circuitos-electricos-8a-ed-compressed-compressed

Circuitos
Eléctricos
OCTAVA EDICIÓN
Richard C. Dorf
Universidad de California
James A. Svoboda
Universidad Clarkson
México • Argentina • Colombia • Chile
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd i
Alfaomega
5/10/11 10:07 AM
Formación
Editec
Al cuidado de la edición:
Luz Ángeles Lomelí Díaz
lalomeli@alfaomega.com.mx
Gerente editorial:
Marcelo Grillo Giannetto
mgrillo@alfaomega.com.mx
Datos catalográficos
Dorf, Richard y Svoboda, James
Circuitos Eléctricos
Octava Edición
Alfaomega Grupo Editor, S.A. de C.V., México
ISBN: 978-607-707-232-4
Formato: 20 x 25.5 cm
Páginas: 908
Circuitos Eléctricos
Richard C. Dorf y James A. Svoboda
ISBN: 978-0-470-52157-1 edición original en inglés “Introduction to Electric Circuits”, 8th Edition,
publicada por John Wiley & Sons, Inc. New Jersey, USA.
Derechos reservados © John Wiley & Sons, Inc.
Octava edición: Alfaomega Grupo Editor, México, junio 2011
© 2011 Alfaomega Grupo Editor, S.A. de C.V.
Pitágoras 1139, Col. Del Valle, 03100, México D.F.
Miembro de la Cámara Nacional de la Industria Editorial Mexicana
Registro No. 2317
Pág. Web: http://www.alfaomega.com.mx
E-mail: atencionalcliente@alfaomega.com.mx
ISBN: 978-607-707-232-4
Derechos reservados:
Esta obra es propiedad intelectual de su autor y los derechos de publicación en lengua española han
sido legalmente transferidos al editor. Prohibida su reproducción parcial o total por cualquier medio
sin permiso por escrito del propietario de los derechos del copyright.
Nota importante:
La información contenida en esta obra tiene un fin exclusivamente didáctico y, por lo tanto, no está
previsto su aprovechamiento a nivel profesional o industrial. Las indicaciones técnicas y programas
incluidos, han sido elaborados con gran cuidado por el autor y reproducidos bajo estrictas normas
de control. ALFAOMEGA GRUPO EDITOR, S.A. de C.V. no será jurídicamente responsable
por: errores u omisiones; daños y perjuicios que se pudieran atribuir al uso de la información
comprendida en este libro, ni por la utilización indebida que pudiera dársele.
Edición autorizada para venta en todo el mundo.
Impreso en México. Printed in Mexico.
Empresas del grupo:
México: Alfaomega Grupo Editor, S.A. de C.V. – Pitágoras 1139, Col. Del Valle, México, D.F. – C.P. 03100.
Tel.: (52-55) 5575-5022 – Fax: (52-55) 5575-2420 / 2490. Sin costo: 01-800-020-4396
E-mail: atencionalcliente@alfaomega.com.mx
Colombia: Alfaomega Colombiana S.A. – Carrera 15 No. 64 A 29, Bogotá, Colombia,
Tel.: (57-1) 2100122 – Fax: (57-1) 6068648 – E-mail: cliente@alfaomega.com.co
Chile: Alfaomega Grupo Editor, S.A. – Dr. La Sierra 1437, Providencia, Santiago, Chile
Tel.: (56-2) 235-4248 – Fax: (56-2) 235-5786 – E-mail: agechile@alfaomega.cl
Argentina: Alfaomega Grupo Editor Argentino, S.A. – Paraguay 1307 P.B. Of. 11, C.P. 1057, Buenos Aires,
Argentina, – Tel./Fax: (54-11) 4811-8352, 4811 7183 y 4811 0887 – E-mail: ventas@alfaomegaeditor.com.ar
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd ii
Circuitos Eléctricos - Dorf
6/9/11 11:25 AM
La naturaleza científica del hombre común
es salir y hacer lo mejor que pueda.
—John Prine
Pero capitán, yo no puedo cambiar las leyes de la física.
—Teniente Comodoro, MontgomeryScott (Scotty), del USS Enterprise
Dedicado a nuestros nietos:
Ian Christopher Boilard, Kyle Everett Schafer, y Graham Henry Schafer
y
Heather Lynn Svoboda, James Hugh Svoboda, Jacob Arthur Leis,
Maxwell Andrew Leis, y Jack Mandlin Leffler
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd iii
Alfaomega
5/10/11 10:07 AM
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd iv
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Los autores
Richard C. Dorf, profesor de Ingeniería Eléctrica y Computacional en la Universidad de California, Davis, da cursos a graduados y estudiantes universitarios de Ingeniería Eléctrica en los campos de circuitos y sistemas de control.
Obtuvo su doctorado en Ingeniería Eléctrica por la U.S. Naval Postgraduate
School, una maestría por la Universidad de Colorado y una Licenciatura en
Ciencias por la Universidad Clarkson. Profundamente comprometido con la
materia de la ingeniería eléctrica y su gran valor para los menesteres sociales y
económicos, ha escrito y disertado a nivel internacional sobre la contribución
y avances de la Ingeniería Eléctrica.
El profesor Dorf tiene una vasta experiencia a nivel educacional e industrial y está activo profesionalmente
en los campos de la robótica, la automatización, los circuitos eléctricos, y las comunicaciones. Ha prestado servicio
como profesor huésped en la Universidad de Edinburgo, Escocia; el Instituto Massachusetts de Tecnología (MIT), la
Universidad Stanford y la Universidad de California en Berkeley.
Miembro del Institute of Electrical and Electronic Engineers y de la American Society for Engineering Education, el Doctor Dorf es ampliamente conocido en el medio por sus obras Sistemas de control modernos, undécima
edición (Prentice Hall, 2008) y The International Encyclopedia of Robotics (Wiley, 1988). El Doctor Dorf es coautor
de Circuits, Devices and Systems (con Ralph Smit), quinta edición (Wiley, 1992). También editó el ampliamente usado Electrical Engineering Handbook, tercera edición (CRC Press e IEEE Press) publicado en 2008. Su más reciente
obra es Technology Ventures, tercera edición (McGraw-Hill 2010).
James A. Svoboda es profesor adjunto de ingeniería eléctrica y computacional
en la Universidad Clarkson, donde da cursos sobre temas de circuitos, electrónica y programación computacional. Obtuvo su doctorado en Ingeniería Eléctrica por la Universidad de Wisconsin en Madison, una maestría por la Universidad de Colorado y una licenciatura en Ciencias del General Motors Institute.
Circuitos para segundo año es uno de los cursos favoritos del profesor
Svoboda. Ha dado este curso a 5 500 estudiantes universitarios en la Universidad Clarkson durante los últimos 30 años. En 1986 recibió el Distinguished
Teaching Award de la Universidad Clarkson.
El profesor Svoboda ha escrito varios artículos en los cuales describe las
ventajas de utilizar nulificadores (nullors) para modelar circuitos eléctricos en
análisis por computadora. Le interesa la manera en que la tecnología afecta la
formación en ingeniería y ha desarrollado algunos paquetes de software para
su uso en los Circuitos para segundo año.
v
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd v
Alfaomega
5/10/11 10:07 AM
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd vi
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Prefacio
El tema central de Circuitos Eléctricos es el concepto de que los circuitos eléctricos forman parte de
la estructura básica de la tecnología moderna. Ante tal tema, nos esforzaremos por demostrar cómo
tanto el análisis y el diseño de circuitos eléctricos están estrechamente enlazados con la habilidad del
ingeniero para diseñar sistemas complejos de electrónica, comunicaciones, cómputo y de control, así
como productos para el consumidor.
ENFOQUE Y ORGANIZACIÓN
Este libro está diseñado para un curso de uno a tres periodos en circuitos eléctricos o análisis de
circuitos lineales, además de que su estructura permite una flexibilidad máxima en su manejo. El
diagrama de flujo de la figura 1 muestra organizaciones de capítulo alternativas que pueden ajustarse
a diferentes perfiles, sin interrumpir la continuidad.
La presentación se acopla a los lectores que van a descubrir los conceptos básicos de los circuitos eléctricos por vez primera y el alcance de este libro es amplio. Los estudiantes deben llegar a este
curso con un conocimiento elemental de cálculo diferencial e integral.
Este libro se esfuerza en preparar al lector para que resuelva problemas reales que involucran
circuitos eléctricos. Por consiguiente, los circuitos se muestran como resultado de invenciones reales
y las respuestas a necesidades reales en la industria, la oficina y el hogar. Aun cuando las herramientas
del análisis de circuitos eléctricos pudieran ser parcialmente abstractas, los circuitos eléctricos son
los bloques de la construcción de la sociedad moderna, actual. El análisis y diseño de los circuitos
eléctricos son habilidades imprescindibles para todos los ingenieros.
N O V E D A D E S E N L A O C TAVA E D I C I Ó N
Se incrementa el uso de PSpice® y MATLAB®
De manera importante, se ha dado una mayor atención al uso de PSpice y MATLAB para la resolución de problemas de circuitos. Empieza con dos nuevos apéndices, uno para presentar PSpice y otro
para MATLAB. Estos apéndices describen brevemente las capacidades de los programas e ilustran
los pasos necesarios para empezar a utilizarlos. A continuación se utilizan PSpice y MATLAB a lo
largo del texto para resolver varios problemas de análisis y diseño de circuitos. Por ejemplo, PSpice
se utiliza en el capítulo 5 para encontrar un circuito equivalente de Thévenin, y en el capítulo 15 para
representar entradas y salidas de circuitos como series de Fourier. MATLAB se usa frecuentemente
para obtener diagramas de entradas y salidas de circuitos que nos ayudan a ver qué nos dicen nuestras
ecuaciones. MATLAB también nos ayuda algo con la larga y tediosa aritmética. Por ejemplo, en el
capítulo 10, MATLAB nos ayuda a hacer la compleja aritmética para analizar circuitos de corriente
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd vii
vii
Alfaomega
5/10/11 10:07 AM
viii
Prefacio
Código
de color
Matrices,
determinantes
E
A
1
2
3
4
VARIABLES
DE CIRCUITOS
ELÉCTRICOS
ELEMENTOS
DE CIRCUITOS
RCIRCUITOS
CRESISTIVOS
MÉTODOS DE
ANÁLISIS DE
CIRCUITOS
RESISTIVOS
Números
compuestos
B, C, D
9
10
11
RESPUESTA TOTAL
DE CIRCUITOS CON
DOS ELEMENTOS DE
ALMACENAMIENTO
DE ENERGÍA
ANÁLISIS
SENOIDAL EN
ESTADO ESTABLE
POTENCIA
DE CA EN
ESTADO ESTABLE
12
CIRCUITOS
TRIFÁSICOS
FIGURA 1 Diagrama de flujo que muestra rutas alternativas a través de los temas de este libro.
alterna (ca), y en el capítulo 14 nos ayuda con la fracción parcial requerida para encontrar las transformaciones inversas de Laplace.
Desde luego, PSpice y MATLAB hacen más que el solo ejecutar los programas. Ponemos especial atención a la interpretación del resultado de estos programas de cómputo y su verificación para
estar seguros de que están correctos. Por lo común, esto se hace en la sección “¿Cómo lo podemos
comprobar...?” que se incluye en cada capítulo. Por ejemplo, la sección 8.9 muestra cómo interpretar y
comprobar una respuesta transitoria de PSpice, y la sección 13.7 muestra cómo interpretar y verificar
una respuesta de frecuencia utilizando MATLAB o PSpice.
Revisiones para mejorar la claridad
Los capítulos 14 y 15, que cubren la transformada de Laplace y la serie y transformada de Fourier,
han sido reescritos a fondo, tanto para mejorar la claridad de la exposición como para aumentar la
cobertura de MATLAB y PSpice. Además, se hicieron revisiones en el texto para mejorar la claridad.
En ocasiones dichas revisiones son menores, pues se refieren a frases o párrafos; otras veces implican
revisiones mayores que implican páginas o secciones enteras.
Más problemas
La octava edición contiene 120 problemas nuevos, con lo que el total supera los 1 350. Esta edición
emplea varios tipos de problemas y van de los sencillos a los que significan un reto, como:
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd viii
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
ix
Prefacio
PSpice
F, G
5
6
7
8
TEOREMAS
DE CIRCUITOS
EL
AMPLIFICADOR
OPERACIONAL
ELEMENTOS QUE
ALMACENAN
ENERGÍA
RESPUESTA
TOTAL DE LOS
CIRCUITOS
RL Y RC
14
TRANSFORMADA DE
LAPLACE
16
CIRCUITOS
DE FILTRADO
13
14
15
RESPUESTA
DE FRECUENCIA
TRANSFORMADA DE
LAPLACE
SERIE Y
TRANSFORMADA DE
FOURIER
6
EL
AMPLIFICADOR
OPERACIONAL
16
CIRCUITOS
DE FILTRO
17
REDES DE
DOS Y TRES
PUERTOS
17
REDES DE
DOS Y TRES
PUERTOS
Leyendas:
Flujo primario
Capítulo
Apéndice
Flujo opcional
• Problemas de análisis directos.
• Análisis de circuitos complejos.
• Problemas sencillos de diseño. (Por ejemplo, dados un circuito y la respuesta especificada,
determinar los valores RLC requeridos.)
• Comparar y contrastar, problemas de multipartes que llaman la atención a las semejanzas o
diferencias entre dos situaciones.
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd ix
Alfaomega
5/10/11 10:07 AM
x
Prefacio
• Problemas de MATLAB y PSpice.
• Problemas de diseño. (Dadas ciertas especificaciones, invente un circuito que las satisfaga.)
• ¿Cómo lo podemos comprobar...? (Verifique si una solución es en verdad la correcta.).
CARACTERÍSTICAS DE EDICIONES ANTERIORES QUE SE
C O N S E R VA N
Introducción
Cada capítulo inicia con una introducción que invita a estudiar el material de ese capítulo.
Ejemplos
Considerando que este libro está orientado a formar expertos en la solución de problemas, hemos incluido más de 260 ejemplos ilustrativos. Incluso, cada ejemplo tiene un título que indica al estudiante
qué es exactamente lo que se ilustra en ese ejemplo en particular.
En algunos ejemplos seleccionados de han incorporado varios métodos de solución de problemas. Estos casos indican a los estudiantes que se pueden emplear múltiples métodos para obtener
soluciones similares o, en algunos casos, que múltiples soluciones pueden ser correctas. Esto ayuda a
los estudiantes a formarse las habilidades de pensar de manera crítica para discernir la mejor opción
entre diversos resultados.
Las secciones Diseño de ejemplos, Método para resolver un problema, y
¿Cómo lo podemos comprobar...?”
Cada capítulo concluye con un ejemplo de diseño que utiliza los métodos de ese capítulo para resolver
un problema de diseño. En el capítulo 1 se presenta un método formal para la solución de problemas
en cinco etapas, y que luego se utiliza en cada uno de los ejemplos de diseño. Un paso importante en
el método de resolución de problemas requiere que usted mismo compruebe sus resultados para verificar que son correctos. En cada capítulo se incluye una sección “¿Cómo lo podemos comprobar...?”
que ilustra cómo se puede comprobar el tipo de resultados obtenidos en ese capítulo para asegurarse
de su exactitud.
Ecuaciones clave y fórmulas
Encontrará que las ecuaciones clave, fórmulas y notas importantes se han destacado en un recuadro
sombreado para ayudarle a identificar con precisión la información de importancia.
Resumen de tablas y figuras
Los procedimientos y métodos desarrollados en este texto se han resumido en determinadas tablas y
figuras clave. Los estudiantes encontrarán que conforman un excelente recurso para la resolución de
problemas.
• Tabla 1.5.1. La convención pasiva.
• Figura 2.7.1 y Tabla 2.7-1. Fuentes dependientes.
• Tabla 3.10-1. Fuentes en serie y en paralelo, elementos en serie y en paralelo. Voltaje y división
de corriente.
• Figura 4.2-3. Nodos de voltajes comparados con corrientes y voltajes de elementos.
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd x
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Prefacio
xi
• Figura 4.5-4. Enlaces de corrientes comparados con corrientes y voltajes de elementos.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Figuras 5.4-3 y 5.4-4. Circuitos equivalentes de Thévenin.
Figura 6.3-1. El amplificador operacional ideal.
Figura 6.5-1. Catálogo de circuitos de amplificadores operacionales de amplio uso.
Tabla 7.8-1. Condensadores e inductores.
Tabla 7.13-2. Condensadores e inductores en serie y en paralelo.
Tabla 8.11-1. Circuitos de primer orden.
Tablas 9.13-1, 2 y 3. Circuitos de segundo orden.
Tabla 10.6-1. Circuitos de CA en el dominio de frecuencia (fasores e impedancias).
Tabla 10.8-1. Voltaje y división de corriente para circuitos de CA.
Tabla 11.5-1. Fórmulas de potencia para circuitos de CA.
Tablas 11.13-1 y 11.13-2. Inductores acoplados y transformadores ideales.
Tabla 13.4-1. Circuitos resonantes.
Tablas 14.2-1 y 14.2-2. Tablas de Transformada de Laplace.
Tabla 14.7-1. Modelos de dominios de elementos de circuitos.
Tabla 15.4-1. Series de Fourier de formas de onda periódicas seleccionadas.
Introducción al procesamiento de señal
El procesamiento de señal es una aplicación importante de los circuitos eléctricos. Este libro lo
presenta de dos maneras. La primera, dos secciones (6.6 y 7.9) describen métodos para diseñar circuitos eléctricos que implementen ecuaciones algebraicas y diferenciales. La segunda, numerosos
ejemplos y problemas a lo largo del libro ilustran el procesamiento de señal. Las señales de entrada y
salida de un circuito eléctrico están identificadas de manera explícita en cada uno de estos ejemplos
y problemas, los cuales investigan la relación entre las señales de entrada y de salida impuesta por
el circuito.
Ejemplos interactivos y ejercicios
Muchos ejemplos a lo largo del libro están marcados como ejemplos interactivos. Esta marca indica
que las versiones computarizadas de ese ejemplo están disponibles en el sitio web de la Editorial
ubicado en http://virtual.alfaomega.com.mx. La figura 2 ilustra la relación entre el ejemplo del libro y
el ejemplo computarizado disponible en el sitio web. La figura 2a muestra un ejemplo del capítulo 3.
El problema presentado por el ejemplo interactivo de la figura 2b es semejante al ejemplo del texto,
pero difiere en algunas formas:
• Los valores de los parámetros del circuito se han vuelto aleatorios.
• Las fuentes independientes y las dependientes pueden haberse invertido.
• La dirección de referencia del voltaje medido puede haberse invertido.
• Se hace una solicitud distinta. En este caso, se pide al estudiante que resuelva el problema del libro
de manera contraria, utilizando el voltaje medido para determinar el valor del parámetro del circuito.
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd xi
Alfaomega
5/24/11 2:07 PM
xii
Prefacio
47
57
Voltímetro
+
12 V +–
+
–
ia
vm
3ia
–
(a)
Ejemplos realizados
R
1.2 V
27 7
Voltímetro
Calculadora
12 V
Problema nuevo
–
+
+
–
ia
+
2 ia
vm
–
Muestra la respuesta
El voltímetro mide el voltaje en voltios.
¿Cuál es el valor de la resistencia, R, en 7?
(b)
Ejemplos realizados
47
27
Amperímetro
Calculadora
Problema nuevo
12 V +–
ia
3ia
im
Muestra la respuesta
El amperímetro mide una corriente en amperios. ¿Cuál
es el valor de la corriente medida por el amperímetro?
(c)
FIGURA 2 (a) El circuito comprende el ejemplo 3.2-5. (b) El ejemplo interactivo correspondiente. (c) Un ejercicio
interactivo correspondiente.
El ejemplo interactivo plantea un problema y luego acepta y comprueba la respuesta del usuario. A los
estudiantes se les proporciona inmediata retroalimentación con respecto a la exactitud de su trabajo.
El ejemplo interactivo selecciona los valores del parámetro de alguna manera aleatoria, proporcionando una aparente provisión infinita de problemas. Este emparejamiento de una solución a un problema
particular con un abasto infinito de problemas similares es una ayuda eficaz para el aprendizaje de los
circuitos eléctricos.
El ejercicio interactivo que muestra la figura 2c considera un circuito similar, pero diferente.
A semejanza del ejemplo interactivo, el ejercicio interactivo plantea un problema y luego acepta y
comprueba la respuesta del usuario. El aprendizaje del estudiante se apoya en mayor forma a partir de
la amplia ayuda en forma de problemas de ejemplo resueltos, disponibles en el ejercicio interactivo
mismo, al cual se accede mediante el botón Worked Example.
Las variaciones de este problema se obtienen con el botón New Problem. También podemos
atisbar la respuesta si utilizamos el botón Show Answer. Los ejemplos y los ejercicios interactivos
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd xii
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Prefacio
xiii
proporcionan cientos de problemas prácticos adicionales, con infinitas variaciones, todos con respuestas que se verifican de inmediato por la computadora.
S U P L E M E N T O S Y M AT E R I A L E N S I T I O W E B
El uso de las computadoras y de la red ha proporcionado una emocionante oportunidad de repensar el
material suplementario. Los suplementos disponibles se han incrementado considerablemente.
Sitio asociado del libro
Se pueden encontrar recursos adicionales para el estudiante y para el maestro en el sitio web de la
Editorial ubicado en http://virtual.alfaomega.com.mx.
Estudiante
• Ejemplos interactivos Los ejemplos y los ejercicios interactivos son poderosos recursos de
apoyo para los estudiantes. Se conformaron como herramientas para asistir a los estudiantes en
habilidades de pericia. Los ejemplos seleccionados del texto e incluidos en la red dan a los estudiantes opciones para navegar a través del problema. Cuando los estudiantes realizan la tarea, han
desarrollado las habilidades adecuadas para completar con éxito sus asignaciones. Es una ayuda
virtual para las tareas.
• MATLAB Tutorial, de Gary Ybarra y Michael Gustafson de la Universidad Duke, construye
ejemplos sobre MATLAB en el texto. Al proporcionar estos ejemplos adicionales, los autores
muestran cómo esta poderosa herramienta se utiliza fácilmente en áreas apropiadas del análisis
de circuitos. Se crearon diez problemas de ejemplo en HTML. Los archivos M para los ejemplos
basados en computadora están disponibles para bajarlos en el sitio web de la Editorial ubicado en
http://virtual.alfaomega.com.mx.
Maestro
• Manual de soluciones
• Diapositivas de PowerPoint
AGRADECIMIENTOS
Estamos muy agradecidos con muchas personas cuyos esfuerzos contribuyeron a la elaboración de
este texto. En especial damos las gracias a nuestro editor asociado Daniel Sayre, al gerente ejecutivo
de marketing Chris Ruel y a Diana Smith, asistente de marketing, por su apoyo y entusiasmo. También agradecemos a Janet Foxman y Dorothy Sinclair de Wiley, y a Heather Johnson de Elm Street
Publishing Services por sus esfuerzos en la producción de este libro. Deseamos agradecer a Lauren
Sapira, Carolyn Weisman y Andre Legaspi por sus importantes contribuciones a este proyecto.
En particular, agradecemos al equipo de revisores que verificaron los problemas y soluciones
para cerciorarse de su certeza.
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd xiii
Alfaomega
5/24/11 2:50 PM
xiv
Prefacio
Revisores de datos
Khalid Al-Olimat, Ohio Northern
University
Lisa Anneberg, Lawrence
Tchnological University
Horace Gordon, University of South Florida
Limachos Kondi, SUNY, Buffalo
Michael Polis, Oakland University
Sannasi Ramanan, Rochester Institute
of Technology
William Robins, University of Minnesota
James Rowland, University of Kansas
Mike Shen, Duke University
Thyagarajan Srinivasan, Wilkes
University
Aaron Still, U.S. Naval Academy
Howard Winert, Johns Hopkins
University
Xiao-Bang Xu, Clemson University
Jiann Shiun Yuan, University of
Central Florida
Revisores
Rehab Abdel-Kader, Georgia Southern
University
Said Ahmed-Zaid, Boise State
University
Farzan Aminian, Trinity University
Constantin Apostoaia, Purdue
University Calumet
Jonathan Bagby, Florida Atlantic University
Carlotta Berry, Tennessee State University
Kiron Bordoloi, University of Louisville
Mauro Caputi, Hofstra University
Edward Collins, Clemson University
Glen Dudevoir, U.S. Militar Academy
Malik Elbuluk, University of Akron
Prasad Enjeti, Texas A&M University
Alieydaghi, University of Maryland
Eastern Shore
Carlos Figueroa, Cabrillo College
Walid Hubbi, New Jersey Institute of Technology
Brian Huggins, Bradley University
Chris Ianello, University of Central Florida
Simone Jarzabek, ITT Technical Institute
James Kawamoto, Mission College
Rasool Kenarangui, University
of Texas Arlington
Jumoke Ladeji-Osias, Morgan State University
Mark Lau, Universidad del Turabo
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd xiv
Seyed Mousavinezhad, Western Michigan
University
Philip Munro, Youngstone State University
Ahmad Nafisi, California Polytechnic State
University
Arnost Neugroschel, University of Florida
Tokunbo Ogunfunmi, Santa Clara University
Gary Perks, California Polytechnic State
University, San Luis Obispo
Owe Petersen, Milwaukee School of Engineering
Ron Pieper, University of Texas
Teodoro Robles, Milwaukee School of Engineering
Pedda Sannuti, Rutgers University
Marcelo Simoes, Colorado School of Mines
Ralph Tanner, Western Michigan University
Tristan Tayag, Texas Christian University
Jean-Claude Thomassian, Central
Michigan University
John Ventura, Christian Brothers University
Annette von Jouanne,
Oregon State University
Ravi Warrier, Kettering University
Gerald Woelfi, Milwaukee School of
Engineering
Hewlon Zimmer, U.S. Merchant
Marine Academy
Circuitos Eléctricos - Dorf
6/9/11 11:26 AM
Contenido
CAPÍTULO 1
Variables de circuitos eléctricos ........................................................................................ 1
1.1
Introducción ............................................................................................................................ 1
1.2
Circuitos eléctricos y corriente ............................................................................................... 1
1.3
Sistemas de unidades .............................................................................................................. 5
1.4
Voltaje ..................................................................................................................................... 7
1.5
Potencia y energía ................................................................................................................... 7
1.6
Análisis y diseño de circuitos ............................................................................................... 11
1.7
¿Cómo lo podemos comprobar . . . ? ....................................................................................... 13
1.8
Ejemplo de diseño — Controlador de válvulas de un motor de propulsión a chorro ........... 14
1.9
Resumen ............................................................................................................................... 15
Problemas ............................................................................................................................. 15
Problemas de diseño ..............................................................................................................19
CAPÍTULO 2
Elementos de circuitos ..................................................................................................... 20
2.1
Introducción .......................................................................................................................... 20
2.2
Ingeniería y modelos lineales ............................................................................................... 20
2.3
Elementos de circuito activos y pasivos ............................................................................... 24
2.4
Resistencias........................................................................................................................... 25
2.5
Fuentes independientes ......................................................................................................... 28
2.6
Voltímetros y amperímetros .................................................................................................. 31
2.7
Fuentes dependientes ............................................................................................................ 33
2.8
Transductores ........................................................................................................................ 37
2.9
Interruptores .......................................................................................................................... 39
2.10
¿Cómo lo podemos comprobar . . . ? ....................................................................................... 41
2.11
Ejemplo de diseño — Sensor de temperatura ....................................................................... 42
2.12
Resumen ............................................................................................................................... 44
Problemas ............................................................................................................................. 44
Problemas de diseño ............................................................................................................. 52
xv
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd xv
Alfaomega
5/10/11 10:07 AM
xvi
Contenido
CAPÍTULO 3
Circuitos resistivos ........................................................................................................... 53
3.1
Introducción .......................................................................................................................... 53
3.2
Leyes de Kirchhoff ............................................................................................................... 53
3.3
Resistores en serie y división de voltaje ............................................................................... 61
3.4
Resistores en paralelo y división de la corriente .................................................................. 66
3.5
Fuentes de voltaje en serie y fuentes de corriente en paralelo .............................................. 72
3.6
Análisis de circuitos .............................................................................................................. 73
3.7
Análisis de circuitos resistivos utilizando MATLAB ........................................................... 78
3.8
¿Cómo lo podemos comprobar . . . ? ....................................................................................... 82
3.9
Ejemplo de diseño — Fuente de voltaje ajustable .................................................................84
3.10
Resumen ............................................................................................................................... 87
Problemas ............................................................................................................................. 88
Problemas de diseño ........................................................................................................... 106
CAPÍTULO 4
Métodos de análisis de circuitos resistivos...................................................................108
4.1
Introducción .........................................................................................................................108
4.2
Análisis de voltajes de nodos de circuitos con fuentes de corriente ....................................109
4.3
Análisis de voltajes de nodos de circuitos con fuentes de corriente y de voltaje ................115
4.4
Análisis de voltajes de nodos con fuentes dependientes .....................................................120
4.5
Análisis de corrientes de enlaces con fuentes de voltaje independientes ............................122
4.6
Análisis de corrientes de enlaces con fuentes de corriente y de voltaje ..............................127
4.7
Análisis de corrientes de enlaces con fuentes dependientes ................................................131
4.8
Comparación entre el método de voltajes de nodos y el método de corrientes de enlaces .....134
4.9
Análisis de corrientes de enlaces utilizando MATLAB .......................................................136
4.10
Uso de PSpice para determinar los voltajes de nodos y las corrientes de enlaces...............138
4.11
¿Cómo lo podemos comprobar . . . ? ......................................................................................140
4.12
Ejemplo de diseño — Despliegue angular del potenciómetro.............................................143
4.13
Resumen ..............................................................................................................................146
Problemas ............................................................................................................................147
Problemas de PSpice............................................................................................................160
Problemas de diseño ............................................................................................................160
CAPÍTULO 5
Teoremas de circuitos......................................................................................................162
5.1
Introducción .........................................................................................................................162
5.2
Transformaciones de fuentes ...............................................................................................162
5.3
Superposición ......................................................................................................................167
5.4
Teorema de Thévenin...........................................................................................................171
5.5
Circuito equivalente de Norton ............................................................................................175
5.6
Transferencia de potencia máxima ......................................................................................179
5.7
Uso de MATLAB para determinar el circuito equivalente de Thévenin .............................182
5.8
Uso de PSpice para determinar el circuito equivalente de Thévenin ..................................185
5.9
¿Cómo lo podemos comprobar . . . ? ......................................................................................188
5.10
Ejemplo de diseño — Puente de indicador de tensión ........................................................189
5.11
Resumen ..............................................................................................................................192
Problemas ............................................................................................................................192
Problemas de PSpice............................................................................................................205
Problemas de diseño ............................................................................................................206
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd xvi
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Contenido
xvii
CAPÍTULO 6
El amplificador operacional ............................................................................................208
6.1
Introducción .........................................................................................................................208
6.2
El amplificador operacional .................................................................................................208
6.3
El amplificador operacional ideal ........................................................................................210
6.4
Análisis nodal de circuitos que contienen amplificadores operacionales ideales................212
6.5
Diseño mediante el uso de amplificadores operacionales ...................................................217
6.6
Circuitos de amplificadores operacionales y ecuaciones algebraicas lineales ....................222
6.7
Características de los amplificadores operacionales prácticos ............................................227
6.8
Análisis de circuitos de amplificadores operacionales mediante el uso de MATLAB ........234
6.9
Análisis de circuitos de amplificadores operacionales mediante el uso de PSpice .............236
6.10
¿Cómo lo podemos comprobar . . . ? ......................................................................................237
6.11
Ejemplo de diseño — Circuito de interfase de transductor .................................................239
6.12
Resumen ..............................................................................................................................241
Problemas ............................................................................................................................242
Problemas de PSpice............................................................................................................255
Problemas de diseño ............................................................................................................256
CAPÍTULO 7
Elementos que almacenan energía ................................................................................257
7.1
Introducción .........................................................................................................................257
7.2
Condensadores .....................................................................................................................258
7.3
Almacenamiento de energía en un condensador..................................................................264
7.4
Condensadores en serie y en paralelo ..................................................................................267
7.5
Inductores ............................................................................................................................269
7.6
Almacenamiento de energía en un inductor ........................................................................274
7.7
Inductores en serie y en paralelo .........................................................................................276
7.8
Condiciones iniciales de los circuitos permanentes.............................................................277
7.9
Circuitos de amplificadores operacionales y ecuaciones diferenciales lineales ..................281
7.10
Uso de MATLAB para trazar el voltaje y la corriente de un condensador
o un inductor ........................................................................................................................281
7.11
¿Cómo lo podemos comprobar . . . ? ......................................................................................287
7.12
Ejemplo de diseño — Integrador e interruptor ...................................................................290
7.13
Resumen ..............................................................................................................................293
Problemas ............................................................................................................................294
Problemas de diseño ............................................................................................................309
CAPÍTULO 8
Respuesta total de los circuitos RL y RC .......................................................................311
8.1 Introducción ................................................................................................................................311
8.2 Circuitos de primer orden ...........................................................................................................311
8.3 Respuesta de un circuito de primer orden a una entrada constante ............................................314
8.4 Conmutación secuencial .............................................................................................................327
8.5 Estabilidad de circuitos de primer orden ....................................................................................329
8.6 Fuente de paso unitario ...............................................................................................................331
8.7 Respuesta de un circuito de primer orden a una fuente no constante .........................................335
8.8 Operadores diferenciales .............................................................................................................340
8.9 Uso de PSpice para analizar circuitos de primer orden ..............................................................342
8.10 ¿Cómo lo podemos comprobar . . . ? ...........................................................................................345
8.11 Ejemplo de diseño — Una computadora y su impresora ..........................................................349
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd xvii
Alfaomega
5/10/11 10:07 AM
xviii
Contenido
8.12
Resumen ..............................................................................................................................352
Problemas ............................................................................................................................353
Problemas de PSpice............................................................................................................366
Problemas de diseño ............................................................................................................367
CAPÍTULO 9
Respuesta total de circuitos con dos elementos de almacenamiento de energía .......368
9.1
Introducción .........................................................................................................................368
9.2
Ecuación diferencial para circuitos con dos elementos de almacenamiento de energía .......369
9.3
Solución de la ecuación diferencial de segundo orden: la respuesta natural .......................373
9.4
Respuesta natural del circuito RLC en paralelo no forzado .................................................376
9.5
Respuesta natural del circuito RLC en paralelo no forzado críticamente amortiguado .......379
9.6
Respuesta natural de un circuito RLC en paralelo no forzado subamortiguado ..................380
9.7
Respuesta forzada de un circuito RLC .................................................................................382
9.8
Respuesta total de un circuito RLC ......................................................................................386
9.9
Método de las variables de estado para el análisis de circuitos ...........................................389
9.10
Raíces en el plano compuesto ..............................................................................................393
9.11
¿Cómo lo podemos comprobar . . . ? ......................................................................................394
9.12
Ejemplo de diseño — Dispositivo de encendido de la bolsa de aire de un automóvil ........397
9.13
Resumen ..............................................................................................................................399
Problemas ............................................................................................................................401
Problemas de PSpice............................................................................................................412
Problemas de diseño ............................................................................................................413
CAPÍTULO 10
Análisis senoidal en estado estable...............................................................................425
10.1
Introducción .........................................................................................................................415
10.2
Fuentes senoidales ...............................................................................................................416
10.3
Respuesta de estado estable de un circuito RL para una función
de forzamiento senoidal .......................................................................................................421
10.4
Función de forzamiento exponencial compuesta .................................................................422
10.5
El fasor .................................................................................................................................426
10.6
Relaciones del fasor para los elementos R, L y C. ...............................................................430
10.7
Impedancia y admitancia .....................................................................................................434
10.8
Leyes de Kirchhoff que utilizan fasores ..............................................................................438
10.9
Análisis del voltaje de nodos y de la corriente de enlaces utilizando fasores .....................443
10.10 Superposición, equivalentes de Thévenin y Norton y transformaciones de fuentes ...........449
10.11 Diagramas de fasores ...........................................................................................................454
10.12 Circuitos de fasores y el amplificador operacional..............................................................455
10.13 La respuesta total .................................................................................................................457
10.14 Uso de MATLAB para el análisis de circuitos en estado estable
con entradas senoidales........................................................................................................464
10.15 Uso de PSpice para analizar circuitos de CA.......................................................................466
10.16 ¿Cómo lo podemos comprobar . . . ? ......................................................................................469
10.17 Ejemplo de diseño — Circuito del amplificador operacional .............................................471
10.18 Resumen ..............................................................................................................................474
Problemas ............................................................................................................................474
Problemas de PSpice............................................................................................................493
Problemas de diseño ............................................................................................................494
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd xviii
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Contenido
xix
CAPÍTULO 11
Potencia de CA de estado estable..................................................................................496
11.1
Introducción .........................................................................................................................496
11.2
Potencia eléctrica .................................................................................................................496
11.3
Potencia instantánea y potencia promedio ...........................................................................497
11.4
Valor efectivo de una forma de onda periódica ...................................................................501
11.5
Potencia compleja ................................................................................................................503
11.6
Factor de potencia ................................................................................................................511
11.7
Principio de superposición de potencia ...............................................................................519
11.8
Teorema de la transferencia de potencia máxima ................................................................522
11.9
Inductores acoplados ...........................................................................................................523
11.10 El transformador ideal .........................................................................................................531
11.11 ¿Cómo lo podemos comprobar . . . ? ......................................................................................536
11.12 Ejemplo de diseño — Transferencia de potencia máxima...................................................538
11.13 Resumen ..............................................................................................................................540
Problemas ............................................................................................................................542
Problemas de PSpice............................................................................................................556
Problemas de diseño ............................................................................................................556
CAPÍTULO 12
Circuitos trifásicos ...........................................................................................................558
12.1
Introducción .........................................................................................................................558
12.2
Voltajes trifásicos .................................................................................................................559
12.3
Circuito Y a Y ......................................................................................................................562
12.4
Fuente y carga conectadas a ⌬ .............................................................................................571
12.5
Circuito Y a ⌬ ......................................................................................................................573
12.6
Circuitos trifásicos balanceados ..........................................................................................576
12.7
Potencias promedio e instantánea en una carga trifásica balanceada ..................................578
12.8
Medición de potencia con dos vatímetros ...........................................................................581
12.9
¿Cómo lo podemos comprobar . . . ? ......................................................................................584
12.10 Ejemplo de diseño — Corrección del factor de potencia ...................................................587
12.11 Resumen ..............................................................................................................................588
Problemas ............................................................................................................................589
Problemas PSpice ................................................................................................................593
Problemas de diseño ............................................................................................................593
CAPÍTULO 13
Respuesta de frecuencia .................................................................................................594
13.1
Introducción .........................................................................................................................594
13.2
Ganancia, cambio de fase y la función de red .....................................................................594
13.3
Diagramas de Bode ..............................................................................................................606
13.4
Circuitos resonantes .............................................................................................................623
13.5
Respuesta de frecuencia de circuitos de amplificadores operacionales...............................630
13.6
Trazo de diagramas de Bode utilizando MATLAB .............................................................632
13.7
Uso de PSpice para trazar un diagrama de respuesta de frecuencia ....................................634
13.8
¿Cómo lo podemos comprobar . . . ? ......................................................................................636
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd xix
Alfaomega
5/10/11 10:07 AM
xx
Contenido
13.9
13.10
Ejemplo de diseño — Sintonizador de radio .......................................................................640
Resumen ..............................................................................................................................642
Problemas ............................................................................................................................643
Problemas de PSpice ...........................................................................................................656
Problemas de diseño ............................................................................................................658
CAPÍTULO 14
Transformada de Laplace ................................................................................................660
14.1
Introducción .........................................................................................................................660
14.2
Transformada de Laplace.....................................................................................................661
14.3
Entradas de pulso .................................................................................................................667
14.4
Transformada inversa de Laplace ........................................................................................671
14.5
Teoremas del valor inicial y final ........................................................................................677
14.6
Solución de ecuaciones diferenciales que describen un circuito .........................................680
14.7
Análisis de circuitos utilizando impedancia y condiciones iniciales ...................................681
14.8
Función de transferencia e impedancia................................................................................692
14.9
Convolución .........................................................................................................................695
14.10 Estabilidad ...........................................................................................................................699
14.11 Expansión de fracción parcial utilizando MATLAB ...........................................................702
14.12 ¿Cómo lo podemos comprobar . . . ? ......................................................................................707
14.13 Ejemplo de diseño — Compuerta de carga del transbordador espacial ..............................710
14.14 Resumen ..............................................................................................................................713
Problemas ............................................................................................................................714
Problemas de PSpice............................................................................................................728
Problemas de diseño ............................................................................................................729
CAPÍTULO 15
Serie y transformada de Fourier .....................................................................................730
15.1
Introducción .........................................................................................................................730
15.2
Serie de Fourier....................................................................................................................731
15.3
Simetría de la función f (t) ...................................................................................................739
15.4
Serie de Fourier de formas de onda seleccionadas ..............................................................744
15.5
Forma exponencial de la serie de Fourier ............................................................................746
15.6
Espectro de Fourier ..............................................................................................................754
15.7
Circuitos y serie de Fourier..................................................................................................758
15.8
Uso de PSpice para determinar la serie de Fourier ..............................................................761
15.9
Transformada de Fourier .....................................................................................................766
15.10 Propiedades de la transformada de Fourier .........................................................................769
15.11 Espectro de las señales.........................................................................................................773
15.12 Convolución y respuesta del circuito ...................................................................................774
15.13 Transformada de Fourier y la transformada de Laplace ......................................................777
15.14 ¿Cómo lo podemos comprobar . . . ? ......................................................................................779
15.15 Ejemplo de diseño — Alimentación de potencia de CD .....................................................781
15.16 Resumen ..............................................................................................................................784
Problemas ............................................................................................................................785
Problemas de PSpice............................................................................................................791
Problemas de diseño ............................................................................................................791
Alfaomega
A01_DORF_1571_8ED_SE_i-xxii.indd xx
Circuitos Eléctricos - Dorf
5/10/11 10:07 AM
Contenido
xxi
CAPÍTULO 16
Circuitos de filtro .............................................................................................................793
16.1
Introducción .........................................................................................................................793
16.2
Filtro eléctrico ......................................................................................................................793
16.3
Filtros ...................................................................................................................................794
16.4
Filtros de segundo orden ......................................................................................................797
16.5
Filtros de alto orden .............................................................................................................805
16.6
Simulación de circuitos de filtro utilizando PSpice .............................................................811
16.7
¿Cómo lo podemos comprobar . . . ? ......................................................................................815
16.8
Ejemplo de diseño — Filtro antiseudónimo ........................................................................817
16.9
Resumen ..............................................................................................................................820
Problemas ............................................................................................................................820
Problemas de PSpice............................................................................................................825
Problemas de diseño ............................................................................................................828
CAPÍTULO 17
Redes de dos y tres puertos ...........................................................................................829
17.1
Introducción .........................................................................................................................829
17.2
Transformación de T a ⌸ y redes de dos puertos y tres terminales .....................................830
17.3
Ecuaciones de redes de dos puertos .....................................................................................832
17.4
Parámetros Z y Y para un circuito con fuentes dependientes...............................................835
17.5
Parámetros híbridos y de transmisión ..................................................................................837
17.6
Relaciones entre parámetros de dos puertos ........................................................................839
17.7
Interconexión de redes de dos puertos .................................................................................841
17.8
¿Cómo lo podemos comprobar . . . ? ......................................................................................844
17.9
Ejemplo de diseño — Amplificador de transistores ............................................................846
17.10 Resumen ..............................................................................................................................848
Problemas ............................................................................................................................848
Problemas de diseño ............................................................................................................852
APÉNDICE A
Inicios con PSpice ............................................................................................................853
APÉNDICE B
MATLAB, matrices y aritmética compuesta ..................................................................860
APÉNDICE C
Fórmulas matemáticas ....................................................................................................871
APÉNDICE D
Código de colores del resistor estándar ........................................................................874
Referencias .......................................................................................................................876
Índice ................................................................................................................................879
Circuitos Eléctricos - Dorf
A01_DORF_1571_8ED_SE_i-xxii.indd xxi
Alfaomega
5/10/11 10:07 AM
A01_DORF_1571_8ED_SE_i-xxii.indd xxii
5/10/11 10:07 AM
Variables
de circuitos eléctricos
CAPÍTULO
EN ESTE CAPÍTULO
1. 1
1. 2
1. 3
1. 4
1. 5
1. 6
1.1
Introducción
Circuitos eléctricos y corriente
Sistemas de unidades
Voltaje
Potencia y energía
Análisis y diseño de circuitos
1. 7
1. 8
1. 9
¿Cómo lo podemos comprobar . . . ?
EJEMPLO DE DISEÑO — Controlador de
válvulas de un motor de propulsión a chorro
Resumen
Problemas
Problemas de diseño
INTRODUCCIÓN
Un circuito consta de elementos eléctricos conectados entre sí. Los ingenieros utilizan los circuitos
eléctricos para resolver problemas de importancia para la sociedad actual. En particular:
1. L
os circuitos eléctricos se usan en la generación, transmisión y consumo de la potencia eléctrica
y la energía.
2. Los circuitos eléctricos se emplean en la codificación, decodificación, almacenamiento,
transmisión y procesamiento de la información.
En este capítulo haremos lo siguiente:
•Representar la corriente y el voltaje de un elemento del circuito eléctrico, prestando particular
atención a la dirección de referencia de la corriente y a la dirección de referencia o polaridad del
voltaje.
• Calcular la potencia y la energía proporcionadas o recibidas por un circuito.
•Utilizar la convención pasiva para determinar si el producto de la corriente y el voltaje de un
elemento de circuito es la potencia proporcionada por ese elemento o la potencia recibida
por el elemento.
•Aplicar notación científica para representar cantidades eléctricas con un amplio margen de
magnitudes.
1.2
CIRCUITOS ELÉCTRICOS Y CORRIENTE
Al comparar con otras fuentes de potencia, las principales características de la electricidad son su
movilidad y su flexibilidad. La energía eléctrica puede ser trasladada a cualquier punto a lo largo
de un conjunto de cables y, dependiendo de las necesidades del usuario, convertida en luz, calor
o movimiento.
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 1
1
Alfaomega
4/12/11 5:15 PM
2
2
Electric Circuit Variables
Variables de circuitos eléctricos
An
or oelectric
network
an interconexión
interconnection
electricaleléctricos
elementsunidos
linked
Un electric
circuito circuit
eléctrico
red eléctrica
es isuna
de of
elementos
together
in
a
closed
path
so
that
an
electric
current
may
flow
continuously.
entre sí en una vía cerrada, de modo que una corriente eléctrica pueda fluir constantemente.
Consider
simple
circuit
consisting
of two
well-known
a battery
and
Considereaun
circuito
sencillo
que conste
de dos
elementoselectrical
eléctricoselements,
bien conocidos,
una
ba-a
resistor,
as
shown
in
Figure
1.2-1.
Each
element
is
represented
by
the
two-terminal
element
tería y una resistencia, como se muestra en la figura 1.2-1. Cada elemento está representado por un
shown
inde
Figure
1.2-2. Elements
are sometimes
called
devices,
terminals
are se
sometimes
elemento
dos terminales
que se muestran
en la figura
1.2-2.
A estosand
elementos
a veces
les llama
called
nodes.
dispositivos, y nodos a las terminales.
Cable
Wire
Batería
Battery
Resistencia
Resistor
a
Cable
Wire
b
FIGURA
eléctrico
de dos terminales
FIGURE 1.2-2
1.2-2Elemento
A general
two-terminal
electrical
aelement
y b generales.
with terminals a and b.
FIGURA
Circuito
FIGURE 1.2-1
1.2-1 A
simplesencillo.
circuit.
La cargamay
puede
enelectric
un circuito
eléctrico.
Corriente
es rate
la velocidad
deofcambio
carga
que
Charge
flowfluir
in an
circuit.
Current
is the time
of change
chargede
past
a given
pasa
en
un
punto
dado.
Carga
es
la
propiedad
intrínseca
de
materia
que
causa
los
fenómenos
eléctripoint. Charge is the intrinsic property of matter responsible for electric phenomena. The quantity of
�19 21.602 3
cos. Laqcantidad
de carga qinseterms
puedeofexpresar
en términos
de cargawhich
en uniselectrón,
decir,
coulombs.
charge
can be expressed
the charge
on one electron,
�1.602 es
� 10
219
18
18
10
culombios.
Poris tanto,
21 culombio
la carga
en 6.24The
3 10current
electrones.
través
electrons.
throughLaa corriente
specifiedaarea
is
Thus,
�1 coulomb
the charge
on 6.24 es
� 10
de
un
área
específica
es
definida
por
la
carga
eléctrica
que
pasa
por
el
área
por
unidad
de
tiempo.
En
defined by the electric charge passing through the area per unit of time. Thus, q is defined as the charge
consecuencia,
q se define
como la carga expresada en culombios (C).
expressed
in coulombs
(C).
Carga esislathe
cantidad
de of
electricidad
de los
eléctricos.
Charge
quantity
electricitycausante
responsible
forfenómenos
electric phenomena.
Entonces,
lo podemos
expresarascomo
Then
we can
express current
i¼
dq
dt
ð1:2-1Þ
(1.2-1)
The unit of current is the ampere (A); an ampere is 1 coulomb per second.
La unidad de corriente es el amperio (A); un amperio es 1 culombio por segundo.
Current
thela time
rate ofdeflow
pastque
a given
point.
Corrienteises
velocidad
flujoofdeelectric
la cargacharge
eléctrica
pasa por
un punto dado.
Note
thatque
throughout
this
we useutilizamos
a lowercase
as q, tocomo
denote
variable
that
is
Observe
a lo largo
dechapter
este capítulo
unaletter,
letra such
minúscula,
q, apara
indicar
una
avariable
function
of
time,
q(t).
We
use
an
uppercase
letter,
such
as
Q,
to
represent
a
constant.
que es una función de tiempo, q(t); y una mayúscula, como Q, para representar una constante.
The
flowdeofcorriente
current isseconventionally
a flow
positive
charges.
This convention
El flujo
representa porrepresented
convenciónascomo
unofflujo
de cargas
positivas.
Esta conwas
initiated
by
Benjamin
Franklin,
the
first
great
American
electrical
scientist.
Of course,
we
vensión la inició Benjamín Franklin, primer gran científico estadounidense de la electricidad.
Desde
now
know
that
charge
flow
in
metal
conductors
results
from
electrons
with
a
negative
charge.
luego, ahora sabemos que la carga que fluye en conductores de metal es el resultado de electrones
Nevertheless,
we will
of current as thela flow
of positive
accepted
con carga negativa.
No conceive
obstante, consideraremos
corriente
como elcharge,
flujo deaccording
una carga to
positiva,
de
convention.
acuerdo con la convención aceptada.
Figure
1.2-3
shows
the notation
that we
usedescribir
to describe
current. There
are two
parts
to
La
figura
1.2-3
muestra
la notación
para
unaacorriente.
Hay dos
partes
para
i1
this
notation:
a
value
(perhaps
represented
by
a
variable
name)
and
an
assigned
direction.
As
i1
esta notación: un valor (quizá representado por un nombre de variable) y una dirección asig-a
a
b
matterAofmanera
vocabulary,
we say thatdecimos
a current
exists
or through
anen
element.
Figure
1.2-3
shows
a
b
nada.
de vocabulario,
que
se dainuna
corriente
o a través
de un
elemento.
i
i22
that
there
are
two
ways
to
assign
the
direction
of
the
current
through
an
element.
The
current
i1
La figura muestra que hay dos maneras de asignar la dirección de la corriente a través del
is
the
rate
of
flow
of
electric
charge
from
terminal
a
to
terminal
b.
On
the
other
hand,
the
FIGURE 1.2-3
1.2-3 La
Current elemento. La corriente i1 es la proporción del flujo de carga de electricidad de la terminal a a
FIGURA
is the
flowlaofcorriente
electric icharge
terminal
b todeterminal
a. The
i1 and
current
in a circuit element.
la
b. Pori2otro
lado,
es el from
flujo de
la carga
electricidad
decurrents
la terminal
b ai2laare
a.
corriente en un circuito.
Alfaomega
M01_DORF_1571_8ED_SE_001-019.indd 2
2
Circuitos Eléctricos - Dorf
4/12/11 5:15 PM
Circuitos eléctricos y corriente
3
i
I
FIGURA 1.2-4 Corriente directa de magnitud I.
t
0
Las corrientes i1 e i2 son semejantes pero diferentes. Tienen el mismo tamaño pero diferentes direcciones. Por lo tanto, i2 es la negativa de i1 y entonces
i1 i2
Siempre se asocia una flecha con una corriente para indicar su dirección. Una descripción completa de
corriente requiere un valor (que puede ser positivo o negativo) y una dirección (indicada por una flecha).
Si la corriente que fluye a través de un elemento es constante, se representa por la constante I,
como se muestra en la figura 1.2-4. Una corriente constante se denomina corriente directa (cd).
Una corriente directa (cd) es una corriente de magnitud constante.
Una corriente que varía con el tiempo i(t) puede tomar varias formas, ya sea de rampa, sinusoide o
exponencial, como se ven en la figura 1.2-5. La corriente sinusoidal se denomina corriente alterna (ca).
i
(A)
i
(A)
i = Mt, t > 0
M
i
(A)
I
1
i = Ie–bt, t > 0
I
0
t (s)
0
i = I sen W t, t > 0
t (s)
t (s)
0
–I
(a)
(b)
(c)
FIGURA 1.2-5 (a) Rampa con una pendiente M. (b) Sinusoide. (c) Exponencial. I es una constante. La corriente i es
cero para t 0.
Si se conoce la carga q, la corriente i se encuentra fácilmente mediante la ecuación 1.2-1. O bien,
si se conoce la corriente i, se puede calcular la carga q. Observe que de la ecuación 1.2-1 obtenemos
Z
q¼
t
1
Z
i dt ¼
t
0
i dt þ qð0Þ
(1.2-2)
donde q(0) es la carga en t 0.
E J E M P L O 1 . 2 - 1 Corriente a partir de una carga
Obtenga la corriente en un elemento cuando la carga entrante sea
q 12t C
donde t es el tiempo en segundos.
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 3
Alfaomega
4/13/11 5:07 PM
1
E1C01_1
11/26/2009
4
4444 4
4 4
4
Electric
Circuit
Variables
Electric
Circuit
Variables
Electric
Circuit
Variables
Electric
Circuit
Variables
Variables
de
circuitos
eléctricos
Electric
Circuit
Variables
Electric
Circuit
Variables
Electric
Circuit
Variables
Solution
Solución
Solution
Solution
Solution
Solution
Solution
Recall
thatque
thelaunit
of charge
is coulombs,
C. Then
thetanto,
current,
from Eq.a1.2-1,
Recuerde
unidad
de carga
es el culombio,
C. Por
la corriente,
partir is
de la ecuación 1.2-1 es
Recall
that
the
unit
of charge
charge
coulombs,
C. Then
Then
the
current,
from
Eq.
1.2-1,
Recall
that
the
unit
of
is
C.
the
current,
from
Eq.
1.2-1,
Recall
that
the
unit
of
isis coulombs,
C.
the
current,
from
Eq.
1.2-1,
isis is
Recall
unit
of
charge
isis coulombs,
coulombs,
C.
Then
current,
from
1.2-1,
isis
Recall
thatthat
thethe
unit
of charge
charge
coulombs,
C. Then
Then
thethe
current,
from
Eq.Eq.
1.2-1,
dq dq
dq
dq
dq
¼iii¼
¼ 12
12
A
¼ dq
¼12
12 AA
¼
iii¼
¼
A
¼
¼
12
¼ dt
¼
AA
dt
dt 12
dt
dt
dt
donde
la unidad
corriente
son
los amperios,
where
the
unitunit
ofdecurrent
current
is amperes,
amperes,
A. A.
where
the
unit
of
current
amperes,
A. A.
where
the
of
current
is
where
the
unit
of
is
A.
where
current
isis amperes,
amperes,
where
thethe
unitunit
of of
current
is amperes,
A. A.
Xm
AXM
M
Po
E
Ej X
X
A
M
E
-22Carga
Charge
from
Current
X
M
Charge
from
Current
A
M
Charge
from
Current
EE
lA
1LLL1.11EE2E...1
-221
a partir
deCurrent
corriente
A
PP
LLLLP
E
from
EE
A
PP
1
22Charge
Charge
from
Current
EeXE
AXpM
PM
E
22..-.-2
2---2
Charge
from
Current
Find
thethe
charge
that
has
entered
the
terminal
ofdean
an
element
from
¼
¼3tt33t¼
the
current
entering
the
Obtenga
la charge
carga
que
hahas
entrado
a la
terminal
unan
elemento
de
5t0t¼
0¼s0a0totssstt5
ss3cuando
lathe
corriente
entrante
althe
Find
the
charge
that
has
entered
the
terminal
of
an
element
from
to
¼
3when
when
the
current
entering
the
Find
that
entered
the
terminal
of
element
from
to
ssswhen
current
entering
Find
the
charge
that
has
entered
the
terminal
of
from
tttt¼
¼
the
current
entering
the
Find
charge
entered
terminal
an
element
from
when
current
entering
Find
thethe
charge
thatthat
hashas
entered
thethe
terminal
of of
anelement
element
from
¼t00¼ss0to
to tto
¼ 3¼ss3when
when
thethe
current
entering
thethe
element
is
as
shown
in
Figure
1.2-6.
elemento
es
como
se
muestra
en
la
figura
1.2-6.
element
is
as
shown
in
Figure
1.2-6.
element
is
as
shown
in
Figure
1.2-6.
element
is
as
shown
in
Figure
1.2-6.
element
is as
shown
Figure
1.2-6.
element
is as
shown
in in
Figure
1.2-6.
(A)i i(A)
(A)
i (A)
(A)
iiii(A)
(A)i (A)
4
4
4 4 4
4
4 4
3
3 33 3
3
3 3
E1C01_1
11/26/2009
2 2
14 2 222 2
2
1 111 1
1
1 1
–1 –1
–1
–1–1
–1
–1 –1
11/26/2009
14
0 000 0
0
0 0
1 111 1
1
1 1
2 222 2
2
2 2
(s)t FIGURA
(s)
1.2-6
Forma
de onda
de corriente
ejemplo
1.2-2.
3 333t (s)
FIGURE
1.2-6
Current
waveform
forExample
Example
1.2-2.
t(s)
(s)FIGURE
3
1.2-6
Current
waveform
for
Example
1.2-2.
ttt3
FIGURE
1.2-6
Current
waveform
fordel
Example
1.2-2.
FIGURE
1.2-6
Current
waveform
for
Example
1.2-2.
1.2-6
Current
waveform
for
1.2-2.
FIGURE
1.2-6
Current
waveform
Example
1.2-2.
3 3
t(s)
(s)t (s)FIGURE
FIGURE
1.2-6
Current
waveform
for for
Example
1.2-2.
Solución
Solution
Solution
Solution
Solution
Solution
Solution
De
la Figure
figura 1.2-6
describir i(t)
i(t) as
como
From
1.2-6,podemos
we can describe
From Figure
Figure 1.2-6,
1.2-6, we
we can
can describe
describe i(t)
i(t) as
as
From
as
From
Figure
1.2-6,
we
can
describe
i(t)
as
From
Figure
1.2-6,
can
describe
14
Electric
Circuit
Variables
From
Figure
1.2-6,
wewe
can
describe
i(t)i(t)
as
14
88
8
8
8
800<
<
<00
tt0<
<00t <
<
<
<t 00< 0
< 10 0t0t <
iiððttÞÞiiiðð¼
¼
<
�ttt11�
ðtttÞÞÞ¼
¼1 1110 <000tt<
<
�11
¼
<
i ðt Þ ¼ : :
1: 0 < t �
� 1� 1
>ttt11>
1 . 8 D E S I G N E X A M:
PttL:
Etttt >
:
>111
>
t t>1
Using
Eq.
1.2-2,
wewe
have
Using
Eq.
1.2-2,
we
have
Using
Eq.
1.2-2,
have
Using
1.2-2,
we
have
Con
laEq.
ecuación
1.2-2,
tenemos
Using
Eq.
1.2-2,
have
Using
Eq.
1.2-2,
wewe
have
ZZZ 111Z
ZZZ 333Z
ZZ 33
ZZ 11 ZZZ 333Z
ZZ 33
3 3
1 1
3 3
Electric Circuit Variables
i
ð
t
Þdt
¼
1
dt
þ
q
ð
3
Þ
�
q
ð
0
Þ
¼
JET
VALVE
CONTROLLER
q
ð
3
Þ
�
q
ð
0
Þ
¼
i
ð
t
Þdt
¼
1
dt
þ
dt
33ÞÞq�
ÞÞ ¼
iiððttÞdt
¼
1 dt
þ t dttttdt
¼
qqðð33qqÞÞðð�
ðð00qqÞÞðð00¼
¼00 iiððttÞdt
Þdt
dt
Þdt
¼ ¼00 1010dt
dt1þ
þ þ11 t1tdt
dt dt
� q�
¼
0 000
0 0
1 11
1
���1101 �����111t222���33322�����333 0 1
���1 ��� tuses
rocket
1
Wire
2���tt3t2 ��a� two-1 1
1
1 . 8 D E SAI small,
G N Eexperimental
X A M P L Espace¼
¼ t¼
t¼
¼
þ1111þ
�
Þ¼
¼
C555C
�� þ
t��� tþ
þ� ¼
¼
þ
¼
ttþ
ððð99911�
111ÞÞÞ55¼
þ
ððð999�
Þ�
C
¼
þ
�
¼
CC
���� 11¼
element circuit, as shown ¼
in ¼
to
t�Figure
¼
1
þ
�
1
Þ
¼
5
C
2
�0022þ���1.8-1,
Jet value
2
2
2
2
1
2
1
�000 þ
2
2
i
+
+
2
0 2 11 111
controller
control
a integration
jet valve from
point
oft0¼
liftoff
at
t3¼s simply
0
Alternatively,
we
note
that
of
i(t)
from
0
to
t
¼
requires
the
calculation
of
the
areaarea
under
Alternatively,
wenote
note
that
integration
ofintegración
i(t)from
from
¼de
tott3t¼
¼
ssimply
simply
requires
the
calculation
ofthe
the
area
under
Alternatively,
we
that
integration
of
i(t)
tt¼
00to
the
calculation
of
under
Alternatively,
we
note
that
integration
of
i(t)
from
tt¼
¼
ss3simply
the
calculation
of
area
under
Alternatively,
we
note
that
integration
of
i(t)
from
to
33simply
simply
requires
calculation
of
the
area
under
De
manera alternativa,
observamos
que
i(t)
de
t ss5
0 arequires
t 5requires
3 s sólo
requiere
calcular
el
área
bajo
Alternatively,
weuntil
note
that
integration
ofesa
i(t)
from
¼t0one
0¼to
to0ttminute.
¼ 3¼
requires
thethe
calculation
ofthe
the
area
under
expiration
of
the
rocket
after
Element
Element
JET
VALVE
CONTROLLER
v
v
the
curve
shown
in
Figure
1.2-6.
Then,
we
have
the
curve
shown
in
Figure
1.2-6.
Then,
we
have
the
curve
shown
in
Figure
1.2-6.
Then,
we
have
1
2
the
curve
shown
in
Figure
1.2-6.
Then,
we
have
the
curve
shown
in
Figure
1.2-6.
Then,
we
have
la
se muestra
en la1.2-6.
figura
1.2-6.
Entonces
tenemos
que1
1
2
thecurva
curveque
shown
in Figure
Then,
have by
The
energy
that
must
bewe
supplied
element
¼qqq11¼
þ11122þ
�22222�
¼22255¼
C555C
¼
þ
�
¼
C
qqq ¼
þ
�
¼
C
¼
þ
�
¼
C
for
the
one-minute
period
is
40
mJ.
Element
1
is
a
¼ 1 þ 2 � 2 ¼ 5 C Wire
A small, experimental space rocket uses a two–
–
to beinselected.
element circuit,battery
as shown
Figure 1.8-1, to
Jet
value
�t/60
i
+
+
Wire
It ispoint
known
i(t)at
¼ tDe
controller
control a jet valve from
of that
liftoff
¼ 0 mA for t � 0,
EXERCISE
1.2-1
Find
the
charge
that
has
entered
an
element
by
time
when
EXERCISE
1.2-1
Find
the
charge
that
has
entered
an
element
by
time
when
EXERCISE
1.2-1
Find
the
charge
that
has
entered
an
by
time
ttt when
(t)
¼
and
the
voltage
across
the
second
element
is
v
EJERCICIO
1.2-1
Obtenga
la
carga
que
ha
entrado
en
un
elemento
ttttcuando
EXERCISE
the
charge
has
entered
an
element
by
time
when
2that
FIGURE
1.8-1
Theelement
circuiten
to
control
EXERCISE
1.2-1
Find
charge
entered
element
by
time
when
until expiration
of22�t/60
the
rocket 1.2-1
after
one Find
minute.
Element
Element
EXERCISE
1.2-1
Find
thethe
charge
thatthat
hashas
entered
an an
bytiempo
time
when
velement
v1t <
2 4t A, t �
2
28t
28t
2
0.
2
�
Assume
q(t)
¼
0
for
0.
i
¼
8t
�
4t
A,
t
�
0.
Assume
q(t)
¼
0
for
t
<
0.
i
¼
�
4t
A,
t
�
0.
Assume
q(t)
¼
0
for
t
<
0.
i
¼
V
for
�
0.
The
maximum
magnitude
of
the
Be
i
5
8t
2
4t
A,
t
0.
Suponga
que
q(t)
5
0
para
que
t
,
0.
2
a
jet
valve
for
a
space
rocket.
tt �t 0.
q(t)
¼
tt<
1 0.
2
�
4t
�Assume
0.
Assume
0 for
<
¼
8t�
The energy ithat
must
be
supplied
by
element
� 4t
4t A,
A, A,
Assume
q(t)1q(t)
¼00¼for
for
<t0.
0.
i¼
¼i8t
8t
8 3Element
38
current,
D,
is
to222 11 is
mA.
Determine the
3limited
888 tmJ.
32t222 C
8
3
for the one-minute
period
is
40
a
Answer:
q
ð
t
Þ
¼
�
3
Answer:
qððtttÞÞÞ¼
¼ ttt 2t�
�22t
2t C
C
Answer:
Answer:
qqððttÞqqconstants
C
–
Respuesta:
Answer:
�
2t
Þð¼
¼ 33¼tt 3333�
�D2t
C BCand describe the required
required
and
battery. –
battery to beAnswer:
selected.
3 �t/60
mA forand
t � 0,the Assumptions Wire
It is knownDescribe
that i(t) ¼the
De Situation
(t)
¼
and the voltage
across
the
second
element
is
v
EJERCICIO
1.2-2
La
carga
total
que
ha
un
circuito
es q(t)
5q(t)
4 4sen
CC
cuando
EXERCISE
1.2-2
The
total
charge
thatthat
hasentrado
entered
circuit
element
is q(t)
q(t)
¼
4¼
3t
when
2total
FIGURE
1.8-1
Theen
to element
control
EXERCISE
1.2-2
The
total
charge
that
has
entered
circuit
element
q(t)
¼sin
sin
3twhen
when
EXERCISE
1.2-2
The
total
charge
has
entered
aaacircuit
element
is
4443tsin
3t
C
EXERCISE
1.2-2
The
charge
that
has
entered
aaacircuit
circuit
is
¼
3t
EXERCISE
1.2-2
The
total
charge
that
has
entered
circuit
element
isis q(t)
sin
3t
CCwhen
when
1. The current
enters
the
plus
terminal
of
the
second
element.
1.2-2
The
total
charge
that
has
entered
circuit
element
is q(t)
¼
4¼sin
sin
3t CC
when
0.
The
maximum
magnitude
of
the
Be�t/60 V forttEXERCISE
tt �
0,
y
q(t)
5
0
cuando
t

0.
Determine
la
corriente
en
este
circuito
para
que
t

0.
a
jet
valve
for
a
space
rocket.
�
0,
and
q(t)
¼
0
when
t
<
0.
Determine
the
current
in
this
circuit
element
for
t
>
0.
�and
0, and
and
q(t)0¼
¼
when
<Determine
0. Determine
Determine
the current
current
in this
this
circuit
element
for
t>
>0.
0.
ttt 0,
�
0,
q(t)
000 when
ttt 0.
<
0.
the
in
circuit
element
for
tt0.
�
q(t)
¼
when
t
<
the
current
in
this
circuit
element
for
t
>
�
0,
and
q(t)
¼
when
<
0.
Determine
the
current
in
this
circuit
element
for
>
0.
t �limited
0, and q(t)
0 when
t < 0. the
Determine the current in this circuit element for t > 0.
current, D, is
1 d¼
Determine
dmA.
2. Thetocurrent
the plus terminal of the first element.
dddleaves
d
Answer:
i
ð
t
Þ
¼
4
sin
3t
¼
12
cos
3t A
A
¼
12
cos
3t
Answer:
i
ð
t
Þ
¼
4
sin
3t
4
sin
3t
¼
12
cos
3t
A
Answer:
i
ð
t
Þ
¼
Respuesta:
sen
Answer:
i
ð
t
Þ
¼
4
sin
3t
¼
12
cos
3t
Answer:
ð¼
tÞand
¼ 4dt
4 sin
¼ required
12
3t
AA
required constants
D and
describe
battery.
Answer:
iðtÞiB
3t 3t
¼the
12
coscos
3t A
dt
dtsin
dt
dt
3. The wires
perfect and have no effect on the circuit (they do not absorb energy).
dtare
Describe the Situation and the Assumptions
Alfaomega
Eléctricos
4. The model of the circuit, as shown in Figure 1.8-1, assumes that the Circuitos
voltage across
the - Dorf
1. The current enters
plus terminal
of the
twothe
elements
is equal;
thatsecond
is, v1 ¼element.
v 2.
�t/60
2. The current 5.
leaves
plus terminal
thev1first
element.
¼ Be
V where B is the initial voltage of the battery that
Thethe
battery
voltage vof1 is
as circuit
it supplies
the valve.
M01_DORF_1571_8ED_SE_001-019.indd
4 will discharge
3. The wires are perfect
and have exponentially
no effect on the
(theyenergy
do nottoabsorb
energy).
4/12/11 5:15 PM
Sistemas de unidades
1.3
5
SISTEMAS DE UNIDADES
Para representar un circuito y sus elementos, debemos definir un sistema que conste de unidades para
las cantidades que ocurran en el circuito. En la reunión general de la Conferencia General de Pesos y
Medidas de 1960, los representantes modernizaron el sistema métrico y crearon el Système International d’Unites, más conocido como Unidades SI.
SI es el Système International d’Unités; o International System of Units.
Las unidades fundamentales, o básicas, del SI se muestran en la tabla 1.3-1. Los símbolos de
unidades que representan nombres propios (de persona) van con inicial mayúscula, los demás no. No
se usan puntos después de los símbolos, y los símbolos no tienen forma plural. Las unidades derivadas
de otras cantidades físicas se obtienen de la combinación de unidades fundamentales. La tabla 1.3-2
muestra las unidades derivadas más comunes junto con sus fórmulas en términos de unidades fundamentales o unidades derivadas anteriores. Se muestran los símbolos para las unidades que los tienen.
Tabla 1.3-1 Unidades base del SI
UNIDAD SI
CANTIDAD
NOMBRE
Longitud
metro
SÍMBOLO
m
Masa
kilogramo
kg
Tiempo
segundo
s
Corriente eléctrica
amperio
A
Temperatura termodinámica
kelvin
Cantidad de sustancia
mol
Intensidad luminosa
candela
K
mol
cd
Tabla 1.3-2 Unidades derivadas del SI
CANTIDAD
NOMBRE DE UNIDAD
FÓRMULA
SÍMBOLO
2
Aceleración – lineal
metro por segundo por segundo
m/s
Velocidad – lineal
metro por segundo
m/s
Frecuencia
hertz
s21
Fuerza
newton
Hz
2
kg  m/s
2
N
Presión o tensión
pascal
N/m
Densidad
kilogramo por metro cúbico
kg/m3
Energía de trabajo
joule (julio)
Nm
J
Potencia
watt (vatio)
J/s
W
Carga eléctrica
culombio
As
C
Potencial eléctrico
voltio
W/A
V
Resistencia eléctrica
ohmio
V/A
V
Conductancia eléctrica
siemens
A/V
S
Capacitancia eléctrica
faradios
C/V
F
Pa
Flujo magnético
weber
Vs
Wb
Inductancia
henry
Wb/A
H
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 5
Alfaomega
4/12/11 5:15 PM
E1C01_1
11/26/2009
6
Variables de circuitos eléctricos
6
6
Electric Circuit Variables
Tabla 1.3-3 Prefijos SI
MÚLTIPLO
Table 1.3-3
SI Prefixes
12
10
MULTIPLE
109
6
12
1010
103
109
1022
106
10323
10
PREFIJO
SÍMBOLO
tera
T
PREFIX
giga
GSYMBOL
mega
tera
kilo
giga
centi
mega
mili
kilo
micro
centi
26
�2
1010
29
�3
1010
10�6212
10
10�9215
10
M
k
c
m
m
nano
milli
pico
micro
n
femto
nano
f
p
T
G
M
k
c
m
m
n
10�12
pico
p
10�15
femto
f
Las unidades básicas como las de longitud en metros (m), de tiempo en segundos (s), y de corriente en amperios (A), se pueden usar para obtener unidades. Así, por ejemplo, tenemos la unidad de
carga (C)
resultado
del such
producto
de corriente
y tiempo
(Ain s).
La unidad
fundamental
para la energía
The
basic units
as length
in meters
(m), time
seconds
(s), and
current in amperes
(A) can
esbeelused
jouleto(J),
la cual
fuerzaunits.
por distancia
N  m. we have the unit for charge (C) derived from the
obtain
the es
derived
Then, foroexample,
La gran
ventajaand
deltime
sistema
quefundamental
incorpora ununit
sistema
decimal
que haya
una relación
product
of current
(A �SI
s).esThe
for energy
is para
the joule
(J), which
is force
entimes
cantidades
mayores
menores con respecto a la unidad básica. Las potencias de 10 se representan
distance
or N � om.
por los prefijos
queof
sethe
muestran
en la
tablait1.3-3.
Un ejemplo
del uso
común
un prefijo
The greatestándar
advantage
SI system
is that
incorporates
a decimal
system
for de
relating
largeresor
elsmaller
centímetro
(cm), que
es 0.01
o lapowers
centésima
parte
un metro.by standard prefixes given in
quantities
to the
basicmetros,
unit. The
of 10
are de
represented
multiplicador
decimal
siempre
debe
acompañar
lasthe
unidades
apropiadas
y nunca
se escribe
TableEl1.3-3.
An example
of the
common
use
of a prefixa is
centimeter
(cm), which
is 0.01
meter.
solo. DeThe
manera
quemultiplier
podemosmust
escribir
2 500
W comothe
2.5appropriate
kW. Del mismo
modo,
podemos
decimal
always
accompany
units and
is never
writtenescribir
by itself.
0.012
como
mA. 2500 W as 2.5 kW. Similarly, we write 0.012 A as 12 mA.
Thus,Awe
may12write
Ejem
. 3 - 1 Unidades del SI
E pXlAoM 1
P L E 1 . 3 - 1 SI Units
Una masa de 150 gramos experimenta una fuerza de 100 newtons. Obtenga la energía o trabajo que se consumió
A mass of 150 grams experiences a force of 100 newtons. Find the energy or work expended if the mass moves 10
si la masa se movió 10 centímetros. Además, obtenga la potencia si la masa completa su movimiento en 1 milicentimeters. Also, find the power if the mass completes its move in 1 millisecond.
segundo.
Solution
Solución
energy
is found ascomo
LaThe
energía
se encuentra
energía
5 fuerza
3�distancia
energy
¼ force
distance5¼100
1003�0.1
0:15
¼10
10JJ
Observe
que we
la distancia
utilizó in
en units
unidades
de metros.
potencia
se encuentra
Note that
used the se
distance
of meters.
The La
power
is found
from a partir de
energy
energía
power
potencia
 ¼ time period
periodo de tiempo
�3
23
s.
Thus,
where
the
time
period
is
10
donde el periodo de tiempo es 10 s. Por consiguiente,
10
power 5
¼ �3 ¼
W5
¼10
10kW
kW
potencia
5 1044 W
10
24
EJERCICIO
tresthree
corrientes
i1 5i45¼mA,
i2 5 0.03
mA, emA,
i3 5and
25 3i 10
A,
EXERCISE 1.3-1
1.3-1 ¿Cuál
Whichdeoflasthe
currents,
45 mA,
i2 ¼ 0.03
1
3 ¼ 25 �
es10
la�4más
A,grande?
is largest?
la más grande.
Respuesta:
Answer: i i3iseslargest.
3
Alfaomega
M01_DORF_1571_8ED_SE_001-019.indd 6
Circuitos Eléctricos - Dorf
4/12/11 5:15 PM
E1C01_1
11/26/2009
7
Potencia
y energía
Power and
and
Energy
Power
Energy
1.4
1.4
1.4
V O LTA J E
VO
OL
LT
TA
AG
GE
E
V
–
7
7
7
vba
+
Las variables básicas en un circuito eléctrico son la corriente y el voltaje. Estas variables descriThe
basic
variables
in
an electrical
electrical
circuit are
aredecurrent
current
and voltage.
voltage.
These
variables
+
vba
The
basic
an
circuit
and
These
variables
ben el
flujovariables
de carga in
a través
de los elementos
un circuito
y la energía
requerida
para que la a––
+
v
ba
describe
the
flow
of
charge
through
the
elements
of
a
circuit
and
the
energy
required
to
describe
the La
flow
of charge
elementspara
of adescribir
circuit and
energy
required
to en esta
carga fluya.
figura
1.4-1 through
muestra the
la notación
un the
voltaje.
Hay
dos partes
+
vab
–b
aa
cause
charge
to flow.
flow.
Figure
1.4-1 shows
shows
the variable
notationdewe
we
use to
to ydescribe
describe
voltage.
b
cause
charge
to
Figure
1.4-1
the
notation
use
aa voltage.
notación:
un valor
(quizá
representado
por una
nombre)
una dirección
asignada. El
There
are
two
parts
to
this
notation:
a
value
(perhaps
represented
by
a
variable
name)
+
v
–
FIGURA
1.4-1
ab
There
areuntwo
partspuede
to thisser
notation:
(perhaps
represented
a variable
name) (1, 2). +
valor de
voltaje
positivoaovalue
negativo.
La dirección
se laby
dan
sus polaridades
vab
–
a través de un
and
anuna
assigned
direction.
The value
value
of
voltage
may be
beocurre
positive
or negative.
negative.
The
and
an
assigned
direction.
The
aa voltage
may
positive
or
The
Como
cuestión
de vocabulario,
se of
dice
que un voltaje
a través
de un elemento.
La Voltaje
FIGURE
1.4-1
Voltage
FIGUREelemento
1.4-1 Voltage
del circuito.
direction
of muestra
voltageque
is given
given
byformas
its polarities
polarities
(þ, el
�).
As aa amatter
matter
ofdevocabulary,
vocabulary,
weEl voltaje
direction
of
aa voltage
is
by
its
(þ,
�).
As
we
figura 1.4-1
hay dos
de marcar
voltaje
travésof
un elemento.
across aa circuit
circuit element.
element.
across
say
that
voltage exists
exists
acrossrequerido
an element.
element.
Figure
1.4-1
showspositiva
that there
there
are
two ways
waysa a la b. Por otra
say
that
aa voltage
across
an
shows
that
two
vba es
proporcional
al trabajo
paraFigure
mover1.4-1
una carga
deare
la terminal
is
proportional
to
the
work
required
to move
move baa
to
label
the
voltage
across
an
element.
The
voltage
v
ba
proportional
the work
required
to
to
labelelthe
voltage
across
an element.
The voltage
vba is
parte,
voltaje
vab es
proporcional
al trabajo
requerido
para
mover unatocarga
positiva
de la terminal
is
proportional
tomisthe
positive
charge
from
terminal
a
to
terminal
b.
On
the
other
hand,
the
voltage
v
ab
is proportional
the
positive
fromvba
terminal
a to terminal
b. On
other hand,
therespecto
voltageavla
se lee como
“el voltaje
de the
la terminal
b con
a”. Delto
a la a. Encharge
ocasiones
ab terminal
as
‘‘the
work
required
to
move
a
positive
charge
from
terminal
b
to
terminal
a.
We
sometimes
read
v
ba
as ‘‘the
work
required
to puede
move aleer
positive
fromenterminal
b to aterminal
a. Weasometimes
vba manera
mo modo,
vab se
comocharge
“el voltaje
la terminal
con respecto
la terminalread
b”. De
can be
bea read
read
asLos
‘‘thevoltajes
voltagev at
at terminal
terminal
voltage
at terminal
terminal
with
respect
toesterminal
terminal
a.’’
Similarly,
ab can
as
‘‘the
voltage
voltage
at
a.’’
Similarly,
vvab
alternativa,
a veces bbsewith
dicerespect
que vbato
el voltaje
que
va de la terminal
a la b.
vba son
ab y
is the
the voltage
voltage
drop from
from
with respect
respect
todiferentes.
terminal Tienen
b.’’ Alternatively,
Alternatively,
we sometimes
sometimes
say that
thatpolaridades.
ba is
drop
aasemejantes
with
terminal
b.’’
we
say
vvba
peroto
la misma magnitud
pero diferentes
Eso significa
que
terminal
a
to
terminal
b.
The
voltages
v
and
v
are
similar
but
different.
They
have
the
same
ab
ba
terminal a to terminal b. The voltages vab and vba are similar but different. They have the same
magnitude but
but different
different polarities.
polarities. This
This means
means
that
vab that
5 2vba
magnitude
b
ab ¼ �vba
Si se considera vba, la terminal b se denominavv“terminal
ab ¼ �vba1” y la terminal a se denomina “terminal 2”.
Por otra parte, cuando nos referimos a vab, la terminal a se denomina la “terminal 1” y la terminal b
When considering
considering vba,, terminal
terminal bb is
is called
called the
the ‘‘þ
‘‘þ terminal’’
terminal’’ and
and terminal
terminal aa is
is called
called the
the ‘‘�
‘‘�
When
es la “terminal 2”. vba
,
terminal
a
is
called
the
‘‘þ
terminal’’
and
terminal.’’
On
the
other
hand,
when
talking
about
v
ab
terminal.’’ On the other hand, when talking about vab, terminal a is called the ‘‘þ terminal’’ and
terminal bb is
is called
called the
the ‘‘�
‘‘� terminal.’’
terminal.’’
terminal
El voltaje que pasa a través de un elemento es el trabajo (energía) que se requiere para mover una
unidad
de carga
positiva
de lawork
terminal
2 arequired
la terminal
1. Laaunidad
de voltaje
es
The
voltage
across
an element
element
is the
the
(energy)
to move
move
unit positive
positive
charge
The
voltage
across
an
is
work (energy)
required to
a unit
charge
el
voltio,
V.
from the
the �
� terminal
terminal to
to the
the þ
þ terminal.
terminal. The
The unit
unit of
of voltage
voltage is
is the
the volt,
volt, V.
V.
from
La ecuación
queacross
pasa athe
través
del elemento
es
The
equationpara
for el
thevoltaje
voltage
element
is
The
equation
for
the
voltage
across the
element
is
dw
¼ dw
vv ¼
dq
dq
ð1:4-1Þ
(1.4-1)
ð1:4-1Þ
where
is
voltage,
w is
is
energy
(or work),
work),
and qq is
isycharge.
charge.
A
charge
ofcarga
coulomb
delivers an
an
energyuna
of
donde vvv is
esvoltage,
el voltaje,
w energy
es la energía
(o trabajo),
q es la A
carga.
Una
de 1 culombio
entrega
where
w
(or
and
charge
of
11 coulomb
delivers
energy
of
1
joule
as
it
moves
through
a
voltage
of
1
volt.
de it1 joule
moverseaavoltage
través de
voltaje de 1 voltio.
1energía
joule as
movesal through
ofun
1 volt.
1.5
1.5
1.5
POTENCIA Y ENERGÍA
P
OW
WE
ER
R A
AN
ND
D E
EN
NE
ER
RG
GY
Y
PO
La potencia y la energía que se entregan a un elemento tienen una gran importancia. Por ejemplo, la
The
power
and
energy
delivered
to an
an element
element
arepuede
of great
great
importance.
For example,
example,
the useful
useful
output
salida
usualand
de energy
una bombilla
eléctrica,
o foco, se
expresar
en términos
de potencia.
Un foco
de
The
power
delivered
to
are
of
importance.
For
the
output
of
an
electric
lightbulb
can
be
expressed
in
terms
of
power.
We
know
that
a
300-watt
bulb
delivers
300anwatts
(vatios)
proporciona
más luz que
de of
100
watts.We know that a 300-watt bulb delivers
of
electric
lightbulb
can be expressed
in uno
terms
power.
more light
light than
than aa 100-watt
100-watt bulb.
bulb.
more
Potencia
periodo
o absorción
de energía.
Power
is es
theeltime
time
rate de
of gasto
expending
or absorbing
absorbing
energy.
Power
is
the
rate
of
expending
or
energy.
Thus,
we
have
the equation
equation
Por lo we
tanto,
tenemos
la ecuación
Thus,
have
the
dw
¼ dw
pp ¼
dt
dt
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 7
ð1:5-1Þ
(1.5-1)
ð1:5-1Þ
Alfaomega
4/12/11 5:15 PM
Electric
Circuit
Variables
Variables
de circuitos
eléctricos
8
8
a
i
+
b
vv
donde
potencia
en watts,
es energía
en joules,
t es tiempo
en segundos.
potencia
where pp es
is power
in watts,
w iswenergy
in joules,
and t isy time
in seconds.
The powerLaassociated
asociada
la carga
fluyean
a través
de is
un elemento es
with the con
charge
flow que
through
element
–
p¼
(a)
i
dw dw dq
¼
�
¼v�i
dt
dq dt
(1.5-2)
ð1:5-2Þ
A partir de la ecuación 1.5-2 vemos que potencia es simplemente el producto del voltaje
From Eq. 1.5-2, we see that the power is simply the product of the voltage across an
a
través
de un elemento por la corriente mientras la corriente fluye a través del elemento. La
–
vv
+
element times the current through the element. The power has units of watts.
potencia tiene unidades de watts.
Two circuit variables are assigned to each element of a circuit: a voltage and a current.
(b)
A cada elemento de un circuito se asignan dos variables de circuito: un voltaje y una
Figure 1.5-1 shows that there are two different ways to arrange the direction of the current
FIGURA
(a) The
La
corriente. La figura 1.5-1 muestra que hay dos diferentes maneras de arreglar la dirección de
FIGURE 1.5-1
1.5-1 (a)
and the polarity of the voltage. In Figure 1.5-1a, the current enters the circuit element at the
convención
pasiva
se
la corriente y la polaridad del voltaje. En la figura 1.5-1a, la corriente entra en el circuito en la
passive convention is
þ terminal of the voltage and exits at the � terminal. In contrast, in Figure 1.5-1b, the
usa
voltaje y
usedpara
for element
terminal 1 del voltaje y sale en la terminal 2. Por el contrario, en la figura 1.5-b, la corriente
current enters the circuit element at the � terminal of the voltage and exits at the þ terminal.
corriente
del
elemento.
voltage and current. (b) entra en el elemento del circuito en la terminal 2 del voltaje y sale en la terminal 1.
First, consider Figure 1.5-1a. When the current enters the circuit element at the þ
(b)
se usaconvention
la
TheNo
passive
Primero veamos la figura 1.5-1a. Cuando la corriente entra en el elemento de circuito en
convención pasiva.
terminal
of the voltage and exits at the � terminal, the voltage and current are said to ‘‘adhere
is not used.
la terminal 1 del voltaje y sale en la terminal 2, se dice que el voltaje y la corriente se “apegan
to the passive convention.’’ In the passive convention, the voltage pushes a positive charge in
a la convención pasiva”. En ella, el voltaje impulsa una carga positiva en la dirección indicada
the direction indicated by the current. Accordingly, the power calculated by multiplying the
por
la corriente.
la potencia
element
voltageSegún
by theesto,
element
currentcalculada al multiplicar el voltaje del elemento por la
corriente del elemento
p ¼pvi5 vi
a
b
is
power absorbed
by por
the element.
(This
power
is also también
called ‘‘the
power received
by the
element’’
esthe
la potencia
absorbida
el elemento.
(Esta
potencia
se denomina
“potencia
recibida
por
and
‘‘the powerLa
dissipated
the element.’’)
The power
absorbed
by anpositiva
elementcomo
can be
either positive
el elemento”).
potencia by
absorbida
por un elemento
puede
ser tanto
negativa,
lo cual
or
negative.deThis
will depend
on they values
of thedelelement
voltage and current.
dependerá
los valores
del voltaje
la corriente
elemento.
Next,
consider Figure
the Aquí
passive
convention
hasla not
been used.
Instead,
the
A continuación,
veamos1.5-1b.
la figuraHere
1.5-1b.
no se
ha utilizado
convención
pasiva.
En camcurrent
enters
the
circuit
element
at
the
�
terminal
of
the
voltage
and
exits
at
the
þ
terminal.
In
this
bio, la corriente entra al elemento de circuito en la terminal 2 del voltaje y sale en la terminal 1. En
case,
the voltage
pushes
a positive
charge
in theendirection
opposite
to the
by the
este caso,
el voltaje
impulsa
una carga
positiva
la dirección
opuesta
a ladirection
indicada indicated
por la corriente.
current.
Accordingly,
when
the
element
voltage
and
current
do
not
adhere
to
the
passive
convention,
Por consiguiente, cuando el voltaje y la corriente del elemento no se apegan a la convención pasiva, la
the
powercalculada
calculated
multiplying
the element
bypor
thelaelement
current
is the power
potencia
al by
multiplicar
el voltaje
de del voltage
elemento
corriente
del elemento
es lasupplied
potencia
by
the
element.
The
power
supplied
by
an
element
can
be
either
positive
or
negative,
depending
on
alimentada por el elemento. La potencia alimentada por un elemento puede ser positiva o negativa,
the
values
of
the
element
voltage
and
current.
dependiendo de los valores del voltaje y la corriente del elemento.
The
power absorbida
absorbed por
by an
element en
and
the power
supplied
same
elementestán
are
La potencia
n elemento
la energía
alimentada
porbyesethat
mismo
elemento
related
by
relacionadas por
powerabsorbida
absorbed5¼2�power
potencia
potenciasupplied
alimentada
The
the se
passive
convention
aredesummarized
in Table
When
the element
voltage and
En larules
tablafor
1.5-1
resumen
las reglas
la convención
pasiva.1.5-1.
Cuando
el voltaje
y la corriente
del
current
adhere
to theapassive
convention,
thelaenergy
bypor
an un
element
can se
bepuede
determined
from
elemento
se apegan
la convención
pasiva,
energíaabsorbed
absorbida
elemento
determinar
Table 1.5-1 Power Absorbed or Supplied by an Element
Tabla 1.5-1 Potencia absorbida o alimentada por un elemento
POWER ABSORBED BY AN ELEMENT
POTENCIA ABSORBIDA POR UN ELEMENTO
a a
i i
+ +
b b
v v
– –
POWER SUPPLIED BY AN ELEMENT
POTENCIA ALIMENTADA POR UN ELEMENTO
i
i
a
a
–
v
b
b
–+
v
+
Because the reference directions of
las direcciones
de referencia
vDado
and ique
adhere
to the passive
de v e i se apegan
a la convención
convention,
the power
pasiva, la potencia
Because the reference directions of
Dado que lasv direcciones
referencia
and i do notde
adhere
to the
de v e i no sepassive
apeganconvention,
a la convención
the power
pasiva, la potencia
p 5 vi
is the power absorbed by the
es la potencia absorbida por el
element.
elemento.
p 5 vi
is the power supplied by the
es la potencia alimentada por el
elemento. element.
p ¼ vi
Alfaomega
M01_DORF_1571_8ED_SE_001-019.indd 8
p ¼ vi
Circuitos Eléctricos - Dorf
4/12/11 5:15 PM
E1C01_1
11/26/2009
9
1.8
DESIGN EXAMPLE
JET VALVE CONTROLLER
Power and Energy
9
Potencia
y energía
Power and
Energy
99
A small, experimental space rocket uses a twoPower andWire
Energy
9
element circuit, as shown in Figure 1.8-1, to
Jet value
i
+
+
Eq.
1.5-1
by rewriting
ascontrol acomo
controller
por la
ecuación
1.5-1 alitreescribirla
jet valve from point of liftoff at t ¼ 0
Eq.
Eq. 1.5-1
1.5-1 by
by rewriting
rewriting it
it as
asuntil expiration ofdwthe
¼ rocket
dt
ð1:5-3Þ
dw 5
ppp dt
dt after one minute.
(1.5-3) Element v2
v1 Element ð1:5-3Þ
dw
¼
1
2
dw
¼
p
dt
ð1:5-3Þ
The
energy
that
must
be
supplied
by
element
1
On
integrating,
we have
Al integrarla
tenemos
On
integrating,
we
have
On integrating, we have for the one-minute period
Z t is 40 mJ. Element 1 is a
–
–
Z
battery to be selected.
w ¼ Z tt p dt �t/60
ð1:5-4Þ
(1.5-4)
Wire
w
¼
pp dt
ð1:5-4Þ
mA for t � 0,
It is known
i(t) ¼
De
�1
w that
¼ �1
dt
ð1:5-4Þ
and the voltage across the
�1 second element is v2(t) ¼
FIGURE 1.8-1 The circuit to control
If the element receives power
onlyVfor
� 0.t0 The
andmaximum
we let t0 magnitude
¼ 0, then of
wethe
have
for tttt �
Be�t/60
Si the
el elemento
sólo
recibe
potencia
para
tt00yand
obtenemos
t
5
0,
entonces
tenemos
a jet valve for a space rocket.
If
element
receives
power
only
for
we
let
t
¼
0,
then
we
have
0
Z we
t
If the element receives power only for t � t0 and
let t00 ¼ 0, then we have
Z
current, D, is limited
to
t 1 mA. Determine the
Z
w ¼ t p dt
ð1:5-5Þ
(1.5-5)
¼
ð1:5-5Þ
required constantsw
Bdt
and describe the required battery.
0 p
wD
¼ and
p
dt
ð1:5-5Þ
0
0
Describe the Situation and the Assumptions
plus terminal
of the second
element.
E1.X AThe
M Pcurrent
L E 1 . enters
5 - 1 the
Electrical
Power
and Energy
E
5
1
Potencia
y
energía
eléctricas
E
X jAeMmPpLlEo 11.
5
1
Electrical
Power
and
Energy
E2.X AThe
M Pcurrent
L E 1 . leaves
5 - 1 the
Electrical
Power
andelement.
Energy
plus terminal
of the first
3. shown
The wires
are perfect
have
do not absorbed
absorb energy).
Let us consider the element
in Figure
1.5-1aand
when
v ¼no4 effect
V andon
i ¼the
10 circuit
A. Find(they
the power
by the
Let
us
consider
the
element
shown
in
Figure
1.5-1a
when
vv ¼
44cuando
V
and
ii ¼
10
A.
Find
the
power
absorbed
by
Consideremos
el
elemento
que
se
muestra
en
la
figura
1.5-1a
v
5
4
V
e
i
5
10
A.
Obtenga
la potencia
Let
us
consider
the
element
shown
in
Figure
1.5-1a
when
¼
V
and
¼
10
A.
Find
the
power
absorbed
by the
the
element and the energy absorbed
over
a
10-s
interval.
4.y laThe
model
of10-s
the interval.
circuit,
as un
shown
in Figure
1.8-1,
assumes that the voltage across the
element
and
the
energy
absorbed
over
a
absorbida
por
el
elemento
energía
absorbida
durante
intervalo
de
10
s.
element and the energy absorbed over a 10-s interval.
two elements is equal; that is, v1 ¼ v2.
Solution
�t/60
Solución
Solution
V where B is the initial voltage of the battery that
5. element
The battery
Solution
The
power absorbed by the
is voltage v1 is v1 ¼ Be
La potenciaabsorbed
absorbida porthe
el elemento
es exponentially as it supplies energy to the valve.
The
is
will discharge
The power
power absorbed by
by the element
element
is
p ¼ vi ¼ 4 � 10 ¼ 40 W
pp 5
vi 5
4  10
10 5 40
40 W
W
¼
p¼
¼ vi
vifrom
¼ 44 t�� ¼
100¼
¼to40
The circuit
t ¼W
60 s.
The energy absorbed by 6.
the element
is operates
La energía absorbida
por el
elemento es
The
the
Zis
Z 10
The energy
energy absorbed
absorbed by
by 7.
the element
element
is
10
The current
D � 1 mA.
Z is limited,
Z so
10
w ¼ Z 10
10 p dt ¼ Z 10 40 dt ¼ 40 � 10 ¼ 400 J
w
w¼
¼ 00 pp dt
dt ¼
¼ 00 40
40 dt
dt ¼
¼ 40
40 �� 10
10 ¼
¼ 400
400 JJ
State the Goal
0
0
Determine the energy supplied by the first element for the one-minute period and then select
the constants D and B. Describe the battery selected.
E X A M P L Generate
E 1 . 5 - 2 a Plan
Electrical Power and the Passive Convention
je
2Electrical
PotenciaPower
eléctrica
la convención
pasiva
E
Pm
L Ep l1o. 51.
and
the
Passive
Convention
E XX AAEM
M P L First,
E 1 . find
5 -- 2
25v1-(t)
Electrical
andythe
the
Passive
and i(t) and Power
then obtain
power,
p1(t), Convention
supplied by the first element. Next,
using p1(t), find the energy supplied for the first 60 s.
Consider the element shown in Figure 1.5-2. The current i and voltage vab adhere to the passive convention, so the
Consider
the
shown
Figure
1.5-2.
The
and
voltage
adhere
to the
passive
convention,
so
Considere
el element
elemento
que
sein
en la figura
1.5-2.iiLa
y el
voltaje
vab se
apegan
a la convención
Consider
the
element
shown
inmuestra
Figure
The current
current
andcorriente
voltage vvi ab
passive
convention,
so the
the
power
absorbed
by this
element
is 1.5-2.
ab adhere to the
GOAL absorbida
EQUATION
NEED
INFORMATION
power
absorbed
by
this
element
is
pasiva,
por
lo
tanto,
la
potencia
por
este
elemento
es
power absorbed by this element is
2 � ð�4Þ ¼ �8 W
power absorbed ¼ i � vab Z¼ 60
potencia
absorbida
5i i� vab
vab¼522� ð�4
1242
5 28
W
v1 and i known except for
The energy
w1 for
the
ÞÞ ¼
power
absorbed
¼
¼
2 �pð�4
¼ �8
�8 W
W (t)
power
absorbed
¼w
i � v¼
ab
1
1
The
current
i
and
voltage
v
do
not
adhere
to
the
passive
convention,
power
supplied
by this
element
is
first
60
s
constants
D and
Bpor este
1 ðt Þ dtso the p
ba
La corriente
y elvoltage
voltaje vvba
no not
se apegan
atolathe
convención
lothe
tanto
la potencia
alimentada
ba do
The
current
iiiand
adhere
passive
convention,
power
supplied
by
this
element
is
0pasiva, porso
The
current
and
voltage
v
do
not
adhere
to
the
passive
convention,
so
the
power
supplied
by
this
element
is
ba
power
supplied
¼
i
�
v
¼
2
�
ð
4
Þ
¼
8
W
ba
elemento es
power
supplied
¼
i
�
v
¼
2
�
ð
4
Þ
¼
8
W
ba
power alimentada
supplied ¼5
i �ivbavab
¼5
2 �2ð4 Þ142
¼ 85W
potencia
8W
As expected
As
expected
Act
on
the
Plan
Como
se esperaba
As
expected
power absorbed ¼ �power supplied
First, wepotencia
need
p1(t),
so we first
calculate alimentada
power
absorbed
�power
absorbida
power
absorbed5¼
¼2potencia
�power� supplied
supplied
��
�
p1 ðtÞ ¼ iv1 ¼ De�t/60 � 10�3 A Be�t/60 V
–
vba = 4 V
+
�t/30
� 10�3 W ¼ DBe�t/30 mW
i=2A –
vba = 4 V
+ ¼ DBe
a
a
a
i=2A
i=2A
–
vba = 4 V
+
+
+
+
vab = –4 V
vab = –4 V
vab = –4 V
–
–
–
b
b
b
FIGURE 1.5-2 The element
FIGURA 1.5-2
El elemento
FIGURE
1.5-2
The
element
considered
in Example
1.5-2.
FIGURE 1.5-2
The element
consideradoinenExample
el ejemplo
1.5-2.
considered
1.5-2.
considered in Example 1.5-2.
Now let us consider an example when the passive convention is not used. Then p ¼ vi is the
Ahoraletveamos
un ejemplo
de cuando nothe
se utiliza laconvention
convención pasiva.
Entonces pp¼5viviisesthe
la
Now
an
let us
usbyconsider
consider
an example
example when
when the passive
passive convention is
is not
not used.
used. Then
Then p ¼ vi is the
powerNow
supplied
the element.
potencia
alimentada
porelement.
el elemento.
power
supplied
by
the
power supplied by the element.
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 9
Alfaomega
4/12/11 5:15 PM
JET VALVE CONTRO
A small, experimental space rocket uses a twoelement circuit, as shown in Figure 1.8-1, to
control a jet valve from point of liftoff at t ¼ 0
until expiration of the rocket after one minute.
10
Variables
de circuitos
eléctricos
v1
10
Electric
Circuit
Variables
10
Electric
Circuit
Variables
The energy that must be supplied by element 1
for the one-minute period is 40 mJ. Element 1 is a
battery
jP
ePm
1.
5 - 3Power,
Potencia,
energía
la convención
pasivato be selected.
E l11o..55
Power,
Energy,
andythe
the
Passive Convention
Convention
EEXXAAEMM
LLEp
--33
Energy,
and
Passive
It is known that i(t) ¼ De�t/60 mA for t � 0,
and the voltage across the second element is v2(t) ¼
FIG
�t V and i ¼ 20e
�t A for t � 0.
2t �t
Consider
the
circuit
shown
inmuestra
Figure1.5-3
1.5-3
with
¼1.5-3
8e�t
Find
theVque
power
supplied
by
for tt �supplied
0.
maximum
magnitude of the
Be
Considerethe
elcircuit
circuito
que se
en lawith
figura
con
v5
V
e ifor
5t 20e
A�t/60
para
0. The
Obtenga
a je
Consider
shown
in
Figure
vv¼
8e
V
and
i ¼8e
20e
A
� 0.2tFind
the
power
by
this
elementalimentada
andthe
theenergy
energy
supplied
bythe
theyelement
element
over
thefirst
firstsecond
second
of
operation.
Weassume
assume
that
andiito
are
current,
is
limited
la potencia
por supplied
este
elemento
la energía
alimentada
por elof
elemento
durante
el D,
primer
segundo
de1 mA. Determine the
this
element
and
by
over
the
operation.
We
that
vvand
are
zero
for tt<
<Supongamos
0.
required constants D and B and describe the required batter
operación.
que v e i son cero para que t , 0.
zero
for
0.
aa
––
Describe the Situation and the Assumptions
ii
bb
++
v
vv
1. The current enters the plus terminal of the second elem
FIGURA
1.5-3
Elemento
fluye
FIGURE 1.5-3
1.5-3 An
An
elementcuya
withcorriente
thecurrent
current
FIGURE
element
with
the
hacia
lainto
terminal
con un signo
voltaje voltage
negativo.
flowing
into
theterminal
terminal
withaade
negative
voltage2.
sign.
flowing
the
with
negative
sign.
The current leaves the plus terminal of the first elemen
3. The wires are perfect and have no effect on the circuit
Solución
Solution
Solution
La potencia
alimentada
The
power supplied
supplied
The
power
isis es
2t
2t
22t
�t B A20e�t
�2t
5vivi
vi ¼
5
5160e
160e�2t
W
¼
¼ ðA8e
8e�t
ð8e
20e�t
¼
160e
W
ppp¼
ÞÞðð20e
ÞÞB¼
W
This
element isisestá
providing
energy to
toenergía
the charge
charge
flowing
through
it.
Este element
elemento
proporcionando
a la carga
quethrough
fluye a it.
través
de él.
This
providing
energy
the
flowing
The
energy
supplied
during
the
first
second
is
La
energía
alimentada
durante
el
primer
segundo
es
The energy supplied during the first second is
ZZ 11 �
ZZ 11
�
�
�2t�
p
dt
¼
160e�2t
dt
w
¼
p dt ¼
160e
dt
w ¼
00
4. The model of the circuit, as shown in Figure 1.8-1, ass
two elements is equal; that is, v1 ¼ v2.
5. The battery voltage v1 is v1 ¼ Be�t/60 V where B is the
will discharge exponentially as it supplies energy to th
6. The circuit operates from t ¼ 0 to t ¼ 60 s.
00
7. The current is limited, so D � 1 mA.
�
�
�
�
�
�
�
�
�
160
e �� 160 �2
�2
¼
e�2 � 1 ¼ 80 1 � e�2
¼ 160
160
¼ 69:2
69:2JState
J
¼
¼
the Goal
�2��0¼ �2
�2 e � 1 ¼ 80 1 � e
�2
0
Determine the energy supplied by the first element for the
the constants D and B. Describe the battery selected.
�
�2t���11
e�2t
Generate a Plan
First, find v1(t) and i(t) and then obtain the power, p1(t), s
4 Energía
un rayo
Energy
in aaen
Thunderbolt
EEXXAAEMMjPPeLLmEEp11l .o.55-1-4.45 -Energy
in
Thunderbolt
using p1(t), find the energy supplied for the first 60 s.
GOAL
EQUATION
4
4
The
average current
current in
in aen
a typical
typical
lightning
thunderbolt
isA,
104duración
A, and
and its
its
typical
duration
0.1 ss (Williams,
(Williams,
La corriente
promedio
un rayo
común es
de 2 3 10is
y�10
su
suele
ser de
0.1 s (Williams,
1988). El
The
average
lightning
thunderbolt
22�
A,
typical
duration
isis 0.1
Z
8
88
The
energy
w1 for
theenergía
V.carga
Determine
thetotal
total
charge
transmitted
tothe
the
1988).
The
voltage
between
theclouds
clouds
and
the10
ground
�10
10 la
voltajeThe
entre
las nubes
y el the
suelo
es deand
53
V. Determine
total transmitida
a la transmitted
tierra
y la
V.
Determine
the
charge
to
1988).
voltage
between
the
ground
isis55�
w
¼
1
first 60 s
earth
and the
the energy
energy released.
released.
liberada.
earth
and
0
Solución
Solution
Solution
La carga
es isis
The
total total
charge
The
total
charge
60
p1 ðtÞ dt
00
00
00
First, we need p1(t), so we first calculate
�
p1 ðtÞ ¼ iv1 ¼ De�t/60 � 10�3 A
¼ DBe�t/30 � 10�3 W ¼ D
00
EJERCICIO1.5-1
1.5-1 Figure
La figura
E 1.5-1
muestra
cuatroelements
elementosidentified
de circuito
EXERCISE
1.5-1
Figure
1.5-1
shows
fourcircuit
circuit
elements
identified
byidentificados
theletters
lettersA,
A,con
B,
EXERCISE
EE1.5-1
shows
four
by
the
B,
lasand
letras
C,
and
D.A, B,C y D.
C,
D.
¿Cuál de
alimenta
12 W?
(a) Which
(a)
Which
oflos
thedispositivos
devices supply
supply
12 W?
W?
(a)
of
the
devices
12
(b) Which
¿Cuál de
absorbe
12 W?
(b)
Which
oflos
thedispositivos
devices absorb
absorb
12 W?
W?
(b)
of
the
devices
12
M01_DORF_1571_8ED_SE_001-019.indd 10
p1(t)
Act on the Plan
ZZ 0:1
ZZ 0:1
0:1
0:1
i
ð
t
Þ
dt
¼
�10
1044 dt
dt ¼
¼ 22�
�10
1033 CC
Q
¼
iðtÞ dt ¼
22�
Q¼
The
total energy
energy
releasedesisis
La energía
total liberada
The
total
released
ZZ 0:1
ZZ 0:1
0:1
0:1 �
�
����
��
i
ð
t
Þ
�
v
ð
t
Þ
dt
¼
�10
1044 55�
¼ 11TJ
TJ
�10
1088 dt
dt ¼
¼ 10
101212 JJ ¼
w
¼
iðtÞ � vðtÞ dt ¼
22�
w¼
Alfaomega
NEED
Circuitos Eléctricos - Dorf
4/12/11 5:15 PM
Análisis y diseño de circuitos
11
(c) ¿Cuál es el valor de la potencia recibida por el dispositivo B?
(d) ¿Cuál es el valor de la potencia entregada por el dispositivo B?
(e) ¿Cuál es el valor de la potencia entregada por el dispositivo D?
+
4V
–
–
2V
+
+
6V
–
–
3V
3A
6A
2A
4A
(A)
(B)
(C)
(D)
+
FIGURA E 1.5-1
Respuestas: (a) B y C, (b) A y D, (c) 212 W, (d) 12 W, (e) 212W
1.6
ANÁLISIS Y DISEÑO DE CIRCUITOS
El análisis y diseño de circuitos eléctricos son las actividades primarias que se describen en este libro,
a la vez que son las habilidades propias de un ingeniero electricista. El análisis de un circuito tiene
que ver con el estudio metódico del circuito dado diseñado para obtener la magnitud y dirección de
una o más variables de circuitos, como una corriente o el voltaje.
El proceso del análisis empieza con una exposición del problema, y por lo común se incluye un
modelo de circuito dado. El objetivo es determinar la magnitud y la dirección de una o más variables
de circuito, y la tarea final es verificar que la solución propuesta sea la correcta. Suele suceder que el
ingeniero identifique primero lo que se conoce y los principios que utilizará para determinar la variable desconocida.
En la figura 1.6-1 se muestra el método que se seguirá a lo largo de este libro para la solución
del problema. Por lo general se da el planteamiento del problema. Entonces el proceso de análisis
se mueve de manera secuencial pasando por las cinco etapas que se muestran en la figura 1.6-1.
En la primera se describen la situación y los supuestos. En la segunda se establecen los objetivos y
Problema
Plantea el problema.
Situación
Describe la situación y
los supuestos.
Objetivo
Establece los objetivos y
requerimientos.
Plan
Acción
Verificación
Incorrecta
Actúa sobre el plan.
Verifica que la solución
propuesta sea la correcta.
Correcta
Solución
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 11
Genera un plan para obtener
una solución del problema.
Se informa sobre la solución.
FIGURA 1.6-1 El método para la solución del problema.
Alfaomega
4/12/11 5:15 PM
Variables de circuitos eléctricos
12
requerimientos, y por lo común se registra la variable de circuito requerida que se ha de determinar.
La tercera etapa es para generar un plan que ayudará a obtener la solución del problema. La cuarta
etapa implica efectuar las actividades que se planearon y hacer un seguimiento a las etapas descritas
en el plan. En la etapa final se verifica que la solución propuesta sea la correcta. Si lo es, se informa
sobre la solución y se registra por escrito o se presenta de viva voz. Si la etapa de verificación indica
que la solución propuesta no es la correcta o es inadecuada, entonces se vuelve a las etapas del plan,
para reformular un plan mejorado y se repiten las etapas 4 y 5.
Para ilustrar este método analítico se propone un ejemplo. En el ejemplo 1.6-1 utilizaremos las
etapas descritas en el método de solución de problemas de la figura 1.6-1.
E j e m p l o 1. 6 - 1 El método formal para la solución del problema
Un experimentador en un laboratorio supone que un elemento está absorbiendo potencia y utiliza un voltímetro y
un amperímetro para medir el voltaje y la corriente como se muestra en al figura 1.6-2. Las mediciones indican que
el voltaje es v 5 112 V y que la corriente es i 5 22 A. Determine si el supuesto del experimentador es correcto.
Describa la situación y los supuestos: estrictamente hablando, el elemento está absorbiendo potencia. El
valor de la potencia absorbida por el elemento puede ser positiva, nula o negativa. Cuando decimos que alguien
“supone que un elemento está absorbiendo potencia” significa que alguien asume que la potencia absorbida por
el elemento es positiva.
Los medidores son ideales. Se conectan al elemento de tal manera que midan el voltaje marcado v y la
corriente marcada como i. Los valores del voltaje y de la corriente aparecen en las lecturas de los medidores.
Establecer los objetivos: es calcular la potencia absorbida por el elemento para determinar si el valor de la
potencia absorbida es positivo.
Generar un plan: es verificar que el voltaje y la corriente del elemento se apegan a la convención pasiva.
Si es así, la potencia absorbida por el dispositivo es p 5 vi. Si no lo es, la potencia absorbida por el dispositivo
es p 5 2vi.
Actuar sobre el plan: refiriéndose a la tabla 1.5-1, podemos ver que el voltaje y la corriente del elemento
se apegan a la convención pasiva. Por lo tanto, la potencia absorbida por el elemento es
p 5 vi 5 12  (22) 5 224 W
El valor de la potencia absorbida no es positivo.
Verificar la solución propuesta: implica invertir las pruebas, como se muestra en la figura 1.6-3. Ahora el
amperímetro mide la corriente i1 en vez de la corriente i, por lo que i1 5 2 A y v 5 12V. Puesto que i1 y v no se
apegan a la convención pasiva, p 5 i1  v 5 24 W es la potencia alimentada por el elemento. Alimentar 24 W es
el equivalente a absorber 224 W, por consiguiente, verificar la solución propuesta.
1 2 . 0
1 2 . 0
Voltímetro
– 2 . 0
Voltímetro
2 . 0 0
Amperímetro
+
v
–
Elemento
Amperímetro
+
i
FIGURA 1.6-2 Un elemento con un voltímetro y un
amperímetro.
Alfaomega
M01_DORF_1571_8ED_SE_001-019.indd 12
v
–
Elemento
i1
FIGURA 1.6.3 El circuito de la figura 1.6-2 probado al revés
con el amperímetro.
Circuitos Eléctricos - Dorf
4/12/11 5:15 PM
¿Cómo lo podemos comprobar . . . ?
13
Diseño es una actividad con un propósito determinado en la cual un diseñador visualiza un
resultado deseado. Es el proceso de originar circuitos y predecir cómo cumplirán tales circuitos su
objetivo. Diseño en ingeniería es el proceso de producir un conjunto de descripciones de un circuito que satisfaga un conjunto de requerimientos de desempeño y exigencias.
El proceso del diseño puede incorporar tres fases: análisis, síntesis y evaluación. La primer tarea
es diagnosticar, definir y preparar, es decir, entender el problema y elaborar un enunciado explícito
de objetivos; la segunda implica encontrar soluciones plausibles, la tercera se refiere a formarse una
opinión sobre la validez de las soluciones en relación con los objetivos y la selección de diversas
alternativas. Se impone un ciclo en el cual la solución sea revisada y mejorada por un nuevo examen
del análisis. Estas tres fases forman parte de un marco de trabajo en la planeación, organización
y desarrollo de los proyectos de diseño.
Diseño es el proceso de la creación de un circuito para cumplir con un conjunto de objetivos.
El proceso de la solución de problemas que se muestra en la figura 1.6-1 se utiliza en algunos ejemplos de diseño incluidos en este capítulo.
1.7
¿ C Ó M O LO P O D E M O S C O M P R O B A R . . . ?
A los ingenieros se les suele solicitar comprobar que la solución de un problema sea la correcta. Por
ejemplo, las soluciones propuestas para problemas de diseño se deben comprobar para confirmar que
se ha cumplido con todas las especificaciones. Además, se deben revisar los resultados de la computadora para protegerse contra errores de captura de datos, así como las exigencias de los comerciantes,
las cuales se deben analizar a fondo.
También a los estudiantes de ingeniería se les pide que verifiquen la exactitud de sus trabajos.
Por ejemplo, tomarse un breve lapso antes de terminar un examen permitiría dar una vista rápida e
identificar esas soluciones que podrían requerir un poco más de aplicación.
Este texto incluye algunos ejemplos que ilustran útiles técnicas para comprobar las soluciones
de determinados problemas que se analizan en el capítulo. Al final de cada capítulo se presentan algunos problemas con los cuales se tiene la oportunidad de practicar las técnicas aprendidas.
E j e m p l o 1.7-1 ¿Cómo podemos comprobar la potencia y la convención pasiva?
El reporte de un laboratorio indica que los valores de v e i medidos por el elemento de circuito mostrado en la figura 1.7-1 son 25 V y 2 A, respectivamente. También establece que
la potencia absorbida por el elemento es de 10 W. ¿Cómo podemos comprobar el valor
reportado de la potencia absorbida por este elemento?
Solución
i
–
v
+
FIGURA 1.7-1 Un
elemento de circuito
con voltaje y
corriente medidos.
¿El elemento de circuito absorbe 210 W o 110 W? El voltaje y la corriente mostrados en
la figura 1.1-7 no se apegan al signo de la convención pasiva. Si nos referimos a la tabla 1.5-1 podemos ver que
el producto de este voltaje y corriente es la potencia alimentada por el elemento, más que la potencia absorbida
por el elemento.
Entonces, la potencia alimentada por el elemento es
p 5 vi 5 1252122 5 210 W
La potencia absorbida y la potencia alimentada por un elemento tienen la misma magnitud pero signos opuestos.
Por lo tanto, hemos verificado que en realidad el elemento de circuito está absorbiendo 10 W.
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 13
Alfaomega
4/12/11 5:15 PM
E1C01_1 11/26/2009
11/26/2009
E1C01_1
14 14
14
1414
14
Electric Circuit Variables
Electric
Circuit
Variables
Variables
de circuitos
eléctricos
Electric
Circuit
Variables
Electric
Circuit
Variables
1414
Electric
Circuit
Variables
1.8 DESIGN EXAMPLE
1 .118. 8 D E
NPNELX
PDLPIELSEE Ñ O
EGIM
OA
DES
EJIS
G
E
XDM
AEM
IG
11
. 8. 8 DD
EE
SS
IG
NNE E
XX
AA
MM
PP
LL
EE
JET VALVE
CONTROLLER
CONTROLADOR DE VÁLVULA
DE UN
JET
VALVE
CONTROLLER
JET
VALVE
CONTROLLER
MOTOR DE
PROPULSIÓN
A
CHORRO
JET
VALVEspace
CONTROLLER
A small,
experimental
rocket uses a twoJET
VALVE
CONTROLLER
Wire
AA
small,
experimental
space
rocket
uses
a twoCable
element
circuit,
as shown in Figure
1.8-1, to Controlador de
Wire
Un
cohete
espacial experimental
pequeño
utiliza
small,
experimental
space
rocket
uses
a twoJet value
Wire
i
+
+
Asmall,
small,
experimental
space
uses
two- of liftoff at t ¼ 0Wire
controller
element
circuit,
as as
shown
in in
Figure
1.8-1,
tolato
válvula
de
motor
Wire
Jet
value
control
a 1.8-1,
jetrocket
valve
from
point
Ade
experimental
space
rocket
uses
a atwoun circuito
dos
elementos,
como
se
ve
en
fielement
circuit,
shown
Figure
Jet value
ii i
++ +
++ + de
propulsión
controller
element
circuit,
as
shown
1.8-1,
control
a jet
valve
from
point
liftoff
at
tun
¼Figure
0¼ of
controller
value Element
until
expiration
the
rocket
one minute.
element
circuit,
asof
shown
inin
Figure
toto after
gura 1.8-1,
para
controlar
la
válvula
de
control
a jet
valve
from
point
of
liftoff
at
t motor
0 1.8-1,
JetJet
value
i
v2
v1+ Element
+
+
i
+
controller
control
ajet
jetvalve
from
point
ofliftoff
liftoff
atvt ¼tElement
¼
0
controller
1
2
until
of
the
rocket
after
one
minute.
Elemento
Elemento
Element
The
energy
mustatvbe
supplied
by
element
1
control
a the
from
point
ofthat
0
de expiration
propulsión
a of
chorro
avalve
partir
del
punto
de
desuntil
expiration
rocket
after
one
minute.
Elementv2v2v
1 1v1 Element
2
11 mJ.
until
expiration
the
rocket
after
one
minute.
The
energy
must
be be
by
element
1 one
1 Element
Element
Element v
for
the
one-minute
period
is 40
1 is22a2
ofofthe
rocket
after
minute.
pegue
en that
tuntil
5
0expiration
hasta
lasupplied
expiración
del
cohete
The
energy
that
must
supplied
by
element
1un
Element
v1v1Element
– v2 2
–
1
Theenergy
energy
must
be
supplied
by
element1 1– –
1
2 2
forfor
thethe
one-minute
period
isthat
40
mJ.
Element
1 is1selected.
battery
to be
The
that
beelemento
supplied
by
minuto
después.
Laperiod
energía
que
el
1ais
debe
one-minute
ismust
40
mJ.
Element
a element
–– –
–
for
the
one-minute
period
Element
1¼is
a �t/60 mA for t –� 0,
battery
to to
befor
selected.
Wire
is40
known
that i(t)
the
one-minute
isItis
40
mJ.
Element
1 is
aDe
alimentar
para
un
lapso
de unperiod
minuto
es
demJ.
40
mJ.
battery
be
selected.
–
– –
�t/60
�t/60
Cable
Wire
battery
to
be
selected.
for
t �t 0,
It is
thatthat
¼
De
Wire
the
voltage
across
the second element
is v2(t) ¼
toi(t)
bei(t)
selected.
El elemento
1 debe
ser
una
batería.
mA
for
�
0,
It known
is battery
known
¼
DeandmA
FIGURE
1.8-1
The circuit to control
�t/60
�t/60
�t/60
2t/60
Wire
mA
t
�
0,
isknown
known
that
i(t)
¼mA
andand
thethe
voltage
the
second
element
is De
vis
Wire
V
for
0.
The
maximum
magnitude
of
the
Beelement
tFIGURA
�
0,
It Itisi(t)
i(t)
¼¼
De
2(t)
FIGURE
1.8-1
The
circuit
to
control
Sevoltage
sabeacross
que
5 second
Dethat
mA
para
0,
yforfor
(t)
¼
across
the
vtt2�
a
jet
valve
for
a
space
rocket.
1.8-1
Circuito
para
controlar
la
FIGURE 1.8-1 The circuit to control
�t/60
(t)
¼
and
voltage
across
the
second
element
v(t)
V voltaje
for
t�
The
maximum
magnitude
the
BeBe
jet
valve
aFIGURE
space
rocket.
2jet
FIGURE
1.8-1
The
to
control
deforunfor
motor
de rocket.
propulsión
a circuit
chorro
current,
D, ofelement
isof
limited
to
1¼valve
mA.
Determine
theThe
and
thethe
voltage
across
the
second
vválvula
que�t/60
el
que
pasa
por
el segundo
elemento
esis ais
V for
t 0.
�
0.
The
maximum
magnitude
the
a
a
space
2
1.8-1
circuit
to
control
�t/60
�t/60
2t/60
V
for
0.
The
maximum
magnitude
thecohete
Be
un
espacial.
current,
to
1t �
Determine
the
a jet
valve
for
a space
rocket.
required
constants
and
describe
the
required
battery.
V
for
�
0.
The
maximum
magnitude
ofBof
the
Be
v2(t) 5D,Be
V
para
ttto
0.
La
magnitud
máxima
current,
D,is
islimited
limited
1mA.
mA.
Determine
theD andpara
a jet
valve
for
a space
rocket.
current,
D,
is
limited
to
1
mA.
Determine
required
constants
DD
and
Blimitada
and
describe
required
battery.
current,
D,and
is
1themA.
Determine
thethe
de la corriente,
D,
está
a 1tomA.
Determine
required
constants
Blimited
and
describe
the
required
battery.
Describe
the
Situation
and the
Assumptions
required
constants
and
required
battery.
required
constants
B Band
describe
thethe
required
battery.
las constantes
requeridas
D yDthe
BDand
yand
describa
ladescribe
batería
requerida.
Describe
the
Situation
and
Assumptions
Describe
the
Situation
and1.the
Assumptions
The
current enters the plus terminal of the second element.
the
Situation
and
the
Assumptions
Describe
the
and
the
Assumptions
Describa
laDescribe
situación
ySituation
los
supuestos
1. 1.
The
current
enters
the
plus
terminal
of
the
second
element.
The
current
enters
the
plus
terminal
of
the
second
element.
2.
The
current
leaves
the
plus
terminal
of the first element.
1.
The
current
enters
the
plus
terminal
of
the
second
element.
1.
The
current
enters
the
plus
terminal
of
the
second
element.
1.
La
corriente
entra
por
la
terminal
positiva
del
segundo
elemento.
2. 2.The
current
leaves
thethe
plus
terminal
of of
thethe
first
element.
The
current
leaves
plus
terminal
first
element.
3. The
wiresterminal
are perfect
andfirst
have
no effect on the circuit (they do not absorb energy).
The
current
leaves
plus
element.
2.2.
The
current
thethe
plus
terminal
ofof
thethe
first element.
2.The
Lawires
corriente
sale
deand
la leaves
terminal
positiva
del
primer
elemento.
3. 3.
are
perfect
have
no
effect
on
the
circuit
(they
do
not
absorb
energy).
The wires are perfect and have no effect on the circuit (they do not
absorb
energy).
4. tienen
Theand
model
ofalguno
theeffect
circuit,
as
shown
in(they
Figure
1.8-1,
assumes
that the voltage across the
3. The
The
wires
are
perfect
have
no
on
the
circuit
do
not
absorb
energy).
3. Los cables
son
precisos
yperfect
no
efecto
sobre
elcircuit
circuito
(no
absorben
3.
wires
are
and
have
no
effect
on
the
(they
do
not
absorb
energy).
4. 4.The
model
of of
thethe
circuit,
as as
shown
in elements
Figure
1.8-1,
assumes
thatthat
the
voltage
across
thethe
two
is
equal;
that
is,
v
¼
v
.
The
model
circuit,
shown
in
Figure
1.8-1,
assumes
the
voltage
across
1
2
energía).
model
of
as shown in Figure 1.8-1, �t/60
assumes
that
voltage
across
twotwo
elements
isThe
equal;
that
is,the
vcircuit,
¼
4.4. The
ofthat
the
that
thethe
voltage
across
thethe
1 circuit,
elements
ismodel
equal;
is,
v1 v¼2.vas
2. shown in Figure 1.8-1, assumes
is v1 ¼supone
Be
V where
B is the
initial voltage of the battery that
5.equal;
The
battery
voltage
v1 1.8-1,
4. El modelotwo
del
circuito,
como
se
muestra
en
la
figura
que
el
voltaje
a
través
two
elements
is
that
is,
v
¼
v
.
1 v 2. 2
elements
equal;
that
is, v1 ¼
�t/60
�t/60
Be
V where
is
voltage
of of
thethe
battery
that
5. 5.The
v1 vis1esvis1is
will
discharge
asvoltage
it supplies
energy
to
the
v1mismo;
¼ Be
V
where
B the
is
initial
battery
thatvalve.
The
battery
voltage
debattery
los
dosvoltage
elementos
el¼
es
decir,
vB�t/60
5
v2the
. initial
1exponentially
�t/60
is
v
¼
Be
V
where
B
is
the
initial
voltage
of
the
battery
that
5. The
The
battery
voltage
v
will
discharge
exponentially
as as
itvsupplies
energy
to
the
valve.
1
1
is
v
¼
Be
V
where
B
is
the
initial
voltage
of
the
battery
that
5.
battery
voltage
2t/60
will
discharge
exponentially
it
supplies
energy
to
the
valve.
1 Be 1
5. El voltaje v1 de la batería es
v1The
5
donde Bfrom
es elt voltaje
inicial
6.
circuitV
operates
¼ 0 to
tthe
¼ 60
s.que se descargará
will
discharge
exponentially
as
it
supplies
energy
to
valve.
will
discharge
exponentially
as
it
supplies
energy
to
the
valve.
decircuit
manera
exponencial
a la válvula.
6. 6.The
operates
from
tal¼tabastecer
0¼to
t ¼t 60
s.energía
The
circuit
operates
from
0 to
¼de
60
s.
7. The
current
istolimited,
so D � 1 mA.
6.
The
circuit
operates
from
t
¼
0
t
¼
6. isThe
circuit
from
s. s.
6.The
Elcurrent
circuito
opera
de so
t operates
5D0�a 1t 5
60 s.t ¼ 0 to t ¼ 6060
7. 7.
limited,
The
current
is limited,
so D �mA.
1 mA.
The
current
limited,
D
1 mA.
mA.
7. La corriente
escurrent
limitada,
por
lo tanto,
D
7.7. The
is is
limited,
soso
DGoal
��
1 1mA.
State
the
State
the
Goal
Determine the energy supplied by the first element for the one-minute period and then select
State the Goal
Establezca
elenergy
objetivo
State
the
Goal bythe
Determine
the
energy
supplied
theconstants
first
element
forfor
one-minute
period
and
then
select
D and
B.theDescribe
the battery
selected.
State
the
Goal
Determine
the
supplied
by
the
first
element
the
one-minute
period
and
then
select
Determine
la
energía
alimentada
por
el
primer
elemento
para
un
periodo
de
un
minuto
y luego
Determine
the
energy
supplied
by
the
first
element
for
the
one-minute
period
and
thenselect
select
thethe
constants
D
and
B.
Describe
the
battery
selected.
Determine
theDescribe
energy supplied
by the
first element for the one-minute period
and
then
constants
D and B.
the battery
selected.
seleccionethe
las
constantes
yand
B.B.Describa
lathe
batería
seleccionada.
theconstants
constantsDDDand
B.Describe
Describe
thebattery
battery
selected.
selected.
Generate a Plan
Generate
Genere aunPlan
First, find v1(t) and i(t) and then obtain the power, p1(t), supplied by the first element. Next,
Generate
a plan
Plan
Generate
a
Plan
First,
find
v
(t)
and
i(t)
and
then
obtain
power,
ppotencia,
supplied
the
first
element.
Primero,
encuentre
v
(t)
e
i(t)
y
luego
lathe
psupplied
alimentada
por
el60primer
ele(t),
find
energy
for
the
first
s.Next,
using
pthe
Generate
a
Plan
1(t),
First, find1 v1(t) and i(t)
obtain
the
power,
p1(t),
supplied
by
the
first
element.
Next,
1 and then
1(t), by
1obtenga
First,
find
vutilizando
and
and
then
obtain
the
power,
supplied
firstelement.
Next,
find
the
energy
supplied
the
first
60la
s.energía
using
p1(t),
mento.
continuación,
p1for
(t),for
encuentre
alimentada
para los
primeros
60element.
s.
1(t)
1(t),
First,
find
venergy
and
i(t)i(t)and
then
obtain
the
p1p(t),
supplied
byby
thethefirst
Next,
find
the
supplied
the
first
60
s.power,
using
pA
1(t)
1(t),
findthethe
energy
supplied
for
the
first
60
s.
usingp1p(t),
1(t),
find
energy
supplied
for
the
first
60
s.
using
GOAL
EQUATION
NEED
INFORMATION
GOAL
OBJETIVO
GOAL
EQUATION
ECUACIÓN
EQUATION
NEED
REQUERIDO
NEED
Z
INFORMATION
INFORMACIÓN
INFORMATION
60
The energy w1EQUATION
for the
v1 and i known except for
NEED
INFORMATION
Z Z
EQUATION
NEED
INFORMATION
6060
w
¼
p
ð
t
Þ
dt
p
(t)
1
60
1
v
TheThe
energy
w
for
the
and
i
known
except
for
first
s
constants
D and B
1
1
1
energy
w
for
the
and
i
known
except
for
v
La energía w1 1para los
v1 1e0 i son conocidas, excepto para
w¼
¼ p1 ðpt1Þðtdt
Þ ZdtZ60 60 p1p(t)
firstfirst
60 60
s s 60The
constants
D and
B
p1(t)
v
energyw1ww
and
i
known
except
for
1 the
1(t)
1 1for
1
constants
D
and
B
for
the
and
i
known
except
for
v
primeros
sTheenergy
las
constantes
D
y
B.
1
0 0
1 ¼
w1w¼
p1pð1tÞðtÞdtdt
first
constants
and
p1p(t)1(t)
first
6060
s s
constants
DD
and
BB
0
GOAL
GOAL
0
Act on the Plan
Actúe
sobre
el
plan
Act
on
the
Plan
First, we need p1(t), so we first calculate
Act on the Plan
En primer
lugar,
sethe
requiere
p1calculate
(t), de modo que hay que calcular antes�
Act
on
the
Plan
First,
we
need
p
(t),
so
we
first
Act
on
Plan
1
First, we need p1(t), so we first calculate
Alfaomega
M01_DORF_1571_8ED_SE_001-019.indd 14
��
�
p�1 ð�t�Þ� ¼ iv1 �¼ � De�t/60 � 10�3 A Be�t/60 V
� first
First,weweneed
needp1p(t),
sowewe
calculate
�first
1(t),
First,
so
calculate
�t/60
�3 �3
�t/60
�t/60
�t/30
p1 ðpt1Þðt¼
iv1iv¼1 ¼DeDe
� 10
VV
Þ ¼
��10
ABeBe
¼ �t/60
DBe
10�3�W
� A�t/60
�� � �t/60
� ¼ DBe�t/30 mW
�3 ��
�3
�t/30
�t/30
�t/60
�3
�t/60
�3
p
ð
t
Þ
¼
iv
¼
De
�
10
A
Be
V
�t/30
�t/30
1W W
¼¼
DBe
¼¼
DBe
p1 ð1tÞ �
¼ 10
De
� 10 mW
A Be
V
1 ¼
�iv10
DBe
mW
DBe
�t/30
�t/30
�3�3 W ¼ DBe
�t/30
DBe
mW
¼¼DBe
��
1010
W ¼ DBe�t/30 mW
Circuitos Eléctricos - Dorf
4/12/11 5:15 PM
E1C01_1
11/26/2009
E1C01_1
11/26/2009
15 151515 15
E1C01_1
11/26/2009
E1C01_1
11/26/2009
E1C01_1
11/26/2009
E1C01_1
11/26/2009
15
E1C01_1
11/26/2009
15
E1C01_1 11/26/2009
15
JET VALVE CONTROLLER
E1C01_1
11/26/2009
15
A11/26/2009
small, experimental
space rocket uses a twoWire
15
ProblemsJet value15
15
element circuit, as shown in Figure 1.8-1, to
Problems
15
Problems
i
+
+ Problems
Problems
Problemas
15
controller 15
Problems
15
control a jet valve from point of liftoff at t ¼ 0
Problems
Problems
15 151515 15
Problems
Problems
Problems
Problems
until
expiration
of
the
rocket
after
one
minute.
Element v
v1 Element
Second,
we
need
to
find
w
for
the
first
60
s
as
1
2
Second,
we
need
to
find
w
for
the
first
60
s
as
Second,
we
need
to
find
w
for
the
first
60
s
as
En
segundo
se
necesita
encontrar
w60
para
los primeros
60 s como
11 that
1
2
1supplied
The
energy
must
be
Second,
we lugar,
need
to
find
w
for
first
s as
Second,
we
need
to
find
wwthe
the
first
60
as
1
11 for
Second,
we
need
to
find
for
the
first
60assssby
aselement 1�3 �t=30���60
Z
Second,
we
need
to
find
w
for
the
first
60
s to
as
Second,
we
need
to
find
w
for
first
60
s60
as
Second,
we
need
w
for
the
first
60
as
60the
Second,
we
need
tofind
find
w
for
the
first
60
sas
Second,
we
need
to
w
the
first
s
1find
60
1to
ZZ
1
60
1for
�
160
�
�
�3
�t=30
for
the
one-minute
period
is
40
mJ.
Element
1
is
a
60
e
DB
�
10
�3
Second,
we
need
find
w
for
the
first
60
s
as
1 DB �
�–60
��60
Z 60��Z
�t/30
�3��
�10
10�3ee�t=30
–
Z 60
�t=30
�t/30
�3
�3
60
w1 ¼
¼
DBe
�10
10�3
dt ¼
¼��DB
����e60
�60
� dt
60�t/30
�3
�60
�� �
�t=30
Z Z60be
e�3
DB
�
10
Z
�60for�����60
��t=30
DB
�
10
¼
DBe
�
10
dt
¼
Z�60Z60
ww
DBe
60
�3
battery
selected.
�3Second,
60�t/30
�t=30
�3
e
DB
�
10
�3
�t=30
�t/30
�3
�t=30
�60 15
�3
�
�t=30
�t=30
�1=30
Problems
Z
we
need
to
find
w
the
first
60
s
�
�
�
�
�
�
�
�
�t/30
�3
�
�
1
e
DB
�
10
�
�
�
�
0
e
DB
�
10
w111 ¼toww
DBe
�
10
dt
¼
e
DB
�
10
60
DBe
�
10
¼
�1=30
�1010 e e� ��000� �� �� DB � 10�3 as
DB
�
�1=30
1010 ¼
�t/30
�t/30
�3 �3
�t/30
�3 dt
�
¼ DBe
DBe
��3
10mA
dt�DB
¼
�t/60
�t/30
�t/30
�3
e�t=30
¼
�
10
dt
¼
�3
�3 � �3
Wire
w1ww
¼
�
10
dt
¼
�2
¼
DBe
�
10
dt
¼
�t/30
�1=30
for
t �1=30
��
0,
isw
known
i(t)
¼
De
¼�30DB
DBe
�
10
dt¼25:9DB
¼
1w
DBe
�
10w
dt
10 DBe
Second,
we
need
first
� to find w1 for the Problems
1¼
1It
�3
�3
00that
�1=30
�2
0
�
�
0
¼
�
10
ð
e
�
1
Þ
¼
10
J
�3
�3
� 60 s as
�2
�
�
¼
DBe
�
10
dt
¼
Z
�1=30
¼0 �30DB
�30DB
�10
10�3ððee�2�3
�111�2
Þ¼
¼ 25:9DB
25:9DB
�
10�1=30
J 0 �3
�1=30
60 � 00
¼
�
Þ
�
10
J
�1=30
�1=30
00 0
�3
0�
0
� �0 DB � 10�3 e�t=30 �60
0 0J
�1=30
�3
�3
�2
�30DB
��3
10
ð�3
e10
�ððelement
Þ¼
�
10�3�3
JFIGURE
(t)
¼�3
and the¼voltage
across
the
second
v225:9DB
¼
�30DB
�
ee1�2
�
11Þis
¼
�
10
025:9DB
�t/30
�3to control
1.8-1
The
circuit
�3�3
Z
�2
�
�3
�3
¼
�30DB
�
10
�
Þ
¼
25:9DB
�
10
J
�2
�3
�2
�2
60 �
w
¼10
DBe
� 10 �3 dt ¼
¼�30DB
�30DB
�
Þ1¼Þ¼
25:9DB
�
10
¼w
��30DB
10�10
ð10
e�
1Þ�
25:9DB
10�
¼
ð1e¼
�
Þ25:9DB
¼�25:9DB
1�
Second,
we need
to Be
find
the
first
60
sðeas
�3J10
�t/60
¼
�30DB
�
ðe�
�
1Þ¼
¼125:9DB
�10
Because
weserequire
require
�for
40
mJ,
¼
�30DB
10
ð10
e�
J J 1JÞ ¼
�
�
1V
Because
we
require
�
40
mJ,
�30DB
10
ðeJ�2
� 10 J w ¼ �1=30
Because
we
www111que
�
40
for
tmJ,
� 0.
The
maximum
magnitude
of�the
�t/30
Puesto
que
requiere
w
40
mJ,
a jet�0valve
for25:9DB
a space rocket.
1
0 � 10�3
DBe
Because
we
require
w
�
40
mJ,
Because
we
require
wwSecond,
40
mJ,
1
we need to find w1 for the first�6060 s as
1
11�
Because
we
require
�
40
mJ,
Z
�3
�3
Because
we
require
�
mJ,
Because
we
require
wrequire
�w
Because
we
w
�
40 mJ,
�t=30 � ¼ �30DB � 10 ðe�2 � 1Þ ¼ 25:9DB 0� 10
current,
D,
is
limited
to
1 �25:9DB
mA.
Determine
�
40
mJ,
Because
we
require
w
40
�
25:9DB
Because
we
require
J
�
40
mJ,
1 40
1w
1
�mJ,
140
160
DB
10�3 ethe
40
�
25:9DB
Because
we40
require
w1 �
40
mJ,
�
�
Z�the
40B�3
�
25:9DB
60 �required ��battery. �
wrequired
DBe�t/30
� 10
dt
¼
�3 �t=30 �60 ¼ �30DB � 10�3 ðe�2 �
40
�
25:9DB
1 ¼
constants
D 40
and
and
describe
40
�
25:9DB
DB � 10 e
40
�
�
25:9DB
40
� 25:9DB
�t/30w01 � 40
�3 mJ,
�1=30
40
�
25:9DB
Next,
select the
the seleccione
limiting
value,
D¼
¼de
1,limitación,
to
get
40get
�25:9DB
25:9DB
Because
we DBe
require
�
0 value,
A
continuación,
el valor
obtener
Next,
select
the
limiting
value,
D
¼
1,
to
get
w
dt ¼
40��10
25:9DB
Next,
select
limiting
D
1,
to
1 D¼5 1, para
� w � 40 mJ,
�31,D�2
�3
Next, Next,
select
the
limiting
value,
D
¼
to
get
select
the
limiting
value,
¼
1,
to
get
�1=30
Because
we
require
Next,
select
the
limiting
value,
D
¼
1,
to
get
¼
�30DB
�
10
ð
e
�
1
Þ
¼
25:9DB
�
10
J
0
1
0
Describe
the
Situation
and
the
Assumptions
Next,
select
limiting
value,
D
¼
1,
to
get
Next,
select
thethe
limiting
value,
D
¼
1,
to
get
Next,
select
the
limiting
value,
D
¼
1,
to
get
40
Next,
select
the
limiting
value,
D
¼
1,
to
get
Next,
select
the
limiting
value,
D
¼
1,
to
get
40
�
25:9DB
�3
�3
40
�2
Next, select
limiting
value,
�30DB
�1,10to get
ðe � 1Þ ¼ 25:9DB � 10 J
¼¼1:54
1:54
VD¼
B�
� the40
40
40
B
�
¼
1:54
V
B
¼
V
40 � 25:9
40
The
the
plus
terminal
of1:54
the second
element.
40 current
mJ, enters
Because we require 1.w1 �
25;
:9
40
40
B � ððð25;
¼ 1:54
BB �
¼
25;
:9
40
:9
ÞÞÞ40
ððð111Next,
ÞÞÞ40
select
theV
limiting
�
¼VV
1:54
V
40 value, D ¼ 1, to get
BB�B�we
¼
B�
1:54
ð25;
:9ððÞ25;
ð¼
1Þ:9
Brequire
�
¼
1:54
ÞÞ1:54
ð�
111:54
ÞV
Because
w
�
¼
1:54
VB �V
40
mJ,
¼
V
1
25;
:9
ð
Þ
¼
1:54
V
Next,
the limiting value, D ¼ 1, to get
ð�
25;
:9
1ðÞterminal
ð40
25;
Þ25;
ðplus
1:9
25;
Þ of
2.
Thebattery
current
the
element.
ð25:9DB
1fin
ÞÞð1of
Thus,
we
select
2-V
battery
soleaves
that
the
magnitude
of
thethe
current
is:9less
less
than
mA. select
Por
lowe
tanto,
seleccionamos
una
batería
de
2ÞðÞð:9
de
que
lafirst
magnitud
la corriente
sea
ð:9
25;
ÞV
1ðaÞ:9
Thus,
we
select
2-V
battery
so
that
the
magnitude
of
the
current
less
than
mA.
40
ð25;
Þð1de
Þthan
Thus,
select
aaa 2-V
so
that
the
magnitude
the
current
isis
111 mA.
Thus, Thus,
we
select
a 2-V aabattery
so thatso
the
magnitude
of theof
current
is �
less
1than
mA.11 mA.
we
select
2-V
battery
that
the
magnitude
the
current
isisthan
less
40
25:9DB
B
�
¼ 1:54 V
Thus,
we
select
2-V
battery
so
that
the
magnitude
of
the
current
less
than
mA.
menor
de
1
mA.
40
Thus,
select
aselect
battery
so
that
the
magnitude
current
is
less
than
1of
mA.
Thus,
wewe
select
a 2-V
battery
so
that
the
magnitude
of
the
less
than
1than
mA.
Thus,
we
a battery
2-V
so
that
the
magnitude
of
the
current
is
less
1current
mA.
Next,
select
the
limiting
value,
Dbattery
¼
1,we
to
get
Thus,
we
select
a2-V
2-V
battery
so
that
the
magnitude
ofcurrent
the
current
isless
less
1mA.
mA.
Thus,
we
select
a2-V
so
that
the
magnitude
ofthe
the
current
is
than
1than
ð25;
:9Þð1energy).
Þis less than 1 mA. B �
3.
The
wires
are
perfect
and
have
noof
effect
on
theis
circuit
(they
do
not
absorb
Thus,
select
a
2-V
battery
so
that
the
magnitude
the
Verify
the
Proposed
Solution
Next,
select
the
limiting
value,
D
¼
1,
to
get
ð
25;
:9Þð1Þ
Verify
the
Proposed
Solution
Verify
the
Proposed
Solution
40
�t/60
Thus,
we
a1.8-1,
2-V
battery
so
thatthe
thevoltage
magnitude
the current is less than 1 m
Verify
the
Proposed
Solution
Verify
the
Solution
Verifique
la
solución
propuesta
4. Proposed
The
model
of
the
circuit,
as
shown
in select
Figure
assumes
that
acrossofthe
�t/60
�t/60
We
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
Verify
the
Proposed
Solution
B
�
¼
1:54
V
We
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
Verify
the
Proposed
Solution
Verify
the
Proposed
Verify
the
Solution
We
must
verify
thatProposed
atSolution
least
40
mJ
is the
supplied
using the
2-V battery. Because
i ¼ e �t/60
mA
and
Verify
theProposed
Proposed
Solution
Verify
the
Solution
�t/60
40Because
�t/60
�t/60
Thus,
we
select
so that the magni
Verify
Solution
ðbatería
25;
:9Proposed
1that
Þusing
We
must
verify
thatenergy
at
least
40
mJ40
is
supplied
the
2-V
battery.
Because
i¼
e�t/60
and
We
must
verify
that
at
least
mJ
is
supplied
using
2-V
battery.
ii¼
emA
mA
and
v2the
.the
two
elements
equal;
Habrá
que
verificar
que
con
una
de
2 is,
V vse
alimenten
al
menos
40
mJ.
Dado
que
�t/60
�t/60
�t/60
�t/60
¼
2e
V,
the
supplied
by
the
battery
is¼
We
must
verify
that
at
least
40
mJ
isÞðusing
supplied
using
2-V
battery.
Because
¼
eV
mA
and a 2-V battery
1is
�t/60
�t/60
B Because
�
¼
1:54
2¼
¼
2e
V,
the
energy
supplied
by
the
battery
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
WeWe
must
verify
that
atthat
least
40
mJ
is
supplied
the
2-V
battery.
i
¼
e
mA
and
We
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
2e
V,
the
energy
supplied
by
the
battery
is
vvv2We
�t/60
2
�t/60
We
must
verify
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
�t/60
2t/60
2t/60
�t/60
ð
25;
:9
Þ
ð
1
Þ
We
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i
¼
e
mA
and
¼
2e
V,
the
energy
supplied
by
the
battery
is
v
¼
2e
V,
the
energy
supplied
by
the
battery
is
v
Verify
the
Proposed
Solution
i
5
e
mA
y
que
v
5
2e
V,
la
energía
alimentada
por
la
batería
es
2
�t/60
2
�t/60
Z
Z
�t/60
�t/60current
¼
2e
V,
the
energy
supplied
by
the
battery
is where
2the
�t/60
2�t/60
Thus,
we
select
athe
battery
that
the
magnitude
of
is Bless
than
1 mA.
60the
60
ZZ2-V
¼
V,
the
energy
supplied
by
the
battery
is6060
V,
energy
supplied
by
the
battery
v2 v¼2v2e
2e
V,
energy
supplied
by
battery
is
vv2e
�t/60
vthe
¼isZZ
Be
V
is battery
the
initial
voltage
ofthe
the Proposed
battery thatSolution
5.
battery
voltage
vthe
¼2e
V,
energy
supplied
by
the
battery
is
�so
�supplied
�is
V,
the
energy
by
battery
isthe
60
2¼
1��
1
60
22e
2v¼
��The
�
�
¼
2e
V,
the
energy
supplied
by
the
is
v
Z
Z
Verify
�t/60
�t/60
�3
�t/30
�3
�
�
�
Z
Z
We
must
verify
that
at
least
40
mJ
is
supplied
using
the
2-V
battery.
Because
i ¼ e�t/60
2
60
60
Z
Z
60
60
�t/60
�t/60
�3 � dt
�t/30
�3
we
select
a¼
2-V
battery
thedt
magnitude
�3
e�t/60
¼
2e
�10
10
2e
�that
10�3
dt
¼
51:8
mJof the current is less than 1 mA.
w¼
¼Z60Z60Z �2e
60discharge
60so�
�� ��Thus,
�����
��Z60
Z¼
Z¼
Z�t/30
Z �t/60
e
dt
2e
�
10
2e
�
10
dt
¼
51:8
mJ
Z
will
exponentially
as
it
supplies
energy
to
the
valve.
Z
e
dt
2e
10
¼
51:8
mJ
ww
�t/60
60
�t/60
�t/60
�3
�t/30
�3
60
60
�t/60
�t/60
�3
�t/30
�3
60
60
60
60
Z
Z
��t/60
�2e
����t/60
� �10
���2e
We
must
verify that
���
�t/60
�t/60
�t/30
�351:8
� �3
0� ¼
�2e��t/60
V,
the
by mJ
the battery
is at least 40 mJ is supplied usin
v2� ¼
w ¼ w0w
e���t/60
10��3�3
dt
¼
��t/30
10�3energy
dt
¼
60
60mJ
ee�t/60
dt
¼
�
2e
�
dt
¼
51:8
�t/30
�t/60
�3
�t/30
�3
�3supplied
00 2e
dt
¼
�
10
2e
�dt
10¼
dt51:8
¼
51:8
mJ
¼
��3
�2e
�2e
�10
�t/60
�3
�3
0dt
�t/60
�t/30
�3
Verify thew Proposed
Solution
�t/60�3
e�t/60
dt
2e
�
10
2e
�
51:8
mJ
ww¼w¼¼0w
e2e
dt
¼
��t/60
10
��t/30
10
¼
51:8
mJ
¼
e�
¼
�
10
2e
�
10
dt
¼
51:8
mJ
¼
�t/60
�t/60
�3
�t/30
dt
�10
10
�
10
dt¼¼
¼51:8
mJ
e eoperates
dtt¼
¼¼
2e
�10
10dt
dtdt
mJ
06.2e
02e
002ecircuit
02e
¼
2e
V,
the
energymJsupplied by the batter
v
e
2e
�
10
2e
�
10
dt
¼
w
¼
Z
Z
0
The
from
¼
0
to
t
¼
60
s.
2
�t/60
60it
60 51:8
Thus,that
we
have
verified
the
solution,
andthe
we
communicate
by recording
recording
the requirerequire0communicate
0 0have
0 2-V
the using
Proposed
0 40 mJVerify
0
0 verified
0 battery.
We must verify
at
is
supplied
Because
i ¼�e� �t/60
mA
and�3 �
0 least
0Solution
Thus,
we
have
verified
the
solution,
we
communicate
by
recording
the
requireThus,
we
the
solution,
and
we
itit�by
0and
0 the
�t/60
�t/30 �t/60 �3
Z
Thus,
we
have
verified
the
solution,
and
we
communicate
it
by
recording
the
requirePor
lo
tanto,
se
ha
verificado
la
solución,
y
se
emite
un
comunicado
registrando
el
requeThus,
we
have
verified
the
solution,
and
we
communicate
it
by
recording
the
require�t/60
60
e
dt
¼
2e
�
10
2e
¼ 51:8
w
¼
ment
forV,
2-V
battery.
Thus,
we
have
verified
the
solution,
and
we40communicate
communicate
itusing
by recording
recording
the requirerequireWe
must
verify
that
at
least
mJ is
supplied
the
2-V
battery.
Because
e�
mAdt
and
� i¼
��10�t/60
� mJ
2eThus,
the
energy
supplied
by
the
battery
iscommunicate
v2 ¼ment
ment
for
aa 2-V
2-V
battery.
Thus,
we
have
verified
the
solution,
and
we
communicate
it
by
recording
the
require7.
The
current
is
limited,
so
D
�
1
mA.
have
verified
the
solution,
and
we
it
by
recording
the
requirefor
awe
battery.
Thus,
we
have
verified
the
solution,
and
we
it
by
the
Thus,
we
have
verified
the
solution,
and
we
communicate
it
by
recording
the
requireThus,
we
have
verified
the
solution,
and
we
communicate
it
by
recording
the
require�t/60
�3
0
0
Thus,
we
have
verified
the
solution,
and
we
communicate
it
by
recording
the
require�t/60
ment for
a
2-V
battery.
rimiento
de
una
batería
de
2
V.
e
dt ¼
2e
�
10
w
¼
ment
for
a
2-V
battery.
ment
for
2-V
battery.
V, the
supplied by the battery is
v2 ¼ 2e
Z energy
ment
aZa2-V
ment
forfor
afor
2-V
battery.
ment
aabattery.
2-V
battery.
60
60
ment
for
afor
2-V
battery.
ment
2-V
0
�battery.
�� �t/60
ment for�3a �2-VZbattery.
Thus, we�3have verified the
solution,
and
we
communicate
it
by
recording
the
�t/60
�t/30
Z 60
dt ¼ 60 � 2e ��� 10 dt ¼ �51:8 mJ
2e
� 10
w¼
State
thee Goal
Thus,
we
have
verified
the
solution,
and
w
ment
for
a
2-V
battery.
�t/60
�t/60
�3
�t/30
�3
0
0 2e
dt one-minute
¼
� 10for the
2e period
� 10 dt
¼ select
51:8 mJ
¼
Determine the energywsupplied
by the firste element
ment for a and
2-Vthen
battery.
0
0
weM
have
the solution,
we communicate
by recording the requirethe
D and B.and
Describe
the battery itselected.
1.9 Thus,
SU
UM
M
AR
Rverified
Yconstants
1.9
S
U
M
Mbattery.
A
R
Y
1.9
S
M
A
Y
Thus,
we
have
verified
the
solution,
and we communicate it by recording the require1.9
R
E
S
U
M
E
N
ment
for
a
2-V
1.9Charge
S Uis
M
M
A
R
YA
1.9
S
U
M
M
RRYY ofof
Charge
the
intrinsic
property
of matter
matter responsible
responsible for
for
Table 1.5-1
1.5-1 summarizes
summarizes the
the use
use of
of the
the passive
passive convention
convention
1.9
S
U
M
M
A
the
intrinsic
property
matter
responsible
for
Table
1.5-1
summarizes
the
use
of
the
passive
convention
Charge
isis
the
intrinsic
property
Table
S
U
M
M
A
R
Y
1.91.9
S
U
M
M
A
R
Y
1.9
S
U
M
M
A
R
Y
S
U
M
A
R
Y
1.9
La
carga
es
la
propiedad
intrínseca
de
la
sustancia
que
hace
La
tabla
1.5-1
resume
la utilización
laof
convención
pasiva
1.9
S
U
M
M
A
R
Y
ment
for
a
2-V
battery.
Charge
is the is
intrinsic
property
of
matter
responsible
for for Table
1.5-1
thesupplied
usethe
of de
the
passive
convention
1.9property
S aU
M
A Rresponsible
Y is the
the The
intrinsic
of M
matter
Table summarizes
1.5-1
summarizes
use
the passive
convention
electricCharge
phenomena.
current
in
circuit
element
when calculating
the
power
or received
by a circuit
E1C01_1
Charge
is the
the
intrinsic
property
of
matter
responsible
forTable
Table
1.5-1
summarizes
the
use
of the
the
passive
convention
Generate
ain
Plan
electric
phenomena.
The
current
in
aofcircuit
circuit
element
the
when
calculating
thesummarizes
power
supplied
or
received
by
acircuit
circuit
electric
phenomena.
The
current
amatter
element
isis elethe
when
calculating
the
power
supplied
or
received
by
aconvention
Charge
is
intrinsic
property
of
responsible
for
1.5-1
use
of
the
passive
convention
Charge
is Charge
the
property
ofproperty
matter
responsible
for
Table
1.5-1
summarizes
thethe
use
of
the
passive
convention
is
intrinsic
of
matter
for
Table
1.5-1
the
use
of
passive
convention
posibles
los
fenómenos
eléctricos.
enresponsible
un
al
calcular
lasummarizes
potencia
alimentada
oof
recibida
por
un
elemento
Charge
isintrinsic
the
intrinsic
property
matter
responsible
Table
1.5-1
summarizes
the
use
of
the
passive
Charge
isthe
the
intrinsic
property
responsible
forfor
Table
1.5-1
summarizes
the
use
the
passive
convention
electric
phenomena.
The
current
inofLa
a matter
circuit
element
is
the
when
calculating
thefor
power
supplied
or
received
by
a circuit
Charge
iscorriente
the
intrinsic
property
ofM
matter
responsible
Table
1.5-1
summarizes
the
of the passive co
electric
phenomena.
The
current
in
a
circuit
element
is
the
when
calculating
the
power
supplied
or
received
by
circuit
rate
of
movement
of
charge
through
element.
The
element.
electric
phenomena.
The
current
in
a
circuit
element
is
the
when
calculating
the
power
supplied
or
received
byaaause
circuit
1.9
S
U
M
A
R
Y
rate
ofelectric
movement
ofThe
charge
the
element.
The
element.
First,
find
vthrough
(t)
i(t)
and
then
obtain
the
power,
p1(t),
supplied
by
the
first
element.
rate
of
movement
of
charge
the
element.
The
element.
electric
phenomena.
The
current
in
aand
isde
the
calculating
the
power
supplied
ororreceived
by
aby
electric
phenomena.
The
current
in
a in
circuit
element
iselement
the
calculating
the
power
supplied
or
received
aNext,
circuit
phenomena.
The
current
in
a element
circuit
is when
thewhen
when
calculating
the
power
supplied
orbyreceived
circuit
1through
mento
de
circuito
es
la
proporción
movimiento
lathe
de
circuito.
electric
phenomena.
The
current
indel
acircuit
circuit
element
isthe
when
calculating
power
supplied
or
received
a by
circuit
electric
phenomena.
current
acircuit
element
is
when
calculating
thethe
power
supplied
received
by
acircuit
circuit
rate
of
movement
of
charge
through
the
element.
The
element.
electric
phenomena.
The
current
in
a
circuit
element
is
the
when
calculating
the
power
supplied
or receivedthe
by
rate
of
movement
of
charge
through
the
element.
The
element.
voltage
across
an element
element
indicates
the
energy
available
The
SIunits
units
(Table1.3-1)
1.3-1)
areS
used
bytoday’s
today’s
engineers
and1.5-1 summarizes
Ude
M
M
Afor
Rengineers
YentreTable
rate
of an
movement
of
charge
through
the
element.
The
element.
Charge
is The
the
intrinsic
property
of1.9
matter
responsible
voltage
across
an
element
indicates
the
energy
available
The
SI
units
(Table
1.3-1)
are
used
by
today’s
engineers
and
voltage
across
indicates
energy
available
The
SI
(Table
are
used
by
and
of
movement
of
charge
through
the
element.
The
element.
raterate
ofrate
movement
ofelemento.
charge
through
the
element.
The
element.
(t),
find
the
energy
supplied
for
the
first
60element.
s.
using
pelement
rate
of
movement
charge
through
the
element.
element.
carga
aof
través
del
Elrate
voltaje
que
pasa
aenergy
través
deThe
Las
unidades
SI (tabla
1.3-1)
son
uso
común
ingeof
movement
of
charge
through
the
element.
element.
rate
movement
ofan
charge
through
the
element.
The
element.
1of
voltage
across
an
element
indicates
the
energy
available
The
SI
units
(Table
1.3-1)
are
used
by
today’s
engineers
and
of
movement
of
charge
through
the
The
element.
voltage
across
indicates
the
available
The
SI
units
(Table
1.3-1)
are
used
by
today’s
engineers
and
to
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
Charge
is
the
intrinsic
property
of
matter
responsible
voltage
across
anindicates
element
indicates
the
energy
available
The
SI
units
(Table
1.3-1)
are
used
by
today’s
engineers
and the power
electric
phenomena.
current
inAlaare
circuit
element
isby
the
when
calculating
tovoltage
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
to
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
voltage
across
element
indicates
the
energy
available
SI
units
(Table
1.3-1)
are
used
by
today’s
engineers
and
across
element
the
energy
available
TheThe
SIThe
units
(Table
1.3-1)
used
by
today’s
engineers
and
voltage
across
an
element
indicates
the
energy
SI
units
(Table
1.3-1)
are
used
today’s
engineers
and
un
elemento
indica
la
energía
disponible
para
que available
laindicates
nieros
yThe
científicos.
usar
prefijos
decimales
(tabla
1.-3)
voltage
across
element
indicates
the
energy
available
SI
units
(Table
1.3-1)
used
by
today’s
engineers
and
across
anan
element
indicates
the
energy
available
The
units
(Table
1.3-1)
areare
used
today’s
engineers
and
1.9 voltage
S
U
M
Aan
Ran
Y
to
cause
charge
to
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
voltage
anhacer
element
the SIenergy
available
SIby
units
(Table
1.3-1)
are
by today’s
engin
to
cause
charge
move
through
the
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
simply
express
electrical
quantities
withThe
a(Table
wide
range
of
Given
the
current,
i,move
andto
voltage,
v,
ofelement.
aacross
circuit
element,
the
electric
phenomena.
The
current
inused
amay
circuit
element
is
toM
cause
charge
to
move
through
the element.
element.
scientists.
Using
decimal
prefixes
1.3-3),
we
may
ratethe
of
movement
of
charge
through
theThe
element.
element.
simply
express
electrical
quantities
with
a
wide
range
of
Given
the
current,
i,
and
voltage,
v,
of
a
circuit
element,
the
simply
express
electrical
quantities
with
a
wide
range
of
Given
the
current,
i,
and
voltage,
v,
of
a
circuit
element,
to
cause
charge
to
move
through
the
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
to
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
to
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
carga
se
mueva
a
través
del
elemento.
podemos
simplificar
la
expresión
de
las
cantidades
eléctrito
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we
may
to is
cause
charge
toi,
move
through
the
scientists.
Using
decimal
prefixes
(Table
1.3-3),
we may
1.9
Uaelement.
M
M
Amove
Rfor
Yelement,
Charge
the
intrinsic
property
ofgiven
matter
responsible
Table
1.5-1
summarizes
theelectrical
use
of
the
passive
GOAL
EQUATION
NEED
INFORMATION
express
electrical
quantities
withUsing
aconvention
wide
of
Given
the
current,
andw,voltage,
v,S
of
circuit
element,
the across
tovoltage,
cause
charge
through
thesimply
element.
scientists.
decimal
prefixes
(Table
1.3-3),
simply
express
quantities
with
aarange
wide
range
of
Given
the
current,
i,i,are
and
v,
of
aa to
circuit
the
magnitudes.
power,
p,
and
energy,
are
by
rate
of
movement
of
charge
through
the
element.
simply
express
electrical
quantities
with
wide
range
of
Given
the
current,
and
voltage,
v,
of
circuit
element,
the
voltage
an
element
indicates
the
energy
available
The
SI
units
(Table
1.3-1)
ar
magnitudes.
power,
p,
and
energy,
w,
given
by
magnitudes.
power,
p,
and
energy,
w,
are
given
by
simply
express
electrical
quantities
with
a by
wide
ofrange
the
current,
i,yi,and
voltage,
v,
aof
circuit
the
simply
express
electrical
quantities
with
awith
wide
of the
Given
the
current,
i,The
and
voltage,
v,v,Charge
of
circuit
element,
the
simply
express
electrical
with
arange
wide
of
Given
the
current,
i,voltage,
and
v,
of
aelement,
circuit
element,
the
Given
Dada
la
corriente
i,
el
voltaje
de
un
elemento
de
circuito,
cas
con
un
amplio
espectro
dequantities
magnitudes.
express
electrical
awide
wide
range
Given
the
current,
i,
and
voltage,
v,
athe
circuit
simply
express
electrical
quantities
with
asummarizes
range
ofof
Given
the
current,
and
v,aof
of
ais
circuit
element,
thethewhen
property
matter
responsible
for
Table
1.5-1
useUsing
of with
the
passive
electric
phenomena.
current
in
avoltage,
circuit
element
the
calculating
the
power
supplied
orquantities
received
arange
circuit
magnitudes.
power,
p,
and
energy,
w,
are
given
by
simply
express
electrical
quantities
a wide
Given
the
current,
i,iselement,
and
v,ofof
asimply
element,
the
magnitudes.
power,
p,
and
energy,
w,
are
given
by
Z
voltage
an except
element
indicates
the
energy
availpc
magnitudes.
power,
p,
and
energy,
w,
are
given
by intrinsic
to voltage,
cause
tocircuit
move
through
the
scientists.
decimal
60 charge
Zthe
The
energy
wgiven
for
and when
i across
known
for
v1 element.
magnitudes.
power,
p,
w,
given
by
magnitudes.
laof
potencia
penergy,
yenergy,
laenergy,
energía
w,
resultan
de
power,
p, power,
and
w,energy,
are
given
by
magnitudes.
p,
w,
are
given
1
tt by
magnitudes.
power,
p,and
and
energy,
w,
are
by
magnitudes.
power,
p,and
w,are
are
given
by
Z
Z
electric
phenomena.
The
current
in
a
circuit
element
is
the
calculating
the
power
supplied
or
received
rate
movement
ofand
charge
through
the
element.
The
element.
magnitudes.
power, p,
and
energy,
w,
are
given
by
t
to
cause
charge
to
move
through
the
element.
simply
express
electrical
qb
Given
the
current,
i,
and
voltage,
v,
of
a
circuit
element,
the
Z t pdt
ZZ t w1 ¼
p1 ðtÞ dt
p1(t)
first
60
s rate
constants
Dtoday’s
and B engineers and
pelement
¼
v
�
i
and
w
¼
t
Z
of
movement
of
charge
through
the
element.
The
element.
Z
voltage across anpp
indicates
the
energy
available
Z
The
SI
units
(Table
1.3-1)
are
used
by
¼
v
�
i
and
w
¼
pdt
Z
Z
¼ v � i and w ¼ t 0t pdt
0 p, and energy,
t
t
tpdt
v, of a circuit element
magnitudes.
power,
Z t w, are given by Given the current, i, and voltage,
p ¼ v p�pthrough
i¼
w
¼ 0w0w
vand
�� iithe
and
¼
¼
and
¼ pdt
pdt
across
anpdt
element indicates
theUsing
energydecimal
available
The
SI units
(Table
1.3-1)
are used
to cause chargep to
element.
scientists.
prefixes
(Table
1.3-3),
wew,may
p move
wand
¼w¼¼pdt
¼
vp¼
� vi¼v�piv�and
w
¼
¼
�and
i voltage
w
¼
pdt
yvvand
p¼
i� iand
w
pdt
power,
p,
and
energy,
are
given
by by today’s eng
0 pdt
0
p¼
i and
w¼
pdt electrical quantities
Z t scientists.
0 v � the
to acause
through
the express
element.
Using
decimal
simply
with a wide
range
of prefixes (Table 1.3-3)
Given the current, i, and voltage, v, of
circuit
0 0 0charge
0 to move
0element,
0
Z t with a wide
� i and
w¼
pdtsimply express electrical quantities
v,pof¼a vcircuit
element,
the
magnitudes.
power, p, and energy, w,
areon
given
by
Act
theGiven
Planthe current, i, and voltage,
0
p ¼ v � i and w ¼
pdt
magnitudes.
power,
p, and
energy,
are given by
First, we need
so we
firstw,
calculate
Z p1(t),
t
p ¼ v � i and w ¼
�
��
�
Z t
pdt
PR
RO
OBB
BLpL
L1EE
EðtM
M
S iv1 ¼ De�t/60
Þ S
¼
� 10�3 A Be�t/60 V
P
R
O
M
S
P
LRR
M
0PPRRO
OBPB
E
M
SiS
BB
L�L
EEM
SS�t/30w ¼ �3pdt
pLEO
¼
vA
and
P
O
M
¼
DBe
� 100 W ZZ¼ 0DBe�t/30 mW
RP
OR
P RPP
O
LBP
E
M
SB
LSE
M
S
Z 0
O
L
E
M
S
RB
O
BLRB
LEO
EM
MS
ZZ
ZMS
PROBLE
0
00
00
Zdt
ZÞÞÞdt
Z 0¼
0dt
¼ZZ 0 000dt
¼ 000
Hint:qqqððð000ÞÞÞZ¼
¼ZZ 0 iiiðððtZ
0 Z
0¼
Section1.2
1.2 Electric
ElectricCircuits
Circuitsand
andCurrent
Current
ttZ
dt
¼
Hint:
¼
Z
¼
Hint:
Z
Z
Z
Section
1.2
Electric
Circuits
and
Current
P
R
O
BZ0¼
L00Edt
0
0
0
0
0
0¼S
Section
0
0
�1
�1
Z 0
Z¼
i
ð
t
Þ
dt
¼
dtM
Hint:
q
ð
0
Þ
¼
i
ð
t
Þ
dt
00dt
Hint:
q
ð
0
Þ
¼
0 00
�1
�1
Section
1.2
Circuits
and Current
Sección
1.2 Electric
Circuitos
y corriente
eléctricos
�1
�1
q
ð
0
Þ
¼
i
ð
t
Þ
dt
¼
dt
¼
Hint:
Section
1.2
Electric
Circuits
and
Current
Sugerencia:
Section
1.2
Electric
Circuits
and
Current
i
ð
t
Þ
dt
¼
0
dt
¼
Hint:
q
ð
0
Þ
¼
Þþidt
00:8
dt
0dt¼00¼
qHint:
ð0ÞHint:
idt
ð¼
t Þ�
dt
¼0¼
ð�1
0ið4t
Þt¼
P0RdtO¼B0L E M S
�5t
ðÞ¼
tdt
Þ�5t
¼Hint:
0dt
0 0¼ 0iðt Þ dt ¼
Þq¼
¼
ð0:8e
ti�1
0dt
Hint:
q¼
ðqq0ððÞ0tÞ¼
�1
PSection
1.2-1
The
total
charge
that
has
entered
acircuit
circuit
elementisis
isqqqHint:
�1
Section
1.2
Electric
Circuits
and
Current
Section
1.2
Electric
Circuits
and
Current
Section
1.2
Electric
Circuits
and
Current
Section
1.2
Electric
Circuits
and
Current
1.2
Electric
Circuits
and
Current
�5t
Answer:
C
for
t
�
�1
�1
q
ð
0
Þ
¼
1.2-1
The
total
charge
that
has
entered
a
circuit
element
PP
1.2-1
The
total
charge
that
has
entered
a
element
Z
Z 0
Section
1.2
Electric
Circuits
and
Current
Answer:
q
ð
t
Þ
¼
4t
þ
0:8e
�
0:8
C
for
t
�
0
�1
�1
�1
qðtÞ�1
¼ 4t
þ 0:8e
� 0:8
C �1
for t � 0
�1 �5t
�5t
�1
�1
�1
�1
�5t
P
1.2-1
The
total
charge
that
has
entered
aelemento
circuit
element
is q Answer:
PP1.2-1
The
total
that
has
entered
aattcircuit
element
is
qq Answer:
�5t
La
carga
total
quecharge
ha00entrado
a¼
un
de0.
circuito
�5t
�1
�1
)
when
t
�
and
q(t)
¼
0
when
<
0.
Deter(t)
¼
1.25(1�e
�5t �
Answer:
q
ð
t
Þ
¼
4t
þ
0:8e
�
0:8
C
for
t
�
0
1.2-1
The
total
charge
that
has
entered
circuit
element
is
q
ð
t
Þ
¼
4t
þ
0:8e
0:8
C
for
t
�
0
)
when
t
�
and
q(t)
0
when
<
Deter(t)
¼
1.25(1�e
Answer:
q
ð
t
Þ
¼
4t
þ
0:8e
�
0:8
C
for
t
�
0
�5t
)charge
when
t
�
0
and
q(t)
¼
0
when
t
<
0.
Deter(t)P1.2-1
¼
1.25(1�e
�5t
�5t
P
The
total
that
has
entered
a
circuit
element
is
q
P 1.2-1
The
total
charge
that
has
entered
a
circuit
element
is
q
P
1.2-1
The
total
charge
that
has
entered
a
circuit
element
is
q
�5t
�5t
�5t
i
ð
t
Þ
dt
¼
Hint:
q
ð
0
Þ
¼
�5t
P
1.2-1
The
total
charge
that
has
entered
a
circuit
element
is
q
P
R
O
B
L
E
M
S
1.2-1
The
total
charge
that
has
entered
a
circuit
element
is
q
25t
P
1.2-3
The
current
in
a
circuit
element
is
i(t)
¼
4
sin
5t
A
Answer:
q
ð
t
Þ
¼
4t
þ
0:8e
�
0:8
C
for
t
�
0
Answer:
qaAnswer:
ðtcircuit
ÞThe
þ
0:8
Celement
for
tCfor
�for
0C
q¼4t
ð0:8e
tÞ4tþ¼
4t
þ
0:8e
�Cpara
0:8
for
t�
0sin�5t
�5t
1.2
Electric
and
Current
Answer:
t¼
Þelement
þ
�
0:8
ti(t)
�
0¼
Answer:
q¼Circuits
ðqt4t
Þðcurrent
0:8e
�
0:8
t�
0¼
Respuesta:
)inwhen
twhen
� 0 Pand
q(t)
¼
0q(t)
when
twhen
< Section
0. Deter(t)
¼ 1.25(1�e
1.2-1
The
total
charge
that
has
entered
is0:8e
1.2-3
The
current
in
circuit
element
i(t)
5t �
))circuit
�
000 and
¼
Deter(t)
¼
1.25(1�e
es
q(t)
5
1.25(12e
) cuando
ttelement
yand
q(t)
5
0 0cuando
ttt<
,
0.
PP
1.2-3
in
aaq�
circuit
440:8e
sin
5t
AA0:8 C�1
mine
the
current
this
circuit
element
for
�
0.
�5t
�5t
Answer:
qðelement
tÞisis
4t
þ
for t � 0
when
�q(t)
q(t)
when
<0.
0.
Deter(t)current
¼�5t
1.25(1�e
�5t
mine
the
current
inwhen
tt�5t
�
0.
mine
the
this
element
for
t0when
�
0.
)in
t)0�
0for
when
<
0.
Deter¼(t)
P
1.2-3
The
current
a Section
circuit
element
is¼i(t)
4i(t)
sincharge
5t
) �5t
when
twhen
�circuit
and
q(t)
¼
0¼when
t¼
<
Deter(t) (t)
¼
1.25(1�e
when
tt �
0q(t)
and
q(t)
¼
00t0.
when
t<
0.
Deter(t)
¼
1.25(1�e
PP
1.2-3
The
current
in
istotal
¼
44 A
sin
A
P¼R
Owhen
LE
M
SDeter)this
and
q(t)
¼
when
t�<
¼
1.25(1�e
t this
�0t �
0and
and
¼
0circuito
tpara
0.0.
(t)
¼1.25(1�e
1.25(1�e
when
t0�
�
0Band
and
i(t)
¼
0in
when
t<
<aa0.
0.circuit
Determine
the¼
total
1.2-3
The
current
in
circuit
element
is
i(t)
¼
sin 5t
5tCurrent
A
1.2
Electric
Circuits
and
mine
the
current
in) when
this
circuit
element
for
t�
0.
)for
when
tt0.Deter�Deter0 and
q(t)
t
<
0.
(t)0circuit
¼
1.25(1�e
when
t
�
0
and
i(t)
¼
0
when
t
<
0.
Determine
the
charge
mine
current
element
for
t<
Determine
lathe
corriente
enin
este
elemento
de
0.
when
t
0
i(t)
¼
0
when
t
Determine
the
total
charge
P
1.2-3
The
current
in
a
circuit
element
is
i(t)
¼
4
sin
5t
A
Z
Z
�5t
P
1.2-3
The
current
in
a
circuit
element
is
i(t)
¼
4
sin
5t
A
P
1.2-3
The
current
in
a
circuit
element
is
i(t)
¼
4
sin
5t
A
�5t
mine
the
current
in
this
circuit
element
t
�
0.
P
1.2-1
The
total
charge
that
has
entered
a
circuit
element
is
q
P
1.2-3
The
current
in
a
circuit
element
i(t)
¼
4
sin
5t
A
P
1.2-3
La
en
un
de
circuito
es
i(t)
5
4
sen
P
1.2-3
The
current
in
a
circuit
element
is
i(t)
¼
4
sin
5t
A
0corriente
0elemento
Answer:
q
ð
t
Þ
¼
þis0:8e
�5t
Answer:
i
ð
t
Þ
¼
6:25e
A
mine
the
current
in
this
circuit
element
for
t
�
0.
when
t
�
0
and
i(t)
¼
0
when
t
<
0.
Determine
the
total
charge
�5t
mine
the
current
in
this
circuit
element
for
t
�
0.
mine
the
current
in
this
circuit
element
for
t
�
0.
P
1.2-3
The
current
in
a
circuit
element
i(t) ¼ 4�
when
t
�
0
and
i(t)
¼
0
when
t
<
0.
Determine
the
total
charge
minethe
this
circuit
element
�
0. circuit element
mine
inin�5t
this
circuit
element
forfort �tin
0.this
that
has
entered
circuit
element
for
t�
�0.
0.
Answer:
iððtcurrent
tÞÞcurrent
¼ 6:25e
6:25e
when
t0.�
�
0aa0circuit
and
i(t)element
¼
0t1.2-1
when
t<
<
0.
Determine
the
total
chargea4t
Answer:
ithe
¼
AA
�5t
mine
the
current
for
t
�
that
has
entered
circuit
element
for
t
�
0.
�5t
that
has
entered
a
for
t
when
t
�
0
and
i(t)
¼
0
when
<
0.
Determine
the
total
charge
25t
Z
Z
when
t
�
0
and
i(t)
¼
when
t
<
0.
Determine
the
total
charge
when
t
0
i(t)
¼
0
when
t
0.
Determine
the
total
charge
P
The
total
charge
that
has
entered
circuit
elemen
)
when
t
�
0
and
q(t)
¼
0
when
t
<
0.
Deter(t)
¼
1.25(1�e
5t
A
cuando
t
0
e
i(t)
5
0
cuando
t
,
0.
Determine
la
carga
when
t
�
0
and
i(t)
¼
0
when
t
<
0.
Determine
the
total
charge
i
ð
t
Þ
dt
0
dt
¼
0
Hint:
q
ð
0
Þ
¼
when
t
�
0
and
i(t)
¼
0
when
t
<
0.
Determine
the
total
charge
�5t
Answer:
i
ð
t
Þ
¼
6:25e
A
0 ¼ 0 when t < 0.
0
Answer:
iiððtt�5t
Þ6.25e
¼
6:25e
Respuesta:
i(t)current
5
A�5t
�5t
Section
1.2
Electric
andA
Current
Zentered
Zwhen
that has
entered
a circuit
element
for
0. tti(t)
t �t 0��5t
and
tot
Answer:
ÞCircuits
¼
6:25e
A
that
has
aa circuit
element
for
�
�5t
�5t
�5t
�5t) A
Z
that
has
entered
circuit
element
for
�0.
0. t P� 1.2-3
TheDetermine
current
inthe
a circ
PAnswer:
1.2-2
inA
circuit
element
isi(t)
i(t)¼
¼4(1�e
4(1�e
�5t
ZaZ
Zelement
Answer:
ðThe
Answer:
iðAnswer:
tÞiThe
¼
i6:25e
ðtÞ6:25e
¼
6:25e
A
�1
�1
00 acircuit
001.25(1�e
Answer:
ðcurrent
tcurrent
Þ6:25e
¼
itðÞti6:25e
Þ¼
¼
AA
�5t
0a
0Hint:
that
has
entered
t0.
�
1.2-2
The
in
circuit
element
i(t)
¼
4(1�e
)AAthat
has
entered
element
for
t0.
�qfor
that
has
entered
a element
circuit
�
0.
0element
and
0t when
(t)
PP
1.2-2
in
aaaA
circuit
element
)the
mine
current
in
this
circuit
element
for
t element
�0for
total
que
ha
entrado
en
un
elemento
de
para
0.
iðtt Þ dt
¼ q(t)for
0¼dt
¼0.0 t < 0. D
ðcircuito
Þ0.
¼
that
has
entered
acircuit
circuit
t0for
�
that
has
entered
element
t0�
0.)0.twhen
�5t
Z circuit
Z¼that
Answer:
iðis
tis
Þ¼
6:25e
A
�5t
ZtZ
Zfor
Section
1.2
Electric
Circuits
and
Current
0
0
has
entered
a
circuit
�
�5t
Z
0
0
Hint:
q
ð
0
Þ
¼
i
ð
Þ
dt
¼
dt
¼
P
1.2-2
The
current
in
a
circuit
element
is
i(t)
¼
4(1�e
)
A
)
A
P
1.2-2
The
current
in
a
circuit
element
is
i(t)
¼
4(1�e
�5t
when telement
� 0 andfor
i(t)t¼�00.when t
0 ¼
0¼ 00
P 1.2-1
The
total
charge
has
a circuit
element
q4(1�e
when
�The
and
i(t)
¼in
<
0.
Determine
theis
total
charge
�5t
La
corriente
en
unin
elemento
de
circuito
esis
5 �5tHint:
P
1.2-2
Hint:
ÞZ¼
¼
dt
¼
00 dt
dt
�5t
A qqðqtððÞ00�5t
P
1.2-2
The
current
in
circuit
element
i(t)
¼
Zcurrent
�1
Þ¼
iiððttZÞÞdt
0for
�5t
Zthe
Z�1
4t0Zþ
0:8e
0:8
t¼
when
tt�
�
00�5t
and
i(t)
¼
00circuit
when
ttelement
<
0.
Determine
the
total
charge
0Z
0C
0Z Z
0 �Z
0�
mine
this circuit
when
tThe
01.2-2
and
i(t)
¼
0that
when
tentered
<
0.
Determine
the
total
charge
A
P
current
awhen
isisi(t)
¼
4(1�e
0 i0ðt�5t
0¼
)i(t)
A)�5t
P 1.2-2
current
in
is element
i(t)
¼
4(1�e
))¼A
P
The
aaelement
circuit
is
i(t)
¼
4(1�e
�1
Zin¼
Z �1
)charge
P1.2-2
1.2-2
The
inacircuit
a 0circuit
element
isi(t)
i(t)
¼
4(1�e
)Answer:
A
P1.2-2
current
in
circuit
element
¼
4(1�e
Hint:
qis
ða0i(t)
Þ¼
Þ dt
0 dt
¼000dt
Hint:
qA
0�1
iið¼
Answer:
icharge
ðA
tÞentered
6:25e
0 00 that �5t
0
�1
when
t25t
�entered
0)The
and
i(t)
¼tatcurrent
0�
when
twhen
<
0.1.2-1
Determine
the
total
) isA
P
The
current
in
acharge
circuit
element
¼
�1
Hint:
qð4(1�e
ð�1
0iÞelement
Þt¼
¼
ðqtttÞÞÞ dt
dt
¼
dt
¼
when
t)cuando
�
0current
i(t)
¼
tt0<
0.
Determine
the
total
has�entered
aforcircuit
P1.2-2
The
total
charge
that
has
circuit
that
has
aand
circuit
element
for
t�
�
0.
when
0
and
q(t)
¼
when
t
<
0.
Deter(t) ¼when
1.25(1�e
4(12e
A
0
e
i(t)
5
0
cuando
t
,
0.
Determine
la
when
t
�
0
and
i(t)
¼
0
when
<
0.
Determine
the
total
Hint:
q
ð
0
Þ
¼
i
ð
t
Þ
dt
¼
0
dt
¼
Answer:
q
ð
t
Þ
¼
4t
þ
0:8e
0:8
C
t � 0elem
Hint:
q
ð
0
Þ
¼
ð
Þ
dt
¼
0
dt
¼
0
Hint:
i
ð
dt
¼
0
dt
¼
0
q
ð
0
Þ
¼
Sugerencia:
that
has
entered
a
circuit
element
for
t
�
0.
Hint:
q
ð
0
Þ
¼
i
ð
t
Þ
dt
¼
0
dt
¼
0
Hint:
q
ð
0
Þ
¼
i
ð
t
Þ
dt
¼
0
dt
¼
0
�1
�1
that
has
entered
a
circuit
element
for
t
0.
�1
�1
t0�
0t �
¼when
<
0.
total
charge
when
twhen
�has
and
i(t)
¼0i(t)
0and
twhen
<0t0.
Determine
thethe
total
charge
t and
�i(t)
i(t)
¼
when
t<
Determine
the
total
charge
0and
i(t)
telement
<
0.Determine
Determine
the
total
charge
when
twhen
�
0and
¼0¼
0when
twhen
<
0.Determine
the
total
charge
1.2-3
The
current
a Detercircuit
element
¼ 4 �5t
sin
5tdtA¼
�1
�5t
Hint:
ð�1
0Þ¼i(t)
¼6:25e
iðt ÞA
0 dt
iqð4(1�e
t�1
Þis
�5t
Z 0¼ 0
Z
that
entered
aentrado
circuit
element
t 00.
�
0.
t1.25(1�e
�
and
¼when
0para
when
<The
0. Determine
the
total
that
has
entered
aa0when
circuit
for
tti(t)
�
0.
�1
�1
�1
�1
�1 Answer:
)0.
tt �t0.
0P
and
q(t)
¼ 0 in
when
t�1
<in
0.charge
(t)
¼for
minethat
thehas
current
in
this
circuit
element
for
t
�
0.
carga
total
que
ha
en
un
elemento
de
circuito
�1
�1
�1
that
has
entered
circuit
element
for
�
)
A
P
1.2-2
current
a
circuit
element
is
i(t)
¼
�1
�1
entered
a acircuit
for
t0.
�
that that
has
a circuit
element
for
tfor
�for
that
has
entered
a element
circuit
element
0.
thatentered
has
entered
acircuit
circuit
element
tfor
�
has
entered
element
t �0.
0.0.t �
P 1.2-3 The
current
in a circuit
i(t)¼¼ �4
when
t �t 0�and
i(t)
¼
0 when t < 0. Determine
the total
charge
that
has
entered
a in
circuit
element
for
0.0 when
Hint:
qelement
ð0Þ ¼element
iðt¼is
Þ 4(1�e
dt
mine
the
current
this
circuit
element
for
t
�
0.
�5t
P
1.2-2
The
current
in
a
circuit
is
i(t)
when
t
�
0
and
i(t)
¼
t
<
0.
Determine
the
total
charge
Answer:
iðtÞEléctricos
¼ 6:25e - Dorf
A
when
the to
that has entered a circuit element for
t � t0.� 0 and i(t) ¼ 0 when t < 0. Determine
�1
Circuitos
�5t has entered a circuit element for twhen
when t < 0. Determine the total c
that
� 0. t � 0 and i(t) ¼ 0Alfaomega
Answer:
i
ð
t
Þ
¼
6:25e
A
�5t
Z
Z
that
has
entered
a
circuit
element
for
t
�
0.
0
0
P 1.2-2 The current in a circuit element is i(t) ¼ 4(1�e ) A
has entered a Zcircuit elementZfor t � 0.
�5t that
Hint:
qð0Þ ¼is i(t) ¼ið4(1�e
t Þ dt ¼
0
0
) A 0 dt ¼ 0
P 1.2-2 The
in a circuit
element
when t � 0 and i(t) ¼ 0 when t < 0. Determine
the current
total charge
�1
�1 Hint: qð0Þ ¼
iðt Þ dt ¼
0 dt ¼ 0
when
that has entered a circuit element for
t � t0.� 0 and i(t) ¼ 0 when t < 0. Determine the total charge
�1
�1
that has entered a circuit element for t � 0.
M01_DORF_1571_8ED_SE_001-019.indd 15
4/12/11 5:16 PM
rate of11/26/2009
movement of
E1C01_1
11/26/2009
16 charge through the element. The
E1C01_1
16
element.
Weenergy
must verify
that at least
supplied
using
battery.engineers
Becauseand
i ¼ e�t/60 mA and
voltage across an element indicates the
available
The40
SI mJ
unitsis(Table
1.3-1)
are the
used2-V
by today’s
�t/60
V, the energy
supplied
by the
battery
is (Table 1.3-3), we may
v2 ¼ 2e
to cause charge to move through the element.
scientists.
Using
decimal
prefixes
Z
Z
simply
express electrical
quantities
with a wide range of
Given the current, i, and voltage, v, of a circuit element, the60 �
60
�� �t/60
�
magnitudes.
power, p, and energy, w, are given by
e
2e�t/60
� 10�3 dt ¼
2e�t/30 � 10�3 dt ¼ 51:8 mJ
w¼
16
16
16
16
Z
0
t
0
Thus, we have verified the solution, and we communicate it by recording the requirep ¼Electric
v � i and
w¼
pdt
Circuit
Variables
Variables
de circuitos
eléctricos
0ment for a 2-V battery.
ElectricCircuit
CircuitVariables
Variables
Electric
PP 1.2-4
current in
element
is
Section
Units
1.2-4 The
La corriente
en auncircuit
elemento
de circuito
es
Sección1.3
1.3 Systems
Sistemasofde
unidades
1.2-4 The
The current
current in
in aa8
circuit element
element isis
Section1.3
1.3 Systems
Systemsof
ofUnits
Units
PP 1.2-4
circuit
Section
PP 1.3-1
current
of 3.2
flowsfluye
through
an
1.3-1 A
Unaconstant
corriente
constante
de mA
3.2 mA
a través
>
80 t < 2
8
>
1.3-1 What
constant
current
of
3.2has
mA
flowsthrough
throughthe
an
PP 1.3-1
AA constant
of
3.2
mA
flows
through
an
<
element.
is¿Cuál
thecurrent
charge
that
passed
0
t
2
>
0
t
<
2
2
2
<
t
<
4
>
de
un
elemento.
es
la
carga
que
ha
pasado
a
través
del
>
P2R O2S
B<U
Lt E<
M
S
<
1.9
M
M
A
R
Y
element.
What
is
the
charge
that
has
passed
through
the
<
iðtÞ ¼ >
element.
What
is
the
charge
that
has
passed
through
the
element
in
the
first
millisecond?
2 42 < t < 844
elemento en el primer milisegundo?
> �1
¼>
iiððttÞÞ¼
Charge
is<the
intrinsic property of
matterin
for
Table 1.5-1 summarizes the use of the passive convention
element
inresponsible
the first
first millisecond?
millisecond?
:
element
�1
>
Z 0
Z 0 3.2the
�1
44<<tt <
88
>
0
8
Answer:
nC
>
>
: electric phenomena. The current inRespuesta:
:
a
circuit
element
is
the
when calculating the power supplied or received by a circuit
3.2
nC
<tt
Answer:
3.2
88<
Answer:
3.2
nC
iðtcharge
ÞnC
dt ¼of 45 nC
0 dtpasses
¼ 0 through a circuit element
Hint: qð0P
Þ 1.3-2
¼
Section 1.2 Electric Circuits and00Current
A
rate
of
movement
of
charge
through
the
element.
The
element.
donde
las
unidades
de
corriente
son
A
y
las
unidades
de
tiempo
�1
�1
where the units of current are A and the units of time are s.
Una
carga
pasa
a través
de
un
elemento
1.3-2
P 1.3-2
1.3-2
A charge
charge
ofde
4545
nCnC
passes
through
ams
circuit
element
PP
A
45
nC
passes
through
circuit
element
aenergy
particular
interval
of
time
that
is 5a(Table
in 1.3-1)
duration.
across
an
element
the
available
The
SI
units
are used by today’s engineers and
�5tof
P 1.2-1 Determine
The
total
that
has entered
aand
circuit
element
qelemento
son
s.
Determine
lacurrent
carga
total
que
ha
entrado
en
un
where
thecharge
unitstotal
of
current
arevoltage
Ahas
and
the
units
ofis
time
areindicates
where
the
units
of
are
A
the
units
of
time
are
s.s. during
the
charge
that
entered
a circuit
element
Answer:
q
ð
t
Þ
¼
4t
þ
0:8e
�
0:8
C
for
t
�
0
de
circuito
durante
un
intervalo
determinado
cuya
duración
during
a
particular
interval
of
time
that
is
5
ms
in
duration.
�5t
during
a
particular
interval
of
time
that
is
5
ms
in
duration.
Determine
the
average
current
in
this
circuit
element
during
to
cause
charge
to
move
through
the
element.
scientists.
Using
decimal
prefixes (Table 1.3-3), we may
and q(t)
¼ has
0has
when
t < 0.
Deter(t) ¼ 1.25(1�e
de
circuito
para
tt � 00.
Determine
the total
total
charge
that
entered
circuit element
element
Determine
the
charge
that
entered
aa circuit
for
t � 0.) when
es
deainterval
5circuit
ms.the
Determine
lacurrent
corriente
promedio
en
este
P 1.2-3v,that
The
current
inaverage
a circuit
element
i(t)circuit
¼ 4 sin
5t
Aelemento
Determine
the
average
current
inis
this
circuit
element
during
Determine
in
this
element
during
of
time.
simply
express
electrical
quantities
with a wide range of
Given
the
current,
i,
and
voltage,
of
element,
the
mine thefor
current
in
this
circuit
element
for
t
�
0.
for
t
�
0.
t � 0.
de
circuito
ese
Respuesta: 8
Answer:
when
t �that
0 by
and
i(t) ¼durante
0of
t <intervalo.
0. Determine
the total charge
that
interval
ofwhen
time.
interval
time.
magnitudes.
power,
p,
and
energy,
w,
are
given
�5t
Answer:
i
¼
9
mA
t<2
Answer:Answer:
iAnswer:
ðtÞ ¼ 6:25e >
80A
8
>
that has entered
a circuit
for t � 0.
<
Respuesta:
9element
mA.
Answer:
¼i995
mA
Answer:
¼
mA
ZZ t iiTen
>
00 � 4 2tt <
2t
< 2t2< 4
>
�5t
>
>
P 1.3-3
billionZ electrons
per second pass through a
0
0
<
where
the
units
of
qðtÞ ¼in<
)
A
P 1.2-2 The current
a 2t
circuit
element
is
i(t)
¼
4(1�e
�t 44 422<<tt<<844
82t��
>
P
1.3-3
Ten
billion
electrons
second
pass current
throughpap
¼
v
�
i
and
w
¼
pdt
P
Ten
billion
electrons
per
second
pass
through
>
particular
circuit
element.
What
is circuito
the
average
inaa
Hint:
q
ð
0
Þ
¼
i
ð
t
Þ
dt
¼
0
dt
¼
0per
A
través
de
un
elemento
de
determinado
P
1.3-3
where
the
units
of
q
ð
t
Þ
¼
where
the
units
of
q
ð
t
Þ
¼
when t � 0 and i(t) ¼ 0:
when
Determine
�ttt< 840.
4<<
<88 the total charge
>
88�
>
tt <
0 �1mil
�1
>
particular
circuit
element.
What isispor
thesegundo.
average ¿Cuál
current
in
>
particular
circuit
element.
What
the
average
current
san
diez
millones
de electrones
es in
la
that
circuit
element?
:0element
:
that has entered a circuit
for
t
�
0.
<tt
00
88<
donde
las
unidades
de
that
circuit
element?
that
circuit
element?
corriente
promedio
en
ese
elemento
de
circuito?
charge are C.
Answer: i ¼ 1.602 nA
carga
chargeson
areC.
C.
charge
are
C.
Answer:ii¼
¼1.602
1.602
nA nA
Answer:
nA
Respuesta:
1.602
P 1.2-5 The total charge q(t), in coulombs, that enters the P
1.3-4 Thei 5charge
flowing in a wire is plotted in Figure
1.2-5
The
total
charge
q(t),
in
coulombs,
that
enters
the
PPP 1.2-5
The
total
charge
q(t),
in
coulombs,
that
enters
the
terminal
of
an
element
is
P
1.3-4
The
charge
flowing in aacurrent.
wire isis plotted
plotted in
in Figure
Figure
1.3-4
The
charge
wire
1.2-5 La carga total q(t), en culombios, que entra en la ter- PP 1.3-4.
Sketch
the
corresponding
fluye en in
un cable
se ha diagramado
en la
1.3-4 La carga que flowing
terminal
of
an
element
is
terminal
of
an
element
is
P
1.3-4.
Sketch
the
corresponding
current.
8
P
1.3-4.
Sketch
the
corresponding
current.
P
R
O
B
L
E
M
S
minal de un elemento es
figura P 1.3-4. Bosqueje la corriente correspondiente.
0
t
<
0
<
8
8
Z 0
Z 0
<0t0� 2
<2t
00
qðtÞ ¼ <
0tt <
�
:
ðt�2Þ 00�
q (t), nC
¼ 32t
2tþ e�2
�2tt �
�
qqððttÞÞ¼
22
i
ð
t
Þ
dt
¼
0 dt ¼ 0
Hint:
q
ð
0
Þ
¼
t
>
Section
1.2
Electric
Circuits
and
Current
:
:
�2ððt�2
t�2ÞÞ
q(t),
(t),nC
nC
�1
�1
qq
(t),
nC
þee�2
>22
33þ
tt >
15
P 1.2-1
The totalfor
charge
a circuit element is q Answer: qðtÞ ¼ 4t þ 0:8e�5t � 0:8 C for t � 0
Find the current i(t) and sketch
its waveform
t � 0.that has entered
15
15
) when
¼ 0 when t < 0. Deter(t)su
¼
Obtenga
la
corriente
i(t)
trace
forma
de �5t
onda
para
tt � 00. and q(t)15
Find the
the current
current
i(t) and
andysketch
sketch
its1.25(1�e
waveform
for
�0.0.
Find
i(t)
its
waveform
for
tt�
P 1.2-3 The current in a circuit element is i(t) ¼ 4 sin 5t A
P 1.2-6 An electroplating bath,
as shown
in Figure
P 1.2-6,
is element for t � 0.
mine
the current
in this
circuit
1.2-6
Anelectroplating
electroplating
bath,onto
asshown
shown
inFigure
Figure
1.2-6,en
PPP1.2-6
An
bath,
as
in
PPkitchen1.2-6,
isis
Una
solución
de electroplastia,
como
se as
muestra
when t � 0 and i(t) ¼ 0 when t < 0. Determine the total charge
used
to plate
silver
uniformly
objects
such
�5t
t, μshas entered a circuit element for t � 0.
Answer:
iðtde
Þfor
¼manera
6:25e
A and
used
to plate
plate
silver
uniformly
onto
objects
such
as
kitchenused
to
silver
uniformly
onto
objects
such
as
kitchenla
figura
P
1.2-6,
utiliza
platinar
uniforme
that
ware
and
plates.
Asecurrent
ofpara
450
A flows
20
minutes,
μs
t,t,t,μs
µs
�5t
Z 0
Z 0
ware
and
plates.
A
current
of
450
A
flows
for
20
minutes,
and
objetos
como
utensilios
de
cocina
y
platos.
Una
corriente
de
ware
and
plates.
A
current
of
450
A
flows
for
20
minutes,
and
2
4
7
each coulomb transports 1.118Pmg
of silver.
is the
1.2-2
The What
current
in aweight
circuit element is i(t) ¼ 4(1�e ) A
2
4
7
each
coulomb
transports
1.118when
mgof
silver.
What
is
the
weight
2
4
7
Hint:
q
ð
0
Þ
¼
i
ð
t
Þ
dt
¼
0 dt ¼ 0
450
A
fluye
durante
minutos,
yoftcada
culombio
transporta
each
coulomb
transports
1.118
mg
silver.
What
is
the
weight
2
4
7
of
silver
deposited
in20
grams?
� 0 and i(t) ¼ 0 when t < 0. Figure
Determine
the total charge
P 1.3-4
�1
�1
of silver
silver
deposited
in grams?
grams?
1.118
mgdeposited
de plata. ¿Cuál
es elthat
peso
enentered
gramosadecircuit
la plata
de- for
of
in
has
element
t
�
0.
Figura
1.3-4
Figure PPP 1.3-4
1.3-4
Figure
positada?
Object to
be
plated
Object
to
Object
Objeto to
que se
beplated
plated
be
va a platinar
i
ii
i
ii
i
i
Silver bar
Silverbar
bar
Silver
Barra de plata
P 1.3-5 The current in a circuit element is plotted in Figure
1.3-5
La
enaaun
elemento
de flowing
circuito
está
P1.3-5.
1.3-5Sketch
Thecorriente
current
in
circuit
element
plotted
indiagraFigure
1.3-5
The
current
in
circuit
element
isis plotted
in
Figure
PP
the corresponding
charge
through
the
mada
en
la
figura
P
1.3-5.
Bosqueje
la
carga
correspondiente
1.3-5.Sketch
Sketch
thecorresponding
correspondingcharge
chargeflowing
flowingthrough
throughthe
the
PP1.3-5.
element
for
t > the
0.
que
fluyefor
a través
element
for
>0.0.del elemento para t . 0.
element
tt >
Bath
Bath
Bath
Baño
i (t), μA
(t), μA
µA
μA
ii (t),
–450
–450
–450
Figure P 1.2-6 An electroplating bath.
Figure PPP 1.2-6
1.2-6 An
Anelectroplating
electroplating
bath.
Figure
1.2-6
bath.
Figura
Baño
de electroplastia.
PP 1.2-7
the charge,
sketch
its waveform
1.2-7 Find
Obtenga
la carga,q(t),
q(t),and
y trace
su forma
de onda when
cuanP1.2-7
1.2-7
Find
the
charge,
q(t),
and
sketch
itswaveform
waveform
when
Pdo
Find
the
q(t),
and
sketch
its
when
la
corriente
quecharge,
entra
en
una
de
un
sea
the
current
entering
a terminal
ofterminal
an
element
is elemento
as shown
in
the
current
entering
a
terminal
of
an
element
is
as
shown
in
the
current
entering
a
terminal
of
an
element
is
as
shown
como se
muestra
en la figura
P 1.2-7.
Suponga
0
Figure
P 1.2-7.
Assume
that q(t)
¼ 0 for
t < 0. que q(t) 5in
Figure
1.2-7. Assume
Assume that
that q(t)
q(t)¼
¼00 for
for tt<
<0.0.
Figure
PP0.
1.2-7.
para
t,
i (A)
i (A)
(A)
ii(A)
3
3
33
2
2
22
1
1
11
80
80
80
t, ms
ms
t,t, ms
140
140
140
–600
–600
–600
Figure P 1.3-5
Figura
1.3-5
Figure PPP 1.3-5
1.3-5
Figure
1
1
11
Figure
1.2-7
Figura PP 1.2-7
Figure PP 1.2-7
1.2-7
Figure
2
2
22
Alfaomega
M01_DORF_1571_8ED_SE_001-019.indd 16
3
3
33
4
4
44
t (s)
t (s)
(s)
tt(s)
PP 1.3-6
current inena un
circuit
element
is plottedestá
in diagraFigure
1.3-6 The
La corriente
circuito
de elemento
P 1.3-6
1.3-6 Determine
The current
currentthe
in atotal
a circuit
circuit
element
plotted
in Figure
Figure
Pmada
in
element
isis
plotted
in
1.3-6.
charge
flows
thea
en The
la figura
P 1.3-6.
Determine
lathat
carga
totalthrough
que
fluye
P 1.3-6.
1.3-6.
Determine
the 300
totaland
charge
that
flows through
through the
the
Ptravés
Determine
the
total
charge
that
flows
circuit
element
between
1200
ms.
del elemento de circuito entre 300 y 1200 ms.
circuit element
element between
between 300
300 and
and 1200
1200 ms.
ms.
circuit
Circuitos Eléctricos - Dorf
4/12/11 5:16 PM
Problems
Problems
17
Problemas
17
720
720
E1C01_1
i (t), nA
i (t), nA
i (t), nA
720
11/26/2009
–720
–720
400
400
17–720
(amp)
i i(amp)
i (amp)
30
30
400
800
800
t, μs
t, μs
t, µs1200
800
1200
1200
17
55
30
5
0
00
10
10 15
15
10
(b)
(b)
15
25
25
(b)
25
(s)
t t(s)
t (s)
1.5-4
(a) Voltage
v(t) and (b) current
i(t)un
for an element.
FiguraPFigure
P1.5-4
1-5-4P
(a)
Voltaje
v(t),and
y (b)
Figure
(a)
Voltage
v(t)
(b) corriente
current i(t)i(t)
forpara
an element.
elemento.
Figure P 1.3-6
FiguraPP1.3-6
1.3-6
Figure
P 1.5-5 An automobile battery is charged with a constant
Section 1.5 Power and Energy
PP 1.5-5
An
battery
is charged
with
a constant
Sección1.5
1.5 Power
Potencia
energía
Section
andyEnergy
1.5-5current
Unaautomobile
batería
se carga
con una
corrien- of the
of 2 Adeforautomóvil
five hours.
The terminal
voltage
P 1.5-1 Figure P 1.5-1 shows four circuit elements identified
current
of
2
A
for
five
hours.
The
terminal
voltage
of the(a) Find
te
constante
de
2
A
durante
cinco
horas.
El
voltaje
terminal
battery is v ¼ 11 þ 0.5t V for t > 0, where t is in hours.
1.5-1 Figure
La figura
1.5-1shows
muestra cuatro
circuito
PP1.5-1
P 1.5-1
circuitelementos
elements de
identified
by the letters
A, B, C, four
and D.
battery
is
v¼
11
þ
0.5t
V1
for0.5t
tto>the
0,
where
t.
is
in hours.
(a)
Find
Problems
de
la
batería
es
v
5
11
V
para
t
0,
donde
t
son
las (b) 17
the
energy
delivered
battery
during
the
five
hours.
If
identificados
por
las
letras
A,
B,
C
y
D.
by the letters A, B, C, and D.
the
energy
delivered
to
the
battery
during
the
five
hours.
(b)
horas. electric
(a) Obtenga
la costs
energía
a la
durante
(a) Which of the devices supply 30 mW?
energy
15 entregada
cents/kWh,
findbatería
the cost
ofIfcharging
(a) Which
¿Cuál de
los
proporciona
30 mW?
(a)
the dispositivos
devices
supply
30
mW?0.03
electric
energy
costs
cents/kWh,
find the tiene
cost of
las cinco
horas.
(b)for15
Sifive
la
energía
uncharging
costo de
(b) of
Which
of the devices
absorb
W?
i (t), nA
the
battery
hours. eléctrica
(b) Which
¿Cuál de
los
dispositivos
absorbe
0.03W?
(b)
of
the
devices
absorb
0.03
W?
the
for fivei (amp)
hours.
15 battery
centavos/kWh,
encuentre
el costo de cargar la batería du(c) What is the value of the power received by device B?
Answer: (b) 1.84 cents
(c) What
¿Cuál
el valor
de la the
potencia
recibida porby
el dispositivo
B? rante cinco
(c)
isesWhat
the
value
power
devicebyB?
720
(d)
is theofvalue
of thereceived
power delivered
device Answer:
B?
(b) horas.
1.84 cents30
(d)
¿Cuál
es
el
valor
de
la
potencia
entregada
por
el
dispositivo
B?
(d) What
the value
the power
by device
(e)isWhat
is theofvalue
of thedelivered
power delivered
by B?
device C?
¿Cuálisesthe
el valor
deof
la potencia
entregada
por by
elt,dispositivo
(e) What
μs
(e)
value
the power
delivered
device C?C? Respuesta: (b) 1.84 centavos.
++
+
10VV
10
10 V
––
–720
mA 3 mA
33mA
(A)
(A)
(A)
Figura Figure
P. 1.5-1P
Figure
P 1.5-1
1.3-6
Figure P 1.5-1
–
++
400
+
5V
55VV 800
––
mA 6 mA
66mA
(B)
(B)
(B)
–
–
–– 66VV
1200
6V
++
+
––
mA 5 mA
55mA
(C)
(C)
(C)
–
15VV
15
15 V
++
mA 2 mA
22mA
+
5
0
10
(D) (D)
(D)
15
25
t (s)
(b)
Figure P 1.5-4 (a) Voltage v(t) and (b) current i(t) for an element.
P 1.5-2Section
Un hornillo
eléctrico
tiene
una corriente constante P 1.5-6P Obtenga
la potencia,
p(t)
alimentada
por
el elemento
1.5-5 Find
An
automobile
battery
is charged
with
a constant
1.5electric
Power
and
1.5-6
the power,
p(t),
supplied
by the
element
shown
P 1.5-2 An
range
hasEnergy
a constant current of 10 A entering P
10 AAn
queelectric
entra range
por lahas
terminal
de current
voltaje of
positiva
con un Pque
1.5-6
Find the
p(t),
supplied
by
the
element
shown
se muestra
enpower,
laAfigura
P 1.5-6
cuando
v(t)
5 4 cosvoltage
3t V
Pde1.5-2
a
constant
10
A
entering
current
of
2
for
five
hours.
The
terminal
of
sin
3t the
the
positive
voltage
terminal
with
a
voltage
of
110
V.
The
range
is
Pde1.5-1
P 1.5-1funciona
shows four
circuit
elements
identified in Figure P 1.5-6 when v(t) ¼ 4 cos 3t V andsin
iðthours.
Þ3t¼ (a) Find
A.
voltaje
110 V.Figure
El hornillo
durante
dos
horas.
(a)
the
positive
voltage
terminal
with
a
voltage
of
110
V.
The
range
is
battery
is
v
¼
11
þ
0.5t
V
for
t
>
0,
where
t
is
in
sen
3t
operated
for two
hours.
(a) D.
Find the charge in coulombs inthat
Figure
P 1.5-6 A.
when
v(t) ¼p(t)
4 cos
3t5V0.5
ands iyðten
Þ ¼t 5 1 s. ObA.12
by
the
letters
A,culombios
B,
C, and
e
i(t)
5
Evalúe
en
t
Encuentre
la
carga
en
que
pasa
a
través
del
horni12
operatedpasses
for two
hours.the
(a)range.
Find the
coulombs
that by the Evaluate
the energy
delivered
to sthe
the five
hours.
(b) If
at t ¼ 0.5
andbattery
at t ¼ 1during
s. Observe
that
the power
12 p(t)
through
(b) charge
Find theinpower
absorbed
p(t) at tenergy
¼ 0.5
scosts
and at
¼
1 as.positive
Observe
that
power
llo. (b)through
Obtenga
potencia
absorbida
por30
el mW?
hornillo. (c)
Si
(a)
Which
of
the (b)
devices
passes
thelaIfrange.
Find supply
the
power
thela Evaluate
electric
15t has
cents/kWh,
find
thethe
oftimes
charging
supplied
by
thisalimentada
element
value
atcost
some
and
range. (c)
electric
energy
costs
12 absorbed
cents per by
kilowatt-hour,
serve
que
la
potencia
por
estos
elementos
tiene
un
by
this element
has
a positive
value at some times and
energía(c)
eléctrica
tiene
un devices
costo
deabsorb
12cents
centavos
por kilovatio- suppliedathe
(b)
Which
ofenergy
the
0.03
W?
range.
If
electric
costs
12
per
kilowatt-hour,
battery
for
five
hours.
negative
value
at
other
times.
determine the cost of operating the range for two hours. avalor
positivo
enatalgunos
momentos y negativo en otros.
value
other times.
hora, determine
del funcionamiento
delhours.
hornillo
du- B?negative
(c)the
What
the
value
of
power
by device
determine
costelisofcosto
operating
thethe
range
for received
two
1
Answer:
(b)
1.84
cents
sin ða þ
Þt þ
ða � bÞtÞ
sin at
Þðcos1 bt
¼ ð(sen(a
P horas.
1.5-3
walker’s
cassette
tapedelivered
player uses
four B?
AA Hint: ð(sen
Sugerencia:
at)(cos
bt)Þ 5
1bb)t
1sin
sen(a2b)t)
rante dos
(d)
What A
is the
value of
the power
by device
2þ bÞt þ sinða � bÞtÞ
ð
sin
ð
a
Hint:
ð
sin
at
Þ
ð
cos
bt
Þ
¼
P 1.5-3batteries
A walker’s
cassette
tape player
usesplayer
four circuit.
AA
inis series
to provide
6 V todelivered
the
The Answer:
(e)
What
the
value
of
the
power
by
device
C?
2
P 1.5-3 in
Un reproductor
portátil
de
utiliza
cuatroThe
pibatteries
to battery
provide
6 Vstore
to sonido
the
player
circuit.
four series
alkaline
cells
a total
of 200
watt-seconds
of
Answer:
Respuesta:
las AA
en+serie
65of
VV 200
al –circuito
reproduc10para
V cells
–proporcionar
+total
–
6ofV
+
–
15 V
+
four
alkaline
battery
store
a
watt-seconds
energy. If the cassette player is drawing a constant 10 mA pðtÞ ¼ 1 sin 6t W; pð0:5Þ ¼ 0:0235 W; pð1Þ ¼ �0:0466 W
1
tor. LasIfcuatro
celdas de
pilasisalcalinas
almacenan
un
total
energy.
the
cassette
player
drawing
a
constant
10
mA
6
= sen
W, p(1)
W W
sin 6t W,
W;p(0.5)
pð0:5=Þ 0.0235
¼ 0:0235
W; =
pð20.0466
1Þ ¼ �0:0466
pp(t)
ðtat
Þ¼
from
the
pack,energía.
how long
will
the cassette
operate
de 200
watts
porbattery
segundo
reproductor
utiliza
6
from
thenormal
battery
pack,
howdelong
will Si
theelcassette
operate
at
i
power?
i
una
constante
de
10
mA
del
paquete
de
pilas,
¿cuánto
tiempo
6 mA
5 mA
2 mA
normal power?3 mA
i
+
P a1.5-4
The current
through and voltage across an element
funciona
una potencia
normal?
+
(A) through and voltage
(B) across an element
(C)
(D)
P 1.5-4vary
Thewith
current
+
time as
shown
in Figure
P 1.5-4.
Sketch
v
P 1.5-4
La
y in
el
voltaje
pasan
a través
de the
un power
vary
with
timecorriente
as1.5-1
shown
Figure
Pque
1.5-4.
Sketch
the
power
v
Figure
P
delivered
to
the
element
for
t
>
0.
What
is
the
total
energy
v
elemento to
varían
al paso del
como isse the
muestra
en la fi–
delivered
the element
for tiempo
t > 0.between
What
total
delivered
to thelaelement
t¼
and energy
t ¼para
25 s? The
–
gura P 1.5-4.
Bosqueje
potencia
entregada
alt0elemento
–
delivered
to
the
element
between
t
¼
0
and
¼
25
s?
The
P
1.5-6
Find
the
power,
p(t),
supplied
by the element shown
P
1.5-2 voltage
An electric
range
hasadhere
a constant
current
of 10 A entering
element
and
current
to the
passive
Figure P 1.5-6 An element.
t . 0. ¿Cuál
es laand
energía
total
entregada
alpassive
elemento
entre tconvention.
50
sin 3t
element
voltage
current
adhere
to the
the
positive
voltage
terminal
with
a voltage
ofconvention.
110 V. The range
is
An
Figure
Pelement.
1.5-6
when v(t) ¼ 4 cos 3t V and iðtÞ ¼
A.
FiguraPin
P1.5-6
1.5-6
Un
elemento.
y t 5 25 s? El voltaje y la corriente del elemento se apegan a Figure
12
operated for
two hours. (a) Find the charge in coulombs that
v (volts)
P
1.5-7
Find
the
power,
p(t),
supplied
by
the
element
shown
la convención
pasiva.
at t ¼p(t),
0.5 p(t),
s andalimentada
at tby
¼ 1 s. Observe
that the power
v (volts)
passes
through 30
the range. (b) Find the power absorbed byPPthe
1.5-7Evaluate
Obtenga
la1.5-6
potencia,
por
elemento
1.5-7
thep(t)
element
shown
inFind
Figure
Ppower,
whensupplied
v(t)
¼
8 sinthe
3t value
V
andelati(t)
¼ 2times
sin 3t A.
supplied
by
this
element
has
a
positive
some
v
(voltios)
range. 30
(c) If electric energy costs 12 cents per kilowatt-hour,
la figura
1.5-6
5 8i(t)
sen¼3t2 V,
i(t)A.5 and
inmostrado
Figure Pen1.5-6
whenP v(t)
¼ cuando
8 sin 3tv(t)
V and
sine 3t
1
a A.
negative value at other times.
30the cost of operating the range for two hours. 2 sen 3tHint:
determine
ðsin atÞðsin1 btÞ ¼ ðcosða � bÞt � cosða þ bÞtÞ
Hint: ðsin atÞðsin btÞ ¼ ðcosða12� bÞt � cosða þ bÞtÞ
sin ða þ
Þt þ
ða �1bÞt
Þ
sin at
Þðcos2bt
¼ ð(cos(a
P 1.5-3 A walker’s
cassette tape player uses four Sugerencia:
AA Hint: ð(sen
at)(sen
bt)Þ 5
2bb)t
2sin
cos(a
b)t)
5
Answer: pðtÞ ¼ 8 � 8cos26t W
batteries5 in series to provide 6 V to the player circuit. The Answer:
Answer: pðtÞ ¼ 8 � 8cos 6t W
5 battery
four alkaline
a total of 200
of P 1.5-8
0 cells store
10 15
25 watt-seconds
Respuesta:
p(t) Find
582
cos 6t W
t (s)
the8 power,
p(t), supplied by the element shown
10 15
25
1 power, p(t), supplied by the element shown
t (s)a constant 10PmA
1.5-8 pinðFind
the
energy. If0 the cassette
player is(a)
drawing
Figure
P 6t
1.5-6.
The
element
voltage
isð1represented
sin
W;
p
ð
0:5
Þ¼
0:0235
W;
Þ ¼ �0:0466asW
t
Þ
¼
potencia,
p(t),
alimentada
porpel
elemen1.5-8
0
10 how
15
25 thet cassette
(s)
(a)long will
6 laThe
inP at
Figure Obtenga
P 1.5-6.
element
voltage
is represented
as
from the battery
pack,
operate
(a)
to mostrado en la figura 1.5-6. El ivoltaje del elemento está
normal power?
P 1.5-4
The current
Circuitos
Eléctricos
- Dorf through and voltage across an element
vary with time as shown in Figure P 1.5-4. Sketch the power
delivered to the element for t > 0. What is the total energy
delivered to the element between t ¼ 0 and t ¼ 25 s? The
element voltage and current adhere to the passive convention.
M01_DORF_1571_8ED_SE_001-019.indd 17
+
Alfaomega
v
–
Figure P 1.5-6 An element.
4/12/11 5:16 PM
t
p ¼ v � i and w ¼
E1C01_1
11/26/2009
pdt
0
18
PROBLEMS
Z 0
Z 0
i
ð
t
Þ
dt
¼
0 dt ¼ 0
Hint:
q
ð
0
Þ
¼
Section 1.2 Electric Circuits and Current
�1
�1
Z 0
Z 0
P 1.2-1 The total charge that has entered aHint:
circuit
Answer:
qð0tÞdt¼¼4t0þ 0:8e�5t � 0:8 C for t � 0
¼
qð0element
Þ ¼ is iqðt Þ dt
Section 1.2 Electric (t)
Circuits
and
Current
22t �5t
sea cero.
LaDeterfigura
P
1.7-1
muestra
un circuito. Todos los volta) when tt � 00 and
q(t)
t < 0.
1.25(1�e
representado por v(t) 5¼4(12e
)V cuando
y v(t)
5¼0 0 when
�1
�1
P �5t
1.2-3 están
The current
in a circuit element
18
Circuit
Variables
jes
y
corrientes
del
elemento
¿están is i(t) ¼ 4 sin 5t A
P 1.2-1
total
that the
has
entered in
a circuit
element
is qporfor
mine
current
this representada
circuit
element
t
�
0.
cuandoThe
t ,Electric
0. Lacharge
corriente
del
elemento
está
Answer: qðtÞ ¼ 4t þ 0:8e
�
0:8
C especificados;
for
t¼�00when pero,
when
t
�
0
and
i(t)
t
<
0.
Determine the total charge
22t �5t
correctos? Justifique su respuesta.
) whenAnswer:
when
(t) i(t)
¼ 1.25(1�e
5 2e A cuando
tt � 00,and
e ii(t)
0 0cuando
0. Deter�5t tt <
ðtq(t)
Þ5
¼¼
6:25e
A,0.
P 1.2-3 The current inthat
a circuit
element
is i(t) ¼element
4 sin 5tfor
A t � 0.
has entered
a circuit
�2tin this circuit
mine
the
current
element
for
t
�
0.
Hint:
Calculate
the
power
absorbed
each
element.
Add up
v(t) ¼ 4(1�ep(t) )V
when
� 0)e22t
andWv(t) ¼ 0 when t < 0. The when
Sugerencia:
Calcule
la potencia
absorbida
por
cada
�5t
Z by
Z elemento.
Respuesta:
5 8(1
2
et22t
t
�
0
and
i(t)
¼
0
when
t
<
0.
Determine
the
total
charge
0
0
)
A
P
1.2-2
The
current
in
a
circuit
element
is
i(t)
¼
4(1�e
of these
powers.
IfHint:
the
sum
zero,
conservation
of cumple
energy
element
is�5t
represented
as i(t) ¼ 2e�2t A when t � 0 thatall
Conjunte
todas
potencias.
lat �
suma
Answer:
iðtÞcurrent
¼ 6:25e
A t � 0 and
qð0ÞisSi
¼
iðt Þes
dtcero,
¼ se
0 dt ¼ 0
has
entered
a estas
circuit
element
for
0.
when
i(t) ¼ 0 desarrolla
when t < 0.3V
Determine
the total
charge
La
batería
de
una
luz
intermitente
y la is
P
1.5-9
�1
�1
and i(t) ¼ 0 when t < 0.
and
the
voltages
and
currents
are
probably
consatisfied
la conservación
de
energía
y
probablemente
los
voltajes
�5t
Z
Z
0
0
) A t � 0.
P 1.2-2
Thea current
in that
abulbo
circuit
element
i(t) ¼
4(1�e
hasesentered
circuit
element
for
corriente
través del
de 200aismA.
¿Cuánta
potencia
y corrientes
estén
Si 0ladt
suma
es posible
correct.
If the
sum
is ¼
not zero,
the
voltages
and
�2t
qð0Þ ¼
iðtcorrectos.
Þ dt
¼ element
0no es cero,
Answer:
pðbulbo?
tÞi(t)
¼¼
8ð10Obtenga
�
e�2ttÞe
W
when
t � 0 el
and
when
<
Determine
the total
absorbe
la 0.energía
absorbida
porcharge
el bulbo Hint:
�1 ybe
�1 elemento no sean los correctos.
que los voltajes
corrientes
currents
cannot
correct.del
thaten
adecircuit
element
fordevelops
t � 0. 3 V, and the current
un entered
periodo
cinco
P has
1.5-9
The battery
of aminutos.
flashlight
– 5V +
–5 A
– 5
V +
–5 A
through
bulb
is 200 mA. What
powerque
is absorbed
the bulb?
investigadores
médicos
estudianby
la hipertenP
1.5-10theLos
Find suelen
the energy
absorbed
by the llamada
bulb in a“electroforesis
five-minute period.
sión
emplear
una técnica
de gel
18
PR
OBLEMS
Variables de circuitos
eléctricos
2D”
para analizar
proteína contenida
una muestra de
teP 1.5-10
Medicalla researchers
studyingenhypertension
often
jido.
imagencalled
de un‘‘2D
“gel”
se puede vertoen
la figura
use a La
technique
geltípico
electrophoresis’’
analyze
the
P
1.5-10a.
protein
content of a tissue sample. An image of a typical ‘‘gel’’
El procedimiento
para la preparación del gel emplea el
is shown
in Figure P1.5-10a.
circuitoThe
eléctrico
que for
se muestra
la figura
P 1.5-10b.
La
procedure
preparingenthe
gel uses
the electric
muestra
consta deinun
gel y1.5-10b.
un filtroThe
de papel
que
contiene
circuit illustrated
Figure
sample
consists
of aprogel
teínas
ionizadas.
fuente ionized
de voltaje
originaAunvoltage
voltajesource
largo
and a filter
paper Una
containing
proteins.
ycauses
constante
de constant
500 V que
pasa a500
través
de la the
muestra.
Dicho
a large,
voltage,
V, across
sample.
The
voltaje
mueve voltage
las proteínas
través del
filtro
large, constant
movesionizadas
the ionizeda proteins
from
thehacia
filter
el
gel. to
Lathe
corriente
encurrent
la muestra
se sample
da por is given by
paper
gel. The
in the
2at
�at
i(t)
mA
iðtÞ 5
¼ 22 1
þ 30e
30e mA
donde
es the
el tiempo
consumido
el inicio of
delthe
procediwhere tt is
time elapsed
since desde
the beginning
miento
y eland
valor
la constante
a es
procedure
thedevalue
of the constant
a is
1
a ¼ 0:85
hr
Determine la
theenergía
energyalimentada
supplied bypor
thelavoltage
when
the
Determine
fuente source
de voltaje
cuangelelpreparation
procedure
lasts 3 hours.
do
procedimiento
de la preparación
del gel dura 3 horas.
+
+
–2 V
–2 V
–
–
2A
2A
2A
2A
+ 1V –
+ 1
V –
3V
+
–
3A 3V
Figura P 1.7-2
+ 500 V –
+ 500 V –
500 V
500 V
Figura P 1.5-10 (a) Imagen de un gel, y (b) el circuito eléctrico
Figure
P 1.5-10
(a) An image
a gel and (b) the electric circuit
que
se usó
en la preparación
delofgel.
used to prepare gel.
Sección 1.7 ¿Cómo lo podemos comprobar . . . ?
Section 1.7 How Can We Check . . . ?
P 1.7-1 La conservación de energía requiere que la suma de
P 1.7-1
Conservation
of energy
requires
that the
sumcircuito
of the
la
potencia
absorbida por
todos los
elementos
en un
power absorbed by all of the elements in a circuit be zero.
Alfaomega
Figure P 1.7-1 shows a circuit. All of the element voltages and
currents are specified. Are these voltage and currents correct?
Justify your answer.
M01_DORF_1571_8ED_SE_001-019.indd 18
2A
+ 4V –
–3 A
–3 V
–
+
–3 V
–3 A
+ 4 V – –3+A
–3 A
P 1.7-3 Excepto por uno, los voltajes y corrientes
que se
muestran
en la figura P 1.7-3 son correctos: la dirección de
Figure
P 1.7-2
referencia de exactamente una de las corrientes del elemento
está
invertida.
Determine
cuál dirección
de referencia
la que
P
1.7-3
The element
currents
and voltages
shown inesFigure
se1.7-3
ha invertido.
P
are correct with one exception: the reference direction
– 3Vcurrents
+
of exactly one of the element
is reversed. Determine
which reference direction has been reversed.
– 1V +
(b)
(b)
3V
–
2A +
3V
2A +
–
3A
2 A + 4 V – 3– A
–
+
–
i (t)
i (t)
5A
5A
3A
3A
P 1.7-2 La conservación de energía requiere que la suma de
la potencia absorbida por todos los elementos en un circuiP 1.7-2 Conservation of energy requires that the sum of the
to sea cero. La figura P 1.7-2 muestra un circuito. Todos los
power absorbed by all of the elements in a circuit be zero.
voltajes y corrientes del elemento están especificados; pero,
Figure P 1.7-2 shows a circuit. All of the element voltages and
¿están correctos? Justifique su respuesta.
currents are specified. Are these voltage and currents correct?
Sugerencia:
Calcule la potencia absorbida por cada elemento.
Justify
your answer.
Conjunte todas estas potencias. Si la suma es cero, se cumple
Hint: Calculate the power absorbed by each element. Add up
con la conservación de energía y probablemente los voltajes
all of these powers. If the sum is zero, conservation of energy
y corrientes estén correctos. Si la suma no es cero, es posible
is satisfied and the voltages and currents are probably
que los voltajes y corrientes del elemento no sean los correctos.
correct. If the sum is not zero, the element voltages and
+ 4V –
3A
currents cannot be correct.
3V
+
–
3V
muestra
sample
+ 4V –
+ 4
V –
Figura P 1.7-1
Figure P 1.7-1
+
(a)
(a)
+
+
3V
3V
–
–
a
+
5V a
–
4A
– 1V +
7A
+
5V
7A
Figura
– P 1.7-3
4A
–3A
– 3V +
b
–
−6V
+
–
−6V
+
– 2V +
c
–3A
b
d
d
2A
2A
−2A
– 2V + –
−8V
+
−2A
–
−8V
+
c –5A
–5A
Circuitos Eléctricos - Dorf
Figure P 1.7-3
4/12/11 5:16 PM
voltage the
across
an element indicates the
energy a
to cause charge to move through
element.
scientists.
to causev, charge
to move
through
simply exp
Given the current, i, and voltage,
of a circuit
element,
the the element.
Given
the by
current, i, and voltage, v, of a magnitude
circuit elem
power, p, and energy, w, are
given
power, p, and
Z energy, w, are given by
t
p ¼ v � i and w ¼
0p
Z
pdt
¼ v � i and w ¼
Problemas de diseño
Problemas de diseño
PD 1-1 Un elemento de circuito en particular está disponible
en tres grados. El grado A garantiza que el elemento puede absorber con seguridad 1/2 W de manera continua. Del mismo
modo, el grado B asegura que se puede absorber 1/4 W sin problemas, y el grado C confirma que puede ser incluso 1/8 W. Por
regla general, los elementos que pueden absorber más potencia
son incluso más costosos y voluminosos.
Se espera que el voltaje que fluye a través de un elemento
pueda ser de 20V y que la corriente sea de 8 mA. Ambas estimaciones tienen una certeza de 25%. La referencia de voltaje y
corriente se apegan a la convención pasiva.
Circuitos Eléctricos - Dorf
M01_DORF_1571_8ED_SE_001-019.indd 19
19
t
pdt
0
PROBLEMS
Especifique el grado de este elemento. La seguridad es la
PROBLEM
consideración más importante, pero no especifique un elemento
Hint: qð0Þ ¼
que seaSection
más costoso
de lo necesario.
1.2 Electric
Circuits and Current
Section
1.2 Electric
Circuitsisand
Theque
total
charge
that
has
a circuit
element
q Current
El voltaje
circula
a través
deentered
un elemento
de circuito
PD 1-2P 1.2-1
Answer: qðtÞ
28t �5t
The
charge
has entered a circuit el
when tt P
�1.2-1
0 when
t <that
Deter1.25(1�e) V )cuando
es v(t) (t)
5¼
20(12e
00 yand
v(t)q(t)
5 0¼total
cuando
t
0.0.La
�5t
28t
)twhen
¼ 1.2-3
0 whenThe
t<
(t)5
¼ 30e
1.25(1�e
mine
in this
circuit
element
forcuando
� 0. tt � 00,and q(t) P
corriente
enthe
estecurrent
elemento
es i(t)
mA
when
t�
� 0.
0 an
mine
the
current
in
this
circuit
element
for
t
e i(t) 5
0
cuando
t

0.
La
corriente
y
el
voltaje
del
elemento
�5t
Answer: iðtÞ ¼ 6:25e
A
that has ente
se apegan a la convención pasiva.
Especifique
la6:25e
potencia
�5t
Answer:
iðtÞ ¼is
Aque
�5t
P 1.2-2 The
current
in a circuit
element
i(t) ¼ 4(1�e
este dispositivo
pueda
ser capaz
de absorber
de manera
segura. ) A
Hint:
qð¼
0Þ4(1
¼
P 1.2-2
The currentthe
in total
a circuit
element
is i(t)
when t � 0 and i(t) ¼ 0 when
t < 0. Determine
charge
Sugerencia:
MATLAB,
un programa
when
t �for
0 and
i(t)
whendit < 0. Determine the to
that hasutilice
entered
a circuit oelement
t �similar,
0. ¼ 0 para
señar el trazo de la potencia.
that has entered a circuit element for t � 0.
Alfaomega
4/12/11 5:16 PM
CAPÍTULO
Elementos
de circuitos
E N E STE CAPÍTULO
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Introducción
Ingeniería y modelos lineales
Elementos de circuito activos y pasivos
Resistencias
Fuentes independientes
Voltímetros y amperímetros
Fuentes dependientes
2.1
2.8
2.9
2.10
2.11
2.12
Transductores
Interruptores
¿Cómo lo podemos comprobar . . . ?
EJEMPLO DE DISEÑO — Sensor de
temperatura
Resumen
Problemas
Problemas de diseño
INTRODUCCIÓN
No es de sorprender que el comportamiento de un circuito eléctrico dependa del comportamiento
individual de los elementos de circuito que comprenden el circuito. Desde luego, diferentes tipos de
elementos de circuitos se comportan de manera diferente. Las ecuaciones que describen el comportamiento de los diversos tipos de elementos de circuito se denominan ecuaciones constitutivas. Con
frecuencia las ecuaciones constitutivas describen una relación entre la corriente y el voltaje del elemento. La ley de Ohm es un ejemplo claro de una ecuación constitutiva.
En este capítulo investigaremos el comportamiento de varios tipos comunes de elementos de
circuito:
•Resistencias
•Fuentes independientes de voltaje y corriente
•Circuitos abiertos y circuitos cerrados
•Voltímetros y amperímetros
•Fuentes dependientes
•Transductores
•Interruptores
2.2
20
I N G E N I E R Í A Y M O D E LO S L I N E A L E S
El arte de la ingeniería es tomar una idea brillante y, con dinero, material, personas expertas y una mirada al entorno, producir algo que el comprador quiere a un precio a su alcance.
Los ingenieros utilizan modelos para representar los elementos de un circuito eléctrico. Un
modelo es una descripción de aquellas propiedades de un dispositivo que se consideran importantes. En ocasiones el modelo constará de una ecuación relacionada con el voltaje y la corriente
del elemento. Aun cuando el modelo sea diferente del dispositivo eléctrico, se puede usar en los
cálculos manuales que predecirán cómo funcionará un circuito compuesto de dispositivos reales. A
veces los ingenieros se topan con un dilema al seleccionar un modelo para un dispositivo. Es fácil
trabajar con modelos sencillos, pero pueden no ser precisos. Los modelos de precisión suelen ser
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 20
Circuitos Eléctricos - Dorf
4/12/11 5:16 PM
E1C02_1
E1C02_1
10/23/2009
10/23/2009
21
21
Ingeniería y modelos
lineales
Engineering
Engineering and
and Linear
Linear Models
Models
Engineering and Linear Models
21
21
21
21
complicados y pesados de usar. El consenso general indica que primero se deben usar los modelos
usually
more
complicated
harder
use.
The
wisdom
suggests
that
simple
models
be
sencillos
y comprobar
susand
resultados
que el uso
de estos
modelos
adecuado.
usually
more
complicated
and
harder to
topara
use. constatar
The conventional
conventional
wisdom
suggests
that es
simple
modelsLos
be
used
first.
The
results
obtained
using
the
models
must
be
checked
to
verify
that
use
of
these
simple
usually
more
complicated
and
harder
to
use.
The
conventional
wisdom
suggests
that
simple
models
be
modelos
se usaránusing
cuando
necesario.
used
first.más
Theprecisos
results obtained
the sea
models
must be checked to verify that use of these simple
models
is
appropriate.
More
accurate
models
are
used
when
necessary.
used
first.
The
results
obtained
using
the
models
must
be
checked
to
verify
that
use
of
these
simple
modelos idealizados
de dispositivos
eléctricos
se definen
de manera muy precisa. Es de
modelsLos
is appropriate.
More accurate
models are
used when
necessary.
idealized
models
of
electric
devices
are
precisely
important
distinguish
models
is appropriate.
More
models
are
used
when defined.
necessary.
The
idealized
models
ofaccurate
electric
devices
are
precisely
defined.
Ityis
issus
important
to
distinguisha
suma The
importancia
hacer
una clara
distinción
entre
los
dispositivos
realesIt
modelos to
idealizados,
between
actual
devices
and
their
idealized
models,
which
we
call
circuit
elements.
The
goal
of
The
idealized
models
of
electric
devices
are
precisely
defined.
It
is
important
to
distinguish
between
and elementos
their idealized
models, which
we call
elements.
The goal
of circuit
circuit
los cualesactual
se lesdevices
denomina
de circuitos.
El objetivo
delcircuit
análisis
de circuitos
es pronosticar
analysis
is
to
predict
the
quantitative
electrical
behavior
of
physical
circuits.
Its
aim
is
to
predict
and
to
between
actual
devices
and
their
idealized
models,
which
we
call
circuit
elements.
The
goal
of circuit
analysis
is to predicteléctrico
the quantitative
electrical
physical
Itses
aim
is to predict
andlos
to
el comportamiento
cuantitativo
de losbehavior
circuitosoffísicos.
Sucircuits.
propósito
prever
y explicar
explain
the
terminal
voltages
and
terminal
currents
of
the
circuit
elements
and
thus
the
overall
analysis
is
to
predict
the
quantitative
electrical
behavior
of
physical
circuits.
Its
aim
is
to
predict
and
to
explain
voltages
and terminal
currents
of the
elements
and thus the
overall
voltajes the
y lasterminal
corrientes
de las terminales
de los
elementos
delcircuit
circuito,
y por consiguiente
el funciooperation
of
the
circuit.
explain
the
terminal
voltages
and
terminal
currents
of
the
circuit
elements
and
thus
the
overall
operation
of
the
circuit.
namiento total del circuito.
Models
of
elements
categorized
in
of
For
it
operation
the
circuit.
i
Models
of circuit
circuit
elements can
can
be
categorized
in aa variety
variety
of ways.
ways.
For example,
example,
it is
is
Los of
modelos
de elementos
de be
circuitos
se pueden
clasificar
de distintas
maneras.
i
important
to
distinguish
linear
models
from
nonlinear
models
because
circuits
that
consist
Models
of
circuit
elements
can
be
categorized
in
a
variety
of
ways.
For
example,
it
is
important
to
distinguish
linear
models
from
nonlinear
models
because
circuits
that
consist
Por ejemplo, es importante distinguir los modelos lineales de los no lineales dado que los + i
–
i
––
+
entirely
circuit
elements
are
easier
to
than
circuits
that
contain
some
important
tolinear
distinguish
models
nonlinear
models
that
consist
v
+
entirely
ofque
linear
circuit
elements
are
easier
to analyze
analyze
thanbecause
circuits circuits
that más
contain
some
circuitosof
constan
porlinear
completo
defrom
elementos
de circuitos
lineales
son
fáciles
de
v
v
–
+
nonlinear
entirely
ofelements.
linear
circuit
elements
are elementos
easier to analyze
than circuits that contain some
nonlinear
elements.
analizar que
los que
contienen
algunos
no lineales.
v
FIGURA 2.2-1
An
or
linear
and
certain
nonlinear
elements.
An
element
oro circuit
circuit
ises
linear
ifsithe
the
element’s
excitation
and
response
satisfy
certain
Un element
elemento
circuitois
linealif
el element’s
estímulo yexcitation
la respuesta
delresponse
elementosatisfy
cumplen
con Elemento
FIGURE
2.2-1
el impulso
FIGUREcon
2.2-1
properties.
Consider
the
element
shown
in
Figure
2.2-1.
Suppose
that
the
excitation
is
An
element
or
circuit
is
linear
if
the
element’s
excitation
and
response
satisfy
certain
properties.
Consider
the
element
shown
in
Figure
2.2-1.
Suppose
that
the
excitation
is the
the de
An
element
with
una
corriente
y
determinadas propiedades. Considere el elemento que se muestra en la figura 2.2-1. Suponga
FIGURE
2.2-1
An
element
withi an
an
, it excitation
current
ii and
the
is
v.
element
is
aa current
properties.
the element
shown
inWhen
Figure
2.2-1.
Suppose
that
the to
excitation
isii1the
v.current
ii and
current
andConsider
theesresponse
response
is the
the
voltage
v.
When
the
element
is subjected
subjected
to
current
que el estímulo
la corriente
i yvoltage
la
respuesta
es elthe
voltaje
v. Cuando
el elemento
está sujeto
An element
with an
1, it respuesta
excitation
current
and
.
Furthermore,
when
the
element
is
subjected
to
a
current
i
,
it
provides
a
response
v
,
current
i
and
the
response
is
the
voltage
v.
When
the
element
is
subjected
to
a
current
i
1
2
1
a
response
v.
excitation
current
i and
.
Furthermore,
when
the
element
is
subjected
to
a
current
i
,
it
provides
a
response
v
a una corriente i1, proporciona
una respuesta v1. Aún más, cuando el elemento está sujeto a2una a response v.
1
a linear
is
necessary
that
the
ii1 þ
i2 result
provides
vv2v..1For
. Furthermore,
whenit
element
is
a current
i2,i ittenga
a response
element,
itthe
is lineal
necessary
thatsubjected
the excitation
excitation
provides
response
corriente aaia2response
, response
da una respuesta
v2. En element,
un elemento
es necesario
que eltoimpulso
1
como v.
2 For a linear
1 þ ii21 result
2
þ
v
.
This
is
usually
called
the
principle
of
superposition.
in
a
response
v
.
For
a
linear
element,
it
is
necessary
that
the
excitation
i
þ
i
result
provides
a
response
v
1
2
2
1
2
þ
v
.
This
is
usually
called
the
principle
of
superposition.
in
a
response
v
resultado una respuesta
1
2 v1 1 v2. A esto se le conoce comúnmente como principio de superposición.
Also,
multiplying
the
input
of
linear
device
by
aa of
constant
have
of
þ v2. This
is
usually
called
the
principle
superposition.
in a response
val1 multiplicar
Also,
multiplying
the la
input
of aade
linear
device
bylineal
constant
must
have the
the
consequence
of
Incluso,
entrada
un dispositivo
por unamust
constante
debeconsequence
tener la consemultiplying
the
output
by
the
same
constant.
For
example,
doubling
the
size
of
the
input
causes
the
size
Also,
multiplying
the
input
of
a
linear
device
by
a
constant
must
have
the
consequence
of
multiplying
the
output
by
the
same
constant.
For
example,
doubling
the
size
of
the
input
causes
the
size
cuencia de multiplicar la salida por la misma constante. Por ejemplo, al duplicar el tamaño de la entraof
the
output
to
double.
This
is
called
the
property
of
homogeneity.
An
element
is
linear
if,
and
only
if,
multiplying
the
output
by
the
same
constant.
For
example,
doubling
the
size
of
the
input
causes
the
size
of
the
output
to
double.
This
is
called
the
property
of
homogeneity.
An
element
is
linear
if,
and
only
if,
da se ocasiona que el de la salida también se duplique. A esto se le llama propiedad de homogeneidad.
the
properties
superposition
and
are
satisfied
for
all
and
responses.
of
output to
is called
the property
of homogeneity.
Anexcitations
element isylinear
if, and only
if,
the
properties
ofdouble.
superposition
and
homogeneity
are
satisfied de
forsuperposición
all
excitations
and
responses.
Unthe
elemento
esof
lineal
si,This
y sólo
si, sehomogeneity
cumplen
las propiedades
homogeneidad
para
the
properties
of
superposition
and
homogeneity
are
satisfied
for
all
excitations
and
responses.
todos los impulsos y respuestas.
A
A linear
linear element
element satisfies
satisfies the
the properties
properties of
of both
both superposition
superposition and
and homogeneity.
homogeneity.
A
linear
element
satisfies
the
properties
of
both
superposition
homogeneity.
Un elemento lineal satisface las propiedades de superposición y and
homogeneidad.
Let
Let us
us restate
restate mathematically
mathematically the
the two
two required
required properties
properties of
of aa linear
linear circuit,
circuit, using
using the
the arrow
arrow
notation
to
imply
the
transition
from
excitation
to
response:
Let
us
restate
mathematically
the
two
required
properties
of
a
linear
circuit,
using
the lineal,
arrow
Declaremos
de
nuevo
matemáticamente
las
dos
propiedades
requeridas
de
un
circuito
notation to imply the transition from excitation to response:
notation
fromsignificar
excitationla
to
response:
ii !
v
utilizandotolaimply
flechathe
de transition
notación para
transición
del
impulso
a
la
respuesta:
!v
ii ! vv
Then
we
state
two
properties
required
follows.
Podemos
establecer
de este
las propiedades
Then
we may
may
state the
the
two modo
properties
required as
asrequeridas.
follows.
Superposition:
Then
we may state the two properties required as follows.
Superposición:
Superposition:
Superposition:
ii1 !
v1
1 ! v1
ii21 !
v
2
2 ! v21
i
!
v
2
then
i
þ
i
!
v
ð2:2-1Þ
1
2
1
entonces
(2.2-1)
then
i1 þ i2 ! v12 þ
þ vv22
ð2:2-1Þ
then
i1 þ i2 ! v1 þ v2
ð2:2-1Þ
Homogeneidad:
Homogeneity:
Homogeneity:
ii !
! vv
Homogeneity:
i
!v
then
ki
ð2:2-2Þ
then
ki !
! kv
kv
ð2:2-2Þ
entonces
(2.2-2)
then
ki ! kv
ð2:2-2Þ
Undevice
dispositivo
que no
los principios
de superposición
ni de homogeneidad
se is
considera
A
that
not
satisfy
the
or
principle
said
be
A
device
that does
does
notsatisface
satisfy either
either
the superposition
superposition
or the
the homogeneity
homogeneity
principle
is
said to
to no
be
lineal.
nonlinear.
A
device
that
does
not
satisfy
either
the
superposition
or
the
homogeneity
principle
is
said
to
be
nonlinear.
nonlinear.
E j e m p l o 2 . 2 - 1 Un dispositivo lineal
E
E XX AA M
MP
PL
LE
E 2
2 .. 2
2 -- 1
1 A
A Linear
Linear Device
Device
E X A M P L E 2 . 2 - 1 A Linear Device
Considere el elemento representado por la relación entre la corriente y el voltaje como
Consider
between
current
Ri
Consider the
the element
element represented
represented by
by the
the relationship
relationshipv 5
between
current and
and voltage
voltage as
as
Consider
the
element
represented
by
the
relationship
between
current
and
voltage as
Determine si este dispositivo es lineal.
vv ¼
Ri
¼ Ri
v ¼ Ri
Determine
whether
this
device
is
linear.
Determine whether this device is linear.
Determine
whether
this device is linear.
Circuitos
Eléctricos
- Dorf
M02_DORF_1571_8ED_SE_020-052.indd 21
Alfaomega
4/12/11 5:17 PM
22
22
22
Circuit
Elements
Circuit
Elements
Elementos
de circuitos
Circuit
Elements
Solution
Solución
Solution
The
response
aa current
i is
Therespuesta
responseato
to
current
ises
La
una
corrienteii111i1is
The
response
to
a current
The
response
aa current
La
una
corrienteiii22i2is
Therespuesta
responseato
to
current
ises
The
response
to
a current
2 is
La suma
de estas
respuestas es
The
The sum
sum of
of these
these responses
responses is
is
The
sum
of
these
responses
is
vvv1 5
¼
Ri
¼ Ri
Ri1111
v11 ¼
Ri
vvv2 5
¼
Ri
¼ Ri
Ri2222
v22 ¼
Ri
vv1 1
v2 5 Ri
Ri2 5
R1i
1 1 Ri
1 1 ii22Þ2
þ
þ vvv222 ¼
¼ Ri
Ri111 þ
þ Ri
Ri222 ¼
¼R
Rðððiii111 þ
þ ii22 ÞÞ
vv111 þ
¼
Ri
þ
¼
R
þ
Because
the
sum
of
the
responses
to
i
and
i
is
equal
to
the
response
, the
of
is
Dado
que
la
suma
de
las
respuestas
a
i
e
i
es
igual
a
la
respuesta
a
1ii1i2þ
, seii2satisface
el principio
de superpo1
2
1
2
Because the
the sum
sum of
of the
the responses
responses to
to ii11 and
and ii22 is
is equal
equal to
to the
the response
responsei1to
to
þ
the principle
principle
of superposition
superposition
is
Because
to
i11 þ
i22,, the
principle
of
superposition
is
satisfied.
Next,
consider
the
principle
of
homogeneity.
Because
sición.
A
continuación
veamos
el
principio
de
homogeneidad.
Dado
que
satisfied. Next,
Next, consider
consider the
the principle
principle of
of homogeneity.
homogeneity. Because
Because
satisfied.
¼
vvv1 5
¼ Ri
Ri1111
Ri
Ri
v111 ¼
we
Tenemos
paraan
impulso iii2i225¼
we have
have for
for
anunexcitation
excitation
¼kiki
ki111
we
have
for
an
excitation
2 ¼ ki1
vvv22 ¼
¼
5
Ri222 5
Rki
¼ Ri
Ri
¼ Rki
Rki11
v22 ¼
Ri
2 ¼ Rki11
Por
consiguiente,
Therefore,
Therefore,
Therefore,
5 kv
kv1
vvv22 ¼
¼ kv
kv111
v22 ¼
satisfies
principle
of
homogeneity.
of
superposition
satisface
el principio
Dado quethe
el element
elementosatisfies
satisfacethe
propiedades
de superposición
hosatisfies the
the
principlede
ofhomogeneidad.
homogeneity. Because
Because
the
element
satisfies
thelasproperties
properties
of both
both
superpositiony and
and
satisfies
the
principle
of
homogeneity.
Because
the
element
satisfies
the
properties
of
both
superposition
and
homogeneity,
it
is
linear.
mogeneidad,
es
lineal.
homogeneity, it
it is
is linear.
linear.
homogeneity,
PLE 2.2-2
Nonlinear
Device
EjXXXe AAAmM
M
Po
LE
E22. 2
. 2- 2
- 2 UnA
Adispositivo
Nonlinear no
Device
EE
p lP
lineal
M
L
Now
us
an
represented
by
the
between
current
Now let
let
us consider
consider
an element
element
represented
byrelación
the relationship
relationship
between
current and
and voltage:
voltage:
Now
let
us
consider
an
element
represented
by
the
relationship
between
current
and
voltage:
Ahora
veamos
un elemento
representado
por la
entre
corriente
y voltaje.
2
vv ¼
i
2
2
¼ ii 2
¼
vv 5
i
Determine
whether
this
device
is
linear.
Determine
whether
this
device
is
linear.
Determine si
whether
this device
linear.
Determine
este dispositivo
es is
lineal.
Solution
Solution
Solución
The
response
aa current
i1 is
Therespuesta
responseato
to
current
ises
The
response
to
a current
La
una
corrienteii11i1is
The
response
aa current
i2 is
Therespuesta
responseato
to
current
ises
The
response
to
a current
La
una
corrienteii22i2is
The
The sum
sum of
of these
these responses
responses is
is
The
sum
of
these
responses
is
La
suma
de
estas
respuestas
es
The
response
ii1 þ
ii2 is
Therespuesta
responseato
to
þ
is
The
response
to
i11 þ
i22 is
La
i1 1
i2 es
Because
Debido
Becausea que
Because
2
vv1 ¼
¼ iii11122
v11 ¼
2
vv2 ¼
¼ iii11122
v22 ¼
2
2
vv1 þ
þ vvv222 ¼
¼ iii11122 þ
þ iii11122
v11 þ
¼
þ
2
2
2
ððii1 þ
þ iii222 ÞÞÞ22 ¼
¼ iii11122 þ
þ 2i
2i111 iii222 þ
þ iii11122
ði11 þ
¼
þ
2i
þ
2
2
ii1 222 þ
þ iii11122 6¼
6¼ ðððiii111 þ
þ iii222ÞÞÞ22
6¼
þ
i11 þ
the
principle
superposition
is
Therefore,
the
is
no
de superposición.
Por lo
tanto, el dispositivo
es lineal.
thesatisface
principleelof
ofprincipio
superposition
is not
not satisfied.
satisfied.
Therefore,
the device
device no
is nonlinear.
nonlinear.
the
principle
of
superposition
is
not
satisfied.
Therefore,
the
device
is
nonlinear.
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 22
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
E1C02_1
10/23/2009
E1C02_1
E1C02_1 10/23/2009
10/23/2009
23
23
23
Ingeniería y modelos lineales
23
Engineering
and
Linear
Models
Engineering
Engineeringand
andLinear
Linear
Models
Models
23
23
23
E j e m p l o 2 . 2 - 3 Un modelo de dispositivo lineal
Model
of
Linear
Device
EEEXXXAAAMMMPPLPLELEE222. .2.22--3-33 AAAModel
Modelof
ofaaaLinear
LinearDevice
Device
Un elemento lineal tiene un voltaje v y una corriente i como se muestra en la figura 2.2-2a. Los valores de la
corriente i y del voltaje v correspondiente se han tabulado como se muestra en la figura 2.2-2b. Represente el eleAAlinear
linear
element
has
voltage
vand
current
shown
Figure
2.2-2a.
Values
the
current
corresponding
Amento
linear
element
element
has
hasvoltage
voltage
vand
andcurrent
current
i ias
iasas
shown
shown
inininFigure
Figure
2.2-2a.
Values
Values
ofofof
the
the
current
current
i iand
iand
and
corresponding
corresponding
por
una ecuación
quevexprese
v como
una
función
de i. 2.2-2a.
Esta
ecuación
es
un
modelo
del
elemento.
Utilice
voltage
v
have
been
tabulated
as
shown
in
Figure
2.2-2b.
Represent
the
element
by
an
equation
that
expresses
voltage
v have
v have
been
tabulated
tabulated
as
shown
shown
in
Figure
Figure
2.2-2b.
2.2-2b.
Represent
the
the
element
element
by
anan
equation
equation
that
that
expresses
expresses
vvas
vasas
elvoltage
modelo
parabeen
pronosticar
elas
valor
de vincorrespondiente
aRepresent
una corriente
de i 5by
100
mA
y el valor
de
i correspona
function
of
i.
This
equation
is
a
model
of
the
element.
Use
the
model
to
predict
the
value
of
v
corresponding
to
adiente
function
a function
ofof
i. i.This
This
equation
equation
isa model
a modelofof
the
theelement.
element.
Use
Use
the
themodel
model
toto
predict
predict
the
thevalue
value
ofofv corresponding
v correspondingtoto
aaa
a un
voltaje
de
v 5 18isV.
current
of
i
¼
100
mA
and
the
value
of
i
corresponding
to
a
voltage
of
v
¼
18
V.
current
currentofofi ¼
i ¼100
100
mA
mAand
andthe
thevalue
valueofofi corresponding
i correspondingtotoa avoltage
voltageofofv v¼¼1818
V.V.
i
+
+v++
i ii
v–vv
–––
(a)
(a)
(a)
(a)
v, V i, mA
v,v,
v,
VVV i,i,mA
i,mA
mA
10
4.5
25
11.25
10
4.5
10
4.5
4.5 10
50
22.5
25
11.25
25
11.25
11.25 25
50
22.5
50
22.5
22.5 50
(b)
(b)
(b)
(b)
v, V
30 v,v,v,
VVV
30
30
30
20
20
20
20
10
10
10
10
10
10
10
10
25
25
25
25
i, mA
i,i,mA
mA
50 i, mA
50
5050
FIGURA 2.2-3 Diagrama del voltaje comparado con la
FIGURA 2.2-2 (a) Elemento de circuito lineal y (b) tabulación
FIGURE
2.2-3
plot
voltage
versus
current
for
the
linear
FIGURE
FIGURE2.2-3
2.2-3AAAplot
plotofofofvoltage
voltageversus
versuscurrent
currentfor
for
the
thelinear
linear
FIGURE
2.2-2
(a)
linear
circuit
element
(b)
FIGURE
FIGURE
2.2-2
2.2-2
(a)
(a)
AAA
linear
linear
circuit
circuit
element
and
and
(b)
(b)
aatabulation
atabulation
tabulation
corriente del elemento lineal de la figura 2.2-2.
de
los valores
correspondientes
a suelement
voltaje
yand
corriente.
element
from
Figure
2.2-2.
element
element
from
from
Figure
Figure
2.2-2.
2.2-2.
corresponding
values
its
voltage
and
current.
ofofof
corresponding
corresponding
values
values
ofofof
its
its
voltage
voltage
and
and
current.
current.
Solución
Solution
Solution
Solution
La
figura 2.2-3 es un diagrama del voltaje v comparado con la corriente i. Los puntos etiquetados por balas repre-
Figure
2.2-3
plot
the
voltage
versus
the
current
The
points
marked
by
dots
represent
corresponding
Figure
Figure2.2-3
2.2-3
isisisaaaplot
plot
ofofofthe
thevoltage
voltageavvlos
vversus
versus
the
thecurrent
current
Thepoints
points
markedby
byde
dots
dots
represent
corresponding
sentan
los
valores
correspondientes
valores
de
v e i i.ai.i.The
partir
de
losmarked
renglones
la represent
tabla
de lacorresponding
figura
2.2-2b.
values
ofof
vand
the
rows
the
table
Figure
2.2-2b.
Because
the
circuit
element
isis
linear,
we
expect
these
values
values
ofel
velemento
vand
and
i ifrom
ifrom
from
the
the
rows
rows
ofof
of
the
the
table
table
ininin
Figure
Figure
2.2-2b.
2.2-2b.
Because
Because
the
circuit
circuit
element
element
linear,
linear,
we
we
expect
these
these
Como
del
circuito
es
lineal,
se
espera
que
esos
puntos the
descansen
en
una is
línea
recta,
yexpect
de
hecho
lo
points
lie
on
aastraight
straight
line,
and
indeed
they
do.
We
can
represent
the
straight
line
by
the
equation
points
pointstoto
tolie
lieon
onarecta
straight
line,
line,
and
andindeed
indeed
they
they
do.We
Wecan
canrepresent
representthe
thestraight
straightline
lineby
bythe
theequation
equation
hacen.
La
línea
se puede
representar
por
lado.
ecuación
mi
¼¼mi
mi
þþbbbb
vvvv¼5
miþ1
donde
mmisis
es
la
pendiente
es
lav-intercept.
intersección
en v. Observando
que laline
línea
rectathrough
pasa
a través
del origen,
vwhen
50
where
the
slope
and
the
v-intercept.
Noticing
that
the
straight
line
passes
through
the
origin,
where
wheremm
isthe
theslope
slopeand
andbybbisbisisthe
the
v-intercept.Noticing
Noticing
that
thatthe
thestraight
straight
linepasses
passes
through
the
theorigin,
origin,
vvv¼¼¼000when
when
i5
0,that
se
ve
b0.We
5
0.are
El
resultado
i¼
we
see
that
We
are
left
with
i cuando
i¼¼0,0,0,we
we
see
see
that
bbbque
¼¼¼0.0.
We
areleft
leftwith
withes
v 5 mi
mi
vvv¼¼¼mi
mi
La pendiente m se puede calcular a partir de los datos de cualquiera de las dos filas de la tabla en la figura 2.2-2b.
The
slope
can
be
calculated
from
the
data
any
two
rows
the
table
Figure
2.2-2b.
For
example:
The
Theslope
slopemmmcan
canbe
becalculated
calculatedfrom
fromthe
thedata
datainininany
anytwo
tworows
rowsofofofthe
thetable
tableinininFigure
Figure2.2-2b.
2.2-2b.For
Forexample:
example:
Por
ejemplo:
11:25
4:5
22:5
11:25
22:5
4:5
11:25
11:25���4:5
4:5
VVV 22:5
22:5���11:25
11:25
VVV
22:5
22:5���4:5
4:5
VVV
0:45
0:45
0:45
¼¼¼0:45
¼¼¼0:45
0:45
¼¼¼0:45
0:45 ; ; ;
0:45 ; ;and
;and
and
y
25
10
mA
50
25
mA
50
10
mA
25
mA
mA
50���10
10
mA
25���10
10
mA 50
50���25
25
mA
50
mA
Consequently,
Consequently,
Consequently,
En
consecuencia,
VVV
VVV
450
0:45
¼¼¼450
450
0:45
mmm¼¼¼0:45
mA
mA
mA
AAA
entonces
and
and
and
v 5 450i
450i
vvv¼¼¼450i
450i
Esta ecuación es un modelo de un elemento lineal. Pronostica que el voltaje v 5 450(0.1) 5 45 V corresponde a la
This
equation
the
linear
element.
predicts
that
the
voltage
450
45
corresponds
the
This
This
equation
equation
isisis
aamodel
amodel
model
ofofof
the
the
linear
linear
element.
element.
ItItIt
predicts
predicts
that
that
the
the
voltage
voltage
vvv¼¼¼450
450
ðð0:1
ð0:1
0:1
ÞÞ¼
Þ¼¼45
45
VVV
corresponds
corresponds
tototo
the
the
corriente i 5 100 mA 5 0.1 A y que la corriente i 5 18>450 5 0.04 A 5 40 mA corresponde al voltaje v 5 18 V.
current
100
mA
0:1
and
that
the
current
18=450
0:04
40
mA
corresponds
the
voltage
current
currenti i¼
i¼¼100
100
mA
mA¼¼¼0:1
0:1
AAAand
andthat
thatthe
thecurrent
currenti i¼
i¼¼18=450
18=450¼¼¼0:04
0:04
AAA¼¼¼40
40
mA
mAcorresponds
correspondstototothe
thevoltage
voltage
18
V.
vvv¼¼¼18
18V.
V.
EJERCICIO 2.2-1 Considere el elemento de circuito que se muestra en la figura E 2.2-1a. En
la figura E 2.2-1b se muestra un diagrama del voltaje v, del elemento, comparado con la corriente i del
elemento. El diagrama es una línea recta que pasa a través del origen y su pendiente tiene un valor m.
EXERCISE
2.2-1
Consider
the
circuit
element
shown
Figure
2.2-1a.
plot
the
EXERCISE
EXERCISE2.2-1
2.2-1 Consider
Considerthe
thecircuit
circuitelement
elementshown
showninininFigure
FigureEEE2.2-1a.
2.2-1a.AAAplot
plotofof
ofthe
the
En consecuencia, v e i se relacionan por
element
voltage,
versus
the
element
current,
shown
Figure
2.2-1b.
The
plot
element
elementvoltage,
voltage,v,v,v,versus
versusthe
theelement
elementcurrent,
current,i,i,i,isisisshown
showninininFigure
FigureEEE2.2-1b.
2.2-1b.The
Theplot
plotisisisaastraight
astraight
straight
v 5with
mi
line
that
passes
through
the
origin
and
has
with
value
m.
Consequently,
related
by
line
linethat
thatpasses
passesthrough
throughthe
theorigin
originand
andhas
hasaaslope
aslope
slope
withvalue
valuem.
m.Consequently,
Consequently,vvand
vand
andi iare
iare
arerelated
relatedby
by
v
¼
mi
Muestre que este dispositivo es lineal.
v v¼¼mimi
Show
that
this
device
linear.
Show
Showthat
thatthis
thisdevice
deviceisisislinear.
linear.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 23
Alfaomega
4/12/11 5:17 PM
Elementos de circuitos
24
v
v
+
v
+
–
v
–
m
m
b
i
i
i
(a)
(a)
(b)
FIGURA E 2.2-1
i
(b)
FIGURA E 2.2-2
EJERCICIO 2.2-2 Considere el elemento de circuito que se muestra en la figura E 2.2-2a.
Un diagrama del voltaje del elemento, v, comparado con la corriente del elemento, i, se muestra en la
figura E 2.2-2b. El diagrama es una línea recta con una intersección en y de valor b y una pendiente
con valor m. En consecuencia, v e i están relacionados por
v = mi + b
Muestre que este dispositivo es no lineal.
2.3
E L E M E N T O S D E C I R C U I T O A C T I V O S Y PA S I V O S
Los elementos de circuito se pueden clasificar en dos categorías, pasivos y activos, al determinar si
absorben o alimentan energía. Se dice que un elemento es pasivo si toda la energía que se le proporcionó por el resto del circuito siempre es no negativa (cero o positiva). Entonces, para un elemento pasivo
con la corriente que fluye a la terminal ⫹ como se muestra en la figura 2.3-1a, esto significa que
Z t
w¼
vi dt 0
(2.3-1)
1
para todos los valores de t.
Un elemento pasivo absorbe energía.
Nodo de
entrada
+
i
Nodo de
salida
+
v
v
–
Nodo de
salida
i
–
Nodo de
entrada
(a)
(b)
FIGURA 2.3-1 (a) El nodo de entrada de la corriente i es el nodo positivo del voltaje v;
(b) el nodo de entrada de la corriente i es el nodo negativo del voltaje v. La corriente fluye
desde el nodo de entrada hasta el nodo de salida.
Se dice que un elemento es activo si es capaz de proporcionar energía. Por consiguiente, un
elemento activo no satisface la ecuación 2.3-1 cuando se representa por la figura 2.3-1a. En otras
palabras, un elemento activo es aquel que es capaz de generar energía. Los elementos activos son
fuentes potenciales de energía, mientras que los elementos pasivos son reductores o absorbedores de
energía. Las baterías y los generadores son ejemplos de elementos activos. Considere el elemento que
se muestra en la figura 2.3-1b. Observe que la corriente fluye hacia la terminal negativa y sale de la
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 24
Circuitos Eléctricos - Dorf
5/24/11 9:48 AM
2.3
ACTIVE AND PASSIVE CIRCUIT ELEMENTS
We may classify circuit elements in two categories, passive and active, by determining whetherResistors
they
25
Resistors
25
Resistors
25
absorb energy or supply energy. An element is said to be passive if the total energy delivered to itResistencias
from
the
rest of the circuit is always nonnegative (zero or positive). Then for a passive element, with the current
terminal.
This element
is said
be
active
if es activo
terminalinto
positiva.
Se
dice
que to
este
elemento
terminal.
element
isbe
said
to
if sithis means that
flowing
þThis
terminal
in be
Figure
Z 2.3-1a,
terminal.
Thisthe
element
is saidastoshown
active
if active
t
Z t
Z
Z
tt
w¼
¼
vi¼dt
dt �
� 00 dt � 0
ð2:3-2Þ ð2:3-2Þ
w
w
vi
ð2:3-1Þ
(2.3-2)
w ¼ �1 vi dt � 0vi
ð2:3-2Þ
for
atalleast
one
value
ofvalue
t.
for one
at of
least
one
para
menos
un
for
all
values
t.valor
for
at
least
value
ofde
t. t. of t.
25
�1
�1
�1
An
active
element
is
capable
of supplying
energy.
Anelement
active
capable
of supplying
Unpassive
elemento
activoelement
escapable
capazis
de
energía.
A
element
absorbs
energy.
An
active
is
ofalimentar
supplying
energy. energy.
Entry
node
+
v
–
i
Exit
node
E X A M PELXEA 2M.P3L-E12 .An
Active
CircuitCircuit
Element
3 -elemento
1Active
An Active
Element
EEj Xe m
p lPoL E2 2
. 3. 3- 1- 1 Un
de circuito
activo
AM
An
Circuit
Element
+
A
circuit
has
an element
represented
by Figure
2.3-1b
where
the
current
is
a constant
5 A and 5the
voltage
is
a
Uncircuit
circuito
tiene
un
representado
por
la
figura
2.3-1b
donde
la is
corriente
de 5the
A yis
el
Avhas
circuit
haselemento
an element
represented
by2.3-1b
Figure
2.3-1b
where
the
current
isesa una
constant
A voltage
and
voltage
is a
A
an element
represented
by Figure
where
the
current
a constant
5 Aconstante
and the
a
constant
6
V.
Find
the
energy
supplied
over
the
time
interval
0
to
T.
voltaje
es
una
constante
de
6
V.
Obtenga
la
energía
alimentada
durante
el
intervalo
de
0
a
T.
6 V.
the supplied
energy supplied
time interval
constantconstant
6 V. Find
theFind
energy
over theover
timethe
interval
0 to T. 0 to T.
i
–
Solution
Solución
Solution
FIGURE 2.3-1 (a) The entry node of the current i is the positive node of the voltage v; (b) the
Solution
Exit
Entry
Because
current enters
terminal,
thelaenergy
by the
is given
byde
Dado quethe
la corriente
entra the
por negative
la terminal
negativa,
energíasupplied
alimentada
por element
el elemento
resulta
node
the
current
the negative
terminal,
thethe
energy
by flows
the element
entry
nodethe
ofenters
the
current
i terminal,
is the negative
of
voltage supplied
v.by
The
current
the is
BecauseBecause
thenode
current
enters
negative
the node
energy
supplied
the
element
isfrom
given
bygiven
(a)
(b)
entry node to the exit node.
Z
Z
by
Z T
Þdtð6¼Þð30T
wðð66¼ÞÞðð55Þdt
5ÞdtJJ¼ 30T J
¼ 30T
T
T
w¼
w¼ 0
0
0
element
is said
be es
active
if an
it isactive
capable
ofelemento
delivering
energy.
an active
element
Por An
lo the
tanto,
elthe
dispositivo
un
generador
oanun
activo;
este
caso
es battery.
una
bateríaviolates
de cd.
Thus,
device
isdevice
a to
generator
or
element,
in this
case
aThus,
dc battery.
Thus,
is
a
generator
or
active
element,
in en
this
a dc
device
is a generator
or an2.3-1a.
active In
element,
in this
dccase
battery.
Eq.Thus,
2.3-1the
when
it is represented
by Figure
other words,
ancase
activea element
is one that is capable of
generating energy. Active elements are potential sources of energy, whereas passive elements are sinks or
absorbers of energy. Examples of active elements include batteries and generators. Consider the element
2.4
R
E
S
S
T
O
RSC
STI O
2.4 R
R2.4
E
S III S
S
T
E
Athat
S Sthe current flows into the negative terminal and out of the positive
R2.3-1b.
EO
SN
IR
R
2.4
E
S
T
S
shown
in
Figure
Note
The
ability
ofability
a material
to resist
the flow
of
charge
is called
itscalled
resistivity,
r. Materials
that are good
La capacidad
de
un of
material
de resistir
elthe
flujo
de of
carga
se denomina
Los
Theof
a material
to
flow
isits
itsresistividad,
resistivity,
r.. Materials
that are good
The
ability
a material
to resist
theresist
flow
of
charge
ischarge
called
resistivity,
r. Materials
thatmateriales
are good
electrical
insulators
have
a
high
value
of
resistivity.
Materials
that
are
good
conductors
ofson
electric
que son electrical
buenos
aisladores
tienen
un alto
valor
de resistividad.
Los
queof
bueinsulators
havevalue
a high
of resistivity.
Materials
are
good conductors
of electric
electrical
insulators
have aeléctricos
high
ofvalue
resistivity.
Materials
that are that
goodmateriales
conductors
electric
current
have
low de
values
ofvalues
resistivity.
Resistivity
values
for
selected
materials
are givenare
in given
Table 2.4-1.
current
lowof
of
resistivity.
Resistivity
values
for selected
materials
Table
nos conductores
electricidad
tienen
bajos valores
de for
resistividad.
En la tabla
los 2.4-1.
current
have
lowhave
values
resistivity.
Resistivity
values
selected
materials
are2.4-1
givense
in registran
Table in
2.4-1.
Copper
is
commonly
used
for
wires because
it permits
current
to cobre
flow relatively
unimpeded.
Silicon
is
Copper
is commonly
used
wires because
permitsel
to
relatively
unimpeded.
valores is
de
resistividad
de for
materiales
selectos.
Por loitcomún
se flow
utiliza
en los cables
porque
Copper
commonly
used
wiresfor
because
it permits
current
tocurrent
flow relatively
unimpeded.
Silicon
isSilicon is
commonly
used
to
provide
resistance
in
semiconductor
electric
circuits.
Polystyrene
is
used
as
an
used to
provide
resistance
in problemas.
semiconductor
electric
Polystyrene
is used
permite commonly
queused
la corriente
fluya
relativamente
sin
El silicio
se circuits.
emplea
para
commonly
to provide
resistance
in semiconductor
electric
circuits.
Polystyrene
isproporcionar
used as
an as an
insulator.
insulator.
resistencia
en circuitos eléctricos semiconductores. Por su parte, el poliestireno se usa como aislante.
insulator.
Resistance
is the physical
property
of an
element
or device
that
impedes
the flow
of current;
Resistencia
propiedad
física property
de
un element
elemento
dispositivo
dethat
impedir
el flujo
de coResistance
is the physical
of an element
orthat
device
impedes
the
flow
of current;
Resistance
isesthelaphysical
property
of an
orodevice
impedes
the flow
of current;
itrriente;
is represented
by
the
symbol
R.
se
representa
con
el
símbolo
R.
it
is
represented
by
the
symbol
R.
it is represented by the symbol R.
Georg
Simon
Ohm
was
able
to show
that
the
current
in
circuit
composed
of a battery
and
Georg Simon
Simon
Ohm
pudo
demostrar
la the
corriente
enin
unaa circuit
circuito
compuesto
unaof
batería
yaa and a
GeorgOhm
Simon
Ohm
was
ableque
to
show
that
the
current
in acomposed
circuit composed
a and
battery
Georg
was
able
to show
that
current
ofde
a battery
conducting
wire
of
uniform
cross-section
could
be
expressed
as
un cableconducting
conductor
uniforme
se expresaría
comobe expressed
wire of uniform
cross-section
as
conducting
wire of seccional
uniform
cross-section
could becould
expressed
as
Av
Av
ð2:4-1Þ ð2:4-1Þ
i ¼ Av
ð2:4-1Þ
i ¼ rL i ¼ rL
(2.4-1)
rL
Table
2.4-1
Resistivities
of Selected
Materials
Tabla
2.4-1Resistivities
Resistividad
materiales
selectos
Table
2.4-1 Resistivities
of Selected
Materials
Table
2.4-1
of de
Selected
Materials
MATERIAL
MATERIAL
MATERIAL
MATERIAL
Poliestireno
Polystyrene
Polystyrene
Polystyrene
Silicio
Silicon
Silicon Silicon
Carbono
Carbon
Carbon Carbon
Aluminio
Aluminum
AluminumAluminum
Cobre
Copper
Copper Copper
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 25
RESISTIVITY
r (OHM.CM)
RESISTIVIDAD
r (OHMCM)
RESISTIVITY
r (OHM.CM)
RESISTIVITY
r (OHM.CM)
18 18
1 3 110� 10
1 � 1018
1 �5 1018
5
2.3 3
10
2.3 � 10
5
2.3 � 105 2.3 � 10
4 3 410�2310�3
�3
4 � 10
4 � 10�3
�6
2.7 3
2.710�2610�6
�6
2.7
2.7 �2610 �6 � 10
1.7 3
1.710� 10�61.7 � 10�6
1.7 � 10
Alfaomega
4/12/11 5:17 PM
26
w¼
Z
T
0
ð6Þð5Þdt ¼ 30T J
Thus, the device is a generator or an active element, in this case a dc battery.
2.4
26
R E S I S T26O
26 R S
Circuit Elements
Circuit
Elementos
de circuitos
Circuit Elements
Elements
26
ircuit Elements
where
is the
cross-sectional
area, r that
the resistivity,
L they length,
and va the
voltage
across the
Aisis
escalled
elAcross-sectional
área
resistividad,
LLlagood
longitud,
v elthe
voltaje
través
del
The ability of a material to resist the flow where
ofdonde
charge
itstransversal,
resistivity,
r. la
Materials
are
area,
the
length,
and
voltage
the
where A
Awire
is the
the
cross-sectional
area,isr
rshown
the resistivity,
resistivity,
L the
the
length,
and vvthe
the constant
voltage across
across
the R as
element.
Ohm,
who
in
Figure
2.4-1,
defined
resistance
elemento
cable.
Ohm,
cuya
aparece
en
la figura
2.4-1, definió
la resistenciaRconselectrical insulators have a high value ofwire
resistivity.
Materials
that
areimagen
goodinconductors
of
electric
element.
Ohm,
who
is
Figure
2.4-1,
defined
where A is the cross-sectional area,
r the
resistivity,
L the
and in
v the
voltage
across
the the
wire
element.
Ohm,
wholength,
is shown
shown
Figure
2.4-1,
defined
the constant
constant resistance
resistance R as
as
current have low values of resistivity. Resistivity
values
for selected materials are given in Table 2.4-1.
tante
R
como
rL
wire element. Ohm, who is shown in Figure 2.4-1, defined the constant resistance
R
as
rL R ¼
ð2:4-2Þ
Copper is commonly used for wires because it permits current to flow relatively unimpeded.
R¼
¼ rLSiliconAis
ð2:4-2Þ
(2.4-2)
R
ð2:4-2Þ
A
rL
commonly used to provide resistance in semiconductor electric circuits. Polystyrene is A
used as an
R¼
ð2:4-2Þ
insulator.
Ohm’s
law,
which
related
the
voltage
and current,
was published
in 1827 as
A
Ohm’s
law,
which
related
the
voltage
and
current,
in
1827
as
La
ley
de
Ohm,
que
relaciona
el
voltaje
y
la
corriente,
publicó en
Ohm’s law, which related the voltage and current, was
wassepublished
published
in 1827
1827como
as
ð2:4-3Þ
Ohm’s law, which related the voltage and current, was published in v1827
as v ¼ Ri
(2.4-3)
ð2:4-3Þ
v¼
¼ Ri
Ri
ð2:4-3Þ
FIGURE
2.4-1
FIGURA
2.4-1
Resistance isFIGURE
the
physical
property
of
an
element
or
device
that
impedes
the
flow
of
current;
2.4-1
The
unit
of resistance
wasthe
named
the
ohm
in
honor
of Ohm
and is
usually abbreviated
by the
v¼
ð2:4-3Þ
FIGURE
2.4-1 Ohm
La unit
unidad
deRi
resistencia
R named
seRdenominó
ohmio
en honor
deand
Ohm
y suele
abreviarse
of
R
in
of
is
abbreviated
by
the
Georg
Simon
Georg
Simon Ohm
The
unit
of resistance
resistance
R was
was
named
the ohm
ohm
in honor
honor
of Ohm
Ohm
and
is usually
usually
abbreviatedcon
by el
the
it is represented
bySimon
the
symbol
R. The
Georg
Ohm
V
(capital
omega)
symbol,
where
1
V
¼
1
V/A.
The
resistance
of
a
10-m
length
of
common
1
Georg
Simon
Ohm
símbolo
V
(letra
omega
mayúscula),
donde
1
V
5
1
V>A.
La
resistencia
de
un
cable
común
V
(capital
omega)
symbol,
where
1
V
¼
1
V/A.
The
resistance
of
a
10-m
length
of
common
(1787-1854)
The unit of
resistance
Rdeterminó
was who
named
ohm in
honor symbol,
of Ohm and
is usually
by the
(1787–1854),
V the
(capital
omega)
where
1 V ¼ abbreviated
1 V/A. The resistance
of a 10-m length of common
(1787–1854),
who
Ohm
(1787–1854),
who
TV
cable
is largo
2 mV.
la leydetermined
de symbol,
Ohm
en Ohm’s
1827.
deVcable
TV
deV/A.
10
de
es de of
2 mV.
TV
is
22 m
mV.
law
V
(capital
omega)
where
1
¼
1
The
resistance
a
10-m
length
of
common
TV
cable
is
mV.
determined
Ohm’s
law
who
determined
Ohm’s
law
elThe
ohmio
An element
that
has a resistance
R is
called
resistor.
Aisresistor
is represented
by the
Georg
Simon
Ohm
was
able
to
show
the
current
in a has
circuit
composed
of
a battery
and resistor.
aaA
Unelement
elemento
resistencia
denomina
Un resistor
se representa
that
aa resistance
R
is
called
aa resistor.
by
inhonor,
1827.
ohm
was thatAn
TV cable
isEn
2 su
mV.
in
1827.
The
ohm
was
An
element
thatcon
hasuna
resistance
RR
is se
called
resistor.
A resistor
resistor
is represented
represented
by the
the
m’sconducting
law
in
1827.
The
ohm
was
fue
elegido
como
la
wire
of
uniform
cross-section
could
be
expressed
as
two-terminal
symbol
shown
in
Figure
2.4-2.
Ohm’s
law,
Eq.
2.4-3,
requires
that
the
i-versus-v
por
el
símbolo
de
dos
terminales
que
se
muestra
en
la
figura
2.4-2.
La
ley
de
Ohm,
ecuatwo-terminal
symbol
shown
in
Figure
2.4-2.
Ohm’s
law,
Eq.
2.4-3,
requires
that
the
i-versus-v
chosen
as
the
unit
of
as
the
unit
An chosen
element
that
hasof
R is called
a resistor.
is represented
by Eq.
the 2.4-3, requires that the i-versus-v
two-terminal
symbol
shown A
in resistor
Figure 2.4-2.
Ohm’s law,
hm was
chosen
as de
the
unit
of a resistance
unidad
resistencia
relationship
beque
linear.
Asin
shown
in
Figure
2.4-3,
resistor
may
become
nonlinear
AvOhm’s
2.4-3,
requiere
la
relación
de i comparada
con v asea
lineal.
Como
se muestra
en laitsoutside its
electrical
resistance
inción
relationship
be
linear.
As
shown
Figure
2.4-3,
aai-versus-v
resistor
may
become
nonlinear
outside
electrical
resistance
in
two-terminal
symbol
shown
in
Figure
2.4-2.
law,
Eq.
2.4-3,
requires
that
the
relationship
be
linear.
As
shown
in
Figure
2.4-3,
resistor
nit of
electrical
resistance
in
eléctrica.
ð2:4-1Þmay become nonlinear outside its
i¼
his
honor.
normal
rated
range
of
operation.
We
will
assume
that
a
resistor
is
linear
unless stated
figura
2.4-3,
un
resistor
puede
llegar
a
convertirse
en
no
lineal
fuera
de
su
rango
proporciohis
honor.
normal
rated
range
of
operation.
We
will
assume
that
a
resistor
is
linear
unless
rL
relationship
be linear. As shown in
Figurerated
2.4-3,range
a resistor
may become
outside
ance in
his honor.
normal
of operation.
Wenonlinear
will assume
thatitsa resistor is linear unless stated
stated
nal
de
operación.
Supondremos
que
un
resistor
es
lineal
si
no
se
establece
otra
cosa.
Por
lo
otherwise.
Thus,
we
will
use
a
linear
model
of
the
resistor
as
represented
by
Ohm’s
law.
otherwise.
Thus,
we
will
use
a
linear
model
of
the
resistor
as
represented
by
Ohm’s
law.
normal rated range of operation.
We will Thus,
assume
resistor
is model
linear of
unless
stated as represented by Ohm’s law.
otherwise.
we that
will ause
a linear
the resistor
tanto,
utilizaremos
un
modelo
lineal
de
resistor
como
el
que
representa
la
ley
de
Ohm.
In
Figure
2.4-4,
the
element
current
and
element
voltage
of
a
resistor
are
labeled.
The
In
Figure
2.4-4,
the
element
current
and
element
voltage
of
a
resistor
are
labeled.
The
otherwise. Thus,
we2.4-1
will use
aInlinear
model
of the
resistor
represented
Ohm’svoltage
law. of a resistor are labeled. The
Figure
2.4-4,
the
elementascurrent
and by
element
Table
Resistivities
of
Selected
Materials
En
la
figura
2.4-4
se
han
etiquetado
la
corriente
y
el
voltaje
de
un
elemento
de
resistor.
La
re­
relationship
between
the
directions
of
this
current
and
voltage
is
important.
The
voltage
direction
relationship
between
the
directions
of
this
current
and
voltage
is
important.
The
voltage
direction
In Figure 2.4-4, the element
current between
and element
voltage ofof athis
resistor
relationship
the directions
currentareandlabeled.
voltageThe
is important. The voltage direction
MATERIAL
RESISTIVITY
(OHM.CM)
lación
lasone
direcciones
deþ
y�.
voltaje
importante.
dirección
delmarked
voltajeþ
indica
flows
the
terminal
to
marks
one
resistor
terminal
and
the
The
current
iiaacurrent
flows
from the
terminal
marked
marks
resistor
terminal
þ other
and the
other
�.es
The
iaLa
relationship between the
directions
ofentre
this
current
and
voltage
isrcorriente
important.
voltage
direction
flows from
from
the
terminal
marked
þ
to the
the þ to the
marks
one
resistor
terminal
þesta
and
the
other
�.elThe
The
current
una
terminal
de
resistor
1
y
la
otra
2.
La
corriente
i
fluye
de
la
terminal
etiquetada
1
a
la
terminal
18
terminal
marked
�.
This
relationship
between
the
current
and
voltage
reference
directions
is
a current
terminal
This
relationship
between
the
current
and reference
voltage reference
is a
flows
from
the
terminal
marked
þ to
thevoltage
marks one resistor terminal
þ andterminal
the other
�. Themarked
current
marked
�. Thisi�.
between
the
and
directions directions
is aa
Polystyrene
1 � 10
a relationship
etiquetada
2.called
Estathe
relación
entre
las
de referencia
de
la when
yelement
el voltaje
es unaand
conconvention
the
passive
convention.
Ohm’s
law
states
that
the
the
5direcciones
convention
called
the
passive
convention.
Ohm’s
law
that
thevoltage
element
voltage
terminal marked �. This
between
current
and
voltage
reference
directions
isstates
acorriente
convention
called
the
passive
convention.
Ohm’s
law
states
that
when
thewhen
element
voltage
and
the and the
Siliconrelationship
2.3
� 10
vención
denominada
convención
pasiva.
La ley
de
Ohm
establece
que cuando la corriente y el voltaje
element
current
adhere
to
passive
convention,
then
�3
convention called the Carbon
passive convention.
Ohm’scurrent
law
states
that
when
the
element
voltage
and
adhere
the
passive
convention,
thenthe
elementelement
current
adhere
to the
the
convention,
then
4 passive
�to
10
del
elemento
se
apegan
a
la
convención
pasiva,
entonces
ð2:4-4Þ
v
¼
Ri
element current adhereAluminum
to the passive convention, then
ð2:4-4Þ ð2:4-4Þ
v ¼ Riaa v ¼ Ria
2.7 � 10�6
v
5
Ri
(2.4-4)
and
i
are
the
same
except
for
the
assigned
direction,
Consider
Figure
2.4-4.
The
element
currents
i
a
ð2:4-4Þ
v
¼
Ri
a
b
�6
a
same
for thedirection,
assigned so
direction, so
Consider
FigureThe
2.4-4.
ia and
theisame
except
forexcept
the assigned
so
Consider
Figure 2.4-4.
element
currents currents
ia and ib are
b are the
Copper
1.7The
� 10 element
Considere
corrientes
ia ethe
ib en
elemento
son
las
mismas
excepto
por
la
dirección
and ib2.4-4.
are theLas
same
except for
assigned
direction,
so
Consider Figure 2.4-4. The element
currentslaiafigura
iiaa el
¼
�i
¼ �ibb ia ¼ �ib
asignada, por lo tanto,
icurrent
The
the
element
voltage
v adhere
the
a ¼ �ib iia and
The element
element
current
element
voltage
toadhere
the passive
passive
convention,
The element
current
ia and
the element
vto
to theconvention,
passive convention,
a and the
ia 5vvoltage
iadhere
b
v
¼
Ri
aa v ¼ Ri
The element current ia and the La
element
voltage
v
adhere
to
the
passive
convention,
v
¼
Ri
corriente ia y el voltaje v del elemento se apegan a la convención
pasiva,
a
v ¼�i
Riba gives
Replacing
iiaa by
Replacing
by
�i
gives
v
5
Ri
b
Replacing ia by �ib gives
a
vv ¼
Replacing ia by �ib gives
Sustituyendo ia por 2ib resulta
¼ �Ri
�Rivbb ¼ �Rib
v ¼ �Ri
v5
There
aa minus
signbin
equation because
the
element
current iibb and
voltage vv do
not
b
There is
isThere
minus
in this
this
because
the2Ri
element
and the
theibelement
element
not v do not
is asign
minus
signequation
in this equation
because
the current
element current
and thevoltage
element do
voltage
adhere
to
the
passive
convention.
We
must
pay
attention
to
the
current
direction
so
that
we
don’t
En
esta
ecuación
hay
un
signo
de
menos
porque
la
corriente
i
y
el
voltaje
v
del
elemento
no
se
apegan
There is a minus sign in this equation
because
the
element
current
i
and
the
element
voltage
v
do
not
bto
adhere
to
the
passive
convention.
We
must
pay
attention
the
current
direction
so
that
we
don’t
b
adhere to the passive convention. We must pay attention to the current direction so that we don’t
this
minus
sign.
a overlook
laWe
convención
pasiva.
Haytoque
especial
atención
a la
adhere to the passive convention.
must
attention
theponer
current
direction
so that
wedirección
don’t de corriente de modo que no
overlook
thispay
minus
sign.
overlook
this
minus
sign.
Ohm’s
law,
Eq.
2.4-3,
can
also
be
written
as
se
pase
por
alto
este
signo
menos.
overlook this minus sign.
Ohm’s Ohm’s
law, Eq.law,
2.4-3,
alsocan
be written
Eq.can
2.4-3,
also be as
written as
de Ohm,
Ohm’s law, Eq. 2.4-3, can also La
be ley
written
as ecuación 2.4-3, también seii puede
¼
ð2:4-5Þ
¼ Gv
Gv escribir
ð2:4-5Þ ð2:4-5Þ
i ¼ Gv
i
5
Gv
(2.4-5)
i
¼
Gv
ð2:4-5Þ
where
where G
G denotes
denotes the
the conductance
conductance in
in siemens
siemens (S)
(S) and
and is
is the
the reciprocal
reciprocal of
of R;
R; that
that is,
is, G
G¼
¼ 1=R.
1=R. Many
Many
G conductancia
denotes
theconductance
conductance
(S)I and
is es
thedecir,
reciprocal
of Muchos
R; that is,
G ¼ 1=R. Many
donde Gwhere
indica
la
en siemens as
(S)in
y siemens
es la
recíproca
de R;
G 5 inverted
1>R.
ingeniedenote
the
with
where G denotes the conductanceengineers
in siemens
(S) and
is theof
of as
R;mhos
that is,
G the
¼ 1=R.
Manywhich
engineers
denote
the units
units
ofreciprocal
conductance
mhos
with
the
I symbol,
symbol,
which is
is an
an inverted omega
omega (mho
(mho is
is
ros
denotan
lasbackward).
unidades
conductancia
como mhos,
el with
símbolo
es una
omega
invertida
(mho
engineers
denotedethe
units ofwe
conductance
as con
mhos
the Iisque
symbol,
which
is an
inverted
omega (mho is
ohm
SI
and
engineers denote the units of conductance
as mhos
with the IHowever,
symbol, which
an inverted
(mhosiemens
ohm spelled
spelled
backward).
However,
we will
willis use
use
SI units
unitsomega
and retain
retain
siemens as
as the
the units
units for
for conductance.
conductance.
es ohm alohm
revés).
Sin embargo, utilizaremos
unidades
del SI
SI units
y se mantendrá
siemensascomo
unidad
However,
and retainelsiemens
the units
forde
conductance.
ohm spelled backward). However, we will
usespelled
SI unitsbackward).
and retain siemens
as we
the will
unitsuse
for conductance.
v
conductancia.
v
v
v
v
– iim
m
––im
– im
R
R
R
R
– im
im
0
ia
Alfaomega
R
ib
iim
m
im
0
R
+
FIGURE 2.4-2 Symbol for a
resistor having a resistance of R
ohms.
0
00
v
–
iia
a
ia
im
+
+
+
R
R
R
v
v
v+
iib
b
ia ib
R
––
–v
ib
–
FIGURE
2.4-2
Symbol
for
FIGURE
2.4-3 A
operating
FIGURE
2.4-4
A
with
FIGURE2.4-2
2.4-2Símbolo
Symbol para
for aaun
FIGURE
A resistor
resistor
operating
FIGURE
2.4-4
A resistor
resistor
withcon
FIGURA
FIGURA
2.4-3 2.4-3
Un resistor
funcionando
FIGURA
2.4-4
Un resistor
resistor
having
a
resistance
of
R
within
its
specified
current
range,
�
element
current
and
element
FIGURE
2.4-2
Symbol
for
a
resistor
having
a una
resistance
R
FIGURE
2.4-3
A
resistor
FIGURE
2.4-4
A resistor with
FIGURE
2.4-3
A resistencia
resistorofoperating
FIGURE
2.4-4
Acurrent
resistor
with
within
specified
range,
� operating
element
current
and element
resistor
que
tenga
dentro
de suits
rango
de corriente
especificado,
corriente
y voltaje
del
elemento.
, can
modeled
by
Ohm’s
law.
voltage.
melement
resistor
having
a resistance
ohms.
deohms.
Rwithin
ohmios.
modelar
por
la
ley de
Ohm. range,
its specified
current
range, �ofimR,iise
currentitsand
element
specified
current
�
element current and element
can be
bewithin
modeled
by
Ohm’s
law.
voltage.
m, puede
im, canohms.
be modeled by Ohm’s law.
voltage.im, can be modeled by Ohm’s law.
voltage.
M02_DORF_1571_8ED_SE_020-052.indd 26
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
E1C02_1
E1C02_1
10/23/2009
10/23/2009
27
27
Resistencias
Resistors
Resistors
Resistors
27
27
27
27
FIGURA
embobinado
unaan
FIGURE
2.4-5
(a)
Wirewound
resistor
with
FIGURE 2.4-5
2.4-5 (a)
(a)Resistor
Wirewound
resistorcon
with
an
FIGURE
2.4-5
(a)
Wirewound
with
an
tapa
al centro
ajustable.
(b)
Resistorresistor
embobinado
adjustable
center
tap.
(b)
Wirewound
resistor
with
a
adjustable
center
tap.
(b)
Wirewound
resistor
with
adjustable
center
tap. (b)
resistor with aa
con
tapa
fija.
Cortesía
de Wirewound
Dale
Electronics.
fixed
tap.
Courtesy
of
Dale
Electronics.
fixed
tap.
Courtesy
of
Dale
Electronics.
fixed tap. Courtesy of Dale Electronics.
FIGURA 2.4-7 Resistor de película de metal
de 1>4 de watt. El cuerpo del resistor tiene
FIGURA 2.4-6 Pequeños chips resistores de película gruesa que se
FIGURE
2.4-7
metal
resistor.
The
FIGURE
2.4-6
Small
thick-film
resistor
chips
used
for
FIGURE
2.4-7
Ade1=4-watt
1=4-watt
metal film
film
resistor.
The body
body
6mmA
largo. Cortesía
de Dale
Electronics.
usan
para circuitos
miniaturizados.
Cortesía
de Corning
Electronics.
FIGURE
2.4-6
Small
thick-film
resistor
chips
used
for
FIGURE 2.4-7
A
1=4-watt
metal
film
resistor.
The
body
FIGURE 2.4-6 Small thick-film resistor chips used for of
the
resistor
is
6
mm
long.
Courtesy
of
Dale
Electronics.
miniaturized
circuits.
Courtesy
of
Corning
Electronics.
of
the
resistor
is
6
mm
long.
Courtesy
of
Dale
Electronics.
miniaturized
circuits.
Courtesy
of
Corning
Electronics.
of the resistor is 6 mm long. Courtesy of Dale Electronics.
miniaturized circuits. Courtesy of Corning Electronics.
La mayoría de resistores discretos caen en una de cuatro categorías básicas: compuesto de carMost
resistors
fall
one
four
carbon
carbon
film,
Most discrete
discrete
resistors
fall into
into
one of
of
four basic
basic categories:
categories:
carbondecomposition,
composition,
carbon
film,
bón, película
de carbón,
película
de metal
y four
embobinado.
Los resistores
compuesto de
carbón
se
Most
discrete
resistors
fall
into
one
of
basic
categories:
carbon
composition,
carbon
film,
metal
film,
or
wirewound.
Carbon
composition
resistors
have
been
in
use
for
nearly
100
years
and
metal
film,
or
wirewound.
Carbon
composition
resistors
have
been
in
use
for
nearly
100
years
and
han
utilizado
más de 100Carbon
años y composition
siguen siendoresistors
populares.
Losbeen
resistores
de carbón
metal
film, orpor
wirewound.
have
in use de
forpelícula
nearly 100
years han
and
are
Carbon
film
resistors
have
supplanted
carbon
composition
resistors
for
many
are still
still popular.
popular.
Carbon
filmde
resistors
havemuchos
supplanted
carbon
composition
resistors
fory mejor
many
desplazado
a los deCarbon
compuesto
carbón para
propósitos
generales
por suresistors
bajo costo
are
still
popular.
film
resistors
have
supplanted
carbon
composition
for
many
general-purpose
uses
because
of
their
lower
cost
and
better
tolerances.
Two
wirewound
resistors
are
general-purpose
uses
because
of
their
lower
cost
and
better
tolerances.
Two
wirewound
resistors
are
tolerancia.
En la figura
2.4-5 seofmuestran
doscost
resistores
de embobinado.
general-purpose
uses because
their lower
and better
tolerances. Two wirewound resistors are
shown
in
Figure
2.4-5.
shown
in
Figure
2.4-5.
resistores
de película gruesa, como los de la figura 2.4-6, se usan en circuitos por su bajo
shownLos
in Figure
2.4-5.
Thick-film
resistors,
as
in
2.4-6,
used
circuits
because
their
low
cost
Thick-film
resistors,Los
as shown
shown
in Figure
Figure
2.4-6, are
aregeneral
used in
inestán
circuits
because of
of
their
low estándar
cost and
and
costo Thick-film
y tamaño pequeño.
resistores
de propósito
disponibles
entheir
valores
resistors,
as
shown
in
Figure
2.4-6,
are
used
in
circuits
because
of
low
cost
and
small
size.
General-purpose
resistors
are
available
in
standard
values
for
tolerances
of
2,
5,
10,
and
20
small
size.
General-purpose
resistors
are
available
in
standard
values
for
tolerances
of
2,
5,
10,
and
20
para
de 2, 5, 10,resistors
y 20 porareciento.
Losinresistores
compuesto
de carbón
smalltolerancias
size. General-purpose
available
standard de
values
for tolerances
of 2,y5,algunos
10, and de
20
percent.
Carbon
composition
resistors
and
some
wirewounds
have
aa color
code
with
three
to
five
percent.
Carbon
composition
resistors
and
some
wirewounds
have
color
code
with
three
to
five
embobinado
tienen
un código resistors
de color con
a cinco
bandas.have
El código
decode
colorwith
es unthree
sistema
de
percent. Carbon
composition
and tres
some
wirewounds
a color
to five
bands.
color
is
of
colors
for
identification
of
the
of
bands. A
A
color code
code
is aaa system
system
of standard
standard
colors adopted
adopted
for
identification
of 2.4-7
the resistance
resistance
of
colores
estándar
adoptado
para identificar
la resistencia
de los for
resistores.
La figura
muestra un
bands.
A
color
code
is
system
of
standard
colors
adopted
identification
of
the
resistance
of
resistors.
Figure
2.4-7
shows
aa metal
film
resistor
with
its
color
bands.
This
is
aa 1=4-watt
resistor,
resistors.
Figure
2.4-7
shows
metal
film
resistor
with
its
color
bands.
This
is
1=4-watt
resistor,
resistor
película
de metal
sus bandas
de colores.
resistor
1>4 is
de awatt,
lo queresistor,
indica
resistors.deFigure
2.4-7
showscon
a metal
film resistor
with Es
its un
color
bands.deThis
1=4-watt
implying
that
be
operated
at
or
below
of
delivered
to
normal
range
of
implying
that it
it should
should
be de
operated
at
ordebajo
below 1=4
1=4lawatt
watt
of power
power
delivered
to it.
it. The
The
normal
rangede
of
que
debe funcionar
a 1>4
watt o at
poror
de
energía
que sedelivered
le suministre.
El rango
normal
implying
that
it
should
be
operated
below
1=4
watt
of
power
to
it.
The
normal
range
of
resistors
is
from
less
than
1
ohm
to
10
megohms.
Typical
values
of
some
commercially
available
resistors
is
from
less
than
1
ohm
to
10
megohms.
Typical
values
of
some
commercially
available
resistores
de menos
de 11ohmio
En el Apéndice
se dan
los valores comunes
resistors isvafrom
less than
ohm toa 10
10 megaohmios.
megohms. Typical
values of D
some
commercially
available
resistors
are
given
Appendix
resistors
areresistores
given in
indisponibles
Appendix D.
D.
de
algunos
comercialmente.
resistors
are
given
in
Appendix
D.
The
power
to
resistor
(when
the
convention
is
is
Thepotencia
power delivered
delivered
to aaaaun
resistor
(when
the passive
passive
convention
is used)
used)
is es
La
transmitida
resistor
(cuando
se
utiliza2convention
la convención
pasiva)
The
power
delivered
to
resistor
(when
the
passive
is
used)
is
�
�
�
�
2
v
v
� v � vv2
ð2:4-6Þ
pp ¼
ð2:4-6Þ
¼ vi
vi ¼
¼ vvv Rv ¼
¼
(2.4-6)
p¼
vi
¼
ð2:4-6Þ
R
R ¼R
R
R
Alternatively,
because
vv ¼
we
can
write
the
for
as
Alternatively,
because dado
¼ iR,
iR,
we
can
write
the equation
equation
for power
power
as
De
manera alternativa,
quewe
v 5can
iR,write
la ecuación
de la potencia
se as
puede
escribir como
Alternatively,
because
v¼
iR,
the
equation
for
power
22
R
ð2:4-7Þ
pp ¼
vi
¼
ð
iR
Þi
¼
i
¼ vi
vi ¼
¼ ððiR
iRÞi
Þi ¼
¼ ii2 R
R
ð2:4-7Þ
(2.4-7)
ð2:4-7Þ
p¼
Thus,
the
power
is
expressed
as
a
nonlinear
function
of
the
current
i
through
the
resistor
or
of
Por
lo tanto,
la potencia
se expresa
una función
nooflineal
de la corriente
i, o the
del voltaje
Thus,
the power
power
is expressed
expressed
as aacomo
nonlinear
function
the current
current
through
resistorv,or
ora través
of the
the
Thus,
the
is
as
nonlinear
function
of the
ii through
the resistor
of
the
voltage
v
across
it.
del
resistor.
voltage
v
across
it.
voltage v across it.
Recall
passive
one
which
the
absorbed
is
Recuerde
definiciónof
pasivo, as
según
es aquel
cuya energía
absorbida
es
Recall the
theladefinition
definition
ofdeaaa elemento
passive element
element
as
one lafor
forcual
which
the energy
energy
absorbed
is always
always
Recall
the
definition
of
passive
element
as
one
for
which
the
energy
absorbed
is
always
nonnegative.
The
equation
for
energy
delivered
to
a
resistor
is
siempre
no negativa.
La ecuación
de la delivered
energía transmitida
a unis resistor es
nonnegative.
The
equation
for
energy
to
a
resistor
nonnegative. The equation for energy delivered
to a ZZresistor
is
ZZ t
Z ttt 2
Z tt
2
(2.4-8)
w
pdt
ii2Rdt
ð2:4-8Þ
pdt ¼
¼
Rdt
ð2:4-8Þ
w¼
¼
ð2:4-8Þ
w
¼
�1 pdt ¼ �1 i Rdt
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 27
�1
�1
�1
�1
Alfaomega
4/12/11 5:17 PM
28
28
28
Circuit
Elements
Elementos
de circuitos
Circuit
Elements
2
Como
i2 siempre
es positiva,
energía
es positiva
el resistor
es uniselemento
Because
is always
always
positive,lathe
the
energysiempre
is always
always
positive yand
and
the resistor
resistor
passivepasivo.
element.
Because
ii2 is
positive,
energy
is
positive
the
is aa passive
element.
Resistenciaisesauna
medida
capacidad
de untoelemento
disipar
potencia de manera
Resistance
measure
of de
an laelement’s
ability
dissipatedepower
irreversibly.
Resistance
irreversible.is a measure of an element’s ability to dissipate power irreversibly.
E X A M P L E 2 . 4 - 1 Power Dissipated
Dissipated by
by aa Resistor
Resistor
E
E jXeAmMpPlLoE 22 .. 44--11 Power
Potencia disipada por
un resistor
Let us
us devise
devise aa model
model for
for aa car
car battery
battery when
when the
the lights
lights are
are left
left on
on and
and the
the engine
engine is
is
Let
Inventemos
unallmodelo
para una
batería
de
automóvil
cuando
las If
luces
han quedado
off.
We
have
experienced
or
seen
a
car
parked
with
its
lights
on.
we
leave
the
car
off. We have
all ha
experienced
or seen aSi
carelparked
with se
itsdeja
lightsasíon.
If we leave
the carla
encendidas
se
apagado
el
un constanttiempo,
for aa period,
period,ythe
the
battery
will run
runmotor.
down or
or go
goautomóvil
dead. An
An auto
auto
batterydurante
is aa 12-V
12-V
for
battery
will
down
dead.
battery
is
constantbatería
funcionando
hastacan
que
baje o by
se aagote.
Laofbatería
deThe
un automóvil
voltageseguirá
source, and
and the
the lightbulb
lightbulb
bese
modeled
resistor
ohms.
circuit is
is
voltage
source,
can
be
modeled
byestar
a resistor
of 66 ohms.
The
circuit
es
una
fuente
constante
de
12-V,
y
el
bulbo
puede
modelado
por
un
resistor
de 6
shown
in
Figure
2.4-8.
Let
us
find
the
current
i,
the
power
p,
and
the
energy
supplied
shown
inEn
Figure
2.4-8.
Letse
usmuestra
find theelcurrent
i, the power p, and the energy supplied
ohmios.
la figura
2.4-8
by the
the battery
battery
for aa four-hour
four-hour
period. circuito. Encuentre la corriente i, la potencia p,
by
for
period.
y la energía alimentada por la batería para un periodo de una hora.
Solution
Solution
Solución
According to Ohm’s law,
law, Eq.
Eq. 2.4-3,
2.4-3, we
we have
have
According
Según la leytodeOhm’s
Ohm, ecuación
2.4-3, tenemos
ii
i
12 VV
12
12 V
+
+–
–+
–
Ω
66 Ω
6Ω
R
R
R
FIGURE 2.4-8
2.4-8 Model
Model of
of a
FIGURE
FIGURA
2.4-8
Modelo dea
car
battery
and
the
car
the
unabattery
bateríaand
de automóvil
y
headlight lamp.
lamp.
headlight
la lámpara.
¼ Ri
Ri
vvv 5
¼
Ri
Becauseque
¼
1212
VVand
and
R5¼
¼666V,V,
V,tenemos
we have
have
¼
A.A.
Puesto
v5
y RR
que
i5
Because
vv ¼
12
V
we
ii ¼
22 2A.
Para
encontrar
la potencia
transmitida
por la batería,
To find
find
the power
power
delivered
by the
the battery,
battery,
we use
use se utiliza
To
the
delivered
by
we
¼ vi
vi ¼
¼ 12
12ðð22ÞÞ ¼
¼ 24
24 W
W
pp ¼
Finally,
the la
energy
delivered
in the
the
four-hour
period
ishoras es
Por
último,
energía
transmitida
en four-hour
un
lapso deperiod
cuatrois
Finally,
the
energy
delivered
in
ZZ tt
w¼
¼
pdt ¼
¼ 24t
24t ¼
¼ 24
24ðð60
60 �
� 60
60 �
� 44ÞÞ ¼
¼ 3:46
3:46 �
� 10
1055 JJ
w
pdt
0
0
Because the battery
battery has aa finite
finite amount
amount of
of stored
stored energy,
energy, itit will
will deliver
deliver this
this energy
energy and
and eventually
eventually be
be unable
unable to
to
Because
Dado quethe
la batería has
tiene una cantidad
finita
de energía almacenada,
transmitirá
esta energía
y acabará
por no
deliver further
further energy
energy without
without recharging.
recharging. We
We then
then say
say the
the battery
battery is
is run
run down
down or
or dead
dead until
until recharged.
recharged. A
A typical
typical
deliver
poder hacerlo más si no se6 recarga. Se dice entonces que la batería se baja o muere hasta que se recargue. Una
in aa fully
fully charged
charged 6condition.
auto battery
battery may
may store 10
106 J in
auto
batería
normal de store
automóvilJpuede
almacenar 10condition.
julios a carga plena.
EXERCISE 2.4-1
2.4-1 Find
Find the
the power
power absorbed
absorbed by
by aa 100-ohm
100-ohm resistor
resistor when
when itit is
is connected
connected directly
directly
EXERCISE
EJERCICIO
2.4-1
Obtenga la potencia absorbida por una resistencia de 100 ohmios cuando
across
a
constant
10-V
source.
across a constant 10-V source.
se conecta directamente a una fuente constante de 10-V.
Answer: 1-W
1-W
Answer:
Respuesta: 1-W
EXERCISE 2.4-2
2.4-2
A voltage
voltage
sourcede
¼
10 cos
cosv tt5
V10
is connected
connected
across
resistor
of 10
10deohms.
ohms.
Find
EJERCICIO
2.4-2
Una fuente
voltaje
cos t V estáacross
conectada
a través
una resisEXERCISE
A
source
vv ¼
10
V
is
aa resistor
of
Find
the power
power
delivered
to
the resistor.
resistor.
tencia
de 10delivered
ohmios. to
Obtenga
la potencia transmitida al resistor.
the
the
2
2
Answer: 10
10 cos
cos tt W
2W
Answer:
Respuesta:
10 cos
tW
2.5
2.5
NUD
DEE
ENP
PTE
EEN
NSD
DIE
ENN
NDT
TES
S
O
UD
RC
ENS
ST E S
PO
EN
ICEE
IIFN
U
R
Some devices
devices
are intended
intended
tocomo
supplypropósito
energy to
tosuministrar
circuit. These
These
devices
are
called A
sources.
Sources are
are
Algunos
dispositivos
tienen
energía
a unare
circuito.
estos dispositivos
Some
are
to
supply
energy
aa circuit.
devices
called
sources.
Sources
categorized
as
being Están
one of
ofclasificados
two types:
types: voltage
voltage
sourcesen
anddos
current
sources.
Figurede
2.5-1a
shows
the
se
les llama as
fuentes.
para funcionar
tipos,sources.
como fuentes
voltaje
y como
categorized
being
one
two
sources
and
current
Figure
2.5-1a
shows
the
symbol that
thatcorriente.
is used
used to
toLa
represent
voltage
source.
The voltage
voltage
of aase
voltage
source
is fuente
specified,
but the
the
fuentes
de
figura 2.5-1a
muestra
el símbolo
con que
representa
una
de voltaje.
symbol
is
represent
aa voltage
source.
The
of
voltage
source
is
specified,
but
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 28
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
Fuentes independientes
29
El voltaje de una fuente de voltaje es específico, pero la corriente la determina el resto del circuito.
Una fuente de voltaje se describe especificando la función v(t), por ejemplo,
vðtÞ ¼ 12 cos 1 000t o vðtÞ ¼ 9
i(t)
+
o vðtÞ ¼ 12 2t
–
Un elemento activo de dos terminales que alimenta energía a un circuito es una fuente de energía. Una
fuente de voltaje independiente proporciona un voltaje específico independiente de la corriente que
fluye a través de él y es independiente de cualquier otra variable de circuito.
Una fuente es un generador de corriente o voltaje capaz de suministrar energía a un circuito.
Una fuente de corriente independiente proporciona una corriente independiente del voltaje que
fluye a través del elemento de fuente y es independiente de cualquier otra variable de circuito. Por lo
tanto, cuando se dice que una fuente es independiente, significa que es independiente de cualquier otra
corriente o voltaje en el circuito.
Una fuente independiente es un generador de voltaje o de corriente que no depende de
otras variables del circuito.
Supongamos que la fuente de voltaje es una batería y
v(t)
(a)
+
i(t)
v(t)
–
(b)
FIGURA 2.5-1
(a) Fuente
de voltaje.
(b) Fuente de
corriente.
v1t2 9 voltios
Se sabe que el voltaje de esta batería es de 9 voltios independientemente del circuito en que se use la
batería. Por el contrario, la corriente de la fuente de voltaje no se conoce y depende del circuito en que
se use la fuente. La corriente podría ser de 6 amperios cuando la fuente de voltaje está conectada a un
circuito y de 6 miliamperios cuando está conectada a otro circuito.
La figura 2.5-1b muestra con qué símbolo se representa una fuente de corriente. La corriente de
una fuente de corriente se especifica, pero el voltaje lo determina el resto del circuito. Una fuente
de corriente se describe especificando la función i(t), por ejemplo,
iðtÞ ¼ 6 sen 500t o iðtÞ ¼ 0:25
o iðtÞ ¼ t þ 8
Una fuente de corriente especificada por i(t) 0.25 miliamperios tendrá una corriente de 0.25
miliamperios en cualquier circuito que se use. El voltaje que fluye a través de esa fuente de corriente
dependerá del circuito en particular.
En los párrafos anteriores se han dejado de lado algunas complejidades para presentar una
descripción sencilla de la manera como funcionan las fuentes. El voltaje a través de una batería de
9 voltios puede no ser en realidad de 9 voltios. Este voltaje depende del tiempo de vida de la batería,
la temperatura, las variaciones en su fabricación, y la corriente de la batería. Aquí conviene hacer
una distinción entre las fuentes reales, como las baterías, y las fuentes sencillas de voltaje y corriente
descritas anteriormente. Sería ideal que las fuentes reales funcionaran como estas fuentes sencillas.
En verdad, la palabra ideal se usa para hacer esta distinción. A las fuentes sencillas descritas en los
párrafos anteriores se les denomina fuentes de voltaje ideales y fuentes de corriente ideales.
El voltaje de una fuente de voltaje ideal se proporciona para una función específica,
digamos v(t). La corriente la determina el resto del circuito.
La corriente de una fuente de corriente ideal se proporciona para una función específica,
digamos i(t). El voltaje lo determina el resto del circuito.
Una fuente ideal es un generador de voltaje o de corriente independiente de la corriente
a través de la fuente de voltaje o del voltaje a través de la fuente de corriente.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 29
Alfaomega
5/24/11 9:49 AM
E1C02_1
10/23/2009period
30 is 40 mJ. Element 1 is a
for the one-minute
o be selected.
–
–
�t/60
E1C02_1
10/23/2009
Wire
battery
to
be
selected.
mA for30t � 0,
s known that i(t) ¼ De
�t/60
mA
for t �
0, The circuit to control Wire
It is known
thatisi(t)
¼ De
¼
oltage across the second
element
v2(t)
FIGURE
1.8-1
(t)
and
the
voltage
across
the
second
element
is
v
for14t � 0. The
maximum
magnitude
of
the
a jet valve2 for¼
a space rocket.
Electric Circuit Variables
FIGURE 1.8-1 The circuit to control
t �Determine
0. The maximum
Be�t/60
D, is limited
to V
1 for
mA.
the magnitude of the
a jet valve for a space rocket.
D, describe
is limitedthetorequired
1 mA.battery.
Determine the
constants Dcurrent,
and B and
1.8 D D
E BS and
IGN
X AtheMrequired
P L E battery.
required constants
and
describe
30
Elementos
deEcircuitos
30
Circuit
Elements
Circuit Elements
e the Situation and30the Assumptions
Describe
the
Situation
theelement.
Assumptions
30
Elements
current enters the plus terminal Circuit
of theand
second
JET
VALVE
CONTROLLER
1. The current enters the plus
element.
EEj terminal
em
p lPoLof
- 1- 1Una
batería
fuente deSource
voltaje
X
AM
M
E2the
X
A
PLE
2. 5. 5second
A
Batterymodelada
Modeledcomo
as a Voltage
current leaves the plus terminal of the first
element.
2. The current
leaves
the plus
the. 5first
E Xterminal
A M P L of
E 2
- 1element.
A Battery
Modeled Wire
as a Voltage Source
small,
experimental
space
rocket
a twowires are perfect andAhave
no effect
the un
circuit
(they
douses
not
absorb
energy).
Veamos
lo queon
hace
ingeniero
que
necesita
analizar
un circuito que
contiene una batería de 9 voltios. ¿En reaelement
circuit,
as que
shown
in
Figure
1.8-1,
to
3. The wires
areesperfect
and
have
noengineer
effect
on
the
circuit
(they
do not
absorb
energy).
value
Consider
the
plight
of
the
who
needs
to
analyze
a
circuit
containing
9-voltJet
battery.
Is it really
necessary
lidad
necesario
este
ingeniero
incluya
la
dependencia
del
voltaje
de ala
en
el tiempo
de vida
de la
i
+ batería
model of the circuit, control
as shown
in
Figure
assumes
that theatvoltage
the +
controller
aengineer
jettemperatura,
valveto1.8-1,
from
point
of
liftoff
tof¼
0 across
for
this
include
the
dependence
battery
voltage
on
the
age
of
the
battery,
the
temperature,
variations
batería,
la
las
variaciones
en
su
fabricación,
y
la
corriente
de
la
batería
en
este
análisis?
Es
de
esperarConsider
plightasofshown
the engineer
who
needsassumes
to analyze
circuit
containing
a 9-volt battery. Is it really necessary
Thethat
model
the
inafter
Figure
thatathe
voltage
across the
lements is4.equal;
is, vof
vthe
1¼
2.circuit,
until
expiration
of and
the the
rocket
one1.8-1,
minute.
Element
Element
in
battery
infuncione
this
analysis?
Hopefully
not.of
expect
act enough
like
se manufacturing,
que
no.
La expectativa
es
que
lacurrent
batería
lovoltage
parecido
aWe
una
fuentevthe
debattery
voltajeto
ideal
de 9 voltios
vbastante
for
this�t/60
engineer
to include
the
dependence
of
battery
on
the
age
the
battery,
1
2 the temperature, variations
two elements
is
equal;
that
is,
v
¼
v
.
1supplied
2
1
2
energy
that
must
be
by
element
1 alto.
¼
Be
V
where
B
is
the
initial
voltage
of
the
battery
that
battery voltage v1 is vThe
de
modo
que
las
diferencias
se
puedan
pasar
por
En
este
caso
se
dice
que
la
batería
está
modelada
como
una
ideal
9-volt
voltage
source
that
the
differences
can
be
ignored.
In
this
case,
it
is
said
that
the
battery
is
modeled
1an
in manufacturing, and the�t/60
battery current in this analysis? Hopefully not. We expect the battery to act enough like
for
the
one-minute
period
isthe
40 valve.
mJ.
Element
1the
is ainitial voltage –of the battery that
is
v
¼
Be
V
where
B
is
5. The battery
voltage
v
discharge exponentially
as
it
supplies
energy
to
fuente
de
voltaje
ideal.
as
an
ideal
voltage
source.
1
1
an ideal 9-volt voltage source that the differences can be ignored. In this case,–it is said that the battery is modeled
battery To
toexponentially
beser
selected.
will discharge
as itconsidere
supplies
energy
to the
valve.
Para
concretos,
unaspecified
batería
especificada
por
el diagrama
voltaje
comparado
con
la corrienspecific,
consider
a battery
by
the plot of
voltage
versus de
current
shown
in Figure
2.5-2a.
This
ideal
voltage
circuit operates from tas
¼ 0anto
t ¼be
60
s. thatsource.
�t/60
Wire
mAwill
forbe
tdiagrama
�v 0,
It se
is
known
i(t)
¼ Devoltage
te que
muestra
en
la
figura
2.5-2a.
Este
indica
que
el
voltaje
de
la
batería
será
v
5
9
voltios
cuando
plot
indicates
that
the
battery
¼
9
volts
when
i
�
10
milliamps.
As
the
current
increases
above
10
To be specific,
consider
battery
specified by the plot of voltage versus current shown in Figure 2.5-2a. This
6. The circuitthe
operates
from
t¼
to t ¼ a60
s.
voltage
across
the0Conforme
second
element
isvolts.
v2(t) ¼
current is limited, so and
Dmilliamps,
� 110
mA.
iplot
miliamperios.
la
corriente
se
incrementa
por
arriba
de
10
miliamperios,
el
voltaje
baja
de
FIGURE
1.8-1
The
circuit
to
control
the
voltage
decreases
from
9
When
i
�
10
milliamps,
the
dependence
of
the
battery
voltage
on
indicates that the battery voltage will be v ¼ 9 volts when i � 10 milliamps. As the current increases above 10
�t/60
Vlimited,
forCuando
t � 0.
The
maximum
magnitude
ofbattery
the can
Be
a jet
valve
for a space
rocket.
7. The current
is
so
D
�
1
mA.
9
voltios.
i
10
miliamperios,
la
dependencia
del
voltaje
de
la
batería
en
la
corriente
de
la
batería
the
battery
current
can
be
ignored
and
the
be
modeled
as
an
ideal
voltage
source.
milliamps, the voltage decreases from 9 volts. When i � 10 milliamps, the dependence of the battery voltage on
current,
D,obviar
is limited
to 1 se
mA.
Determine
the
sevolts
puede
y lacan
batería
puede
modelar
comocan
una be
fuente
de voltaje
e Goal
the
battery
current
be ignored
and
the battery
modeled
as anideal.
ideal voltage source.
v,
v,
volts
required
constants
D
and
B
and
describe
the
required
battery.
v, the
voltios
State
the Goal
e the energy
supplied
by
first element for the one-minute period and then
select
v, volts
9
9
the
energy
supplied
by theand
firstthe
element
for the one-minute
period and then select
ants D andDetermine
B. Describe
the
battery
selected.
9 the
ii
Describe
Situation
Assumptions
i
the constants D 9and B. Describe the battery selected.
i
1. The current enters the plus terminal of the second element.
v=
=9
9V
V
v
e a Plan
+
+
R
R
v = first
9 –V
– element.
+
2.a Plan
The current
leavespthe
plus terminal of
the
Battery
R
d v1(t) and Generate
i(t) and then
obtain
the power,
Next,
1(t), supplied by the
vBattery
= first
9 V element.
–
+
Batería
FIGURA 2.5-2 (a) Diagrama de un voltaje de
R
–
findsupplied
v
(t)
and
i(t)
and
then
obtain
the
power,
p
(t),
supplied
by
the
element.
Next,
t), find theFirst,
energy
for
the
first
60
s.
1
1
Battery
FIGURE
2.5-2
(a) A
A con
plotlaof
of
batterydevoltage
voltage
versus
3. The wires are perfect
and
have no effect on the circuit (theyfirst
do
not
absorb
energy).
FIGURE
2.5-2
(a)
plot
battery
versus
batería
comparado
corriente
la batería.
i,
mA
10
i,
mA
10
for
the
first
60
s.
using p1(t), find the energy supplied
battery
current.
(b) está
The modelada
battery is
is modeled
modeled
asfuente
an de
battery
current.
(b)
The
battery
as
an
10 i, mA
(b)
La
batería
como
una
FIGURE
2.5-2 across
(a) A plot
4. EQUATION
The model of(a)
the
circuit,
1.8-1,(b)
assumes
that
the voltage
the of battery voltage versus
(a)
i, mAas shown in Figure
10
(a)
(b)
independent
voltage source.
source.
NEED
INFORMATION(b)
voltaje
independiente.
independent
voltage
battery
current.
(b) The battery is modeled as an
¼
v
.
two
elements
is
equal;
that
is,
v
1
2 NEED
GOAL
EQUATION
Z 60Suponga que
(a) hay
(b) deINFORMATION
source. como se muestra en la
una
resistencia
conectada
a except
través
lasindependent
terminales
de la batería,
v1 �t/60
y w1 for the
andacross
i known
forthe
Suppose a resistor
is connected
the terminals
of
battery as voltage
shown
in Figure 2.5-2b. The battery
Z
is
v
¼
Be
V
where
B
is
the
initial
voltage
of
the
battery
that
5.
The
battery
voltage
v
w
¼
p
ð
t
Þ
dt
p
(t)
60
figura
2.5-2b.
La
corriente
de
la
batería
será
1
1
1
1
constants
D
and
B
1
v
The energy current
w1 forSuppose
the
and
i
known
except
for
1
will
be
a resistor is connected across the terminals of the battery as shown in Figure 2.5-2b. The battery
0
will
discharge
to thev valve.
w1exponentially
¼
p1 ðtÞ as
dtit supplies
p1(t) energyconstants
first 60 s current
D
and
B
ð2:5-1Þ
i¼
(2.5-1)
will be
0
v
R
6. The circuit operates from t ¼ 0 to t ¼ 60 s.
ð2:5-1Þ
i ¼complica
La relación
entre between
v e i que se
muestra
en lainfigura
2.5-2a
esta ecuación.
Tal complicación
se puede
pasar
The
relationship
v and
i shown
Figure
2.5-2a
this equation.
This complication
can
be
R complicates
he Plan
7.
The
current
is
limited,
so
D
�
1
mA.
por
alto
sin
problema
cuando
i
10
miliamperios.
Cuando
la
batería
se
modela
como
una
fuente
de
voltaje
ideal
safely
ignored
when
i
�
10
milliamps.
When
the
battery
is
modeled
as
an
ideal
9-volt
voltage
source,
the
voltage
Plan
need p1(t),Act
so on
we the
first
calculate
The
relationship between v and i shown in Figure 2.5-2a complicates this equation. This complication can be
de 9pvoltios,
la
corriente
de�milliamps.
la
voltaje
source
current
is
given
� so �t/60
� fuente de�When
First, we need
(t),
wewhen
first
safely
ignored
icalculate
� by
10
theresulta
batterydeis modeled as an ideal 9-volt voltage source, the voltage
p1 ðtÞ State
¼ iv1 1the
¼ De
� 10�3 A � Be�t/60 V
9
Goal
�
�
source �t/30
current is�3given by �t/30
ð2:5-2Þ
i ¼V�
�t/60
�3
�t/60
(2.5-2)
p
ð
t
Þ
¼
iv
¼
De
�
10
A
Be
¼ DBe the
�1energy
10 Wsupplied
¼
mW
1 DBeby the
Determine
first element for the R
9one-minute period and then select
�t/30
�t/30i ¼
ð2:5-2Þ
¼estas
�
10�3
W es¼importante.
DBe
mW
the
Dentre
and
B.DBe
Describe
the
battery
selected.
Laconstants
distinción
dos
La
que implica
relación v 2shown
i que se
The
distinction
between
theseecuaciones
two
equations
is important.
Eq. 2.5-1,2.5-1,
involving
the v�ilarelationship
in
R ecuación
muestra
en
la
figura
2.5-2a,
es
más
exacta
pero
también
más
complicada.
La
ecuación
2.5-2
es
más
sencilla
pero
Figure
2.5-2a,
is
more
accurate
but
also
more
complicated.
Equation
2.5-2
is
simpler
but
may
be
inaccurate.
The distinction between these two equations is important. Eq. 2.5-1, involving the v�i relationship shown in
puedeSuppose
ser
errónea.
that
R ¼accurate
1000 ohms.
Equation
gives theEquation
current 2.5-2
of theisideal
voltage
Figure
2.5-2a,
is more
but also
more 2.5-2
complicated.
simpler
but source:
may be inaccurate.
Generate
a Plan
Suponga
que
R
5
1000
ohmios.
La
ecuación
2.5-2
da
la
corriente
de
la
fuente
de
voltaje
ideal:
9
Suppose
thati(t)
R¼
2.5-2
gives
the current
of
theelement.
ideal voltage
source:
First, find
v1(t) and
and1000
thenohms.
obtainEquation
the power,
p
(t),
supplied
by
the
first
Next,
1
ð2:5-3Þ
i ¼ 9 ¼ 9 mA
601000
s.9  9 mA
using p1(t), find the energy supplied for the first
(2.5-3)
ii 
ð2:5-3Þ
¼ 000 ¼ 9 mA
Because this current is less than 10 milliamps, the 1ideal
1000 voltage source is a good model for the battery, and it is
Como esta corriente
menor battery
que 10 current
miliamperios,
la fuente de voltaje ideal es un buen modelo para la batería,
GOAL
EQUATION
NEED
reasonable
expectes
9 milliamps.
Because thistocurrent
isthat
lessthe
than
10 milliamps,isthe
ideal voltage sourceINFORMATION
is a good model for the battery, and it is
y es razonable
esperar
que
la
corriente
de
la
batería
sea
deEq.
9 miliamperios.
Suppose,
instead,
that
R battery
¼Z 600
ohms.
Once
again,
2.5-2 gives the current of the ideal voltage source:
reasonable
to
expect
that
the
current
is
9
milliamps.
60
The energy
w1 forpor
the el contrario, que R 5 600 ohmios. Una vezv1más,
andla iecuación
known 2.5-2
exceptda for
Suponga,
la corriente de la fuente
instead, that
again,
2.5-2
gives
the current
of the ideal voltage source:
w1 R¼¼ 600p1ohms.
ðtÞ dtOnce
p19(t) ¼Eq.
first
sSuppose,
constants
D and
B
15
mA
ð2:5-4Þ
i
¼
de 60
voltaje
ideal:
0
600
9
(2.5-4)
ð2:5-4Þ
¼ 15 mA
i¼
Because this current is greater than 10 milliamps, the600
ideal voltage source is not a good model for the battery. In this
case,
isthis
reasonable
that
battery current
different
from ideal
thenot
current
the modelo
ideal
Dado
esta
corriente
esexpect
mayor
que
10
miliamperios,
laisfuente
de voltaje
no
es unfor
buen
para lasource.
batería.
Because
current
istogreater
than
10the
milliamps,
the ideal
voltage
source
is
a good
model
for thevoltage
battery.
In this
Act
onitque
the
Plan
En
este
caso,
es
razonable
esperar
que
la
corriente
de
la
batería
sea
diferente
de
la
corriente
de
la
fuente
de
voltaje
ideal.
case,
it
is
reasonable
to
expect
that
the
battery
current
is
different
from
the
current
for
the
ideal
voltage
source.
First, we need p1(t), so we first calculate
� �t/60
�
�
�
p1 ðtÞ ¼ iv1 ¼ De
� 10�3 A Be�t/60 V
Engineers
frequently
face
a trade-off when
�t/30
�t/30 selecting a model for a device. Simple models
¼ DBe
�ingenieros
10�3 W ¼seDBe
mW
Suele
suceder
que
los
enfrenten
a unmodels
dilemaare
cuando
tienen
que
seleccionarand
un
are easy
to
work
with
but
may
not
be
accurate.
Accurate
usually
more
complicated
Engineers
frequently Los
facemodelos
a trade-off
when
selecting
ademodel
forpero
a device.
Simple
models
modelo
para
un
dispositivo.
sencillos
son
fáciles
trabajar
pueden
no
ser
exactos.
harder
totouse.
The
conventional
wisdom
suggests
that models
simple are
models
be used first.
The results
are
work
with
but por
maylonot
be accurate.
Accurate
complicated
and
Loseasy
modelos
precisos
son
común
más complicados
y difíciles
deusually
usar. Elmore
sentido
común sugiere
obtained
using
the
models
must
be
checked
to
verify
that
use
of
these
simple
models
is
harder
to
use.
The
conventional
wisdom
suggests
that
simple
models
be
used
first.
The
results
que primero se usen los modelos sencillos. Los resultados obtenidos al usar estos modelos se deben
appropriate.
More
accurate
models
are
used
when
necessary.
obtained using the models must be checked to verify that use of these simple models is
Alfaomega
Circuitos Eléctricos - Dorf
appropriate. More accurate models are used when necessary.
M02_DORF_1571_8ED_SE_020-052.indd 30
4/12/11 5:17 PM
E1C02_1
E1C02_1
10/23/2009
10/23/2009
31
31
Voltímetros y amperímetros
31
comprobar para cerciorarse que es adecuado utilizar estos modelos sencillos. Los modelos más preciVoltmeters
and
i(t) =
0 Ammeters
sos se usarán cuando sea necesario.
Voltmeters and Ammeters
31
+
El cortocircuito y el circuito abierto son casos especiales de fuentes ideales. Un cortocircuito es
v(t)
The
short
circuit
and
open
circuit
are
special
cases
of
ideal
sources.
A
short
circuit
is an ideal
una fuente de voltaje ideal que tiene v(t) 5 0. La corriente en un cortocircuito está determinada por
–
voltage
source
having
v(t)
¼
0.
The
current
in
a
short
circuit
is
determined
by
the
rest
of
the
circuit.
An
The
short
circuit
and
open
circuit
are
special
cases
of
ideal
sources.
A
short
circuit
is
an
ideal
el resto del circuito. Un circuito abierto es una fuente de corriente ideal que tiene i(t) 5 0. El voltaje i(t) = 0
+
open
circuit
is
an
ideal
current
source
having
i(t)
¼
0.
The
voltage
across
an
open
circuit
is
determined
voltage source
having
v(t)
¼
0.
The
current
in
a
short
circuit
is
determined
by
the
rest
of
the
circuit.
An
a través de un circuito abierto está determinado por el resto del circuito. La figura 2.5-3 muestra los
v
+
the rest
of¼
the0.circuit.
2.5-3 an
shows
symbols
usedlatopotencia
representabsorthe short
circuit and the open
open circuit
is an ideal
sourcebyhaving
The voltage
across
openthe
circuit
is determined
símbolos
con current
que se representan
eli(t)
cortocircuito
yFigure
el circuito
abierto.
Observe
que
v(t)
(a)
circuit.
Notice
that
the
power
absorbed
by
each
of
these
devices
is
zero.
by the rest
of
the
circuit.
Figure
2.5-3
shows
the
symbols
used
to
represent
the
short
circuit
and
the
open
bida por cada uno de estos dispositivos es cero.
–
and
short
circuits
be added
to a circuit
without disturbing
circuit. Notice Los
that circuitos
the powerabierto
absorbed
byOpen
each
of these
devices
iscan
zero.
y corto
se pueden
agregar
a un
circuito
sin alterar
las corrientes
y los volta-the branch currents and
voltages
of
all
the
other
devices
in
the
circuit.
Figure
2.6-3
shows
how this
Open
and
short
circuits
can
be
added
to
a
circuit
without
disturbing
the
branch
currents
and
jes de las derivaciones de todos los demás dispositivos en el circuito. La figura 2.6-3 muestra cómo
se can be done. Figure
(a
2.6-3a
shows
an
example
circuit.
In
Figure
2.6-3b
an
open
circuit
and
a
short
circuit+have been added
voltages puede
of all hacer
the other
devices
in
the
circuit.
Figure
2.6-3
shows
how
this
can
be
done.
Figure
esto. La figura 2.6-3a presenta un circuito de ejemplo. En la figura 2.6-3b se han agregado(a)
this
example
The
connected
between
twoentre
nodes
of the
circuit. In
2.6-3a shows
an exampleycircuit.
In to
Figure
2.6-3b
ancircuit.
open circuit
and circuit
a short
circuit
have
been
un cortocircuito
un circuito
abierto
a este
circuito
de open
ejemplo.
El was
circuito
abierto
seadded
conectó
dos
i(t)
v(t)original
=0
contrast,
the
short
circuit
was
added
by
cutting
a
wire
and
inserting
the
short
circuit.
Adding
open
to this example
circuit.
The
open
circuit
was
connected
between
two
nodes
of
the
original
circuit.
In
nodos del circuito original. Por el contrario, el cortocircuito se agregó al cortar un cable e insertar el
–
+
circuits
and
short
circuits
to
a
network
in
this
way
does
not
change
the
network.
contrast,cortocircuito.
the short circuit
was
added
by
cutting
a
wire
and
inserting
the
short
circuit.
Adding
open
Agregar cortocircuitos y circuitos abiertos a una red de esta manera no modifica la red.
+
circuits
circuits
candescribir
also be described
as especiales
special cases
circuits and short
to a network
inOpen
this way
doesand
notshort
change
the network.
Loscircuits
cortocircuitos
y los circuitos
abiertos
también
se pueden
como casos
de of resistors. A resistor v(t) = 0
v(t)
=
0
with
resistance
R
¼
0
(G
¼
1)
is
a
short
circuit.
A
resistor
with
conductance
¼ 0 (b)
(R ¼ 1) is an
Open
circuits
and
short
circuits
can
also
be
described
as
special
cases
of
resistors.
A
resistor
–
resistores. Un resistor con una resistencia R 5 0(G 5 ) es un cortocircuito. Un resistor con conduc- Gi(t)
open
circuit.
with resistance
R
¼
0
(G
¼
1)
is
a
short
circuit.
A
resistor
with
conductance
G
¼
0
(R
¼
1)
is
an
–
tancia G 5 0 (R 5 ) es un circuito abierto.
FIGURA 2.5-3
open circuit.
(a) Circuito
2.6
(b
abierto. (b)
VO
E REST R
AO
NS
D AMMETERS
2.6 V O LT Í M E T R2.6
OS Y
A LMTPMEERTÍ M
VOLTMETERS AND AMMETERS
(b)Cortocircuito.
Las mediciones de corriente
y voltaje se hacen
lectores
directosare
(análogos)
o medidores
digitales,
FIGURE
2.5-3 or digital meters,
Measurements
of dc con
current
and voltage
made with
direct-reading
(analog)
como
se
muestra
en
la
figura
2.6-1.
Un
medidor
de
lectura
directa
tiene
un
apuntador
cuya
desviación
(a)
Open
circuit.
as shown
in Figure
2.6-1. A direct-reading
an indicating pointer whose angular
Measurements of dc current and voltage
are made
with direct-reading
(analog) ormeter
digitalhas
meters,
an­ginular
depende
de la
de la
variable
que
está
midiendo.
medidor
despliega
uncircuit.
(b)A
Short
deflection
depends
on
theanmagnitude
ofpointer
theUn
variable
it isdigital
measuring.
digital
meter displays a set
as shown
Figure
2.6-1.
A magnitud
direct-reading
meter
has
indicating
whose
angular
conjunto
de
dígitos
que
indican
el
valor
de
la
variable
medida.
of of
digits
indicating
measuredAvariable
value.displays a set
deflection depends on the magnitude
the variable
it isthe
measuring.
digital meter
Para medir
un voltajevariable
o corriente
se conecta
un medidor
a unacircuito
unas
To value.
measure
a voltage
or current,
meter ismediante
connected
to terminales
a circuit, using terminals called
of digits indicating
the measured
llamadas
probadores.
Estos
probadores
tienen
colores
codificados
para
indicar
la
dirección
de refe- of the variable being
probes.
These isprobes
are color
coded to
indicate
the reference
To measure a voltage or current,
a meter
connected
to a circuit,
using
terminals
called direction
rencia
de
la
variable
que
se
va
a
medir.
Los
probadores
métricos
suelen
presentar
los
colores
rojo voltmeter
y
measured.
Frequently,
meter probes
are colored
red and being
black. An ideal
measures the
probes. These probes are color coded
to indicate
the reference
direction
of the variable
negro.
Un
voltímetro
ideal
mide
los
voltajes
del
probador
rojo
al
negro.
La
terminal
roja
es
positiva,
voltage
from the
the black
probe.
The red measures
terminal isthe
the positive terminal, and the black
measured. Frequently, meter probes
are colored
redred
andtoblack.
An ideal
voltmeter
y la negra
es negativa
(verprobe.
figuraThe
2.6-2b).
terminal
is red
the terminal
negative isterminal
(see Figure
2.6-2b).
voltage from
the red
to the black
the positive
terminal,
and the black
Un
amperímetro
ideal
mide
la ideal
corriente
que fluye
a través
sus terminales,
como se its
muestra
en as shown in Figure
An
ammeter
measures
thedecurrent
flowing through
terminals,
terminal is the negative terminal (see Figure
2.6-2b).
la
figura
2.6-2a
y
tiene
voltaje
cero,
v
,
a
través
de
sus
terminales.
Un
voltímetro
ideal
mide
el
voltaje
m
, across itsasterminals.
ideal voltmeter measures the voltage
and has
zero through
voltage, its
vmterminals,
An ideal ammeter measures2.6-2a
the current
flowing
shown inAn
Figure
a través
de sus
terminales,
como its
se
enAn
la
2.6-2b
y tiene
una corriente
terminal,
im,
its muestra
terminals,
as figura
shownvoltmeter
in Figure
2.6-2b,
and voltage
hasde
terminal
current,
im, equal
terminals.
ideal
measures
the
2.6-2a and
has zero
voltage,
vm, across
(a) to zero.
igual
a
cero.
Los
instrumentos
de
medición
útiles
sólo
se
aproximan
a
las
condiciones
ideales.
Para
measuring
only
approximate
the to
ideal
conditions. For a practical ammeter,
zero.
across its terminals, as shown inPractical
Figure 2.6-2b,
andinstruments
has terminal
current,
im, equal
amperímetro
útil el voltaje
a travésacross
de sus
suele
ser
de tan
pequeño.the
Del
the voltage
itsterminales
terminals
is usually
negligibly
Similarly,
current into a voltmeter
Practicalunmeasuring
instruments
only
approximate
the
ideal
conditions.
Forinsignificante
a practicalsmall.
ammeter,
mismo
modo,
la
corriente
en
un
voltímetro
suele
ser
ínfima.
is usually
negligible.
the voltage across its terminals is usually
negligibly
small. Similarly, the current into a voltmeter
(a)
Los voltímetros ideales actúan
circuitos
amperímetros
actúan
Idealcomo
voltmeters
actabiertos,
like openy los
circuits,
and idealideales
ammeters
act como
like short circuits. In other
is usually negligible.
cortocircuitos.
Es
decir,
el
modelo
de
un
voltímetro
ideal
es
un
circuito
abierto,
y
el
de
un
amperímetro
words,
the model
of an
ideal voltmeter
an open
circuit,
and the model of an ideal ammeter is a
Ideal voltmeters act like open
circuits,
and ideal
ammeters
act like isshort
circuits.
In other
ideal
es unofcortocircuito.
Considere
circuito
dethe
la circuit
figura
2.6-3a
luego
abierto
short is
circuit.
Consider
of Figure
2.6-3a
and
thenunadd
circuit with a voltage v and
words, the
model
an ideal voltmeter
anelopen
circuit,
and
the model
ofy an
idealagregue
ammeter
iscircuito
aan open
con
un
voltaje
v
y
un
cortocircuito
con
una
corriente
i
como
se
muestra
en
la
figura
2.6-3b.
En
la
figura
short circuit
current
i asopen
shown
in Figure
InvFigure
the open circuit has been
short circuit. Consider the circuit ofaFigure
2.6-3awith
and athen
add an
circuit
with a2.6-3b.
voltage
and 2.6-3c,
2.6-3c,with
el circuito
ha
sido
reemplazado
un the
voltímetro
y el
cortocircuito
ha by
sido
un The voltmeter will
replaced
by a voltmeter,
and
short circuit
hascircuit
been replaced
an por
ammeter.
a short circuit
a currentabierto
i as shown
in Figure
2.6-3b. por
In Figure
2.6-3c,
the
open
haslo
been
FIGURE 2
(a) Open ci
(b) Short ci
(a
measure
labeledbyv an
in ammeter.
Figure 2.6-3b
whereas the
replaced by a voltmeter, and the short
circuitthe
hasvoltage
been replaced
The voltmeter
willammeter will measure the current
i. Notice
that Figure
2.6-3c could
be obtained
Figure 2.6-3a by adding(b)
a voltmeter
measure the voltage labeled v in labeled
Figure 2.6-3b
whereas
the ammeter
will measure
the from
current
Voltímetro
labeled i. Notice that Figure 2.6-3c could be obtained from Figure 2.6-3a by adding
a voltmeter
im = 0
Amperímetro
Elemento i
vm = 0
Ammeter Element
+
Element
(a)
i
–
i
Ammeter
i
im = 0
+
FIGURA 2.6-2 (a) Amperímetro ideal. (b) Voltímetro+ideal.
+
Circuitos Eléctricos - Dorf
(a)
vm = 0
–
+
v
Voltmeter
i
i
Elemento
v
–
(b)
vm = 0
(a)
–
–
Element
i
FIGURA 2.6-1
(a) Medidor de
(b)Voltmeter
lectura directa
(análogo). (b)
FIGURE 2.6-1
Medidor digital.
0 directim
(a)= A
+
v
–
reading (analog)
meter.
(b) A digital
i
i
Element
meter.
Alfaomega
(b)
(b)
FIGURE 2.6-2 (a) Ideal ammeter. (b) Ideal voltmeter.
FIGURE 2.6-2 (a) Ideal ammeter. (b) Ideal voltmeter.
M02_DORF_1571_8ED_SE_020-052.indd 31
4/12/11 5:17 PM
(b
FIGURE 2
(a) A direct
reading (an
meter.
(b) A digita
meter.
32
Elementos de circuitos
50 Ω
+
–
20 Ω
10 Ω
50 Ω
10 Ω
Cortocircuito
+ v –
Circuito abierto
2 voltios
20 Ω
60 Ω
i
+
–
2 voltios
60 Ω
(a)
(b)
Voltímetro
+
v
Amperímetro
–
i
50 Ω
10 Ω
+
–
2 voltios
60 Ω
20 Ω
(c)
FIGURA 2.6-3 (a) Un circuito de ejemplo, (b) con un circuito abierto y un cortocircuito agregados. (c) El circuito
abierto ha sido reemplazado por un voltímetro y el cortocircuito por un amperímetro.
amperímetro. El voltímetro medirá el voltaje etiquetado v en la figura 2.6-3b en tanto que el amperímetro medirá la corriente etiquetada i. Observe que la figura 2.6-3c podría obtenerse de la figura 2.6-3a al
agregar un voltímetro y un amperímetro. Idealmente, agregar el voltímetro y el amperímetro de esta manera no debe alterar el circuito. Una interpretación más de la figura 2.6-3 es útil. La figura 2.6-3b podría
formarse con la figura 2.6-3c si se reemplazan el voltímetro y el amperímetro por sus modelos (ideales).
La dirección de referencia es una parte importante del voltaje o la corriente de un elemento.
Las figuras 2.6-4 y 2.6-5 indican que se debe poner bastante atención a las direcciones de referencia
cuando se mide un voltaje o la corriente de un elemento. La figura 2.6-4a muestra un voltímetro. Los
voltímetros tienen probadores bicolores. Este color indica la dirección de referencia del voltaje que se
ha de medir. En las figuras 2.6-4b y la figura 2.6-4c se ha utilizado un voltímetro para medir el voltaje
a través de la resistencia de 6 kV. Cuando el voltímetro se conecta al circuito, como se muestra en la
Voltímetro
+
v
– 3 . 6
Voltímetro
Voltímetro
+
–
5 kΩ
12 V
(a)
+ 3 . 6
+
–
va –
–
6 kΩ
10 k Ω
(b)
4 kΩ
5 kΩ
12 V
+
–
vb +
6 kΩ
10 k Ω
4 kΩ
(c)
FIGURA 2.6-4 (a) Correspondencia entre los probadores del código de color del voltímetro y la dirección de referencia
del voltaje medido. En (b), el signo 1 de va está a la izquierda, en tanto que en (c), el signo 1 de vb está a la derecha.
El probador coloreado se muestra en gris. En el laboratorio este probador será rojo. Al referirnos al probador de color
se dirá que es el “probador rojo”.
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 32
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
Fuentes dependientes
Amperímetro
+ 1 . 2
– 1 . 2
Amperímetro
Amperímetro
33
i
6 kΩ
12 V
(a)
+
–
6 kΩ
ia
(b)
4 kΩ
12 V
+
–
ib
4 kΩ
(c)
FIGURA 2.6-5 (a) Correspondencia entre los probadores del código de color del amperímetro y la dirección de
referencia de la corriente medida. En (b) la corriente ia está dirigida a la derecha, en tanto que en (c) la corriente ib está
dirigida a la izquierda. El probador coloreado se muestra aquí en gris. En el laboratorio será rojo. Nos referiremos al
probador de color como el “probador rojo”.
figura 2.6-4b, el voltímetro mide va, con 1 a la izquierda, en el probador rojo. Cuando se intercambian
las pruebas del voltímetro como se muestra en la figura 2.6-4c, el voltímetro mide vb, con el signo 1
a la derecha, de nuevo en el probador rojo. Observe que vb 5 2va.
La figura 2.6-5a muestra un amperímetro. Los amperímetros tienen probadores de código de
dos colores. Esta codificación de color indica la dirección de referencia de la corriente que se va a
medir. En las figura 2.6-5b y c, el amperímetro se utiliza para medir la corriente en el resistor de 6 kV.
Cuando se conecta el amperímetro al circuito como se muestra en la figura 2.6-5b, el amperímetro
mide ia, direccionada del probador rojo al probador negro. Cuando los probadores del amperímetro están intercambiados como se muestra en la figura 2.6-5c, el amperímetro mide ib, de nuevo
direccionado del probador rojo al negro. Observe ib 5 2ia.
2.7
FUENTES DEPENDIENTES
Las fuentes dependientes modelan la situación en la cual el voltaje o la corriente de un elemento de circuito es proporcional al voltaje o corriente del segundo elemento de circuito. (En contraste, un resistor
es un elemento de circuito en el cual el voltaje del elemento es proporcional a la corriente del mismo elemento.) Las fuentes dependientes se emplean para modelar dispositivos electrónicos como transistores o
amplificadores. Por ejemplo, el voltaje de salida de un amplificador es proporcional al voltaje de entrada
de ese amplificador, de modo que un amplificador puede ser modelado por una fuente dependiente.
La figura 2.7-1a muestra un circuito que incluye una fuente dependiente. El símbolo de diamante representa una fuente dependiente. Los signos más y menos dentro del diamante identifican la
fuente dependiente como una fuente de voltaje e indican la polaridad de referencia del voltaje del elemento. La etiqueta “5i ” representa el voltaje de esta fuente dependiente. Este voltaje es un producto
de dos factores, 5 e i. El segundo factor, i, indica que el voltaje de esta fuente dependiente es controlado
por la corriente, i, en el resistor de 18 V. El primer factor, 5, es la ganancia de esta fuente dependiente,
la cual es la proporción del voltaje controlado, 5i, para la corriente predominante, i. Esta ganancia
tiene unidades de V>A u V. Dado que esta fuente dependiente es una fuente de voltaje y como una corriente controla el voltaje, la fuente dependiente se denomina fuente de voltaje de corriente controlada
(CCVS, por sus siglas en inglés).
La figura 2.7-2b muestra el circuito de 2.7-1a, desde un punto de vista diferente. En la figura
2.7-1b, se ha insertado un cortocircuito en serie con el resistor de 18 V. Ante esto se considera la
corriente predominante i como la corriente en un cortocircuito más que como la corriente en el resistor de 18 V en sí. De este modo, se puede tratar siempre la corriente predominante de una fuente
dependiente como la corriente en un cortocircuito. En esta sección usaremos este segundo punto de
vista para clasificar las fuentes dependientes.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 33
Alfaomega
4/12/11 5:17 PM
34
Elementos de circuitos
i
i
18 Ω
18 Ω
+
–
12 Ω
24 V
+
–
5i
+
–
12 Ω
24 V
(a)
+
v
5i
(b)
+
–
v
–
18 Ω
18 Ω
+
–
+
–
12 Ω
24 V
(c)
0.2 v
+
–
24 V
12 Ω
0.2 v
(d)
FIGURA 2.7-1 La corriente predominante de una fuente dependiente mostrada como (a) la corriente en un elemento, y
(b) la corriente en un cortocircuito en serie con ese elemento. El voltaje predominante de una fuente dependiente se muestra
como (c) el voltaje a través de un elemento y (d ) el voltaje a través de un circuito abierto en paralelo con ese elemento.
La figura 2.7-1c muestra un circuito que incluye una fuente dependiente, representada por el
símbolo del diamante. La flecha dentro del diamante identifica la fuente dependiente como una fuente
de corriente e indica la dirección de referencia de la corriente del elemento. La etiqueta “0.2 v” representa la corriente de esta fuente dependiente. Esta corriente es producto de dos factores, 0.2 y v. El
segundo factor, v, indica que la corriente de esta fuente dependiente está controlada por el voltaje, v, a
través del resistor de 18 V. El primer factor, 0.2, es la ganancia de esta fuente dependiente, la cual es
a su vez la proporción de fuente controlada, 0.2 v, para el voltaje predominante, v. Esta ganancia tiene
unidades de A/V. Dado que esta fuente dependiente es una fuente de corriente y como un voltaje controla la corriente, la fuente dependiente se denomina fuente de corriente de voltaje controlado (VCCS
por sus siglas en inglés).
La figura 2.7-1d muestra el circuito de la figura 2.7-1c, desde un punto de vista diferente. En la
figura 2.7-1d, se ha agregado un circuito abierto en paralelo con el resistor de 18 V. Ahora consideramos
el voltaje predominante v como el voltaje a través de un circuito abierto de la figura 2.7-1, en vez del
voltaje a través del mismo resistor de 18 V. De este modo, se puede tratar siempre el voltaje predominante de una fuente dependiente como el voltaje a través de un circuito abierto.
Estamos listos para clasificar las fuentes dependientes. Cada fuente dependiente consta de dos
partes: la parte predominante y la parte controlada. La parte predominante es o bien un circuito abierto
o un cortocircuito. La parte controlada puede ser una fuete de voltaje o una fuente de corriente. Hay
cuatro tipos de fuente dependiente que corresponden a las cuatro formas de seleccionar una parte controladora y una parte controlada. Estas cuatro fuentes dependientes se denominan (* por sus siglas en
inglés) fuente de voltaje controlada por voltaje (VCVS*), fuente de voltaje controlada por corriente
(CCVS*), fuente de corriente controlada por voltaje (VCCS*), y fuente de corriente controlada por
corriente (CCCS*). En la tabla 2.7-1 se muestran los símbolos que representan las fuentes dependientes.
Considere la CCVS de la tabla 2.7-1. El elemento predominante es un cortocircuito. La corriente
y el voltaje del elemento que predomina están indicados como ic y vc. El voltaje a través de un cortocircuito es cero, por lo tanto, vc 5 0. La corriente de cortocircuito, ic es la señal de control de esa fuente
dependiente. El elemento controlado es una fuente de voltaje. La corriente y el voltaje del elemento
controlado se representan por id y vd, respectivamente. El voltaje controlado vd es controlado por ic:
vd 5 ric
La constante r se denomina la ganancia de CCVS. La corriente id, como la corriente de toda fuente de
voltaje, está determinada por el resto del circuito.
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 34
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
Fuentes dependientes
35
Tabla 2.7-1 Fuentes dependientes
DESCRIPCIÓN
Fuente de voltaje controlada por corriente (CCVS)
r es la ganancia de la CCVS.
r tiene unidades de voltios/amperios
SÍMBOLO
+
+
vc +
=
vc +
=
vc –=
vc –=
–
–
0
0
0
0
Fuente de voltaje controlada por voltaje (VCVS)
b es la ganancia de la VCVS.
b tiene unidades de voltios/voltios
Fuente de corriente controlada por voltaje (VCCS)
g es la ganancia de la VCCS.
g tiene unidades de amperios/voltios
Fuente de corriente controlada por corriente (CCCS)
d es la ganancia de la CCCS
d tiene unidades de amperios/amperios
+
+
vc +
=
vc +
=
vc –=
vc –=
–
–
0
0
0
0
+
+
–
+
–
+
–
–
id
id
id
id v
d
vd
vd
vd
=
=
=
=
ric
ric
ric
ric
+
+
–
+
–
+
–
–
id
id
id
id v
d
vd
vd
vd
=
=
=
=
bvc
bvc
bvc
bvc
+
+
v+d
v+d
v–d
v–d
–
–
id
id
id
id
=
=
=
=
gvc
gvc
gvc
gvc
+
+
v+d
v+d
v–d
v–d
–
–
id
id
id
id
=
=
=
=
dic
dic
dic
dic
ic
ic
ic
ic
ic
ic
+ ic
+ ic
v+c
vc
–
v+
–c
v–c
–
=
=
=
=
0
0
0
0
ic
ic
+ ic
+ ic
v+c
vc
–
v+
–c
v–c
–
=
=
=
=
0
0
0
0
ic
ic
ic
ic
A continuación, considere la VCVS de la tabla 2-7-1. El elemento predominante es un circuito
abierto. La corriente en un circuito abierto es cero, por lo tanto, ic 5 0. El voltaje del circuito abierto, vc,
es la señal controladora de esta fuente dependiente. El elemento controlado es una fuente de voltaje.
El voltaje vd es controlado por vc:
vd 5 bvc
La constante b se denomina ganancia de VCVS. La corriente id está determinada por el resto del
circuito.
El elemento controlador de la VCCS que se muestra en la tabla 2.7-1 es un circuito abierto.
La corriente en este circuito abierto es ic 5 0. El voltaje del circuito abierto, vc, es la señal contro­
la­do­ra de esta fuente dependiente. El elemento controlado es una fuente de corriente. La corriente
id está controlada por vc:
id 5 gvc
La constante g se denomina ganancia de VCCS. El voltaje vd, como el voltaje a través de toda fuente
de corriente, está determinado por el resto del circuito.
El elemento controlador de la CCCS que se muestra en la tabla 2.7-1 es un cortocircuito. El
voltaje a través de este circuito abierto es vc 5 0. La corriente del cortocircuito, ic, es la señal controladora de esta fuente dependiente. El elemento controlado es una fuente de corriente. La corriente id
está controlada por ic:
id 5 dic
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 35
Alfaomega
4/12/11 5:17 PM
36
Elementos de circuitos
ic
ib
b
c
vbe
b
c
+
r
–
e
e
(b)
(a)
RB
RB
vin
RC
vo
ic
ib
+
+
+
–
gmvbe
+
–
vin
vbe
+
r
–
RC
vo
–
–
(c)
gmvbe
(d)
FIGURA 2.7-2 (a) Símbolo para un transistor. (b) Modelo del transistor. (c) Amplificador de transistor. (d ) Modelo del
amplificador de transistor.
La constante d se denomina ganancia de CCCS. El voltaje, vd como el voltaje a través de toda fuente
de corriente, está determinado por el resto del circuito.
La figura 2.7-2 ilustra el uso de las fuentes dependientes para modelar dispositivos electrónicos.
En ciertas circunstancias, el comportamiento del transistor que se muestra en la figura 2.7-2a se puede
representar utilizando el modelo que se muestra en la figura 2.7-2b. Este modelo consta de una fuente
dependiente y un resistor. El elemento controlador de la fuente dependiente es un circuito abierto a
través del resistor. El voltaje controlador es vbe. La ganancia de la fuente dependiente es gm. La fuente
dependiente se utiliza en este modelo para representar una propiedad del transistor, específicamente,
que la corriente ic es proporcional al voltaje vbe, es decir,
ic 5 gmvbe
donde gm tiene unidades de amperios/voltios. Las figuras 2.7-2c y d ilustran la utilidad de este modelo.
La figura 2.7-2d se obtiene de la figura 2.7-2c al reemplazar el transistor por el modelo de transistor.
E j e m p l o 2 . 7- 1
Fuentes de poder y dependientes
Determine la potencia absorbida por la VCVS de la figura 2.7-3.
Solución
La VCVS consta de un circuito abierto y una fuente de voltaje controlado. En un circuito abierto no hay corriente,
por lo tanto, no hay absorción de potencia en un circuito abierto.
El voltaje, vc, a través del circuito abierto es la señal controladora de la VCVS. El voltímetro mide vc para que
El voltaje de la fuente de voltaje controlado es
vc 5 2 V
vd 5 2 vc 5 4 V
El amperímetro mide la corriente en la fuente de voltaje controlado para que
id 5 1.5 A
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 36
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
Transductores
Transducers
+
+ 2.
2. 0
0 0
0
Voltímetro
Voltmeter
+
+
+
+ 1.
1. 5
5 0
0
Amperímetro
Ammeter
––
vvc
c
2
2Ω
Ω
+
12
12 VV –+–
37
37
++ ––
0.5
0.5 A
A
iid
d
vvd =
2v
d = 2vcc
4Ω
Ω
4
FIGURE2.7-3
2.7-3UnAcircuito
circuitque
containing
a VCVS.
FIGURA
contiene una
VCVS.The
Los meters
2.0
indicate that
the voltage
the controlling
element
is vc ¼es
medidores
indican
que el of
voltaje
del elemento
controlador
and
that the
current
of the controlled
element
is id ¼ 1.5
vcvolts
5 2.0
voltios
y que
la corriente
del elemento
controlado
esamperes.
id 5 1.5 amperios.
La
elemento,
id, yvoltage,
el voltaje,
la convención
pasiva.
Por lo tanto,
Thecorriente
elementdel
current,
id, and
vd, vadhere
to the apassive
convention.
Therefore,
d, se apegan
p ¼ id vd ¼ ð1:5Þð4Þ ¼ 6 W
is the
power absorbed
es
la potencia
absorbidaby
porthe
la VCVS.
VCVS.
EJERCICIO
2.7-1FindObtenga
la potencia
por lainCCCS
enEla2.7-1.
figura E 2.7-1.
EXERCISE 2.7-1
the power
absorbedabsorbida
by the CCCS
Figure
– 1. 2 0
– 1. 2 0
Amperímetro
Ammeter
2Ω
2Ω
12 V +–+
12 V –
+ 2 4. 0
+ 2 4. 0
Voltímetro
Voltmeter
2Ω
2Ω
ic
ic
4Ω
4Ω
id = 4ic
id = 4ic
+
+
vd
vd
–
–
FIGURA
circuitocontaining
que contiene
una CCCS.
Los medidores
indican
que la corriente
del elemento
controlador
FIGURE EE2.7-1
2.7-1 Un
A circuit
a CCCS.
The meters
indicate that
the current
of the controlling
element
is ic ¼
es
ic 5amperes
21.2 amperios
que
el voltaje
delcontrolled
elementoelement
controlado
24 voltios.
�1.2
and that ythe
voltage
of the
is vdes¼vd245volts.
Sugerencia:
El elemento
controlador
esta fuente
dependiente
un cortocircuito.
Elacross
voltajea ashort
traHint: The controlling
element
of this de
dependent
source
is a shortescircuit.
The voltage
vés
de un
cortocircuito
Porabsorbed
consiguiente,
la controlling
potencia absorbida
el elemento
controlador
circuit
is zero.
Hence, es
thecero.
power
by the
elementpor
is zero.
How much
power es
is
cero.
¿Cuánta
potencia
absorbe
el elemento controlado?
absorbed
by the
controlled
element?
Answer: �115.2 watts are absorbed by the CCCS. (The CCCS delivers þ115.2 watts to the rest of
Respuesta: La CCCS absorbe 2115.2 watts (la CCCS entrega 1 115.2 watts al resto del circuito).
the circuit.)
2.8
2.8
TRANSDUCTORES
TRANSDUCERS
Los transductores son dispositivos que convierten cantidades físicas en cantidades eléctricas. Esta
sección
describe
dos transductores:
potenciómetros
sensores
dequantities.
temperatura.
potenciómeTransducers
are devices
that convert physical
quantitiesy to
electrical
ThisLos
section
describes
tros
posición en resistencia
y los sensores
temperatura convierten
la temperatura
two convierten
transducers:lapotentiometers
and temperature
sensors.de
Potentiometers
convert position
to resisten
corriente.
ance,
and temperature sensors convert temperature to current.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 37
Alfaomega
4/12/11 5:17 PM
Describe
Describe
the the
Situation
Situation
and
and
the
the
Assumptions
Assumptions
mAmA
for tfor
� 0,
t � 0,
It isItknown
is known
that that
i(t) ¼
i(t)De
¼�t/60
De�t/60
38
38
Circuit
38Elements
Circuit Elements
Circuit Elements
Circuit Elements
Elementos de circuitos
W
1. The
1. The
current
current
enters
theacross
plus
theacross
plus
terminal
of the
ofsecond
the second
element. FIGURE
(t)v¼
and
and
the
voltage
theenters
voltage
the terminal
second
the second
element
element
is velement.
2is
2(t) ¼ FIGURE
1.8-11.8
T
�t/60�t/60
V
for
V
t
for
�
0.
t
�
The
0.
The
maximum
maximum
magnitude
magnitude
of
the
of
the (1
Be
Be
a jet
avalve
jetp valve
for afor
sp
(1
–
a)R
– a)R
pVoltmeter
2. The
2.Voltmeter
The
current
current
leaves
leaves
the plus
the plus
terminal
terminal
of the
offirst
the first
element.
element.
D, D,
is limited
is limited
to 1to
1vmmA.
Determine
the the
– current,
– Determine
+ vm current,
+ mA.
+ the
3. The
3. The
wires
wires
are perfect
are
perfect
andDhave
and
no
effect
noand
effect
on
onthe
circuit
thethe
circuit
(they(they
do battery.
not
do absorb
not absor
en+
required
required
constants
constants
and
Dhave
and
B and
B
describe
describe
required
required
battery.
4. The
4. The
model
model
of the
ofthe
circuit,
thethe
circuit,
as shown
as aR
shown
in Figure
in vthe
Figure
1.8-1,
1.8-1,
assumes
assumes
the
the volta
vam
aRpvoltage
Describe
Describe
Situation
and
the
Assumptions
I that that
m Assumptions
p and
I
RIp Situation
two two
elements
elements
is equal;
is equal;
that that
is, v1is,¼vv12¼
. v 2.
Figure 2.8-1a shows
symbol
for the The
potentiometer.
The
potentiometer
is
Figurethe
2.8-1a
shows
symbol
for
the
potentiometer.
The
potentiometer
is
a element.
The
current
current
enters
enters
the
plus
the plus
terminal
terminal
ofa the
of second
the second
element.
Figure 2.8-1a shows the symbol for1.the1.
potentiometer.
The
potentiometer
�t/60�t/60 – is a
resistor
having
a
third
contact,
called
the
wiper,
that
slides
along
the
resistor.
Two
Figure
2.8-1a
shows
the
symbol
for
the
potentiometer.
The
potentiometer
is
a
resistor
having
a
third
contact,
called
the
wiper,
that
slides
along
the
resistor.
Two
¼vBe
¼
Be
V where
V where
B is B
theisinitial
the initial
voltage
voltage
of the
ofbt–
The
5. the
The
battery
battery
voltage
voltage
v1 isvvalong
figura a2.8-1a
muestra5.called
el símbolo
para
elthat
potenciómetro.
potenciómetro
11is
1 El
(1 – a)R
(1 resistor
– a)Rp Lahaving
third contact,
slides
the
Two
2.wiper,
2.
The
The
current
current
leaves
leaves
the
plus
theresistor.
plus
terminal
terminal
of the
of first
the first
element.
element.
(1 – a)Rpp
and
a,
are
needed
to
describe
the
potentiometer.
The
parameter
R
parameters,
R
resistor
having
a
third
contact,
called
the
wiper,
that
slides
along
the
resistor.
Two
and
a,
are
needed
to
describe
the
potentiometer.
The
parameter
R
parameters,
R
will
will
discharge
discharge
exponentially
exponentially
as
it
as
supplies
it
supplies
energy
energy
to
the
to
valve.
the
valve.
es un resistor que
tiene un tercer
contacto, llamado cursor, que se desliza a lo largo
p
p
p
FIGURE
2.8-2 (a)pA circuit contai
p
(1 – a)Rp (1 – a)Rp parameters, Rp and a, are needed to describe the potentiometer. The parameter R
specifies
theSepotentiometer
resistance
(R3.
>
0).para
The
parameter
aand
represents
the
a,
are needed
describe
the
The
parameter
R
parameters,
Rp and
specifies
the
potentiometer
resistance
(R
>
0).
The
parameter
a effect
represents
the
3.
The
The
wires
wires
are
perfect
are
perfect
and
have
have
no
effect
no
on
the
on
the
circuit
(they
do not
do
del
resistor.
necesitan
dos
parámetros,
R
y
a,
describir
el
potenciómetro.
El
pto
p circuit
p potentiometer.
Rp
p
Rp
potentiometer.
(b)
An(they
equivalent
ci
specifies the potentiometer resistance
(Rp circuit
>operates
0). operates
The from
parameter
the
6.in The
6. range
The
circuit
from
t ¼ 0t to
¼aa0trepresents
¼
to60
t¼
s.60as.¼
Rp
wiper
position
and
takes
values
the
0
�
a
�
1.
The
values
¼
0
and
1
specifies
the
potentiometer
resistance
(R
>
0).
The
parameter
a
represents
the
wiper
position
and
takes
values
in
the
range
0
�
a
�
1.
The
values
a
¼
0
and
a
¼
1
(a)
(a)
(b)
(b)
parámetro
R
especifica
la
resistencia
del
potenciómetro
(R
.
0).
El
parámetro
a
p
a
model
of
the
potentiometer.
R
p
p
p
wiper position and takes values in the range
� The
amodel
� 1.
The
values
a ¼as0 shown
and
a¼
Rp
4. 04.
The
model
of the
of
circuit,
the circuit,
as shown
in1Figure
in Figure
1.8-1,
1.8-1,
assumes
assumes
thattha
th
aR
aRp
correspond
toposición
the extreme
positions
of
the
wiper.
wiper
position
andcursor
in
the
range
0limited,
�the
a�
1.�The
values
¼ 0 and a ¼ 1
correspond
totakes
the
extreme
positions
wiper.
7. yvalues
The
7.
The
current
current
is
limited,
iselof
so
D0so
Da1 �
mA.
1 mA.
representa
lato
del
toma
valores
en
rango
1.
Los avalores
aRpp
correspond
the
extreme
positions
of
the
wiper.
two
two
elements
elements
is
equal;
is
equal;
that
that
is,
v
is,
v
¼
v
¼
.
v
.
1
1
2
2
aRp a 5 0 Figure
2.8-1b
shows
model
for the
potentiometer
that
consists
of
to
thethe
extreme
positions
of
the for
wiper.
Figure
2.8-1b
shows
a model
thattwo
consists of two
y acorrespond
5 1Solving
corresponden
aaaangle
las
posiciones
del
cursor.
aRp
for
gives
Solving
forthe
thepotentiometer
angle
gives of
Figure
2.8-1b
shows
model
for theextremas
potentiometer
that
consists
two
�t/60
�t/60
resistors.
The
resistances
of
these
resistors
depend
on
the
potentiometer
parameters
Figure
2.8-1b
shows
a
model
for
the
potentiometer
that
consists
two
resistors.
The
resistances
of
these
resistors
depend
on
the
potentiometer
¼vBe
where
Vparameters
where
B isBthe
is initial
the initial
volta
v
5.
The
battery
voltage
vque
La figura
2.8-1b muestra
unresistors
modelo
para
elThe
potenciómetro
deBe
dosV of
State
State
the 5.
the
Goal
Goal
1 isvconsta
1v1isparameters
1¼
resistors.
The resistances
of these
depend
onbattery
thevoltage
potentiometer
360
360
and
a.
R
resistors.
The
resistances
of
these
resistors
depend
on
the
potentiometer
parameters
and
a.
R
v
v
u
¼
u
¼
will
will
discharge
discharge
exponentially
exponentially
as
it
as
supplies
it
supplies
energy
energy
to
the
to
valve.
the
valve.
resistores.
Las
resistencias
de
estos
resistores
dependen
de
los
parámetros
R
y
a
del
p
Determine
Determine
the
energy
the
energy
supplied
supplied
by
the
by
first
the
first
element
element
for
the
for
one-minute
the
one-minute
period
period
and
p
(a)
m
m
(a) (b)
(b)
p
Rp and a.
Ito the angular
Rp position
(a)
(b)
Frequently,
position of
the
wiper
corresponds
to corresponds
the angular
of
a position of a Rp I
Rp and a. theFrequently,
the
wiper
potenciómetro.
thethe
constants
theposition
constants
Dofand
D and
B.
Describe
B.
Describe
the battery
theposition
battery
selected.
selected.
(a)
(b)
Frequently,
the
position
of
the
wiper
corresponds
to
the
angular
of
a
(a)
(b)
6.
6.
The
The
circuit
circuit
operates
operates
from
from
t
¼
0
t
¼
to
0
t
¼
to
t ¼s.60 s.
� the
�
FIGURE 2.8-1 (a)
The symbol
FIGURE
2.8-1 (a) The
symbol
shaft
connected
to the
potentiometer.
Suppose
uSuppose
isAn
the
angle
inu10
degrees
and
01del
�
uan
�60
Frequently,
the
of
wiper
corresponds
angular
position
of of
shaft
connected
to
the
potentiometer.
Suppose
isto
the
angle
in
degrees
0a �vofmu163
� 4.53
Suppose
R
10position
kV
and
I ¼the
1 mA.
angle
of
163
kV
would
and
I¼
cause
mA.
output
Anand
angle
¼
would
V. Acau
me
En ocasiones,
la potentiometer.
posición
del
cursor
corresponde
aRlap ¼
posición
angular
eje
p¼
FIGURE 2.8-1 (a) The symbol
shaft
connected
to
the
Suppose
u
is
the
angle
in
degrees
and
0
�
u
�
and
(b)
a
model
for
the
andSímbolo
(b) a (a)
model
for
the
� current
�0 � u �
FIGURE
2.8-1
The
symbol
FIGURA
2.8-1
(a)
y
360.
Then,
shaft
connected
to
the
potentiometer.
Suppose
u
is
the
angle
in
degrees
and
360.
Then,
7.
7.
The
The
current
is
limited,
is
limited,
so
D
so
�
D
1
�
mA.
1
mA.
.
.
7.83
V
would
indicate
that
u
¼
282
7.83
V
would
indicate
that
u
¼
282
conectado
al
potenciómetro.
Suponga
que
es
el
ángulo
en
grados
y
que
0
360.
and (b) a model for the
360. Then,
Generate
Generate
a Plan
a Plan
potentiometer.
potentiometer.
a model for the
(b)
modeloand
del(b)
potenciómetro.
360. Then,
potentiometer.
E1C01_1
E1C01_1
11/26/2009
11/26/2009 14 14 Entonces,
u find find
First,
First,
v
(t)
v
and
i(t) and
i(t) and
thenthen
obtain
obtain
the power,
the power,
p1(t),p1supplied
(t), supplied
by the
by first
the firs
ele
1
1(t) uand
potentiometer.
a¼ u
aState
¼
State
the
the
Goal
Goal
u
a using
¼ 360
(t), find
the
the
energy
supplied
supplied
for the
for first
the first
60 s.60 s.
using
p1(t),p1find
360energy
360 a ¼Determine
the the
energy
energy
supplied
supplied
by the
by the
firstfirst
element
element
for the
for the
one-minute
one-min
360Determine
Temperature
sensors,
as Describe
the
Temperature
AD590
sensors,
such
byasAnalog
the AD590
Devi
the the
constants
constants
Dsuch
and
D and
B.
B.
Describe
the manufactured
the
battery
battery
selected.
selected.
GOALGOAL
EQUATION
EQUATION
NEEDNEED
INFORMAT
INFOR
sources having current proportional
sources
to absolute
havingtemperature.
current proportional
Figure 2.8-3a
to absolute
showste
Z
Z
EeXmApMl oP L 2E .28E.-8
- 1MCircuito
Circuit
X1A
P Potentiometer
L E to
2 .represent
8del
- 1The
Potentiometer
60
60 2.8-3b
E jE
potenciómetro
v1 and
v1 Figure
The
energy
energy
wCircuit
for
w1 the
for the Circuit
and
i known
i 2.8kn
the
to represent
Figure
the temperature
shows thesensor.
circuit
model
of
1 temperature
X A M P L E 2 . 8 - 1 Potentiometer
Generate
Generate
a Plan
awsensor.
Plan
¼w1 ¼
poperate
ðtÞproperly,
dt p1(t) pthe
E X A M P L E 2 . 8 -sensor.
1 first
Potentiometer
Circuit
1 sensor.
1(t) sensor
first
60
s temperature
constants
constants
D and
D vB
and
1 ðt Þpthe
1dt
Fors 60
the
sensor
to
For
temperature
branch
to
voltage
operate
First,
First,
findfind
v (t)v and
(t) and
i(t) i(t)
and and
thenthen
obtain
obtain
the the
power,
power,
p (t),
p supplied
(t), supplied
by tpm
38
38
14
14
I
Electric
Electric
Circuit
Circuit
Variables
Variables
Rp
1
1
0
0
1
1
(t),
find
find
the gives
the
energy
energy
supplied
forof
for
thethe
the
first
first
60 s.60 s.
using
using
pby
p1an
1(t),
Figure
2.8-2a
shows
a2.8-2a
circuit
in which
the voltage
measured
by the
meter
gives
indication
of
the
angular
Figure
shows
aencircuit
inelwhich
themedido
voltage
measured
the
meter
ansupplied
indication
angular
La
figura
2.8-2a
muestra
un circuito
el
voltaje
instrumento
muestra
indiFigure
2.8-2a
shows
a circuit
in which
thecual
voltage
measured
by por
the el
meter
gives anmedidor
indication
of theuna
angular
position
of
the
shaft.
In
Figure
2.8-2b,
the
current
source,
the
potentiometer,
and
the
voltmeter
have
been
Figure
2.8-2a
shows
a
circuit
in
which
the
voltage
measured
by
the
meter
gives
an
indication
of
the
angular
position
of
the
shaft.
In
Figure
2.8-2b,
the
current
source,
the
potentiometer,
and
the
voltmeter
have
been
cación
posición
del eje.
En lathe
figura
2.8-2b,
la fuente
de the
corriente,
el and
potenciómetro
y el have
voltímetro
positiondeofla the
shaft.angular
In Figure
2.8-2b,
current
source,
the Act
potentiometer,
the voltmeter
been
Act
on
on
the
PlanPlan
1
.
8
1
.
8
D
E
D
S
E
I
G
S
I
N
G
E
N
X
E
A
X
M
A
P
M
L
P
E
L
E
GOAL
GOAL
EQUATION
EQUATION
NEED
NEED
replaced
by
models
of
these
devices.
Analysis
of
Figure
2.8-2b
yields
position
of
the
shaft.
In
Figure
2.8-2b,
the
current
source,
the
potentiometer,
and
the
voltmeter
have
been
replaced
by
models
of
these
devices.
Analysis
of
Figure
2.8-2b
yields
han
sido reemplazados
modelos
de estos
dispositivos.
análisis
deneed
la figura
como
resultado
replaced
by models ofpor
these
devices.
Analysis
of FigureEl2.8-2b
yields
First,First,
we
we
need
p1(t),pda
(t),
we
so first
we
first
calculate
calculate
1so
Z
Z
replaced by models of these devices. Analysis of RFigure
2.8-2b
yields
I
R
I
p
p energy
60
� �t/60
� 60�t/60
�� ��t/60
� �t/60
� �v1 vand
energy
w1 the
for the
1
u R IaThe
vm ¼ Rp Ia ¼ RvpmI ¼
�3 �3
u w1p for
¼ The
ðtÞ iv¼1 ¼iv
¼
10
A p1Be
Vconstant
¼ Rp Ia ¼ 360 u pRp Ifirst360
vm
w11 De
¼
w1 De
¼ �
p110
ðtp�
Þ1 ðdt
tA
Þ dtBe
1 ðt Þp1¼
(t)pV
1(t)
first
60
s
60
s
cons
u
¼
R
Ia
¼
v
360
�3
�3
�t/30
�t/30
�t/30
�t/30
m
p
0 � 010
JET
JET
VALVE
VALVE
CONTROLLER
¼ DBe
¼ DBe � 10
W ¼
WDBe
¼ DBe mWmW
360CONTROLLER
II
I
I
Rp
R
Rpp
I
Rp
A small,
A small,
experimental
experimental
rocket
rocket
usesuses
a twoa two(1 ––space
a)Rpspace
Voltímetro
ontothe
on the
Plan
Plan Wire Wire
(1
a)R
(1 – a)Rp ActAct
Voltmeter
Voltmeter
p in Figure
element
element
circuit,
circuit,
as
shown
as
shown
in
Figure
1.8-1,
1.8-1,
to
(1
–
a)R
Jet value
Jet value
Voltmeter
p
i first
+ vvm ––Voltmeter v
+ calculate
+ controller
First,
First,
we we
needneed
p1(t),
p+1(t),
so+ we
soi we
first
calculate
(1 – a)R+p
+
controller
m
a+ jetamvalve
jet –valve
fromfrom
point
point
of liftoff
at t ¼
at 0t ¼+ 0
– control
+ vm control
+of liftoff
� ��t/60�t/60 �3 �3�� ���t/60�t/60�
+ one one
–
vexpiration
+ until
m expiration
until
of the
of the
rocket
rocket
afterafter
minute.
p1 ðtpÞ1 ð¼tÞ Element
iv
¼1 Element
¼
iv1 ¼
De
� 10A A
Be Be V
+ minute.
v2 De
v2 � 10
v1 Element
v1 Element
aR
Ithat
�3 �3
m element
�t/30
�t/30�t/30
1 1
2 DBe
2 �t/30
p supplied
vvby
The
The
energy
energy
that
must
must
be
supplied
be
by
element
1
vm1
aR
I
aR
I
m
¼
DBe
¼
�
10
�
10
W
¼
W
DBe
¼
DBe
mWm
p
p
Rp
vm
aRp
I
for the
for one-minute
the one-minute
period
period
40aRmJ.
Element
Element
1 is a
v1m is a
I is 40ismJ.
p
–
–
–
–
battery
battery
to be
to selected.
be selected.
––
–
–
Wire Wire
mA mA
for t–for
� 0,
t � 0,
It isItknown
is known
that that
i(t) ¼
i(t)De
¼�t/60
De�t/60
FIGURA 2.8-2 (a) Circuito que contiene un
FIGURE 2.8-2 (a)
A circuit2.8-2
containing
a
FIGURE
(a) A circuit
containing a
(t)v¼
¼(b)
and and
the voltage
the voltage
across
across
the second
the second
element
element
is v2is
FIGURE
2.8-2
(a)
A circuit
containing
a circuit
2(t)
FIGURE
FIGURE
1.8-1
1.8-1
The circuit
The
to control
to control
potenciómetro.
Circuito
equivalente
que
contiene
potentiometer.
(b)potentiometer.
An
equivalent
circuit
containing
(b)
An containing
equivalent
�t/60�t/60
FIGURE
2.8-2
(a)
A
circuit
acircuit containing
(a)
(b)
� 0.
t �The
0. The
maximum
maximum
magnitude
magnitude
of
the
of
the
Be (a)
Be V forVtfor
potentiometer.
(b)
An
equivalent
circuit
containing
a
jet
a
valve
jet
valve
for
a
for
space
a
space
rocket.
rocket.
un
modelo
del
potenciómetro.
(a)
(b)
a(b)
model ofpotentiometer.
the potentiometer.
a model(b)
of An
the equivalent
potentiometer.
circuit containing
(a)
a model the
of the
current,
D, is
D, limited
is limited
to(b)
1to mA.
1 mA.
Determine
Determine
thepotentiometer.
(a)current,
(b)
a model of the potentiometer.
required
required
constants
constants
D and
D and
B and
B and
describe
describe
the required
the required
battery.
battery.
Solving for the
angle gives
Solving
for
the angle gives
Despejando
el ángulo
obtiene
Solving for the
angle se
gives
Solving for the angle
gives
Describe
Describe
the the
Situation
Situation
and
the the
Assumptions
Assumptions
360and
360
u ¼ 360 vm
vm
u¼
360
v
uthe
¼ plus
1. 1.
TheThe
current
current
enters
enters
the
plus
terminal
terminal
of
the second
element.
element.
I
R
m
Iof second
Rthe
vm p
Rpp I u ¼
I
R
�
p an
� of
Suppose Rp ¼R10
kV
1 mA.
An
angle
of
163An
cause
output
of first
velement.
¼element.
4.53 V.
reading
2.
The
leaves
the
plus
the163°
plus
terminal
terminal
the
of first
the
Suppose
RpIy¼2.
10
and
Icurrent
¼Un
1leaves
mA.
angle
of podría
163
would
cause
output
ofAvvmmeter
¼ 4.53
4.53
V.
A of
meter reading of
Suponga
10and
kV
que
IkV
5 The
1current
mA.
ángulo
de
ocasionar
salida
de
V. Una
man
� would
m5
p5
Suppose Rque
an output
of vuna
p ¼ 10 kV and I ¼ 1 mA. An
m ¼ 4.53 V. A meter reading of
� angle of 163� would cause
�
7.83 V de
would
indicate
that
u indicaría
¼ 282
Suppose
Rpde
¼7.83
10
kV
and
I ¼� .1that
mA.u An
angle. of 163 would cause an output of vm ¼ 4.53 V. A meter reading of
7.83
V
would
indicate
¼ 282°.
282
lectura
medidor
V
que
5
. wires
7.83 V would indicate that u3.¼3.
282The
The
wires
are perfect
are perfect
and and
havehave
no effect
no effect
on the
on circuit
the circuit
(they
(they
do not
do absorb
not absorb
energy).
energy).
7.83 V would indicate that u ¼ 282� .
4. 4.
TheThe
model
model
of the
of circuit,
the circuit,
as shown
as shown
in Figure
in Figure
1.8-1,
1.8-1,
assumes
assumes
that that
the voltage
the voltage
across
across
the the
two
elements
elements
is equal;
is equal;
that
that
is, vis,
vv12¼
. v 2.
Los sensorestwo
de temperatura,
como
el AD590
manufacturado
por Analog Devices, son fuentes
1¼
Temperature
sensors,
such sensors,
asproporcional
the AD590
manufactured
byabsoluta.
AnalogLa
Devices,
are current
de corriente
que tienen
una corriente
la temperatura
figura
2.8-3a
mues- are current
Temperature
such con
as�t/60
the
AD590 manufactured
by Analog
Devices,
Temperature
sensors,
such
as the
Analog
Devices,
v1 1is¼vBe
Be�t/60
V where
V where
Bby
is Bthe
is initial
the initial
voltage
voltage
ofare
the
ofcurrent
battery
the battery
that that
5. 5.
The
The
battery
battery
voltage
voltage
v1 isvAD590
1 ¼manufactured
sources
havingTemperature
current
proportional
absolute
temperature.
Figure
2.8-3a
the
symbol
used
tra el símbolo
con
el que
se representa
un
sensor
deAD590
temperatura.
La figura
2.8-3b
muestra
elshows
modelo
sensors,
as the
manufactured
byshows
Analog
Devices,
are
sources
having
currenttosuch
proportional
to
absolute
temperature.
Figure
2.8-3a
thecurrent
symbol used
sources having current
proportional
to
absoluteastemperature.
Figure
willwill
discharge
discharge
exponentially
exponentially
itas
supplies
it supplies
energy
energy
to2.8-3a
the
to valve.
theshows
valve.the symbol used
to
the
temperature
Figure
2.8-3b
shows
the circuit
model
ofadecuadamente,
the
temperature
delrepresent
circuito
del
decurrent
temperatura.
Para
que
el
sensor
de
temperatura
funcione
el
sources
having
proportional
to
absolute
temperature.
Figure
2.8-3a
shows
the of
symbol
used
tosensor
represent
thesensor.
temperature
sensor.
Figure
2.8-3b
shows
the
circuit
model
the temperature
to represent the temperature sensor. Figure 2.8-3b shows the circuit model of the temperature
sensor.
For
the
temperature
sensor
to
operate
properly,
theshows
vmodel
mustvoltage
satisfy
the
6.
6.
Thevthe
The
circuit
circuit
operates
operates
from
tsensor
¼Figure
0t ¼
to 0tto
¼
to2.8-3b
60
t ¼s.60
s. branch
voltaje de
debe
satisfacer
lafrom
condición
toderivación
represent
temperature
sensor.
thevoltage
circuit
of
the vtemperature
sensor.
For
the
temperature
operate
properly,
the branch
must satisfy the
sensor. For the temperature sensor to operate properly, the branch voltage v must satisfy the
sensor. For the temperature sensor to operate properly, the branch voltage v must satisfy the
7. 7.
TheThe
current
current
is limited,
is limited,
so Dso�Dv1 �
mA.
1 mA.
4 voltios
30
voltios
Alfaomega
State
State
the the
Goal
Goal
Circuitos Eléctricos - Dorf
Determine
Determine
the energy
the energy
supplied
supplied
by the
by first
the first
element
element
for the
for one-minute
the one-minute
period
period
and and
thenthen
select
select
the the
constants
constants
D and
D and
B. Describe
B. Describe
the the
battery
battery
selected.
selected.
M02_DORF_1571_8ED_SE_020-052.indd 38
4/12/11 5:17 PM
Generate
Generate
a Plan
a Plan
(1 – a)Rp
Voltmeter
+
39
+
condition
I
Switches
–
vm
I
Rp
vm
aRp
i(t)
+
4 volts � v � 30 volts
Interruptores
39
v(t)
–
AD590
When this condition is satisfied, the current, i, in microamps, is numerically equal to
Cuando
se satisface
condición,
corriente,
en microamperios,
es numéricamente
igualaa la –
FIGURE
2.8-2
(a) A that
circuit
containing
the
temperature
T, inesta
degrees
Kelvin.laThe
phrasei,numerically
equal indicates
the
current
temperatura
T, en have
grados
La frase
igual
indica
que la
y la temperatupotentiometer.
(b)corriente
An
equivalent
circuit containing+
and
temperature
theKelvin.
same value
butnuméricamente
different units.
This
relationship
can
be expressed
(a)
(b)Esta relación
a model
of theexpresar
potentiometer.
ra
tienen
el
mismo
valor
pero
diferentes
unidades.
se
puede
como
as
v(t)
(a)
Solving for the angle gives
i ¼i 5
k �kT T
mA
360
a constant
associated
the
sensor.
where
constante
asociadawith
conu el
donde k ¼ 1 � , una
vm
¼sensor.
K
Rp I
i(t)
AD590
–
+
(a)
v(t)
Suppose
Rp ¼ 10 kV 2.8-1
and2.8-1
I ¼ 1For
mA.
An
angle
of 163
would
an2.8-2,
output
of vm2.8-2,
¼the
4.53
V. Avoltage,
meter
reading of
EJERCICIO
Para
el circuito
del
potenciómetro
de
la figura
calcule
la
medición
EXERCISE
the
potentiometer
circuit
ofcause
Figure
calculate
meter
�
– +
.
7.83 Vvdewould
indicate
that
u
¼
282
�
voltaje
v
,
cuando
5
45°.
R
5
20
kV,
e
I
5
2
mA.
, when u m¼ 45 , R ¼ 20 kV,p and I ¼ 2 mA.
�
m
i(t) = kT
p
Respuesta:
5V
m5
5V
Answer:
vm v¼
v(t)
i(t) = kT
–
(b)
EJERCICIO
2.8.2 sensors,
El voltaje
y laas
corriente
de un manufactured
sensor de temperatura
AD590
de la figura
Temperature
such
the AD590
by Analog
Devices,
are current
EXERCISE
The
voltage and current
of an AD590
temperature
sensor of Figure 2.8-3 FIGURE 2.8-3
2.8-3 son 10 V 2.8-2
y 280 mA,
respectivamente.
Determine
la temperatura
medida.
sources having current proportional to absolute temperature. Figure 2.8-3a shows the symbol
used(b) and
(a) The symbol
are 10 V and 280 mA, respectively. Determine the measured temperature.
to
represent
the
temperature
sensor.
Figure
2.8-3b
shows
the
circuit
model
of
the
temperature
Respuesta: T 5 280 °K, o aproximadamente 6.85 °C.
(b) a FIGURA
model for 2.8-3
the
Answer:
T ¼ 280
approximately
6.85�to
C operate properly, the branch voltage v must satisfy(a)the
sensor.
For� K,
theortemperature
sensor
Símbolo y
(b) modelo del
potenciómetro.
2.9
2.9
INTERRUPTORES
SWITCHES
Los interruptores
dosstates:
estados
distintos:
abierto
y cerrado.
Idealmente,
interruptor
Switches
have twotienen
distinct
open
and closed.
Ideally,
a switch
acts as aunshort
circuit
actúa
como
un
cortocircuito
cuando
está
cerrado
y
como
circuito
abierto
cuando
está
abierto.
when it is closed and as an open circuit when it is open.
Las figuras
2.9-2show
muestran
varios
de interruptor.
cadathe
caso
se when
indicathe
el tiempo
Figures
2.9-12.9-1
and y2.9-2
several
typestipos
of switches.
In eachEncase,
time
switch
cuando
el
interruptor
cambia
de
estado.
Veamos
primero
los
interruptores
unipolares
de
una
changes state is indicated. Consider first the single-pole, single-throw (SPST) switches shown in acción
Figure
(SPST,The
porswitch
sus siglas
en inglés)
se muestran
la figura
El state,
interruptor
de laclosed,
figura at
2.9-1a
2.9-1.
in Figure
2.9-1aque
is initially
open.en
This
switch2.9-1.
changes
becoming
time
está
abierto
inicialmente.
Este
interruptor
cambia
de
estado,
a
cerrado,
en
el
tiempo
t
5
0
s.
Cuando
t ¼ 0 s. When this switch is modeled as an ideal switch, it is treated like an open circuit when t < 0 s and
este ainterruptor
se when
modela
un interruptor
ideal,
se le state
trata instantaneously.
como un circuitoThe
abierto
cuando
t,
like
short circuit
t >como
0 s. The
ideal switch
changes
switch
in Figure
0
s
y
como
cortocircuito
cuando
t
.
0
s.
El
circuito
ideal
cambia
de
estado
de
manera
instantánea.
El
2.9-1b is initially closed. This switch changes state, becoming open, at time t ¼ 0 s.
interruptor
de
la
figura
2.9-1b
está
cerrado
inicialmente.
Este
interruptor
cambia
de
estado,
a
abierto,
Next, consider the single-pole, double-throw (SPDT) switch shown in Figure 2.9-1a. This SPDT
en el tiempo
t 5two
0 s.SPST switches, one between terminals c and a, another between terminals c and b.
switch
acts like
A
continuación,
veremos
el interruptor
unipolar
doble
acción
(SPDT,
porbsus
siglasAt
en tinglés)
Before t ¼ 0 s, the switch
between
c and a is closed
anddethe
switch
between
c and
is open.
¼ 0 s,
que
se
muestra
en
la
figura
2.9-1a.
Este
interruptor
SPDT
funciona
como
dos
interruptores
SPST,
uno
both switches change state; that is, the switch between a and c opens, and the switch between c and
b
entre
las
terminales
c
y
a,
y
otro
entre
las
terminales
c
y
b.
Antes
de
t
5
0
s,
el
interruptor
que
está
encloses. Once again, the ideal switches are modeled as open circuits when they are open and as short
tre c y awhen
está cerrado
y el
que está entre c y b está abierto. En t 5 0 s, ambos interruptores cambian de
circuits
they are
closed.
estado,Inessome
decir,
el
que
está
a y ca difference
está abierto,whether
y el quethe
está
entrebetween
c y b está
cerrado.
Unabefore,
vez más,
applications, entre
it makes
switch
c and
b closes
or
los
interruptores
ideales
se
modelan
como
circuitos
abiertos
cuando
están
abiertos
y
como
cortocircuitos
after, the switch between c and a opens. Different symbols are used to represent these two types of
cuando están cerrados.
a
En algunas aplicaciones es importante si el interruptor entre cay b cierra antes o después,
y que
c
c
el interruptor entre c y a abra. Se utilizan diferentes símbolos para brepresentar estos dosbtipos de int=0
t=0
t=0
t=0
terruptor unipolares,
de doble acción.
El interruptor abrir antes de cerrar está fabricado de tal manera
Initially open
Initially closed
(a)
(b)
FIGURE 2.9-1t =SPST
switches. (a) Initially
0
t = 0 open and (b)
abierto
Inicialmente cerrado
initiallyInicialmente
closed.
(a)
(b)
FIGURA 2.9-1 Interruptores SPST. (a) Inicialmente
abierto y (b) Inicialmente cerrado.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 39
Break before make
c
(a)
t=0
a
b
Make before break
c (b)
t=0
a
b
FIGURE 2.9-2 SPDT switches. (a) Break before make
Abrir
antes
de cerrar
Cerrar antes de abrir
and (b)
make
before
break.
(a)
(b)
FIGURA 2.9-2 Interruptores SPDT. (a) Abrir antes de
cerrar y (b) Cerrar antes de abrir
Alfaomega
4/12/11 5:17 PM
40
40
Circuit Elements
Elementos de circuitos
single-pole,
switch.
Theel break-before-make
is manufactured
that the
switch
que
entre c y double-throw
b cierre después
de que
interruptor entre c yswitch
a se abra.
En la figura so
2.9-2a
se muesbetween
c
and
b
closes
after
the
switch
between
c
and
a
opens.
The
symbol
for
the
break-before-make
tra el símbolo del interruptor de abrir antes de cerrar. El interruptor de abrir antes de cerrar está
switch is de
shown
in Figure
The make-before-break
switch
so thatentre
the switch
fabricado
tal manera
que2.9-2a.
el interruptor
entre c y b cierre
antesisdemanufactured
que el interruptor
cya
between
c
and
b
closes
before
the
switch
between
c
and
a
opens.
The
symbol
for
the
make-beforeabra. El símbolo del interruptor cerrar antes de abrir se muestra en la figura 2.9-2b. Recuerde que
switchdel
is shown
in Figure
Remember:
switchque
transition
fromdeterminal
to terminal b
labreak
transición
interruptor
de la2.9-2b.
terminal
a a la b sethe
supone
se efectúa
maneraainstantánea.
is
assumed
to
take
place
instantaneously.
This
instantaneous
transition
is
an
accurate
model
when es
the
Esta transición instantánea es un modelo preciso cuando la transición real de abrir antes de cerrar
actual
make-before-break
transition
is
very
fast
compared
to
the
circuit
time
response.
muy rápida comparada con la respuesta del tiempo del circuito.
E j e m p l o 2 . 9 - 1 Interruptores
E X A M P L E 2 . 9 - 1 Switches
La figura 2.9-3 ilustra el uso de circuitos abiertos y cortos para modelar interruptores ideales. En la figura 2.9-3a
2.9-3
thetres
useinterruptores.
of open and short
for modeling
idealelswitches.
In Figure
2.9-3a,sido
a circuit
seFigure
muestra
un illustrates
circuito con
En lacircuits
figura 2.9-3b
se muestra
circuito como
si hubiera
mocontaining
shown.
In Figure
2.9-3b, the
is shown
as itde
would
been
modeled
before
0 s.
delado
antesthree
de t 5switches
0 s. Losisdos
interruptores
unipolares
de circuit
una acción
cambian
estado
el tiempo
t 5 0t ¼
s. La
The two
single-pole,
single-throw
switches
change
at time cuando
t ¼ 0 s. Figure
2.9-3c
it would
figura
2.9-3c
muestra el
circuito como
si hubiera
sidostate
modelado
el tiempo
estáshows
entre 0the
s ycircuit
2 s. Elasinterrupbeunipolar
modeledde
when
theacción
time iscambia
between
s and 2en
s. el
The
single-pole,
switch
changes
at time
t ¼si2
tor
doble
de0estado
tiempo
t 5 2s.double-throw
La figura 2.9.-3d
muestra
el state
circuito
como
s. Figure
2.9-3d
showsdespués
the circuit
hubiera
sido
modelado
de 2 as
s. it would be modeled after 2 s.
5 kΩ
5 kΩ
t =t =
22
s s
kΩ
4 4kΩ
1212
kΩkΩ
kΩ
55kΩ
t =t =0 0s s
+
1212
V V +– –
++
12VV ++–
6 6V V – – 12
–
88
kΩkΩ
5 kΩ
5 kΩ
66 VV ––
(c)
(c)
kΩ
55kΩ
kΩ
4 4kΩ
10kΩ
kΩ
10
1212
kΩkΩ
++
kΩ
88kΩ
(a)
(a)
+
1212
V V +– –
10 kΩ
kΩ
10
12kΩ
kΩ
12
10kΩ
kΩ
10
= s0 s
t =t 0
kΩ
44 kΩ
12kΩ
kΩ
12
+
12VV ++–
6 6V V +– – 12
–
88
kΩkΩ
kΩ
44 kΩ
(b)
(b)
10 kΩ
kΩ
10
kΩ
88kΩ
66 VV
(d)
(d)
+
+–
–
FIGURE2.9-3
2.9-3
FIGURA
(a)Un
A circuit
(a)
circuito
containing
con
varios
several switches.
interruptores.
(b)ElThe
(b)
circuito
equivalent circuit
equivalente
for tt � 00 s.
s.
para
(c)ElThe
(c)
circuito
equivalent circuit
equivalente
para
0 for
, t0,
s. 2 s.
< 2t <
(d)
circuito
(d El
) The
equivalente
para
equivalent circuit
t.
for2t s.
> 2 s.
EJERCICIO 2.9-1 ¿Cuál es el valor de la corriente i en la figura E 2.9-a en el tiempo t 5 4 s?
EXERCISE
is the
value ofinterruptores
the current iestán
in Figure
E 2.9-1 at time t ¼ 4 s?
Respuesta:
i 5 02.9-1
amperiosWhat
en t 5
4 s (ambos
abiertos).
Answer: i ¼ 0 amperes at t ¼ 4 s (both switches are open).
EJERCICIO 2.9-2 ¿Cuál es el valor del voltaje v en la figura E 2.9-2 en el tiempo t 5 4 s?
¿En
t 5 6 s? 2.9-2 What is the value of the voltage v in Figure E 2.9-2 at time t ¼ 4 s? At t ¼ 6 s?
EXERCISE
Respuesta: v 5 6 voltios en t 5 4 s, y v 5 0 voltios en t 5 6 s.
Answer: v ¼ 6 volts at t ¼ 4 s, and v ¼ 0 volts at t ¼ 6 s.
t t==55ss
t t==33ss
+
33kΩ
kΩ
+
12
12VV ––
66VV +–+
–
ii
FIGURA E 2.9-1 Circuito con dos interruptores SPST.
FIGURE E 2.9-1 A circuit with two SPST switches.
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 40
t=5s
t=5s
2 mA
2 mA
+
+
v
3 kΩ
v
3 kΩ
–
–
i
i
FIGURA E. 2-9-2 Circuito con un interruptor SPDT cerrar
FIGURE
antes
de abrir. E 2.9-2 A circuit with a make-before-break
SPDT switch.
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
¿Cómo lo podemos comprobar . . . ?
2.10
41
¿ C Ó M O LO P O D E M O S C O M P R O B A R . . . ?
Frecuentemente a los ingenieros se les solicita comprobar que la solución de un problema sea la correcta. Por ejemplo, las soluciones propuestas para problemas de diseño se deben comprobar para confirmar
que se ha cumplido con todas las especificaciones. Además, se deben revisar los resultados de la computadora para protegerse contra errores de captura de datos, así como las exigencias de los comerciantes,
las cuales se deben analizar a fondo.
También a los estudiantes de ingeniería se les pide que verifiquen la exactitud de sus trabajos.
Por ejemplo, tomarse un breve lapso antes de terminar un examen permitiría dar una vista rápida e
identificar esas soluciones que podrían requerir un poco más de aplicación.
El ejemplo siguiente ilustra técnicas útiles para comprobar las soluciones a los diversos problemas analizados en este capítulo.
E j e m p l o 2 . 10 - 1
¿Cómo comprobar los valores de voltajes y corrientes?
Los medidores del circuito de la figura 2.10-1 indican que v1 5 24 V, v2 5 8 V y que i 5 1 A. ¿Cómo podemos
comprobar que los valores de v1, v2 e i se han medido correctamente? Verifiquemos los valores de v1, v2 e i de
dos maneras:
(a) Verifique que los valores dados para ambos resistores cumplan con la ley de Ohm.
(b)Verifique que la potencia alimentada por la fuente de voltaje sea igual a la potencia absorbida por los resistores.
– 4 . 0
Voltímetro
1 . 0 0
Amperímetro
–
v1
4Ω
12 V
+
–
+
8 . 0 0
i
+
8Ω
Voltímetro
v2
–
FIGURA 2.10-1 Circuito con medidores.
Solución
(a)Considere el resistor de 8 V. La corriente i fluye a través de este resistor de arriba hacia abajo. Por lo tanto,
la corriente i y el voltaje v2 se apegan a la convención pasiva. Además, la ley de Ohm requiere que v2 5 8i.
Los valores v2 5 8 V e i 5 1 A satisfacen esta ecuación.
A continuación, considere el resistor 4 V. La corriente i fluye de izquierda a derecha a través de este
resistor. Por lo tanto, la corriente i y el voltaje v1 se apegan a la convención pasiva. Además, la ley de Ohm
requiere que v1 5 4(2i). Los valores v1 5 24 V e i 5 1 A satisfacen esta ecuación.
Por lo tanto, se satisface la ley de Ohm.
(b)La corriente i fluye de arriba hacia abajo a través de la fuente de voltaje. Por lo tanto, la corriente i y el
voltaje de 12 V no se apegan a la convención pasiva. En consecuencia, 12i 5 12(1) 5 12 W es la potencia
alimentada por la fuente de voltaje. La potencia absorbida por el resistor de 4 V es 4i2 5 4(12) 5 4 W, y
la potencia absorbida por el resistor de 8 V es 8i2 5 8(12) 5 8 W. La potencia alimentada por la fuente de
voltaje es en realidad igual a la potencia absorbida por los resistores.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 41
Alfaomega
4/12/11 5:17 PM
42
Elementos de circuitos
2 . 11
E J E M P LO D E D I S E Ñ O
SENSOR DE TEMPERATURA
Las corrientes se pueden medir fácilmente utilizando amperímetros. Un sensor de temperatura, como el AD590 de Analog Devices, se puede usar para medir la temperatura al convertir
la temperatura en corriente. La figura 2.11-1 muestra un símbolo para representar un sensor
de temperatura. Para que este sensor funcione adecuadamente, el voltaje v debe cumplir satisfactoriamente la condición
4 voltios v 30 voltios
i(t)
+
v(t)
AD590
–
FIGURA 2.11-1
Un sensor de temperatura.
Cumplida esta condición, la corriente i, en mA, es numéricamente igual a la temperatura
T, en °K. La frase numéricamente igual indica que las dos variables tienen el mismo valor
pero unidades diferentes.
mA
i ¼ k T donde k ¼ 1 K
El objetivo es diseñar un circuito utilizando el AD590 para medir la temperatura de un
depósito de agua. Además del AD590 y un amperímetro, hay varios proveedores de energía
disponibles y algunos resistores estándar de 2%. Los proveedores de energía son fuentes de
voltaje. También se cuenta con alimentadores de energía con voltajes de 10, 12, 15, 18 o
24 voltios.
Describa la situación y los supuestos
Para que el transductor de temperatura funcione adecuadamente, su voltaje del elemento debe
estar entre 4 y 30 voltios. Para establecer este voltaje se utilizarán los alimentadores de energía y
los resistores. Se usará un amperímetro para medir la corriente en el transductor de temperatura.
El circuito debe poder medir temperaturas en el rango de los 0 °C a 100 °C porque a
estas temperaturas el estado del agua es líquido. Recuerde que la temperatura en °C equivale
a la temperatura en °K menos 273°.
Establezca el objetivo
Utilice los alimentadores de energía y los resistores para hacer que el voltaje, v, del transductor de temperatura esté entre los 4 y los 30 voltios.
Utilice un amperímetro para medir la corriente, i, en el transductor de temperatura.
Genere un plan
Modele el alimentador de energía como una fuente de voltaje ideal y el transductor de temperatura como una fuente de corriente ideal. El circuito que se muestra en la figura 2.11-2a
hace que el voltaje a través del transductor de temperatura sea igual al voltaje del alimento
de energía. Dado que todos los alimentadores de energía disponibles tienen voltajes entre 4 y
30 voltios, se puede usar cualquiera de ellos. Observe que no se necesitan los resistores.
En la figura 2.11-2b se ha agregado un cortocircuito de manera que no altere la red. En
la figura 2.11-2c, este cortocircuito ha sido reemplazado con un amperímetro (ideal). Como el
amperímetro medirá la corriente en el transductor de temperatura, la lectura del amperímetro
será numéricamente igual a la temperatura en °K.
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 42
Circuitos Eléctricos - Dorf
5/24/11 9:49 AM
E1C02_1
E1C02_1
E1C02_1 10/23/2009
10/23/2009
10/23/2009
E1C02_1
10/23/2009
43
43
43
43
Ejemplo de diseño
Design
Design
Design
Example
Example
Example
Design
Example
+
–
++++
––––
+
+
+
Amperímetro
+v(t)
+++
+v(t)
+++
+v(t)
+++
Ammeter
Ammeter
Ammeter
Ammeter
+
–
++++
––––
v(t)
v(t)
v(t)
v(t)
–
i(t)
––––
i(t)
i(t)
i(t)
i(t)
v(t)
v(t)
v(t)
v(t)
–
––Corto––
circuito
Short
Short
Short
Short
i(t)
+
–
++++
––––
v(t)
v(t)
v(t)
v(t)
–
––––
i(t)
i(t)
i(t)
i(t)
i(t)
i(t)
i(t)
i(t)
i(t)
circuit
circuit
circuit
circuit
(a)
43
43
43
43
43
(b)
(c)
FIGURA 2.11-2
con un sensor de temperatura.
(b) Agregando un
(a)
(a)
(a) (a) Medición de la temperatura
(b)
(b)
(b)
(c)
(c)
(c)
(a)
(b)
(c)
cortocircuito. (c) El cortocircuito ha sido reemplazado por un amperímetro.
FIGURE
FIGURE
FIGURE
FIGURE2.11-2
2.11-2
2.11-2
2.11-2(a)
(a)
(a)
(a)Measuring
Measuring
Measuring
Measuringtemperature
temperature
temperature
temperaturewith
with
with
withaaatemperature
atemperature
temperature
temperaturesensor.
sensor.
sensor.
sensor.(b)
(b)
(b)
(b)Adding
Adding
Adding
Addingaaashort
ashort
short
shortcircuit.
circuit.
circuit.
circuit.(c)
(c)
(c)
(c)
Aun
cuando
cualquiera
de los alimentadores de energía es adecuado para satisfacer las
Replacing
Replacing
Replacing
the
the
the
short
short
short
circuit
circuit
circuit
by
by
by
an
an
an
ammeter.
ammeter.
ammeter.
Replacing
the
short
circuit
by
an
ammeter.
especificaciones, una ventaja podría ser la elección de un alimentador de energía en particular.
Although
Although
Although
any
any
any
of
of
of
the
the
the
available
available
available
power
power
power
supplies
supplies
supplies
isisisis
adequate
adequate
adequate
to
to
to
meet
meet
meet
the
the
the
specifications,
specifications,
specifications,
there
there
there
Although
any
of
the
available
power
supplies
adequate
to
meet
the
specifications,
there
Por ejemplo, es razonable seleccionar el alimentador de energía que haga que el transductor
may
may
may
still
still
still
be
be
be
an
an
an
advantage
advantage
advantage
to
to
to
choosing
choosing
choosing
aaaparticular
aparticular
particular
power
power
power
supply.
supply.
supply.
For
For
For
example,
example,
example,
itititit
isisisis
reasonable
reasonable
reasonable
may
still
be
an
advantage
to
choosing
particular
power
supply.
For
example,
reasonable
absorba la menor cantidad posible de potencia.
to
to
tochoose
choose
choosethe
the
thepower
power
powersupply
supply
supplythat
that
thatcauses
causes
causesthe
the
thetransducer
transducer
transducerto
to
toabsorb
absorb
absorbas
as
aslittle
little
littlepower
power
poweras
as
aspossible.
possible.
possible.
to
choose
the
power
supply
that
causes
the
transducer
to
absorb
as
little
power
as
possible.
Actúe sobre el plan
Act
Act
Act
Acton
on
on
onthe
the
the
thePlan
Plan
Plan
Plan
La potencia absorbida por el transductor es
The
The
The
Thepower
power
power
powerabsorbed
absorbed
absorbed
absorbedby
by
by
bythe
the
the
thetransducer
transducer
transducer
transducerisisisis
p5vi
ppp¼
p¼
¼
vvv�v�i�Elegir
i�ii
¼
donde v es el voltaje del alimentador de energía.
un v lo más pequeño posible, en este
where
where
where
vvvis
v10
isisisthe
the
thepower
power
power
supply
supply
supply
voltage.
Choosing
Choosing
Choosing
vvvas
vas
aspor
small
small
small
as
aspossible,
possible,
possible,10
10
10volts
volts
voltsin
in
inthis
this
thiscase,
case,
case,
where
the
power
supply
voltage.
Choosing
as
small
as
possible,
10
volts
in
this
case,
caso
de
voltios,
hace
quevoltage.
lavoltage.
potencia
absorbida
el as
transductor
de
temperatura
sea
lo
makes
makes
makes
the
the
the
power
power
power
absorbed
absorbed
absorbed
by
by
the
the
the
temperature
temperature
temperature
transducer
transducer
transducer
as
as
as
small
small
small
as
as
as
possible.
possible.
possible.
Figure
Figure
Figure
2.11-3a
2.11-3a
2.11-3a
makes
the
power
absorbed
by
the
temperature
transducer
as
small
as
possible.
Figure
2.11-3a
más
pequeña
posible.
La by
figura
2.11-3a
muestra
el diseño
final.
La
figura 2.11-3b
muestra
shows
shows
shows
the
the
the
final
final
final
design.
design.
Figure
Figure
Figure
2.11-3b
2.11-3b
2.11-3b
shows
shows
aaagraph
agraph
graph
that
that
that
can
can
can
be
be
be
used
used
used
to
to
to
find
find
find
the
the
the
temperature
temperature
temperature
shows
the
final
design.
Figure
2.11-3b
shows
graph
that
can
be
used
to
find
the
temperature
una
gráfica
quedesign.
se
puede
utilizar
parashows
encontrar
la temperatura
que
corresponda
a cualquier
corresponding
corresponding
corresponding
to
to
toany
any
anyammeter
ammeter
ammetercurrent.
current.
current.
corresponding
to
any
ammeter
current.
corriente
del amperímetro.
Verify
Verify
Verifythe
the
the
Proposed
Proposed
Proposed
Solution
Solution
Verify
the
Proposed
Solution
Verifique
la
soluciónSolution
propuesta
����
Let’s
Let’s
Let’stry
try
tryan
an
anexample.
example.
example.
Suppose
Suppose
Suppose
the
the
thetemperature
temperature
temperature
of
of
ofthe
the
the
water
water
waterisisisis80.6
80.6
80.6
F.
F.
F.
This
This
This
temperature
temperature
temperature
isisisis
Let’s
try
an
example.
Suppose
the
temperature
of
the
water
80.6
F.
This
temperature
Hagamos
la
prueba
con
un ejemplo.
Suponga que
la
temperatura
del
agua
es
80.6 °F. Esta
����
����
or
or300
300
300
K.
K.The
The
The
current
current
in
inthe
the
the
temperature
temperature
temperature
sensor
sensorwill
will
will
be
be
be
equal
equal
equalto
to
to27
27
27CC
CCor
or
300
K.
The
current
in
the
temperature
sensor
will
be
equal
to
27
temperatura
es
igual
aK.
27
°C
ocurrent
300
°K.in
La
corriente
en el sensor
sensor
de
temperatura
será
mA
mA
mA ����
mA
iii¼
i¼
¼ 1111���� 300
300
300KKKK¼
¼
¼
300
300
300
mA
mA
mA
¼
300
¼
300
mA
KKKK
Next,
Next,
Next,
suppose
suppose
suppose
that
that
that
the
the
the
ammeter
ammeter
ammeter
in
in
Figure
Figure
Figure
2.11-3a
2.11-3a
2.11-3a
reads
reads
300
300
300
mA.
mA.
mA.
AAAA
sensor
sensor
sensor
current
current
current
of
of
of
300
300
300
A
continuación,
suponga
que elin
amperímetro
dereads
la
figura
2.11-3a
lee
300
mA. Una
coNext,
suppose
that
the
ammeter
in
Figure
2.11-3a
reads
300
mA.
sensor
current
of
300
mA
mA
mAcorresponds
corresponds
corresponds
to
to
toade
aatemperature
atemperature
temperature
of
of
of
rriente
en el sensor
300
mA corresponde
a una temperatura de
mA
corresponds
to
temperature
of
300
mA
300
300
mA
mA
300
mA
����
����
����
¼¼
300
300
¼
¼
TTTT¼
¼
300
KKK¼
¼¼
27
2727
CCC
¼
¼¼
80:6
80:6
80:6
FFF
¼
mA
mA ¼ 300 K ¼ 27 C ¼ 80:6 F
mA
mA
1111����
KKKK
La
gráfica
de
la
figura
2.11-3b
indica que
la
de unof
sensor
de
300
mA
corresponde
The
The
graph
graph
in
Figure
2.11-3b
2.11-3b
indicates
indicates
that
that
aaasensor
sensor
current
of
300
mA
does
does
correspond
correspond
to
to
The
graphin
inFigure
Figure
2.11-3b
indicates
that
acorriente
sensorcurrent
current
of300
300mA
mA
does
correspond
toaaaaa
The
graph
in
Figure
2.11-3b
indicates
that
sensor
current
of
300
mA
does
correspond
to
����27 °C.
una
temperatura
de
C.
C.
C.
temperature
temperature
of
of
27
temperature
of27
27
C.
temperature
of
27
Este
ejemplo
que
el
circuito
funcionando
de manera adecuada.
This
This
Thisexample
example
examplemuestra
shows
shows
showsthat
that
thatthe
the
the
circuit
circuit
circuitisestá
isisisworking
working
working
properly.
properly.
properly.
This
example
shows
that
the
circuit
working
properly.
Amperímetro
Ammeter
Ammeter
Ammeter
Ammeter
10 V ++–+++
10
10
10
VVVV ––––
10
i(t)
i(t)
i(t)
i(t)
i(t)
(a)
Temperatura, °C
Temperature,
Temperature,
Temperature,
°C
°C
°C
Temperature,
°C
100
100
100
100
100
0
0000
273
373
273
273
273
273
373
373
373
373
µA
Lectura del amperímetro,
Ammeter
Ammeter
Ammeter
reading,
reading,
reading,
μμA
μAA
Ammeter
reading,
μA
(b)
(a)
(a)de un circuito que mide la temperatura con un sensor
(b)
(b)
(b) de temperatura.
(a)
(b)
FIGURA 2.11-3 (a) Diseño(a)
final
(b) Gráfica de la temperatura comparada con la corriente del amperímetro.
FIGURE
FIGURE
FIGURE2.11-3
2.11-3
2.11-3(a)
(a)
(a)
Final
Final
Final
design
design
design
of
of
of
aaacircuit
acircuit
circuit
that
that
that
measures
measures
measures
temperature
temperature
temperature
with
with
with
aaatemperature
atemperature
temperature
sensor.
sensor.
sensor.
(b)
(b)
(b)
FIGURE
2.11-3
(a)
Final
design
of
circuit
that
measures
temperature
with
temperature
sensor.
(b)
Graph
Graph
Graph
Graph
of
of
of
of
temperature
temperature
temperature
temperature
versus
versus
versus
versus
ammeter
ammeter
ammeter
ammeter
current.
current.
current.
current.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 43
Alfaomega
4/12/11 5:17 PM
Circuit Elements
44
Circuit Elements
Elementos de circuitos
44
44
2.12
2.12 S U M M A R Y
A dependent
source
provides
a current
(or a voltage) that is
2.12usesSmodels,
UMM
A Rcircuit
Y elements,
engineer
called
tofuente
repre-dependiente
Una
proporciona
una
corriente
(o un volR E S U MThe
EN
dependent
provides
a current
(or aThe
voltage) that is
dependent
onAanother
variable
elsewhere
the circuit.
The engineer
uses
called
circuit
elements,
to repretaje)
quewe
es dependiente
de otra
variable
ensource
cualquier
pun- in
the devices
thatelementos
make
upmodels,
a circuit.
In this
book,
El ingeniero utilizasent
modelos,
llamados
de
circuidependent
on
another
variable
elsewhere
in
the
circuit. The
constitutive
equations
of
dependent
sources
are
summarized
todevices.
del circuito.
En la we
tabla 2.7-1 se resumen las ecuaciones
theintegra
devices
that
make
circuit.
In A
this book,
onlysent
linear
elements
linear
models
to, para representarconsider
el dispositivo
que
un or
circuito.
Enup a of
constitutive
equations
of
dependent
sources
are
summarized
in
Table
2.7-1.
constitutivas
de
fuentes
dependientes.
consider
only
linear
elements
models of devices. A
device is linear
if it satisfies
theo properties
oflinear
both superpoeste libro sólo se consideran
elementos
lineales
modelos or
in
Tableand
2.7-1.
short circuit
open
are special cases of
El cortocircuito
yThe
el circuito
abierto
son
casos circuit
especiales
device is es
linear
if si
it satisfies
of both superposition Un
anddispositivo
homogeneity.
lineales de dispositivos.
lineal
satisfacethe properties
The short
circuit
and
circuit
aresource
special cases of
independent
sources.
A short
circuit
isopen
andeideal
voltage
de fuentes
Un cortocircuito
es una
fuente
sitionybetween
and
homogeneity.
relationship
the reference directions
of theindependientes.
las propiedades deThe
la superposición
la homogeneidad.
independent
sources.
A
short
circuit
is
an
ideal
voltage source
having
v(t)
¼
0.
The
current
in
a
short
circuit
is
determined
by
voltajedirections
ideal
relationship
between
of thev(t) 5 0. La corriente en un cortocircurrent
and The
voltage
of a circuit
elementtheis reference
important.
The que tiene
La relación entre las
direcciones
de referencia
de la
corriente
having
v(t)
¼
0.
The
current
in
a
short
circuit
is
determined by
the
rest
of
the
circuit.
An
open
circuit
is
an
ideal
current
source
cuito
está
determinada
por
el
resto
del
circuito.
Un
circuito
current
and
voltage
of
a
circuit
element
is
important.
The
polarity
marks one
terminal þ La
and the other �. The
y el voltaje de unvoltage
elemento
de circuito
es importante.
rest
of the
An
open
circuit
an ideal current source
¼ the
0. The
voltage
across
is is
determined
abierto
es other
una fuente
corriente
ideal
quecircuit.
tiene an
i(t)open
5 0.circuit
voltage
marks
terminal
þ and
the
�.having
Thede i(t)
element
andpolarity
current
adhere
to El
the
passive
convenpolaridad del voltaje
marcavoltage
una
terminal
1 y la
otra one
2.
having
i(t)
¼
0.
The
voltage
across
an opencan
circuit
byun
thecircuito
rest of the
circuit.
Open
circuits
andpor
short circuits
alsois determined
Eltovoltaje
a través
de
abierto
está
determinado
element
voltage
and
current
adhere
theþpassive
convention
the current
directed
the terminal
marked
to
voltaje y la corriente
delifelemento
se is
apegan
a lafrom
convención
by
the
rest
of
the
circuit.
Open
circuits
and
short
circuits can also
be
described
as
special
cases
of
resistors.
A
resistor
with
el
resto
del
circuito.
Los
circuitos
abiertos
y
los
cortocircuitos
tion
the�.current
is directed
these
terminal
marked
pasiva si la corriente
dirige de
laifterminal
marcada
comofrom the terminal marked þ to
be
described
as
special
cases
of
resistors.
A
resistor with
resistance
R
¼
0
(G
¼
1)
is
a
short
circuit.
A
resistor
with
también
se
pueden
considerar
casos
especiales
de
resistores.
the
terminal
marked
�.
Resistors
are
widely
used
as
circuit
elements.
When
the
+ a la terminal marcada como 2.
resistance
0) es
(G
¼ open
1) iscircuit.
a short circuit. A resistor with
conductance
G
(R R¼
is an
Un elements.
resistor con
una
resistencia
R¼
5 00 (G
5¼1)
un
cortoResistors
are (owidely
used
as passive
circuit
When
the
Con gran frecuencia
se utilizan
resistores
resistencias)
resistor
voltage
and current
adhere
to the
convenconductance
G5
¼the
(R
¼
circuit.
Ancon
ideal
measures
current
through
its
Un resistor
unaammeter
conductancia
G
00 (R
5 1)flowing
) is an open
como elementos de
circuito.
Cuando
voltajeand
y la
resisresistor
voltage
current
adherecircuito.
to the passive
convention,
resistors
obey el
Ohm’s
law;
the
voltage
across
the
An
ideal
ammeter
measures
the
current
flowing
through its
terminals
and
has
zero
voltage
across
its
terminals.
An
ideal
es
un
circuito
abierto.
tencia del resistorterminals
se apegan
athe
la resistor
convención
pasiva,
resistorsis obey
Ohm’s
the into
voltage
oftion,
related
to los
thelaw;
current
the across the
has
zero
across itsand
terminals.
An ideal
voltmeter
theand
voltage
across
its terminals
has
Un
amperímetro
ideal
laterminals
corriente
que
fluye
a voltage
través
resistores obedecen
a la ley
de
Ohm;
a través
terminals
of¼voltaje
theRi.resistor
is related
to the
into
themide measures
positive
terminal
as vel
The
power
delivered
tocurrent
a
voltmeter
measures
voltage
its terminals
and has
current
equalato
zero. Ideal
act like
open
sus terminales
ytotiene
voltaje
través
dethe
susvoltmeters
ter- across
de las terminales del
resistor se
la corriente
as v ¼ enRi. Thedepower
deliveredterminal
a cero
resistance
is positive
prelaciona
¼ i2R terminal
¼con
v2=R
watts.
2
2
terminal
current
equal
to
zero.
Ideal
voltmeters
act
like
open
circuits,
and
ideal
ammeters
act
like
short
circuits.
minales.
Un
voltímetro
ideal
mide
el
voltaje
a
través
de
la terminal positivaAn
como
v 5 resistance
Ri. Lasource
potencia
a watts.
is pprovides
¼transmitida
i R ¼a vcurrent
=R
independent
or a voltage
circuits,
and
ideal
ammeters
act
like
short
circuits.
Transducers
are
devices
that
convert
physical
quantities,
sus
terminales
y
tiene
una
corriente
terminal
igual
a
cero.
>R
watts.
una resistencia es pindependent
5 i2R 5 v2An
independent
source provides
a current
of other
circuit variables.
The voltage
of anor a voltage
Transducers
areto devices
that quantity
convert physical
such
rotational
position,
anabiertos,
electrical
such as quantities,
voltímetros
como
circuitos
Una fuente independiente
proporciona
unaofcorriente
o un but
independent
other
circuit
variables.
The is
voltageideales
of anasfuncionan
independent
voltage source
is
specified,
theLos
current
such
as rotational
position,
an electrical
quantity such as
y loscurrent
amperímetros
ideales
cortocircuitos.
voltage.
this
chapter,
we describe
two to
transducers:
potenvoltaje independiente
otras variables
circuito.
voltaindependent
voltage
is specified,
but the current
is Incomo
not.deConversely,
thedecurrent
ofElsource
an
independent
Los
transductores
son
dispositivos
que
convierten
cantidavoltage.
In
this
chapter,
we
describe
two
transducers:
potentiometers
and
temperature
sensors.
je de una fuente desource
voltajeisindependiente
es específico,
pero
not. Conversely,
current
of The
an voltages
independent current
specified
whereas
thethe
voltage
is not.
des
físicas,
como
la
posición
de
rotación,
en
cantidad
eléctritiometers
and
temperature
sensors.
Switches
are
widely
used
in
circuits
to
connect
and
disla corriente no lo of
es.independent
Por el contrario,
la
corriente
de
una
source
is specified
the voltage
is not. The voltages
voltage
sources whereas
and currents
of independent
unindependent
voltaje.
En esteelements
capítulo
se describen
dos
transSwitches
are
widely
used
in circuits
connect
and disconnect
and
circuits.
They
can
also be to
used
to
fuente de corrientecurrent
independiente
es específica
aun
of independent
voltage
sources
and ca,
currents
of
sources
are
frequently
usedcuando
as the inputs
to como
electric
ductores:
los
potenciómetros
y
los
sensores
de
temperatura.
connect
elements
and
circuits. They can also be used to
create
discontinuous
voltages
or
currents.
el voltaje no lo sea.
Los voltajes
de fuentes
in- used as the inputs to electric
current
sourcesde
arevoltaje
frequently
circuits.
Los interruptores se utilizan ampliamente
en los circuitos
para or currents.
create discontinuous
voltages
dependientes y las corrientes de
fuentes de corriente indecircuits.
conectar y desconectar elementos y circuitos. También se puependientes se suelen utilizar como las entradas a circuitos
den utilizar para crear voltajes o corrientes discontinuos.
eléctricos.
PROBLEMAS
PROBLEMS
PROBLEMS
thatpara
the model
is linear.
(b) Use
Sectiony2.2
Engineering
and Linear Models
Sección 2.2 Ingeniería
modelos
lineales
(b) Utilice el modelo
pronosticar
el valor
de vthe
quemodel
corres-to predict the value
that
the
model
is
linear.
(b)
Use (c)
theUse
model
predict the value
of
v
corresponding
to
a
current
of
i
¼
40
mA.
theto
model
Section
2.2
Engineering
and
Linear
Models
a la corriente
de i 5 40 mA. (c) Utilice el modelo parar
P 2.2-1
elementv yhas
v and
currentponda
i as shown
in
P 2.2-1 Un elemento
tiene An
un voltaje
unavoltage
corriente
i como
of
v
corresponding
to
a
current
of
i
¼
40
mA.
(c)
Use the model
to
predict
the
value
of
i
corresponding
to
a
voltage
of
v
¼
3
V.
pronosticar
el
valor
de
i
que
corresponda
a
un
voltaje
de
v
5
3
V.
P Los
2.2-1
An element
has voltage
and current i as shown in
Figure
P 2.2-1a.
Values
of
current
ii yand vcorresponding
se muestra en la figura
P 2.2-1a.
valores
de the
la corriente
to
predict
the
value
of
i
corresponding
to
a
voltage
Hint:
Plot
the
data.
We
expect
the
data
points
to
lie
on
a of v ¼ 3 V.
P tabulated
2.2-1a.
Values
of the
current
and corresponding
asseshown
in Figure
Pi 2.2-1b.
Sugerencia:
Diagrame los datos. Se espera que los puntos de
el correspondientevoltage
voltaje vv have
seFigure
hanbeen
tabulado
como
muestra
Hint:
Plot
theundata.
Weoflineal
expect
the data
line.recta.
Obtain
a linear
model
the element
by points
repre- to lie on a
voltage
v have
been
shownseinubiquen
Figure en
Pstraight
2.2-1b.
Determine
whether
the
element
istabulated
linear. as datos
una línea
Obtenga
modelo
en la figura P 2.2-1b.
Determine
si el elemento
es lineal.
straight
line.
Obtain
a linear model of the element by representing that
line
anecuación.
equation.
Determine whether the element isdel
linear.
elemento al representar
la línea
recta
porby
una
v, V
i, A
senting that straight line by an equation.
v, V
i, iA
i
–3+
–4
v
0
12–
+
v
–
32
60
(a)
Figura P 2.2-1
–3
–2
0
2
4
6
(a)
(b)
Figure P 2.2-1
Figure P 2.2-1
+
v
–
–3i
–4
0
12
32
60
(b)
(a)
–3
–2
0
2
4
6
v, V
i, A
–3
–4
0
12
32
60
–3
–2
0
2
4
6
(b)
Figura P 2.2-2
i
+
i
+
v, V
v
–3.6
2.4–
6.0
v
–
(a)
i
i, A
–30
20
50
+
v, V
v –3.6
2.4
–
6.0
(b)
(a)
i, A
–30 v, V
20 –3.6
50 2.4
6.0
(b)
(a)
Figure P 2.2-2
P 2.2-3
Un elemento lineal tiene
un voltaje
P 2.2-2 Un elemento
linealAtiene
voltaje has
v y voltage
una corriente
P 2.2-2
linearunelement
v and current
i as shown
Figure
P 2.2-2v y una corriente
i, A
–30
20
50
(b)
2.2-3
A linear
element
voltage
v and
como
muestra
la figura
2.2-3a.
Los has
valores
de la
co-current i as shown
i como se muestrainenFigure
la figura
2.2-2a.
valores
de lahas
coPP 2.2-2
ALos
linear
v andsecurrent
iP
asen
shown
P 2.2-2a.
Values
of element
the current
ivoltage
and icorresponding
P 2.2-3
Ahan
linear
element
hasse
v and current i as shown
in Figure Pvoltaje
2.2-3a.
Values
of the
current
i voltage
and corresponding
rrienteP ii2.2-2b.
yand
el correspondiente
v se
tabulado
como
rriente i y el correspondiente
voltaje
tabulado
como
in Figure
P han
2.2-2a.
Values
of in
theFigure
current
corresponding
voltage v have
beenv se
tabulated
as
shown
in Figure
P el
2.2-3a.
Values
theFigure
currentP i2.2-3b.
and corresponding
muestra
P 2.2-3b.
Represente
elemento
por of
una
se muestra en la figura
2.2-2b.the
Represente
el elemento
con una
v have
been tabulated
as
shown
in
voltage
v have
tabulated
shown invenFigure
Pvoltage
2.2-2b.
Represent
element
by
anbeen
equation
that as
expresses
asla afigura
ecuación
que
exprese
función
de
Esta
ecuación
es as
ecuación que exprese
v como
función
de
Esta
ecuación
voltage
v have
been
tabulated
shown inv Figure
Represent
the
element
byi. an
equation
that
expresses
as a P 2.2-3b.
thei.iselement
by
aneselement.
equation
thatVerify
expresses
vv como
as a una
function
of una
i.Represent
This
equation
a model
of the
(a)
un of
modelo
del elemento.
(a) Verifique
que the
el modelo
seabylineal.
un modelo del elemento. (a) Verifique
modelo
es lineal.
Represent
element
an equation that expresses v as a
functionque
of i.elThis
equation
is a model
the element.
(a) Verify
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 44
Circuitos Eléctricos - Dorf
4/12/11 5:17 PM
Problems
Problems
Problems
Problems
45
45
45 45
Section 2.4
2.4 Resistors
Resistors
function of
of i.
i. This
This equation
equation is
is aa model
model of
of the
the element.
element. Section
function
Problemas
45
2.4 2.4
Resistors
function
of
This
equation
is (b)
aismodel
the
element.
Section
Resistors
function
i.model
This
a model
of to
the
element.Section
(a) Verify
Verify
thati.ofthe
the
model
isequation
linear.
(b)
Use
the of
model
to
predict
(a)
that
is
linear.
Use
the
model
predict
P 2.4-1
2.4-1 A
A current
current source
source and
and aa resistor
resistor are
are connected
connected in
in series
series
P
(a)
that
the model
is linear.
(b) Use
theii ¼
model
to (c)
predict
the Verify
value
of
corresponding
to linear.
current
of
¼
mA.
(c)
Use
(a) Verify
that
the model
is
(b)of
Use
the
model
toUse
predictP
the
value
of
vv corresponding
to
aa current
66 mA.
2.4-1
A current
source
and and
aP
are connected
in series
in
the
circuit
shown
in source
Figure
Presistor
2.4-1.
Elements
connected
in
P 2.4-1
current
a resistor
are connected
in in
series
the
circuit
shown
in
Figure
2.4-1.
Elements
connected
(b) Use
el of
modelo
para
pronosticar
el
valor
dei of
v que
corresponSección
2.4A
Resistencias
the
value
v
corresponding
to
a
current
of
¼
6
mA.
(c)
Use
the
value
of
v
corresponding
to
a
current
i
¼
6
mA.
(c)
Usein
model
to
predict
the
value
of
i
corresponding
to
a
voltage
the
model
to
predict
the
value
of
i
corresponding
to
a
voltage
in
the
circuit
shown
in
Figure
P
2.4-1.
Elements
connected
in
in
the
circuit
shown
in
Figure
P
2.4-1.
Elements
connected
series
have
the
same
current,
so
i
¼
i
in
this
circuit.
Suppose
s
have the same current, so i ¼ is in this circuit. Suppose in
da vamodel
una
corriente
de the
i 5 value
6 mA.
Utilice
el modelo
the
to
of(c)i of
corresponding
to apara
voltage
the
model
to predict
the
value
i corresponding
to
aprovoltageseries
of
¼ 12
12
V.predict
P 2.4-1
fuente
de7 corriente
y un
están
conectaof
v
¼
V.
series
same
so i so
¼
is¼
inresistor
circuit.
Suppose
series
have
theR
current,
isthis
in this
circuit.
Suppose
¼ 33Una
Atheand
and
Rsame
¼current,
V.
Calculate
the
voltage
across
the
that
i
s
¼
A
¼
7 V.
Calculate
the
voltage
vv across
the
is have
nosticar
el
of
vof¼ v12
V.
¼valor
12 V.de i que corresponda a un voltaje de v 5 12 V. that
dos
en
serie
en
el
circuito
que
se
muestra
en
la
figura
2.4-1.
Athe
R ¼R7absorbed
V.7Calculate
theresistor.
voltage
v across
the the
that
is ¼and
3and
Apower
and
¼
V. Calculate
the voltage
vP across
that
is3 ¼
Hint: Plot
Plot the
the data.
data. We
We expect
expect the
the data
data points
points to
to lie
lie on
on aa resistor
resistor
and
the
power
absorbed
by the
the
resistor.
Hint:
by
Los resistor
elementos
conectados
en serie
la resistor.
misma corriente,
Sugerencia:
Diagrame
losWe
datos.
Se of
espera
quepoints
lostopuntos
Hint:
Plot
the
data.data.
expect
the
data
points
lie
on
and
the
absorbed
bytienen
the
straight
line.
Obtain
linear
model
thedata
element
by
repreHint:
Plot
the
expect
torepreliedeaon aresistor
the
power
by resistor.
the
straight
line.
Obtain
aaWe
linear
model
ofthe
the
element
by
Answer:
¼and
21power
V and
and
theabsorbed
resistor
absorbs
63que
W.i 5 3 A y
Answer:
vv ¼
21
V
the
resistor
W.
por lo tanto,
en
este
circuito
i 5 is. absorbs
Suponga63
datos
se that
ubiquen
en
una
línea
recta.
Obtenga
un element
modelo
lineal
s
straight
line.
Obtain
a
linear
model
of
the
element
by
represtraight
line.
Obtain
a
linear
model
of
the
by
representing
straight
line
by
an
equation.
senting that straight line by an equation.
Answer:
v
¼
21
V
and
the
resistor
absorbs
63
W.
Answer:
v
¼
21
V
and
the
resistor
absorbs
R 5 7 V. Calcule el voltaje v a través del resistor 63
y laW.
potencia
del elemento
al representar
línea
por una ecuación.
senting
that that
straight
line line
bylaan
equation.
senting
straight
by
an recta
equation.
i
i
+
absorbida por el resistor.
+
ii
ii
+
+
+
+
v
v
v
v
––
––
i
is
i
+
v
–
i
i +
v
R ––v
–
+
+
v
R
v
R
Respuesta: v 5 21iissV y el resistor absorbe
63 W.
v, V
V
i, mA
mA
v,
i,
v, V
V v, V i,
i, mA
mAi, mA
v,
12
3.078
12
3.078
12
3.078
12
3.078
20 12
5.133.078 20
5.13
20 20
5.135.13 20
50
12.825
5.13
50
12.825
50 50
12.825
12.825
12.825 50
is
R
i
Figure P
P 2.4-1
2.4-1
is
Figure
Figure
P 2.4-1
Figure
P 2.4-1
R
v
–
v
–
P 2.4-2
2.4-2 A
A current
current source
source and
and aa resistor
resistor are
are connected
connected in
in series
series
P
P
2.4-2
current
source
and and
aP
are connected
in series
in
the
circuit
shown
in source
Figure
Presistor
2.4-1.
Elements
connected
in
P 2.4-2
A
current
a resistor
are connected
in in
series
Figura
PA
2.4-1
in
the
circuit
shown
in
Figure
2.4-1.
Elements
connected
in
the
circuit
shown
in
Figure
P
2.4-1.
Elements
connected
in
series
have
the
same
current,
so
i
¼
i
in
this
circuit.
Suppose
Figure
P
2.2-3
in
the
circuit
shown
in
Figure
P
2.4-1.
Elements
connected
Figure
P 2.2-3
2.2-3
s in this circuit. Suppose in
Figura P
series
have
the
same
current,
so
i
¼
i
s
P 2.4-2
Una
de corriente
uniresistor
conectados
Figure
P 2.2-3
Figure
P 2.2-3
series
thefuente
same
so iyso
¼
inisthis
circuit.
have
thevv same
in están
this
circuit.
that series
¼have
mA
and
¼current,
48 current,
V. Calculate
Calculate
the
resistance
RSuppose
andSuppose
the
s¼
that
ii ¼
33enmA
and
¼
48
V.
the
resistance
R
and
the
en
serie
el
circuito
que
se
muestra
en
la
figura
P
2.4-1.
Los
P
2.2-4
Un
elemento
está
representado
por
la
relación
entre
P
2.2-4
An
element
is
represented
by
the
relation
between
that
i ¼absorbed
3i ¼
mA
andby
v¼
48
V.
Calculate
the
resistance
R
and
the
that
3 mA
and
v
¼
48
V.
Calculate
the
resistance
R
and
the
power
absorbed
by
the
resistor.
P 2.2-4 An element is represented by the relation between power
theenresistor.
elementos
conectados
serie tienen la misma corriente, por lo
P
2.2-4
Anyvoltage
element
represented
by the
between
la corriente
el voltaje
P 2.2-4
An
element
is represented
by relation
the relation
betweenpower
current
and
voltage
asiscomo
absorbed
by the
resistor.
power
absorbed
by
the resistor.
current
and
as
P
2.4-3
A
voltage
source
and
a
resistor
are
connected
in
que i 5are
3 mA
y v 5 48 in
V.
tanto,
en este
circuitosource
i 5 is. Suponga
P
2.4-3
A voltage
and a resistor
connected
current
and and
voltage
as asv ¼ 3i þ 5
current
voltage
= 3i
P
2.4-3
voltage
source
andin
a Figure
resistor
connected
in in
vv ¼
3i 1
þ5
P 2.4-3
A circuit
voltage
a resistor
are
connected
parallel
inAresistencia
the
circuit
shown
inand
Figure
P are
2.4-3.
Elements
Calcule
la
R shown
ysource
la potencia
absorbida
por
elElements
resistor.
parallel
in
the
P
2.4-3.
v ¼ v3i¼þ3i5 þ 5
parallel
in
the
circuit
shown
in
Figure
P
2.4-3.
Elements
parallel
in
the
circuit
shown
in
Figure
P
2.4-3.
Elements
connected
in
parallel
have
the
same
voltage,
so
v
¼
v
in
Determine
whether
the
element
is
linear.
s
connected
in fuente
parallel
same
voltage,
v ¼ vs en
in
Determine
whether
the element
is linear.
Determine si
el elemento
es lineal.
P 2.4-3 Una
de have
voltajethe
y un
resistor
están so
conectados
connected
in
parallel
have
the
same
voltage,
so
v
¼
v
in
Determine
whether
the element
is linear.
connected
in
parallel
have
the
same
voltage,
so
v
¼
Determine
whether
the element
is linear.
¼
10
V
and
R
¼
5
V.
Calculate
the
this
circuit.
Suppose
that
v
s
s
10 V and
¼ 5 V.
Calculate
thevs in
circuit.
vs ¼
P 2.2-5
2.2-5 The
The circuit
circuit shown
shown in
in Figure
Figure P
P 2.2-5
2.2-5 consists
consists of
of aa this
paralelo
en elSuppose
circuito that
que se
muestra
en laRfigura
P 2.4-3.
Los eleP
¼v10
R ¼R5 ¼
V.by
the the
this
circuit.
Suppose
that
vs the
current
icircuit.
in the
the
resistor
and
the
power
absorbed
by
theCalculate
resistor.
¼V10and
Vabsorbed
and
5Calculate
V.
this
Suppose
that
El
circuito
que
se
muestra
en
laPConsider
figura
Pconsists
2.2-5
consta
P
2.2-5
spower
current
i
in
resistor
and
the
resistor.
P
2.2-5
The
circuit
shown
in
Figure
2.2-5
of
a
current
source,
a
resistor,
and
element
A.
three
cases.
P
2.2-5
The
circuit
shown
in
Figure
P
2.2-5
consists
of
a
mentos conectados en paralelo tienen el mismo voltaje, por lo tanto,
current
source,de
a resistor,
and
element
A.
Consider
three
cases.
current
i
in
the
resistor
and
the
power
absorbed
by
the
resistor.
de
una
fuente
corriente,
un
resistor
y
un
elemento
A.
Concurrent
i
in
the
resistor
and
the
power
absorbed
by
the
resistor.
current
source,
a resistor,
and and
element
A. Consider
threethree
cases.
current
source,
a resistor,
element
A. Consider
cases.Answer:
Answer:
¼ 22vA
A
and
the resistor
resistor
absorbs
20
W.
que vabsorbs
y RW.
5 5 V. Calcule
en este circuito
5 and
vs. Suponga
ii ¼
the
s 5 10 V20
sidere tres casos.
Answer:
i¼
A 2resistor
and
theyresistor
absorbs
20 W.
Answer:
i2¼
A and
the
resistor
absorbs
20
la corriente
i en
el
la potencia
absorbida
porW.
el resistor.
(a)
(a)
(a) (a)
(a)
0.4 A
A
0.4
0.4 A
A
0.4
0.4 A
Figure
P 2.2-5
2.2-5
Figura P
Figure
P
2.2-5
Figure
P 2.2-5
Figure
P 2.2-5
(b)
(b)
(b) (b)
(b)
+
+
+
v+
10 Ω
Ω
v
10
v
10 Ω
Ω10 Ω v
10
−
−
−
−
A+ ii
A
Av Aii
A
−
ii
+ W.
+
Respuesta: i 5 2 A y el resistor absorbe
i
i20
i
vs
v
s
vs
+
+
–
–
+
vs
–
vs
+
–
+
–
R
R
Ri
R
+
v
v
R ––v
+
–
v
+
v
–
Figure P
P 2.4-3
2.4-3
–
Figure
Figure
P
2.4-3
Figure
P 2.4-3
(a)
When
element
A
is
a
40-V
resistor,
described
by
i
¼
v
/
40,
(a)
Cuando
el elemento
es un
resistordescribed
de 40 V,by
descrito
(a) When
element
A is a A
40-V
resistor,
i ¼ v / por
40,
P
2.4-4P 2.4-3
A voltage
voltage source
source and
and aa resistor
resistor are
are connected
connected in
in
Figura
(a) then
When
element
Ais
a 40-V
resistor,
described
by i by
¼ vi /¼40,
(a)
element
A
iscircuito
a 40-V
resistor,
described
v / 40,P
2.4-4
A
then
the
circuit
isisrepresented
represented
byrepresenta
i 5 When
v>40,
entonces
el
se
por
the
circuit
by
P
2.4-4
voltage
source
andin
a Figure
resistor
connected
in in
parallel
inA the
the
circuit
shown
inand
P are
2.4-3.
Elements
P 2.4-4
A circuit
voltage
source
a resistor
are
connected
thenthen
the circuit
is represented
by by
the circuit
is represented
parallel
shown
P
2.4-3.
Elements
P 2.4-4 in
Unathe
fuente
de voltaje
y in
un Figure
resistor están
conectados
en
vv þ vv
parallel
in
circuit
shown
Figure
P
2.4-3.
Elements
parallel
in
the
circuit
shown
in
Figure
P
2.4-3.
Elements
connected
in
parallel
have
the
same
voltage,
so
v
¼
v
in
0:4
¼
s
connected
parallel que
havese the
sameenvoltage,
¼ vsLos
in
0:4 ¼ 10
v þv40
v v
paralelo en in
el circuito
muestra
la figurasoP v2.4-3.
connected
in
parallel
have
the
same
voltage,
so
v
¼
v
in
0:4 ¼
connected
in
parallel
have
the
same
voltage,
so
v
¼
v in
þ
0:410¼þ 40
¼
24
V
and
i
¼
3
A.
Calculate
the
this
circuit.
Suppose
that
v
s
s
V andeli ¼
3 A. voltaje,
Calculate
this
circuit.conectados
Suppose that
vs ¼ 24 tienen
10 i.1040
elementos
en paralelo
mismo
porthe
los
40 that the above
Determine
the
values of
of
and
Notice
i¼
A.3Calculate
the the
this
circuit.
vs ¼absorbed
24and
Vby
A. Calculate
this
circuit.
that
v24
Determine the
los valores
devvvand
e i. i.Observe
que the
la ecuación
resistance
RSuppose
andSuppose
the that
power
absorbed
byand
the3i ¼
resistor.
Determine
values
Notice
that
above
s ¼V
resistance
R
and
the
power
the
resistor.
tanto, en este circuito v 5 vs. Suponga que vs 5 24 V e i 5 3A.
Determine
theaauna
values
of
vof
and
i. Notice
that that
the above
Determine
the
values
v and
i. Notice
the above resistance
equation
has
unique
solution.
Rvoltage
and
thesource
power
absorbed
by the
resistor.
anterior
tiene
solución
única.
resistance
R and
the power
absorbed
by are
the connected
resistor. in
equation
has
unique
solution.
P
2.4-5
Aresistencia
and
two
resistors
Calcule
laA
R y la and
potencia
absorbidaare
porconnected
el resistor.in
P
2.4-5
voltage source
two
resistors
equation
has
a
unique
solution.
(b)
When
element
A
is
a
nonlinear
resistor
described
by
equation
hasAa is
unique
(b)When
Cuando
el elemento
A aesnonlinear
unsolution.
resistorresistor
no linealdescribed
descrito por
(b)
element
by P
2.4-5
A voltage
source
and and
two
resistors
connected
in in
P 2.4-5
A voltage
source
two
resistors
are Elements
connected
parallel
in
the
circuit
shown
in Figure
Figure
P are
2.4-5.
Elements
parallel
in
the
circuit
shown
in
P
2.4-5.
(b) (b)
When
element
A
is
a
nonlinear
resistor
described
by
element
A
is
a
nonlinear
resistor
described
by
ii ¼
¼ When
vv22=2,
=2,
then
the
circuit
is
represented
by
P 2.4-5 Una fuente de voltaje y dos resistores están conecthen the circuit
is represented
bypor
entonces
el circuito
se representa
parallel
in
the
circuit
shown
in
Figure
P
2.4-5.
Elements
parallel
in
the
circuit
shown
in
Figure
P
2.4-5.
Elements
2
2
connected
in
parallel
have
the
same
voltage,
so
v
¼
v
and
1
s
connected
in parallel
have
the same
so ven
i ¼ iv ¼=2,v then
the circuit
is represented
by by
1 ¼
s and
=2, then
the circuit
is represented
tados en paralelo
en el
circuito
que voltage,
se muestra
la vfigura
2
in parallel
havehave
thethat
same
so
vso
¼v1V,
vs¼and
and
v2 ¼
¼connected
in this
this
circuit.
Suppose
that
¼voltage,
150
V, R
R1 ¼
¼
50
V,
in parallel
thevvsame
voltage,
vs and
1 50
vv þ vv 22
vconnected
vvss in
circuit.
Suppose
150
V,
22.4-5.
ss ¼
1
P
Los
elementos
conectados
en
paralelo
tienen
el
mismo
0:4
¼
2
0:4 ¼ 10
v þvv2 v
vR
¼
v
in
this
circuit.
Suppose
that
v
¼
150
V,
R
¼
50
V,
and
¼
25
V.
Calculate
the
current
in
each
resistor
and
the
power
R
v
¼
v
in
this
circuit.
Suppose
that
v
¼
150
V,
R
¼
50
V, and
22 ¼ 225
s V.s Calculate the current in each
s
1
sresistor and
1 power
the
2
10
2
voltaje, por lo tanto, en este circuito v1 5 vs y v2 5 vs. Suponga
0:4 ¼
0:4 ¼þ þ
the current
in each
resistor
and and
the power
R
¼V.
25
V.
Calculate
the current
in each
resistor
the power
R25
10i. 10
absorbed
byCalculate
each
resistor.
2¼
2case, there are two absorbed
by
each
Determine
the
values
ofdevv and
and
In2this
this
que
vs2 5 150
V, R1resistor.
5 50 V, y R2 5 25 V. Calcule la corriente
Determine the
los values
valoresof
v e i.i.
En
estecase,
caso,there
hay are
dostwo
so- absorbed
Determine
In
by
each
resistor.
absorbed
by
each
resistor.
Determine
values
of
v of
and
i. Nonlinear
InLos
case,
there
are
two
Determine
the
values
v and
i.this
Incircuitos
this
case,
are twoHint:
solutions
ofthe
theecuación
above
equation.
Nonlinear
circuits
exhibit
Hint:
Notice
theyreference
reference
directions
ofpor
thecada
resistor
currents.
en cada
resistor
la potencia
absorbidaof
resistor.
luciones
de
la
anterior.
nothere
lineales
solutions
of
the
above
equation.
circuits
exhibit
Notice
the
directions
the
resistor
currents.
solutions
ofunthe
above
equation.
Nonlinear
circuits
exhibit
more
complicated
behavior
than
linear
circuits.
Notice
the
reference
directions
of
the
currents.
solutions
of
thebehavior
above
equation.
Nonlinear
circuits
exhibitHint:
Hint:
Notice
the
reference
directions
of resistor
the450
resistor
currents.
presentan
comportamiento
más
complejo
que
los
cirmore
complicated
than
linear
circuits.
Answer:
i
¼
3
A
and
i
¼
�6
A.
R
absorbs
W
and
R2
Sugerencia:
direcciones
referencia
de and
las coAnswer:
i11 ¼Observe
3 A andlas
i22 ¼
�6 A. R11de
absorbs
450 W
R
2
more
complicated
behavior
than
linear
circuits.
more
complicated
behavior
than
linear
circuits.
(c)
When
element
A
is
a
nonlinear
resistor
described
by
i
¼
cuitos lineales.
(c) When
element A is a nonlinear resistor described by i ¼ absorbs
Answer:
i
¼
3
A
and
i
¼
�6
A.
R
absorbs
450
W
and
R
Answer:
i
¼
3
A
and
i
¼
�6
A.
R
absorbs
450
W
and
absorbs
900
W.
1
2
1
2 R2
rrientes de losW.
1resistores.
2
1
(c) When
element
A isAaAis
nonlinear
resistor
described
by ipor
¼ i ¼absorbs 900
(c) When
a
nonlinear
resistor
described
by
22unelement
v
(c)
Cuando
elemento
es
un
resistor
no
lineal,
descrito
900
W.
v
absorbs
900
W.
0:8
þ
,
then
the
circuit
is
described
by
0:8 þ v222 , then
the circuit is described by
Respuesta: i1 5 3 e i2 5 26i A. R1 absorbe
450 W y R2 absorbe
v 2 , entonces
i
0:8
the circuit
is described
by bypor
0:82þ, then
is described
el circuito
lo describe
i 5þ
i11
+ i22
+
2 then the circuit
900 W.
+
vv
vv 22
i2 i2 +
i 1 i1
+
+
+
þ
0:8
þ
0:4
¼
2
+
v
v
2
+
v
0:4 ¼ 10
R1 v11 R
R2
v þv0:8 þ v2 v
vss ––
v22 +
R
1
2
0:4 ¼
þþ
0:82þ
0:410¼þ 0:8
vs +–vs + R
i122 R2v–2 v2
i11 R1v–1 vR
–
–+
–+
10no10solution.
2 This
2 result usually
Show that
that this
this equation
equation has
has
Show
no solution. This
result usually
– –
– –
+
v
v
v
R
R
Show
that
this
equation
has
no
solution.
This
result
usually
s –
Showque
thisecuación
equation
hastiene
no
result
usually
1
2
indicates
modeling
problem.
At solution.
least
oneThis
of
the
three
1
2
Muestre
esta
no
solución.
Este
resultaindicates
aathat
modeling
problem.
At
least
one
of
the
three
–
–
indicates
ainmodeling
problem.
At
least
oneAl
ofaccurately.
the
elements
the
circuit
hasproblem.
notdebeen
been
modeled
accurately.
Figure P
P 2.4-5
2.4-5
indicates
a modeling
Atmodeled
least
one
of three
theuno
three Figure
do suele
indicar
un problema
modelado.
menos
elements
in
the
circuit
has
not
P 2.4-5
Figure
P 2.4-5
elements
the
not
beenbeen
modeled
elements
in circuit
the circuit
not
accurately.Figure
de los
tresinelementos
enhas
el has
circuito
no
hamodeled
sidoaccurately.
modelado
Figura P 2.4-5
con certeza.
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 45
Alfaomega
4/12/11 5:18 PM
<<Nota:
ecuacion
= if; vs =
4646
46
46
Circuit
Elements
Circuit
Elements
Circuit Elements
Elementos de circuitos
P P2.4-6
2.4-6A Acurrent
currentsource
sourceand
andtwo
tworesistors
resistorsareareconnected
connectedinin
P 2.4-6
2.4-6 A current
source
and two
resistors
are connected
in
P
fuente
deshown
corriente
yFigure
dos
resistores
están
conecseries
circuit
P P2.4-6.
Elements
seriesinUna
inthethe
circuit
showninin
Figure
2.4-6.
Elements
series eninserie
the encircuit
shown
in
Figure Pen 2.4-6.
Elements
tados
el
circuito
que
se
muestra
la
figura
P
2.4-6.
connected
connectedininseries
serieshave
havethethesame
samecurrent,
current,sosoi1 i1¼¼is iand
s and
¼ is and
connected
in series
have the
sametienen
current,
so i1 corriente,
Los
elementos
conectados
en serie
la
misma
i2 i¼
is iin
this
circuit.
Suppose
that
is i¼
2525
mA,
R1R¼
4 V,
and
this
circuit.
Suppose
that
mA,
4 V,
and
2¼
s in
s¼
1¼
i2 ¼lois tanto,
in thisen
circuit.
Suppose
ies ¼
25imA,
R1 ¼ 4que
V, iand
por
este circuito
i1 that
5 isacross
i2 5
. Suponga
5the
feach
fthe
R2R¼
8 8V.V.Calculate
thethevoltage
each
resistor
Calculate
voltage
across
resistorand
and
2 ¼
R2 mA,
¼ 8 V.
Calculate
voltage
across each
resistor
and the
25
Rabsorbed
yeach
Rthe
8resistor.
V. Calcule
el voltaje
a través
de
1 5 4 V,
2 5resistor.
power
byby
powerabsorbed
each
powerresistor
absorbed
each resistor.
cada
y laby
potencia
absorbida por cada resistor.
Hint:
Hint:Notice
Noticethethereference
referencedirections
directionsofofthetheresistor
resistorvoltages.
voltages.
Hint: Notice Observe
the reference
directions de
of referencia
the resistordevoltages.
Sugerencia:
las direcciones
los vol– – v1v1+ +
tajes de los resistores.
v
is is
is
if
–
–
i1 i1
i
i11
+
v11 +
R1R i2 i2
R 1 i2
R11 iR2 R
2
R 2
R22
++
+
v+2v2
v
v–22 –
–
–
P P2.4-10
2.4-10The
Thevoltage
voltagesource
sourceshown
shownininFigure
FigureP P2.4-10
2.4-10is isanan
P
2.4-10
voltagedesource
shown
in Figure
P en
2.4-10
is an
P
2.4-10 The
La
fuente
voltaje
que
se
muestra
la figura
aa
adjustable
dcdc
voltage
source.
InIn
other
words,
thethe
voltage
vs viss is
adjustable
voltage
source.
other
words,
voltage
adjustable
dcuna
voltage
source.
In other
words,dethe
voltage
vs is
Pconstant
2.4-10
es
fuente
de
voltaje
ajustable
cd.
En
otras
pa-a
voltage,
butbut
thethe
value
ofof
that
constant
can
bebe
adjusted.
constant
voltage,
value
that
constant
can
adjusted.
constant
the
ofconstante,
that constant
be
adjusted.
labras,
elvoltage,
voltaje
vbut
un value
voltaje
perocan
el
valor
de esa
s es
The
tabulated
data
were
collected
asas
follows.
The
voltage,
vs,v ,
The
tabulated
data
were
collected
follows.
The
voltage,
The
tabulated
data
were
collected
as
follows.
The
voltage,
vs, s
constante
puede
ajustar.
La
tabulación
deacross
los the
datos
capturawas
tose
some
value,
and
thethe
voltages
across
resistor,
vava
wassetset
to
some
value,
and
voltages
the
resistor,
was
set to some
value,
andalgún
the voltages
acrossalthe
resistor,
dos
es
sigue.
Se dio
tipo de
valor
voltaje,
yvaa
and
vbv,como
measured
and
recorded.
Next,
thethe
value
ofof
vvssv,was
and
, were
measured
and
recorded.
Next,
value
bwere
s was
and
vb, wereameasured
recorded.
themidió
valueyof
vs was
los
voltajes
través
deland
resistor,
va the
y the
vNext,
, se
les
registró.
bresistors
changed,
and
thethevoltages
across
were
changed,
and
voltages
across
resistors
weremeasured
measured
changed,
and theelvoltages
across
the resistors
were measured
A
continuación,
valor
de
v
se
cambió,
y
los
voltajes
a
s
again
and
recorded.
This
procedure
was
repeated
several
again and recorded. This procedure was repeated través
several
again
recorded.
This procedure
was
several
de
los and
resistores
se midieron
denot
nuevo
y se repeated
registraron.
Este
times.
(The
ofofvs vwere
Determine
thethe
times.
(Thevalues
values
notrecorded.)
recorded.)
Determine
s were
times.
(The values
of vs varias
were not
recorded.)
Determine
the
procedimiento
se
repitió
veces.
(No
se
registraron
los
value
valueofofthetheresistance,
resistance,R.R.
value
valoresofdethe
vs.)resistance,
DetermineR.el valor de la resistencia, R.
+ + va v – –
+ va a–
vs v + +
vs s +–– –
Figure
FigureP P2.4-6
2.4-6
Figure P
P 2.4-6
2.4-6
Figura
P
Un
calentador
eléctrico
está
conectado
a una
fuente
de
P 2.4-7
AnAn
electric
totoa aconstant
250-V
P2.4-7
2.4-7
electricheater
heateris isconnected
connected
constant
250-V
P 2.4-7
An electric
heater1 is000
connected
to este
a constant
250-V
250
V constante
y absorbe
W. Luego,
calentador
se
source
and
1000
this
source
andabsorbs
absorbs
1000W.W.Subsequently,
Subsequently,
thisheater
heateris isconconsource and
absorbs
W.VSubsequently,
thispotencia
heater isabsorconconecta
una
fuente1000
de
220
constante.
¿Qué
nected
a aconstant
220-V
source.
What
power
does
nectedtoato
constant
220-V
source.
What
power
doesit itabsorb
absorb
nected
tofuente
a constant
220-V
source.laWhat
powerdel
does
it absorb
be
de la
desource?
220
V?What
¿Cuál
resistencia
calentador?
from
thethe
220-V
resistance
from
220-V
source?
Whatisesisthe
the
resistanceof
ofthe
theheater?
heater?
from the 220-V source? What is the resistance of the heater?
Sugerencia:
Modele
el
calentador
como
una
resistencia.
Hint:
Hint:Model
Modelthetheelectric
electricheater
heaterasasa aresistor.
resistor.
Hint: Model the electric heater as a resistor.
P
2.4-8
El
equipo
portátil
de
alumbrado
para
mina
se
loPP
2.4-8
The
portable
lighting
equipment
forfor
auna
mine
is is
located
2.4-8 The portable lighting equipment
a mine
located
P 2.4-8
Themetros
portable
lighting
equipment
fordeacd.
mine
isluces
located
caliza
ameters
100
fuente
de
alimento
Las
dea a
100
from
source.
The
lights
use
100meters
fromitsde
itsdcladcsupply
supply
source.
Themine
mine
lights
use
100
meters
from
its de
dc 5supply
source. The
mine
lights
use a
la
mina
usan
un
total
kW
y
funcionan
a
120
V
cd.
Determitotal
ofof
5 kW
and
operate
at at
120
VV
dc.dc.
Determine
thethe
required
total
5 kW
and
operate
120
Determine
required
total
5 kW
and operate
at 120
V dc. Determine
theserequired
ne
el of
área
seccional
requerida
de
cables
de used
cobre
usaron
cross-sectional
area
ofofthethecopper
wires
thethe
cross-sectional
area
copper
wires
usedtoque
toconnect
connect
cross-sectional
area
ofdethe
copperdewires
used
to requiere
connect que
the
para
conectar
lamine
fuente
lathat
mina
sipower
se
source
toto
thethe
mine
lights
iflas
weluces
require
thethe
lost
inin
thethe
source
lights
if
we
require
that
power
lost
source
to the
mine lights
if we require
that the
la
potencia
perdida
en el
cableado
sea menor
a opower
igual alost
5%indethe
la
copper
copperwires
wiresbebeless
lessthan
thanororequal
equaltoto5 5percent
percentofofthethepower
power
copper wires
be less
than
or equal
tomina.
5 percent of the power
potencia
requerida
por
las
luces
de
la
required
requiredbybythethemine
minelights.
lights.
required by the
mine el
lights.
Sugerencia:
Modele
equipo de iluminación y el cableado
Hint:
Model
both
thethe
lighting
equipment
and
thethe
wire
asas
resistors.
Hint:
Model
both
lighting
equipment
and
wire
resistors.
como
resistencias.
Hint: Model
both the lighting equipment and the wire as resistors.
P
2.4-9
The
resistance
of
a
practical
resistor
depends
onon
P
2.4-9
The
resistance
of
a
practical
resistor
depends
P
unaresistor
útil resistor
dependedepends
de la resisP 2.4-9
2.4-9 La
Theresistencia
resistancedeof
practical
on
the
nominal
resistance
and
the
resistance
tolerance
as
follows:
the
nominal
resistance
and
the
resistance
tolerance
as
follows:
tencia
nominal
y de la tolerancia
de la resistencia
como
sigue:
the nominal
resistance
the resistance
as follows:
and
tolerance
t t t t ��
Rnom
RR
��
Rnom
R 1�
Rnom1 þ
1 �t
1 þt
Rnomnom1 � 100
100� R � Rnom 1 þ 100
100
100
100
donde
RRnom
es
la
resistencia
nominal
y and
t and
es tlaist tolerancia
de la
is is
thethe
nominal
where
nominalresistance
resistance
isthetheresistance
resistance
where
Rnom
nom
is the nominal
resistance
andPor
t isejemplo,
the resistance
where Rnomexpresada
resistencia
porcentaje.
retolerance
For
toleranceexpressed
expressedasen
asa percentage.
aunpercentage.
Forexample,
example,a 100-V,
a un
100-V,
tolerance
expressed
as a percentage.
example,
a 100-V,
sistor
de 2%
de resistencia
de
100
V,For
tendrá
una
2 2percent
resistor
will
a aresistance
given
bybyresistencia
percent
resistor
willhave
have
resistance
given
2 percent
dada
por resistor will have a resistance given by
9898
VV
��
RR
��
102
VV
102
98 V � R � 102 V
The
circuit
shown
inin
Figure
PP
2.4-9
has
one
input,
vs,vsand
one
The
circuit
shown
Figure
2.4-9
has
one
input,
, entraand
one
El
se muestra
figura
2.4-9
tiene
Thecircuito
circuit que
shown
in FigureenPla2.4-9
hasP one
input,
vuna
, and
one
output,
vov. oThe
ofofthis
circuit
is isgiven
byby s
output,
.salida,
Thegain
gain
this
circuit
given
da,
v
,
y
una
v
.
La
ganancia
de
este
circuito
la
da
output,
s vo. The gain oof this circuit is given by
vovo
R2R2
gain
¼¼
vo ¼
ganancia
gain
¼ R2
R
þ
R2R
gain ¼ vs v¼
R
1
1þ
vs s R1 þ
R2 2
Determine
the
range
of
possible
values
ofof
thethe
gain
R1Ris is
Determine
elthe
rango
deof
los
valores
posibles
degain
lawhen
ganancia
Determine
range
possible
values
when
Determine the range of possible values of the gain when R1 is1
thetheresistance
2 2percent
resistor
isV
cuando
R1 es laofresistencia
de
2%
de un
resistorand
de R100
ythe
resistance
ofa a100-V,
100-V,
percent
resistor
and
isthe
2R2
the resistance of a 100-V, 2 percent resistor and R2 is the
resistance
ofofa 400-V,
5 5percent
thethe
gain
R
es la resistencia
de 5%
de un resistor.
resistor
deExpress
400 V.
Exprese
a 400-V,
percent
resistor.Express
gaininin
2resistance
resistance
ofen
a 400-V,
5 percent
resistor. Express
in
la
ganancia
términos
deand
una
nominalthe
y gain
de una
terms
ofofa anominal
gain
a gain
terms
nominal
gainand
a ganancia
gaintolerance.
tolerance.
terms
of
a
nominal
gain
and
a
gain
tolerance.
tolerancia de ganancia.
v v ++ +
– –
vfs s+
v
s –
–
Figure
Figura
PP 2.4-9
Figure
P2.4-9
2.4-9
Figure P 2.4-9
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 46
RR
R11 1
R
1
++
+
vbvb
vb
– –
–
Figure
FigureP P2.4-10
2.4-10
Figure
P 2.4-10
2.4-10
Figura P
Section
2.5
Sources
Section
2.5Independent
Independent
Sources
Sección
2.5
Fuentes
independientes
Section
2.5
Independent
Sources
PP
2.5-1
AA
current
source
and
a voltage
source
areare
connected
2.5-1Una
current
source
and
ay voltage
source
connected
P 2.5-1
2.5-1
fuentesource
de
corriente
una fuente
voltaje
están
P
A current
and a voltage
sourcedeare
connected
inin
parallel
with
a resistor
asas
shown
inin
Figure
P 2.5-1.
AllAll
ofof
thethe
parallel
with
a
resistor
shown
Figure
P
2.5-1.
conectados
en paralelo
un resistor,
comoP se
muestra
la
in
parallel with
a resistorcon
as shown
in Figure
2.5-1.
All ofenthe
elements
connected
inin
parallel
have
thethe
same
voltage,
vs vins in
this
elements
connected
parallel
have
same
voltage,
this
figura P 2.5-1.
En este
todosthe
lossame
elementos
conectados
elements
connected
in circuito
parallel have
voltage,
vs in this
1515V,V,is i¼
5 5V.V.
circuit.
Suppose
that
vs v¼
¼3 3A,A,and
andRvR¼
¼
circuit.
Suppose
that
s ¼
sSuponga
en
paralelo
tienen
el
mismo
voltaje,
v
.
que
=
15
V,
circuit. Suppose that vs ¼ 15 V, is s ¼ 3 A, and R s¼ 5 V.
the
current
i iininthe
thethepower
(a)Calculate
Calculate
current
theresistor
resistorand
and
power
3 A,
y R 5the
5 the
V.current
(a)
Calcule
corriente
el
resistor
y la
i(a)
f 5 Calculate
(a)
i in lathe
resistori en
and
the power
absorbed
by
the
resistor.
(b)
Change
the
current
source
current
absorbed
by
the
resistor.
(b)
Change
the
current
source
current
potencia absorbida
por el(b)
resistor.
(b)the
Cambie
la source
corriente
de la
absorbed
by the resistor.
Change
current
current
toto
is i¼¼
5A
and
recalculate
the
current,
i, in
thethe
resistor
and
thethe
5A
and
recalculate
current,
i, in
and
fuente
a is 5 5 the
A the
ycurrent,
calcule
de
laresistor
corriente,
to
is ¼s de
5 Acorriente
and recalculate
i,
innuevo
the resistor
and thei,
power
absorbed
by
the
resistor.
power
absorbed
by
the
resistor.
en el resistor
y laby
potencia
absorbida por el resistor.
power
absorbed
the resistor.
Answer:
Answer:i ¼
i ¼3 3A Aand
andthetheresistor
resistorabsorbs
absorbs4545WWboth
bothwhen
when
3A
Respuesta:
y el the
resistor
absorbe
45 W
is 5
Answer:
i ¼i 5
3 3AAand
resistor
absorbs
45cuando
W both
when
is i¼
3 3A Aand
5 5A.A.
andwhen
whenis i¼
s ¼
s ¼
5 5 when
A. is ¼ 5 A.
is and
icuando
s ¼ 3 A
i i
i
is i
is s
vs v + +
vs s+– –
–
PP
2.5-2
AA
current
and
a voltage
source
connected
2.5-2Una
current
source
and
ay voltage
source
are
connected
fuentesource
de
corriente
una fuente
deare
voltaje
están
P 2.5-2 A current source and a voltage source are connected
inin
series
with
resistor
shown
inin
Figure
2.5-2.
All
oflaof
the
series
with
a resistor
shown
Figure
P
2.5-2.
All
conectados
en aserie
con as
unas
resistor,
como
seP muestra
en
fi-the
in series with a resistor as shown in Figure P 2.5-2. All of the
elements
connected
inin
series
have
thethe
same
current,
this
elements
connected
series
have
same
, in
this
gura
P 2.5-2.
Todos los
elementos
conectados
encurrent,
serie itienen
la
s,isin
elements connected in series have the same current, is, in this
is i¼
3 3A,
circuit.
Suppose
vs v¼
misma
corriente,
isthat
, en
este
Suponga
que
vand
5V.
¼1010V,V,
A,and
R¼V,
¼5is5V.
circuit.
Suppose
that
scircuito.
s ¼
s 5R10
circuit. Suppose that vs ¼ 10 V, is ¼ 3 A, and R ¼ 5 V.
thethe
voltage
v vacross
thethe
3(a)A,
yCalculate
R 5 5 V.
(a)
Calcule
elacross
voltaje
vresistor
aresistor
travésand
del
resistor
y
(a)Calculate
voltage
andthe
thepower
power
(a) Calculate the voltage v across the resistor and the power
absorbed
byby
thethe
resistor.
the
voltage
source
voltage
la
potencia
absorbida
por(b)(b)
elChange
resistor.
(b)
Cambie
la corriente
absorbed
resistor.
Change
the
voltage
source
voltage
absorbed by the resistor. (b) Change the voltage source voltage
totovlas vfuente
¼¼5 5V de
recalculate
resistor
de
voltaje
a vs 5 the
5 the
Vvoltage,
y voltage,
calculev,de
nuevo
elthe
voltaje,
Vand
and
recalculate
v,across
acrossthe
resistor
to vs ¼s 5 V and recalculate
the voltage, v, across the resistor
v,
a través
del resistor
y labypotencia
absorbida por el resistor.
and
thethepower
absorbed
and
power
absorbed
bythetheresistor.
resistor.
and the power absorbed by the resistor.
–– –
–
RR
R22 2
R
2
RR
R
Figure
FigureP P2.5-1
2.5-1
Figura P
Figure
P 2.5-1
2.5-1
+ ++
+
i i
ii
4040
ΩΩ
40 Ω

RR
R
va,vaV, V vbv, bV, V
vb, V
va, V
11.75
11.75 7.05
7.05
11.75
7.5
4.5
7.5 7.05
4.5
7.5
4.5
5.625
5.625 3.375
3.375
5.625
1010 3.375
66
10
6
4.375
4.375 2.625
2.625
4.375 2.625
++
+
+
vovo
vs
v
–o –
––
iis i
iss s
vvs v
vss s
++
+
+
R R vv v
R
v
R
–– –
–
Figure
P 2.5-2
Figure
P2.5-2
2.5-2
Figura
P
Figure P 2.5-2
Circuitos Eléctricos - Dorf
4/12/11 5:18 PM
Problemas
47
P 2.5-3 La fuente de corriente y la fuente de voltaje en el
circuito que se muestra en la figura P 2.5-3 están conectadas
en paralelo, por lo que tienen el mismo voltaje, vs. La fuente
de corriente y la fuente de voltaje también están conectadas
en serie, de modo que tienen la misma corriente, is. Suponga
que vs ⫽ 12 V, e i ⫽ 3 A. Calcule la energía alimentada por
cada fuente.
al producto del voltaje de la batería y la capacidad de la batería.
Por lo común la capacidad se da con las unidades de amperios hora (Ah). Una batería nueva de 12 V con una capacidad
de 800 mAh está conectada a la carga que jala una corriente de
25 mA. (a) ¿Cuánto tiempo le tomará a la carga descargar la
batería? (b) ¿Cuánta energía se alimentará a la carga durante el
tiempo requerido para descargar la batería?
Respuesta: La fuente de voltaje alimenta ⫺36 W, y la fuente
de corriente alimenta 36 W.
i
+
is
is
+
vs
–
–
v
R
vs +
–
batería
carga
Figura P 2.5-6
Figura P 2.5-3
P 2.5-4 La fuente de corriente y la fuente de voltaje en el
circuito que se muestra en la figura P 2.5-4 están conectadas en
paralelo por lo que tienen el mismo voltaje, vs. La fuente de corriente y la fuente de voltaje también están conectadas en serie
de modo que tiene la misma corriente is. Suponga que vs ⫽ 12 V,
e i ⫽ 2 A. Calcule la energía alimentada por cada fuente.
is
+
vs
–
Sección 2.6 Voltímetros y amperímetros
P 2.6-1 Para el circuito de la figura P 2.6-1:
(a) ¿Cuál es el valor de la resistencia R?
(b) ¿Cuánta energía transmite la fuente de voltaje?
vs +
+ 5 . 0
– . 5 0
Voltímetro
Amperímetro
–
is
Figura P 2.5-4
P .2.5-5
(a) Encuentre la potencia alimentada por la fuente de voltaje que
se muestra en la figura P 2.5-5 cuando para t 0 tenemos
v ¼ 2 cos t V
y
i ¼ 10 cos t mA
(b) Determine la energía alimentada por esta fuente de voltaje
para el periodo 0 t 1 s.
12 V
+
–
R
1
2
A
Figura P 2.6-1
P 2.6-2 La fuente de corriente en la figura P 2.6-2 alimenta
40 W. ¿Qué valores leen los medidores de la figura P 2.6-2?
i
Amperímetro
+
–
v
Voltímetro
Figura P 2.5-5
46
P 2.5-6 La figura 2.5-6 muestra una batería conectada a una
carga. La carga en la figura 2.5-6 podría representar las luces
de un automóvil, una cámara digital o un teléfono celular. La
energía alimentada por la batería a la carga está dada por
Z t2
vi dt
w¼
t1
Cuando el voltaje de la batería es constante y la resistencia de
la carga es fija, la corriente de la batería será constante y
w ¼ viðt2 t1 Þ
La capacidad de la batería es el producto de la corriente de la
batería y el tiempo requerido para que la batería se descargue.
En consecuencia, la energía almacenada en la batería es igual
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 47
+
–
+
v
–
i
12 V
2A
Figura P 2.6-2
P 2.6-3 Se ha modelado un voltímetro ideal como un circuito
abierto. Un modelo más real de un voltímetro es una resistencia grande. La figura P 2.6-3a muestra un circuito con un
voltímetro que mide el voltaje vm. En la figura P 2.6-3b, el voltímetro ha sido reemplazado por el modelo de un voltímetro
ideal, un circuito abierto. Idealmente, en el resistor de 100 ⍀
no hay corriente y el voltímetro mide vmi ⫽ 12 V, el valor ideal
Alfaomega
5/24/11 9:50 AM
E1C02_1
10/23/2009
48
A dependent source provides a current (or a voltage) that is
Y
o repreA dependent
provides
a current (or a voltage) that is
dependent
ontoanother
elsewhere source
in the circuit.
The
els, called
elements,
repre- variable
ook,
we circuit
dependent
on another
variable elsewhere in the circuit. The
constitutive
equations
sources
are summarized
make
In this book,
we of dependent
vices.up
A a circuit.
constitutive equations of dependent sources are summarized
Tableof2.7-1.
ments
or linear in
models
devices. A
uperpoin circuit
Table 2.7-1.
The
short
circuit
and
open
are special cases of
sfies the properties of both
4848superpoCircuit
Elements
Elements
short
and open
independent
ACircuit
short The
circuit
iscircuitos
ancircuit
ideal voltage
sourcecircuit are special cases of
48 sources.Elementos
de
y.
s of the
480. The current
Circuit
Elements
independent
sources.
A
short
circuit
is an ideal voltage source
having
v(t)
¼
in
a
short
circuit
is
determined
by
een The
the reference directions of the
ant.
having
v(t)
¼
0.
The
current
in
a
short
circuit
determined
byel
the
rest
of
the
circuit.
An
open
circuit
is
an
ideal
current
source
ideal
value
of
v
.
In
Figure
P
2.6-3c,
the
voltmeter
is
modeled
bybyis (a)
(a)(a)Express
theerror
that
de
v
.
En
la
figura
P
2.6-3c,
el
voltímetro
está
modelado
por
Exprese
de mediciónerror
que
ocurre
cuando
Rwhen
V¼
m
a�.circuit
The ofmvm. In Figure P 2.6-3c, the voltmeter is modeled
The element is important.
m value
m 5R10
ideal
Express
themeasurement
measurement
error
thatoccurs
occurswhen
Rm¼
the
rest
of
the
circuit.
An
open
circuit
is
an
ideal
current
source
having
i(t)
¼
0.
The
voltage
across
an
open
circuit
is
determined
.
Now
the
voltage
measured
by
the
voltmeter
is
.
the
resistance
R
10
V
as
a
percent
of
i
la
resistencia
R
.
Ahora
el
voltaje
medido
por
el
voltímetro
es
como
un
porcentaje
de
i
.
m
mi
one
terminal
þ
and
the
other
�.
The
convenideal
value ofmRvmm..Now
In Figure
P 2.6-3c,
voltmeter
modeledisby (a) 10
Express
measurement
the
voltage
measured
by theisvoltmeter
the
resistance
V as athe
percent
of imimi. error that occurs when Rm ¼
the
i(t)
¼short
voltage
an
open circuit
is(b)
determined
thepassive
rest ofthe
the
circuit. Open
circuits
and
circuits
also
voltage
canacross
(b)
Determine
the
maximum
value
totoensure
(b)
Determine
el
valor
máximo
de RofmofRrequerido
para
asegum
urrent
tobythe
convenked þ adhere
to
R0.mRThe
Now the
measured
by
the
voltmeter
is
.value
resistance
Rm.having
10
V
as
a
percent
of imi
Determine
the
maximum
Rmrequired
required
ensure
vmvthe
12
by
rest
of themcircuit.
Openwith
circuits and short circuits
can
also
be described
as þ
special
ofm¼ ¼
resistors.
A resistor
12
that
the
measurement
isvalue
smaller
than
555percent.
rarse
de
que
elthe
error
de error
medición
es of
menor
de
por
ciento.
ected from the terminal
marked
to cases
(b)
Determine
maximum
R
required
to
ensure
that
the
measurement
error
is
smaller
than
percent.
RmRþ þ
100
m
Rm100
m as
special12cases
with
resistance R ¼ 0 (G ¼ 1) isbea described
withof resistors. A resistorthat
vshort
.hen the
m ¼ circuit. A resistor
the
measurement
error
is
smaller
than
5
percent.
þ
100
!!
1,1,
the
anan
ideal
and
Because
mbecomes
m
imiim
RRbecomes
¼
0 (G
¼
1)
is
aun
short
circuit.
G que
¼ R0the
isvoltmeter
an
open
circuit.
R
,resistance
el
voltímetro
se
convierte
envoltmeter,
voltímeBecause
R(R
the
voltmeter
ideal
voltmeter,
and A resistor with
m ¼ 1)
m
sed as circuit conductance
elements.Dado
When
convenm
vammeter
!
v
¼
l2
V
.
When
R
<
1,
the
voltmeter
is
not
ideal,
and
mvBecause
mi
m
conductance
G
¼
0
(R
¼
1)
is
an
open
circuit.
An
ideal
measures
the
current
flowing
through
its
tro
ideal,
y
v
v
5
12
V.
Cuando
R
,
,
el
voltímetro
R
!
1,
the
voltmeter
becomes
an
ideal
voltmeter,
and
! vmi ¼m
is not ideal, and
ml2 V .miWhen Rm < 1, the voltmeter
m
im
rrent the
adhere to the passivemconvenross
Amperímetro
Ammeter
<
v
.
The
difference
between
v
and
v
is
a
measurement
v
Ammeter
mvv
mi
mvthe
mi
ammeter
measures
current
flowing
es
ideal,
. ideal
La diferencia
entre
vvmmiideal
yisvthe
es
unideal,
error
and
across
its<terminals.
An
!
¼vvoltage
l2,
VvAn
.mi
When
Rm
1,
voltmeter
not
and through its
<
vvzero
. yThe
difference
between
and
aismeasurement
mi
m
mi
mmhas
mi
m
hm’sthe
law; theterminals
voltageno
across
the
into
error
caused
by
the
fact
that
the
voltmeter
is
not
ideal.
Ammeter
de
medición
causado
por
el
hecho
de
que
el
voltímetro
sea
terminals
and
has
zero
voltage
across
its
terminals.
An
ideal
voltmeter
measures
the
voltage
across
its
terminals
and
has
<
v
.
The
difference
between
v
and
v
is
a
measurement
v
error
caused
by
the
fact
that
the
voltmeter
is
not
ideal.
m
mi
m
mi
or
to the current into the
ed istorelated
a
1 k
2A
1 kΩ
2A
1
kΩ
2
A
no
ideal.
voltmeter
measures
theoccurs
voltage
across
its900
terminals and has
terminal
current
equal
tothe
zero.
Ideal
voltmeters
act
like
error
caused
by
the
fact that
the
voltmeter
isopen
not
ideal.
(a)(a)
Express
measurement
error
that
occurs
when
RmR¼ ¼
900
¼ Ri. The power
delivered
to
a the
Express
measurement
error
that
when
m
1 kΩ
terminal
equal
to
zero. when
Ideal R
voltmeters
act like open 2 A
ideal
act
like
shorterror
circuits.
V
as
aammeters
percent
of
vmi
. .current
¼voltage
v2=R watts. circuits, and
(a)
Express
the
measurement
that
occurs
¼
900
V
as
a
percent
of
v
que ocurre cuando Rmm 5
(a)Exprese el error de medición
mi
circuits,
and
ideal
ammeters
act
like
short
circuits.
Transducers
are
devices
that
convert
physical
quantities,
(b)
Determine
the
minimum
value
of
R
required
to
ensure
that
(a)
ee provides
a current or (b)
a 900
voltage
of an
(a)
VVascomo
a percent
of vmi. value
Determine
the
minimum
un porcentaje
de vmiof. mRm required to ensure that
(a)
Transducers
are
devices
that
quantities,
such
rotational
position,
to
anerror
electrical
quantity
such
asconvert
. physical
theetermine
measurement
error
is issmaller
2 required
percent
of
vasegumi
ircuit is
variables.
Theas voltage
of
an el the
urrent
(b)
Determine
minimum
value
of
R
to
ensure
that
.
the
measurement
smaller
than
2
percent
of
v
(b)
D
valor
mínimo
de Rthan
requerido
para
m
mi
(a)
mi
imi = 2 A
such as rotational
position,
to an electrical
quantity such as
voltage.
In this
chapter,
we describe
two
potenurce
is specified,
but the
current
current
thedeis
measurement
is transducers:
smaller
2 percent
of vvmi
rarse
que el error error
de medición
esthan
menor
de 2% de
imii = 2 A
mi..
2A
voltage.
In
this
chapter,
we
describe
two
transducers:
potentiometers
and
temperature
sensors.
mi
=
current of an independent current
voltages
imi = 2 A
100
Ω Ωand
tiometers
temperature
sensors.
arevoltages
widely used in
circuits
to connect
and
disVoltmeter
100
reas
the voltageSwitches
is not. The
pendent
Voltmeter
Switches
are
widely
used
in
circuits
to
connect
and
disconnect
elements
and
circuits.
They
can
also
be
used
to
100
Ω
sources
electric and currents of independent
100  +
Voltmeter
1 k
Voltímetro
2A
+
connect
elements
and circuits.
They can also be used to
discontinuous
or currents.
1 kΩ
quently used ascreate
the inputs
to electric voltages
2 A2 A
+
1 kΩ
v
1212
V V – +create discontinuous
+
voltages or currents.
+mvm
–
+
+
12 12
V V– –
v−m−vm
− −
2A
(a)(a)
(a)(a)
100
ΩΩ
100
100

S
100 Ω
2A
+ +
2 A2 A
PROBLEMS
+
+
=
V V
vmiv 12
1212
V
=+12
+ +
= 12
V
vmimi
2A
12 V V – –
that the model is linear. (b) –Use
to12
predict
the value
+ the model
=
V
v
mi
−
12 V
–
−
− is(c)
that
linear.
(b) model
Use the model to predict the value
of v corresponding
to a current
of the
i ¼ model
40 mA.
Use the
ring
and
Models
shown
in Linear
− (b)to a current of i ¼ 40 mA. (c) Use the model
of v corresponding
1 kΩ
(b)
(b)
(b)
i
(b)
im im
m
im
1 k
1 kΩ
1 kΩ
R
RmRm
m
Rm
1 kΩ
(c)
(c)
(c)
(c)
to predict
value
of iin
corresponding to a voltage
(b) of v ¼ 3 V.
(b)
Figura
P
2.6-4
Figure
P 2.6-4
ssponding
voltage v and
currentthe
i as
shown
2.6-4
to
predict
the
value
of to
i corresponding
to a voltageFigure
of v ¼ P
3 V.
(b)
Hint:i and
Plotcorresponding
the data. We expect100
thedata points
lie on a
sP of
the current
2.2-1b.
Figure
P
2.6-4
P 2.6-5
El
voltímetro
de la figura P P2.6-5a
mide
el voltaje
100
Ω Ωthe element
P
2.6-5
The
measures
thea
Plot
data. Webyexpect
lie on
a voltmeter ininFigure
straight
line. Obtain
a linearHint:
model
of
repre- the data points
bulated as shown
in Figure
P 2.2-1b.
100
Pto
2.6-5
P2.6-5a
2.6-5a
través
de la The
fuentevoltmeter
de corriente.Figure
La figura
P 2.6-5bmeasures
muestra elthe
voltage
across
current
source.
Figure
P P2.6-5b
shows
line.Ω Obtain++ a linear model of the element
representing that straight line by straight
an equation.
element is linear.
100
P by2.6-5
Thethe
Figure
Py 2.6-5a
measures
the
voltage
across
the
current
source.
Figure
2.6-5b
shows
circuito
después
devoltmeter
eliminar
elinvoltímetro
etiquetar
el voltaje
+ by an equation.
the
circuit
after
removing
the
voltmeter
and
labeling
thethe
+senting thatRstraight
v
line
m
m
voltage
across
the
current
source.
Figure
P
2.6-5b
shows
the
circuit
after
removing
the
voltmeter
and
labeling
12 V +–
medido
por
el
voltímetro
como
v
.
Incluso,
los
demás
volv
R
m
v, V
i, A
m
m
+
vm
1212
Vi V – +
Rm
voltage
measured
bybythethevoltmeter
asasvmv. Also,
thetheother
they circuit
after
the
the
voltage
measured
voltmeter
Also,
other
–
−
m. and
tajes
corrientes
delremoving
elemento
estánvoltmeter
etiquetados
enlabeling
la figura
–3
–3
+
Rm − −vim
element
voltages
and
currents
are
labeled
in
Figure
PP
2.6-5b.
12
V
voltage voltages
measured
bycurrents
the voltmeter
as vinm.Figure
Also,
the
other
element
and
are
labeled
2.6-5b.
+
–
P
2.6-5b.
–2
–4
v, V
i, A (c)
−
element voltages and currents are labeled in Figure P 2.6-5b.
+
v
0
0
(c)
(c)
v, V
i, A
–30
–3.6
E1C02_12 10/23/2009
48
v
12
Figura
P
2-6-3
2525
ΩΩ
–
25

(c)
20
2.4
–30
–3.6
PP
2.6-3
1C02_1 32
10/23/2009
48 Figure
4
Figure
2.6-3
Voltmeter
–
Voltímetro
P 2.6-4 Un amperímetro
cor25 Ω
Voltmeter
50 está modelado como
6.0 ideal
2.4 un20
6
60
Figure
2.6-3 más real de un amperímetro
tocircuito.
Un P
modelo
es
una
reVoltmeter
50
6.0
(a)
(b) Pis 2.6-4a
(b)
+
P P2.6-4
AnAnideal
ammeter
asasa ashort
circuit.
AA
sistencia
La
figura
muestra
unshort
circuito
con
1212
V V+– +
2.6-4pequeña.
ideal
ammeter
ismodeled
modeled
circuit.
2
A
12
V
2A
2A
– –
(a)
(b)
un
amperímetro
que
la
corriente
. En
laafigura
2.6-4b,
more
realistic
model
ofmide
an
ammeter
a small
resistance.
P PA
+
P
2.6-4
An
ideal
ammeter
is ismodeled
asresistance.
shortFigure
circuit.
more
realistic
model
of
an
ammeter
is
aimsmall
Figure
Figure P 2.2-2
12 V –
2A
el
amperímetro
sido
reemplazado
por
elmeasures
modelo
dethe
un
am- P
2.6-4a
a circuit
with
an
ammeter
that
the
current
as shown
moreshows
realistic
model
of
ammeter
is
athat
small
resistance.
Figure
2.6-4a
shows
aha
circuit
with
an
ammeter
measures
current
Figure
Pan
2.2-2
perímetro
ideal,
un
cortocircuito.
Idealmente,
a
través
del
rei
.
In
Figure
P
2.6-4b,
the
ammeter
is
replaced
by
the
model
of
an
mi 2.6-4a
A linear
voltage
v and
current
ithat
as shown
nt has voltagePv2.2-3
and current
shown
esponding
showsPhas
a2.6-4b,
circuit
with
an ammeter
measures
the current
Inelement
Figure
the
ammeter
is replaced
by the model
of an
mi.as
(a)(a)
48sistor
Circuit
Elements
1ammeter,
kV
no aP
hay
voltaje,
ylinear
eli Ideally,
amperímetro
mide
imimodel
5
2across
A,
(a)
ideal
short
circuit.
Ideally,
ishas
no
voltage
P short
2.2-3
Aammeter
element
voltage
vacross
and
i as shown
in Figure
2.2-3a.
Values
of
the
current
and
corresponding
ues
of the current
i andPcorresponding
P 2.2-2b.
im.ammeter,
In
Figure
the
isthere
replaced
by
the
ofcurrent
an
ideal
a2.6-4b,
circuit.
there
is
no
voltage
48
Circuit
Elements
el
valor
ideal
de
i
.
En
la
figura
P
2.6-4c,
el
amperímetro
está
the
1-kV
resistor,
and
the
ammeter
measures
i
¼
2
A,
the
ideal
(a)
m
mi
ina and
Figure
P 2.2-3a.
Values
of
current
iideal
and corresponding
voltage
v have
tabulated
as the
shown
inIdeally,
Figure
P 2.2-3b.
bulated
in Figure
Pbeen
2.2-2b.
es v as as
a shown
ideal
ammeter,
short
circuit.
there
no
the
1-kV
resistor,
ammeter
measures
iis
¼ 2voltage
A, the across
mithe
2525
Ω ΩiiRR i
25

. the
Ahora,
la corriente
medida
por
modelado
por
la
resistencia
R
value
of
i
.
In
Figure
P
2.6-4c,
ammeter
is
modeled
by
the
m
m
voltage
v
have
been
tabulated
as
shown
in
Figure
P
2.2-3b.
Represent
the
element
by
an
equation
that
expresses
v
as
a
by
an equation
that expresses
v
as
a
(a) Verify
the
1-kV
resistor,
and
the
ammeter
measures
i
¼
2
A,
the
ideal
value
of
i
.
In
Figure
P
2.6-4c,
the
ammeter
is
modeled
by
the
ideal value ofmvm. In Figure P 2.6-3c, the voltmeter mi
is modeled
(a) Express the measurement error thatR occurs when Rm ¼
el amperímetro
es
25 Ω iR
..P
Now
the
current
measured
by
the
ammeter
is
Rm
Represent
the
element
by
an
equation
that
expresses
on is a modelideal
of the
element.
(a)
Verify
value
of
i
In
Figure
P
2.6-4c,
the
ammeter
is
modeled
by
the
Now
the
current
measured
by
the
ammeter
is
value
ofresistance
vresistance
.
In
Figure
2.6-3c,
the
voltmeter
is
modeled
by
(a)
Express
the
error
when Rm ¼
.
Now
the
voltage
measured
by
the
voltmeter
is
. occurs
the
R
10measurement
Vv asasaapercent
of ithat
mm
+
m
mi+
+vvRRvR–– –
++
+
11000
by the
the current
measured
ammeter
is
000
by
the
voltmeter
is
.
the resistance Rresistance
10
V
as
a
percent
of
i
(b)
Determine
the
maximum
value
of
R required
to ensure
m. Nowmeasured
m. Now theRvoltage
mi
Rm1000 2
+
+ vR – m v

+
vmiim
+
¼
2
12
V
m¼
2
A
+ 5 percent.
12
m
–of
v
12
V
¼
2
i
(b)
Determine
the
maximum
value
R
required
to
ensure
2
A
that
the
measurement
error
is
smaller
than
v
12
V
1
000

R
m
2
A
m
m
–
m
m
–
1000
þ
R
Rm
Rm 1000
þ 100
1000
þ mRm
+ i
vm ¼
2
im ¼12
that the measurement error 12
is smaller
V – ifs isthan 25 Apercent.
–– –vm
R
þvoltmeter
100
m amperímetro
1000
þ
R
m
Como
R
0,
el
se
convierte
en
un
ideal,
e
i
0,
the
ammeter
becomes
an
ideal
ammeter,
and
i
!
AsAs
RmR!
Because
Rm
!
1,
the
becomes
an
ideal
voltmeter,
and
m m
m
im is
m ! 0, the ammeter becomes an ideal ammeter, and im !
–
i¼
5
A.
Rm
.
0,ammeter
el
no
esimideal,
Because Rmvm!
1,
the
voltmeter
an
ideal
voltmeter,
and
imi!
2mi
A.
When
Rmammeter
>>
0,
the
ammeter
is is
not
ideal,
and
imiie. !
v¼
l2
VCuando
. When
R
<
1,
theamperímetro
voltmeter
is
not
ideal,
and
mi
m
im
(b)
!
0,
the
becomes
an
ideal
ammeter,
and
As
R
imi
2m¼2A.
When
Rbecomes
0,
the
not
ideal,
and
i<
m
m <im
mi.
(b)
(b)
Ammeter
,
i
.
La
diferencia
entre
i
e
i
es
un
error
de
medición
vm ! vmi ¼vmiThe
l2
V
.
When
R
<
1,
the
voltmeter
is
not
ideal,
and
difference
between
i
and
i
is
a
measurement
error
v
.
The
difference
between
v
and
v
is
a
measurement
m
m<
mi
m
mi
m
mi
mi
m
mi
imi ¼difference
2 A. Whenbetween
Rm > 0, the
ammeter
ideal, and im error
< imi.
The
im and
imi isis anot
measurement
(b)
PP
2.6-5
Ammeter
causado
por
hecho
de
que
amperímetro
sea
no ideal. error Figure
Figura
2.6-5
difference
between
vthat
and
vammeter
is a measurement
vm < vmi. The
caused
by
fact
that
ammeter
is not
error
caused
byel
the
fact
the
voltmeter
is
not
ideal.
Figure
2.6-5
mthe
mi
The
difference
between
iel
isnot
aideal.
measurement
caused
bythe
the
fact
that
the
ideal.
m and imi is
1 kΩ
error caused(a)byExpress
the factby
that
the
voltmeter
isammeter
not
ideal.
Figure P2 A
2.6-5
caused
the
fact
that the
is notwhen
ideal.
the
measurement
error
that
occurs
Rm ¼ 900
1 kΩ
2A
Alfaomega
Circuitos Eléctricos - Dorf
(a) Express theVmeasurement
that
as a percent error
of vmi
. occurs when Rm ¼ 900
V as a(b)
percent
of vmithe
. minimum value of Rm required to ensure that
Determine
(a)
(b) Determine the minimum
value
of Rismsmaller
requiredthan
to ensure
that of vmi.
measurement
error
2 percent
(a)
the measurement error is smaller than 2 percent of vmi.
M02_DORF_1571_8ED_SE_020-052.indd 48
imi = 2 A
imi = 2 A
4/12/11 5:18 PM
Problems
Problems
Problemas
Problems
Dado
que
Given
that
Given
that
Given
that
and
and
yand
que
2 2 .. . 0
00 0
00
2
Ammeter
Ammeter
Ammeter
Amperímetro
vvRRþ
vvmmand
and
iiRR¼
iis¼
¼22A22A
1212
¼v¼
vRR
þvþ
and
�i�
¼i¼
AA
1212

12¼



A
m
R
Rv
m
m
RiiR
fs ii
f 2
y �
vvm
ss ¼
25i
¼¼25i
25i
RR
vvRRvR¼
R
8 8 .. . 0
000
00
8
Voltmeter
Voltmeter
Voltmeter
Voltímetro
i
RR iiaa a
R
(a)
elthe
valor
del
medido
por el by
medidor.
(a)Determine
Determinethe
value
thevoltage
voltage
measured
themeter.
meter.
(a)
Determine
the
value
ofofvoltaje
the
voltage
measured
the
meter.
(a)
Determine
value
of
the
measured
bybythe
(b)
D
etermine la
potencia
alimentada
por
cada
elemento.
(b)
Determine
the
power
supplied
by
each
element.
(b)
Determine
the
power
supplied
by
each
element.
(b) Determine the power supplied by each element.
P
amperímetro
la figura
P2.6-6a
mide la
corriente
2.6-6El
The
ammeterin
Figure
2.6-6a
measures
the
current
PP2.6-6
2.6-6
The
ammeter
ininde
Figure
PP2.6-6a
2.6-6a
measures
the
current
P
2.6-6
The
ammeter
Figure
P
measures
the
current
en
la
fuente
de
voltaje.
La
figura
2.6-6b
muestra
el
circuito
in
the
voltage
source.
Figure
P
2.6-6b
shows
the
circuit
after
in the
the voltage
voltage source.
source. Figure
Figure P
P 2.6-6b
2.6-6b shows
shows the
the circuit
circuit
after
in
after
luego
de
eliminar
el
amperímetro
y
etiquetar
la
corriente
meremoving
the
ammeter
and
labeling
the
current
measured
removing the
the ammeter
ammeter and
and labeling
labeling the
the current
current measured
measured by
byby
removing
dida
por
el amperímetro
como the
im. Incluso,
los demás
voltajesand
y
.
Also,
other
element
voltages
the
ammeter
as
i
m
.
Also,
the
other
element
voltages
and
the
ammeter
as
i
m. Also, the other element voltages and
the ammeter
as im
corrientes
del
elemento
están
etiquetados
en
la
figura
P
2.6-6b.
currentsare
arelabeled
labeledin
FigureP
2.6-6b.
currents
are
labeled
ininFigure
Figure
PP2.6-6b.
2.6-6b.
currents
Dado
que
Given
that
Given
that
Given
that
and
and
yand
que
4949
49
49
vfsvf 
i ¼¼iiRiRand
¼
vs¼12
¼
vvRvvRRv


12
V
VVV
R¼v
yand
þþiim
and
1212
mm¼
22 2þ
R
R ¼ vss ¼ 12 V
25i
¼¼25i
25i
RR
vvRRvR¼
R
(a)
Determine
the
value
of
the
current
measuredby
themeter.
meter.
(a) Determine
Determine the
the value
value of
of the
the current
current measured
measured
bybythe
the
meter.
(a)
(a)
Determine
el
valor
de la
corriente
medida
por el medidor.
(b)
Determine
the
power
supplied
by
each
element.
(b)
Determine
the
power
supplied
by
each
element.
(b) Determine the power supplied by each element.
(b)Determine la potencia alimentada por cada elemento.
Ammeter
Ammeter
Ammeter
Amperímetro
+ + v vs
+
– v ss
––
ri
rr iiaa a
++
+
––
–
++
+
v
vvbb b
–
––
FigureP
2.7-1
Figura
Figure
PP2.7-1
2.7-1
Figure
P
2.7-1
P 2.7-2 El amperímetro en el circuito que se muestra en la
2.7-2The
Theammeter
ammeterin
thecircuit
circuitshown
shownin
FigureP
2.7-2
PP2.7-2
2.7-2
The
ammeter
ininithe
the
ininFigure
Figure
PP2.7-2
2.7-2
P
figura
P 2.7-2
indica que
2 A, y shown
el voltímetro
indica
que
a 5circuit
indicates
that
i
¼
2
A,
and
the
voltmeter
indicates
that
a
indicates
that
i
¼
2
A,
and
the
voltmeter
indicates
that
v
¼¼
indicates
iaa ¼ 2 el
A,valor
and the
voltmeter
indicates
that vbbvb¼
vb 5 8 V. that
Determine
de g,
la ganancia
de la VCCS.
8
V.
Determine
the
value
of
g,
the
gain
of
the
VCCS.
8
V.
Determine
the
value
of
g,
the
gain
of
the
VCCS.
8 V. Determine the value of g, the gain of the VCCS.
Respuesta:
g¼5
0.25
A/V
Answer:gg g¼
0.25
A/V
Answer:
¼
0.25
A/V
Answer:
0.25
A/V
2 2 .. . 0
00 0
00
2
Ammeter
Ammeter
Amperímetro
Ammeter
8 8 .. . 0
00 0
00
8
Voltmeter
Voltmeter
Voltímetro
Voltmeter
i
iiaa a
RR
11
R
1
gv vb
b
gg v
b
vsvs ++ +–
v
s –
–
RR
22
R
2
++
+
vbvb
v
b
–
––
Figura
FigureP
2.7-2
Figure
PP2.7-2
2.7-2
Figure
P
2.7-2
+
+
12V V ++
12
––
12
12
VV ––
25ΩΩ
25
25
Ω
25

P
2.7-3
ElThe
amperímetro
en
elcircuit
circuito
que in
se Figure
muestra
en2.7-3
la
2.7-3The
ammetersin
the
circuit
shown
FigureP
PP2.7-3
2.7-3
The
ammeters
inin
the
shown
PP2.7-3
2.7-3
P
ammeters
the
circuit
shown
ininFigure
figura
P that
2.7-3
indica
que
iai 5
32
AA.Determine
e Determine
ib 5 8 A.the
Determine
eld,
indicate
that
i
¼
32
A
and
i
¼
8
the
value
of
a
b
indicate
i
¼
32
A
and
¼
8
A.
value
of
d,
indicate that iaa ¼ 32 A and ibb ¼ 8 A. Determine the value of d,
valor
de d,
ganancia
thegain
gain
the
CCCS.de CCCS.
the
gain
ofoflathe
the
CCCS.
the
of
CCCS.
Answer:dd d¼
Answer:
¼
A/A
Respuesta:
d¼5
4A/A
A/A
Answer:
44 4A/A
2 2A
AA
2
2
A
(a)
(a)
(a)
(a)
i
iimm
m
+ im
+
12V V ++
12
––
12
12
VV ––
25ΩΩ
25
25
Ω
25

i
RR
iiiR
R
++
+
+
vRvR
vvRR
––
––
2 2A
AA
2
2
A
++
+
+
vsvs
vvss
––
––
(b)
(b)
(b)
(b)
FigurePP2.6-6
2.6-6
Figure
Figure
P 2.6-6
2.6-6
Figura
P
Sección
2.7
Fuentes
dependientes
Section2.7
2.7Dependent
Dependent
Sources
Section
2.7
Dependent
Sources
Section
Sources
2.7-1El
The
ammeterin
the
circuit
shown
FigureP
P
2.7-1
amperímetro
elcircuit
circuito
que in
seinFigure
muestra
en2.7-1
la
PP2.7-1
2.7-1
The
ammeter
ininen
the
circuit
shown
in
Figure
PP2.7-1
2.7-1
P
The
ammeter
the
shown
indicates
thatiiindica
ia¼
¼22 2A,
A,and
the
voltmeter
indicates
thatvvque
vb¼
¼
figura
P 2.7-1
que
iand
2 A,
y el voltímetro
indica
indicates
that
¼
A,
and
the
voltmeter
indicates
that
¼
a 5
a
b
indicates
that
the
voltmeter
indicates
that
a
b
V.
Determine
the
value
of
r,
the
gain
of
the
CCVS.
v88b8V.
5
8
V.
Determine
el
valor
de
r,
la
ganancia
de
la
CCVS.
V. Determine
Determine the
the value
value of
of r,
r, the
the gain
gain of
of the
the CCVS.
CCVS.
Answer:
r
¼
4
V/A
Answer:
r
¼
4
V/A
Respuesta:
r
5
4
V/A
Answer: r ¼ 4 V/A
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 49
3 2 . . 00
3
3 2
2 . 0
Ammeter
Ammeter
Amperímetro
Ammeter
iiaaia
di
d
d iibb b
R
R
R
iibbib
8 . 0 00
8
8 .. 0
0 0
Ammeter
Ammeter
Amperímetro
Ammeter
v +
v
vss s +–+ –
–
Figura
FigureP
2.7-3
Figure
PP2.7-3
2.7-3
Figure
P
2.7-3
PP2.7-4
Los
voltímetros
en the
el circuito
que seinmuestra
en2.7-4
la
2.7-4The
Thevoltmeters
voltmetersin
circuitshown
shown
FigureP
P 2.7-4
2.7-4
The
voltmeters
ininthe
the circuit
circuit
shown
in Figure
Figure
PP2.7-4
2.7-4
P
in
figura
P
2.7-4
indican
que
v
5
2
V
y
que
v
5
8
V.
Determine
b
indicatethat
thatvvava¼
andvvabvb¼
Determine
thevalue
valueof
indicate
that
¼¼22 2V
VVand
and
¼¼88 8V.
V.V.Determine
Determine
the
value
ofofb,
b,b,
indicate
the
a ganancia
elthe
valor
deof
b, the
la
deb la VCVS.
gain
VCVS.
the
gain
of
the
VCVS.
the gain of the VCVS.
Answer:bb b¼
Respuesta:
b¼5
4V/V
V/V
Answer:
¼
V/V
Answer:
44 4V/V
Alfaomega
4/12/11 5:18 PM
Elementos de circuitos
50
2 . 0 0
240 6
Amperímetro
+
8 . 0 0
250 mA
10 6
Voltímetro
+
va
+
–
vs
b va
+
+
–
–10 V
–
50 mA
+
Figura P 2.7-7
vb
P 2.7-8 El circuito que se muestra en la figura 2.7-8 contiene
una fuente dependiente. Determine el valor de la ganancia k
de esa fuente dependiente.
R
+
–
k va
–
200 mA
–
va
–
Figura P 2.7-4
200 6
P 2.7-5 En la figura 2.7-5 se muestran los valores de la corriente y el voltaje de cada elemento del circuito.
Determine los valores de la resistencia, R, y de la ganancia de la fuente dependiente, A.
ia
+ 10 V –
+
+
–
20 V
k ia
20 6
10 V
–
450 mA
2V
ia = 0.5 A
4V
Figura P 2.7-8
A ia = 2 V
+
2A
10 V
R
+
–
P 2.7-9 El circuito que se muestra en la figura 2.7-9 contiene
una fuente dependiente. La ganancia de esa fuente dependiente es
–
3.5 A
1.5 A
+
–
14 V
4A
2.5 A
k ¼ 25
12 V
V
A
Determine el valor del voltaje vb.
Figura P 2.7-5
120 6
+
P 2.7-6 Encuentre la potencia alimentada por la VCCS en la
figura P 2.7-6.
Respuesta: La VCCS alimentó 17.6 watts (pero absorbió
⫺17.6 watts).
250 mA
56
–1 V
k ia
–
ia
+
–
+
vb
–
50 mA
Figura P 2.7-9
– 2. 0 0
Voltímetro
+ 2. 2 0
+
Voltímetro
–
vc
26
+
0.2 6
–15.8 V
+
–
6.9 6
id = 4vc
vd
–
Figura P 2.7-6
P 2.7-7 El circuito que se muestra en la figura P 2.7-7 contiene una fuente dependiente. Determine el valor de la ganancia
k de esa fuente dependiente.
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 50
P 2.7-10 El circuito que se muestra en la figura P 2.7-10
contiene una fuente dependiente. La ganancia de esa fuente
dependiente es
mA
A
k ¼ 90
¼ 0:09
V
V
Determine el valor de la corriente ib.
Circuitos Eléctricos - Dorf
5/24/11 9:50 AM
Problemas
+
+
–
va
Determine el valor de v para cada uno de los casos siguientes.
50 mA
100 6
–
10 V
51
+
k va
10 6
5V
–
ib
(a)
(b)
(c)
(d)
El interruptor está cerrado y Rs ⫽ 0 (un cortocircuito).
El interruptor está cerrado y Rs ⫽ 5 ⍀.
El interruptor está cerrado y Rs ⫽ 1 (un cortocircuito)
El interruptor está cerrado y Rs ⫽ 10 k⍀.
Rs
Figura P 2.7-10
Sección 2.8 Transductores
100 6
12 V
P 2.8-1 Para el circuito del potenciómetro de la figura 2.8-2,
la corriente de la fuente de corriente y la resistencia del potenciómetro son 1.1 mA y 100 k⍀, respectivamente. Calcule
el ángulo requerido, u, de modo que el voltaje medido sea
de 23 V.
P 2.8-2 Un sensor AD590 tiene una constante asociada k ⫽
1 mA
. El sensor tiene un voltaje v ⫽ 20 V. La corriente medida,
K
i(t), como se muestra en la figura 2.8-3, es 4 mA ⬍ i ⬍ 13 mA
en una fijación en laboratorio. Encuentre el rango de la temperatura medida.
Sección 2.9 Interruptores
v
100 6
12 V
(a)
v
(b)
Figura P 2.9-3
Sección 2.10 ¿Cómo lo podemos comprobar . . . ?
P 2.10-1 El circuito que se muestra en la figura P 2.10-1 se
utiliza para probar la CCVS. Su compañero de laboratorio
argumenta que esta medición muestra que la ganancia de la
CVVS es ⫺20 V/A en vez de ⫹ 20 V/A. ¿Está de acuerdo?
Justifique su respuesta.
P 2.9-1 Determine la corriente, i, en t ⫽ 1 s y en t ⫽ 4 s para
el circuito de la figura P 2.9-1.
– 2 . 0
4 0 . 0
Amperímetro
Voltímetro
t=2s
R
t=3s
15 V
+
–
5 k6
+
–
10 V
i
P 2.9-2 Determine el voltaje, v, en t ⫽ 1 s y en t ⫽ 4 s para el
circuito que se muestra en la figura P 2.9-1.
t=3s
+
5 k6
vs
is
vo
is
CCVS
= 20
V
A
+
vo
–
Figura P 2.10-1
Figura P 2.9-1
1 mA
+
–
t=2s
v
–
2 mA
P 2.10-2 El circuito de la figura P 2.10-2 se utiliza para medir la corriente en el resistor. Una vez conocida esta corriente,
se puede calcular la resistencia como R ¼ vis . El circuito se
construye utilizando una fuente de voltaje con vs ⫽ 12 V y
un resistor de 1/2 W de 25 ⍀. Luego de una voluta de humo y un
olor desagradable, el amperímetro indica que i ⫽ 0 A. El resistor debe estar mal. Se tienen más resistores de 1>2 W de 25 ⍀.
¿Debería probar con otro resistor? Justifique su respuesta.
0 . 0 0
Figura P 2.9-2
P 2.9-3 Idealmente, un interruptor abierto se modela como
un circuito abierto y un interruptor cerrado se modela como un
circuito cerrado. En la realidad, un interruptor abierto se modela como una resistencia grande, y un interruptor cerrado se
modela como una resistencia pequeña.
La figura P 2.9-3a muestra un circuito con un interruptor. En la figura 2.9-3b, el interruptor ha sido reemplazado por
una resistencia. En la figura P 2.9-3b, el voltaje v lo da
100
v¼
12
Rs þ 100
Circuitos Eléctricos - Dorf
M02_DORF_1571_8ED_SE_020-052.indd 51
Amperímetro
R
+
–
i
vs
Figura P 2.10-2
Sugerencia: Los resistores de 1>2 W son capaces de disipar
una potencia de 1>2 W. Pero pueden fallar si se requiere que
disipen más de 1>2 W de potencia.
Alfaomega
5/24/11 9:51 AM
Elementos de circuitos
52
Problemas de diseño
PD 2-1 Especifique la resistencia R en la figura PD 2-1 de
modo que se cumpla con las dos siguientes condiciones:
Sugerencia: No hay una garantía de que siempre se cumpla con
esas especificaciones.
1.
i . 40 mA.
2.
La potencia absorbida por el resistor es menor de 0.5 W.
PD 2-3 A los resistores se les da una potencia nominal. Por
ejemplo, hay resistores de 1>8 W, 1>4 W, 1>2 W, y de 1 W. Un
resistor de 1>2 W es capaz de disipar indefinidamente con seguridad 1>2 W de potencia. Los resistores con potencia nominal
más grande son más costosos y voluminosos que los de menor
potencia nominal. Una buena práctica de ingeniería requiere
que las potencias nominales de los resistores se especifiquen lo
más grande posible, pero no más de lo necesario.
Considere el circuito que se muestra en la figura PD 2-3.
Los valores de las resistencias son
i
10 V
+
–
R
Figura PD 2-1
R1 5 1 000 V, R2 5 2 000 V, y R3 5 4 000 V
PD 2-2 Especifique la resistencia R en la figura PD 2-2 de
modo que se cumpla con las dos siguientes condiciones:
1.
v . 40 V.
2.
La potencia absorbida por el resistor es menor de 15 W.
El valor de la corriente de la fuente de corriente es
is 5 30 mA
Especifique la potencia nominal para cada resistor.
2A
R
R1
R2
R3
ir = is
+
v
–
is
Figura PD 2-2
Alfaomega
M02_DORF_1571_8ED_SE_020-052.indd 52
Figura DP 2-3
Circuitos Eléctricos - Dorf
4/12/11 5:19 PM
CAPÍTULO
Circuitos
resistivos
EN ESTE CAPÍTULO
3.1
3.2
3.3
3.4
3.5
3.6
3.1
Introducción
Leyes de Kirchhoff
Resistores en serie y división de voltaje
Resistores en paralelo y división de corriente
Fuentes de voltaje en serie y fuentes de
corriente en paralelo
Análisis de circuitos
3.7
3.8
3.9
3.10
nálisis de circuitos resistivos utilizando
A
MATLAB
¿Cómo lo podemos comprobar . . . ?
EJEMPLO DE DISEÑO — Fuente de
voltaje ajustable
Resumen
Problemas
Problemas de diseño
INTRODUCCIÓN
En este capítulo haremos lo siguiente:
•Escribir ecuaciones utilizando las leyes de Kirchhoff.
No es de sorprender que el comportamiento de un circuito eléctrico esté determinado por los
tipos de elementos que comprenden el circuito y por aquellos elementos con los que está conectado.
Las ecuaciones constitutivas describen los elementos en sí mismos, y las leyes de Kirchhoff describen
la manera en que los elementos están conectados entre sí para conformar el circuito.
•Analizar circuitos eléctricos sencillos, utilizando solamente las leyes de Kirchhoff y las ecuaciones
constitutivas de los elementos del circuito.
• Analizar dos configuraciones de circuito muy comunes: los resistores en serie y los resistores en
paralelo.
Veremos que los resistores en serie actúan como un “divisor de voltaje”, y los resistores en serie
lo hacen como un “divisor de corriente”. Además, los resistores en serie y los resistores en paralelo
nos proporcionan nuestros primeros ejemplos de un “circuito equivalente”. La figura 3.1-1 ilustra este
importante concepto. En este punto, un circuito se ha dividido en dos partes, A y B. El reemplazo de B
por un circuito equivalente, Beq, no modifica la corriente o el voltaje de ningún elemento del circuito
en la parte A. En este sentido, Beq es equivalente a B. Veremos cómo obtener un circuito equivalente
cuando la parte B consta de resistores en serie y resistores en paralelo.
•Determinar circuitos equivalentes para fuentes de voltaje en serie y fuentes de corriente en paralelo.
•Determinar la resistencia equivalente a un circuito resistivo.
A veces, los circuitos que están constituidos totalmente por resistores se pueden reducir a un
resistor equivalente único al reemplazar repetidamente resistores en serie y/o en paralelo por resistores
equivalentes.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 53
53
Alfaomega
4/12/11 5:20 PM
54
Circuitos resistivos
3.2
LEYES DE KIRCHHOFF
Un circuito eléctrico consta de elementos de circuito que están conectados entre sí. Los lugares en que
los elementos están conectados entre sí se llaman nodos. La figura 3.2-1a muestra un circuito eléctrico que consta de seis elementos conectados entre sí en cuatro nodos. Es una práctica muy común
trazar circuitos utilizando líneas rectas y posicionar los elementos horizontal o verticalmente, como
se muestra en la figura 3.2-1b.
FIGURA 3.1-1 Reemplazar B
por un circuito equivalente Beq
no modifica la corriente ni el
voltaje de ningún elemento de
circuito en A.
El circuito se muestra de nuevo en la figura 3.2-1c, esta vez remarcando los nodos. Observe que
al trazar de nuevo el circuito, mediante líneas rectas y elementos horizontales y verticales, ha cambiado la forma de representar los nodos. En la figura 3.2-1a los nodos están representados como puntos.
En las figuras 3.2-1b,c, los nodos están representados tanto por puntos como por segmentos de línea.
FIGURA 3.2-1 (a) Un circuito eléctrico.
(b) El mismo circuito, pero con un trazo
nuevo, utilizando líneas rectas y elementos
horizontales y verticales. (c) El circuito
después de etiquetar los nodos y los
elementos.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 54
Circuitos Eléctricos - Dorf
4/12/11 5:20 PM
Leyes de Kirchhoff
55
El mismo circuito puede trazarse de varias formas. Un dibujo de un circuito puede verse muy
diferente de otro trazo del mismo circuito. ¿Cómo podemos discernir cuándo dos trazos de circuitos
representan el mismo circuito? De manera informal, decimos que dos trazos de circuitos representan
el mismo circuito si los elementos correspondientes están conectados a sus nodos correspondientes.
De manera más formal, decimos que los trazos de los circuitos A y B representan el mismo circuito
cuando se cumplen las tres condiciones siguientes.
1. H
ay una correspondencia de uno a uno entre los nodos del trazo A y los nodos del trazo B. (Una
correspondencia de uno a uno es igual a un emparejamiento. En esta correspondencia de uno
a uno, cada nodo en el trazo A empata exactamente con un nodo del trazo B y viceversa. La
posición de los nodos no es importante.).
2. Hay una correspondencia uno a uno entre los elementos del trazo A y los del B.
3. Los elementos correspondientes están conectados a los nodos correspondientes.
Ejemplo 3.2-1
Trazos diferentes del mismo circuito
La figura 3.2-2 muestra cuatro trazos de circuitos. ¿Cuál de éstos representa el mismo circuito trazado en la figura
3.2-1c?
FIGURA 3.2-2 Cuatro trazos de circuitos.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 55
Alfaomega
4/12/11 5:20 PM
E1C03_1
11/25/2009
56
56
56
56
Circuitos Circuits
resistivos
Resistive
Resistive Circuits
Solution
Solución
Solution
The circuit
drawing shown inenFigure
3.2-2a
has five cinco
nodes, labeled
r, s, t, u, and
v.
The
circuit
drawing
in Figure
El
de circuito
la figura
3.2-2a
etiquetados
r, s, v.
t, uThe
y v.circuit
El trazo
de circuito
de la
Thetrazo
circuit
drawingmostrado
shown in Figure
3.2-2a
hastiene
five nodes,nodos,
labeled
r, s, t, u, and
drawing
in Figure
3.2-1c
has
four
nodes.
Because
the
two
drawings
have
different
numbers
of
nodes,
there
cannot
be
a
one-to-one
figura
3.2-1c
tiene
cuatro
nodos.
Dado
que
los
dos
trazos
tienen
diferente
cantidad
de
nodos,
no
puede
haber
una
corres3.2-1c has four nodes. Because the two drawings have different numbers of nodes, there cannot be a one-to-one
correspondence
thelos
nodes
these los
drawings
represent different
pondencia
de unobetween
a uno entre
nodosofdethe
lostwo
dos drawings.
trazos. PorHence,
consiguiente,
trazos representan
circuitos circuits.
diferentes.
correspondence between the nodes of the two drawings. Hence, these drawings represent different circuits.
The
circuit
drawingmostrado
shown inenFigure
3.2-2b
hastiene
four cuatro
nodes nodos
and sixy elements,
the same
numbers
of nodes
and
El
trazo
de circuito
la figura
3.2-2b,
seis elementos,
el mismo
número
de nodos
The circuit drawing shown in Figure 3.2-2b has four nodes and six elements, the same numbers of nodes and
as theque
circuit
drawing
in Figure
The
nodesLos
in Figure
3.2-2b
have3.2-2b
been labeled
in the samedeway
as the
yelements
de elementos
el trazo
del circuito
de3.2-1c.
la figura
3.2-1c.
nodos de
la figura
se han etiquetado
la misma
elements as the circuit drawing in Figure 3.2-1c. The nodes in Figure 3.2-2b have been labeled in the same way as the
manera
que losnodes
nodosincorrespondientes
de la
figura 3.2-1c.
ejemplo,
el nodo
c en la figura
3.2-2b
al
corresponding
Figure 3.2-1c. For
example,
node cPor
in Figure
3.2-2b
corresponds
to node
c in corresponde
Figure 3.2-1c.
corresponding nodes in Figure 3.2-1c. For example, node c in Figure 3.2-2b corresponds to node c in Figure 3.2-1c.
nodo
c de la figura
3.2-1c.
Los have
elementos
de la figura
han
de la misma
manera
que los3.2-1c.
elementos
The elements
in Figure
3.2-2b
been labeled
in the3.2-2b
same se
way
asetiquetado
the corresponding
elements
in Figure
For
The elements in Figure
3.2-2b have been
labeled
theelemento
same way as the
corresponding
elements inalFigure
3.2-1c.
correspondientes
la figura
ejemplo,into
el
la figura
3.2-2b
corresponde
elemento
deFor
la
example, elementde5 in
Figure3.2-1c.
3.2-2b Por
corresponds
element 5 5inen
Figure
3.2-1c.
Corresponding
elements
are 5indeed
example,
element
5
in
Figure
3.2-2b
corresponds
to
element
5
in
Figure
3.2-1c.
Corresponding
elements
are
indeed
figura
3.2-1c.
Los elementosnodes.
correspondientes,
están2 is
conectados
nodos
correspondientes.
Por3.2-2b
ejemplo,
el
connected
to corresponding
For example,pues,
element
connectedatolosnodes
a and
b, in both Figure
and in
connected2toestá
corresponding
For aexample,
element
2 is connected
to nodesEn
a and b, in both Figure
3.2-2b
and iny
elemento
conectado anodes.
losFigure
nodos
y b, en
y 3.2-1c.
las figuras
3.2-2b
Figure 3.2-1c. Consequently,
3.2-2b
andambas
Figurefiguras
3.2-1c3.2-2b
represent
the sameconsecuencia,
circuit.
Figure
3.2-1c.
Consequently,
Figure
3.2-2b
and
Figure
3.2-1c
represent
the
same
circuit.
3.2-1c The
representan
el mismoshown
circuito.
circuit drawing
in Figure 3.2-2c has four nodes and six elements, the same number of nodes and
Thetrazo
circuit drawing
shown
in Figureen3.2-2c
has3.2-2c
four nodes
and
six nodos
elements,
theelementos,
same number
of nodes
and
El
circuito
que
seinmuestra
la figura
tiene
cuatro
y seis
el mismo
número
elements
as thedel
circuit
drawing
Figure 3.2-1c.
The nodes
and
elements
in Figure
3.2-2c
have been
labeled
in the
elements
aselementos
the circuitque
drawing
in Figure
3.2-1c.
The
nodes3.2-1c.
and elements
in Figure
3.2-2cde
have
been labeled
in han
the
de
nodos
y
el
trazo
del
circuito
de
la
figura
Los
nodos
y
elementos
la
figura
3.2-2c
se
same way as the corresponding nodes and elements in Figure 3.2-1c. Corresponding elements are indeed
same way de
aslathe
corresponding
nodes
and yelements
in Figure
3.2-1c.3.2-1c.
Corresponding
elements
are indeed
etiquetado
misma
manera
que
los
nodos
los
elementos
de
la
figura
Los
elementos
correspondientes,
connected to corresponding nodes. Therefore, Figure 3.2-2c and Figure 3.2-1c represent the same circuit.
connected
to
corresponding
nodes.
Therefore, Figure
3.2-2c and Figure
3.2-1c
represent
therepresentan
same circuit.
pues,
están
conectados
a losshown
nodos
correspondientes.
Enfour
consecuencia,
las elements,
figuras
3.2-2c
y 3.2-1c
el misThe circuit
drawing
in
Figure 3.2-2d has
nodes and six
the same
numbers
of nodes
and
The circuit drawing shown in Figure 3.2-2d has four nodes and six elements, the same numbers of nodes and
mo circuito.
elements as the circuit drawing in Figure 3.2-1c. However, the nodes and elements of Figure 3.2-2d cannot be
elements
as the
drawing
in Figure
thecuatro
nodesnodos
and elements
of Figure
3.2-2d número
cannot be
El trazo
delcircuit
circuito
que se muestra
en 3.2-1c.
la figuraHowever,
3.2-2d tiene
y seis elementos,
el mismo
de
labeled so that corresponding elements of Figure 3.2-1c are connected to corresponding nodes. (For example, in
nodos
que el trazo del
circuitoof
de Figure
la figura3.2-1c
3.2-1c.are
Sinconnected
embargo, los
nodos y los elementos
la figura
3.2-2d
labeledy elementos
so that corresponding
elements
to corresponding
nodes. de
(For
example,
in
Figure 3.2-1c, three elements are connected between the same pair of nodes, a and b. That does not happen in
no
se han
podidothree
etiquetar
de modo
los elementos
correspondientes
figura a3.2-1c
estén
conectados
los nodos
Figure
3.2-1c,
elements
are que
connected
between
the same pairdeoflanodes,
and b.
That
does not ahappen
in
Figure 3.2-2d.) Consequently, Figure 3.2-2d and Figure 3.2-1c represent different circuits.
correspondientes.
(Por ejemplo, en
la figura
3.2-1c,
elementos
conectados
entre
el mismo par de nodos, a y
Figure 3.2-2d.) Consequently,
Figure
3.2-2d
and tres
Figure
3.2-1c están
represent
different
circuits.
b, lo que no sucede en la figura 3.2-2d.) En consecuencia, las figuras 3.2-2d y 3.2-1c representan circuitos diferentes.
FIGURE 3.2-3 Gustav
FIGURE 3.2-3 Gustav
Robert Kirchhoff (1824FIGURA
3.2-3 Gustav
Robert Kirchhoff
(18241887). Kirchhoff stated
Robert
1887). Kirchhoff
Kirchhoff stated
two laws in 1847
(1824-1887).
En 1847,
two laws in 1847
regarding the current and
Kirchhoff
regarding estableció
the current and
voltage
inrespecto
an electrical
dos
leyesin
de
voltage
an electrical
circuit.
Courtesy
of
la
corriente
y el voltaje
circuit. Courtesy of
theun
Smithsonian
en
circuito eléctrico.
the Smithsonian
Cortesía
de la Institución
Institution.
Institution.
Smithsonian.
In 1847, Gustav Robert Kirchhoff, a professor at the University of Berlin, formulated
In 1847, Gustav Robert Kirchhoff, a professor at the University of Berlin, formulated
two important laws that provide the foundation for analysis of electric circuits. These laws are
two important
laws that
provide
the foundation
for la
analysis
of electric
circuits.
These
laws are
En 1847,
Gustav
Robert
Kirchhoff,
profesor
universidad
de Berlín,
formuló
referred
to as Kirchhoff’s
current
law (KCL)
anddeKirchhoff’s
voltage
law (KVL)
indos
his imporhonor.
referred
to que
as Kirchhoff’s
current
law (KCL)
and
Kirchhoff’s
voltage law
(KVL)Estas
in hisleyes
honor.
tantes
leyes
constituyen
los
fundamentos
del
análisis
de
los
circuitos
eléctricos.
se
Kirchhoff’s laws are a consequence of conservation of charge and conservation of energy.
Kirchhoff’s
laws are
a consequence
of conservation
of
charge
and
conservation
of yenergy.
refieren
en
su
honor
como
ley
de
la
corriente
de
Kirchhoff
(KCL,
por
sus
siglas
en
inglés),
ley
del
Gustav Robert Kirchhoff is pictured in Figure 3.2-3.
GustavdeRobert
Kirchhoff
is pictured
in
Figure 3.2-3.
voltaje
Kirchhoff
(KVL). Las
Kirchhoff
son consecuencia de la carga y la conservación
Kirchhoff’s
current
law leyes
statesde
that
the algebraic
sum of the currents entering any node is
Kirchhoff’s
current
law states
that the
algebraic
sum en
of the
currents
entering any node is
de la energía.
La imagen
de Gustav
Robert
Kirchhoff
aparece
la figura
3.2-3.
identically zero for all instants of time.
La ley zero
de lafor
corriente
de Kirchhoff
identically
all instants
of time.establece que la suma algebraica de las corrientes que
entran en cualquier nodo es idéntica a cero en todo momento.
Kirchhoff’s current law (KCL): The algebraic sum of the currents into a node at
Kirchhoff’s current law (KCL): The algebraic sum of the currents into a node at
Ley
la corriente
anyde
instant
is zero.de Kirchhoff (KCL): La suma algebraica de las corrientes en un
any es
instant
is todo
zero.instante.
nodo
cero en
The phrase algebraic sum indicates that we must take reference directions into account as
The
phrase
sum
indicates
we tomar
must take
reference
directions into
account as
La
frase
sumaalgebraic
algebraica
indica
que sethat
deben
en cuenta
las direcciones
de referencia
al
we add up the currents of elements connected to a particular node. One way to take
we addlasup
the currents
elements
connected
a particular
node.
One
waydetotomar
take
agregar
corrientes
de los of
elementos
conectados
a untonodo
en particular.
Una
manera
reference directions into account is to use a plus sign when the current is directed away from
enreference
cuenta lasdirections
direcciones
deaccount
referencia
esuse
utilizar
unsign
signo
más the
cuando
la corriente
se dirige
into
is to
a plus
when
current
is directed
away hacia
from
the node and a minus sign when the current is directed toward the node. For example,
fuera
del nodo,
signo sign
menoswhen
cuando
corrienteissedirected
dirige hacia
el nodo.
ejemplo,
considethe node
andyaunminus
thelacurrent
toward
the Por
node.
For example,
consider
the circuit shown
in Figure
3.2-1c.
Four
elements
of
this
circuit—elements
1, 2,
3,
reconsider
el circuito
en la figura
3.2-1c.
Cuatro
elementos
de of
este
circuito
—los elementos
1, 3,
2,
themostrado
circuit shown
in Figure
3.2-1c.
Four
elements
this
circuit—elements
1, 2,
and
4—are
connected
to
node
a.
By
Kirchhoff’s
current
law,
the
algebraic
sum
of
the
3and
y 4—
están
conectados
al
nodo
a.
Según
la
ley
de
la
corriente
de
Kirchhoff,
la
suma
algebraica
4—are connected to node a. By Kirchhoff’s current law, the algebraic sum of the
i4 must be izero.
Currents
i andcero.
i are directed
away
from
node
a, so
we
element currents
i , i , i3, and
corrientes
i2 efrom
i3 senode
dirigen
hacia
las corrientes
los ielementos
1, i2, iCurrents
3 e i4 debe
, and
i22ser
and i33 areLas
directed
away
a, so
we
elementdecurrents
i11, i22, i3de
4 must be zero.
and
i
.
In
contrast,
currents
i
and
i
are
directed
toward
node
a,
so
we
will
will usefuera
a plus
sign
for
i
2
3
1
4
las corrientes
i1 e
delsign
nodo
lo icual
signo más
para
2 e idirected
3. Por el contrario,
and
. Inusaremos
contrast, un
currents
i1 and
i iare
toward node
a, so we will
will use a plus
fora,ipor
i43. usaremos
The KCLel equation
for
nodei41 ae iof
Figure
3.2-1c
is KCL para el nodo
use
a minusalsign
for
i1 2and
iuse
nodo
a,
por
lo
que
signo
menos
para
.
La
ecuación
de
la
4 se adirigen
4
minus sign for i1 and i4. The KCL equation for node a of Figure 3.2-1c is
a de la figura 3.2-1c es
�i þ i þ i � i ¼ 0
(3.2-1)
�i11 þ i22 þ i33 � i44 ¼ 0
(3.2-1)
An
waypara
of obtaining
algebraic
sum de
of the
is es
to set
the sumque
of all
Otraalternate
alternativa
obtener lathe
suma
algebraica
las currents
corrientesinto
en aunnode
nodo
establecer
la
An alternate way of obtaining the algebraic sum of the currents into a node is to set the sum of all
the
currents
directed
away from
the
nodedel
equal
toes
theigual
sumaof
thede
currents
directed
toward
node.
suma
de todas
las corrientes
que se
alejan
nodo
la all
suma
todas las
corrientes
que that
se dirigen
the currents directed away from the node equal to the sum of all the currents directed toward that node.
a ese nodo.
esta técnica
encontramos
que laequation
ecuaciónfor
de la
KCL
el nodo
de la figura
3.2-1c es
Using
this Con
technique,
we find
that the KCL
node
a para
of Figure
3.2-1c
is
Using this technique, we find that the KCL equation for node a of Figure 3.2-1c is
(3.2-2)
i2 þ i3 ¼ i1 þ i4
(3.2-2)
i2 þ i3 ¼ i1 þ i4
Desde luego,
ecuaciones
3.2-1
3.2-2 son equivalentes.
Clearly,
Eqs.las
3.2-1
and 3.2-2
arey equivalent.
Clearly, Eqs. 3.2-1 and 3.2-2 are equivalent.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 56
Circuitos Eléctricos - Dorf
4/12/11 5:20 PM
E1C03_1
E1C03_1
11/25/2009
11/25/2009
57
57
Kirchhoff’s Laws
Leyes de Kirchhoff
Kirchhoff’s Laws
Kirchhoff’s Laws
57
57
57
57
Similarly,
the Kirchhoff’s
node b ofpara
Figure
3.2-1c
Del mismo modo,
la ecuacióncurrent
de la leylaw
de laequation
corrientefor
de Kirchhoff
el nodo
b de is
la figura 3.2-1c es
i1 ¼
i2 þ i3 þ
Similarly, the Kirchhoff’s current law
equation
fori6 node b of Figure 3.2-1c is
Similarly, the Kirchhoff’s current law equation for node b of Figure 3.2-1c is
Antes de
establecervoltage
la ley
del
voltaje
necesitamos
de un
Before
weque
canpodamos
state Kirchhoff’s
law,
we
need
the definition
of a loop.laAdefinición
loop is a closed
i1 ¼ i2 þ i3 þ de
i Kirchhoff,
ies
i2 þ
i3 cerrada
þ i66 a través de un circuito que no encuentra
1 ¼
circuito
cerrado
(loop). that
Un circuito
cerrado
una
ruta
path
through
a
circuit
does
not
encounter
any
intermediate
node
more
than
once.
For
example,
we can state
Kirchhoff’s
voltage
law, we
the definition
of aaloop.
loop 3.2-1c,
is a closed
ningúnBefore
nodo
intermedio
más
de unawe
vez.
Por
ejemplo,
si need
empezamos
ennode
el nodo
dethrough
laA
Before
weacan
state
Kirchhoff’s
voltage
law,
we
need
the definition
of
a loop.
Afigura
loopelement
is a closed
starting
at node
in Figure
3.2-1c,
can
move
through
element
4node
to
c, then
5nos
to
path
through
a
circuit
that
does
not
encounter
any
intermediate
more
than
once.
For
example,
podemos
mover
aelement
travésthat
del
elemento
al nodo
c,any
luego
proseguimos
a back
través
del
elemento
5have
hasta
elclosed
nodo
path
through
a circuit
does
notb,4encounter
intermediate
node
more
than
once.
For
example,
node
d,
through
6
to
node
and
finally
through
element
3
to
node
a.
We
a
starting
at node por
a inelFigure
3.2-1c,
canbmove
through element
4 to node
c, then
through
5 to
d, continuamos
elemento
6 alwe
nodo
y,the
finalmente,
por el elemento
3 de
vuelta
al nodoelement
a. Tenemos
starting
at node
a in Figure
3.2-1c,
we
canofmove
through
element
4 to node
then
through
element
5 to
path,
and
we did
not
encounter
any
intermediate
nodes—b,
c, node
or d—more
than
once.
node
d,
through
element
6
to
node
b,
and
finally
through
element
3
back
to
a.
We
have
a
closed
una ruta
cerrada,elements
yelement
no nos encontramos
ninguno
deloop.
los nodos
intermedios
(b,node
c1,
o d)
de
una
Por
node
d, through
6 to4,node
b, 6con
and
finally through
element
3 back
to
a.más
We
have
avez.
closed
Consequently,
3,
5,
and
comprise
a
Similarly,
elements
4,
5,
and
6
comprise
path,
and welosdid
not encounter
any
of the intermediate
nodes—b,
c,mismo
or d—more
than
once.
consiguiente,
elementos
3,
4,
5
y
6
comprenden
un
circuito
cerrado.
Del
modo,
los
elementos
path,
did shown
not encounter
any
of Elements
the intermediate
nodes—b,
or d—more
thancircuit.
once.
a loop and
of thewe
circuit
in4,Figure
3.2-1c.
1 and
3 comprise
yetc,another
of 6this
Consequently,
elements
5, and
6 comprise
a loop.
Similarly,
elements
4, loop
5, and
comprise1
1, 4, 5 y 6 comprenden
un3,
circuito
cerrado
del circuito
que se
muestra en
la figura1,
elementos
Consequently,
elements
3,
4,
5, and
6 comprise
a 2,
loop.
Similarly,
elements
1,3.2-1c.
4, 5, Los
and
6 comprise
The
circuit
has
three
other
loops:
elements
1
and
elements
2
and
3,
and
elements
2,
4,
5,
and
6.
ay loop
of the circuit
shown in
Figuremás
3.2-1c.
Elements
andcircuito
3 comprise
yet another
loop ofcerrados:
this circuit.
3 conforman
otro circuito
cerrado
de este
circuito.11 El
tiene otros
tres circuitos
los
a loop
of the
circuit
showntoin
Figure
3.2-1c.
Elements
and 3 comprise
yet another
loop of this circuit.
We
are
now
ready
state
Kirchhoff’s
voltage
law.
The
circuit1 has
other loops:
1 and 2,
2, elements
3, and elements 2, 4, 5, and 6.
elementos
y 2, three
los elementos
2 y 3 elements
y los elementos
5 y 6. 22 and
The
circuit has
three
other loops:
elements
1 and 2, 4,elements
and 3, and elements 2, 4, 5, and 6.
We
are
now
ready
to
state
Kirchhoff’s
voltage
law.
Ya estamos preparados para establecer la ley del voltaje de Kirchhoff.
We are now ready to state Kirchhoff’s voltage law.
Kirchhoff’s voltage law (KVL): The algebraic sum of the voltages around any loop in a
Ley delisvoltaje
de Kirchhoff
la suma algebraica de los voltajes en torno a cualquier
circuit
identically
zero
for (KVL):
all time.
Kirchhoff’s
voltage
law
(KVL):
The algebraic
the voltages around any loop in a
circuito
cerrado
en
un
circuito
es
idéntica
a cero ensum
todo of
momento.
Kirchhoff’s voltage law (KVL): The algebraic
sum
of
the voltages around any loop in a
circuit is identically zero for all time.
circuit is identically zero for all time.
La frase
suma
algebraica
que that
se debe
tenertake
en cuenta
la into
polaridad
al agregar
losup
voltajes
de los
The
phrase
algebraic
sumindica
indicates
we must
polarity
account
as we add
the voltages
elementos
que
comprenden
un
circuito
cerrado.
Unapolarity
forma de
tomar
en cuenta
la polaridad
es
moverse
of
elements
that
comprise
a
loop.
One
way
to
take
into
account
is
to
move
around
the
loop
in
The phrase algebraic sum indicates that we must take polarity into account as we add up the voltages
The
phrase
algebraic
sum
indicates
that de
we
must
take polarity
intomientras
account
weWe
addwrite
up
voltages
en torno
al circuito
cerrado
en elobserving
sentido
las
manecillas
del
reloj
seasobservan
lasthe
polaridades
the
clockwise
direction
while
the
polarities
of
the
element
voltages.
the
voltage
of elements that comprise a loop. One way to take polarity into account is to move around the loop in
de elements
losa plus
voltajes
elemento.
El voltaje
se
escribe
con
unpolarity
signo
cuando
encontramos
el signo
2 the
de
of
thatdel
comprise
a loop.
One
take
polarity
intomás
account
is to
move
around
the loop
in
with
sign
when
wewhile
encounter
theway
þthe
oftothe
voltage
before
the
�.
In We
contrast,
the
clockwise
direction
observing
polarities
of the element
voltages.
writewe
thewrite
voltage
la
polaridad
del
voltaje
antes
del
signo
1.
Por
el
contrario,
el
voltaje
se
escribe
con
un
signo
menos
cuando
the
clockwise
direction
while
observing
the
polarities
of
the
element
voltages.
We
write
the
voltage
voltage
with
a minus
when we encounter
the
� of the
voltage
polarity
before
the þ. For
with
a plus
sign
whensign
wedeencounter
the þ
of the
voltage
polarity
before
the
�.
In contrast,
weelexample,
write
the
encontramos
el signo
2
la polaridad
delofvoltaje
antes3,polarity
del
signo
1.
Porthe
ejemplo,
considere
circuito
with
a plus
sign
when
we encounter
the
þ
the
voltage
before
�. In
contrast,
we
write
the
consider
the
circuit
shown
in
Figure
3.2-1c.
Elements
4,
5,
and
6
comprise
a
loop
of
the
circuit.
By
voltage
with
a
minus
sign
when
we
encounter
the
�
of
the
voltage
polarity
before
the
þ.
For
example,
que
se
muestra
en
la
figura
3.2-1c.
Los
elementos
3,
4,
5
y
6,
comprenden
un
circuito
cerrado
del
circuito.
voltage
with
a
minus
sign
when
we
encounter
the
�
of
the
voltage
polarity
before
the
þ.
For
example,
,
v
,
v
,
and
v
must
be
zero.
As
Kirchhoff’s
voltage
law,
the
algebraic
sum
of
the
element
voltages
v
3
4
5
6
consider
the del
circuit
shown
in Figurela3.2-1c.
Elements de
3, 4,
and 6 comprise
a loop vof, the
circuit.
By
Según
la ley
voltaje
de Kirchhoff,
sumadirection,
algebraica
los5,
deþlos
elementos
v�,
v6 of
debe
4, vthe
5y�
consider
the
circuit
shown
inthe
Figure
3.2-1c.
Elements
3,encounter
4,
5,voltajes
and 6the
comprise
a loop the
of3 the
circuit.
By
before
v5
we
move
around
the
loop
in
clockwise
we
of
v
4
v6 must
be zero. As
Kirchhoff’s
voltage law,
the algebraic
sum
of theenelement
voltages
v3, v4, v5, and
ser cero.the
Alþ,
movernos
en
torno
althe
circuito
cerrado
el sentido
de las
manecillas
del reloj,
encontramos
el
v6 must
be
zero.sign
As
Kirchhoff’s
voltage
law,
the algebraic
sum
of �
theofelement
voltages
v3, v4, v5, and
before
the
�
of
v
before
þ,
and
the
v
before
the
þ.
Consequently,
we
use
a
minus
6
3
the �,
the
� of
vy5
we
move
around
thedel
loop
in the
clockwise
direction,
wedelencounter
the
þ of v24 before
antes
signo
2;
el
signo
2
de
v
antes
signo
1;
el
signo
de
v
antes
del
signo
1
signo
1
de
v
4
5
6
before
the
�,
the
�
of
v
we
move
around
the
loop
in
the
clockwise
direction,
we
encounter
the
þ
of
v
4 Figure 3.2-1c is
v5, þ,
and v6�and
plus sign
for
v4. The
for this
loop of
for v3, the
before
of va6 before
theEn
þ, and
the �KCL
of v3equation
before the
Consequently,
use
a minus
sign5
signo 1.
un þ.
signo
menos para vwe
y un signo
el signothe
2 þ,
de vthe
3 antes
3, vuse
5 y va6,minus
before
the
� ofdel
v6 before
the þ,consecuencia,
and
the
��ofvutilizamos
v3�
before
the
þ.
Consequently,
we
sign
v
�
v
v
¼
0
5KCL6cerrado
3 de for
, 4and
plus
.4 The
equation
this loop
of es
Figure 3.2-1c is
for
3, v5v
6 and a de
. La vvecuación
la sign
KCL for
paravv4este
circuito
la figura
3.2-1c
másvvpara
,
v
,
and
and
a
plus
sign
for
.
The
KCL
equation
for
this
loop
of
Figure
3.2-1c
is
for
3
5
6
4
Similarly, the Kirchhoff’s voltage law vequation
thev loop
� v �for
v �
¼ 0 consisting of elements 1, 4, 5, and 6 is
v44 � v55 � v66 � v33 ¼ 0
� v5 �de
v6Kirchhoff
þ
¼ 0consisting
v4 voltaje
Del mismothe
modo,
la ecuación
de la ley
para el circuito
cerrado que
Similarly,
Kirchhoff’s
voltage
lawdel
equation
for
thev1loop
of elements
1, 4,consta
5, andde6los
is
Similarly,
the4, Kirchhoff’s
voltage law equation for the loop consisting of elements 1, 4, 5, and 6 is
elementos
1,
5
y
6
es
The Kirchhoff’s voltage law equation vfor
consisting
�the
v loop
�v þ
v ¼ 0 of elements 1 and 2 is
v44 � v55 � v66 þ v11 ¼ 0
The
Kirchhoff’s
voltage
law de
equation
forpara
the
of
2 is 1 y 2 es
þ
v1consisting
¼ cerrado
0
�v2loop
La ecuación
de la ley
del voltaje
Kirchhoff
circuito
queelements
consta de 11losand
elementos
The
Kirchhoff’s
voltage
law equation
for the el
loop
consisting
of
elements
and
2 is
�v2 þ v1 ¼ 0
�v2 þ v1 ¼ 0
E jXeAmMpPlLoE 33 ..22--22 Kirchhoff’s
Laws
E
Leyes de Kirchhoff
E X A M P L E 3 . 2 - 2 Kirchhoff’s Laws
E X A M P L E 3 . 2 - 2 Kirchhoff’s Laws
IENJ TE EMRPALCOT I N
V ET EERXAACMTPI VL O
E
INTERACTIVE EXAMPLE
INTERACTIVE EXAMPLE
Considere
el circuito
se muestra
en 3.2-4a.
la figuraDetermine
3.2-4a. Determine
la energía
alimentada
porCeland
elemento
C y lareceived
energía
Consider the
circuit que
shown
in Figure
the power
supplied
by element
the power
recibida
por
el
elemento
D.
by element
Consider
theD.
circuit shown in Figure 3.2-4a. Determine the power supplied by element C and the power received
Consider the circuit shown in Figure 3.2-4a. Determine the power supplied by element C and the power received
by
element D.
Solución
by
element D.
Solution
La
figura 3.2-4a proporciona un valor para la corriente en el elemento C pero no para el voltaje, v, a través del elemento
Figure
3.2-4ayprovides
a value
for the current
in en
element
C but
not for
the voltage,
v, across element
voltage
la figura
3.2-4a
se apegan
a la convención
pasiva, C.
porThe
lo tanto,
el
C.
El voltaje
la corriente
del elemento
C dados
Solution
Solution
and
current
of
element
C
given
in
Figure
3.2-4a
adhere
to
the
passive
convention,
so
the
product
of
this
voltage
and
La
figura
3.2-4a
proporciona
un
valor
para
producto
del
voltaje
y
la
corriente
es
la
potencia
recibida
por
el
elemento
C.
Figure 3.2-4a provides a value for the current in element C but not for the voltage, v, across element C. The voltage
Figure
3.2-4a
provides
a value
for
the current
inlaelement
C i,but
not
for
thefor
voltage,
v, across
C.D
The
voltage
current
is the
byD,
element
C.3.2-4a
Figure
3.2-4atoprovides
value
theEl
voltage
element
but
notand
for
el
voltaje
a través
delreceived
elemento
pero
no
para
corriente,
elaelemento
D.
voltaje
y laelement
corriente
elemento
and
current
ofpower
element
C given
in
Figure
adhere
the en
passive
convention,
so theacross
product
of thisdel
voltage
and
current
of
element
C
given
in
Figure
3.2-4a
adhere
to
the
passive
convention,
so
the
product
of
this
voltage
and
the
current,
i,
in
element
D.
The
voltage
and
current
of
element
D
given
in
Figure
3.2-4a
do
not
adhere
to
the
passive
figurareceived
3.24a noby
se element
apegan aC.laFigure
convención
poralo
tanto,
producto
voltaje
y la corriente
la
D dadosisen
current
thelapower
3.2-4apasiva,
provides
value
forelthe
voltagedel
across
element
D but notesfor
current
isalimentada
theso
power
received
bythis
element
C. Figure
3.2-4aisprovides
a value
for thebyvoltage
across
element D but not for
potencia
por
el
elemento
D.
convention,
the
product
of
voltage
and
current
the
power
supplied
element
D.
the current, i, in element D. The voltage and current of element D given in Figure 3.2-4a do not adhere to the passive
the current,
i, into
element
D. The
voltage
of
Dthe
given
Figure
3.2-4a
not
to
the passive
Necesitamos
determinar
voltaje,and
acurrent
través
delelement
elemento
C ycurrent,
lain
corriente,
i, en eldo
elemento
Usaremos
las
We
need
theel
voltage,
v,v,across
element
i, inbyelement
D.
Weadhere
willD.
use
Kirchhoff’s
convention,
so thedetermine
product
of
this
voltage
and
current
is Ctheand
power
supplied
element
D.
convention,
so thevalues
product
of
this i.
voltage
andvidentify
is the
power
supplied
by circuit
element
D.
leyes
dedetermine
Kirchhoff
para
determinar
los
valores
ecurrent
i. Primero,
identificamos
yofetiquetamos
los
nodos del
circuito3.2-4b.
como
laws to
of
v
and
First,
we
and
label
the
nodes
the
as
shown
in
Figure
We need to determine the voltage, v, across element C and the current, i, in element D. We will use Kirchhoff’s
We need
determine
the voltage, v, across element C and the current, i, in element D. We will use Kirchhoff’s
se muestra
en latofigura
3.2-4b.
laws to determine values of v and i. First, we identify and label the nodes of the circuit as shown in Figure 3.2-4b.
laws to determine values of v and i. First, we identify and label the nodes of the circuit as shown in Figure 3.2-4b.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 57
Alfaomega
4/12/11 5:20 PM
E1C03_1
E1C03_1
11/25/2009
11/25/2009
58
58
58
58
58
58
58
Resistive Circuits
Resistive Circuits
Circuits
Resistive
Circuitos resistivos
Resistive Circuits
Resistive Circuits
FIGURA 3.2-4
en Example
el
FIGURE
3.2-4 (a)
(a) El
Thecircuito
circuitconsiderado
considered in
FIGURE3.2-2,
3.2-4y (a)
(a) el
The
circuittrazado
considered
in Example
Example
ejemplo
(b)
circuito
de
nuevo
FIGURE
3.2-4
The
circuit
considered
in
3.2-2
and (b)
the (a)
circuit
to emphasize
FIGURE
3.2-4
Theredrawn
circuit considered
in the
Example
3.2-2destacar
and (b)
(b)
the
circuit
redrawn
to emphasize
emphasize
the
para
los (a)
nodos.
3.2-2
and
the
circuit
to
nodes.
FIGURE
3.2-4
Theredrawn
circuit considered
in the
Example
3.2-2
and
(b)
the
circuit
redrawn
to
emphasize
the
nodes.
nodes.
3.2-2
and
(b)
the
circuit
redrawn
to
emphasize
the
nodes. que consta de los elementos C, D y B para
Aplique la ley del voltaje de Kirchhoff (KVL) al circuito cerrado
nodes.
Apply Kirchhoff’s voltage law (KVL) to the loop consisting
of elements C, D, and B to get
obtener
Apply Kirchhoff’s voltage
voltage law (KVL)
(KVL) to
to the loop
loop consisting
consisting of
of elements
elements C, D,
D, and B
B to get
get
Apply
Apply Kirchhoff’s
Kirchhoff’s voltage law
law (KVL)
to the
the
of V
elements C,
C, D, and
and B to
to get
Þ � loop
6 ¼ 0consisting
) v ¼ �2
�v � ð�4
�v �
� ðð�4
�4
� loop
¼ 00consisting
) vv ¼
¼ �2
V
Apply Kirchhoff’s voltage law (KVL)
to the
of V
elements
C, D, and B to get
�v
ÞÞÞ �
666 ¼
)
�v
� 3.2-4b
0 es
)
¼El�2
�2
Vand current
El valor
corriente
elemento
en�lað�4
figura
7 vA.
voltaje
y la corriente
del elemento
dados
The
valuedeoflathe
currenten
in el
element
C inC
Figure
3.2-4b
is¼
7 A.
The
voltage
of element
C given inCFigure
The value
value of
of the
the current
current in
in element
element C
C in
in�v
Figure
3.2-4b
is¼
A.
Thevvoltage
voltage
and current
current of
of element
element C
C given
given in
in Figure
Figure
� ð�4
Þ � 6is
0 )
¼ �2 V
The
Figure
3.2-4b
777 A.
The
and
3.2-4b
adhere
tocurrent
the
so pasiva,
en
lavalue
figura
3.2-4b
se passive
apegan
aconvention,
la convención
por
lo tanto,
The
of
the
in
element
C
in
Figure
3.2-4b
is
A.
The
voltage
and
current
of
element
C
given
in
Figure
3.2-4b adhere
adhere to
to the
the passive convention,
convention,
so
3.2-4b
pC so
¼ v3.2-4b
ð7Þ ¼ (is�72)(7)
¼ voltage
�14 W and current of element C given in Figure
The value
of thetocurrent
in element
C in Figure
A. The
3.2-4b
adhere
the passive
passive
convention,
pCso
¼ vvðð77ÞÞ ¼
¼ (( �
� 2)(7)
2)(7)
¼ �14
�14 W
W
p
¼
¼
C
3.2-4b
adherereceived
to the passive
convention,
pC so
¼ vð7Þelement
¼ ( � 2)(7)
¼ �1414
W W.
is
the power
by element
C. Therefore,
C supplies
is
the
power
received
by
element
C.
Therefore,
element
C
supplies
14
es the
la potencia
recibida
por el elemento
C.
eltoelemento
alimenta 14 W.
pCPor
¼
vconsiguiente,
ð7Þelement
¼
� 2)(7)
¼get
�1414
WCW.
is
power
received
by
C.
C
Kirchhoff’s
current
law
(KCL)
at (node
is the Next,
powerapply
received
by element
element
C. Therefore,
Therefore,
element
Cbbsupplies
supplies
14 W.
W.
Next,
apply
Kirchhoff’s
current
law
(KCL)
at
node
to
get
A
continuación,
aplique
la
ley
de
la
corriente
de
Kirchhoff
(KCL)
al
nodo
b para obtener
Next,
apply
Kirchhoff’s
current
law
(KCL)
at
node
b
to
get
is the Next,
powerapply
received
by element
C. Therefore,
supplies
Kirchhoff’s
current
law
to iget
7 þ(KCL)
ð�10element
Þ at
þ node
i ¼ 0Cb)
¼ 3 14
A W.
7
þ
ð
�10
Þ
þ
i
¼
0
)
i
¼
3
A
Next, apply Kirchhoff’s current law
(KCL)
at
node0 b)
to iget
7þ
ðð�10
þ
33 A
þFigure
�10ÞÞ 3.2-4b
þ ii ¼
¼ is
0 �4
) V.
i¼
¼The
Avoltage and current of element D given in
The value of the voltage across element D7in
The
value
of
the
voltage
across
element
D
in
Figure
3.2-4b
is
�4
V.
The
voltage
and ycurrent
current
of element
element
D given
given in
7
þ
ð
�10
Þ
þ
i
¼
0
)
i
¼
3 V.
Avoltage
The
value
of
the
voltage
across
element
D
in
Figure
3.2-4b
is
�4
V.
The
and
of
D
El
valor
del
voltaje
a
través
del
elemento
D
en
la
figura
3.2-4b
es
24
El voltaje
la corriente
elemento
D
Figure
3.2-4b
do
not
adhere
to
the
passive
convention,
so
the
power
by current
element
is del
given
The
value
of thedovoltage
acrosstoelement
D in Figure
3.2-4bsois the
�4 V.
The supplied
voltage
and
ofD
element
D by
given in
in
Figure
3.2-4b
not
adhere
the
passive
convention,
power
supplied
by
element
D
is
given
by
Figure
3.2-4b
do
not
adhere
to
the
passive
convention,
so
the
power
supplied
by
element
D
is
given
by
dado
en
la
figura
3.2-4b
no
se
apegan
a
la
convención
pasiva,
por
lo
tanto,
la
energía
alimentada
por
el
elemento
The
value
of
the
voltage
across
element
D
in
Figure
3.2-4b
is
�4
V.
The
voltage
and
current
of
element
D
given
in
Figure 3.2-4b do not adhere to the passive
supplied by element D is given by
pD ¼convention,
ð�4Þi ¼ð�4so
Þð3the
Þ ¼power
�12 W
¼convention,
�4Þi ¼
¼ðð�4
�4so
Þð3the
¼power
�12 W
W
D
es dada
por do not adhere to the passive
Figure
3.2-4b
supplied by element D is given by
pppDD ¼
ððð�4Þi
Þð3
ÞÞÞ ¼
�12
¼
�4Þi
¼
ð
�4
Þð3
¼
�12
W
Therefore, element D receives 12 W. D
Therefore,
element
D
receives
12
W.
p ¼ ð�4Þi ¼ð�4Þð3Þ ¼ �12 W
Therefore,
Therefore, element
element D
D receives
receives 12
12 W.
W. D
Therefore,
element
D
receives
12
W.
Por consiguiente, el elemento D recibe 12 W.
E X A M P L E 3 . 2 - 3 Ohm’s and Kirchhoff’s Laws
E X A eMmPP pLL lEEo33 .3
. 22 -- 33- 3 Ohm’s
Ohm’s
and
Kirchhoff’s
Laws
E
Kirchhoff’s
Laws
Leyesand
de Ohm
y Kirchhoff
E XXEAAj M
M P L E 3 . 2. 2
- 3 Ohm’s
and
Kirchhoff’s
Laws
E X A M P L E 3 . 2 - 3 Ohm’s and Kirchhoff’s Laws
Consider the circuit shown in Figure 3.2-5. Notice that the passive convention was used to assign reference
Consider the
the
circuitde
shown
in Figure
3.2-5. Notice
Notice
that
the passive
convention
was used
used to
to direcciones
assign reference
reference
Consider
circuit
shown
in
3.2-5.
the
convention
was
assign
Considere
eltocircuito
la figura
3.2-5.
Observe
queThis
lathat
convención
pasiva
se utilizó
asignar
deand
redirections
the resistor
voltages
and
currents.
anticipates
using
Ohm’spara
law.used
Findtoeach
current
Consider
the
circuit
shown
in Figure
Figure
3.2-5.
Notice
that
the passive
passive
convention
was
assign
reference
directions
to
the
resistor
voltages
and
currents.
This
anticipates
using
Ohm’s
law.
Find
each
current
and
directions
to
the
resistor
voltages
and
currents.
This
anticipates
using
Ohm’s
law.
Find
each
current
and
ferencia
a
los
voltajes
y
las
corrientes
de
los
resistores.
Esto
se
anticipa
al
uso
de
las
leyes
de
Ohm.
Encuentre
cada
Consider
the
circuit
shown
in
Figure
3.2-5.
Notice
that
the
passive
convention
was
used
to
assign
reference
V, v2 ¼ and
�10currents.
V, i3 ¼ 2This
A, and
R3 ¼ 1 V.
Also,
determine
the resistance
R 2. and
each
voltage
when
R1 ¼ 8voltages
directions
to the
resistor
anticipates
using
Ohm’s
law. Find
each current
¼ 8 V,
V, v2 ¼
¼ �10
V,5ii3210
¼ 22 V,
A,i and
and
RA3 ¼
¼R11 V.
V.
Also,
determine
the resistance
resistance
R2. R .
each voltage
voltage
when
R1 ¼
¼
A,
R
Also,
determine
the
R
each
when
R
corriente
y cada
voltaje
8�10
V,currents.
vV,
5
2
y
5
1
V.
Además,
determine
la
resistencia
3¼
directions
to the
resistor
This
anticipates
using
Ohm’s
law.
Find
each
current
2 i3
3
3
2
¼ 88voltages
V, Rvv122 5
¼ and
�10
V,
¼
2
A,
and
R
1
V.
Also,
determine
the
resistance
R22.. and
each
voltage
when
R11 cuando
3
3
each voltage when R1 ¼ 8 V, v2 ¼ �10 V, i3 ¼ 2 A, and R3 ¼ 1 V. Also, determine the resistance R 2.
Solution
Solution
Solución
Solution
The
sum of the currents entering node a is
Solution
Thesuma
sum de
of las
thecorrientes
currents entering
node
a is a es
La
que entran
al nodo
The
i1 � i2 � i3 ¼ 0
Solution
The sum
sum of
of the
the currents
currents entering
entering node
node aa is
is
� ii22 �
� ii33 ¼
¼ 00
iii11 �
The sum of the currents entering node a is
1 � i2 � i3 ¼ 0
i1 � i2 � i3 ¼ 0
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 58
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
E1C03_1 11/25/2009
11/25/2009
E1C03_1
E1C03_1
E1C03_1
11/25/2009
E1C03_1 11/25/2009
11/25/2009
5959
5959
59
Leyes
deLaws
Kirchhoff
Kirchhoff’s
Laws
Kirchhoff’s
Kirchhoff’s
Laws
Kirchhoff’s
Laws 59
Kirchhoff’s
Laws
Kirchhoff’s
Kirchhoff’s
Laws
Kirchhoff’sLaws
Laws
59
59
59
5959
59
59
59
Utilizando
la ley
defor
para
R3, encontramos
que
Using
Ohm’s
law
R
we
find
that
Using
Ohm’s
law
for
,Ohm
we
that
3,find
3for
Using
Ohm’s
lawRfor
R
wefind
find
that
Using
Ohm’s
law
R
that
3,,33,we
Using
Ohm’s
law
for
R
we
find
that
Using
Ohm’s
law
for
R
,
we
find
that
3
3
v
v
¼
R
i3 ¼
1(2)
¼
R
i
¼
2¼
V¼
Using Ohm’s law for R3, we find that 3 v3v333¼
31(2)
3¼
3R¼
R
¼1(2)
1(2)
¼222VV
V
3 i3i33i33¼
vv3v3¼¼
R
¼
1(2)
¼
222VV
R
i
¼
1(2)
¼
3
3
3
3
¼
R
i
¼
1(2)
¼
V
3
3
3
Kirchhoff’s
voltage
law
for
the
bottom
loop
incorporating
v
Kirchhoff’s
voltage
law
for
the
bottom
loop
incorporating
v
,
v
,
,
v
,
La
ley del voltaje
delaw
Kirchhoff
el circuito
cerrado inferior
que
3 incorpora
1 v3v,1
Kirchhoff’s
voltage
law
forthe
thepara
bottom
loopincorporating
incorporating
Kirchhoff’s
voltage
for
bottom
loop
1 11,v3v,33,
Kirchhoff’s
voltage
for
Kirchhoff’s
voltage
law
for
the
bottom
loop
incorporating
and
source
is
and and
the
source
isde law
vKirchhoff’s
vthe
y 10-V
la10-V
fuente
10
es the
voltage
law
thebottom
bottomloop
loopincorporating
incorporatingvv1v,11,,vv3v,33,,
1, 10-V
3the
and
the
10-V
source
isV for
source
is
and
and
the
10-V
source
andthe
the10-V
10-Vsource
sourceisis
is
¼
�10
�10�10
þ
v1þþ
þvv3v0333¼¼
¼000
�10
þvv1v3111þþ
vv3v3¼¼
000
�10
þþ
vv1v1þþ
�10
þ
¼
�10
þ
1
3
Therefore;
v
Therefore;
v
¼
10
�
¼
10
�
v
¼
V¼
Por
consiguiente,
1 v v11¼¼10
Therefore;
103��vv3v8333¼
¼888VV
V
Therefore;
1 1¼ 10
Therefore;
v
�
v
888VV
Therefore;
v1v11 ¼
10
�
v3v33¼¼
Therefore;
¼
10
�
¼
V
Ohm’s
law
for
the
resistor
R
Ohm’s
law
for
the
resistor
R
is
is
1
1
Ohm’s
law
forthe
theresistor
resistor
Ohm’s
for
RR
La
ley law
de
Ohm
para
el resistor
Ris1 es
1 11is
Ohm’s
Ohm’s
law
for
the
resistor
Ohm’slaw
lawfor
forthe
theresistor
resistorRR
R1 11isis
is
v1 ¼vvvR11¼
1 i1
1 i¼
1RR
¼
R
1 i1i11i11
RR
vv11v11¼¼
1 11ii1
¼
R
¼
=R
¼
or oror
i1 ¼i vii111¼
=R
¼
1 8=8
18=8
11 A
11¼
1 ¼
¼v1v1v=R
¼8=8
8=8
¼111AA
A
oorbien
¼
1 =R
1 11¼
oror
i11ii11¼¼
vv1v=R
8=8
¼¼
111AA
¼
8=8
1 1¼
11=R
or
¼
=R
¼
8=8
¼
A
1
1
Next,
apply
Kirchhoff’s
current
law
at
node
a
to
get
Next,
apply
Kirchhoff’s
current
law
at
node
a
to
get
Next,
applyKirchhoff’s
Kirchhoff’s
current
law
at nodeade
a to
get
Next,
apply
at
get
FIGURE
3.2-5
Circuit
with
two
FIGURE
3.2-53.2-5
Circuit
with with
two
FIGURE
3.2-5
Circuit
with
two
A
continuación,
aplique current
lacurrent
ley delaw
la corriente
Kirchhoff
al nodo a para obtener
FIGURE
3.2-5
Circuit
withtwo
two
Next,
apply
current
law
atatnode
node
aaato
toto
get
FIGURE
Circuit
Next,
apply
Kirchhoff’s
law
get
Next,
applyKirchhoff’s
Kirchhoff’s
current
law
node
to
get
FIGURE
3.2-5
Circuit
with
two
i2at
¼
i
�
i
¼
1
�
2
¼
�1
A
¼node
i
�
i
¼
1
�
2
¼
�1
A
FIGURE
3.2-5
Circuit
with
two
2
1
3
1
3
constant-voltage
sources.
constant-voltage
sources.
2
1
3
constant-voltage
sources.
FIGURE
3.2-5
Circuit
with
two
¼i1i1��i3i3¼¼11��22¼¼�1
�1AA
ii2i2¼
constant-voltage
sources.
constant-voltage
sources.
FIGURA
3.2-5
Circuito
con
¼
�1
2ii2¼
constant-voltage
sources.
constant-voltage
sources.
¼i1ii11��
�i3ii33¼¼
¼111��
�222¼¼
¼�1
�1AA
A
constant-voltage
sources.
We
can
now
find
the
resistance
R
We We
can
now
find
the
R2 from
dos fuentes de
voltaje constante.
2 from
from 2R por
Wecan
canya
now
findresistance
the
resistance
R
now
find
the
resistance
R
2 22from
Ahora
podemos
encontrar
la
resistencia
from
We
can
now
find
the
resistance
R
from
We
can
now
find
the
resistance
R
2
2
2
v
v
¼
R
i
¼
R
i
We can now find the resistance R2 from
2 v v22¼
2¼
2RRi22 i22
vv22v22¼¼
RR22 i222i 2
2 ¼ R22 i22
R
¼
v
=i
¼
�10=�1
10
or oror
R
¼
v
=i
¼
�10=�1
¼ 10¼¼
V10
2
2
2
2
2
2
22 =i
orbien
R
�10=�1
¼
10VV
V
R
2 22¼¼v2v=i
2 22¼¼�10=�1
oor
oror
RR
�10=�1
10
22=i
R2 22¼¼
¼vv2v=i
=i2 22¼¼
¼�10=�1
�10=�1¼¼
¼10
10VV
V
E
j e m p l o 3 . 2 - 4 Leyes de Ohm
X
44 Ohm’s
Ohm’s
and
E X AEE
PA
E P
3LLLLE.EEE233
-3.4..222- --4
Ohm’s
andand
A
M
EM
XA
ALMM
M
P
Ohm’s
and
XX
PP
y Kirchhoff
EE
XXAAMMPPLLEE 33. .22- -44 Ohm’s
and
Ohm’s
and
E X A M P L E 3 . 2 -Kirchhoff’s
4 Kirchhoff’s
Ohm’s
and
Kirchhoff’s
Laws
Laws
Kirchhoff’s
Laws
Laws
Kirchhoff’s
Laws
Kirchhoff’s
Kirchhoff’s Laws
Laws
EJEMPLO INTERACTIVO
IIIN
I N T EI R
AT TTC
IAVA
EM
N
RR
A
CC
VV
XX
A
M
PP
N
ETR
A
IV
EPX
A
M
N
ETEER
CEC
T TTEITVIIX
EAEEEM
E EEX
ALA
M
PP
L LLELEEE
IN
II N
NTTTEEERRRAA
ACCCTTTI VII VVEEEEEEXXXAA
AMM
MPPPL LEL EE
Determine el valor de la corriente, en amperios, medido por el amperímetro en la figura 3.2-6a.
Determine
the
value
the
current,
amps,
measured
by
the
ammeter
Figure
3.2-6a.
Determine
the value
of the
in amps,
measured
by the
in Figure
3.2-6a.
Determine
thevalue
value
ofcurrent,
thecurrent,
current,
inamps,
amps,
measured
byammeter
theammeter
ammeter
inFigure
Figure
3.2-6a.
Determine
the
ofof
the
inin
measured
by
the
inin
3.2-6a.
Solución
Determine
Determine
the
value
the
current,
amps,
measured
by
the
ammeter
Figure
3.2-6a.
Determinethe
thevalue
valueofof
ofthe
thecurrent,
current,inin
inamps,
amps,measured
measuredby
bythe
theammeter
ammeterinin
inFigure
Figure3.2-6a.
3.2-6a.
Solution
Solution
Un
amperímetro ideal es equivalente a un cortocircuito. La corriente medida por el amperímetro es la corriente en
Solution
Solution
Solution
Solution
An
ideal
ammeter
is
equivalent
to
aashort
short
circuit.
The
current
measured
by
the
ammeter
the
current
in
the
short
An ideal
ammeter
is equivalent
to
a to
short
circuit.
current
measured
by the
is theisis
inelthe
el
cortocircuito.
La
figura
3.2-6b
elThe
circuito
después
de reemplazar
amperímetro
por
cortocircuito
Solution
Anideal
ideal
ammeter
isequivalent
equivalent
toamuestra
short
circuit.
Thecurrent
current
measured
byammeter
theel
ammeter
iscurrent
thecurrent
current
inshort
theshort
short
An
ammeter
is
circuit.
The
measured
by
the
ammeter
the
in
the
An
ideal
ammeter
isis
equivalent
toto
acircuit
circuit.
The
current
measured
by
the
ammeter
isis
the
current
inin
the
short
An
ideal
ammeter
equivalent
aashort
short
circuit.
The
current
measured
by
the
ammeter
the
current
the
short
circuit.
Figure
3.2-6b
shows
the
after
replacing
the
ammeter
by
the
equivalent
short
circuit.
circuit.
Figure
3.2-6b
shows
the
circuit
after
replacing
the
ammeter
by
the
equivalent
short
circuit.
equivalente.
An
ideal
ammeter
is
equivalent
to
short
circuit.
The
current
measured
by
the
ammeter
is
the
current
in
the
short
circuit.Figure
Figure3.2-6b
3.2-6bshows
showsthe
thecircuit
circuitafter
afterreplacing
replacingthe
theammeter
ammeterbybythe
theequivalent
equivalentshort
shortcircuit.
circuit.
circuit.
circuit.
Figure
3.2-6b
shows
the
after
the
ammeter
the
equivalent
short
circuit.
circuit.
Figure
3.2-6b
shows
the
circuit
after
replacing
the
ammeter
by
the
equivalent
short
circuit.
The
circuit
been
redrawn
in
Figure
3.2-7
to
label
the
nodes
of
the
circuit.
This
circuit
consists
of
The The
circuit
has
been
redrawn
incircuit
Figure
3.2-7
to label
the
nodes
ofby
the
circuit.
This
circuit
consists
of aconsta
El
circuito
sehas
habeen
trazado
de
nuevo
en
la replacing
figura
3.2-7
para
etiquetar
los
nodos
del This
circuito.
El circuito
circuit.
Figure
3.2-6b
shows
the
circuit
after
replacing
ammeter
by
equivalent
short
circuit.
The
circuit
has
been
redrawn
inFigure
Figure
3.2-7
tolabel
label
the
nodes
ofthe
the
circuit.
This
circuit
consists
ofaaa
circuit
has
redrawn
in
3.2-7
to
the
nodes
of
the
circuit.
circuit
consists
of
The
circuit
has
been
redrawn
in
Figure
3.2-7
to
label
the
nodes
of
the
circuit.
This
circuit
consists
ofof
aa
The
circuit
has
been
redrawn
in
Figure
3.2-7
to
label
the
nodes
of
the
circuit.
This
circuit
consists
voltage
source,
a
dependent
current
source,
two
resistors,
and
two
short
circuits.
One
of
the
short
circuits
is
the
voltage
source,
a
dependent
current
source,
two
resistors,
and
two
short
circuits.
One
of
the
short
circuits
is
the
The
circuit
has
been
redrawn
in
Figure
3.2-7
to
label
the
nodes
of
the
circuit.
This
circuit
consists
of
de
una
fuente
de
voltaje,
una
fuente
de
corriente
dependiente,
dos
resistores
y
dos
cortocircuitos.
Uno
de
los
corvoltagesource,
source,a adependent
dependentcurrent
currentsource,
source,two
tworesistors,
resistors,and
andtwo
twoshort
shortcircuits.
circuits.One
Oneofofthe
theshort
shortcircuits
circuitsisisthe
thea
voltage
voltage
source,
a
dependent
current
source,
two
resistors,
and
two
short
circuits.
One
of
the
short
circuits
is
the
voltage
source,
a
dependent
current
source,
two
resistors,
and
two
short
circuits.
One
of
the
short
circuits
is
the
controlling
element
the
CCCS,
and
the
other
short
circuit
is
aamodel
model
the
ammeter.
controlling
element
the
and and
the
other
short
circuit
isand
a is
model
of the
voltage
source,
aelemento
dependent
current
source,
two
resistors,
two
short
circuits.
One of the short circuits is the
tocircuitos
es
elof
deother
la
CCCS
y circuit
el
otro
es
modelo
del
controlling
element
ofCCCS,
thepredominante
CCCS,
and
the
other
short
circuit
is
model
ofammeter.
theamperímetro.
ammeter.
controlling
element
ofof
the
CCCS,
the
short
aun
ofof
the
ammeter.
controlling
controlling
element
the
CCCS,
and
the
other
short
circuit
model
the
ammeter.
controllingelement
elementofof
ofthe
theCCCS,
CCCS,and
andthe
theother
othershort
shortcircuit
circuitisis
isaaamodel
modelofof
ofthe
theammeter.
ammeter.
Amperímetro
FIGURA 3.2.7 El circuito de la figura 3.2-6 después de etiquetar
FIGURA 3.2-6 (a) Circuito con fuente dependiente y un
FIGURE
3.2-6
(a)
A
circuit
with
dependent
source
and
an
FIGURE
3.2-6
(a)
A
circuit
with
dependent
source
and
an
FIGURE
3.2-6
(a)
A
circuit
with
dependent
source
and
an
FIGURE3.2-6
3.2-6
(a)circuito
circuit
withdependent
dependent
source
andanan los nodos y algunas corrientes y voltajes del elemento.
FIGURE
AAcircuit
with
and
amperímetro.
(b)(a)
El
equivalente
despuéssource
de
sustituir
FIGURE
3.2-6
(a)
AAcircuit
with
dependent
and
FIGURE
3.2-6
(a)
circuit
with
dependent
source
and
an
FIGURE
3.2-7
The
circuit
Figure
3.2-6
after
labeling
the
FIGURE
3.2-73.2-7
The The
circuit
of Figure
3.2-63.2-6
after
labeling
the the
ammeter.
(b)
The
equivalent
circuit
after
replacing
the
ammeter
ammeter.
(b) The
equivalent
after
replacing
the source
ammeter
FIGURE
3.2-7
The
circuit
of
Figure
3.2-6
after
labeling
the
ammeter.
(b)
The
equivalent
circuit
after
replacing
the
ammeter
FIGURE
3.2-6
(a)
circuit
with
dependent
source
andan
an FIGURE
FIGURE
3.2-7
Thecircuit
circuit
ofFigure
Figure
3.2-6after
afterlabeling
labeling
the
ofof
ammeter.
(b)
The
equivalent
circuit
after
replacing
theammeter
ammeter
ammeter.
(b)
The
equivalent
circuit
after
replacing
the
el
amperímetro
por
unAcircuit
cortocircuito.
FIGURE
3.2-7
The
circuit
ofof
Figure
3.2-6
after
labeling
the
FIGURE
3.2-7
The
circuit
Figure
3.2-6
after
labeling
the
ammeter.
(b)
The
equivalent
circuit
after
replacing
the
ammeter
ammeter.
(b)
The
equivalent
circuit
after
replacing
the
ammeter
nodes
and
some
element
currents
and
voltages.
nodes
and
some
element
currents
and
voltages.
by
a
short
circuit.
by a short
circuit.
nodes
and
some
element
currents
and
voltages.
FIGURE
3.2-7
The
circuit
of
Figure
3.2-6
after
labeling
the
short
circuit.
ammeter.
(b)
The equivalent circuit after replacing the ammeter nodes
nodesand
andsome
someelement
elementcurrents
currentsand
andvoltages.
voltages.
bya aashort
shortcircuit.
circuit.
byby
nodes
nodes
and
some
element
currents
and
voltages.
by
by
short
circuit.
nodesand
andsome
someelement
elementcurrents
currentsand
andvoltages.
voltages.
bya aashort
shortcircuit.
circuit.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 59
Alfaomega
4/12/11 5:21 PM
E1C03_1
E1C03_1
11/25/2009
11/25/2009
60
60
60
60
60
Circuitos resistivos
Resistive
Resistive Circuits
Circuits
Aplicando dos veces la KCL, una en el nodo d y otra en el nodo a, se muestra que la corriente en la fuente
de voltaje
y la corriente
en el resistor
de 4 V equivalen
ambas shows
a ia. Estas
corrientes
están
etiquetadas
en la figura
Applying
that
Applying KCL
KCL twice,
twice, once
once at
at node
node dd and
and again
again at
at node
node a,
a, shows
that the
the current
current in
in the
the voltage
voltage source
source and
and the
the
3.2-7. Aplicando
una
vez más
la KCL
en to
el inodo
c,
se
muestra
que
la
corriente
en
el
resistor
de
2
V
es
igual
a iat.
.
These
currents
are
labeled
in
Figure
3.2-7.
Applying
KCL
again,
current
in
the
4-V
resistor
are
both
equal
at
current in the 4-V resistor are both equal to iaa. These currents are labeled in Figure 3.2-7. Applying KCL again, m
Esta corriente
está etiquetada
en lathe
figura 3.2-7.
.
This
current
is
labeled
in
Figure
3.2-7.
node
.
This
current
is
labeled
in
Figure
3.2-7.
node c,
c, shows
shows that
that the
the current
current in
in the 2-V
2-V resistor
resistor is
is equal
equal to
to iim
m
A continuación,
la leyusdethat
Ohm nos
dice que elthe
voltaje resistor
a través
del resistor
de 4that
V es igual
a 4ia y quethe
el
Next,
and that the
the voltage
voltage across
across the
Next, Ohm’s
Ohm’s law
law tells
tells us that the
the voltage
voltage across
across the 4-V
4-V resistor is
is equal
equal to
to 4i
4iaa and
voltaje
a través
del resistor
de
2
V
es
igual
a
2i
.
Ambas
tensiones
están
etiquetadas
en
la
figura
3.2-7.
m
.
Both
of
these
voltages
are
labeled
in
Figure
3.2-7.
2-V
resistor
is
equal
to
2i
. Both of these voltages are labeled in Figure 3.2-7.
2-V resistor is equal to 2im
Aplicando la KCL
enmel nodo b, da
Applying
Applying KCL
KCL at
at node
node bb gives
gives
¼0
�i
� 3i
3iaa �
� iim
�iaa �
m ¼0
Applying
KVL
to
closed
path
a-b-c-e-d-a
gives
Aplicando
la
KVL
a
la
ruta
cerrada
a-b-c-e-d-a
resulta
en
Applying KVL to closed path a-b-c-e-d-a gives
�
� 1 �
�
1i
þ 2im � 12 ¼ 3im � 12
00 ¼
�
12
¼
�4
�
m
i
¼ �4i
�4iaa þ
þ 2i
2im
�
12
¼
�4
�
m
44 m þ 2im � 12 ¼ 3im � 12
Finally,
solving
this
gives
Finalmente,
al resolver
esta ecuación
Finally,
solving
this equation
equation
gives da por resultado
iim
¼ 4A
m ¼ 4A
E j e m p l o 3 . 2 - 5 Leyes de Ohm
y Kirchhoff
E
E XX AA M
MP
PL
LE
E 3
3 .. 2
2 -- 5
5 Ohm’s
Ohm’s and
and
Kirchhoff’s
Kirchhoff’s Laws
Laws
EJEMPLO INTERACTIVO
II N
N TT EE R
RA
AC
C TT II V
V EE EE X
XA
AM
MP
P LL EE
Determine el valor del voltaje, en voltios, medido por el voltímetro en la figura 3.2-8a.
Determine
measured
Determine the
the value
value of
of the
the voltage,
voltage, in
in volts,
volts,
measured by
by the
the voltmeter
voltmeter in
in Figure
Figure 3.2-8a.
3.2-8a.
Voltímetro
FIGURA 3.2-8 (a) Un circuito con fuente dependiente y un
FIGURA 3.2-9 Circuito de la figura 3.2-8b después de etiquetar
voltímetro. (b) El circuito equivalente después de reemplazar
los nodos y algunas corrientes y voltajes del elemento.
el voltímetro por un circuito abierto.
FIGURE
FIGURE 3.2-8
3.2-8 (a)
(a) A
A circuit
circuit with
with dependent
dependent source
source and
and aa
Solución
voltmeter.
(b)
FIGURE
voltmeter.
(b) The
The equivalent
equivalent circuit
circuit after
after replacing
replacing the
the
FIGURE 3.2-9
3.2-9 The
The circuit
circuit of
of Figure
Figure 3.2-8b
3.2-8b after
after labeling
labeling the
the
voltmeter
by
circuit.
Un voltímetro
ideal
equivale a un circuito abierto. El voltaje
medido
por
el
voltímetro
es
el
voltaje
a
través del
nodes
and
some
element
currents
and
voltages.
voltmeter
by an
an open
open
circuit.
nodes and some element currents and voltages.
circuito abierto. La figura 3.2-8b muestra el circuito después de reemplazar el voltímetro por el circuito abierto
Solution
equivalente.
Solution
An
is
to
an
circuit.
voltage
measured
by
the
is
voltage
across
the
Envoltmeter
la figura 3.2-9
aparece el
trazar
para etiquetar
del circuito.
circuito
consta
An ideal
ideal
voltmeter
is equivalent
equivalent
to circuito
an open
open vuelto
circuit.aThe
The
voltage
measuredlos
bynodos
the voltmeter
voltmeter
is the
theEste
voltage
across
the
open
circuit.
Figure
3.2-8b
shows
the
circuit
after
replacing
the
voltmeter
by
the
equivalent
open
circuit.
de
una
fuente
de
voltaje,
una
fuente
de
voltaje
dependiente,
dos
resistores,
un
cortocircuito
y
un
circuito
abierto.
open circuit. Figure 3.2-8b shows the circuit after replacing the voltmeter by the equivalent open circuit.
The
been
in
to
nodes
of
This
circuit
consists
El cortocircuito
el elemento
predominante
la CCVS
y elthe
circuito
un modelo
voltímetro.
The circuit
circuiteshas
has
been redrawn
redrawn
in Figure
Figurede3.2-9
3.2-9
to label
label
the
nodesabierto
of the
theescircuit.
circuit.
This del
circuit
consists of
of aa
voltage
source,
a
dependent
voltage
source,
two
resistors,
a
short
circuit,
and
an
open
circuit.
The
short
circuit
Aplicando
la
KCL
dos
veces,
una
en
el
nodo
d
y
otra
en
el
nodo
a,
se
muestra
que
la
corriente
en
la
fuente
voltage source, a dependent voltage source, two resistors, a short circuit, and an open circuit. The short circuit is
is
the
controlling
element
of
the
CCVS,
circuit
aa model
of
voltmeter.
de voltaje
y la corriente
4 Vthe
sonopen
iguales
a ia. is
Estas
corrientes
etiquetadas en la figura 3.2-9.
the
controlling
element en
of el
theresistor
CCVS,deand
and
the
open
circuit
is
model
of the
theestán
voltmeter.
Applying
KCL
once
dd and
at
a,
shows
current
and
Aplicando
la KCL
unatwice,
vez más,
nodo
c, seagain
muestra
que la
el resistor
dethe
5 Vvoltage
es igualsource
a la corriente
Applying
KCL
twice,
onceenat
atelnode
node
and
again
at node
node
a, corriente
shows that
thatenthe
the
current in
in
the
voltage
source
and the
the
.
These
currents
are
labeled
in
Figure
3.2-9.
Applying
KCL
again,
current
in
the
4-V
resistor
are
both
equal
to
i
at
current in the 4-V resistor are both equal to iaa. These currents are labeled in Figure 3.2-9. Applying KCL again, at
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 60
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
Series
Resistors and
Voltage
Division
Series
Voltage
Series Resistors
Resistors
and
Voltage
Division
Resistores
en serie and
y división
deDivision
voltaje
6161
61
node
c,c,
that
current
inin
5-V
resistor
equal
toto
current
ininthe
circuit,
that
is,is,
This
node
shows
that
the
current
the
5-V
resistor
the
the
open
circuit,
that
zero.
This
node
c,shows
showsabierto,
thatthe
thees
current
inthe
theEsta
5-Vcorriente
resistorisis
is
equal
tothe
the
current
theopen
open
circuit,
thatnos
is,zero.
zero.que
This
en
el circuito
decir, cero.
se equal
etiquetó
en lacurrent
figurain
3.2-9.
La ley
de Ohm
dice
el
current
is
labeled
in
Figure
3.2-9.
Ohm’s
law
tells
us
that
the
voltage
across
the
5-V
resistor
is
also
equal
to
zero.
current
is
labeled
in
Figure
3.2-9.
Ohm’s
law
tells
us
that
the
voltage
across
the
5-V
resistor
is
also
equal
to
zero.
current
is
labeled
in
Figure
3.2-9.
Ohm’s
law
tells
us
that
the
voltage
across
the
5-V
resistor
is
also
equal
to
zero.
voltaje a través del resistor de 5 V también es igual a cero. A continuación, aplicando la KVL a la ruta cerrada
¼¼3i3ia. .
Next,
KVL
totothe
closed
Next,
applying
path
b-c-f-e-b
gives
vvm
Next,applying
applying
KVL
the
closedpath
pathb-c-f-e-b
b-c-f-e-bgives
givesvm
b-c-f-e-b
resultaKVL
vm 5to
3iathe
. closed
m ¼ 3iaa.
Applying
KVL
to
the
closed
path
a-b-e-d-a
gives
Applying
to
path
a-b-e-d-a
gives
Applying
KVL
to the
the
closed
path a-b-e-d-a
a-b-e-d-aresulta
gives
AplicandoKVL
la KVL
a la closed
ruta cerrad
�4i
�4i
12
�4ia aþþ
þ3i3i
3ia a��
�12
12¼¼
¼000
a
por lo que
soso
so
Finally
Finally
Finalmente
Finally
a
iaiia¼¼
�12
¼�12
�12AA
A
a
vm
¼¼3i3ia ¼¼33ð�12
vvm
ð�12Þ Þ¼¼�36
�36VV
m ¼ 3iaa ¼ 3ð�12Þ ¼ �36 V
EXERCISE
3.2-1
Determine
the
ofof
EE
EJERCICIO
3.2-1
Determine
los valores
vand
la
figura
E3.2-1.
3.2-1.
6
EXERCISE
3.2-1
Determine
the
values
,,i4ii,44i,,i36,ii,66i,,4v,2vvi,226,,v, 4vvv,442,,,and
vv666inen
in
Figure
4 yvv
EXERCISE
3.2-1
Determine
thevalues
values
ofi3ii,33de
and
inFigure
Figure
E3.2-1.
3.2-1.
Answer:
i3ii3¼¼
¼¼i333A,
¼¼44A,
¼¼�3
V,
v4vv4¼
V,
v6vv6¼
Answer:
�3
A,
A,
A,
�3
V,
¼
V,
¼
4
6
2
Respuesta:
i�3
5 A,
23
23
v�6
26
Answer:
A,i4iiA,
A,3i6iiA,
A,v42vvA,
¼�6
�6
¼666vV6V
V5 6 V
3 �3
45
2 5V,
4 5V,
3¼
4¼
6 ¼i64 5
2 ¼v�3
4V,
6V,
FIGURE
E 3.2-1
FIGURE
3.2-1
FIGURE
E
3.2-1
FIGURA E
E 3.2-1
3.3
SSEE
RRI IIEE
SSRR
SS
NNDD
VVOO
TT
GGEE
DDI IIVV
I SSI IIOO
3.3
E
S
IISS
O
N
3.3
RER
ESE
SIS
STET
TON
ORR
RS
SEAA
A
OLIL
LV
TA
A
ON
N JE
3.3 S
RE
ER
S IE
SS
TO
RN
I EDYV
D
IA
SG
I ÓEND
DVEII S
VO
LTA
Let
usus
circuit,
asas
inin
3.3-1.
InIn
anticipation
ofof
using
Let
consider
aasingle-loop
single-loop
circuit,
shown
Figure
3.3-1.
anticipation
using
Ohm’s
Let
usconsider
consideraahora
single-loop
circuit,
asshown
shown
inFigure
Figure
3.3-1.
Inse
anticipation
of
usingOhm’s
Ohm’s
Consideremos
un circuito
de circuito
cerrado
único
como
muestra
en
la figura
3.3-1.
law,
the
passive
convention
has
been
used
to
assign
reference
directions
to
resistor
voltages
law,
the
passive
convention
has
been
used
to
assign
reference
directions
to
resistor
voltages
law,
the
passive
convention
has
been
used
to
assign
reference
directions
to
resistor
voltages
Anticipándonos
al
uso
de
la
ley
de
Ohm,
la
convención
pasiva
se
ha
utilizado
para
asignar
and
currents.
and
currents.
and
currents.de referencia a voltajes y corrientes.
direcciones
The
connection
ofof
inin
3.3-1
isis
toto
connection
because
The
connection
resistors
Figure
3.3-1
said
be
aaseries
series
connection
because
The
connection
ofresistors
resistors
inFigure
Figure
3.3-1
issaid
said
tobe
beauna
series
connection
because
Se dice
que
la conexión
decurrent.
resistores
en
la figura
3.3-1
es
conexión
en we
serie
porque
all
the
elements
carry
the
same
To
identify
a
pair
of
series
elements,
look
for
all
the
elements
carry
the
same
current.
To
identify
a
pair
of
series
elements,
we
look
for
all
the
elements
carry
the
same
current.
To
identify
a
pair
of
series
elements,
we
look
for
todos
los
elementos
transportan
la
misma
corriente.
Para
identificar
un
par
de
elementos
en
two
elements
connected
to
a
single
node
that
has
no
other
elements
connected
to
it.
Notice,
two
elements
connected
to
a
single
node
that
has
no
other
elements
connected
to
it.
Notice,
two
elements
connected
to
a
single
node
that
has
no
other
elements
connected
to
it.
Notice,
serie,
buscamos
dos
elementos
conectados
a
un
nodo
único
que
no
tenga
otros
elementos
connected
toto
node
bbband
other
for
example,
that
resistors
RR1 1and
andRR2 2are
are
both
connected
node
and
that
no
other
for
example,
that
resistors
areboth
both
connected
node
andthat
thatno
no
other
for
example,
that
resistors
1 and R2
conectados
a sí.
Observe,
porRejemplo,
que
los
resistores
R
yito
R, ,2so
están
conectados
al nodo
1¼
¼
both
resistors
have
the
circuit
elements
are
connected
to
node
b.
Consequently,
i
1
2
i
so
both
resistors
have
the
circuit
elements
are
connected
to
node
b.
Consequently,
i
1
2
¼
i
,
so
both
resistors
have
the
circuit
elements
are
connected
to
node
b.
Consequently,
i
1
2
b y que
ningún
otros
elemento
del shows
circuito
lo está.
En consecuencia,
i1 also
5also
i2,connected
de
modo que
and
R
are
in
same
current.
A
similar
argument
that
resistors
R
2
3
and
R
are
connected
in
same
current.
A
similar
argument
shows
that
resistors
R
2
3
and
R
are
also
connected
in
same
current.
A
similar
argument
shows
that
resistors
R
2
3
ambosNoticing
resistores
tiene
laismisma
corriente.
Un argumento
parecido
muestra
que
los
resistores
connected
in
series
with
both
R
and
R
,
we
say
that
all
three
series.
that
R
2
1
3
is
connected
in
series
with
both
R
and
R
,
we
say
that
all
three
series.
Noticing
that
R
2
1
3
is
connected
in
series
with
both
R
and
R
,
we
say
that
all
three
series.
Noticing
that
R
2
1
R2 y R3 también
están conectados
en
serie.
Al observar
que
R2 está is3conectado
en serie For
con
resistors
are
inin
The
order
ofof
resistors
not
resistors
are
connected
series.
The
order
series
resistors
is
not
important.
For
resistors
areconnected
connected
inseries.
series.
Theestán
order
ofseries
seriesen
resistors
is
notimportant.
important.
For
R
y
R
,
decimos
que
los
tres
resistores
conectados
serie.
El
orden
de
los
resistores
1
3 the
example,
voltages
and
currents
of
the
three
resistors
in
Figure
3.3-1
will
not
change
ifif
example,
the
voltages
and
currents
of
the
three
resistors
in
Figure
3.3-1
will
not
change
example,
the
voltages
and
currents
of
the
three
resistors
in
Figure
3.3-1
will
not
change
en interchange
serie
no es significativo.
Por
ejemplo,
los
voltajes y corrientes de los tres resistores en laif
and
R
.
we
the
positions
R
2
3
and
R
.
we
interchange
the
positions
R
and R33.
we
interchange
positions si
R22intercambiamos
figura
3.3-1KCL
no
sethe
las posiciones
de obtain
R2 y R3.
Using
atmodificarán
each
node
ofof
the
inin
3.3-1,
Using
KCL
at
each
node
the
circuit
Figure
3.3-1,
we
Using
KCL
at
each
node
of
thecircuit
circuit
inFigure
Figure
3.3-1,we
weobtain
obtain
Al aplicar la KCL en cada nodo
del
circuito
de la figura
3.3-1
obtenemos
aaa: :: isiis ¼¼
i1ii1
¼
s i1
1
a : ibs : 
b : i1ii11 ¼¼
¼ i2ii22
b : ic1b: :
i
2
c : i2ii22 ¼¼
¼ i3ii3
c : id2c: :
i¼
3 i 3
i
3
d : i ¼ si
d : i3d :i33is ¼ iss
Consequently,
isi ¼¼i1ii1¼¼
i2i ¼¼i3i
Consequently,
Consequently,
En consecuencia,
is  iiss1 ¼
 i12 ¼
 ii223 ¼ i33
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 61
s
s
FIGURE
3.3-1
FIGURE
3.3-1
FIGURE3.3-1
3.3-1
FIGURA
Single-loop
circuit
with
a
Single-loop
circuit
with
Single-loop
circuitcerrado
with aa
Circuito
de circuito
voltage
source
vsv.s.
voltage
source
voltage
vs. de
único
consource
una fuente
voltaje vs.
Alfaomega
4/12/11 5:21 PM
The circuit shown in Figure 5.3-1 has one output, vo , and three inputs, v1 , i2 , and v3 . (As expected, the inputs are
voltages of independent voltage sources and the currents of independent current sources.) Express the output as a
linear combination of the inputs.
Solution
62
62
Resistive
Let’s
analyzeCircuits
the circuit using node equations. Label the node voltage at the top node of the current source and
Circuitos resistivos
identify the supernode corresponding to the horizontal voltage source as shown in Figure 5.3-2.
Apply To
KCL to the supernode
get around the loop to obtain
i1,i we
usetoKVL
Paradetermine
determinar
1, nos valemos de la KVL en torno al circuito cerrado para obtener
v1 � ðv3 þ vo Þ
v
þ vv22 
þ vv33þ�i2vvss¼
¼ 00o
vv11 
10
40
voltage
resistor
R1. Using
forpara
each
resistor,
where,
for
example,
donde,
ejemplo,
voltaje
a través
delthe
resistor
R1. Aplicando
lawe
leyhave
delaw
Ohm
cada
resistor,
1v1esiselthe
Multiply
both por
sides
of thisvequation
by
40
toacross
eliminate
the fractions.
ThenOhm’s
v1R
¼vvs4v
5vvvoss
R

R

3þ
22ii332�
11 
R�11ii1ð1v
þ
R22vii2o2Þ
þþR
R40i
¼o 00 )
) R
R11vii111 þ
þ 40i
R22ii211 �
þ vR
R322ii¼
¼
s 
Despejando
i tenemos
Solving for
40 Ωi11, we have
v3
v3
40 Ω
vvss
v3 + vo
¼
ii11 
+–
R1 
þR
R2 
þR
R3
+
R
1
2
3
+ v
i
+ v
vo
1voltage across
10 Ω
i2como
Thus,
the
nth
is vn and
canvnbe
obtained
as
1
10 Ω
Por
lo–the
tanto,
el voltaje2a través
delresistor
enésimoRnresistor
Rn es
y se
obtener
– puede
+–
–
¼ ii11R
Rnn 
¼
vvnn 
FIGURE 5.3-1 The linear circuit for Example 5.3-1.
Por
voltaje
a través
resistorRR
For ejemplo,
example,elthe
voltage
acrossdelresistor
2 2ises
Rnn
vvssR
R11 
þR
R22 
þR
R33
R
+
vo
–
FIGURE 5.3-2 A supernode.
R
R22
vv2 
vs
2 ¼ R1  R2  R3 vs
R 1 þ R2 þ R3
Por lo tanto, el voltaje a través de la combinación de resistores en serie se divide entre los resistores
Thus, the voltage across the series combination of resistors is divided up between the individual
individuales de una manera predecible. El circuito demuestra el principio de la división de voltaje, y
resistors in a predictable way. This circuit demonstrates the principle of voltage division, and the
al circuito se le denomina divisor de voltaje.
circuit is called a voltage divider.
En general, podemos representar el principio del divisor de voltaje por la ecuación
In general, we may represent the voltage divider principle by the equation
Rn
Rn
vn 
vs
vn ¼ R 1 R 2
R N vs
R1 þ R2 þ � � � þ R N
donde vn es el voltaje a través del enésimo resistor de resistores N conectados en serie.
where vn is the voltage across the nth resistor of N resistors connected in series.
Podemos reemplazar los resistores en serie por un resistor equivalente, el cual se ilustra en la
We can replace series resistors by an equivalent resistor. This is illustrated in Figure 3.3-2. The
figura 3.3-2. Los resistores R1, R2 y R3 en la figura 3.3-2a están reemplazados por un único resistor Rs
series resistors R1, R2, and R3 in Figure 3.3-2a are replaced by a single, equivalent resistor Rs in Figure
equivalente, en la figura 3.3-2b. Se dice que Rs debe ser equivalente para los resistores en serie R1, R2
3.3-2b. Rs is said to be equivalent to the series resistors R1, R2, and R3 when replacing R1, R2, and R3 by Rs
y R3, cuando
al reemplazarlos por R no se modifica la corriente o el voltaje de ningún elemento del
does not change the current or voltages of any other element of the circuit. In this case, there is only one
circuito. En este caso sólo hay un elemento en el circuito, que es la fuente de voltaje. Debemos elegir
other element in the circuit, the voltage source. We must choose the value of the resistance Rs so that
el valor de la resistencia Rs para que al reemplazar R1, R2 y R3 por Rs no se modifique la corriente
de
replacing R , R2, and R3 by Rs will not change the current of the voltage source. In Figure 3.3-2a, we have
la fuente de1 voltaje.
En la figura 3.3-2a tenemos
vvss
¼
iiss 
R1 
þR
R2 
þR
R3
R
1
2
3
In Figure
3.3-2b,
have
En
la figura
3.3-2bwe
tenemos
vvss
iiss 
¼R
Rss
Dado
quethe
la corriente
de la fuente
demust
voltaje
ser lainmisma
en amboswe
circuitos,
requiere que
Because
voltage source
current
be debe
the same
both circuits,
require se
that
Rs ¼ R1 þ R 2 þ R3
s
f
s
FIGURE 3.3-2
FIGURA 3.3-2
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 62
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
Series Resistors
Resistors and
and Voltage
Voltage Division
Division
Series
Resistores
en serie and
y división
deDivision
voltaje
Series Resistors
Voltage
63
63
63
63
In general,
general,
the
series connection
connection
ofNN
Nresistores
resistors having
having
resistances
R. .22. R.. .. es
is equivalent
equivalent
to the
the
In
series
of
resistors
resistances
.. RR
to
En
general,the
la conexión
en serie de
con resistencias
R RR
, 1R
equivalente
al resistor
1,, R
N is
N
In
general,
the having
series
connection
of N resistors having resistances 1R1,2R2 . N. . RN is equivalent to the
single
resistor
having
resistance
single
resistor
resistance
único cuya resistencia es
single resistor having resistance
¼ RR11 þ
þ RR22 þ
þ �� �� �� þ
þ RRNN
RRss ¼
R s ¼ R 1 þ R2 þ � � � þ R N
Replacing series
series
resistors
by
an equivalent
equivalent
resistor
does not
not change
change
the current
current
or voltage
voltage
offuente
any other
other
Replacing
resistors
an
resistor
does
the
or
of
any
Reemplazar
resistores
en by
serie
por
un resistor
equivalente
no modifica
la corriente
de la
de
Replacing
series
resistors
by an equivalent
resistor does not change the current or voltage of any other
elementdeof
ofningún
the circuit.
circuit.
element
the
voltaje
otro elemento
del circuito.
element
of the
circuit.
Next,
let’s
calculate
the power
power
absorbed
by the
the series
series
resistors
in en
Figure
3.3-2a:
Next,
let’s
calculate
the
absorbed
by
in
Figure
A
continuación,
calculemos
la potencia
absorbida
por losresistors
resistores
serie3.3-2a:
de la figura 3.3-2a:
Next, let’s calculate the power absorbed by the series resistors in Figure 3.3-2a:
¼ iiss22RR11 þ
þ iiss22RR22 þ
þ iiss22RR33
pp ¼
p ¼ is 2 R1 þ is 2 R2 þ is 2 R3
Doing
littlede
algebra
gives
Con
unaapoco
álgebragives
resulta
Doing
little
algebra
Doing a little algebra gives
(R11 þ
þ RR22 þ
þ RR33)) ¼
¼ iiss22RRss
¼ iiss22(R
pp ¼
p ¼ is 2 (R1 þ R2 þ R3 ) ¼ is 2 Rs
lo
cual is
esequal
igual ato
potencia
el
equivalente
figura 3.3-2b.
3.3-2b.
which
is
equal
tolathe
the
powerabsorbida
absorbedpor
by the
theresistor
equivalent
resistordein
inlaFigure
Figure
3.3-2b. Concluiremos
We conclude
conclude
which
power
absorbed
by
equivalent
resistor
We
which
is power
equal to
the power
by the
equivalent
in Figure
3.3-2b.
Weequivalent
que
potencia
absorbida
por
los resistores
en is
serie
es igual
la
potencia
absorbida
por
elconclude
resistor
that la
the
power
absorbed
by absorbed
series
resistors
is
equal
to resistor
thea power
power
absorbed
by the
the
equivalent
that
the
absorbed
by
series
resistors
equal
to
the
absorbed
by
that
the power absorbed by series resistors is equal to the power absorbed by the equivalent
resistor.
equivalente.
resistor.
resistor.
MP
P LL E
E 3
3 .. 33 -- 11 Voltage
Voltage Divider
Divider
EE XX AA M
EEjXeAmMpPl LoE 33..33--11 Divisor
voltaje
Voltagede
Divider
Let us consider the circuit shown in Figure 3.3-3 and determine the resistance R22 required so that the voltage
Consideremos
circuito
se
muestra
en la3.3-3
figuraand
y determine
la resistencia
parathe
quevoltage
el volLet
us consider
circuit
shown
in voltage
Figure
determine
the resistance
R2 R
required
so that
2 requerida
bethe
1=4
of que
the
source
when
R3.3-3
across
R22 will el
11 ¼ 9 V. Determine the current i when vss ¼ 12 V.
taje
a
través
de
R
sea
de
1>4
de
la
fuente
de
voltaje
cuando
R
5
9
V.
Determine
la
corriente
i
cuando
v
5
across R2 will bes 1=4 of the source voltage when R1 ¼ 9 V.1 Determine the current i when vs ¼ 12 V.s 12 V.
s
Solution
Solution
Solución
Solution
The voltage
voltage across
across resistor
resistor RR2 will
will be
be
The
FIGURE 3.3-3
3.3-3 Voltage
Voltage divider
divider circuit
circuit with
with R
R11 ¼
¼ 99 V.
V.
FIGURE
FIGURE
3.3-3Circuito
Voltagededivider
with con
R1 ¼R91 5
V. 9 V.
FIGURA
3.3-3
divisorcircuit
de voltaje
2
El voltaje
a través
resistor
The
voltage
acrossdel
resistor
R2Rwill
be
2 será
Because we desire v22=vss ¼ 1=4, we have
Because
desire
¼ 1=4, we have
Porque sewe
desea
quevv2=v
2 >svs 5 1>4, tenemos
or
or
o bien
or
RR22
¼ R
vv22 ¼
R2 vvss
R
¼ 11 þ2 R22 vvss
vv22 
R
R11 þ
R
R22
R22
1
¼1
R2 ¼
þ RR22 ¼ 44
RR11 þ
R1 þ R 2 4
¼ 3R
3R22
RR11 ¼
R1 ¼ 3R2
Because R11 ¼
¼ 99 V,
V, we
we require
require that RR22 ¼
¼ 3 V. Using
Using KVL
KVL around
around the
the loop,
loop, we
we have
have
Because
Dado queR
se require
requierethat
queRR22¼5333V.
V.Using
Utilizando
KVL en
circuito
Because
RR1 1¼59 9V,V,we
that
V.
KVL laaround
thetorno
loop,alwe
have cerrado, tenemos
�v
þ vvv11 
þ vvv22 
¼ 000
�v
þ
þ
¼
v ss 
�vss þ v11 þ v22 ¼ 0
¼ iiiR
þ iiiR
or
¼
RR11 
þ
RR22
oorbien
vvvsss 
or
vs ¼ i R11 þ i R22
12
vvs
12
Therefore,
¼ vvsss 
¼ 12
¼ 11 A
A
(3.3-1)
Therefore,
¼
¼
¼
(3.3-1)
12 
iii
Por
lo
tanto,
(3.3-1)
R
þ
R
9
þ33 ¼ 11AA
Therefore,
i ¼ R111 R222 ¼ 9 
(3.3-1)
R 1 þ R2 9 þ 3
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 63
Alfaomega
4/12/11 5:21 PM
E1C03_1
E1C03_1
E1C03_1
11/25/2009
11/25/2009
11/25/2009
64
64
64
64
64
64
64
Circuitos resistivos
Resistive Circuits
Circuits
Resistive
Resistive Circuits
EEjXe AmM
p lPo
en serie
L E33
3. 3. 3- 2
- 2 Resistores
Series Resistors
Resistors
E
E XX AA M
MP
P LL E
E 3 .. 3
3 -- 22 Series
Series Resistors
Para el circuito de la figura 3.3-4a, encuentre la corriente medida por el amperímetro. Luego muestre que la poFor the
the circuit
circuit of
of Figure
Figure 3.3-4a,
3.3-4a, find
find the
the current measured
measured by
by the
the ammeter.
ammeter. Then
Then show
show that the
the power absorbed
absorbed by
For
tencia
los 3.3-4a,
dos resistores
igual ameasured
la alimentada
la fuente.
For theabsorbida
circuit of por
Figure
find theescurrent
current
by thepor
ammeter.
Then show that
that the power
power absorbed by
by
the
two
resistors
is
equal
to
that
supplied
by
the
source.
the
the two
two resistors
resistors is
is equal
equal to
to that
that supplied
supplied by
by the
the source.
source.
Amperímetro
FIGURA 3.3-4 (a) Un circuito con resistores en serie. (b) El circuito, luego de que el amperímetro ideal ha sido reemplazado por
FIGURE
3.3-4 equivalente,
(a) A
A circuitycontaining
containing
series resistors.
resistors.
(b) The
The
circuit
after
the ideal
ideal ammeter
ammeter
has
been replaced
replaced by
the equivalent
equivalent
el
cortocircuito
se le ha agregado
una etiqueta
paracircuit
indicarafter
la corriente
porhas
el been
amperímetro,
FIGURE
3.3-4
series
(b)
the
m
FIGURE
3.3-4 (a)
(a) A circuit
circuit containing
series resistors.
(b) The
circuit
after
the idealmedida
ammeter
has
been replacediby
by. the
the equivalent
short
circuit,
and
a
label
has
been
added
to
indicate
the
current
measured
by
the
ammeter,
i
.
m
short
short circuit,
circuit, and
and aa label
label has
has been
been added
added to
to indicate
indicate the
the current
current measured
measured by
by the
the ammeter,
ammeter, iim..
m
Solución
Solution
Solution
Solution
La
figura 3.3-4b muestra el circuito después de que el amperímetro ideal ha sido reemplazado por el cortocircuito
Figure 3.3-4b shows the circuit after the ideal ammeter has been replaced by the equivalent short circuit and a
Figure
shows
circuit
after
the
ammeter
been
by
the
equivalent
short
and
Figure 3.3-4b
3.3-4b
shows
the
circuit una
afteretiqueta
the ideal
ideal
ammeter
has
been replaced
replaced
bypor
theel
equivalent
shorti circuit
circuit
and aa
equivalente
y se
le hathe
agregado
para
indicarhas
la corriente
medida
amperímetro,
. Aplicando
Applying KVL
KVL gives
gives m
label has
has been
been added
added to
to indicate
indicate the
the current
current measured
measured by
by the
the ammeter,
ammeter, iim
m.. Applying
label
label
hasresulta
been added to indicate the current measured by the ammeter, im. Applying KVL gives
la
KVL
15 þ
þ 5im
þ 10i
10im ¼
¼ 00
15
15 þ 5i
5imm þ
þ 10im
m ¼ 0
La corriente
medida porbyelthe
amperímetro
The
current measured
measured
ammeter is
ises
The
The current
current measured by
by the
the ammeter
ammeter is
15
15 ¼
¼�
� 15
A
¼ �1 A
m ¼
iiim
�1 A
5þ
þ 10
10 ¼ �1
m ¼ �5
5 þ 10
negative?
Why¿Por
can’t
wenojust
just
divide
the source
source
voltage
by de
thelaequivalent
equivalent
resistance?
Recall
that when
when
(Why
is iim
(¿Por
imnegative?
es negativa?
qué
se divide
puede
sólo
dividirvoltage
el voltaje
fuente entre
la resistencia
equivalente?
Why
can’t
we
the
by
resistance?
Recall
that
(Why
is
negative?
Why
can’t
we
just
divide
the
source
voltage
by the
the equivalent
resistance?
Recall
that when
(Why qué
is im
m
we
use
Ohm’s
law,
the
voltage
and
current
must
adhere
to
the
passive
convention.
In
this
case,
the
current
Recuerde
que
cuando
se
utiliza
la
ley
de
Ohm,
el
voltaje
y
la
corriente
se
deben
apegar
a
la
convención
pasiva.
En
we
we use
use Ohm’s
Ohm’s law,
law, the
the voltage
voltage and
and current
current must
must adhere
adhere to
to the
the passive
passive convention.
convention. In
In this
this case,
case, the
the current
current
calculated
by
dividing
the
source
voltage
by
the
equivalent
resistance
does
not
have
the
same
reference
direction
este
caso,
la
corriente
calculada
al
dividir
el
voltaje
de
la
fuente
entre
la
resistencia
equivalente
no
tiene
la
misma
calculated
calculated by
by dividing
dividing the
the source
source voltage
voltage by
by the
the equivalent
equivalent resistance
resistance does
does not
not have
have the
the same
same reference
reference direction
direction
so we
we
need aa minus
minus
sign.)
as
im,, so
dirección
de referencia
que sign.)
isign.)
need
as
m, por lo que se necesita un signo menos.)
,
so
we
need
a
minus
as iim
m
The
total power
power
absorbed
by
the
two
resistors
ises
La
total absorbed
absorbidaby
porthe
lostwo
dosresistors
resistoresis
The
total
Thepotencia
total power
absorbed
by
the
two
resistors
is2
�� ��
2
2
22 ¼ 15 �1222 � ¼ 15 W
p
¼
5i
þ
10i
m
m
2
ppRR ¼
5i
þ
10i
¼
15
1 ¼
m
m
¼ 15
15 W
W
R ¼ 5im þ 10im ¼ 15 1
The
power
supplied
by
the
source
is
The
power
supplied
by
the
source
is
La
alimentada
fuenteises
Thepotencia
power supplied
by por
the la
source
¼ �v
�vss iim
¼ �15
�15ðð�1
�1ÞÞ ¼
¼ 15
15 W
W
m ¼
pppss ¼
¼ �15
pss ¼ �vvss iim
15ð1�11Þ2 ¼
 15
15 W
W
m
Thus, the
the power
power supplied
supplied by
by the
the source
source is
is equal
equal to
to that
that absorbed
absorbed by
by the
the series connection
connection of
of resistors.
resistors.
Thus,
Thus,
the
power
supplied
by
the
source
is
equal
to
that
absorbed
by the series
series
resistors.
Por lo tanto, la potencia alimentada por la fuente es igual a la absorbida
por la connection
conexión enofserie
de resistores.
E jEeXmApMl o
delDivider
divisor de
voltaje
PL
L3
E .3
33.-33-- 3
3 Diseño
Voltage
Design
E
E XX AA M
MP
P LE
E 3 .. 3
3 - 3 Voltage
Voltage Divider
Divider Design
Design
La entrada
divisor
de divider
voltaje in
deFigure
la figura
3.3-5
es el
voltaje,v v, sof
, de
fuente source.
de voltaje.
salidais es
voltaje
The
input to
toalthe
the
voltage
3.3-5
is the
the
voltage,
thelavoltage
voltage
TheLa
output
theelvoltage,
voltage,
The
input
divider
in
3.3-5
is
voltage,
vvsss,, of
the
The
output
is
Themedido
input topor
theelvoltage
voltage
divider
in Figure
Figure
3.3-5 de
is the
voltage,
of especifique
the voltage source.
source.
Thede
output
is the
the voltage,
voltímetro.
Diseñe
elthe
divisor
voltaje;
es
decir,
los
valores
las resistencias,
measured
by
the
voltmeter.
Design
voltage
divider;
that
is,
specify
values
of
the
resistances,
R
and
R22,, Rto
to1
vvvooo,,, measured
by
voltmeter.
Design
the
by the
the las
voltmeter.
Design
the voltage
voltage divider;
divider; that
that is,
is, specify
specify values
values of
of the
the resistances,
resistances, R
R111 and
and R
R2, to
yvoR, measured
satisfacer
siguientes
especificaciones.
2 para
satisfy
both
of
these
specifications.
satisfy
specifications.
satisfy both
both of
of these
these
Especificación
Los specifications.
voltajes
entradavoltages
y salidaare
están
relacionados
por vvso. 5 0.8 vs.
¼ 0.8
0.8
Specification
1:1 The
The
input
anddeoutput
output
related
by vvoo ¼
Specification
1:
input
voltages
are
related
by
0.8 vde
vss.. 1 mW de potencia cuando la entrada
Specification 1:2 The
input and
and
output de
voltages
are
related
by vno
o ¼más
Especificación
Se requiere
la
fuente
voltajeto
para
alimentar
Specification
2:
The
voltage
source
is
required
supply
no
more
than
1 mW of
of power when
when the input
input to the
the
Specification
2:
The
voltage
source
is
required
to
supply
no
more
than
Specification
2: Thesea
voltage
source
is required to supply no more than 11 mW
mW of power
power when the
the input to
to the
al
divisordivider
de voltaje
v
5
20
V.
s
¼
20
V.
voltage
is
v
voltage
¼ 20
20 V.
V.
voltage divider
divider is
is vvsss ¼
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 64
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
Resistores en serie y división de voltaje
65
Voltímetro
vs
vo
Divisor de voltaje
FIGURA 3.3-5 Divisor de voltaje.
Solución
Analizaremos cada especificación para ver a qué se refiere respecto de los valores del resistor.
Especificación 1: Los voltajes de entrada y salida del divisor de voltaje se relacionan por
R2
vs
vo R1 R2
Por lo que la especificación 1 requiere
R2
¼ 0:8 ) R2 ¼ 4R1
R 1 þ R2
Especificación 2: La potencia alimentada por la fuente de voltaje es resultado de
vs
vs 2
vs ps i s v s R1 R2
R1 R2
Por lo que la especificación 2 requiere
202
R 1 þ R2
La combinación de estos resultados da
0:001 ) R1 þ R2 400 103 ¼ 400 kV
5R1 400 kV
La solución no es única. Una solución es
R1 ¼ 100 kV y R2 ¼ 400 kV
EJERCICIO 3.3-1 Determine el voltaje medido por el voltímetro en el circuito mostrado
en la figura E 3.3-1a.
Sugerencia: La figura E 3.3-1b muestra el circuito luego de que el voltímetro ideal ha sido reemplazado por el circuito abierto equivalente y se ha agregado una etiqueta para indicar el voltaje medido
por el voltímetro, vm.
Respuesta: vm ⫽ 2 V.
Voltímetro
FIGURA E 3.3-1 (a) Divisor de voltaje. (b) El divisor de voltaje después de que el voltímetro ideal ha sido reemplazado
por el circuito abierto equivalente y se ha agregado una etiqueta para indicar el voltaje medido por el voltímetro, vm.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 65
Alfaomega
5/24/11 10:02 AM
E1C03_1
E1C03_1 11/25/2009
11/25/2009
66
6666
6666
Circuitos resistivos
Resistive
ResistiveCircuits
Circuits
EJERCICIO 3.3-2 Determine el voltaje medido por el voltímetro en el circuito que se muestra
en la figura3.3-2
E3.3-2
3.3.2a.Determine
EXERCISE
EXERCISE
Determinethe
thevoltage
voltagemeasured
measuredbybythe
thevoltmeter
voltmeterininthe
thecircuit
circuitshown
showninin
Figure
FigureEE3.3-2a.
3.3-2a.
Voltímetro
FIGURA E 3.3-2 (a) Un divisor de voltaje. (b) El divisor de voltaje luego de que el voltímetro ideal ha sido reemplazado
por el circuito abierto equivalente y se ha agregado una etiqueta para indicar el voltaje medido por el voltímetro, vm.
FIGURE
FIGUREE E3.3-2
3.3-2(a)(a)AAvoltage
voltagedivider.
divider.(b)(b)The
Thevoltage
voltagedivider
dividerafter
afterthe
theideal
idealvoltmeter
voltmeterhas
hasbeen
beenreplaced
replacedbybythe
the
Sugerencia:
Lacircuit
figura
muestra
eltocircuito
después
de
que el by
voltímetro
idealvmvha
equivalent
equivalentopen
opencircuit
and
andaElabel
a 3.3-2b
labelhas
hasbeen
beenadded
added
toindicate
indicatethe
thevoltage
voltage
measured
measured
bythe
thevoltmeter,
voltmeter,
.m. sido reem-
plazado por el circuito abierto equivalente y se ha agregado una etiqueta para indicar el voltaje medido
Hint:
Hint:
FigureEE3.3-2b
3.3-2b
showsthe
thecircuit
circuitafter
afterthe
theideal
idealvoltmeter
voltmeterhas
hasbeen
beenreplaced
replacedbybythe
theequivalent
equivalent
por
elFigure
voltímetro,
vm. shows
open
opencircuit
circuitand
anda alabel
labelhas
hasbeen
beenadded
addedtotoindicate
indicatethe
thevoltage
voltagemeasured
measuredbybythe
thevoltmeter,
voltmeter,vmv.m.
Respuesta: vm 5 22 V
Answer:
Answer:vmvm¼¼�2
�2VV
3.4
E S I S T O R E S E N PA R A L E L O Y
R
DIVISIÓN DE LA CORRIENTE
3.4
3.4 PPAARRAALLLLEELLRREESSI S
I STTOORRSSAANNDDCCUURRRREENNTTDDI V
I VI S
I SI O
I ONN
Los elementos de circuito, como los resistores, están conectados en paralelo cuando el voltaje a través
Circuit
Circuitelements,
elements,such
suchasasresistors,
resistors,are
areconnected
connectedininparallel
parallelwhen
whenthe
thevoltage
voltageacross
acrosseach
eachelement
elementisis
de cada elemento es idéntico. Los resistores de la figura 3.4-1 están conectados en paralelo. Observe,
identical.
identical.The
Theresistors
resistorsininFigure
Figure3.4-1
3.4-1are
areconnected
connectedininparallel.
parallel.Notice,
Notice,for
forexample,
example,that
thatresistors
resistorsRR
11
por ejemplo, que los resistores R1 y R2 están conectados cada uno al nodo a y al nodo b. En consecuencia,
and
andRR
each
eachconnected
connectedtotoboth
bothnode
nodea aand
andnode
nodeb.b.Consequently,
Consequently,v1v¼
v2v,2so
, soboth
bothresistors
resistorshave
havethe
the
2 are
2 are
1¼
v1 5 v2, de modo que ambos resistores tienen el mismo voltaje. Un argumento semejante muestra que los
andRR
arealso
alsoconnected
connectedininparallel.
parallel.
same
samevoltage.
voltage.AAsimilar
similarargument
argumentshows
showsthat
thatresistors
resistorsRR
2 2and
3 3are
resistores R2 y R3 también están conectados en paralelo. Observando
que R2 está conectado en paralelo
is
is
connected
connected
in
in
parallel
parallel
with
with
both
both
R
R
and
and
R
R
,
,
we
we
say
say
that
that
all
all
three
three
resistors
resistors
are
are
Noticing
Noticingthat
thatRR
22
11
3
con los resistores
R1 y R3, decimos que estos tres resistores
están 3conectados
en paralelo. El orden de los
connected
connectedininparallel.
parallel.
The
Theorder
orderofofparallel
parallelresistors
resistorsisisnot
notimportant.
important.For
Forexample,
example,the
thevoltages
voltagesand
and
resistores en paralelo no es importante. Por ejemplo, los voltajes y las corrientes de los tres resistores de
andRR
currents
currentsofofthe
thethree
threeresistors
resistorsininFigure
Figure3.4-1
3.4-1will
willnot
notchange
changeififwe
weinterchange
interchangethe
thepositions
positionsRRand
3.3.
la figura 3.4-1 no se modificarán si se intercambian las posiciones de los resistores R2 y R3. 2 2
The
Thedefining
definingcharacteristic
characteristicofofparallel
parallelelements
elementsisisthat
thatthey
theyhave
havethe
thesame
samevoltage.
voltage.To
Toidentify
identify
La característica determinante de los elementos en paralelo es que tienen el mismo voltaje. Para
a apair
pairofofparallel
parallelelements,
elements,we
welook
lookfor
fortwo
twoelements
elementsconnected
connectedbetween
betweenthe
thesame
samepair
pairofofnodes.
nodes.
identificar un par de elementos en paralelo se deben buscar dos elementos entre el mismo par de nodos.
Consider
Considerthe
thecircuit
circuitwith
withtwo
tworesistors
resistorsand
anda acurrent
currentsource
sourceshown
shownininFigure
Figure3.4-2.
3.4-2.Note
Notethat
that
Considere el circuito con dos resistores y una fuente de corriente que se muestra en la figura
both
bothresistors
resistorsare
areconnected
connectedtototerminals
terminalsa aand
andb band
andthat
thatthe
thevoltage
voltagev vappears
appearsacross
acrosseach
eachparallel
parallel
3.4-2. Observe que ambos resistores están conectados a las terminales a y b y que el voltaje v aparece
element.
element.InInanticipation
anticipationofofusing
usingOhm’s
Ohm’slaw,
law,the
thepassive
passiveconvention
conventionisisused
usedtotoassign
assignreference
reference
a través de cada elemento en paralelo. Antes de utilizar la ley de Ohm, se aplica la convención pasiva
directions
directionstotothe
theresistor
resistorvoltages
voltagesand
andcurrents.
currents.We
Wemay
maywrite
writeKCL
KCLatatnode
nodea a(or
(oratatnode
nodeb)b)totoobtain
obtain
para asignar direcciones de referencia a los voltajes y corrientes de resistores. Podríamos escribir la
�i1i1��i2i2¼¼0 0
is i�
sobtener
KCL en el nodo a (o en el b, inclusive) para
oror
o bien
However,
However,
from
from
Ohm’s
Ohm’s
lawOhm
Sin
embargo,
por
la leylaw
de
Alfaomega
is 2 i1 2 i2 5 0
is is¼¼i1i1þþi2i2
is 5 i1 1 i2
vv
vv
e i2i2¼¼
and
and
i1i1¼¼
RR
RR
11
22
FIGURA 3.4-1 Un circuito con resistores en paralelo
FIGURE
FIGURE3.4-1
3.4-1AAcircuit
circuitwith
withparallel
parallelresistors.
resistors.
M03_DORF_1571_8ED_SE_053-107.indd 66
FIGURA 3.4-2 Circuito en paralelo con una fuente de corriente.
FIGURE
FIGURE3.4-2
3.4-2Parallel
Parallelcircuit
circuitwith
witha current
a currentsource.
source.
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
Parallel Resistors
Resistors and
and Current
Current Division
Division
Parallel
Parallel
Parallel Resistors
Resistors and
and Current
Current Division
Division
Parallel
Resistors
and Current
Current
Division
Parallel
Resistors
and
Division
Resistores
en paralelo
y división
de la coriente
Parallel
Resistors
and
Current
Division
67
67
67
67
67
67
67
Then
Then
Then
Then
Entonces
Then
Then
Then
vv
vv
¼ vvvv þ
þ vvvv
(3.4-1)
iiiiissss ¼
(3.4-1)
(3.4-1)
¼
þ
¼
þ
(3.4-1)
R
R
v1111 þ
v2222
i ¼
¼R
þR
(3.4-1)
RR
RR
(3.4-1)
iisss ¼
þ
(3.4-1)
R
R
R11 R
R22
R
1
2 resistance R. We may therefore rewrite Eq.
Recall that
that
weya
defined
conductance
G as
asthe
the
inverse
of
Recall
we
defined
conductance
G
inverse
of
resistance
We
therefore
rewrite
Eq.
Recall
that
we
defined
conductance
G
as
the
inverse
of
resistance
R.
We
may
therefore
rewrite
Eq.
Recuerde
que
se
definió
la
conductancia
G
como
la
inversa
de laR.
resistencia
Por lo tanto,
podeRecall
that
we
defined
conductance
G
as
the
inverse
of
resistance
R.
We may
may R.
therefore
rewrite
Eq.
Recall
that
we
defined
conductance
G
as
the
inverse
of
resistance
R.
We
may
therefore
rewrite
Eq.
Recall
that we
we
defined
conductance
G as
as the
the inverse
inverse of
of resistance
resistance
R.
We
may
therefore
rewrite
Eq.
3.4-1escribir
as
3.4-1
as
Recall
that
defined
conductance
G
R.
We
may
therefore
rewrite
Eq.
3.4-1
as
mos
la
ecuación
3.4-1
como
3.4-1
as
3.4-1 as
as
3.4-1
3.4-1
as
¼G
G11vvvvþ
þG
G22vvvv ¼
¼ ððððG
G11 þ
þG
G22Þv
Þv
(3.4-2)
(3.4-2)
iiiiissss ¼
¼
G
G
G
(3.4-2)
¼
G
þ
G
¼
G
þ
G
Þv
(3.4-2)
11 þ
22 ¼
11 þ
22Þv
¼
G
vþ
þ
G
v¼
¼
ðG
G
þ
G
Þv
(3.4-2)
1v
2v
1þ
2 Þv
¼
G
G
ð
G
G
(3.4-2)
iiiss ¼
1
2
1
2
G1 v þ G2 v ¼ ðG1 þ G2 Þv
(3.4-2)
s
as shown
shown
in
Thus,
theel
equivalent
circuit for
forpara
this
parallel
circuit
is aaaa conductance
conductance
Gpp,,,, as
in
Thus,
the
equivalent
circuit
this
parallel
circuit
is
G
Por
tanto,
circuito equivalente
este
circuito
en paralelo
es una conductancia
Gp, como
shown
in
Thus,
the
equivalent
circuit
for
this
parallel
circuit
is
conductance
G
as
shown
in
Thus,
the
equivalent
circuit
for
this
parallel
circuit
is
conductance
G
pp as
,
as
shown
in
Thus,
the
equivalent
circuit
for
this
parallel
circuit
is
a
conductance
G
as shown
shown in
in
Thus,
the
equivalent
circuit
for this
this parallel
parallel circuit
circuit is
is aa conductance
conductance G
Gpp,, as
Figure
3.4-3,
where
Figure
3.4-3,
where
Thus,
the
equivalent
for
se
muestra
en la
figuracircuit
3.4-3, donde
Figure
3.4-3,
where
Figure
3.4-3,
where
p
f
Figure
3.4-3,
where
Figure
3.4-3,
where
Figure 3.4-3, where
G
¼
G
þ
G
p
1
2
G
¼
G
þ
G
p
1
2
G
Gpp ¼
¼G
G11 þ
þG
G22
G
¼
G
þ
G
G
pp ¼ G 11 þ G 22
G
p ¼ G1 þ G2
Theresistencia
equivalentequivalente
resistance for
for the
the
two-resistor
circuit
is found
found
from en
The
equivalent
resistance
two-resistor
circuit
is
from
FIGURE 3.4-3
3.4-3
La
para
el circuito
de doble
resistor
se encuentra
FIGURA
3.4-3
The
equivalent
resistance
for
the
two-resistor
circuit
is
found
from
FIGURE
The
equivalent
resistance
for
the
two-resistor
circuit
is
found
from
FIGURE
FIGURE 3.4-3
3.4-3
The
equivalent
resistance
for
the
two-resistor
circuit
is
found
from
FIGURE
3.4-3
The
equivalent
resistance
for
the
two-resistor
circuit
is
found
from
Equivalent
circuitfor
foraaa
FIGURE
3.4-3
Circuito
equivalente
Equivalent
circuit
The
equivalent
resistance
for
the
two-resistor
circuit
is
found
from
Equivalent
circuit
FIGURE
Equivalent3.4-3
circuitfor
forpara
11 1111
Equivalent
circuit
for
aa
Gpp ¼
¼ 1111 þ
þ 11
parallel
circuit.
un
circuito
en
paralelo.
Equivalent
circuit
for
G
parallel
circuit.
G
¼
þ
G
¼
þ
parallel
circuit.
pp ¼ R
Equivalent
circuit for aa
parallel
circuit.
R
R
1
1
1
2
R
G
þ
parallel circuit.
circuit.
1 þR
2
Gpp ¼
¼R
parallel
G
RR1111 þ
RR2222
parallel
circuit.
p
R
R
R
R
Because
G
¼
1=R
,
we
have
1
2
Because
G
¼
1=R
,
we
have
p
p
p
Because
have
Because
GGpppp¼
¼51=R
1=R
we
have
pp,, ,we
Dado
queG
1>R
tenemos
Because
G
¼
1=R
we
have
p¼
p,,p we
Because
G
1=R
have
p
p
Because
G
11 1111
1111
p ¼ 1=Rp, we have
¼ 1111 þ
þ 11
1 ¼
¼
1
¼
þ
R
R
R
11111 þ
122
1pppp ¼
RR
R
¼R
þ
R
R
þ
R
R
Rpp ¼ R
R11 þ R
R2222
R
R
R p R1 R 2
or
or
or
or
or
or
o
bien
or
R22
R11R
RR
R
¼ R
(3.4-3)
Rpp ¼
(3.4-3)
R
RR2222
R1111R
¼
(3.4-3)
R
R
¼
(3.4-3)
R
p
p
R1R1 1þ
þ
R22
RR
¼R
(3.4-3)
Rpp ¼
2R
R
þ
R
(3.4-3)
R
R
þ
1
2
1
2
(3.4-3)
Rp ¼ R
R þ
þR
R
R111 þ
R222
Note
that
the
total
conductance,
G
,
increases
as
additional
parallel
elements
are
added
and
that
the
Note
that
the
total
conductance,
G
,
increases
as
additional
parallel
elements
are
added
and
that
p
p
Note
that
the
total
conductance,
G
,
increases
as
additional
parallel
elements
are
added
and
that
the
Note that
that the
the total
total conductance,
conductance, G
Gpp,, increases
increases as
as additional
additional parallel
parallel elements
elements are
are added
added and
and that
that the
the
Note
the
p, increases
Note
that
the total
total
conductance,
G
asadded.
additional parallel
parallel elements
elements are
are added
added and
and that
that the
the
,
declines
as
each
resistor
is
added.
total
resistance,
R
p
total
resistance,
R
,
declines
as
each
resistor
is
p
Note
that
the
conductance,
G
,
increases
as
additional
p
,
declines
as
each
resistor
is
added.
total
resistance,
R
declines total,
as each
each
is added.
added.
total resistance,
resistance,
pGp,resistor
Observe
que la conductancia
se incrementa
conforme se agregan los elementos adicionales
pp,, declines
as
resistor
is
total
RR
p, declines
total
resistance,
R
as
each
resistor
is
added.
Thecircuit
circuit
shown
inFigure
Figure
3.4-2
isdisminuye
called
acurrent
current
divider
circuit
because
itdivides
dividesthe
thesource
source
p
The
shown
in
3.4-2
is
called
aaadded.
total
resistance,
R
, declines
as
each
resistor
is
The
circuit
shown
in
Figure
3.4-2
is
called
current
divider
circuit
because
it
divides
the
source
The
circuit
shown
in
Figure
3.4-2
is
called
a
current
divider
circuit
because
it
divides
the
source
pla
en paralelo,
y
que
resistencia
total,
R
,
a
cadadivider
resistorcircuit
que sebecause
agrega.it
p
The
circuit
shown
in
Figure
3.4-2
is
called
a
current
divider
circuit
because
it
divides
the
source
The
circuit
shown
in
Figure
3.4-2
is
called
a
current
divider
circuit
because
it
divides
the
source
current.
Note
that
current.
that
The
circuit
Figureen
3.4-2
is called
a current
dividercircuito
circuit because
it divides
the porque
source
current.
Note
that
current.
Note
thatshown
El Note
circuito
que se in
muestra
la figura
3.4-2
se denomina
divisor de
corriente
current.
Note
that
current.
Note
that
current.
Note
that
divide la corriente de la fuente. Observe que ii1 ¼
¼G
G11vvvv
(3.4-4)
(3.4-4)
G
(3.4-4)
iii111 ¼
¼
G
(3.4-4)
11
¼
G
v
(3.4-4)
1v
¼
G
(3.4-4)
ii11 ¼
1
(3.4-4)
G1 v
1
¼(G
(G11 þ
þG
G22)v,
)v, we
we solve
solve for
for v,
v,
obtaining
Also, because
because iiiiss¼
obtaining
Also,
(G
G
we
solve
for
v,
obtaining
Also,
because
¼
(G
þ
G
)v,
we
solve
for
v,
obtaining
Also,
because
ss¼
11 þ
22)v,
¼ (G
(G11 þ
þG
G22)v,
)v, we
we solve
solve for
for v,
v, obtaining
obtaining
Also, because
because iiss ¼
Also,
)v,Gwe
for v,
obtaining
Also,
because
Además,
dado ique
is 5
(GG
se obtiene
para
s ¼ (G
1 þ
1 21
2)v,solve
iiviissss
¼
(3.4-5)
vvvv ¼
(3.4-5)
iss
(3.4-5)
¼ GG1 þ
(3.4-5)
þ
G22
iiiþ
v¼
¼
(3.4-5)
ss G
111 þ
G
G
¼
(3.4-5)
G
G
22
vvv 
¼
(3.4-5)
G11 
þG
G22
G
þ
G
G
G
þ
G
11obtain
22
Substituting vvvv from
from Eq.
Eq. 3.4-5
3.4-5 into
into Eq.
Eq. 3.4-4,
3.4-4, we
we
Substituting
obtain
Substituting
from
Eq.
3.4-5
into
Eq.
3.4-4,
we
obtain
Substituting
from
Eq.
3.4-5
into
Eq.
3.4-4,
we
obtain
Substituting
v
from
Eq.
3.4-5
into
Eq.
3.4-4,
we
obtain
Substituting
from laEq.
Eq.
3.4-5 into
into
Eq.obtenemos
3.4-4, we
we obtain
obtain
Al sustituir vvvdesde
ecuación
3.4-5,
Substituting
from
3.4-5
Eq.
3.4-4,
G11iiiiss
G
G
11 ss
¼ G
(3.4-6)
(3.4-6)
iiii1111 ¼
G
i
¼
(3.4-6)
G
¼
(3.4-6)
G
11 iissG 2
G
þ
1þ
¼G
(3.4-6)
11 issG
11G
222
G
þ
G
¼
(3.4-6)
iiii111 
G
þ
G
1
¼
(3.4-6)
G11 
þG
G22
1
G
þ
G
G
G
11 þ G 22
G22iiiiss
G
G
22 ss
Similarly;
¼ G
Similarly;
iiii2222 ¼
i
G
Gþ
Similarly;
¼
G
Similarly;
¼
222iisssG2
G
1þ
Similarly;
ii222 ¼
¼
2 isG
G
11G
G
þ
G
Similarly;
i
G
þ
G2
1þ
Similarly;
i
¼
G
G
2
G111 þ
G
G22222
G
Del mismo modo,
G
þ
G
1
2 the current i in terms of two resistances as
Notethat
thatwe
wemay
mayuse
useG
G22¼
¼1=R
1=R22and
andG
G11¼
¼1=R
1=R11to
toobtain
obtain
Note
the
Note
that
we
may
use
G
1=R
G
1=R
obtain
the
current
in
terms
of
two
resistances
as
Note
that
we
may
use
G
¼
1=R
and
G
¼
1=R
to
obtain
the current
current iiii2222in
in terms
terms of
of two
two resistances
resistances as
as
22¼
22 and
11¼
11 to
Note
that
we
may
use
G
¼
1=R
and
G
¼
1=R
to
obtain
the
current
in
terms
of
two
resistances
as
2 ¼ 1=R2
1 to obtain the
Note
thatque
we la
may
use G
Gde
and
G
¼ 1=R
1=R
current
in
terms
of
two
resistances
as
follows:
2
1
follows:
Observe
manera
G2 G
51111>R
Gto
1>R1,the
para
obtener
la terms
corriente
i2 enresistances
términos de
Note
that
we
may
use
1=R22 and
¼
obtain
current
ii222 in
of
two
as
follows:
2 y1
15
follows:
2 ¼utilizar
follows:
follows:
dos resistencias como sigue:
follows:
R11iiss
RR
¼ R
iiii2222 ¼
R1111iiiiissss
¼
R
R
¼
R
R22
R
isR
¼ RR1111þ
i ¼
1þ
þ
R
þ
¼R
iii222 
R11 
þ
RR2222
R
þ
R
R
R
R 1 þ R2
Thecurrent
currentof
ofthe
thesource
sourcedivides
dividesbetween
betweenconductances
conductancesG
G11and
andG
G22in
inproportion
proportionto
totheir
theirconductance
conductance
The
The
current
of
the
source
divides
between
conductances
G
G
proportion
to
their
conductance
The
current
of
the
source
divides
between
conductances
G
and
G
in
proportion
to
their
conductance
11 and
22 in
The
current of
of
the
source
divides
between
conductances
G
and
G
in
proportion
to
their
conductance
1G
2 in
The
current
the
source
divides
between
conductances
G
and
G
proportion
to
their
conductance
values.
La
corriente
de
la
fuente
se
divide
entre
las
conductancias
y
G
en
proporción
con
sus
valores de
1
2
values.
1 G22in proportion to their conductance
The
current of the source divides between conductances G1 and
values.
values.
values.
values.
conductancia.
values.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 67
Alfaomega
4/12/11 5:21 PM
68
68
68
68
68
68
68
68
68
68
Resistive
Circuits
Resistive
Circuits
Resistive
Circuits
Resistive
Circuits
ResistiveCircuits
Circuits
Resistive
Circuits
Resistive
Circuits
Resistive
Resistive Circuits
Circuitos resistivos
FIGURE
3.4-4
FIGURE
3.4-4
FIGURE
FIGURE3.4-4
3.4-4
FIGURE
3.4-4
FIGURE
3.4-4
Set
of
NNparallel
parallel
Set
of
parallel
Set
of
N
FIGURE
3.4-4
Setof
ofN
Nparallel
parallel
Set
ofof
NN
parallel
Set
parallel
FIGURA
3.4-4
Set
conductances
conductances
conductances
Set
of N parallel
conductances
conductances
Conjunto
de N
conductances
conductances
with
current
with
current
with
aaaaacurrent
conductances
with
current
with
current
conductancias
with
a
current
with
a
current
source
source
source
iisssii..ss.con
with
a current
source
paralelas
source
issi.s..
source
source
is. de
una fuente
corriente is.
Let
us
consider
the
more
general
case
of
current
division
with
set
of
N
parallel
conductors
as
Let
Letus
usconsider
considerthe
themore
moregeneral
generalcase
caseof
ofcurrent
currentdivision
divisionwith
withaaaaaset
setof
ofN
Nparallel
parallelconductors
conductorsas
as
Let
us
consider
the
more
general
case
of
current
division
with
set
of
NN
parallel
conductors
as
Let
us
consider
the
more
general
case
of
current
division
with
set
of
parallel
conductors
as
shown
in
Figure
3.4-4.
The
KCL
gives
shown
in
Figure
3.4-4.
The
KCL
gives
Let
us
consider
the
more
general
case
of
current
division
with
a
set
of
N
parallel
conductors
as
Consideremos
el caso
común
shown
inFigure
Figure3.4-4.
3.4-4.
Themás
KCL
givesde división de corriente con un conjunto de conductores N en
shown
inin
The
KCL
gives
shown
Figure
3.4-4.
The
KCL
gives
shown
Figure
The
gives
¼
þ
þ
þ
þ
(3.4-7)
iis3.4-4.
paraleloincomo
se 3.4-4.
muestra
en KCL
la figura
La
da
¼
þ
þ
þ
þ
(3.4-7)
s¼
þ
þ
(3.4-7)
¼iii111ii1i111þ
þii22i2i2KCL
þiii333ii3i333þ
þ������������þ
þiiiNNNiiNiNNN
(3.4-7)
2þ
¼
þ
þ
(3.4-7)
(3.4-7)
iissssiiss¼
1 þi22i2þ
3 þ� �� �� �þ
N
iis ¼ i1 þ i2 þ i3 þ � � � þ iN
(3.4-7)
for
which
for
which
for
forwhich
which
for
which
for
parawhich
la cual
for
which
¼
G
(3.4-8)
(3.4-8)
iinnnininn¼
¼G
Gnnnnnnnnvvvvv
(3.4-8)
GG
(3.4-8)
¼
(3.4-8)
nin ¼
¼
G
v
(3.4-8)
i
n
n
for
n
¼
1,
.
.
.
,
N.
We
may
write
Eq.
3.4-7
as
for
n
¼
1,
.
.
.
,
N.
We
may
write
Eq.
3.4-7
as
fornnnque
¼1,1,
1,n 5
N.
We
maywrite
write
Eq.3.4-7
3.4-7
as
for
¼¼
. ... ..1,
. ..,.,,N.
We
may
Eq.
as
for
We
may
write
Eq.
3.4-7
as
para
. N.
. , N.
Podemos
escribir
la ecuación
3.4-7 como
for n ¼ 1, . . . , N. We may write
Eq.
3.4-7
as
¼
(G
þ
GG
þ
GG
þ
þ
GG
)v
(3.4-9)
¼
(G
þ
þ
þ
þ
)v
(3.4-9)
11111þ
22222þ
¼
(G
G
G
þ
G
)v
(3.4-9)
iiiisssiisiss¼
¼
(G
þ
G
þG
G3333333þ
þ������������þ
þG
GNNNNNNN)v
)v
(3.4-9)
(3.4-9)
(G111þþGG222þþ
G
þ
G
(3.4-9)
(3.4-9)
ss s ¼(G
3 þ� �� �� �þ
N )v
(3.4-9)
is ¼ (G1 þ G2 þ G3 þ � � � þ GN )v
Therefore,
Therefore,
Therefore,
Therefore,
Por
consiguiente,
Therefore,
Therefore,
NN
Therefore,
N
X
N
X
N
NN
X
N
X
X
(3.4-10)
iiisisis ¼
¼
v
GG
¼
v
(3.4-10)
(3.4-10)
N
v
G
¼
v
Gnnnnnnnn
(3.4-10)
X
s
(3.4-10)
G
(3.4-10)
G
(3.4-10)
ississ¼¼vvn¼1
n¼1
n¼1
G
(3.4-10)
is ¼ v n¼1
n¼1
n¼1
n
n¼1
n¼1
Because
¼
GG
v,
we
may
obtain
from
Eq.
3.4-10
and
substitute
itit
in
Eq.
3.4-8,
obtaining
Dado
queiinniniinn¼
5G
Gnnnnv,
v,we
podríamos
obtener
v porEq.
lan¼1
ecuación
3.4-10
y sustituirla
en
ecuación
3.4-8,
Because
¼
we
may
obtain
from
Eq.
3.4-10
and
substitute
in
Eq.
3.4-8,
obtaining
nv,
Because
may
obtain
vvvvvfrom
from
3.4-10
and
substitute
it
in
Eq.
3.4-8,
obtaining
Because
¼
G
v,
we
may
obtain
from
Eq.
3.4-10and
and
substitute
itin
inEq.
Eq.la
3.4-8,
obtaining
n
Because
innn¼¼
G
v,
we
may
obtain
from
Eq.
3.4-10
and
substitute
it
in
Eq.
3.4-8,
obtaining
Because
i
G
v,
we
may
obtain
v
Eq.
3.4-10
substitute
it
3.4-8,
obtaining
n
n
nn
obteniendo
Because in ¼ Gnv, we may obtain v from Eq.GG
and substitute it in Eq. 3.4-8, obtaining
nnisisis
Gnnnn3.4-10
s
GG
iinn ¼
¼
(3.4-11)
¼
(3.4-11)
nnississ
(3.4-11)
¼P
(3.4-11)
N
N
N
¼
(3.4-11)
i
G
P
iinnnniinn¼
(3.4-11)
N
Nn s
N
P
P
N
N
P
in ¼ P
(3.4-11)
G
n
G
nn
n
G
N
n
G
P
GGnnn
n¼1
n¼1
n¼1
n¼1
n¼1G n
n¼1
n¼1
n¼1
n¼1
such
that
Recall
that
the
equivalent
circuit,
Figure
3.4-12,
has
an
equivalent
conductance
G
that
Recall
the
equivalent
circuit,
3.4-12,
has
an
equivalent
conductance
G
such
thatcomo
Recallthat
that
the
equivalent
circuit,Figure
Figure
3.4-12,
has
anuna
equivalent
conductance
Gpppppppsuch
that
Recall
that
the
circuit,
Figure
an
equivalent
conductance
GG
such
that
Recall
that
the
equivalent
circuit,
Figure
3.4-12,
has
an
equivalent
conductance
Recuerde
que
elequivalent
circuito
equivalente,
figura3.4-12,
3.4-12,has
tiene
conductancia
equivalente
Gp, casi
p such
N
N
such
that
Recall that the equivalent circuit, Figure 3.4-12, X
has
an
equivalent
conductance
G
N
X
p
X
NN
X
N
NN
X
X
G
¼
G
(3.4-12)
G
(3.4-12)
N G
Gpppppppp¼
¼X
Gnnnnnnnn
(3.4-12)
(3.4-12)
GG
¼¼
GG
(3.4-12)
(3.4-12)
n¼1
n¼1
n¼1
Gp ¼ n¼1
G
(3.4-12)
n¼1
n¼1
n
n¼1
n¼1
n¼1
Therefore,
Therefore,
Por
consiguiente,
Therefore,
Therefore,
Therefore,
G
Therefore,
Gnnnnnnnniiissssiisisss
GG
iinn ¼
¼
(3.4-13)
G
¼
(3.4-13)
(3.4-13)
¼G
(3.4-13)
(3.4-13)
¼
(3.4-13)
GGG
iinnnniinn¼
(3.4-13)
npppips
G
G
p
in ¼ GGppp
(3.4-13)
Gp NNconductances.
the
basic
equation
for
the
current
divider
with
conductances.
Of
course,
3.4-12
can
be
which
is
the
basic
equation
for
the
current
divider
with
Of
course,
Eq.
3.4-12
can
be
lawhich
cualis
esthe
labasic
ecuación
básica
las
conductancias
del
divisor
de corriente
con Eq.
NEq.
conductancias.
which
is
the
basicequation
equation
forpara
thecurrent
current
dividerwith
with
Nconductances.
conductances.
Ofcourse,
course,
Eq.
3.4-12can
canbe
be
which
isis
the
basic
equation
for
the
current
divider
with
NN
conductances.
Of
course,
Eq.
3.4-12
can
be
which
is
the
basic
equation
for
the
current
divider
with
conductances.
Of
course,
Eq.
3.4-12
can
be
which
for
the
divider
N
Of
3.4-12
rewritten
as
rewritten
as
which
is
the
basic
equation
for
the
current
divider
with
N
conductances.
Of
course,
Eq.
3.4-12
can
be
Desde
luego,
la
ecuación
3.4-12
se
puede
reescribir
como
rewritten
as
rewritten
as
rewrittenas
as
rewritten
rewritten as
NN
N
X
X
N
N 11
11 X
NN
X
X
N
X
1111
1111 ¼¼
(3.4-14)
(3.4-14)
N
(3.4-14)
¼
(3.4-14)
X
¼
(3.4-14)
¼
(3.4-14)
1R
1R
p
R
R
p
p
nnnnnn
n¼1
R
Rppppp¼ n¼1
n¼1
R
RR
R
(3.4-14)
n¼1
R
R
n¼1 nn
n¼1
n¼1
n¼1
Rp n¼1 Rn
EEeXXXXXXmAAAAAApM
M
PPL
LLE
EE 3
34
44
11
Parallel
Resistors
M
Parallel
Resistors
P
E
M
P
L
E
4
1
Parallel
Resistors
M
PL
LE
E.33
3...-...4
4------1
1 Resistores
ParallelResistors
Resistors
EE
jE
lP
oP
3
1
en
paralelo
M
P
LL
EE
EXXAAM
M
3
4
1
Parallel
Resistors
Parallel
E X A M P L E 3 . 4 - 1 Parallel Resistors
For
the
circuit
Figure
3.4-5,
find
(a)
the
current
each
For
the
circuit
in
Figure
3.4-5,
find
(a)
the
current
in
each
Forthe
the
circuitin
in
Figure
3.4-5,
findencuentre
(a)the
thecurrent
current
in
each en
Para
el circuito
enFigure
la figura
3.4-5,
(a) lain
corriente
For
the
circuit
inin
Figure
3.4-5,
find
(a)
the
current
inin
each
For
the
circuit
in
Figure
3.4-5,
find
(a)
the
current
in
each
For
circuit
3.4-5,
find
(a)
each
branch,
(b)
the
equivalent
circuit,
and
(c)
the
voltage
v.
The
branch,
(b)
the
equivalent
circuit,
and
(c)
the
voltage
v.
The
For
the
circuit
in
Figure
3.4-5,
find
(a)
the
current
in
each
branch,
(b)
the
equivalent
circuit,
and
(c)
the
voltage
v.
Thev. Los
cada
extensión,
(b)
el
circuito
equivalente
y
(c)
el
voltaje
branch,
(b)
the
equivalent
circuit,
and
(c)
the
voltage
v.
The
branch,(b)
(b)the
theequivalent
equivalentcircuit,
circuit,and
and(c)
(c)the
thevoltage
voltagev.v.The
The
branch,
resistors
are
resistors
are
branch,
(b)
the
equivalent
circuit,
and
(c)
the
voltage
v.
The
resistors
are
resistores
son
resistors
are
resistorsare
are
resistors
11
11
11
resistors are
1111V;
1111V;
1111V
¼
¼
¼
V;
R
V;
R
R
R
R
R
111111¼
222222¼
333333¼
¼
¼
¼
V;
R
V;
R
V
R
¼
¼
¼
V;
R
V;
R
VV
R
FIGURE
3.4-5
Parallel
circuit
for
Example
3.3-2.
FIGURE
3.4-5
Parallel
circuit
for
Example
3.3-2.
V; RR2¼¼414 V;
V; RR3¼¼818 V
V
FIGURE
3.4-5
Parallel
circuit
for
Example
3.3-2.
RR11¼¼2122V;
FIGURE
3.4-5Parallel
Parallel
circuitfor
for
Example
3.3-2.
FIGURE
3.4-5
Parallel
circuit
for
Example
3.3-2.
FIGURA
3.4-5 circuit
Circuito
enExample
paralelo3.3-2.
para
el
FIGURE
3.4-5
Parallel
circuit
for
Example
3.3-2.
FIGURE
3.4-5
R1 ¼ 22 V; R22 ¼ 444 V; R33 ¼ 888 V
FIGURE
3.4-5
Parallel
circuit
for
Example
3.3-2.
2
4
8
ejemplo 3.3-2.
Solution
Solution
Solución
Solution
Solution
Solution
The
current
divider
follows
the
equation
The
current
the
equation
Solution
El
divisor
dedivider
corriente
sigue la
ecuación
The
current
dividerfollows
follows
the
equation
The
current
divider
follows
the
equation
The
current
divider
follows
the
equation
The current divider follows the equation
The current divider follows the G
equation
G
Gnnnnnnnniissssiisisss
GG
¼
¼
¼
iiinnniininn¼
¼
G
¼
G
n
npppips
GG
G
ppp
G
inn ¼ G
Gppcircuito
por
lois
que
esto
sensato
encontrar
el
equivalente,
como se
so
it
is
wise
to
find
the
equivalent
circuit,
as
shown
in
so
it
is
wise
to
find
the
equivalent
circuit,
as
shown
in
so
it
wise
find
the
equivalent
circuit,
as
shown
in
soitit
itisis
iswise
wisetoto
tofind
findthe
theequivalent
equivalentcircuit,
circuit,as
asshown
shownin
in
so
wise
find
the
equivalent
circuit,
as
shown
in
so
muestra
en
la
figura
3.4-6,
con
su
conductancia
equivalente
Gp.
.
We
have
Figure
3.4-6,
with
its
equivalent
conductance
G
..We
We
have
Figure
3.4-6,
with
its
equivalent
conductance
G
so
it is3.4-6,
wise with
to
find
the
equivalent
circuit, G
as
in
have
Figure
its
equivalent
conductance
Wehave
have
Figure
3.4-6,
with
itsequivalent
equivalent
conductance
Gpppppp.p.shown
We
have
Figure
3.4-6,
with
its
equivalent
conductance
Figure
3.4-6,
with
its
conductance
GG
p .We
Tenemos
FIGURE
3.4-6
Equivalent
circuit
for
the
parallel
circuit
FIGURE
3.4-6
Equivalent
circuit
for
the
parallel
circuit
FIGURE
3.4-6
Equivalent
circuit
for
the
parallel
circuit
Figure 3.4-6, with
NN its equivalent conductance Gp. We have
FIGURE3.4-6
3.4-6Equivalent
Equivalentcircuit
circuitfor
forthe
theparallel
parallelcircuit
circuit
FIGURE
3.4-6
Equivalent
circuit
for
the
parallel
circuit
N
X
FIGURE
3.4-6
Equivalent
circuit
for
the
parallel
circuit
N
X
FIGURE
N
NN
X
of
Figure
3.4-5.
N
of
Figure
3.4-5.
X
X
of
Figure
3.4-5.
FIGURE
3.4-6
Equivalent
circuit
for
the
parallel
circuit
¼
G
¼
G
þ
G
þ
G
¼
2
þ
4
þ
8
¼
14
S
G
of
Figure
3.4-5.
¼
G
¼
G
þ
G
þ
G
¼
2
þ
4
þ
8
¼
14
S
G
of
Figure
3.4-5.
p
n
1
2
3
N G
FIGURA 3.4-6 Circuito equivalente para el
Figure3.4-5.
3.4-5.
¼¼
¼¼
GG
þþ
GG
þþ
GG
¼¼
þþ
þþ
¼¼
14
SS
GG
¼X
Gnnnnnnn¼
¼G
G1111111þ
þG
G2222222þ
þG
G3333333¼
¼2222þ
þ4444þ
þ8888¼
¼14
14S
S ofofFigure
Gpppppp¼
14
GG
G
of Figure 3.4-5.
n¼1
n¼1
circuito en paralelo de la figura 3.4-5.
G
¼
G
þ
G
þ
G
¼
2
þ
4
þ
8
¼
14
S
Gpp ¼ n¼1
n¼1
n¼1
n
1
2
3
n¼1
n¼1
n¼1
n¼1
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 68
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
E1C03_1
11/25/200969
E1C03_1
11/25/2009
69
Parallel Resistors and Current Division
69
Resistores
en paralelo
y división
de laDivision
coriente
Parallel
Resistors
and
Current
Parallel
Resistors
and
Current
Division
Parallel
Resistors
and
Current
Division
Parallel
Resistors
and Current
Division
69
6969
69
69
Recall that the units for conductance are siemens (S). Then
G
2
1 is siemens
Recuerde
que
las
unidades
para la conductancia
son
(S).
Recall
that
forfor
areare
(S).
Recall
thatthe
theunits
units
forconductance
conductance
aresiemens
siemens
(S).
Then
i1 ¼(S).
¼Then
(28)
¼ Entonces
4A
Recall
that
the
units
conductance
siemens
(S).
Then
Recall
that the
units
for conductance
are siemens
Then
G
14
p
is11 iiss 2 22
G1G
G
is
2 ¼ (28)
i G
¼1¼
¼
¼¼
(28)
¼4 A
A
(28)
44 A
i1 ¼1 ii11 ¼
(28)
4A
G¼
1414
Gpp14¼
14 ¼
Gp pG
G2 is 4(28)
¼
¼ 8A
Similarly,
i2 ¼
Del
mismo modo,
G
14
G2G
is22piiss 4(28)
G
4(28)
is 4(28)
¼2¼
¼¼
¼¼
Similarly,
i G
¼ 4(28)
¼8 A
A
Similarly,
88 A
Similarly,
i2 ¼2 ii22 ¼
Similarly,
G¼
14¼
G
148 A
pG
p
14
Gp G3p i14
s
and
i3 ¼
¼ 16 A
G
p
G3G
is33 iiss
G
is
and
i G
¼3¼
¼¼1616
AA
yand
and and
i3 ¼3 ii33 ¼
Because in ¼ Gnv, we have
G¼
Gpp16¼A 16 A
Gp pG
i1
4
G5nG
have
Because
i ii¼
¼
Gv,G
v,nwe
we
have
Because
Dado
v,we
tenemos
nin¼
nv,
have
Because
v¼
¼ ¼ 2V
n
n
¼nG
have
Because
in que
nv, we
G
i1 ii111 4 442
iv1 ¼
4¼¼
v¼
v
¼
¼2 V
2V
¼ 2¼V
v¼ G
¼G1 ¼
2 ¼ 2V
G1 1G
21 22
ApMl PoL E
Parallel Resistors
EE
j eXm
3 .34.-42- 2 Resistores
en paralelo
E
X
A
M
P
L
E
3
.
4
2
Parallel
Resistors
E
X
A
M
P
L
E
3
.
4
2
Parallel
Resistors
2 Parallel
Resistors
E X AEMXPALME P3L.E43- 2. 4 -Parallel
Resistors
V ET EERXAACMT PI VL O
E
EI NJ TE EMRPALCOT II N
I NIITN
P LPEL E
NETTREEAR
RCA
ATC
CI TVIEV E XEAXM
AM
INTER
ACTIV
E TEIXVAE ME PX LAEM P L E
For the
circuit ofdeFigure
3.4-7a,
findencuentre
the voltageelmeasured
by the por
voltmeter.
Then show
thatmuestre
the power
by
Para
el circuito
la figura
3.4-7a,
voltaje medido
le voltímetro.
Luego
queabsorbed
la potencia
the
two
resistors
is
equal
to
that
supplied
by
the
source.
absorbida
poroflos
dos resistores
esthe
igual
a la alimentada
por
fuente.
For
thethe
circuit
Figure
3.4-7a,
find
voltage
measured
by
thela
voltmeter.
Then
show
that
thethe
power
absorbed
byby
For
the
circuit
of
Figure
3.4-7a,
find
the
voltage
measured
by
the
voltmeter.
Then
show
that
the
power
absorbed
For
circuit
of
Figure
3.4-7a,
find
the
voltage
measured
the
voltmeter.
that
power
absorbed
For the
circuit
of Figure
3.4-7a,
find the
voltage
measured
by theby
voltmeter.
ThenThen
showshow
that the
power
absorbed
by by
thethe
two
resistors
is is
equal
toto
that
supplied
byby
thethe
source.
the
two
resistors
is
equal
to
that
supplied
by
the
source.
two
resistors
equal
that
supplied
source.
the two resistors is equal to that supplied by the source.
Voltímetro
FIGURA 3.4-7 (a) Un circuito que contiene resistores
FIGURE
A circuit
containing
parallel
resistors.
en paralelo.3.4-7
(b) El(a)
circuito
después
de que el
voltímetro
(b)
The
after the ideal
has
been replaced by
ideal
hacircuit
sido reemplazado
porvoltmeter
el circuito
abierto
FIGURE
3.4-7
(a)(a)
containing
resistors.
FIGURE
3.4-7
(a)Aagregado
Acircuit
circuit
containing
parallel
resistors.
FIGURE
A
circuit
containing
parallel
resistors.
the
equivalent
open
circuit
and
a label
hasparallel
been
added
to
equivalente
y se
una
etiqueta
para
indicar
el
FIGURE
3.4-7 3.4-7
(a)
Ahacircuit
containing
parallel
resistors.
(b)(b)
The
circuit
after
the
ideal
voltmeter
has
been
replaced
byby
(b)
The
circuit
after
the
ideal
voltmeter
has
been
replaced
by
The
circuit
after
the
ideal
voltmeter
has
been
replaced
The
indicate
the
voltage
measured
by the
vm. (c)
voltaje
porideal
el
voltímetro,
vm. voltmeter,
(c)
Elreplaced
circuito
después
(b) The
circuitmedido
after
the
voltmeter
has
been
by
thethe
equivalent
open
circuit
and
a label
has
been
added
to
the
equivalent
open
circuit
and
a han
label
has
been
added
toan
equivalent
open
circuit
and
a
label
has
been
added
to
de
que
los
resistores
en
paralelo
sido
reemplazados
por
circuit
after
the
parallel
resistors
have
been
replaced
by
the equivalent open circuit and a label has been added to
(c)
The
indicate
thethe
voltage
measured
byby
thethe
voltmeter,
vmvv. m
(c)
The
indicate
the
voltage
measured
the
voltmeter,
m
.. (c)
The
indicate
voltage
measured
voltmeter,
unathe
resistencia
equivalente.
equivalent
resistance.
The
indicate
voltage
measured
by theby
voltmeter,
vm. (c)
circuit
after
the
parallel
resistors
have
been
replaced
by
an
circuit
after
the
parallel
resistors
have
been
replaced
by
an
after
the parallel
resistors
have replaced
been replaced
circuitcircuit
after the
parallel
resistors
have been
by an by an
equivalent
resistance.
equivalent
resistance.
equivalent
resistance.
equivalent resistance.
Solution
Solución
Figure
3.4-7b
shows
the circuit
afterdespués
the ideal
has been
replaced
byreemplazado
the equivalent
circuit,abierto
and a
La
figura
3.4-7b
muestra
el circuito
de voltmeter
que el voltímetro
ideal
ha sido
poropen
el circuito
Solution
Solution
Solution
label has been added to indicate the voltage measured by the voltmeter, v . The two resistors are connected in
m el voltímetro, vm. Los dos resistores
equivalente,
y shows
se hathe
agregado
una
etiqueta
para
indicar el
voltaje
medido
por
Figure
3.4-7b
shows
circuit
after
the
ideal
voltmeter
has
been
replaced
by
the
equivalent
open
circuit,
and
a aa
Figure
3.4-7b
the
circuit
after
the
ideal
voltmeter
has
been
replaced
by
the
equivalent
open
circuit,
and
Figure
3.4-7b
shows
the
circuit
after
ideal
voltmeter
has
been
replaced
by
the
equivalent
open
circuit,
Figure
3.4-7b
shows
the
after
the ideal
voltmeter
haspor
been
replaced
the
equivalent
circuit,
andde
aand
parallel
and
can
becircuit
replaced
with
athe
single
equivalent
resistor.
The by
resistance
of
thisopen
equivalent
resistor
is
están
conectados
en
paralelo
y
pueden
ser
sustituidos
un
resistor
equivalente
único.
La are
resistencia
este
.
The
two
resistors
connected
in
label
has
been
added
to
indicate
the
voltage
measured
by
the
voltmeter,
v
.
The
two
resistors
are
connected
in
label
has
been
added
to
indicate
the
voltage
measured
by
the
voltmeter,
v
m
m
.
The
two
resistors
are
connected
in
has been
added
to indicate
the voltage
measured
byvoltmeter,
the voltmeter,
v
m
.
The
two
resistors
are
connected
in
labellabel
has
been
added
to
indicate
the
voltage
measured
by
the
v
calculated
as
m
resistor
equivalente
sereplaced
calculawith
como
parallel
and
a aasingle
equivalent
resistor.
The
ofof
resistor
is is
parallel
andcan
canbe
bereplaced
with
single
equivalent
resistor.
Theresistance
resistance
ofthis
thisequivalent
equivalent
resistor
parallel
and
can
be
replaced
single
equivalent
resistor.
resistance
this
equivalent
resistor
parallel
and can
be
replaced
with with
a single
equivalent
resistor.
The The
resistance
of this
equivalent
resistor
is is
calculated
as
40
�
10
calculated
as
calculated
calculated
as as
¼8V
þ 10
4040
� 10
40
�
10
40
�
10
40 � 10
¼¼
¼8 8V
8V
V
¼resistors
8V
40
þþ
10
Figure
3.4-7c
shows
the circuit
afterluego
the
have
been replaced
byreemplazados
the equivalentpor
resistor.
The
40
þ
10
La
figura
3.4-7c
muestra
el circuito
que
resistores
en paralelo
han sido
el resistor
40
10
40parallel
þ
10los
do not adhere
to the
current in theLaequivalent
resistor
is 250equivalente
mA, directed
upward.
This
and
the voltage
vm corriente
equivalente.
corriente en
el resistor
es de
250 mA,
en current
dirección
ascendente.
Esta
y el voltaje
Figure
3.4-7c
shows
thethe
circuit
after
thethe
parallel
resistors
have
been
replaced
byby
thethe
equivalent
resistor.
The
Figure
3.4-7c
shows
the
circuit
after
the
parallel
resistors
have
been
replaced
by
the
equivalent
resistor.
The
Figure
3.4-7c
shows
circuit
parallel
resistors
been
equivalent
resistor.
The
Figure
shows
the circuit
after
the
resistors
have have
been
replaced
equivalent
resistor.
The
convention.
The
current
in after
the
equivalent
can
alsoreplaced
be by
expressed
as
�250
mA,
directed
vpassive
apegan
a la convención
pasiva.
Laparallel
corriente
en resistance
la
resistencia
equivalente
sethe
puede
expresar
como
2250
mA,
m no se 3.4-7c
dodo
not
adhere
toto
thethe
current
inin
thethe
equivalent
resistor
is is
250
mA,
directed
upward.
This
current
and
thethe
voltage
vmvvm
do
not
adhere
to
current
in
the
equivalent
resistor
is
250
mA,
directed
upward.
This
current
and
the
voltage
m
not
adhere
current
equivalent
resistor
250
mA,
directed
upward.
This
current
and
voltage
current in the equivalent resistor is 250 mA, directed upward. This current and the voltage vm do not adhere to the the
passive
convention.
The
current
ininthethe
resistance
can
bebe
asas
mA,
directed
passive
convention.
The
current
theequivalent
equivalent
resistance
canalso
also
beexpressed
expressed
as�250
�250
mA,
directed
passive
convention.
current
equivalent
resistance
can
also
expressed
�250
directed
passive
convention.
The The
current
in thein
equivalent
resistance
can also
be expressed
as �250
mA, mA,
directed
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 69
Alfaomega
4/12/11 5:21 PM
Circuitos resistivos
70
con dirección descendente. Esta corriente y el voltaje vm se apegan a la convención pasiva. De la ley de Ohm surge
vm ¼ 8ð0:25Þ ¼ 2 V
El voltaje vm en la figura 3.4-7b es igual al voltaje vm en la figura 3.4-7c. Esto es consecuencia de la equivalencia del resistor de 8 ⍀ para la combinación en paralelo de los resistores de 40-⍀ y 10-⍀. Si vemos la figura
3.4-7b observaremos que la potencia absorbida por los resistores es
v m 2 v m 2 22 22
þ
¼
þ
¼ 0:1 þ 0:4 ¼ 0:5 W
40
10
40 10
y la corriente de la fuente de corriente se apegan a la convención pasiva, por lo que
pR ¼
El voltaje vm
ps ¼ vm (0:25) ¼ ð2Þð0:25Þ ¼ 0:5 W
es la potencia recibida por la fuente de corriente. La fuente de corriente alimenta 0.5 W.
Entonces, la potencia absorbida por los dos resistores es igual a la alimentada por la fuente.
EJEMPLO 3.4-3
Diseño del divisor de corriente
La entrada a la corriente del divisor de corriente en la figura 3.4-8 es la corriente, is, de la fuente de corriente.
La salida es la corriente, io, medida por el amperímetro. Especifique los valores de las resistencias R1 y R2 para
satisfacer las dos siguientes especificaciones:
o
Amperímetro
s
s
FIGURA 3.4-8 Circuito de divisor de corriente.
Divisor de corriente
Especificación 1: Las corrientes de entrada y salida se relacionan por io ⫽ 0.8 is.
Especificación 2: La fuente de corriente se requiere para alimentar no más de10 mW de potencia cuando la entrada al divisor de corriente es is ⫽ 2 mA.
Solución
Analicemos cada especificación para ver a qué se refiere respecto de los valores del resistor.
Especificación 1: Las corrientes de entrada y salida del divisor de corriente se relacionan por
io ¼
R2
i
R 1 þ R2 s
Por lo que la especificación 1 requiere
R2
¼ 0:8
R1 þ R 2
)
R2 ¼ 4R1
Especificación 2: La potencia alimentada por la fuente de corriente resulta de
ps is vs is is
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 70
R 1 R2
R 1 R2
is 2
R1 R2
R1 R2
Circuitos Eléctricos - Dorf
5/24/11 10:03 AM
Parallel
Parallel
Resistors
Resistors
and
and
Current
Current
Division
Parallel
Parallel
Parallel
Resistors
Resistors
Resistors
and
and
and
Current
Current
Current
Division
Division
Division
Resistores
en paralelo
y división
de laDivision
coriente
Parallel
Resistors
and
Current
Division
71
71
71
71
71
71
71
So
So
specification
22222requires
2requires
requires2 requiere
So
So
So
specification
specification
specification
requires
requires
requires
So
specification
Por
lospecification
que la especificación
�
�
��
�
��
��
�
��
RRR
RRR
R
R
RR
1R
21R
1RR
21RR2
222222 R
1R
11R
2R
222
1
2
)
) R1111R2222 �
0:01
0:01
�
�
ð
0:002
ð
0:002
Þ
Þ
2500
2500
)
)
)
0:01
0:01
0:01��
�
�ðð0:002
ðð0:002
0:002
0:002
ÞÞÞÞ
��
�
��
2500
2500
2500
22500
500
)
0:01
RRRR
R
þ
R
þ
R
R
R
þ
R
þ
þ
þ
R
R
R
R
R
R
þ
þ
þ
RRR
1R
1
2
2
1
1
2R
1111þ R2222
2R
R1111þþ
2222
Combining
Combining
these
these
results
results
gives
Combining
Combining
Combining
these
these
these
results
results
results
gives
gives
gives
La
combinación
de
estosgives
resultados
es
Combining
these
results
gives
RRRR
R
ð4R
4R
44444
ð4R
4R
ðð4R
Þ ÞÞÞ
1R
22Þ
1ð
111ð14R
222Þ2�
2500
2500
) 4RRRR
�
2500
2500
) RRR
�
3125
3125
V
V
22500
500 )
22500
500 )
33125
125
V
��
�
��
2500
2500
2500
)
)
)
�
��
2500
2500
)
)
)
�
��
3125
3125
VV
V
1RR
1�
1RR
1�
)
þ
4R
þ
4R
RRRR
R
5555551111 � 2500 ) R1111 � 3125 V
þ
þ4R
4R
4R
1R
1þ
222222
1111þ
4R
La
solución
no
es
launique.
única.
Una
solución
es
The
The
solution
solution
is
is
not
unique.
One
One
solution
solution
is
The
The
The
solution
solution
solution
isis
is
isnot
not
not
not
unique.
unique.
unique.
One
One
One
solution
solution
solution
isis
is
isis
The
solution
not
unique.
One
solution
yandRRR
¼
333kV
3kV
kVand
¼
12
12
kV
kV
RRRR
¼¼
¼¼
3kV
kV
and
and
and
¼¼
¼¼
12
12
12
kV
kV
kV
1R
2R
1R
2R
1111¼ 3 kV and R
2222¼ 12 kV
EXERCISE
EXERCISE
3.4-1
3.4-1AAAAAA
resistor
resistor
network
network
consisting
consisting
of
of
parallel
parallel
resistors
resistors
is
shown
shown
in
aaaaapackage
apackage
package
EXERCISE
EXERCISE
EXERCISE
3.4-1
3.4-1
3.4-1
resistor
resistor
resistor
network
network
network
consisting
consisting
consisting
of
of
of
parallel
parallel
parallel
resistors
resistors
resistors
isis
is
isis
shown
shown
shown
inin
in
inin
package
package
package
EXERCISE
3.4-1
resistor
network
consisting
of
parallel
resistors
shown
EJERCICIO
3.4-1
En
la
figura
E
3.4-1a
se
muestra
una
red
de
resistores
que
consta
de
used
used
for
for
printed
printed
circuit
circuit
board
board
electronics
electronics
in
in
Figure
Figure
E
3.4-1a.
E
3.4-1a.
This
This
package
package
is
only
is
only
2
cm
2
cm
�
�
0.7
0.7
cm,
cm,
used
used
usedfor
for
forprinted
printed
printedcircuit
circuit
circuitboard
board
boardelectronics
electronics
electronicsinin
in
inFigure
Figure
FigureEEEE3.4-1a.
3.4-1a.
3.4-1a.This
This
Thispackage
package
packageisis
is
isonly
only
only2222cm
cm
cm��
�
�0.7
0.7
0.7
cm,
cm,
cm,
used
for
printed
circuit
board
electronics
Figure
3.4-1a.
This
package
only
cm
0.7
cm,
resistores
paralelos
un
paquete
utilizado
para circuitos
impresos
de as
tableros
electrónicos.
Este
paand
and
each
each
resistor
resistor
is
1kV.
kV.
The
The
circuit
circuit
is
is
connected
to
to
use
four
four
resistors
resistors
as
shown
shown
in
in
Figure
EEEEE3.4-1b.
E3.4-1b.
3.4-1b.
and
and
and
each
each
each
resistor
resistor
resistor
isis
is
is1is
1en
kV.
kV.
kV.
The
The
The
circuit
circuit
circuit
isis
is
isconnected
connected
connected
connected
to
to
touse
use
use
use
four
four
four
resistors
resistors
resistors
as
as
as
shown
shown
shown
in
in
inFigure
Figure
Figure
Figure
3.4-1b.
3.4-1b.
3.4-1b.
and
each
resistor
111kV.
The
circuit
connected
to
use
four
resistors
as
shown
in
Figure
quete
mide
sólo
2
3
0.7
cm,
y
cada
resistor
es
de
1
kV.
El
circuito
está
conectado
para
utilizar
cuatro
¼
1
¼
mA.
1
mA.
Find
Find
the
the
equivalent
equivalent
circuit
circuit
for
for
this
this
network.
network.
Determine
Determine
the
the
current
current
in
each
in
each
resistor
resistor
when
when
i
i
¼
1111mA.
mA.
mA.
Find
Find
Findthe
the
theequivalent
equivalent
equivalentcircuit
circuit
circuitfor
for
forthis
this
thisnetwork.
network.
network.Determine
Determine
Determinethe
the
thecurrent
current
currentin
in
ineach
each
eachresistor
resistor
resistorwhen
when
whenissiiis¼
s¼
s¼
mA.
Find
the
equivalent
circuit
for
this
network.
Determine
the
current
in
each
resistor
when
resistores, como se muestra en la figura E 3.4-1b. Encuentre el circuito equivalente paras esta red.
Determine la corriente en cada resistor cuando is 5 1 mA.
FIGURE
FIGURE
EEE
E3.4-1
3.4-1
FIGURE
FIGURE
FIGURE
E
E3.4-1
3.4-1
3.4-1
3.4-1
FIGURE
(a)
(a)
A
parallel
Aparallel
parallel
resistor
resistor
(a)
(a)
(a)
AA
A
A
parallel
parallel
parallel
resistor
resistor
resistor
(a)
resistor
FIGURA
ECourtesy
3.4-1of
network.
network.
Courtesy
network.
network.
network.
Courtesy
Courtesy
Courtesy
ofof
of
ofof
network.
Courtesy
(a)
Red
de
resistores
Dale
Dale
Electronics.
Electronics.
Dale
Dale
Dale
Electronics.
Electronics.
Electronics.
Dale Electronics.
en
paralelo.
Cortesía
(b)
(b)
The
The
connected
connected
(b)
(b)
(b)
The
The
The
connected
connected
connected
(b)
The
connected
de
Dale
Electronics.
circuit
circuit
uses
uses
four
four
circuit
circuit
circuit
uses
uses
uses
four
four
four
circuit
uses
four
(b)
El circuito
resistors
resistors
where
where
RRRR
¼
resistors
resistors
resistors
where
where
where
R¼R
¼
¼¼
resistors
where
¼
utiliza
11conectado
kV.
1
kV.
1
1
kV.
kV.
kV.
1 kV.
cuatro resistores
cuando R 5 1 kV.
Answer:
Answer:
RRRR
¼
250
¼
250
V
Answer:
Answer:
Answer:
¼p¼
¼
250
250
250
VV
V
VV
pR
pR
Answer:
250
ppp¼
Respuesta: Rp 5 250 V
EXERCISE
EXERCISE
3.4-2
3.4-2Determine
Determine
the
the
current
current
measured
measured
by
by
the
the
ammeter
ammeter
in
the
the
circuit
circuit
shown
shown
in
EXERCISE
EXERCISE
EXERCISE
3.4-2
3.4-2
3.4-2
Determine
Determine
Determine
the
the
the
current
current
current
measured
measured
measured
by
by
by
the
the
the
ammeter
ammeter
ammeter
inin
in
inin
the
the
the
circuit
circuit
circuit
shown
shown
shown
inin
in
inin
EXERCISE
3.4-2
Determine
the
current
measured
by
the
ammeter
the
circuit
shown
EJERCICIO
3.4-2
Determine
la
corriente
medida
por
el
amperímetro
en
el
circuito
que
se
Figure
Figure
E
3.4-2a.
E
3.4-2a.
Figure
Figure
FigureEEEE3.4-2a.
3.4-2a.
3.4-2a.
Figure
3.4-2a.
muestra en la figura E 3.4-2a.
Amperímetro
FIGURE
FIGURE
EEE
3.4-2
3.4-2
(a)
(a)
A
A
current
divider.
divider.
(b)
(b)
The
The
FIGURE
FIGURE
FIGURE
E
EEE
3.4-2
3.4-2
3.4-2
(a)
(a)
(a)
ADivisor
A
Acurrent
current
current
current
divider.
divider.
(b)
(b)
(b)
The
The
FIGURE
3.4-2
(a)
A
current
divider.
(b)
The
FIGURA
3.4-2
(a)
de divider.
corriente.
(b)The
El
current
current
divider
divider
after
after
the
the
ideal
ideal
ammeter
ammeter
has
been
been
current
current
current
divider
divider
divider
after
after
after
the
the
the
ideal
ideal
ideal
ammeter
ammeter
ammeter
has
has
has
been
been
been
current
divider
after
the
ideal
ammeter
has
been
divisor
de corriente
después
de
que has
el
amperímetro
replaced
replaced
by
by
the
the
equivalent
equivalent
short
short
circuit
and
and
aaaalabel
alabel
label
replaced
replaced
replaced
by
by
by
the
the
the
equivalent
equivalent
equivalent
short
short
short
circuit
circuit
circuit
and
and
and
alabel
label
replaced
by
the
equivalent
short
circuit
and
label
ideal ha
sido
reemplazado
porcircuit
el
cortocircuito
has
has
been
been
added
added
to
indicate
to
indicate
the
the
current
current
measured
measured
by
by
has
has
has
been
been
been
added
added
added
to
to
to
indicate
indicate
indicate
the
the
the
current
current
current
measured
measured
measured
by
by
has
been addedy to
the current
measured
by
equivalente
se indicate
ha agregado
una etiqueta
paraby
the
the
ammeter,
ammeter,
iimm
.. . medida por el amperímetro, i .
the
the
the
ammeter,
ammeter,
ammeter,
indicar
la corriente
the
ammeter,
iii.m.mi.m
m
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 71
m
Alfaomega
4/12/11 5:21 PM
E1C03_1
11/25/2009
72
72
72
72
72
72
Resistive Circuits
Resistive
Resistive Circuits
Circuits
Circuitos resistivos
Resistive Circuits
Hint:
E
the
circuit
the
ammeter
has
been
replaced
the
equivalent
Hint: Figure
Figure La
E 3.4-2b
3.4-2b
shows
themuestra
circuit after
after
the ideal
ideal
ammeter
has el
been
replaced by
by
the ha
equivalent
Sugerencia:
figura shows
E 3.4-2b
el circuito
después
de que
amperímetro
ideal
sido reshort
circuit,
and
a
label
has
been
added
to
indicate
the
current
measured
by
the
ammeter,
iimm..
Hint:
Figure
E
3.4-2b
shows
the
circuit
after
the
ideal
ammeter
has
been
replaced
by
the
short
circuit,
and
a
label
has
been
added
to
indicate
the
current
measured
by
the
ammeter,
emplazado por el cortocircuito equivalente, y se ha agregado una etiqueta para indicar laequivalent
corriente
m
short
circuit,
a label hasi been
added to indicate the current measured by the ammeter, im.
Answer:
imm el
¼and
�1
A
medida
por
amperímetro,
.
m
Answer: im ¼ �1 A
Answer: im ¼
A A
Respuesta:
im �1
5 21
3.5
3.5
3.5
S
E
S
N
S
ER
R
ETS
SEV
VSO
ODL
L ET
TA
A
G
ELTA
SO
OJU
U
REC
CNE
ES
SS A
A
NID
DE___Y
F U
E IINE
VG
OE
ER
ER
_____F___U
____E___N
____T___E
___S
____________________________
P
A
R
A
L
L
E
L
C
U
R
R
E
N
T
S
O
U
R
C
E
S
S
E
R
I
E
S
V
O
L
T
A
G
E
S
O
U
R
C
E
S
A
N
D
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
______________________________
D
E
C
O
R
R
I
E
N
T
E
E
N
PA
R
A
L
E
L
O
PARALLEL CURRENT SOURCES
P A R A L L E L C U R R E N T S O U R C E S ________________________________________________________
Voltage
sources
connected
in series
are equivalenta una
to a singledevoltage
source.
The
voltage
of
the
Las
fuentes
de voltaje
conectadas
en serie
voltaje source.
única. ElThe
voltaje
de laof
fuenVoltage
sources
connected
in series
are equivalen
equivalent to afuente
single voltage
voltage
the
equivalent
voltage
source
is
equal
to
the
algebraic
sum
of
voltages
of
the
series
voltage
sources.
te
de voltaje
equivalente
a la
algebraica
deof
voltajes
dethe
lassource.
fuentesvoltage
de voltaje
enof
serie.
Voltage
sources
connected
in
series
are algebraic
equivalent
to
alossingle
voltage
The
voltage
the
equivalent
voltage
sourceesisigual
equal
to suma
the
sum
voltages
of
series
sources.
Consider
the
circuit is
shown
intoFigure
3.5-1a.3.5-1a.
Notice
that
the que
currents
of bothvoltage
voltage
sources
are
Considere
el
circuito
mostrado
en
la
figura
Observe
las
corrientes
de
ambas
fuentes
equivalent
voltage
source
equal
the
algebraic
sum
of
voltages
of
the
series
sources.
Consider the circuit shown in Figure 3.5-1a. Notice that the currents of both voltage sources are
,
to
be
equal.
Accordingly,
define
the
current,
i
de
voltaje
son iguales.
En shown
consecuencia,
queNotice
la corriente,
, sea
ss, 3.5-1a.
Consider
the circuit
in Figure
that theiscurrents
of both voltage sources are
equal.
Accordingly,
define
the current,
idefina
s to be
(3.5-1)
equal. Accordingly, define the current, is, toiiissbe
a
b
(3.5-1)
¼
i
¼
i
a
b
(3.5-1)
s ¼ ia ¼ ib
(3.5-1)
A
continuación,
quevvssel
, tovoltaje,
be vs, sea is ¼ ia ¼ ib
Next,
define
voltage,
Next,
define the
the defina
voltage,
s, to be
v
¼
v
þ
v
(3.5-2)
v
(3.5-2)
Next, define the voltage, vs, to be
vssss ¼ vaaa þ vbbb
(3.5-2)
Con
lasKCL,
leyesKVL,
KCL,and
KVL
y de law,
Ohm,wepodemos
representar
el circuito
de la
figura
por
las
Using
Ohm’s
in Figure
3.5-1a
by3.5-1a
the equations
vcan
va þ
vb the circuit
(3.5-2)
s ¼represent
Using KCL, KVL, and Ohm’s law, we can
represent
the circuit in Figure 3.5-1a by the equations
ecuaciones
Using KCL, KVL, and Ohm’s law, we can
vv11 þ i the circuit in Figure 3.5-1a by the equations
(3.5-3)
i ¼represent
1
ss
iccc ¼ R
(3.5-3)
v1111 þ is
R
(3.5-3)
ic ¼ 2 þ is
vv21
R
2 þ i33
iiss ¼
(3.5-4)
(3.5-4)
s ¼ R
v222 þ i3
is ¼ R2 þ i3
(3.5-4)
v R¼ v
(3.5-5)
vccc ¼2 v111
(3.5-5)
(3.5-5)
v v¼ ¼
v vþ v
(3.5-6)
(3.5-6)
v111 ¼c vsss þ1 v222
v1v22¼¼vsi33þR33v2
(3.5-6)
(3.5-7)
(3.5-7)
v2 ¼ i3 R3
vequations
(3.5-7)
2 ¼ i3 R3 result from applying KCL, KVL, and Ohm’s
vs¼
where
donde
5iiaaaia¼
1vvvbbb.b. .These
Estas same
mismas
ecuaciones
el applying
resultadoKCL,
de la KVL,
aplicación
de las
¼
¼5iibbbiand
and
¼ vvvaaaa þ
þ
These
same
equations
resultson
from
and Ohm’s
where iiissss ¼
b y vvsss 5
s¼
iiaa ¼
iibb and
vvss ¼
vaa þ
vv5bb,, ithen
in
law
to
circuit
Figure
3.5-1b.
If
iaKVL
¼ ib and
vOhm
þcircuito
vb. These
equations
result
applying
KCL,
KVL,
and
Ohm’s
where
ithe
leyes
yin
al
la
figura
3.5-1b.
is from
ibthe
y vcircuits
va 1shown
vb, entonces
los
s ¼ circuit
s ¼ va3.5-1b.
¼same
the
circuits
shown
in Figures
Figures
law
toKCL,
the
inde
Figure
If iissde
a5
s5
a¼
b and
s ¼ vSi
a þ
b then
3.5-1a
and
3.5-1b
equivalent
because
both
represented
by
the
same
equations.
ithey
¼ ibyare
and
vs ¼son
va þ
vb, then
circuits
shownestán
in Figures
law to the
circuit
inare
Figure
If is ¼
circuitos
que
se muestran
en3.5-1b.
las figuras
3.5-1a
3.5-1b
equivalentes
porque
reprea
3.5-1a
and
3.5-1b
are
equivalent
because
they
are
both
represented
bythe
the
sameambos
equations.
¼
2are
V, both
R22 ¼represented
6 V, R33 ¼ 3 by
V, the
vaa ¼same
1 V, equations.
and vbb ¼ 3 V. The
example,
suppose
that
11 ¼
3.5-1aFor
and
3.5-1b
are
equivalent
sentados
por
las mismas
ecuaciones.
¼ 44 A,
A, R
Rthey
For
example,
suppose
that iicccbecause
1 ¼ 2 V, R2 ¼ 6 V, R3 ¼ 3 V, va ¼ 1 V, and vb ¼ 3 V. The
equations
describing
the
circuit
in
Figure
3.5-1a
become
¼Figure
44 A,
3 V,
va v¼a 15V,
and
For example,
suppose
that iin
icc 5
Por
ejemplo,
suponga
que
A, RR3.5-1a
522V,
V,RR2 2¼56 6V,V,RR
3 V,
1V
y vvbb ¼
5 33 V.
V. The
Las
11 ¼
3¼
equations
describing
the circuit
become
35
vbecome
1
equations describing
the circuit
in Figure
3.5-1a3.5-1a
ecuaciones
que describen
el circuito
de la figura
son
1
v1 iss
(3.5-8)
44 ¼
þ is
(3.5-8)
¼ v21 þ
2
4 ¼ v2 þ is
(3.5-8)
v222 þ i33
(3.5-9)
iiss ¼
(3.5-9)
s ¼ v62 þ i3
6
(3.5-9)
is ¼ þ i3
6
s
s
FIGURE 3.5-1 (a)
(a) A circuit
circuit containing
FIGURE
FIGURE 3.5-1
3.5-1 (a) A
A circuit containing
containing
voltage
sources
connected
in
seriesde
and
FIGURA
3.5-1
(a)
Circuito
con in
fuentes
voltaje
voltage
sources
connected
and
FIGURE
3.5-1 connected
(a) A circuit
containing
voltage
sources
in series
series
and
(b)
an
equivalent
circuit.
conectadas
en
serie
y
(b)
circuito
equivalente.
(b)
an
circuit.
voltage
sources connected
(b)
an equivalent
equivalent
circuit. in series and
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 72
(b) an equivalent circuit.
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
Circuit Analysis
Circuit Analysis
Circuit Analysis
Table 3.5-1 Parallel and Series Voltage and Current Sources
Table 3.5-1 Parallel and Series Voltage and Current Sources
Table 3.5-1 Parallel and Series Voltage and Current Sources
CIRCUIT
CIRCUIT
CIRCUIT
EQUIVALENT CIRCUIT
EQUIVALENT CIRCUIT
EQUIVALENT CIRCUIT
CIRCUIT
CIRCUIT
CIRCUIT
73
73
73
EQUIVALENT CIRCUIT
EQUIVALENT CIRCUIT
EQUIVALENT CIRCUIT
Análisis de circuitos
73
Tabla 3.5-1 Fuentes de voltaje y corriente en paralelo y en serie
CIRCUITO
CIRCUITO EQUIVALENTE
CIRCUITO
CIRCUITO EQUIVALENTE
No permitido
No permitido
vvc ¼
(3.5-10)
¼ vv1
(3.5-10)
vcc ¼ v11
(3.5-10)
(3.5-10)
vv1 ¼
4
þ
v
(3.5-11)
2
¼4þv
(3.5-11)
(3.5-11)
v11 ¼ 4 þ v22
(3.5-11)
(3.5-12)
vv2 ¼
(3.5-12)
¼ 3i
3i3
(3.5-12)
v22 ¼ 3i33
(3.5-12)
The
solution
to
set
of
is
v2 ¼
and A,
vc ¼v26 5
V. 2V
Eqs.y
La
este
de ecuaciones
v11 A,
5 i63 ¼
V,0.66
is 5A,1A,
i3 25V,0.66
s¼
Thesolución
solutionpara
to this
this
set conjunto
of equations
equations
is vv11 ¼
¼ 66 V,
V, iies
s ¼ 1 A, i3 ¼ 0.66 A, v2 ¼ 2 V, and vc ¼ 6 V. Eqs.
The
solution
to
this
set
of
equations
is
v
¼
6
V,
i
¼
1
A,
i
¼
0.66
A,
v
¼
2
V,
and
v
¼
6
V.
Eqs.
1
s
3
2
c
3.5-8
to
3.5-12
also
describe
the
circuit
in
Figure
3.5-1b.
Thus,
v
¼
6
V,
i
¼
1
A,
i
¼
0.66
A,
v3.5-8
V.3.5-12
Las ecuaciones
3.5-8the
a 3.5-12
el circuito
Por
lo
c 5 6to
also describe
circuit también
in Figuredescriben
3.5-1b. Thus,
v11 ¼ 6deV,laissfigura
¼ 1 A,3.5-1b.
i33 ¼ 0.66
A,
¼
6
V,
i
¼
1
A,
i
¼
0.66
A,
3.5-8
to
3.5-12
also
describe
the
circuit
in
Figure
3.5-1b.
Thus,
v
1
s
3
¼
2
V,
and
v
¼
6
V
in
both
circuits.
Replacing
series
voltage
sources
by
a
single,
equivalent
v
tanto,
6 V,vcis¼56 1VA,
iboth
5 0.66
A, v2Replacing
5 2 V y vcseries
56V
en ambos
circuitos.
Al reemplazar
las
1 5and
3
in
circuits.
voltage
sources
by
a
single,
equivalent
v22 ¼ 2vV,
and
vccdoes
¼en
6 serie
V
in both
circuits.
Replacing
series
voltage
sources
byof
a the
single, el
equivalent
vvoltage
2 ¼ 2 V,
source
not
the
voltage
or
of
elements
fuentes
de
voltaje
por una
única,
la fuente
de
voltaje
equivalente
no
modifica
voltaje o
voltage
source
does
not change
change
the
voltage
or current
current
of other
other
elements
of
the circuit.
circuit.
voltage
source
does
not
change
the
voltage
or
current
of
other
elements
of
the
circuit.
Figure
3.5-2a
shows
a
circuit
containing
parallel
current
sources.
The
circuit
la corriente
otros elementos
del circuito.
Figurede3.5-2a
shows a circuit
containing parallel current sources. The circuit in
in Figure
Figure
Figure
3.5-2a
shows
a circuit
containing
parallelsources
current
sources.
Theequivalent
circuit
in current
Figure
3.5-2b
obtained
by
replacing
parallel
current
by
3.5-2a
circuito
con fuentes
corriente
paralelo.
El circuito
de la
3.5-2bLais
isfigura
obtained
by muestra
replacingunthese
these
parallel
current de
sources
by aaensingle,
single,
equivalent
current
3.5-2b
is
obtained
by
replacing
these
parallel
current
sources
by
a
single,
equivalent
current
source.
The
current
of
the
equivalent
current
source
is
equal
to
the
algebraic
sum
of
the
currents
figura
se obtiene
estas source
fuentesisdeequal
corriente
paralelosum
porofuna
de
source.3.5-2b
The current
of thereemplazando
equivalent current
to theen
algebraic
thefuente
currents
source.
The
current
of the
equivalent
current
equal
to theequivalente
algebraic sum
of theacurrents
of
current
sources.
corriente
equivalente
La corriente
de lasource
fuenteisde
corriente
es igual
la suma
of the
the parallel
parallel
currentúnica.
sources.
of theWe
parallel
current
sources.
not
allowed
connect
independent
current
sources
algebraica
de las
las fuentes
de corriente
en paralelo.
We are
are
notcorrientes
allowed to
tode
connect
independent
current
sources in
in series.
series. Series
Series elements
elements have
have
We
are
not
allowed
to
connect
independent
current
sources
in
series.
Series
elements
have
the
same
current.
This
restriction
prevents
series
current
sources
from
being
independent.
No nos
está permitido
conectarprevents
fuentes deseries
corriente
independientes
en serie.
elementos
the same
current.
This restriction
current
sources from
beingLos
independent.
the
same
current.
This
restriction
prevents
series
current
sources
from
being
independent.
Similarly,
we
are
not
allowed
to
connect
independent
voltage
sources
in
parallel.
en
serie tienen
la misma
corriente.
Esta restricción
evitavoltage
que lassources
fuentes in
de parallel.
corriente en serie se
Similarly,
we are
not allowed
to connect
independent
Similarly,
we
are not allowed
to connect
independent
voltage
sources
in parallel.
Table
3.5-1
the
series
connections
of
and
voltage
tornen
independientes.
Del mismo
modo, and
tampoco
nos
permite
conectar
voltaje
Table
3.5-1 summarizes
summarizes
the parallel
parallel
and
series se
connections
of current
current
andfuentes
voltagedesources.
sources.
Table
3.5-1
summarizes
the
parallel
and
series
connections
of
current
and
voltage sources.
independientes en paralelo.
La tabla 3.5-1 resume las conexiones en paralelo y en serie de las fuentes de corriente y de
voltaje.
3.6 C I R C U I T A N A L Y S I S
3.6
3.6
C
C II R
RC
CU
U II T
T A
AN
NA
AL
LY
YS
S II S
S
In
In this
this section,
section, we
we consider
consider the
the analysis
analysis of
of aa circuit
circuit by
by replacing
replacing aa set
set of
of resistors
resistors with
with an
an
In this Asection,
the
circuit
by replacing
a set of resistors with an
3.6
Nresistance,
Á L we
I S Iconsider
Sthus
DE
C Ianalysis
R CtheUnetwork
I of
T Oa S
equivalent
reducing
to a form
easily analyzed.
equivalent resistance, thus reducing the network to a form easily analyzed.
equivalent
resistance,
thus reducing
the network Note
to a formiteasily
analyzed.
Consider
Consider the
the circuit
circuit shown
shown in
in Figure
Figure 3.6-1.
3.6-1. Note that
that it includes
includes aa set
set of
of resistors
resistors that
that is
is in
in
theset
circuit
shown
in
Figure
3.6-1.
Note
that
it includes
athe
set output
of resistors
thatvois, so
in
En
estaConsider
sección
consideramos
el análisis
de parallel.
un circuito
mediante
el find
reemplazo
de un
conjunto
de
series
and
another
of
resistors
that
is
in
It
is
desired
to
voltage
series and another set of resistors that is in parallel. It is desired to find the output voltage vo, so
series
andto
another
set
ofcircuit
resistors
thatequivalent
is in con
parallel.
It is
desired
find
the3.6-2.
output
vo, so
resistores
con
una resistencia
equivalente,
locircuit
que
seshown
reduceinto
laFigure
red para
tener voltage
así una forma
we
wish
reduce
the
to
the
we wish to reduce the circuit to the equivalent circuit shown in Figure 3.6-2.
we wish
to reduce
más
fácil de
analizar.the circuit to the equivalent circuit shown in Figure 3.6-2.
Considere el circuito que se muestra en la figura 3.6-1. Observe que incluye un conjunto
de resistores que está en serie y otro que está en paralelo. Como lo que se pretende es encontrar
el voltaje de salida vo, habrá que reducir el circuito al circuito equivalente que se muestra en la
figura 3.6-2.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 73
FIGURE 3.5-2
FIGURE 3.5-2
FIGURE
(a)
A circuit3.5-2
(a) A circuit
(a) A circuitparallel
containing
containing parallel
containing
parallel
current
sources
and (b)
current sources and (b)
current
sourcescircuit.
and (b)
an
equivalent
an equivalent circuit.
an equivalent circuit.
FIGURA 3.5-2
(a) Circuito con
fuentes de corriente
en paralelo y (b)
circuito equivalente.
Alfaomega
4/12/11 5:21 PM
74
74 74
74
74
74
74
Resistive
Circuits
Resistive
Circuits
Resistive
Circuits
Circuitos
resistivos
Resistive
ResistiveCircuits
Circuits
Resistive
Circuits
s
o
o
s
FIGURE
3.6-2
Equivalent
circuit
for
the
circuit
of
3.6-2
Equivalent
circuit
for for
the the
circuit
of of
FIGURE
3.6-2
Equivalent
circuit
circuit
FIGURE
3.6-1
Circuit
with
setconjunto
of series
series
resistors
anden FIGURE
FIGURE
3.6-1
Circuit
withwith
a setaun
of
series
resistors
andand
FIGURA
3.6-1
Circuito
con
de
resistores
FIGURA
equivalente
para
el circuito
FIGURE
3.6-1
Circuit
set
of
resistors
FIGURE
FIGURE3.6-2
3.6-2
3.6-2Circuito
Equivalent
Equivalent
circuit
circuit for
for the
the
circuit
circuit of
of
FIGURE
3.6-2
Equivalent
circuit
for
the
circuit
of
FIGURE
FIGURE3.6-1
3.6-1Circuit
Circuitwith
withaaaaset
setof
ofseries
seriesresistors
resistorsand
and Figure
FIGURE
3.6-1
Circuit
with
set
of
series
resistors
and
Figure
3.6-2.
3.6-2.
Figure
3.6-2.
a of
setparallel
of
parallel
resistors.en paralelo.
a setserie
resistors.
y otro
de
resistores
de
la figura
3.6-1.
set
of
parallel
resistors.
Figure
Figure
3.6-2.
3.6-2.
Figure
3.6-2.
aaaaset
setof
ofparallel
parallelresistors.
resistors.
set
of
parallel
resistors.
We
note
that
the
equivalent
series
resistance
is es
WeWe
note
thatthat
thethe
equivalent
series
resistance
is serie
Observamos
resistencia
equivalente
en
note
equivalent
series
resistance
is
We
Wenote
notethat
thatque
the
thela
equivalent
equivalent
series
series
resistance
resistance
is
is
We
note
that
the
equivalent
series
resistance
is
Rss R¼
¼1 R
R
þ2 R
R
þ3 R
R
Rs R
¼
þ11 Rþ
þ22 Rþ
RRRsss¼
¼RRR111þ
þRRR222þ
þRRR33333
¼
þ
þ
and
the
equivalent
parallel
resistance
is es
andy
the
equivalent
parallel
resistance
is is
quethe
la equivalent
resistencia
equivalente
en paralelo
and
parallel
resistance
and
andthe
theequivalent
equivalentparallel
parallelresistance
resistanceis
is
and
the
equivalent
parallel
resistance
is
1 1
¼ 1111
R
¼pp ¼
Rp R
¼p G
RRRppp¼
¼
Gpp
G
G
Gppp
G
where
G
¼
G
þ5 G
G
þ6 G
G66
where
Gp G¼pp G
þ44 G
þG
4G
where
¼
þ
þ
donde
where
where
G
Gppp¼
¼G
G444þ
þG
G55555þ
þG
G666
where
G
¼
G
þ
G
þ
G
Then,
using
the
voltage
divider
principle,
with
Figure
3.6-2,
we
have
Then,
using
the
voltage
divider
principle,
with
Figure
3.6-2,
we
have
Then,
using
the
voltage
divider
principle,
with
Figure
3.6-2,
we
have
Entonces,
utilizando
el
principio
del
divisor
de
voltaje,
con
la
figura
3.6-2, tenemos
Then,
Then, using
usingthe
thevoltage
voltagedivider
dividerprinciple,
principle,with
withFigure
Figure3.6-2,
3.6-2, we
wehave
have
Then,
using
the
voltage
divider
principle,
with
Figure
3.6-2,
we
have
Rpp
Rp R
v ¼
¼ RRRpppvs vvss
vo v¼
vvvooooo ¼
¼s þ
¼
Rss Rþ
þp R
R vvvsss
R
R
RR
Rsssþ
þRRRppppp
þ
Replacing
the
series
resistors
by
the
equivalent
resistor
Rss R
did
not
change
the
current
orelvoltage
voltage
of
Replacing
thethe
series
resistors
thethe
equivalent
resistor
R R
did
not
change
thethe
current
or voltage
of de
Replacing
series
resistors
by
equivalent
resistor
did
not
change
current
or
of
Reemplazar
losseries
resistores
enby
serie
por
el
resistor
equivalente
modifica
corriente
voltaje
s no
Replacing
Replacingthe
the
series
resistors
resistors
by
bythe
theequivalent
equivalent
resistor
resistors RR
Rsssdid
did
not
not
change
changela
the
the
current
currentoor
orvoltage
voltage
of
of
Replacing
the
series
resistors
by
the
equivalent
resistor
did
not
change
the
current
or
voltage
of
did
not
change.
Also,
the
voltage
v
across
the
any
other
circuit
element.
In
particular,
the
voltage
v
did
not
change.
Also,
the
voltage
v
across
the
anyningún
other
circuit
element.
In
particular,
the
voltage
v
o voo v
did
not
change.
Also,
the
voltage
across
the
any
other
circuit
element.
In
particular,
the
voltage
otro
elemento
del circuito.
En particular,
el ovoltaje
cambia.Also,
Además,
el voltaje
o no
o a través
didvnot
not
change.
change.
Also,
the
thevoltage
voltage
vvvoooacross
across
the
the
any
anyother
other
circuit
circuit
element.
element.
In
Inparticular,
particular,
the
thevoltage
voltage
vvvvooooodid
did
not
change.
Also,
the
voltage
across
the
any
other
circuit
element.
In
particular,
the
voltage
isRequal
equal
to the
the
voltage
across
each
ofcada
the
parallel
resistors.
Consequently,
the
equivalent
resistor
R
is equal
to the
voltage
across
each
of
the
parallel
resistors.
Consequently,
thethe
equivalent
resistor
Rp R
is
to
voltage
across
each
of
the
parallel
resistors.
Consequently,
equivalent
resistor
del
resistor
equivalente
alvoltage
voltaje
que
pasa
porof
uno
de
los
resistores
en paralelo.
En
p es igual
isequal
equal
to
tothe
the
voltage
across
across
each
each
ofthe
theparallel
parallel
resistors.
resistors.
Consequently,
Consequently,
the
the
equivalent
equivalent
resistor
resistor
RRRpppppis
is
equal
to
the
voltage
across
each
of
the
parallel
resistors.
Consequently,
the
equivalent
resistor
in
Figure
3.6-2
is
equal
to
the
voltage
v
in
Figure
3.6-1.
We
can
analyze
the
simple
voltage
v
in
Figure
3.6-2
is
equal
to
the
voltage
v
in
Figure
3.6-1.
We
can
analyze
the
simple
voltage
v
o voltaje
o voo in Figure
o val
3.6-2
is
equal
to
the
voltage
in
Figure
3.6-1.
We
can
analyze
the
simple
voltage
consecuencia,
el
voltaje
v
de
la
figura
3.6-2
es
igual
v
de
la
figura
3.6-1.
Podemos
analizar
o
o is
o 3.6-1.
in Figure
Figure 3.6-2
3.6-2
is equal
equal to
to the
the voltage
voltage vvvooo in
in Figure
Figure
3.6-1. We
We can
can analyze
analyze the
the simple
simple
voltage
voltage vvvooo in
in
Figure
3.6-2
is
equal
to
the
voltage
in
Figure
3.6-1.
We
can
analyze
the
simple
voltage
and
know
that
the
voltage
v the
involtaje
the
more
circuit
in Figure
Figure
3.6-2
tofigura
find
the
value
ofencontrar
the
voltage
voo and
know
thatthat
thethe
voltage
vque
more
circuit
in Figure
3.6-2
to la
find
thethe
value
of
the
voltage
velo and
o in
know
voltage
in
the
more
circuit
in
3.6-2
to
find
value
of
the
voltage
el
circuito
sencillo
de
3.6-2
para
voltaje
saber
vo
o y voltage
anddel
know
know
that
thatvthe
the
voltage
vvvvoooooel
in
inthe
themore
more
circuit
circuit
in
inFigure
Figure
3.6-2
3.6-2
to
tofind
findthe
the
value
value
of
ofthe
thevoltage
voltage
vvvvalor
voooand
and
know
that
the
voltage
in
the
more
circuit
in
Figure
3.6-2
to
find
the
value
of
the
voltage
complicated
circuit
shown
in
Figure
3.6-1
has
the
same
value.
complicated
circuit
shown
in
Figure
3.6-1
has
the
same
value.
complicated
circuit
shown
in
Figure
3.6-1
has
the
same
value.
en
el
circuito
más
complejo
de
la
figura
3.6-1
tiene
el
mismo
valor.
complicated
complicated circuit
circuit shown
shown in
in Figure
Figure 3.6-1
3.6-1 has
has the
the same
same value.
value.
complicated
circuit
shown
in
Figure
3.6-1
has
the
same
value.
APM
ML P
P
E. 3
Series
and
Parallel
Resistors
EEXEjAeXXMm
E LL3E
6 .- 6
1 - 1Series
and
Parallel
pMlPP
serie Resistors
y Resistors
en
paralelo
EEEXXXAAAAM
Po
LLLEEE33
...66-61
---111 Resistores
Series
Series and
andenParallel
Parallel
Resistors
M
3.36
Series
and
Parallel
Resistors
Consider
the
circuit
shown
in Figure
Figure
3.6-3.
Find
the
current
i11 when
when i cuando
Consider
thethe
circuit
shown
in Figure
Find
thethe
current
i1lawhen
Consider
circuit
shown
in
3.6-3.
Find
current
Considere
el
circuito
mostrado
en
la3.6-3.
figura
3.6-3.
Consider
Considerthe
the
circuit
circuit shown
shownin
inFigure
Figure
3.6-3.
3.6-3.
Find
FindEncuentre
the
thecurrent
current
iii1icorriente
when 1
Consider
the
circuit
shown
in
Figure
3.6-3.
Find
the
current
when
11 when
R
¼V22 V
Vandand
andR2 R
R
¼3 R
R
¼V88 V
V
R4 R
¼44 2¼
¼22 R¼
¼33 8¼
RRR ¼
¼222V
V and
and R
R¼
¼RRR ¼
¼888V
V
¼
V
and
R
¼
¼
V
444
y
222
333
s
s
FIGURE
3.6-3
(a) Circuit
Circuit
for Example
Example
3.6-1.
(b) Partially
Partially
reduced
circuit
for Example
Example
3.6-1.
FIGURE
3.6-3
(a) (a)
Circuit
for for
Example
3.6-1.
(b) (b)
Partially
reduced
circuit
for for
Example
3.6-1.
FIGURE
3.6-3
3.6-1.
reduced
circuit
3.6-1.
FIGURE
FIGURE3.6-3
3.6-3(a)
(a)Circuit
Circuitfor
forExample
Example3.6-1.
3.6-1.(b)
(b)Partially
Partiallyreduced
reducedcircuit
circuitfor
forExample
Example3.6-1.
3.6-1.
FIGURE
3.6-3
(a)
Circuit
for
Example
3.6-1.
(b)
Partially
reduced
circuit
for
Example
3.6-1.
FIGURA 3.6-3 (a) Circuito para el ejemplo 3.6-1. (b) Circuito reducido parcialmente para el ejemplo 3.6-1.
Solution
Solution
Solución
Solution
Solution
Solution
Because
objective
to find
i , we
attempt
to reduce
circuit
so that
resistor
in parallel
with
Because
thethe
objective
is tois find
i , we
willwill
attempt
to reduce
thethe
circuit
so that
thethe
3-V3-V
resistor
is inis parallel
with
1i ,i11intentaremos
Because
the
objective
is
to
find
, we
we
will
attempt
to
reduce
the
circuit
so
that
the
3-V
resistor
is
in
parallel
with
Como
el the
objetivo
es encontrar
reducir
el circuito
de
manera
quethe
el resistor
de 3-V
esté
en paralelo
Because
Because
theobjective
objective
is
isto
tofind
find
wewill
willattempt
attempt
to
toreduce
reduce
the
thecircuit
circuit
so
sothat
that
the
3-V
3-Vresistor
resistor
is
isin
inparallel
parallel
with
with
Because
the
objective
is
to
find
will
attempt
to
reduce
the
circuit
so
that
the
3-V
resistor
is
in
parallel
with
1ii1i11,,,we
Then
we
can
use
the
current
divider
principle
to obtain
obtain
i11.. Because
Because
R22obtener
and
R
one
resistor
and
the
current
source
iss.. Then
we
cancan
use
thethe
current
divider
principle
todivisor
obtain
i1.corriente
Because
R
and
R3 R
onecon
resistor
andand
the
current
source
is. Then
2R
we
use
current
divider
principle
to
i
and
one
resistor
the
current
source
i
un
resistor
y
la
fuente
de
corriente
i
.
Luego
podemos
aplicar
el
principio
del
de
para
Then
wecan
canuse
usethe
thecurrent
currentdivider
dividerprinciple
principleto
toobtain
obtainii1i11...Because
BecauseRRR222and
andRRR33333
one
oneresistor
resistorand
andthe
thecurrent
currentsource
sourceiisis.s..Then
Then
we
can
use
the
current
divider
principle
to
obtain
Because
and
one
resistor
and
the
current
source
s we
are
in
parallel,
we
find
an
equivalent
resistance
as
areiare
in
parallel,
we
find
an
equivalent
resistance
as
in
parallel,
we
find
an
equivalent
resistance
as
que Rwe
y find
R
en paralelo,
encontramos
are
are
in
inparallel,
parallel,
find
an
anequivalent
equivalent
resistance
resistance
as
as una resistencia equivalente como
are
in
parallel,
we
find
an
equivalent
resistance
as
1. Puesto
2we
3 están
R322R
R
R2 RR
R222RRR3333¼
R¼
¼ RR
¼V44 V
V
Rp1R
3 4
p1 ¼
¼
p1
RRRp1
¼2 R
¼444V
V
¼
¼
V
R
þ3 R
R ¼
R
þ22 Rþ
p1¼
p1
RRR222þ
þRRR33333
þ
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 74
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
E1C03_111/25/2009
11/25/2009 75 75
E1C03_1
Circuit
Analysis
Análisis
de circuitos
Circuit
Analysis
Circuit
Analysis
Circuit
Analysis
75
75
75
75 75
s
FIGURE 3.6-4
3.6-4 Equivalent
Equivalent circuit
circuit for
for Figure
Figure 3.6-3.
3.6-3.
FIGURE
FIGURA
3.6-4
Circuito
equivalente
para
la3.6-3.
figura
3.6-3.
FIGURE
3.6-4
Equivalent
circuit
for
Figure
3.6-3.
FIGURE
3.6-4
Equivalent
circuit
for Figure
Este
equivalente
está
conectado
seriewith
conR
Al agregar
Rp1Rp1
a Rto
un resistor
en serie
This resistor
equivalent
resistor is
is
connected
in en
series
with
RR444...Then
Then
adding
to4,R
Rtenemos
have
series equivalente
equivalent resistor
resistor
p1
4, we have
This
equivalent
resistor
in
adding
aaa series
equivalent
4 . Then
p1R
44,, we
This
equivalent
resistor
is connected
connected
in series
series
with
adding
Rto
have
series
equivalent
resistor
This
equivalent
resistor
is connected
in series
with
R4.RThen
adding
Rp1R
we
have
a series
equivalent
resistor
4
p1 to
4 we
4,R
Rs ¼ R
R4 þ R
Rp1 ¼
¼ 2þ
þ 4 ¼ 66 V
V
R
p1 2¼þ2
Rss R¼
¼4 þ
R44Rþ
þp1R¼
24þ¼44 6¼
¼V6 V
Rs ¼
p1
Now the
the
R
is
in parallel
parallel
withcon
three
resistors
as shown
in Figure
Figure
3.6-3b.
However,
we wish
wish
to obtain
obtain the
the
s resistor
Ahora
resistor
está
paralelo
losresistors
tres resistores
como
muestra
en
laHowever,
figurawe
3.6-3b.
Sin
el
Now
R
is
in
with
three
as
in
3.6-3b.
we
to
ss resistor
sin
Now
R
resistor
is
in en
parallel
with
three
resistors
as shown
shown
inise
Figure
3.6-3b.
However,
we
wish
toembargo,
obtain
Now
thethe
Rels resistor
isRas
parallel
three
resistors
as we
shown
in
Figure
3.6-3b.
However,
wish
to obtain
thethe
scircuit
.
Therefore,
we
combine
the
9-V
resistor,
the
equivalent
shown
inwith
Figure
3.6-4
so se
that
canen
find
1
objetivo
es
obtener
el
circuito
equivalente
que
muestra
la
figura
3.6-4
para
que
podamos
encontrar
i
.
Por
.
Therefore,
we
combine
the
9-V
resistor,
the
equivalent
circuit
as
shown
in
Figure
3.6-4
so
that
we
can
find
i
1
1 the
. Therefore,
we
combine
the
resistor,
equivalent
circuit
as
shown
in the
Figure
3.6-4
so that
find
i113.6-3b
combine
the
9-V9-V
resistor,
the
equivalent
circuit
as shown
in Figure
3.6-4
that
we we
cancan
i1. Therefore,
shown
to
right
of9so
terminals
a-b
infind
Figure
intowe
one
parallel
equivalent
conductance
18-V resistor,
resistor,
and
Rss shown
consiguiente,
combinamos
el
resistor
de
V,
el
de
18-V
y
el
R
que
aparece
a
la
derecha
de
las
terminales
a-b
en
to
the
right
of
terminals
a-b
in
Figure
3.6-3b
into
one
parallel
equivalent
conductance
18-V
and
R
s
s
to the
right
of terminals
in Figure
3.6-3b
parallel
equivalent
conductance
18-V
resistor,
Rs shown
to the
right
of terminals
a-ba-b
in Figure
3.6-3b
intointo
oneone
parallel
equivalent
conductance
18-V
resistor,
and
R
s shown
. Thus,
Thus,
weand
find
Gp2
p2
la
figura
3.63b
en
una
conductancia
equivalente
en
paralelo,
G
.
Entonces,
encontramos
.
we
find
G
p2
. Thus,
Thus,
we we
findfind
Gp2G. p2
p2
1
1 11
1 11 11 11
1
1
1
11 ¼ 3 V
¼1þ
þ 1 1¼
¼1þ
þ 1 1¼
¼ 1 SS )
) R
Rp2
¼
Gp2
p2 1¼
p2 ¼
1þ 1 1þ
1þ 1 1þ
1 1 ¼
3V
p2
p2
9
18
R
9
18
6
3
G
¼
þ
þ
¼
þ
þ
¼
S
)
R
¼
G¼
s
p2 3¼V3 V
þ
þ
¼
þ
þ
¼
S
)
R
¼
¼
Gp2G
p2
p2 p2
9918 18
G
p2
18Rs R
Rss 9 9918 18
186 66 3 33
G
9
G
p2
p2
Then, using
using the
the current
current divider
divider principle,
principle,
Then,
Then,
using
current
divider
Then,
using
thethe
current
divider
principle,
Luego,
aplicando
el principio
delprinciple,
divisor de corriente,
G1iis
1 s
¼1 iG
G
s 1 is
¼
iii11 G
Gp
¼
i1 ¼
1
G
Gp Gpp
1 11 22
where
Gp ¼
¼G
G1 þ
þG
Gp2 1¼
¼ 111þ
þ 1 2¼
¼2
where
G
p
1
p2
3 þ¼33 ¼
3
where
¼1 G
where
Gp G
¼p G
þ1Gþp2G¼
p2 ¼þ3
3
3
3 3 3 33
donde
Therefore,
Therefore,
Por
consiguiente,
Therefore,
Therefore,
1=3
1
¼ 1=3
¼ 11 iiss
1=3 iisss 1¼
¼
iii111 1=3
s
2=3
i
¼
¼
i
i
¼
i1 ¼
1
s s
s 2 is
2=3
2=32=3
2 22
E XXj eAAmM
MpP
PlL
LoE
E 3
6 ---22
2 Equivalent
Equivalent
Resistance
. 66
Resistencia Resistance
equivalente
E
3
Equivalent
Resistance
E XE
AXMAPML P
E L3E . 3
6 ..- 6
2 - 2Equivalent
Resistance
El
figura3.6-5a
3.6-5acontains
contienean
unohmmeter.
ohmímetro.An
Éste
es un instrumento
que mide
ohmios laresistance
resistencia.
Thecircuito
circuitde
in laFigure
ohmmeter
is an
an instrument
instrument
thatenmeasures
measures
in
circuit
in
3.6-5a
contains
an
ohmmeter.
An
ohmmeter
is
that
resistance
The
circuit
in Figure
Figure
3.6-5a
contains
an
ohmmeter.
An
ohmmeter
is resistor
an
instrument
measures
resistance
in
TheThe
circuit
in Figure
3.6-5a
contains
an
ohmmeter.
Ancircuito
ohmmeter
isthe
an
instrument
that
measures
resistance
in in
El
ohmímetro
medirá
lawill
resistencia
equivalente
del
de of
resistores
conectado
athat
sus
terminales.
Determine
ohms.
The
ohmmeter
measure
the
equivalent
resistance
circuit
connected
to
its
terminals.
ohms.
ohmmeter
will
measure
the
equivalent
resistance
of
the
resistor
circuit
connected
to
terminals.
ohms.
The
ohmmeter
will
measure
the
equivalent
resistance
of 3.6-5a.
the
resistor
circuit
connected
to its
its
terminals.
ohms.
TheThe
ohmmeter
will
measure
theby
equivalent
resistance
of the
resistor
circuit
connected
to its
terminals.
la
resistencia
medida
por
elmeasured
ohmímetro
en
laohmmeter
figura
3.6-5a.
Determine
the
resistance
the
in Figure
Figure
Determine
the
resistance
measured
by
the
ohmmeter
in
3.6-5a.
Determine
resistance
measured
ohmmeter
in Figure
3.6-5a.
Determine
thethe
resistance
measured
by by
thethe
ohmmeter
in Figure
3.6-5a.
Solución
Solution
Solution
Funcionando de izquierda a derecha, el resistor de 30-V está en paralelo con el resistor de 60-V. La resistencia
Solution
Solution
Working from left to right, the 30-V resistor is parallel to the 60-V resistor. The equivalent resistance is
Working
from
to
30-V
resistor
is
to
60-V
resistor.
equivalent
resistance
equivalente
es
Working
from
left
to right,
right,
the
30-V
resistor
is parallel
parallel
to the
the
60-V
resistor.
The
equivalent
resistance
is
Working
from
leftleft
to right,
thethe
30-V
resistor
is parallel
to the
60-V
resistor.
TheThe
equivalent
resistance
is is
60
�
30
¼ 20
20 V
V
30 ¼
60 60
�60
30�� 30
60 þ
þ¼
30 20
¼V
20 V
6030
þ 30
30
60 60
þ
In
Figure
3.6-5b,
the
parallel
combination
of
the
30-V
and
60-V de
resistors
has
been
replaced
with
the equivalent
equivalent
In
Figure
3.6-5b,
the
parallel
combination
of
the
30-V
and
60-V
resistors
has
been
replaced
with
the
EnFigure
la figura
3.6-5b,
laparallel
combinación
en paralelo
de
los resistores
30-V yhas
debeen
60 Vreplaced
ha sido with
reemplazada
por el
In
3.6-5b,
the
combination
ofare
the
30-V
60-V
resistors
equivalent
In Figure
3.6-5b,
the
parallel
combination
of the
30-V
andand
60-V
resistors
has been
replaced with
thethe
equivalent
20-V
resistor.
Now
the
two
20-V
resistors
in
series.
20-V
resistor.
Now
the
two
20-V
resistors
are
in
series.
resistor
de
20-V.
Ahora
los
dos
resistores
de
20-V
están
en
serie.
20-V
resistor.
Now
two
20-V
in series.
20-V
resistor.
Now
thethe
two
20-V
resistors
areare
in series.
The
equivalent
resistance
is resistors
The
equivalent
resistance
is
La
resistencia
equivalente
es
equivalent
resistance
TheThe
equivalent
resistance
is is
20 þ
þ 20 ¼
¼ 40
40 V
2020
þ 20
20
¼V
40 V
V
20 20
þ
¼ 40
In Figure
Figure
3.6-5c,
the
series
combination
ofde
thelostwo
two
20-V
resistors
has been
been
replaced
with the
the
equivalent
40-V
En
la figura
3.6-5c,
la series
combinación
en serie
dos20-V
resistores
de 20-V
ha sido
reemplazada
porequivalent
el resistor 40-V
equiIn
3.6-5c,
the
combination
of
the
resistors
has
replaced
with
In
Figure
3.6-5c,
series
combination
of to
the
two
20-V
resistors
has
been
replaced
with
equivalent
40-V
In Figure
3.6-5c,
thethe
series
combination
of the
two
20-V
resistors
has
been
replaced
with
thethe
equivalent
40-V
resistor.
Now
the
40-V
resistor
is
parallel
the
10-V
resistor.
The
equivalent
resistance
is
valente de
40-V.
Ahora,
el resistor
de
40-V to
está
en10-V
paralelo
con elThe
resistor
de 10-V.
La resistencia
equivalente es
resistor.
Now
the
40-V
resistor
is
parallel
the
resistor.
equivalent
resistance
is
resistor.
Now
40-V
resistor
is parallel
to the
10-V
resistor.
equivalent
resistance
resistor.
Now
thethe
40-V
resistor
is parallel
to the
10-V
resistor.
TheThe
equivalent
resistance
is is
40 �� 10
10
40
¼
8
V
40 �40
10� 10 ¼ 8 V
40 þ
þ¼
10 8¼V8 V
4010
þ 10
10
40 40
þ
In
Figure
3.6-5d
the
parallel
combination
of
the
40-V
and 10-V
10-V
resistors
has been
been
replaced
with the
the
equivalent
EnFigure
la figura
3.6-5d
la parallel
combinación
en paralelo
de los
resistores
de 40-V
yhas
20-V
ha
sido
reemplazada
porequivalent
el resistor
In
3.6-5d
combination
of
40-V
and
resistors
has
replaced
with
In
Figure
3.6-5d
the
parallel
combination
of the
the
40-V
and
10-V
resistors
has
been
replaced
with
equivalent
In Figure
3.6-5d
thethe
parallel
combination
of the
40-V
and
10-V
resistors
been
replaced
with
thethe
equivalent
8-V
resistor.
Thus,
the
ohmmeter
measures
a
resistance
equal
to
8
V.
equivalente
de
8-V.
Por
consiguiente,
el
ohmímetro
mide
una
resistencia
igual
a
8
V.
8-V
resistor.
Thus,
the
ohmmeter
measures
a
resistance
equal
to
8
V.
resistor.
Thus,
ohmmeter
measures
a resistance
equal
8 V.
8-V8-V
resistor.
Thus,
thethe
ohmmeter
measures
a resistance
equal
to 8toV.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 75
Alfaomega
4/12/11 5:21 PM
Circuitos resistivos
76
Ohmímetro
Ohmímetro
Ohmímetro
Ohmímetro
FIGURA 3.6-5
Ejemplo 3.6.3
Análisis de circuitos utilizando resistencias equivalentes
Determine los valores de i3, v4, i5 y v6 en el circuito que se muestra en la figura 3.6-6.
Solución
El circuito de la figura 3.6-7 se obtuvo a partir del circuito que se muestra en la figura 3.6-6 al reemplazar las
combinaciones en serie y en paralelo de las resistencias por resistencias equivalentes. Podemos utilizar el circuito
equivalente para resolver este problema en tres pasos:
1. D
etermine los valores de las resistencias R1, R2 y R3 de la figura 3.6-7 que conforman el circuito de la figura
3.6-7 equivalente al circuito de la figura 3.6-6.
FIGURA 3.6-6 El circuito considerado en el ejemplo 3.6-3.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 76
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
Circuit
Circuit
CircuitAnalysis
Analysis
Analysis
Circuit
Analysis
Circuit
Circuit
Analysis
Análisis
de Analysis
circuitos
77
77
77
77
77
77
77
FIGURE
FIGURE
FIGURE3.6-7
3.6-7
3.6-7An
An
Anequivalent
equivalent
equivalentcircuit
circuit
circuitfor
for
forthe
the
the
FIGURE
3.6-7
An
equivalent
circuit
for
the
FIGURE
3.6-7
An
equivalent
circuit
for
FIGURE
3.6-7
An
equivalent
circuit
forthe
the
FIGURE
FIGURA
3.6-83.6-8
FIGURA
3.6-7
Circuito
equivalente
FIGURE
FIGURE
3.6-8
3.6-8
circuit
in
Figure
3.3-6.
FIGURE
3.6-8
circuit
circuit
in
in
Figure
Figure
3.3-6.
3.3-6.
FIGURE
FIGURE3.6-8
3.6-8
circuit
in
Figure
3.3-6.
circuit
in
Figure
3.3-6.
circuit
in
Figure
3.3-6.
para el circuito de la figura 3.3-6.
2.
2.
2.
2.
2.
2.
2.
3.
3.
3.
3.
3.
3.
3.
Determine
in
Figure
3.6-7.
Determine
Determinethe
the
thevalues
values
valuesof
of
ofvvvvvv1v111,,,1,,,,vvvvvv2v222,,,2,,e,and
and
and
in
infigura
Figure
Figure
3.6-7.
3.6-7.
Determine
los
valores
de
i eniiiiiila
3.6-7.
Determine
the
values
of
and
in
Figure
3.6-7.
Determine
the
values
of
in
Figure
3.6-7.
Determine
the
values
of
and
in
Figure
3.6-7.
11 22 and
iiiiin
Because
circuits
are
equivalent,
vvv11vv,,1,1,vv,v2v2v,,2,2,and
C
omo losthe
circuitos
son
losvalues
valores
de
eand
i de
laFigure
figura
3.6-6
son
iguales
los
valores
devvvvv1v11,,,1,,,
and
in
in
Figure
Figure3.6-6
3.6-6
3.6-6are
are
areequal
equal
equalto
to
toathe
the
thevalues
values
valuesof
of
of
Because
Because
the
the
circuits
circuits
are
areequivalentes,
equivalent,
equivalent,the
the
the
values
valuesof
of
of
and
in
Figure
3.6-6
are
equal
to
the
values
of
Because
the
circuits
are
equivalent,
the
values
of
ii in
Figure
3.6-6
are
equal
to
the
values
of
Because
the
circuits
are
equivalent,
the
values
of
vv111,, vv222,, and
and
in
Figure
3.6-6
are
equal
to
the
values
of
Because
the
circuits
are
equivalent,
the
values
of
11
vvvv2v21,,2,,,and
i
in
Figure
3.6-7.
Use
voltage
and
current
division
to
determine
the
values
of
i
,
v
,
i
,
and
v
vand
la
figura3.6-7.
3.6-7.Use
Utilice
la división
de voltaje
y deto
corriente
parathe
determinar
los
de
i3v,v6v6v64in
,in
and
and
in
inFigure
Figure
Figure
3.6-7.
3.6-7.
Use
Use
voltage
voltage
and
andcurrent
current
current
division
division
to
todetermine
determine
determine
the
the
values
valuesof
of
of
,vvvv4v44,,4,,,iii5i5i5,,5,,,and
and
and
in
in
Figure
3.6-7.
Use
voltage
and
current
division
to
determine
the
values
of
and
in
2 e iiiiiide
vv222,, and
in
voltage
and
division
values
iiii333i33,,3,,valores
and
in
Figure
3.6-7.
Use
voltage
and
current
division
to
determine
the
values
of
in
44 55 and vv66
6 in
Figure
3.6-6.
i
y
v
de
la
figura
3.6-6.
Figure
Figure
3.6-6.
3.6-6.
Figure
3.6-6.
5
6
Figure
Figure 3.6-6.
3.6-6.
Paso1:1:Figure
La figura
3.6-8a
muestra
los resistors
tres resistores
en la
partecircuit
alta del
circuito3.6-6.
de la We
figura
3.6-6.the
Vemos
Step
3.6-8a
shows
three
at
of
in
see
6-V
Step
Step1:
1:
1:Figure
Figure
Figure
3.6-8a
3.6-8a
shows
showsthe
the
the
three
three
resistors
resistors
at
atthe
the
thetop
top
top
of
ofthe
the
thecircuit
circuit
circuit
in
inFigure
Figure
Figure
3.6-6.
3.6-6.
We
We
see
seethat
that
thatthe
the
theserie
6-V
6-V
Step
1:
Figure
3.6-8a
shows
the
three
resistors
at
the
top
of
the
circuit
in
Figure
3.6-6.
We
see
that
the
6-V
Step
3.6-8a
shows
the
three
resistors
at
the
top
of
the
in
Figure
3.6-6.
We
see
that
6-V
Step
1:
Figure
3.6-8a
shows
the
three
resistors
at
the
top
of
the
circuit
in
Figure
3.6-6.
We
see
that
the
6-V
que
el
resistor
de
6-V
está
conectado
en
serie
con
el
resistor
de
18-V.
En
la
figura
3.6-8b,
estos
resistores
en
resistor
is
connected
in
series
with
the
18-V
resistor.
In
Figure
3.6-8b,
these
series
resistors
have
been
replaced
by
resistor
resistor
is
is
connected
connected
in
in
series
series
with
with
the
the
18-V
18-V
resistor.
resistor.
In
In
Figure
Figure
3.6-8b,
3.6-8b,
these
these
series
series
resistors
resistors
have
have
been
been
replaced
replaced
by
by
resistor
is
connected
in
series
with
the
18-V
resistor.
In
Figure
3.6-8b,
these
series
resistors
have
been
replaced
by
resistor
connected
the
Figure
3.6-8b,
these
resistors
have
been
by
resistor
connectedin
inseries
series
with
the18-V
18-Vresistor.
resistor.
In
Figure
3.6-8b,
theseseries
series
resistors
have
beenreplaced
replaced
by
han
sidoisisreemplazados
por elwith
resistor
equivalente
deIn
24-V.
Ahora
el resistor
de
24-V
está
conectado
en
paralelo
the
equivalent
24-V
resistor.
Now
the
24-V
resistor
is
connected
in
parallel
with
the
12-V
resistor.
Replacing
the
the
equivalent
equivalent
24-V
24-V
resistor.
resistor.
Now
Now
the
the
24-V
24-V
resistor
resistor
is
is
connected
connected
in
in
parallel
parallel
with
with
the
the
12-V
12-V
resistor.
resistor.
Replacing
Replacing
the
equivalent
24-V
resistor.
Now
the
24-V
resistor
is
connected
in
parallel
with
the
12-V
resistor.
Replacing
the
resistor.
Now
resistor
in
the equivalent
equivalent
24-V
resistor.
Now the
the
24-V
resistor
is connected
connected
in parallel
parallel with
with the
the 12-V
12-V
resistor.
Replacing
con
el resistor 24-V
de
12-V.
Reemplazar
los24-V
resistores
en is
serie
por una
resistencia
equivalente
noresistor.
modificaReplacing
el voltaje
series
series
seriesresistors
resistors
resistorsby
by
byan
an
anequivalent
equivalent
equivalentresistance
resistance
resistancedoes
does
doesnot
not
notchange
change
changethe
the
thevoltage
voltage
voltageor
or
orcurrent
current
currentin
in
inany
any
anyother
other
otherelement
element
elementof
of
ofthe
the
the
series
resistors
by
an
equivalent
resistance
does
not
change
the
voltage
or
current
in
any
other
element
of
the
series
resistors
by
an
equivalent
resistance
does
not
change
the
voltage
or
current
in
any
other
element
of
the
series
resistors
by
an
equivalent
resistance
does
not
change
the
voltage
or
current
in
any
other
element
of
the
o
la
corriente
en
ningún
otro
elemento
del
circuito.
En
particular,
v
,
el
voltaje
a
través
del
resistor
de
12-V,
no
1
,
the
voltage
across
the
12-V
resistor,
does
not
change
when
the
series
resistors
are
circuit.
In
particular,
v
1
,
,
the
the
voltage
voltage
across
across
the
the
12-V
12-V
resistor,
resistor,
does
does
not
not
change
change
when
when
the
the
series
series
resistors
resistors
are
are
circuit.
circuit.
In
In
particular,
particular,
v
v
1
1
,
the
voltage
across
the
12-V
resistor,
does
not
change
when
the
series
resistors
are
circuit.
In
particular,
v
across
the
does
change
the
series
are
circuit.
vv111,, the
the voltage
voltage
across
the 12-V
12-V resistor,
resistor,
does not
not
change when
when
the
series resistors
resistors
are
circuit. In
In particular,
particular,
cambia
cuando
los
resistores
en
serie
son
reemplazados
por
el
resistor
equivalente.
Por
el
contrario,
v
no
es
un
4
replaced
not
replaced
replacedby
by
bythe
the
theequivalent
equivalent
equivalentresistor.
resistor.
resistor.In
In
Incontrast,
contrast,
contrast,vvvvv4v444is
isis
not
notan
an
anelement
element
elementvoltage
voltage
voltageof
of
ofthe
the
thecircuit
circuit
circuitshown
shown
shownin
in
inFigure
Figure
Figure3.6-8b.
3.6-8b.
3.6-8b.
replaced
by
the
equivalent
resistor.
In
contrast,
not
an
element
voltage
of
the
circuit
shown
in
Figure
3.6-8b.
replaced
by
the
equivalent
resistor.
In
contrast,
an
element
voltage
of
the
circuit
shown
in
Figure
3.6-8b.
replaced
by
the
equivalent
resistor.
In
contrast,
isisnot
not
an
element
voltage
of
the
circuit
shown
in
Figure
3.6-8b.
voltaje
del
elemento
del
circuito
mostrado
en
la
3.6-8b.
4figura
4is
In
In
InFigure
Figure
Figure3.6-8c,
3.6-8c,
3.6-8c,the
the
theparallel
parallel
parallelresistors
resistors
resistorshave
have
havebeen
been
beenreplaced
replaced
replacedby
by
bythe
the
theequivalent
equivalent
equivalent8-V
8-V
8-Vresistor.
resistor.
resistor.The
The
Thevoltage
voltage
voltageacross
across
across
In
Figure
3.6-8c,
the
parallel
resistors
have
been
replaced
by
the
equivalent
8-V
resistor.
The
voltage
across
In
Figure
3.6-8c,
the
parallel
resistors
have
been
replaced
by
the
equivalent
resistor.
The
voltage
across
In
Figure
3.6-8c,
the
parallel
resistors
have
been
replaced
by
the
equivalent
8-V
resistor.
The
voltage
across
En
la figura
3.6-8c,
los
resistores
en paralelo
han
sido
reemplazados
por el8-V
resistor
equivalente
de 8-V.
El
in
this
case.
In
summary,
the
the
equivalent
resistor
is
equal
to
the
voltage
across
each
of
the
parallel
resistors,
v
1
in
in
this
this
case.
case.
In
In
summary,
summary,
the
the
the
the
equivalent
equivalent
resistor
resistor
is
is
equal
equal
to
to
the
the
voltage
voltage
across
across
each
each
of
of
the
the
parallel
parallel
resistors,
resistors,
v
v
1
1
in
this
case.
In
summary,
the
the
equivalent
resistor
is
equal
to
the
voltage
across
each
of
the
parallel
resistors,
v
1
this
In
summary,
the
the
resistor
isisequal
to
across
each
the
resistors,
vvlos
inresistores
thiscase.
case.en
Inparalelo,
summary,
the
theequivalent
equivalent
resistor
equal
tothe
thevoltage
voltage
across
eachof
of
theparallel
parallel
resistors,
voltaje
a
través
del
resistor
equivalente
es
igual
al
voltaje
a
través
de
cada
uno
de
v
en
11in
1
given
resistance
1 in
in
inFigure
Figure
Figure3.6-7
3.6-7
3.6-7is
isis
given
givenby
by
by
resistance
resistanceR
RR
in
Figure
3.6-7
given
by
resistance
Figure
3.6-7
is
resistance
in
Figure
3.6-7
isisgiven
given
by
resistance
RR11111in
este
caso. R
En
resumen,
la
resistencia
R1by
de la figura 3.6-7 resulta de
R
1 ¼
RR
¼
¼12
12
12kkkkkkððððð666ð666þ
þ
þ18
18
18ÞÞÞÞÞÞ¼
¼
¼888888V
V
V
¼
12
þ
18
¼
V
R
12
þ
18
¼
V
RR11111¼
¼
12
þ
18
¼
V
Similarly,
the
resistances
R
and
R
in
Figure
3.6-7
are
given
by
2
3
Similarly,
Similarly,
the
the
resistances
resistances
R
R
and
and
R
R
in
in
Figure
Figure
3.6-7
3.6-7
are
are
given
given
by
by
Similarly,
the
resistances
and
Figure
3.6-7
are
given
by
Similarly,
resistances
RR
3.6-7
given
by
Similarly,
the
resistances
R22222and
and RR
R333233in
inRFigure
Figure
3.6-7 are
are
given
by de
Del
mismothe
modo,
las resistencias
R
yin
3.6-7
resultan
3 de la figura
¼
12
þ
ð
20
k
5
Þ
¼
16
R
22 2¼
¼
12
12
þ
þ
ð
20
ð
20
k
k
5
5
Þ
Þ
¼
¼
16
16V
V
V
R
R
¼
12
þ
ð
20
k
5
Þ
¼
16
V
R
V
RR222 ¼
¼12
12þ
þðð20
20kk55ÞÞ¼
¼16
16
V
¼
8
k
ð
2
þ
6
Þ
¼
4
V
R
3
¼
¼88888kkkkkðððð22ð222þ
þ
þ66666ÞÞÞÞÞ¼
¼
¼44444V
V
V
RR
¼
þ
¼
V
R
¼
þ
¼
V
RR33333¼
Step
2:
Apply
KVL
to
the
circuit
of
Figure
3.6-7
to
get
Step
Step2:
2:
2:Apply
Apply
ApplyKVL
KVL
KVL
to
tothe
the
the
circuit
circuitof
of
ofFigure
Figure
Figure
3.6-7
3.6-7
to
topara
get
get obtener
Step
2:
Apply
KVL
to
the
circuit
of
Figure
3.6-7
to
get
Paso
2:
Aplique
la KVL
al circuito
de
la
figura
3.6-7
Step
to
circuit
3.6-7
to
get
Step
2:
Apply
KVL
to
the
circuit
of
Figure
3.6-7
to
get
18
18
18
18
18
18
18
18
18
18
18
18
¼
¼
R
R
8i
�
18
¼
0000 )
iiii¼
R
11 ii1iiþ
22 ii2iiþ
33 ii3iiþ
¼
¼
¼
¼0:5
0:5
0:5A
AA
A
þ
þ
R
R
þ
þ
R
R
þ
þ
8i
8i
�
�
18
18
¼
¼
)
)
¼
¼
R
R
¼
þ
R
þ
R
þ
8i
�
18
¼
)
¼
¼
0:5
R
1
2
3
¼
0:5
A
RR11iiþ
¼
¼
0:5
A
þRR22iiþ
þRR33iiþ
þ8i
8i�
�18
18¼
¼00 )
) ii¼
¼R
þ
R
þ
R
þ
8
8
þ
16
þ
1
2
3
þ
þ
R
R
þ
þ
R
R
þ
þ
8
8
8
8
þ
þ
16
16
þ
þ444444þ
þ
þ888888¼
R
R
þ
þ
þ
þ
16
þ
þ
þ
RR
þRR
R22222þ
þRR
R33333þ
þ888 888þ
þ16
16þ
þ
þ
R11111þ
Next,
Ohm’s
Next,
Next,
Ohm’s
Ohm’slaw
law
law
gives
gives
A
continuación
la gives
ley
de Ohm proporciona
Next,
Ohm’s
law
gives
Next,
Ohm’s
law
gives
Next,
Ohm’s
law
gives
vvv1v1 1¼
1i ¼
3i ¼
¼
¼R
RR
¼
¼888888ððððð0:5
0:5
ð0:5
0:5ÞÞÞÞÞÞ¼
¼
¼444444V
VV
V and
and
and
¼
¼R
RR
¼
¼444444ððððð0:5
0:5
ð0:5
0:5ÞÞÞÞÞÞ¼
¼
¼222222V
VV
V
y vvvvv2v222¼
¼
¼
¼
and
¼
¼
¼
vv111 ¼
R
0:5
¼
V
and
R
0:5
¼
V
¼
RR1111ii1iii¼
¼
0:5
¼
V
and
¼
RR3333ii3iii¼
¼
0:5
¼
V
22 ¼
,
v
,
and
i
in
Figure
3.6-6
are
equal
to
the
values
of
v
, v , and
iiiiin
Step
3:
The
values
of
v
Paso
3:
Los
valores
devvv1v1v1,,11,,vv,v2v2v2,,22,,and
eand
i dein
la
figura
3.6-6
son
iguales
losvalues
valores
de
eand
i de
laFigure
figura3.6-7.
3.6-7.
inFigure
Figure
Figure3.6-6
3.6-6
3.6-6are
are
areequal
equal
equalto
to
toathe
the
the
values
valuesof
of
of
and
in
in
Figure
Figure
3.6-7.
3.6-7.
Step
Step3:
3:
3:The
The
Thevalues
values
valuesof
of
of
and
in
Figure
3.6-6
are
equal
to
the
values
of
and
in
Figure
3.6-7.
Step
3:
The
values
of
vvvv11v111v,,1,,1,vv,vv22v222v,,2,,2,and
ii in
Figure
3.6-7.
Step
and iiiiiin
in
Figure
3.6-6
are
equal
to
the
values
of
and
in
Figure
3.6-7.
Step
3:
The
values
of
v11, v22, and
Returning
our
attention
to
Figure
3.6-6,
and
paying
attention
to
reference
directions,
we
can
determine
the
values
Si
volvemos
la
atención
a
la
figura
3.6-6
y
vemos
con
más
detenimiento
las
direcciones
de
referencia,
podemos
Returning
Returning
our
our
attention
attention
to
to
Figure
Figure
3.6-6,
3.6-6,
and
and
paying
paying
attention
attention
to
to
reference
reference
directions,
directions,
we
we
can
can
determine
determine
the
the
values
values
Returning
our
attention
to
Figure
3.6-6,
and
paying
attention
to
reference
directions,
we
can
determine
the
values
Returning
Returningour
ourattention
attentionto
toFigure
Figure3.6-6,
3.6-6,and
andpaying
payingattention
attentionto
toreference
referencedirections,
directions,we
wecan
candetermine
determinethe
thevalues
values
vvv66v6using
current
division,
law:
of
determinar
valores
de ivoltage
, v4, i5 ydivision,
v6 mediante
la división
deand
voltajes,
la división
,and
and
and
using
using
voltage
voltage
division,
division,
current
current
division,
division,
and
andOhm’s
Ohm’s
Ohm’s
law:
law: de la corriente y la ley de Ohm:
of
ofiiiii33i3,,,3,,,vvvvv44v4,,,4,,,iiii5i5i5,,,5,,los
3voltage
and
using
division,
current
division,
and
Ohm’s
law:
of
vv666 using
voltage
division,
current
division,
and
Ohm’s
law:
of
and
using
voltage
division,
current
division,
and
Ohm’s
law:
of
33 44 55 and
8888
1111
88
11ð0:5
iii33i3 ¼
iiii¼
¼
¼
¼
¼
0:5
ð0:5
0:5ÞÞÞÞÞÞ¼
¼
¼0:25
0:25
0:25A
AA
A
¼
¼
¼
0:25
3
¼
0:25
A
ii33 ¼
¼ 8888þ
¼2222ðððð0:5
0:5
¼
0:25
A
þ
þððððð222ð222þ
þ
þ666666ÞÞÞÞÞÞii¼
þ
þ
88þ
þ
22
þ
þ
18
18
18 v ¼ � 333333ð4Þ ¼ �3 V
18
18
18
�
vvv4v4 4 ¼
¼
¼
�
�
vv1v1 1¼
¼�
�
�4 ðððð44ð444ÞÞÞÞÞ¼
¼
¼�3
�3V
VV
V
¼
�
¼
�
¼
�3
�
vv444 ¼
¼ �6666þ
¼
�
�3
V
þ
þ18
18
18vv111 ¼
4444� ¼�3
þ
18
�
66þ
18
4
þ
18
�
��
��
�
�
�
�
5555
1111�
5
1
5
1
iii55i5 ¼
i
¼
�
ð0:5
�
¼
¼
�
�
i
i
¼
¼
�
�
0:5
ð0:5
0:5ÞÞÞÞÞÞ¼
¼
¼�0:1
�0:1
�0:1A
AA
A
¼
�
i¼
�
¼
�0:1
ii555 ¼
¼
�0:1
A
¼�
�20
¼�
� 5555 ðððð0:5
0:5
¼
�0:1
A
20
20þ
þ
þ555555ii¼
20
þ
20
þ
55
20
þ
¼
vvv6v6 6 ¼
¼
¼ððððð20
20
ð20
20kkkkkk555555Þi
ÞiÞi
¼
¼444444ððððð0:5
0:5
ð0:5
0:5ÞÞÞÞÞÞ¼
¼
¼222222V
VV
V
¼
¼
¼
20
Þi
0:5
¼
V
vv666 ¼
¼
20
ÞiÞi¼
¼
0:5
¼
V
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 77
Alfaomega
4/12/11 5:21 PM
78
78
Resistive Circuits
Circuitos resistivos
InEngeneral,
may find
the equivalent
resistance
for a portion
of aparte
circuit
consisting
of
general,we
podemos
encontrar
la resistencia
equivalente
para una
de un
circuito only
que sólo
resistors
and
then
replace
that
portion
of
the
circuit
with
the
equivalent
resistance.
For
example,
conste de resistores y luego reemplazar esa parte del circuito con la resistencia equivalente. Por ejemconsider
the circuit
shown que
in Figure
3.6-9.en
The
circuit
(a) is equivalent
theessingle
56 V
plo, considere
el circuito
se muestra
la resistive
figura 3.6-9.
El in
circuito
resistivo ento(a)
equivalente
.
We
say
that
R
is
‘‘the
equivalent
resistor
in
(b).
Let’s
denote
the
equivalent
resistance
as
R
al resistor de 56 Ω único en (b). Indiquemos ahora la resistencia
equivalente como
Req. Decimos que
eq
eq
resistance
lookingequivalente
into the circuit
Figure 3.6-9(a)
from
terminals
3.6-9(c)
shows
a
Req es “laseen
resistencia
queofaparece
dentro del
circuito
de laa-b.’’
figuraFigure
3.6-9(a)
a partir
de las
notation
used
to
indicate
the
equivalent
resistance.
Equivalent
resistance
is
an
important
concept
that
terminales a-b”. La figura 3.6-9(c) muestra una notación que se usa para indicar la resistencia equioccurs
in aLavariety
of situations
and has
a variety
of importante
names. ‘‘Input
‘‘output
resistance,’’
valente.
resistencia
equivalente
es un
concepto
que resistance,’’
se presenta en
diversas
situaciones
‘‘Thevenin
resistance,’’
and
‘‘Norton
resistance’’
are
some
names
used
for
equivalent
y tiene varios nombres. “Resistencia de entrada”, “resistencia de salida”, “resistencia de resistance.
Thevenin”, y
“resistencia de Norton”, son algunos de los nombres con que se conoce la resistencia equivalente.
FIGURE 3.6-9 The resistive circuit in (a) is equivalent to the single resistor in (b). The notation used to indicate the
FIGURA 3.6-9 El circuito resistivo en (a) es equivalente al resistor único de (b). La notación utilizada para indicar la
equivalent resistance is shown in (c).
resistencia equivalente se muestra en (c).
EXERCISE
3.6-1
Determine
the resistance
measured
bypor
theelohmmeter
Figure
E 3.6-1.
EJERCICIO
3.6-1
Determine
la resistencia
medida
ohmímetroinen
la figura
E 3.6-1.
Ohmímetro
FIGURE
EE
3.6-1
FIGURA
3.6-1
Answer:
Respuesta:
(30 þ 30) � 30
þ 30 ¼ 50 V
(30 þ 30) þ 30
3.7 AANNAÁLLYI ZSIIN
S GDREEC
3.7
SIIR
SC
TU
I VI T
E OCSI RRCEUSI ITSST I V O S
_
_
_
_
_
_
_
_
_
_
_
_
_________________________________________________________________________________
U
T
I
L
I
Z
A
N
D
O
M
AT
L
A
B
USING MATLAB
Podemos
analizar
circuitos
sencillos
mediante
la escritura
despeje deWe
unuse
conjunto
de ecuaciones.
We
can analyze
simple
circuits
by writing
and solving
a set ofyequations.
Kirchhoff’s
law and
Empleamos
la
ley
de
Kirchhoff
y
las
ecuaciones
de
elementos,
por
ejemplo,
la
ley
de
Ohm,example
para esthe element equations, for instance, Ohm’s law, to write these equations. As the following
cribir
estas
ecuaciones.
Como
ilustra
el
ejemplo
siguiente,
MATLAB
proporciona
una
manera
más
illustrates, MATLAB provides a convenient way to solve the equations describing an electric circuit.
cómoda de despejar las ecuaciones que describen un circuito eléctrico.
E
circuitos
sencillos
E jXeAmMpPlLoE 33..77 -11 MATLAB
MATLAB para
for Simple
Circuits
Determinethe
los values
valoresofdethe
losresistor
voltajesvoltages
y corrientes
los resistores
el circuito
en la figura 3.7-1.
Determine
and de
currents
for the para
circuit
shown inmostrado
Figure 3.7-1.
FIGURA 3.7-1 Circuito considerado en el ejemplo 3.7-1.
FIGURE 3.7-1 The circuit considered in Example 3.7-1.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 78
Circuitos Eléctricos - Dorf
4/12/11 5:21 PM
E1C03_1
11/25/2009
79
Analyzing
Resistive
Circuits
UsingMATLAB
MATLAB
Analyzing
Resistive
Circuits
Using
MATLAB
Análisis
de circuitos
resistivos
utilizando
MATLAB
Analyzing
Resistive
Circuits
Using
Analyzing
Resistive
Circuits
UsingMATLAB
MATLAB
Analyzing
Resistive
Circuits
Using
79
79
79
79
79
FIGURA
3.7-2
El
de
la Figure
figura
del etiquetado
de voltages
los
voltajes
las corrientes.
FIGURE3.7-2
3.7-2The
Thecircuito
circuitfrom
from
Figure3.7-1
3.7-1después
afterlabeling
labeling
theresistor
resistor
voltages
andycurrents.
currents.
FIGURE
3.7-2
The
circuit
from
Figure
3.7-1
after
labeling
the
resistor
voltages
and
currents.
FIGURE
circuit
3.7-1
after
the
and
FIGURE3.7-2
3.7-2The
Thecircuit
circuitfrom
fromFigure
Figure3.7-1
3.7-1after
afterlabeling
labelingthe
theresistor
resistorvoltages
voltagesand
andcurrents.
currents.
FIGURE
Solución
Etiquetemos,
Solution
Solution
Solution entonces, los voltajes y corrientes de los resistores. Antes de utilizar la ley de Ohm, etiquetaremos
Solution
el
voltaje
la corriente
de cada and
resistor
para que
se apeguen of
aoflausing
convención
pasiva.
una
de lasand
vaLet’s
labelythe
the
resistor voltages
voltages
and currents.
currents.
In anticipation
anticipation
using
Ohm’s law,
law,
we(Seleccione
will label
label the
the
voltage
and
Let’s
label
resistor
In
Ohm’s
we
will
voltage
Let’s
label
the
voltages
and
currents.
In
anticipation
of
using
Ohm’s
law,
we
will
label
the
voltage
and
Let’slabel
labelthe
the resistor
resistorvoltages
voltagesand
andcurrents.
currents.In
Inanticipation
anticipationof
ofusing
usingOhm’s
Ohm’slaw,
law,we
wewill
willlabel
labelthe
thevoltage
voltageand
and
Let’s
riables,
yaeach
sea resistor
la
corriente
o el voltaje
del
resistor,
y etiquete
la dirección
de
referencia
como
desee.
Etiquete
la
current
of
each
resistor
to
adhere
to
the
passive
convention.
(Pick
one
of
the
variables—the
resistor
current
or
the
current
of
resistor
to
adhere
to
the
passive
convention.
(Pick
one
of
the
variables—the
resistor
current
or
the
current
of
each
resistor
to
adhere
to
the
passive
convention.
(Pick
one
of
the
variables—the
resistor
current
or
the
current
ofvoltage—and
each
resistor label
todeadhere
tovariable
the passive
convention.
(Pick
one
ofLabel
the variables—the
resistor
current
orother
the
dirección
de
referencia
la
otra
para
que
se
apegue
a
la
convención
pasiva
con
la
primera
variable.)
resistor
the
reference
direction
however
you
like.
the
reference
direction
of
the
resistor
voltage—and
label
the
reference
direction
however
you
like.
Label
the
reference
direction
of
the
other
resistorvoltage—and
voltage—andlabel
labelthe
thereference
referencedirection
directionhowever
howeveryou
youlike.
like.Label
Labelthe
thereference
referencedirection
directionof
ofthe
theother
other
resistor
La
figurato
3.7-2
muestra
elpassive
circuito
etiquetado.with
variable
to
adhere
to
the
passive
convention
with
the
first
variable.)
Figure
3.7-2
shows
the
labeled
circuit.
variable
adhere
to
the
convention
the
first
variable.)
Figure
3.7-2
shows
the
labeled
circuit.
variable
to
adhere
to
the
passive
convention
with
the
first
variable.)
Figure
3.7-2
shows
the
labeled
circuit.
variable
to adhere
touse
theKirchhoff’s
passive convention
the KCL
first variable.)
Figure
3.7-2
shows
the
circuit.
A
continuación,
leyes
dewith
Kirchhoff.
aplicamos
la KCL
al nodo
allabeled
cualand
están
Next,
we
will
laws.
First,
apply
to
the
node
at
which
the
current
source
and
the
40-V,
Next,
we
will
use
Kirchhoff’s
laws.
First,
apply
KCL
to
the
node
at
which
the
current
source
the
40-V,
Next,
wewill
willuse
useaplicaremos
Kirchhoff’slas
laws.
First,
applyKCL
KCLPrimero
tothe
thenode
node
atwhich
which
thecurrent
current
source
and
theconec40-V,
Next,
we
Kirchhoff’s
laws.
First,
apply
to
at
the
source
and
the
40-V,
tados
la
fuente
de
corriente
y
los
resistores
de
40-V,
48-V
y
80-V
entre
sí,
para
escribir
48-V,
and
80-V
resistors
are
connected
together
to
write
48-V,
and
80-V
resistors
are
connected
together
to
write
48-V,
and
80-V
resistors
are
connected
together
to
write
48-V, and 80-V resistors are connected together to write
(3.7-1)
(3.7-1)
þii55¼
¼ 0:5
0:5
þ
þ
(3.7-1)
ii22þ
0:5þ
þii4ii4444
(3.7-1)
(3.7-1)
i2i22þþi5i55 ¼¼0:5
Luego
aplicamos
al nodo
en el que
resistores
de 48-V
y 32-V
están
conectados
entre
sí
para escribir
Next,
apply
KCL
toKCL
the
node
at which
which
the
48-V
and
32-V
resistors
are
connected
together
to write
write
Next,
apply
KCL
to
the
node
at
the
48-V
and
32-V
resistors
are
connected
together
to
Next, apply
apply
KCLla
the node
node
which
thelos
48-V
and 32-V
32-V
resistors
are connected
connected
together
write
Next,
KCL
toto the
atat which
the
48-V
and
resistors
are
together
toto write
(3.7-2)
(3.7-2)
¼ ii66
(3.7-2)
ii55¼
(3.7-2)
(3.7-2)
ii55 ¼¼ii66
5
6
Aplicamos
latoKVL
al circuito
cerrado
quevoltage
consta de
la fuente
de 40-V
voltajeand
y los
resistores
deto40-V
y 80-V para
Apply
KVL
the
loop
consisting
of
the
source
and
the
80-V
resistors
write
Apply
Apply KVL
KVL to
to the
the loop
loop consisting
consisting of
of the
the voltage
voltage source
source and
and the
the 40-V
40-V and
and 80-V
80-V resistors
resistors to
to write
write
Apply
escribirKVL to the loop consisting of the voltage source and the 40-V and 80-V resistors to write
12¼¼vv22þþvv44
(3.7-3)
12
(3.7-3)
(3.7-3)
(3.7-3)
12¼
¼vv222þþvv444
12
(3.7-3)
Apply
KVL
to
the
loop
consisting
of
the
48-V,
32-V,
and
80-V
resistors
to
write
Apply
KVL
to
the
loop
consisting
of
the
48-V,
32-V,
and
80-V
resistors
to
write
Aplicamos
KVL
al
circuito
cerrado
que
consta
de
los and
resistores
de
48-V, to
32-V,
y 80-V para escribir
Apply KVL
KVLlato
to
the loop
loop
consisting
of the
the 48-V,
48-V, 32-V,
32-V,
and 80-V
80-V resistors
resistors
to write
write
Apply
the
consisting
of
v
þ
v
þ
v
¼
0
(3.7-4)
v 44þ v 55þ v 66¼ 0
(3.7-4)
(3.7-4)
vv44þþvv55þþvv66 ¼¼00
(3.7-4)
4
5
6
Apply Ohm’s
Ohm’s
lawdeto
toOhm
the resistors.
resistors.
Aplicamos
la ley
a los resistores.
Apply
law
the
Apply
Ohm’s
law
the
resistors.
Apply
Ohm’s
law
totothe
resistors.
(3.7-5)
¼ 40
40
¼
80
¼
48
¼
32
(3.7-5)
¼
80
¼
48
¼
32
(3.7-5)
vv22¼
40ii2ii2222;;;; vvvv44444¼
¼80
80ii4ii4444;;;; vvvv55555¼
¼48
48ii5ii5555;;;; vvvv66666¼
¼32
32ii6ii6666
(3.7-5)
(3.7-5)
vv222¼¼40
Podemos
utilizar
las ecuaciones
de la
de Ohm
para
eliminar
las variables
que voltages.
representen
a losso
Wecan
canuse
use
theOhm’s
Ohm’s
lawequations
equations
toley
eliminate
the
variables
representing
resistor
voltages.Doing
Doing
sovoltajes
enablesdel
us
We
the
law
to
eliminate
the
variables
representing
resistor
enables
us
Wecan
canuse
usethe
theOhm’s
Ohm’slaw
lawequations
equationsto
toeliminate
eliminatethe
thevariables
variablesrepresenting
representingresistor
resistorvoltages.
voltages.Doing
Doingso
soenables
enablesus
us
We
resistor.
Hacerlo
nos
permite
reescribir
la
ecuación
3.7-3
como
to
rewrite
Eq.
3.7-3
as:
to rewrite
Eq.
3.7-3
as:
rewrite Eq.
Eq. 3.7-3
3.7-3 as:
as:
toto rewrite
þ80
80iii444
(3.7-6)
12¼
¼40
40iii222þ
(3.7-6)
(3.7-6)
12
80
(3.7-6)
12
40
i44
(3.7-6)
12
¼¼40
i22þþ80
Similarly,we
wecan
canrewrite
rewriteEq.
Eq.3.7-4
3.7-4as
as
Similarly,
Del
mismowe
modo,
podemos
reescribir
Similarly,
we
can rewrite
rewrite Eq.
Eq.
3.7-4 as
asla ecuación 3.7-4 como
Similarly,
can
3.7-4
80ii4i44þ
þ48
48ii5i55þ
þ32
32ii6i66¼
¼00
(3.7-7)
80
(3.7-7)
(3.7-7)
80
48
32
(3.7-7)
80
i44þþ48
i55þþ32
i66 ¼¼00
(3.7-7)
fromEq.
Eq.
3.7-6
asfollows
follows
Next,
useEq.
Eq.3.7-2
3.7-2
eliminate
3.7-6
as
Next,
use
totoeliminate
i i66from
A
continuación,
usamos
la ecuación
3.7-2
para3.7-6
eliminar
i6 de la ecuación 3.7-6 como sigue
from
Eq.
3.7-6
as follows
follows
Next,
use Eq.
Eq. 3.7-2
3.7-2
eliminate
Eq.
as
Next,
use
toto eliminate
i6i66 from
þ48
48ii5i55þ
þ32
32ii5i55¼
¼00 )
) 80
80ii4i44þ
þ80
80ii5i55¼
¼00 )
) ii4i44¼
¼�i
�i555
(3.7-8)
80ii4i44þ
(3.7-8)
80
(3.7-8)
48
32
)
80
80
)
�i
80
(3.7-8)
i55þþ32
i55 ¼¼00 )
80
i44þþ80
i55 ¼¼00 )
i44 ¼¼�i
(3.7-8)
80
i44þþ48
55
from
Eq.3.7-1.
3.7-1.
UseEq.
Eq.3.7-8
3.7-8
eliminate
Usamos
la
ecuación
3.7-8 para
eliminar
i3.7-1.
Eq.
Use
totoeliminate
i i55from
5 de la ecuación 3.7-1.
from
Eq. 3.7-1.
Use Eq.
Eq. 3.7-8
3.7-8
eliminate
Eq.
Use
toto eliminate
i5i55 from
0:5þ
þii44 )
) iii222¼
¼0:5
0:5þ
þ22ii44
(3.7-9)
(3.7-9)
ii22��ii44¼¼0:5
(3.7-9)
0:5
)
0:5
(3.7-9)
i2i22��i4i44 ¼¼0:5
þþi4i44 )
i22 ¼¼0:5
þþ22i4i44
(3.7-9)
UseEq.
Eq.3.7-9
3.7-9
eliminate
from
Eq.3.7-6.
3.7-6.
Solve
theresulting
resulting
equationto
todetermine
determinethe
thevalue
valueof
of i22. .
Usamos
la
ecuación
3.7-9 para
eliminar
i3.7-6.
ecuación
3.7-6.
Despejamos
la
ecuación
Use
totoeliminate
i i44from
Eq.
the
equation
4 de laSolve
Use Eq.
Eq. 3.7-9
3.7-9
eliminate
from
Eq. 3.7-6.
Solve
the resulting
resulting
equation to
to determine
determineresultante
the value
valuepara
of ii2i2determi2. .
Use
toto eliminate
i4i44 from
Eq.
Solve
the
equation
the
of
�
�
nar el valor de i2.
�� i � 0:5��
12
þ
20
� ii222�
0:5�
12
þ
20
12þ
þ20
20 ¼ 0:4 A
2 � 0:5 ¼
12
¼
40
¼
80
þ
80
�
20
)
¼
(3.7-10)
12
¼
40
80
þ
80
�
20
)
¼
¼ 0:4
AA
(3.7-10)
12¼
¼40
40iii2i2222þ
¼80
80ii2ii2222�
þ80
80 i2 �220:5 ¼
�20
20 )
) ii2ii2222¼
¼12 80
0:4A
(3.7-10)
80
12
(3.7-10)
¼¼0:4
(3.7-10)
2
80
2
80
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 79
Alfaomega
4/12/11 5:22 PM
80
8080
80
Resistive Circuits
Resistive
ResistiveCircuits
Circuits
Resistive
Circuitos Circuits
resistivos
Now we are ready to calculate the values of the rest of the resistor voltages and currents as follows:
Now
Now
we
weare
areready
readytotocalculate
calculatelos
the
thevalues
valuesofofthe
therest
rest
ofthe
the
resistor
resistor
voltages
voltagesand
andlos
currents
currents
asasfollows:
follows:
Ya
estamos
calcular
resto
deof
los
voltajes
y corrientes
de
resistores
como sigue
Now
we arelistos
readypara
to calculate
thevalores
values del
of the
rest
of
the
resistor
voltages and
currents
as follows:
i2 � 0:5 0:4 � 0:5
0:5¼ 0:4
0:4��0:5
0:5¼ �0:05 A;
i4 ¼ i2i2��0:5
i4i4¼¼i2 �2 0:5 ¼¼0:4 �
�0:05A;
A;
2 0:5 ¼¼�0:05
i4 ¼ 22 ¼
¼ �0:05 A;
2
2
i62¼ i5 ¼ �i42¼ 0:05 A;
�i4 4¼¼0:05
0:05A;
A;
i6i6¼¼i5i5¼¼�i
¼ ii52 ¼
¼ Þ0:05
A; V;
i6 40
¼ �i
404ð0:4
¼ 1:6
v2 ¼
40i2i2¼¼40
40ð0:4
ð0:4Þ Þ¼¼1:6
1:6V;
V;
v2v2¼¼40
ð0:4Þ Þ¼¼1:6
v4v2¼¼8040i4 i2¼¼8040
ð�0:05
�4V;V;
v4v4¼¼80
80i4i4¼¼80
80ð�0:05
ð�0:05Þ Þ¼¼�4
�4V;
V;
vv45¼¼80
�4V;
V;
48i4i5¼¼80
48ðð�0:05
0:05Þ Þ¼¼2:4
48i i ¼¼48
48ð0:05
ð0:05Þ Þ¼¼2:4
2:4V;
V;
v5v5¼¼48
48 i5 5 ¼ 48ð0:05Þ ¼ 2:4 V;
v5v¼
and
6 ¼ 325 i6 ¼ 32ð0:05Þ ¼ 1:6 V:
and
and
v6v6¼¼32
32i i ¼¼32
32ð0:05
ð0:05Þ Þ¼¼1:6
1:6V:
V:
yand
v6 ¼ 32 i666 ¼ 32ð0:05Þ ¼ 1:6 V:
Solución 1Solution
con MATLAB
MATLAB
1
MATLAB
MATLAB
Solution
Solutionmuestra
11
El
álgebra
anterior
quethat
estethis
circuito
secan
puede
MATLAB
Solution
1 shows
The
preceding
algebra
circuit
be representar
representedpor
by estas
theseecuaciones:
equations:
The
Thepreceding
precedingalgebra
algebrashows
showsthat
thatthis
thiscircuit
circuitcan
canbeberepresented
representedby
bythese
theseequations:
equations:
The preceding algebra shows that this circuit can be represented by these equations:
i2 � 0:5
0:5; i6 ¼ i5 ¼ �i4 ; v2 ¼ 40 i2 ; v4 ¼ 80 i4 ;
12 ¼ 80 i2 � 20; i4 ¼ i2i2��0:5
20;i i ¼¼i2 �2 0:5; i; i ¼¼i i ¼¼�i
�i; v; v ¼¼40
40i i; ;vv ¼¼80
80i i; ;
12
12¼¼80
80i i ��20;
12 ¼ 80 i222� 20; i444 ¼ 22 ; i666 ¼ i555 ¼ �i444; v222 ¼ 40 i222; v444 ¼ 80 i444;
v5 ¼2 48 i5 ; and v6 ¼ 32 i6
v5v5¼¼48
48i i; ;and
and
32i i
y vv ¼¼32
v5 ¼ 48 i555; and v666 ¼ 32 i666
These equations can be solved consecutively, using MATLAB as shown in Figure 3.7-3.
These
Theseequations
equationscan
canpueden
bebesolved
solved
consecutively,
consecutively,
usingMATLAB
MATLAB
asasshown
shown
ininFigure
Figure
3.7-3.
3.7-3.
Estas
se
despejar
de manerausing
consecutiva
utilizando
MATLAB
como
muestra la figura 3.7-3.
Theseecuaciones
equations can
be solved
consecutively,
using
MATLAB
as shown
in Figure
3.7-3.
FIGURE 3.7-3 Consecutive equations.
FIGURE
FIGURE3.7-3
3.7-3Consecutive
Consecutiveequations.
equations.
FIGURE
3.7-3 Consecutive
equations.
FIGURA
3.7-3 Ecuaciones
consecutivas.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 80
FIGURE 3.7-4 Simultaneous equations.
FIGURE
FIGURE3.7-4
3.7-4Simultaneous
Simultaneousequations.
equations.
FIGURE
3.7-4 Simultaneous
equations.
FIGURA
3.7-4 Ecuaciones
simultáneas.
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Analyzing Resistive Circuits Using MATLAB
Analyzing
Resistive
Circuits
Using
MATLAB
Analyzing
AnalyzingResistive
ResistiveCircuits
CircuitsUsing
UsingMATLAB
MATLAB
Analyzing
Resistive
Circuits
Using
MATLAB
Análisis de circuitos resistivos utilizando MATLAB
81
81
81
81
81
81
MATLAB Solution 2
MATLAB
Solution
MATLAB
MATLABSolution
Solution2222
MATLAB
Solución
2Solution
con
MATLAB
We
can avoid
some
algebra if we are willing to solve simultaneous equations.
We
can
avoid
some
algebra
we
are
willing
to
solve
simultaneous
equations.
We
Wecan
canavoid
avoid
some
some
algebra
algebra
weare
arewilling
willing
to
tosolve
solvesimultaneous
simultaneouslas
equations.
equations.
We
can
avoid
some
algebra
ifififwe
we
are
willing
to
solve
simultaneous
equations.
Podemos
pasar
por
alto
algo if
de
álgebra
si nos
inclinamos
ecuaciones
simultáneas.
After
applying
Kirchhoff’s
laws
and
then
using
thea despejar
Ohm’s law
equations to
eliminate the variables
After
applying
Kirchhoff’s
laws
and
then
using
the
Ohm’s
law
equations
to
eliminate
the
variables
After
After
applying
applying
Kirchhoff’s
Kirchhoff’s
laws
laws
and
and
then
then
using
using
the
the
Ohm’s
Ohm’s
law
lawequations
equations
to
eliminate
eliminate
thevariables
variables
After
applying
Kirchhoff’s
laws
and
then
using
the
Ohm’s
law
equations
to
eliminate
the
variables
Después
de aplicar
las leyes
de Kirchhoff
de utilizar
las ecuaciones
de lato
ley
de Ohm the
para
eliminar
representing
resistor
voltages,
we have
Eqs 3.7-1,y luego
2, 6, and
7:
representing
resistor
voltages,
we
have
Eqs
3.7-1,
2,
6,
and
7:
representing
representing
resistor
resistor
voltages,
voltages,
we
we
have
have
Eqs
Eqs
3.7-1,
3.7-1,
2,
2,
6,
6,
and
and
7:
7:
representing
resistor
voltages,
we
have
Eqs
3.7-1,
2,
6,
and
7:
las variables que representan los voltajes de los resistores, tenemos las ecuaciones 3.7-1, 2, 6 y 7:
i2 þ i5 ¼ 0:5 þ i4 ; i5 ¼ i6 ; 12 ¼ 40 i2 þ 80 i4 ;
þ
¼
0:5
þ
¼
12
¼
40
þ
80
þiii55i55¼
¼0:5
0:5þ
þiii44i4;4;;; iii55i55¼
¼iii66i6;6;;12
;12
12¼
¼40
40iii22i22þ
þ80
80iii44i4;4;;;
iii22i22þ
þ
¼
0:5
þ
¼
¼
40
þ
80
y
and
80 i4 þ 48 i5 þ 32 i6 ¼ 0
and
80
þ
48
þ
32
¼
and
and
80
80iii44i44þ
þ48
48iii55i55þ
þ32
32iii66i66¼
¼0000
and
80
þ
48
þ
32
¼
This set of four simultaneous equations in i2, i4, i5, and i6 can be written as a single matrix equation.
This
set
of
four
simultaneous
equations
in
can
be
written
as
single
matrix
equation.
Este
de cuatro
ecuaciones
simultáneas
,and
i53e2
se
escribir
como
una matrix
ecuación
de matriz única.
This
Thisconjunto
set
setof
offour
four
simultaneous
simultaneous
equations
equations
in
iniii22i2,2,,en
,iii44i4,4,,i,i2ii55,i5,5,i,4,and
and
iii6i6i666can
can
be
be
written
as
aaasingle
single
matrixequation.
equation.
2
3puede
2written
3 as
This
set
of
four
simultaneous
equations
in
and
can
be
written
2
33
332
22 3
33as a single matrix equation.
2
2
12 �1 1
0 3
i222 3
0:5
2
2
3
3
2
3
�1
i22i22 6 0:5
0:5
�1 1111 0007
0 6 ii7
0:5
�1
6 0 1111 0�1
7
i6467
00:5
660000 0000 1 1111 �1
7
7
6
77
6
66
7
6
6
7
7
7
7
6
6
i
�1
07
6
6
7
7
6
7
¼
(3.7-11)
i
i
�1
�1
0
07
4
�1
0
44 47
656i5
66 80 0
77¼
776
6
77
46
54
4
5
7
6
6
7
¼
(3.7-11)
6
6
7
7
6
7
i
40
0
12
¼
(3.7-11)
(3.7-11)
¼
(3.7-11)
4
4
5
5
4
55
4
4
4
5
5
5
5
4
4
5
i
40
80
0
0
12
4
4440
5
5
4
5
i
i
40
80
80
0
0
0
0
12
12
5
i
0 40 8080 48 0 32 0 i6 555
0 12
80
48
32
0000 80
80 48
48 32
32 iii66i66
0000
80
48
32
We can write this equation as
We
can
write
this
equation
as
We
Wecan
canwrite
write
this
this
equation
equation
as
as como
We
can
write
this
equation
as
Podemos
escribir
esta
ecuación
Ai ¼ B
(3.7-12)
Ai
¼
BB
(3.7-12)
Ai
Ai¼
¼BB
(3.7-12)
(3.7-12)
Ai
5
B
Ai
¼
(3.7-12)
where
donde
where
where
where
2 3
2
3
2
3
where
33
22 3
33
2
33
2
2
2
0:5
12 �1 1
0 3
i222 3
2
3
2
3
2
3
0:5
1
�1
1
0
i
0:5
0:5
1
�1
1
0
1
�1
1
0
i
i
2
22 2
6 i4 i7
6 00:5
7
6 0 1 0�1 1 1 �107
66ii7
77 B ¼ 6 6
66 0007
77
660000 0000 1111 �1
77
7;7
6
6
i44i447
�1
6
7
6
7
6
7
and
A¼6
i¼6
07
�1
�1
6565
77and
6
77
66 80 0
77
4
4
4
57
6
7
6 5
6
7
and
B
¼
¼
;
i
¼
6
7
6
7
6
7
i
12
40
0
y
B
¼
and
B
¼
AAAA¼
¼6
;
i
¼
;
i
¼
and
B
¼
¼
;
i
¼
4
5
4
55
4
5
4
4
4
12
40
80
44iii55i555
55
4412
5
4440
55
125
40 80
80 0000 0000 5
12
40
80
i6
0
0 80 48 32
80
48
32
iii66i66
0000
0000 80
80 48
48 32
32
80
48
32
This matrix equation can be solved using MATLAB as shown in Figure 3.7-4. After entering matrices A and B,
This
matrix
equation
can
be
solved
using
MATLAB
as
shown
in
Figure
3.7-4.
After
entering
matrices
AAand
and
B,
Thisecuación
matrixequation
equation
can
be
solved
usingMATLAB
MATLABas
asshown
shownin
in
Figure
3.7-4.After
After
entering
matrices
andB,
B,
This
matrix
can
solved
using
Figure
A
This
matrix
equation
can
be
solved
using
MATLAB
as
shown
in
Figure
3.7-4.
After
entering
matrices
A
and
B,
de matriz
sebe
puede
despejar
utilizando
MATLAB
como
se3.7-4.
muestra
en entering
la
figura matrices
3.7-4.
Después
de
theEsta
statement
the
statement
thestatement
statement
the
the
statement
introducir
las matrices A y B, la expresión
i ¼ A=B
¼
A=B
¼A=B
A=B
iiii¼
¼
A=B
tells MATLAB to calculate i by solving Eq 3.7-12.
tells
MATLAB
to
calculate
by
solving
Eq
3.7-12.
tellsMATLAB
MATLAB
tocalculate
calculate
solvingEq
Eq
3.7-12.
tells
to
iiiiby
solving
tells
MATLAB
to
calculate
by
solving
Eq
3.7-12.
indica
a MATLAB
que
calcule
iby
despejando
la3.7-12.
ecuación 3.7-12.
A Un
circuit
consisting
n elementstiene
has n ncurrents
and nyvoltages.
A set
of equationsde
representing
that
circuito
con nofof
elementos
corrientes
voltajes.
Un
ecuaciones
que
AAcircuit
circuit
consisting
of
elements
has
currents
and
voltages.
AA
set
of
equations
representing
that
A
circuit
consisting
consisting
ofnnnelements
elementshas
hasnnnncurrents
currentsand
andnnnnnvoltages.
voltages.
A
set
setconjunto
of
ofequations
equationsrepresenting
representingthat
that
A
circuit
consisting
of
elements
has
currents
and
voltages.
A
set
equations
representing
that
circuit
could
have
as manypodría
as n2n
unknowns.
We
can
reduce
the
number
ofof
unknowns
by
labeling
the
representen
ese
circuito
tener
incógnitas
de
hasta
2
enésimas.
Podemos
reducir
la
cantidad
de
circuit
could
have
as
many
as
2n
unknowns.
We
can
reduce
the
number
of
unknowns
by
labeling
the
circuit
circuitcould
couldhave
haveas
asmany
manyas
as2n
2nunknowns.
unknowns.We
Wecan
canreduce
reducethe
thenumber
numberof
ofunknowns
unknownsby
bylabeling
labelingthe
the
circuit
could
have
as
many
as
2n
unknowns.
We
can
reduce
the
number
of
unknowns
by
labeling
the
currents
and voltages
carefully.
For
example,
suppose
two of ythe
circuit
elements
are connected
in series.
incógnitas
si
etiquetamos
con
gran
cuidado
las
corrientes
los
voltajes.
Por
ejemplo,
suponga
que
dos
currents
and
voltages
carefully.
For
example,
suppose
two
of
the
circuit
elements
are
connected
in
series.
currents
currentsand
andvoltages
voltagescarefully.
carefully.For
Forexample,
example,suppose
supposetwo
twoof
ofthe
thecircuit
circuitelements
elementsare
areconnected
connectedin
inseries.
series.
currents
and
voltages
carefully.
For
example,
suppose
two
of
the
circuit
elements
are
connected
in
series.
We
can
choose
the reference
directions
for the currents
in those
elements
that they are
equal
and use
de
los
elementos
del
circuito
están conectados
en serie.
las so
direcciones
deare
referencia
para
We
can
choose
the
reference
directions
for
the
currents
in
those
elements
so
that
they
are
equal
and
use
We
We
can
can
choose
choosethe
the
reference
reference
directions
directions
for
forthe
thecurrents
currents
in
inElegimos
those
thoseelements
elements
so
sothat
thatthey
they
are
equal
equaland
and
use
use
We
can
choose
the
reference
directions
for
the
currents
in
those
elements
so
that
they
are
and
one
variable
to represent
both
currents.
Table
3.7-1
presents
some
guidelines
that
will
help
usequal
reduce
theuse
las
corrientes
en
esos
elementos
de
modo
que
sean
iguales
y
utilicen
una
variable
para
representar
one
variable
to
represent
both
currents.
Table
3.7-1
presents
some
guidelines
that
will
help
us
reduce
the
one
onevariable
variableto
torepresent
representboth
bothcurrents.
currents.Table
Table3.7-1
3.7-1presents
presentssome
someguidelines
guidelinesthat
thatwill
willhelp
helpus
usreduce
reducethe
the
one
variable
to
represent
both
currents.
Table
3.7-1
presents
some
guidelines
that
will
help
us
reduce
the
number
of
unknownsLaintabla
the set
of presenta
equationsalgunos
describing
a given circuit.
ambas
corrientes.
3.7-1
lineamientos
nos ayudarán a reducir el número
number
of
unknowns
in
the
set
of
equations
describing
given
circuit.
number
number
of
ofunknowns
unknownsin
inthe
the
set
setof
ofequations
equationsdescribing
describing
aaaagiven
givenque
circuit.
circuit.
number
of
unknowns
in
the
set
of
equations
describing
given
circuit.
de incógnitas en el conjunto de ecuaciones que describen un circuito dado.
Table 3.7-1 Guidelines for Labeling Circuit Variables
Table
3.7-1
Guidelines
for
Labeling
Circuit
Variables
Table
Table3.7-1
3.7-1 Lineamientos
Guidelines
Guidelinesfor
forpara
Labeling
Labeling
Circuit
CircuitVariables
Variables
Table
3.7-1
Guidelines
for
Labeling
Circuit
Variables
Tabla
3.7-1
el etiquetado
de variables de circuito
CIRCUIT FEATURE
CIRCUIT
FEATURE
CIRCUIT
CIRCUITFEATURE
FEATURE
CARACTERÍSTICA
CIRCUIT
FEATURE
Resistors
DEL CIRCUITO
Resistors
Resistors
Resistors
Resistors
Resistores
Series elements
Series
elements
Series
Serieselements
elements
Series
elements
Elementos en serie
Parallel elements
Parallel
elements
Parallel
Parallelelements
elements
Parallel
elements
Elementos en paralelo
Ideal Voltmeter
Ideal
Voltmeter
Ideal
IdealVoltmeter
Voltmeter
Ideal
Voltmeter
Voltímetro ideal
Ideal Ammeter
Ideal
Ammeter
Ideal
IdealAmmeter
Ammeter
Ideal
Ammeter
Amperímetro ideal
GUIDELINE
GUIDELINE
GUIDELINE
GUIDELINE
GUIDELINE
Label
the voltage and current of each resistor to adhere to the passive convention. Use
LINEAMIENTO
Label
the
voltage
and
current
of
each
resistor
to
adhere
to
the
passive
convention.
Use
Label
Label
the
the
voltage
voltage
and
andeither
current
current
of
of
each
eachresistor
resistor
to
toadhere
adhere
to
tothe
thepassive
passiveconvention.
convention.Use
Use
Label
the
voltage
and
current
of
each
resistor
to
adhere
to
the
passive
convention.
Use
Ohm’s
law
to
eliminate
the
current
or voltage
variable.
Etiquete
el
voltaje
y la corriente
de
cada
resistor
para variable.
que
se apeguen a la convención pasiva.
Ohm’s
law
to
eliminate
either
the
current
or
voltage
variable.
Ohm’s
Ohm’slaw
law
to
toeliminate
eliminate
either
eitherthe
thecurrent
current
or
orvoltage
voltage
variable.
Ohm’s
law
to
eliminate
either
the
current
or
voltage
variable.
Label
reference
for series
elements
so de
thatlatheir
currents
equal. Use one
Use
lathe
ley
dereference
Ohm directions
para
eliminar
la variable
ya sea
corriente
o elare
voltaje.
Label
the
directions
for
series
elements
so
that
their
currents
are
equal.
Use
one
Label
Labelthe
the
reference
reference
directions
directions
forseries
series
serieselements.
elements
elementsso
sothat
thattheir
theircurrents
currentsare
areequal.
equal.Use
Useone
one
Label
the
reference
directions
for
series
elements
so
that
their
currents
are
equal.
Use
one
variable
to
represent
the
currentsfor
of
variable
to
represent
the
currents
of
series
elements.
Etiquete
las
direcciones
de
referencia
para
elementos
en
serie
de
modo
que
sus
corrientes
variable
variable
to
to
represent
represent
the
the
currents
currents
of
of
series
series
elements.
elements.
variable to represent the currents of series elements.
Label
the reference
directions
for parallel
elements so
that
their voltages
are equal. en
Useserie.
one
sean
iguales.
Utilice
una
variable
para
representar
lasso
corrientes
de
los elementos
Label
the
reference
directions
for
parallel
elements
so
that
their
voltages
are
equal.
Use
one
Label
Label
the
the
reference
reference
directions
directions
for
parallel
parallel
elements
elements
sothat
thattheir
their
voltages
voltages
are
areequal.
equal.Use
Useone
one
Label
the
reference
directions
for
parallel
elements
so
that
their
voltages
are
equal.
Use
one
variable
to
represent
the
voltagesfor
of
parallel
elements.
variable
to
represent
the
voltages
of
parallel
elements.
Etiquete
las
devoltages
referencia
para
elementos
en paralelo de modo que sus voltajes
variable
variable
to
todirecciones
represent
representthe
the
voltages
of
ofparallel
parallel
elements.
elements.
variable
to
represent
the
voltages
of
parallel
elements.
Replace
each (ideal)
voltmeter
by an
open
circuit. Label
the voltage
the openencircuit
sean
iguales.
Utilice
una
variable
para
representar
las
corrientes
de across
losacross
elementos
paralelo.
Replace
each
(ideal)
voltmeter
by
an
open
circuit.
Label
the
voltage
across
the
open
circuit
Replace
(ideal)
voltmeter
voltmeter
by
byan
anopen
opencircuit.
circuit.
Label
Label
the
thevoltage
voltage
across
the
theopen
open
circuit
circuit
Replace
each
(ideal)
voltmeter
by
an
open
circuit.
Label
the
voltage
across
the
open
circuit
to Replace
be
equaleach
toeach
the(ideal)
voltmeter
voltage.
to
be
equal
to
the
voltmeter
voltage.
to
beequal
equal
to
the
voltmeter
voltmeter
voltage.
voltage.
Reemplace
cada
voltímetro
(ideal)
por un circuito abierto. Etiquete el voltaje a través
totobe
be
equal
totothe
the
voltmeter
voltage.
Replace
eachabierto
(ideal) para
ammeter
by aigual
shortalcircuit.
Label
the current in the short circuit to be
del
circuito
que sea
voltaje
del
voltímetro.
Replace
each
(ideal)
ammeter
by
short
circuit.
Label
the
current
in
the
short
circuit
to
be
Replace
Replace
each(ideal)
(ideal)
ammeter
ammeter
by
byaaaashort
short
circuit.
circuit.
Label
Label
the
thecurrent
currentin
inthe
theshort
shortcircuit
circuitto
tobe
be
Replace
each
(ideal)
ammeter
by
short
circuit.
Label
the
current
in
the
short
circuit
to
be
equal
to theeach
ammeter
current.
equal
to
the
ammeter
current.
equal
equalto
tothe
the
ammeter
ammeter
current.
current.
Reemplace
cada
amperímetro
(ideal) por un cortocircuito. Etiquete la corriente en el
equal
to
the
ammeter
current.
cortocircuito para que sea igual a la corriente del amperímetro.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 81
Alfaomega
4/12/11 5:22 PM
82 82
828282
Resistive
Resistive
Circuits
Circuits
Circuitos
resistivos
Resistive
Resistive
Circuits
Circuits
3.8 ¿ C Ó M O L O P O D E M O S C O M P R O B A R . . . ?
3.83.8
3.8 H
HO
O
HW
W
OW
W
CA
A
CN
N
AN
N
WW
W
EE
E
CH
H
CE
E
HC
C
EK
K
CK
K.. .... .... ??.. ??
3.8
H
O
C
C
A
W
E
C
C
H
E
C
AEngineers
los ingenieros
se les suele
solicitar
comprobar
que
la
solución
un
problema
la correct.
correcta.
Por
Engineers
areare
are
frequently
frequently
called
called
upon
upon
to
to
check
check
that
that
solution
solution
to de
to
problem
problem
is indeed
indeed
issea
indeed
correct.
For
For
Engineers
Engineers
are
frequently
frequently
called
called
upon
upon
to
to
check
check
that
that
aadiseño
solution
aa solution
to
to
aa problem
aa problem
is
is
indeed
correct.
correct.
For
For
ejemplo,
las
soluciones
propuestas
para
problemas
de
se
deben
comprobar
para
confirmar
que
se
example,
example,
proposed
proposed
solutions
solutions
to
to
design
design
problems
problems
must
must
be
be
checked
checked
to
to
confirm
confirm
that
that
all
all
of
of
the
the
example,
proposed
proposed
solutions
solutions
to to
design
design
problems
problems
must
be be
checked
checked
toresultados
to
confirm
confirm
that
that
all all
of of
thethe
haexample,
cumplido
con
todas
lassatisfied.
especificaciones.
Además,
semust
deben
revisar
los
de
la
computadora
specifications
specifications
have
have
been
been
satisfied.
In
In
addition,
addition,
computer
computer
output
output
must
must
be
be
reviewed
reviewed
to
to
guard
guard
against
against
specifications
specifications
have
have
been
been
satisfied.
satisfied.
In
In
addition,
addition,
computer
computer
output
output
must
be be
reviewed
reviewed
to
guard
guard
against
against
para
protegerse
contra
errores
de
captura
de datos,
así
como
lasmust
exigencias
de los to
comerciantes,
las
data-entry
data-entry
errors,
errors,
andand
and
claims
claims
made
made
by by
by
vendors
vendors
must
must
be be
be
examined
examined
critically.
critically.
data-entry
data-entry
errors,
errors,
and
claims
claims
made
made
by
vendors
vendors
must
must
be
examined
examined
critically.
critically.
cuales
se
deben
analizar
a
fondo.
Engineering
Engineering
students
students
areare
are
also
also
asked
asked
to to
to
check
check
thethe
the
correctness
correctness
of of
of
their
their
work.
work.
ForFor
For
example,
example,
Engineering
Engineering
students
students
are
also
asked
asked
to
check
check
the
correctness
correctness
of
their
their
work.
work.
For
example,
example,
También
aa los
estudiantes
dealso
ingeniería
se
les
pide
que
verifiquen
la
exactitud
de
sus
trabajos.
occasionally
occasionally
just
just
little
a
little
time
time
remains
remains
at
the
at
the
end
end
of
of
an
an
exam.
exam.
It
is
It
useful
is
useful
to
be
to
be
able
able
to
quickly
to
quickly
identify
identify
occasionally
occasionally
just
just
a little
a little
time
time
remains
remains
atantes
the
at the
end
of
an
of an
exam.
exam.
isIt useful
is useful
to be
to be
able
able
quickly
to quickly
identify
Por
ejemplo,
tomarse
un
breve
lapso
deend
terminar
un Itexamen
permitiría
dartouna
vistaidentify
rápida
e
those
those
solutions
solutions
that
that
need
need
more
more
work.
work.
those
those
solutions
solutions
that
that
need
need
more
more
work.
work. requerir un poco más de aplicación.
identificar
esas
soluciones
que
podrían
The
The
following
following
example
example
illustrates
illustrates
techniques
techniques
useful
useful
forfor
for
checking
checking
thethe
the
solutions
solutions
of of
of
thethe
the
sortsort
sort
of of
of
The
following
following
example
example
illustrates
illustrates
techniques
techniques
useful
useful
for
checking
checking
the
solutions
solutions
of
the
sort
of
ElThe
ejemplo
siguiente
ilustra
técnicas
útiles para
comprobar
las soluciones
a los diversos
probleproblem
problem
discussed
discussed
in
in
this
this
chapter.
chapter.
problem
problem
discussed
discussed
in
thiscapítulo.
this
chapter.
chapter.
mas
analizados
enineste
EmXXE
E
A
Xo
M
AP
MLPEL 3
E .38.¿Cómo
811-- 1
1How
How
How
Can
Can
We
We
Check
Check
Voltage
Voltage
and
and
Current
Current
Values?
E j eE
pA
lX
3.8-1
podemos
comprobar
los
valores
del
voltaje
yValues?
la
corriente?
MA P
MLPEL 3
E .3
8.--8
How
Can
Can
We
We
Check
Check
Voltage
Voltage
and
and
Current
Current
Values?
Values?
The
The
circuit
circuit
shown
shown
inmuestra
in
Figure
Figure
3.8-1a
3.8-1a
was
was
analyzed
analyzed
by
by
writing
writing
andand
and
solving
solving
set
set
of of
of
simultaneous
simultaneous
equations:
equations:
ElThe
circuito
que
sein
en
la
figura
3.8-1a
se by
analizó
escribiendo
y despejando
un
conjuntoequations:
de
ecuaciones
The
circuit
circuit
shown
shown
in
Figure
Figure
3.8-1a
3.8-1a
was
was
analyzed
analyzed
by
writing
writing
and
solving
solving
aa set
aa set
of
simultaneous
simultaneous
equations:
simultáneas:
vv¼22 vvþ22 iþ
vv5 vv¼55 i¼4 þ
12 12
12
¼ v¼
v¼2 þ
vþ2 4i
þ34i
4i
; i34; ¼
v¼5 4i
4i
¼34i
and
5i
þ45i
5i4
3 ; iv35; ¼
3 ; and5 ¼
þ
iþ
ii44 5i
i¼
12
¼
þ
ii44 ¼
;; and
þ
2 v2 4i
3 ; i34 ; ¼
3 ; iv35; v
5 ¼34i
3 ; yand
4þ
4 4
5
5
5 5
22 22
The
The
computer
computer
Mathcad
Mathcad
(Mathcad
(Mathcad
User’s
User’s
Guide,
Guide,
1991)
1991)
was
was
used
used
tosesolve
solve
toempleó
solve
thethe
the
equations
equations
as shown
as
shown
shown
in Figure
Figure
in Figure
Figure
3.83.8Para
resolver
lasMathcad
ecuaciones
que User’s
se
muestran
en1991)
la1991)
figura
3.8-1b
la
computadora
Mathcad
(Mathcad
The
The
computer
computer
Mathcad
(Mathcad
(Mathcad
User’s
Guide,
Guide,
was
was
used
used
to
to
solve
the
equations
equations
as
as
shown
in
in
3.83.81b.
1b.
It
was
It
was
determined
determined
that
that
User’s
Guide,
1991).
Se
determinó
que
1b.1b.
It was
It was
determined
determined
thatthat
�60
¼ �60
V; V;
V;
¼
18
¼ 18
18
A; A;
A;
¼
6¼A;
A;
A;
and
and
v¼5 72
72
¼ 72
V: V:
V:
v¼2 �60
yand
V;
ii33 ¼
ii33 18
¼
A;
ii44 ¼
ii44 6¼
66 A;
and
vv55 v¼
V:
vv22 v¼
2 ¼ �60
5 ¼ 72
¿Cómo
podemos
comprobar
quecurrents
estasand
corrientes
y voltajes
son correctos?
How
can
we
check
that
these
and
voltages
are
correct?
How
cancan
wewe
check
thatthat
these
currents
voltages
areare
correct?
How
How
can
we
check
check
that
these
these
currents
currents
andand
voltages
voltages
are
correct?
correct?
FIGURA
3.8-1
decircuit
ejemplo
(b)(b)
análisis
por
computadora
utilizando
FIGURE
FIGURE
3.8-1
3.8-1
(a)(a)
(a)
AnCircuito
An
example
example
circuit
andyand
and
(b)
computer
computer
analysis
analysis
using
using
Mathcad.
Mathcad.Mathcad.
FIGURE
FIGURE
3.8-1
3.8-1
(a)
(a)
An
An
example
example
circuit
circuit
and
(b)
(b)
computer
computer
analysis
analysis
using
using
Mathcad.
Mathcad.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 82
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
E1C03_1
11/25/2009
83
How
Can
We
Check. .. .. ?. ?
How
Can
We
Check
How
Can
We
Check
...?
How
Can
We
Check
Can
WeWe
Check
. . ... ...?..... ????
How
Can
We
Check
¿Cómo How
lo How
podemos
comprobar
Can
Check
How Can We Check . . . ?
8383
83
83
83 83
83
83
Solution
Solution
Solution
Solución
Solution
Solution
Solution
Thecurrent
currenti2 ican
canbebecalculated
calculatedfrom
fromv2v, 2i,3,i3i,4,i4and
, andv5vin
ina acouple
coupleofofdifferent
differentways.
ways.First,
First,Ohm’s
Ohm’slaw
lawgives
gives
The
The
current
iii2222 se
can
be
calculated
from
v , i , i44y,, and
aa couple
of
different
ways.
First,
gives
La
corriente
puede
calcularfrom
desde
v5venvvvin555unin
de maneras
distintas.
Primero,
laOhm’s
ley law
de law
Ohm
da
The
current
can
be
calculated
from
and
in
couple
of
different
ways.
First,
Ohm’s
law
gives
Solution
The
current
i2 can
be
calculated
v2,vvvv2i22223,,,,,iiii3i3333,4,,,,i4iiiand
a par
couple
of of
different
ways.
First,
Ohm’s
gives
The
current
can
be
calculated
from
and
in
couple
of
different
ways.
First,
Ohm’s
law
gives
5 v5
4,, and
5 in
The
current
ii22 can
be
calculated
from
aa couple
different
ways.
First,
Ohm’s
law
gives
4
The current i2 can be calculated from v2, i3, i4v, vand
v�60
5 in a couple of different ways. First, Ohm’s law gives
2v2 �60
�12
A
22 ¼ �60
�60
v
2 ¼
i2 ii¼
¼
¼¼�12
AA
�60
v
�60
v
�12
522 ¼
5¼ ¼
2 ¼
¼ 5�60
¼
�12
A
¼52 v¼
i2 ¼
�12
A
ii222 ¼
¼
¼
�12
A
5
5
5 v552 ¼5�60
55 ¼ �12 A
i2 ¼
Next,applying
applyingKCL
KCLat atnode
nodeb bgives
gives
Next,
5
5
Next,
applying
KCL
at
node
b gives
Next,
applying
KCL
at
node
gives
A
continuación,
aplicando
lab KCL
en el nodo b resulta
Next,
applying
KCL
at at
node
gives
Next,
applying
KCL
node
bb gives
Next, applying KCL at node b gives i2 i¼
¼i3 iþ
þ
18
þ
¼2424
A
4 ¼
i4 ii¼
1818
þþ
6 66¼¼
AA
ii333 þ
24
ii222 ¼
4 ¼
¼
þ
¼
18
þ
¼
24
A
i3 þ
1818
þ 6þ
2424
AA
i2 ¼
¼
þ
¼
18
þ¼
¼
24
A
4 ¼
ii33 iþ
iii444 ¼
666 ¼
ii22 ¼
¼so
ithe
þ values
i4values
¼ 18calculated
þcalculated
6 ¼ 24 A
2 so
3 the
Clearly,i2 icannot
cannotbebeboth
both�12
�12and
and2424A,iA,
forv2v, 2i,3,i3i,4,i4and
, andv5vcannot
cannotbebecorrect.
correct.
Clearly,
forfor
Clearly,
ii222 cannot
be
both
�12
and
24
A,
so
the
values
calculated
vv22,, ii33,, ii44,, and
v555 cannot
be
correct.
Clearly,
cannot
be
both
�12
and
24
A,
so
the
values
calculated
for
and
cannot
be
correct.
Clearly,
i
cannot
be
both
�12
and
24
A,
so
the
values
calculated
for
vcalculados
i3,,sign
, ii33i4,,error
, ii44and
v5vinvvv2the
cannot
correct.
Clearly,
i
cannot
be
both
�12
and
24
A,
so
the
values
calculated
for
,para
and
cannot
be
correct.
i3por
,4values
i4lo
, and
v, 5we
, valores
wefind
find
avsign
error
Checking
used
tocalculate
calculate
v, 2i,A,
2 the
2,v
2 the
2
Desde
luego,
i2 equations
no puede
ser
nitode
212
ni A,
dev24
quecalculated
los
,55 ithe
, KCL
i4 KCL
ybe
v5be
noequation
pueden
Clearly,
i
cannot
be
both
�12
and
24
so
the
for
,
and
cannot
correct.
,
i
,
and
v
a
in
equation
Checking
equations
used
2
2
3
2
3
5
ii33,, values
i ,, and
vv55,, we
find
aav ,sign
in
KCL
Checking
equations
used
to
calculate
v ,,the
and
we
find
sign
error
in
the
KCL
equation
Checking
the
equations
used
to
calculate
Clearly,
i2 the
cannot
be b.
both
�12
and
24should
A,
so
calculated
, encontramos
i4error
, and
v5 the
cannot
beequation
correct.
,vvv2222iusaron
vcalcular
find
in in
the
KCL
equation
Checking
the
equations
used
to
calculate
v2se
,3, be
i33i,,4, iii4444and
, and
and
v55we
, we
we
find
a2ysign
sign
error
in
the
KCL
equation
Checking
the
equations
used
to
calculate
corresponding
tonode
node
b.This
This
equation
should
5,v
ser
correctos.
Alequations
verificar
las
ecuaciones
que
para
v2find
, for
i3a, isign
vi53,error
unKCL
error
de
signo
,
i
,
,
a
error
the
equation
Checking
the
used
to
calculate
corresponding
to
equation
be
4
corresponding
to
node
b.
This
equation
should
corresponding
to
node
b.
This
equation
should
be
, be
ib.
i4, and
v5, we
find asersign error in the KCL equation
Checking
thetode
equations
used
to
calculate
v2be
corresponding
node
b. b.
This
equation
should
corresponding
to
node
b.
This
equation
should
be
3, Esta
en
la ecuación
la
KCL
correspondiente
al nodo
ecuación
debería
corresponding
to
node
This
equation
should
be
v2vv2
corresponding to node b. This equation should be
22 �
v�
4 ¼
i3 i3
i4 ii¼
v
2 v5
¼
22 �
4 ¼5
�3 iii333
�
i
i4 ¼
�
ii444 ¼
5
5 v5552 � i3 to be
¼
i4calculated
Aftermaking
makingthis
thiscorrection,
correction,v2v, 2i,3,i3i,4,i4and
, andv5vare
3 be
5 are
After
calculated
to
After
making
this
correction,
v
,
i
,
i
,
and
v
are
calculated
to
be
5 to
2
3
4
5
After
making
this
correction,
, and
are
calculated
to
be
4
5
After
making
thisthis
correction,
v2,vvvi2223,,,, iiii3334,,,, iiiand
v5 vvvare
calculated
be be
After
making
this
correction,
and
are
calculated
to
be
4,, and
5 are
After
making
correction,
calculated
to
4
5
Luego
de
haber
hecho
esta
corrección,
se
calcula
que
v
,
i
,
i
y
v
sean
2
3
4
5
After making this correction,
,
i
,
i
,
and
v
are
calculated
to
¼
7:5
V;
i
¼
1:125
A;
i
¼
0:375
A;
4:5
V
v
2
3
4
5
2 7:5 V; i3 ¼
4 0:375be
5 ¼
1:125
A;A;i4 i¼
A;
v5vv¼
4:5
VV
v2v¼
7:5
V;
ii333 ¼
1:125
0:375
A;
¼
4:5
22 ¼
44 ¼
¼
7:5
V;
¼
1:125
A;
i
¼
0:375
A;
¼
4:5
V
v
7:57:5
V;V;
i3 ¼
1:125
A;A;
i4 ¼
0:375
A;A;
v5 vv¼5555 ¼
4:54:5
VV
v2 v¼22 ¼
i33 ¼
1:125
i44 ¼
0:375
v2 ¼ 7:5 V; i3 ¼ 1:125 A; i4 ¼ 0:375 A; v5 ¼ 4:5 V
Now
Now
Now
Now
Now
Now
Ahora
Now
Now
7:5
v2vv2 7:5
¼1:5
1:5
A
22 ¼ 7:5
7:5¼¼
v¼
2 ¼
AA
i2 ii¼
v
7:5
2
2
7:5
v
¼
1:5
¼
2
5
5 ¼
2
¼
1:5
A
¼
1:51:5
AA
i2 ¼
¼57:5
¼
1:5
A
¼5 v¼
552 ¼
55¼ ¼
iii222 ¼
5 5 ¼5 5 ¼ 1:5 A
i2 ¼
5
5
1:125
þ
0:375¼¼1:5
1:5
A
yand
2 ¼
3þ
4 ¼
and
i2 ii¼
i3 iiþ
i4 ii¼
1:125
þþ
0:375
AA
and
1:125
0:375
¼
1:5
2 ¼
3þ
4 ¼
and
¼
þ
¼
1:125
þ
0:375
¼
1:5
A
andand
i2 ¼
i3 þ
1:125
þ 0:375
¼¼
1:51:5
AA
and
¼
þ
¼
1:125
þ 0:375
0:375
¼
1:5
A
4 ¼
iii222 ¼
iii333 iþ
iii444 ¼
1:125
þ
and se esperaba, esto ya concuerda. i2 ¼ i3 þ i4 ¼ 1:125 þ 0:375 ¼ 1:5 A
Como
Thischecks
checksasasweweexpected.
expected.
This
This
checks
as
we
expected.
This
checks
as
we
expected.
Como
una
comprobación
adicional,
v3. law
Primero,
la ley de Ohm da
This
checks
as
wewe
expected.
This
checks
as
we
expected.
.considere
First,Ohm’s
Ohm’s
lawgives
gives
Asanan
additional
check,consider
consider
This
checks
as
expected.
As
additional
check,
v3vv. 3First,
.. First,
Ohm’s
law
gives
As
an
additional
check,
consider
3
First,
Ohm’s
law
gives
As
an
additional
check,
consider
v
ThisAschecks
as
we
expected.
3
.
First,
Ohm’s
law
gives
an
additional
check,
consider
v
.
First,
Ohm’s
law
gives
As
an
additional
check,
consider
v
As an additional check, consider3 v33. First, Ohm’s law gives
First,
law¼gives
As an additional check, consider vv33v.¼
¼4i4i
¼4(1:125)
4(1:125)
¼4:54:5
V
3Ohm’s
VV
3 ¼
4i
¼
4(1:125)
¼
4:5
v3 ¼
¼
4i¼333 ¼
¼
4(1:125)
¼
4:5
V
4i34i
4(1:125)
¼
4:5
V
v3 vv¼3333 ¼
4(1:125)
¼
4:5
V
3
4ithe
¼voltage
4(1:125)
¼ de
4:5
v3of¼
A
continuación,
aplicando
laloop
KVL
al circuito
que source
consta
laVthe
fuente
deand
voltaje
los
resistores
3voltage
Next,
applyingKVL
KVL
tothetheloop
consisting
ofcerrado
source
and
the4-V
4-V
and5-V
5-Vyresistors
resistors
givesde 4-V y
Next,
applying
toto
consisting
the
and
gives
Next,
applying
KVL
the
loop
consisting
of
the
voltage
source
and
the
4-V
and
5-V
resistors
gives
Next,
applying
KVL
to
the
loop
consisting
of
the
voltage
source
and
the
4-V
and
5-V
resistors
gives
Next,
applying
to to
thethe
loop
consisting
of of
thethe
voltage
source
andand
thethe
4-V
andand
5-V
resistors
gives
5-V,
nos
da KVL
Next,
applying
KVL
loop
consisting
voltage
source
4-V
5-V
resistors
gives
Next, applying KVL to the loop consisting
of��
the
source
and
the 4-V and 5-V resistors gives
12
¼1212
�
7:5¼¼4:5
4:5
V
3 ¼
2voltage
1212
v2vv¼
��
7:5
VV
v3vv¼
¼
�
¼
12
7:5
¼
4:5
¼
12
�
¼
12
�
7:5
¼
4:5
V
1212
� v�
1212
� 7:5
¼
4:5
V
v3 vvv¼3333 ¼
¼
12
�2 vvv¼2222 ¼
¼
12
�
7:5
¼
4:5
V
� 7:5 ¼ 4:5 V
�
v22-V
¼2-V
12and
�
7:5
¼resistors
4:5
V gives
v3 ¼ 12
Finally,applying
applyingKVL
KVLtotothetheloop
loopconsisting
consisting
of
the
and
4-V
resistors
gives
Finally,
of
the
4-V
Finally,
applying
KVL
to
the
loop
consisting
of
the
2-V
and
4-V
resistors
gives
Finalmente,
aplicando
KVL
al circuito
cerrado
consta
de4-V
los
resistores
de 2-V y 4-V, resulta
Finally,
applying
KVL
to
the
loop
consisting
of
the
2-V
and
4-V
resistors
gives
Finally,
applying
KVL
tolato
the
loop
consisting
of of
theque
2-V
andand
4-V
resistors
gives
Finally,
applying
KVL
the
loop
consisting
the
2-V
resistors
gives
Finally, applying KVL to the loop consistingvof
and
¼v5v2-V
4:5
V4-V resistors gives
v¼
3the
5 ¼
4:5
VV
3v
¼
v¼
¼
4:5
¼
¼
4:5
V
v5 vv¼5555 ¼
4:54:5
VV
v3 vv¼3333 ¼
v5 indicating
¼indicating
4:5 V that
vother,
3 ¼
Theresults
resultsofofthese
thesecalculations
calculationsagree
agreewith
witheach
each
other,
that
The
The
results
of
these
calculations
agree
with
each
other,
indicating
that
The
results
of
these
calculations
agree
with
each
other,
indicating
that
Los
resultados
de
estos
cálculos
concuerdan
entre
sí,
lo
que
indicathat
que
The
results
of of
these
calculations
agree
with
each
other,
indicating
The
results
these
calculations
agree
with
each
other,
indicating
that
The results of these calculations
agree
with
each
other,
indicating
that
7:5
V;
1:125
A;i4 i¼
¼0:375
0:375
A;
4:5
V
2 ¼
3 ¼
5 ¼
7:5
V;V;
i3 ii¼
1:125
A;A;
A;A;
v5vv¼
4:5
VV
v2vv¼
¼
7:5
1:125
ii444 ¼
0:375
¼
4:5
¼
7:5
V;
iii333 ¼
¼
1:125
A;
¼
0:375
A;
¼
4:5
V
7:57:5
V;V;
i3 ¼
1:125
A;
i
¼
0:375
A;
v5 vvv¼5555 ¼
4:54:5
VV
v2 vvv¼2222 ¼
¼
7:5
V;
¼
1:125
A;
i
¼
0:375
A;
¼
4:5
V
3 ¼ 1:125 A;4 i44 ¼ 0:375 A;
v2 ¼ 7:5 V; i3 ¼ 1:125 A; i4 ¼ 0:375 A; v5 ¼ 4:5 V
arethethecorrect
correctvalues.
values.
areare
the
correct
values.
are
the
correct
values.
areare
thethe
correct
values.
correct
values.
son
los
valores
correctos.
are the correct values.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 83
Alfaomega
4/12/11 5:22 PM
84
84
Resistive Circuits
Circuitos resistivos
33 .. 99
DE EJ S
E XDAEMD
P ILSEE Ñ O
E IMGPNL O
FUENTE
DE VOLTAJE
AJUSTABLE
ADJUSTABLE
VOLTAGE
SOURCE
Se requiere que un circuito proporcione un voltaje ajustable. Las especificaciones para este
A circuitson
is required
to provide an adjustable voltage. The specifications for this circuit are
circuito
que:
that:
1. El voltaje debe poderse ajustar a cualquier valor entre 25 V y 15 V. No deberá haber
1. It
should be de
possible
adjust the
voltage accidental
to any value
�5 Vde
and
þ5margen.
V. It should
posibilidad
que setoobtenga
de manera
unbetween
voltaje fuera
ese
not be possible accidentally to obtain a voltage outside this range.
2.
La corriente
de carga
ser insignificante.
2. The
load current
willdebe
be negligible.
3.
El circuito
utilizar
la little
menorpower
potencia
posible.
3. The
circuitdebe
should
use as
as possible.
Los
disponibles
The componentes
available components
are:son:
1. Potentiometers:
values of
10 kV,
kV, andde
5010kV
1.
Potenciómetros: resistance
hay en existencia
valores
de 20
resistencia
V,are
20 in
V ystock
50 V.
2. A
standard estándar
2 percentde
resistors
values
V (vea
and 1apén­
MV
2.
Unlarge
granassortment
surtido deof
resistores
2% conhaving
valores
entrebetween
10 V y 10
1V
(see
Appendix
D)
dice D).
3. Two power supplies (voltage sources): one 12-V supply and one �12-V supply, both
3. rated
Dos alimentadores
de potencia (fuentes de voltaje): uno de 12 V y otro de 212 V, ambos
at 100 mA (maximum)
clasificados a 100 mA (máximo).
Describe the Situation and the Assumptions
Describa
lashows
situación
y los supuestos
Figure 3.9-1
the situation.
The voltage v is the adjustable voltage. The circuit that uses
La
muestra
situación.
El voltaje
v es el voltaje
the figura
output3.9-1
of the
circuitlabeing
designed
is frequently
called ajustable.
the load. Al
In circuito
this case,que
theutiliza
load
la
salidaisdel
circuito que
current
negligible,
so se
i ¼va
0. a diseñar se le llama a veces la carga. En este caso, la corriente
de carga no es significativa: por lo tanto, i 5 0.
Corriente de carga
Circuito
que se ha
de diseñar
Circuito
de carga
FIGURE 3.9-1 The circuit being
FIGURA 3.9.1 El circuito que se
designed provides an adjustable
va a diseñar proporciona un voltaje
voltage, v, to the load circuit.
ajustable, v, a la carga del circuito.
State the Goal
Establezca
el objetivo
A circuit
providing
the adjustable
voltage
Un
circuito
que proporciona
el voltaje
ajustable
�5V � v � þ5V
se
debe
utilizando
losavailable
componentes
disponibles.
must
bediseñar
designed
using the
components.
Genere
unaplan
Generate
Plan
Haga
Make las
thesiguientes
followingobservaciones:
observations.
1. La adaptabilidad de un potenciómetro se puede utilizar para obtener un voltaje v ajustable.
1. The adjustability of a potentiometer can be used to obtain an adjustable voltage v.
2.
Se deben
utilizar
los dos
alimentadores
de energía
de modovoltage
que el can
voltaje
ajustable
pueda
2. Both
power
supplies
must
be used so that
the adjustable
have
both positive
contar
con losvalues.
dos valores, positivo y negativo.
and
negative
3. The
terminals of
potentiometer
be conectar
connected
directly to athe
supplies
3.
Las terminales
delthe
potenciómetro
nocannot
se deben
directamente
lospower
suministros
de
because
the voltage
is not allowed
to be vassea
large
asde
1212
V Voro�12
V. V.
energía porque
no se vpermite
que el voltaje
tanto
de 212
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 84
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Design Example
Designde
Example
Ejemplo
diseño
85
85
85
These observations suggest the circuit shown in Figure 3.9-2a. The circuit in Figure 3.9-2b is
Theseobservaciones
observations
the model
circuit
shown
Figure
3.9-2a.
The
in
FigureEl3.9-2b
is
obtained
by using thesuggest
simplest
eachinque
component
in Figure
3.9-2a.
Estas
nos
recuerdan
el for
circuito
se muestra
en
la circuit
figura
3.9-2a.
circuito
obtained
by
using
the
simplest
model
for
each
component
in
Figure
3.9-2a.
de la figura 3.9-2b se obtiene utilizando el modelo más sencillo para cada componente de la
figura 3.9-2a.
Circuito
de carga
FIGURE 3.9-2 (a) A proposed circuit for producing the variable voltage, v, and (b) the equivalent circuit
FIGURE
3.9-2 (a) Aisproposed
after
the potentiometer
modeled.circuit for producing the variable voltage, v, and (b) the equivalent circuit
FIGURA
3.9-2 (a) Propuesta
de un circuito para producir el voltaje variable, v, y (b) el circuito equivalente
after the potentiometer
is modeled.
después de que el potenciómetro ha sido modelado.
To complete the design, values need to be specified for R1, R2, and Rp. Then several
To
complete
the design,
values need made,
to be specified
for R , R , and R . Then several
results Para
need
to be checked
and se
adjustments
if necessary.
completar
el diseño
necesitan especificar
los valores1 de 2R1, R2 y pRp. Después, se
results need to be checked and adjustments made, if necessary.
requiere que cada resultado se compruebe y se le hagan los ajustes que fueren necesarios.
1. Can the voltage v be adjusted to any value in the range �5 V to þ5V?
1. ¿El
Canvoltaje
the voltage
v be adjusted
any value
in en
theelrange
V toVþ5V?
1.
v se puede
ajustar a to
cualquier
valor
rango�5
de 25
a 15 V?
2. Are the voltage source currents less than 100 mA? This condition must be satisfied if the
2.
Are
the
voltage
source
currents
less
than
100
mA?
This
condition
must
be satisfied
the
supplies are
modeled
as ideal
2. power
¿Las corrientes
de to
la be
fuente
de voltaje
son voltage
menoressources.
de 100 mA? Se debe
satisfacerif esta
power
supplies
are
to
be
modeled
as
ideal
voltage
sources.
condición si se han de modelar los alimentadores de potencia como fuentes de voltaje ideales.
3. Is it possible to reduce the power absorbed by R1, R2, and Rp?
3.
Is it puede
possible
to reduce
the power
absorbed
R1, R2, and Rp?
3. ¿Se
reducir
la potencia
absorbida
por Rby
1, R2 y Rp?
Act on the Plan
Act
onsobre
the
Plan
ItActúe
seems
likely
R1 and R2 will have the same value, so let R1 ¼ R2 ¼ R. Then it is
elthat
plan
It seems likely
Rtuvieran
Rel
theinpor
same
value,
soRlet5RR1 ¼. Entonces
R2 ¼ R. Then
it is
1 and3.9-2b
2 will
convenient
to redraw
as have
shown
Figure
3.9-3.
Pareciera
que
R1that
y R2Figure
mismo
valor,
lo que
R1 5
es conve2
p
convenient
to
redraw
Figure
3.9-2b
as
shown
in
Figure
3.9-3.
niente hacer un nuevo dibujo de la figura 3.9-2b como se muestra en la figura 3.9-3.
FIGURE 3.9-3 The circuit after setting R1 ¼ R2 ¼ R.
FIGURA
3.9-3 El 3.9-3
circuito,
de haber
establecido
1 5 R2 5 R.
FIGURE
Thedespués
circuit after
setting
R1 ¼ R2 ¼RR.
Applying KVL to the outside loop yields
Aplicar
la KVL
KVL al
cerrado
Applying
to circuito
the outside
loopexterior
yields da como resultado
�12 þ Ria þ aRp ia þ (1 � a)Rp ia þ Ria � 12 ¼ 0
�12 þ Ria þ aRp ia þ (1 � a)Rp ia þ Ria � 12 ¼ 0
24
ia ¼
24
2R
i a ¼ þ Rp
2R þ Rp
Next, applying KVL to the left loop gives
A
continuación,
aplicando
KVL
al circuito
Next,
applying KVL
to thelaleft
loop
gives cerrado de la izquierda resulta
v ¼ 12 � (R þ aRp )ia
v ¼ 12 � (R þ aRp )ia
Substituting for ia gives
Substituting for
ia gives
Sustituyendo
ia nos
da
�
�
24 �R þ aRp �
v ¼ 12 � 24 R þ aRp
v ¼ 12 � 2R þ Rp
2R þ Rp
sopor lo que
so
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 85
Alfaomega
4/12/11 5:22 PM
E1C03_1
E1C03_1
11/25/2009
11/25/2009
86
86
E1C03_1
E1C03_1
E1C03_1
E1C03_1
11/25/2009
11/25/2009
11/25/2009
11/25/2009
86
86
86
86
p
available values.
Then,
R ¼ 14 kV
86
86
86
86
86
86
86
86
Verify the Proposed Solution
Circuitos Circuits
resistivos
Resistive
Resistive
AsCircuits
a check,
Resistive Circuits
notice that when a ¼ 1,
�
�
14,000 þ 20,000
Resistive Circuits
Circuits
Resistive
When
a
¼
0,
v
must
be
5
V,
so
Resistive
Circuits
v
¼
12
�
24 ¼ �5
Cuando
a
5
0,
v
debe
ser
5
V,
por
lo
tanto,
Resistive
Circuits
When
a ¼ 0, v must be 5 V, so
28,000 k þ 20,000
When a ¼ 0, v must be 5 V, so
24R
12 � 24R
as
required.
specification
When
¼ 0,
0,The
must
be 55 V,
V, so
sothat 55 ¼
24RRp
When
aa ¼
vv must
be
¼
12
�
2R
When a ¼ 0, v must be 5 V, so
5 ¼ 12 � 2R þ
þ Rp
þ
Rp
24R
�5 V � v2R
�24R
5V
Solving for R
R resulta
gives
5
¼
12
�
24R
Despejando
5
¼
12
�
Solving for R gives
2R
þR
Rpp
5 ¼by
12 the
� 2R
þ
Solving
R givesThe power absorbed
has
beenfor
satisfied.
three
is
2R
þ
Rresistances
pp
R
¼
0:7R
pp
R
5
0.7R
R
¼
0:7R
Solving for
for R
R gives
gives
Solving
R ¼ 0:7Rpp 242
2is selected
Solving for
Rpotentiometer
gives
Suppose
the
resistance
toseleccionado
be Rp ¼ 20 kV,
theRmiddle
of the
Suponga
que
la
resistencia
del
potenciómetro
como
el three
valor
(2R
þ Rsep )ha
¼
p
¼
i
p 5 20 kV,
a
Suppose the potentiometer resistance is R
selected
be þ
RpR¼
20 kV,
the middle
of the
three
R
¼ 0:7R
0:7Rto
p 2R
¼
p
Suppose the
potentiometer
resistance
is R
selected
to
be
R
¼
20
kV,
the
middle
of
the
three
p
available
values.
p
intermedio
de
los
tres
valores
disponibles.
¼ 0:7Rpp
available values.
available
values.
Then,
Entonces,
Suppose
the
potentiometer
resistance
is
selected
to
be
R
¼
20
kV,
the
middle
of
the
three
Suppose
the potentiometer resistance is selected
to be Rppp ¼ 20 kV, the middle of the three
so
p ¼ 12 mW
Then,
Suppose
the
potentiometer resistance is selected
to be Rp ¼ 20 kV, the middle of the three
Then,
available
values.
available
values.
R
¼
14
kV
R
5
available
values.
Notice
that
this
power
can
be
reduced
by
choosing
R ¼ 14 kV Rp to be as large as possible, 50 kV in
Then,
Then,
R ¼a 14
kVvalue of R:
Then,
this case.
Changing Rp to 50 kV requires
new
R ¼ 14
14 kV
kV
R
Verify
the
R¼
¼
R ¼ 0:7
� 14
Rp kV
¼ 35 kV
Verifique
laProposed
solución Solution
propuesta
Verify
the
Proposed
Solution
Verify
thecomprobación,
Proposed
As
a check,
notice that Solution
when
a ¼ 1,
A
guisa
de
observe
que
cuando
a
5
1,
As
a
check,
notice
that
when
a
¼
1,
Because
�
As a check,
notice that Solution
when a ¼ 1,�
Verify
the Proposed
Proposed
Solution
� 14,000
�
Verify
the
þ
20,000
�
�
�
�
Verify
the
Proposed
Solution
�
þ 20,000 �
v ¼þ
12
24 ¼ 35,000
�5
As aa check,
check, notice
notice that
that
when
a 50,000
¼�
1, 14,000
35,000
As
when
a
¼
1,
14,000
þ
20,000
v
¼
12
�
24
¼
�5
20,000
�5 V ¼
12 �that when
24 �kk vþ
12 � �24 ¼ �5
24 ¼ 5 V
As a check,
notice
a ¼�1,�
v ¼þ
12
�28,000
þ�20,000
28,000
70,000
50,000
20,000�
14,000kþ
þþ20,000
20,000
� 28,000
� 70,000 þ 50,000
14,000
as required. The specification
that
¼ 12
12
� 14,000 þ 20,000 24
24 ¼ �5
�5
vv ¼
�
as required.
The that
specification
that
Como
se requería.
La especificación
que kk þ
28,000
þ 20,000
20,000 24 ¼
the
specification
v ¼ 12
� de
¼ �5
28,000
as required.
The specification
that
28,000
�5 V �k vþ�20,000
5V
�5 V
V�
� vv �
� 55 V
V
as required.
required. The
The specification
specification that
that �5
as
�5
V
�
v
�
5V
as required.
The specification
has
been satisfied.
The power that
absorbed by the three resistances is
hashabeen
been
satisfied.
The
power
absorbed
by
thevtres
three
resistances
is now
�5
Vlas
�
� 5resistencias
5V
Vresistances
has
satisfied.
power
absorbed
by
the
se
satisfecho.
LaThe
potencia
absorbida
por
es is
�5
V
�
�
has been
satisfied.
The
power
absorbed
thevv three
three
is
2
�5by
V�
� 5 Vresistances
24
2
2
2
24
2
has been
been satisfied.
satisfied. The
The power
power absorbed
absorbed
by
the
is
þthe
Rp three
)three
¼ resistances
p ¼ ia 2(2R
24
has
by
is
24
Rpthree
) ¼ ¼resistances
p¼¼ ia 2 (2R
Rp
has been satisfied. The power pabsorbed
byþ
is
5þ
p ¼50,000
ia (2R
þthe
Rp ) ¼ 2R
2Rresistances
þ2mW
Rp
2
þ
70,000
2R24
þ2Rp
(2R
þ
R
¼ 24
¼ iiaa2222(2R
242
so
pþ
¼R
12
p))mW
¼
pp ¼
p
a
p
Finally,
the power supply current
is
so
p
¼
12
mW
2R
þ
Rpp
¼ 2R þ R
p ¼ ia (2Rp þ
p ) mW
so
¼ R12
þ
Rbe
pp as large as possible, 50 kV in
por
lo tanto,
p 5choosing
12 mW2R
Notice
that this power can be reduced by
R
to
Notice
that this power can be reduced byp24
choosing
Rpp to be as large as possible, 50 kV in
so
¼
12
mW
so
¼a 12
mW
Notice
thatChanging
this power
bei kV
reduced
bypp choosing
toofmA
beR:as large as possible, 50 kV in
¼ requires
¼Rp0:2
this
case.
R can
to 50
new
value
so
12
mW
this case. Changing Rpp to 50 akV 50,000
requires
new
value
of R:
þ¼aa70,000
this
case.
Changing
R
to
50
kV
requires
new
value
of
Observe
que
esta
potencia
se
puede
reducir
si
se
elige
que
R
sea
másas
posible,
50 in
V
p
Notice
that
this
power
can
be
reduced
by
choosing
R
to
be
as lo
large
asgrande
possible,
50 kV
kV
in
p R:
p to be
Notice that this power can be reduced
by choosing
as
large
possible,
50
R ¼the
0:7
� Rp ¼sources
35R
kV
p
p to be
Notice
that
this
power
can
be
reduced
by
choosing
R
as
large
as
possible,
50
kV
in
which
iscaso.
well
below
the
100
mA
that
voltage
are
able
to
supply.
The
design
is
p
R
¼
0:7
�
R
¼
35
kV
en
este
Cambiar
R
a
50
V
requiere
un
nuevo
valor
de
R:
p
this
case.
Changing
R
to
50
kV
requires
a
new
value
of
R:
this case. Changing Rppppto 50 kV R
requires
aR
new
R:
¼ 0:7 �
35 kVof
p ¼ value
this case. Changing Rp to 50 kV requires
a new
value
of R:
complete.
Because
Because
R¼
¼�
0:7 �
�R
Rp ¼ 35
35 kV
kV
�
�
�
R
0:7
Because
R ¼�
0:7 � Rppp ¼
¼ 35 kV�
�35,000 þ 50,000
�
35,000
�
�
�
�24 ¼ 5 V
35,000
þ
50,000
35,000
�5 V ¼ 12 � 35,000 þ 50,000 24 � v � 12 �
Because
Porque
Because
35,000
�5
V
¼
12
�
24
�
v
�
12
�
24 ¼ 5 V
70,000
þ
50,000
70,000
þ
50,000
Because
�5 V ¼ 12 � �
�70,000 þ 50,000�
�24 � v � 12 � �
�70,000 þ 50,000�
�24 ¼ 5 V
70,000 þ
þ
50,000
35,000
þ 50,000
50,000�
35,000
�35,000
� 70,000
�
35,000
the specification
�5 V
V¼
¼ 12
12that
� 35,000 þ 50,000 24
24 � vv �
� 12
12 �
�
24 ¼
¼ 55 V
V
35,000
�5
�
the specification
70,000 þ
þ 50,000
50,000 24 �
70,000 þ
þ 50,000
50,000 24
�5 V ¼ 12that
� 70,000
� v � 12 � 70,000
24 ¼ 5 V
the specification
that
70,000 þ 50,000�5 V � v � 5 V
70,000 þ 50,000
�5 V � v � 5 V
the specification
specification that
that
the
�5
V
�
v
�
5
V
la
especificación
de
que
the specification
has
been satisfied.that
The power absorbed by the three resistances is now
has been satisfied. The power absorbed
thev three
is now
�5by
V�
�
� 55 V
Vresistances
�5
V
v three
�
has been satisfied. The power absorbed
the
resistances is now
2 v � 5V
�5by
V
�
24 2
24the
2porthree
has been
been
satisfied.
The power
power
absorbed
by
the
three
resistances
is now
now
¼resistances
5 resistencias
mW
pabsorbed
¼ absorbida
has
satisfied.
The
by
is
se
satisfecho.
Ahora
potencia
las tres
es
24the
¼resistances
5 mW
pabsorbed
¼ 50,000
70,000
hashabeen
satisfied.
The lapower
by
three
is now
¼ 5 mW
p ¼ 50,000 þ
þ2270,000
50,00024
þ2270,000
24
Finally, the power supply current
¼ 55 mW
mW
¼is
pp ¼
Finally, the power supply current
is50,00024
þ
70,000 ¼
¼ 5 mW
p
¼
Finally, the power supply current is50,000 þ 70,000
50,00024
þ 70,000
24
ia ¼ is
¼ 0:2 mA
Finally, the
the power
power supply
supply current
current
is
24
Finally,
ia ¼ is
0:2 mA
þ 70,000
Finalmente,
corriente
delcurrent
suministro
de
energía
es ¼
Finally, the la
power
supply
ia ¼ 50,000
¼ 0:2 mA
50,000 þ 70,000
50,00024
þ 70,000
24
which is well below the 100 mA
voltage sources
are able to supply. The design is
¼that the
¼ 0:2
0:2 mA
mA
iiaaa ¼
¼
which is well below the 100 mA
the 24
voltage
are able to supply. The design is
50,000
þ
70,000sources
ia ¼ that
¼ 0:2 mA
which is well below the 100 mA
that
the þ
voltage
sources
are able to supply. The design is
50,000
70,000
complete.
50,000 þ 70,000
complete.
complete.
which is
is well below
below the
the 100 mA
mA that
that the
the voltage
voltage sources
sources are able
able to
to supply. The
The design
design is
which
which
is well
well
the 100
100
thatmA
the que
voltage
sourcesdeare
are
able son
to supply.
supply.
design is
is
la
cual está
bienbelow
por debajo
de mA
los 100
las fuentes
voltaje
capacesThe
de alimentar.
complete.
complete.
complete.
El
diseño está completo.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 86
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Resumen
3.10
RESUMEN
La ley de la corriente de Kirchhoff (KCL) establece que la
suma algebraica de las corrientes que entran en un nodo es
cero. La ley del voltaje de Kirchhoff (KVL) establece que la
suma algebraica de los voltajes en torno a un circuito cerrado (loop) es cero.
Se pueden analizar circuitos eléctricos sencillos utilizando
solamente las leyes de Kirchhoff y las ecuaciones constitutivas de los elementos del circuito.
Los resistores en serie actúan como un “divisor de voltaje”,
los resistores en paralelo funcionan como un “divisor de corriente”. Las primeras dos hileras de la tabla 3.10-1 resumen
estas ecuaciones tan importantes.
87
Los resistores en serie equivalen a un “resistor equivalente”
único. Del mismo modo, los resistores en paralelo equivalen
a un “resistor equivalente” único. Las dos primeras hileras
de la tabla 3.10-1 resumen tan importantes ecuaciones.
Las fuentes de voltaje en serie equivalen a una “fuente de
voltaje equivalente” única. Del mismo modo, las fuentes
de corriente en paralelo equivalen a una “corriente equivalente” única. Las dos últimas hileras de la tabla 3.10-1 resumen tan importantes ecuaciones.
En ocasiones, los circuitos que constan por completo de resistores se pueden reducir a un resistor equivalente único al
reemplazar de manera repetida los resistores en serie o en
paralelo por resistores equivalentes.
Tabla 3.10-1 Circuitos equivalentes para elementos en serie y en paralelo
Resistores en serie
Circuito
Circuito
y
Resistores en paralelo
Circuito
y
Circuito
e
Fuentes de voltaje
en serie
Circuito
y
Circuito
y
Fuentes de corriente
en paralelo
Circuito
Circuito
e
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 87
Alfaomega
4/12/11 5:22 PM
88
Circuitos resistivos
PROBLEMAS
Sección 3.2 Leyes de Kirchhoff
P 3.2-1 Considere el circuito que se muestra en la figura 3.2-1.
Determine los valores de la potencia alimentada por la extensión B y la potencia alimentada por la extensión F.
Figura P 3.2-4
P 3.2-5 Determine la potencia absorbida por cada resistor en
el circuito que se muestra en la figura P 3.2-5.
Respuesta: El resistor de 4 V absorbe 16 W, el de 6 V absorbe
24 W y el de 8 V absorbe 8 W.
Figura P 3.2-1
P 3.2-2 Determine los valores de i2, i4, v2, v3 y v6 en la figura
P 3.3-2.
Figura P 3.2-5
P 3.2-6 Determine la potencia alimentada por cada fuente de
corriente en el circuito de la figura P 3.2-6.
Respuesta: La fuente de corriente de 2-mA alimenta 6 mW y
la fuente de corriente de 1-mA alimenta 27 mW.
Figura P 3.2-2
P 3.2-3 Considere el circuito que se muestra en la figura P 3.2-3.
(a)Suponga que R1 5 8 V y R2 5 4 V. Encuentre la corriente
i y el voltaje v.
(b)Suponga, en cambio, que i 5 2.25 A y v 5 42 V. Determine las resistencias R1 y R2.
(c)Suponga, en cambio, que la fuente de voltaje alimenta
24 W de potencia y que la fuente de corriente alimenta
9 W de potencia. Determine la corriente i, el voltaje v y las
resistencias R1 y R2.
Figura P 3.2-6
P 3.2-7 Determine la potencia alimentada por cada fuente de
voltaje en el circuito de la figura P 3.2-7.
Respuesta: La fuente de voltaje de 2 V alimenta 2 mW y la
fuente de voltaje de 3-V alimenta 26 mW.
Figura P 3.2-3
P 3.2-4 Determine la potencia absorbida por cada resistor en
el circuito que se muestra en la figura P 3.2-4.
Respuesta: El resistor de 4-V absorbe 100 W, el de 6-V absorbe 24 W y el de 8-V absorbe 72 W.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 88
Figura P 3.2-7
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Problemas
89
P 3.2-8 ¿Cuál es el valor de la resistencia R en la figura P 3.2-8?
Sugerencia: Suponga un amperímetro ideal, pues equivale a
un cortocircuito.
Respuesta: R 5 4 V
Amperímetro
Figura P 3.2-11
Figura P 3.2-8
P 3.2-9 El voltímetro de la figura P 3.2-9 mide el valor del
voltaje a través de la fuente de corriente a 56 V. ¿Cuál es el
valor de la resistencia R?
P 3.2-12 Determine la potencia recibida por cada resistor en
el circuito mostrado en la figura P 3.2-12.
Sugerencia: Suponga un voltímetro ideal, el cual es equivalente a un circuito abierto.
Respuesta: R 5 10 V
Voltímetro
Figura P 3.2-12
Figura P 3.2-9
P 3.2-10 Determine los valores de las resistencias R1 y R2 en
la figura P 3.2-10.
Voltímetro
Voltímetro
P 3.2-13 Determine el voltaje y la corriente de cada uno de
los elementos de circuito en el circuito que se muestra en la
figura P 3.2-13
Sugerencia: Necesitará especificar las direcciones de referencia para los voltajes y las corrientes. Hay más de una manera
de hacerlo, y sus respuestas dependerán de las direcciones de
referencia que elija.
Figura P 3.2-13
Figura P 3.2-10
P 3.2-11 El circuito que se muestra en la figura P 3.2-11 consta de cinco fuentes de voltaje y cuatro fuentes de corriente.
Exprese la potencia alimentada por cada fuente en términos de
los voltajes de las fuentes de voltaje y de las corrientes de las
fuentes de corriente.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 89
P 3.2-14 Determine el voltaje y la corriente de cada uno de
los elementos de circuito en el circuito de la figura P 3.2-14.
Sugerencia: Necesitará especificar las direcciones de referencia para los voltajes y corrientes de los elementos. Hay más
de una manera de hacerlo, y sus respuestas dependerán de las
direcciones de referencia que elija.
Alfaomega
4/12/11 5:22 PM
90
Circuitos resistivos
P 3.2-19 La fuente de voltaje en la figura P 3.2-19 alimenta
3.6 W de potencia. La fuente de corriente alimenta 4.8 W. Determine los valores de las resistencias R1 y R2.
Figura P 3.2-19
P 3.2-20 Determine la corriente i en la figura 3.2-20.
Figura P 3.2-14
Respuesta: i 5 4 A
P 3.2-15 Determine el valor de la corriente medida por el
contador en la figura P 3.2-15.
Amperímetro
Figura P 3.2-20
P 3.2-21 Determine el valor de la corriente im en la figura
P 3.2-21a.
Figura P 3.2-15
P 3.2-16 Determine el valor de la corriente medida por el
contador en la figura P 3.2-16.
Amperímetro
Figura P 3.2-16
P 3.2-17 Determine el valor del voltaje medido por el contador en la figura P 3.2-17.
Voltímetro
Figura P 3.2-21 (a) Un circuito que contiene una VCCS. (b)
El circuito después de haber etiquetado los nodos y algunas
corrientes y voltajes de elementos.
Figura P 3.2-17
P 3.2-18 Determine el valor de la corriente medida por el
contador en la figura P 3.2-18.
Sugerencia: Aplicar la KVL a la ruta cerrada a-b-d-c-a en la
figura P 3.2-21a para determinar va. Luego aplique la KCL en
el nodo b para encontrar im.
Respuesta: im 5 9 A
Voltímetro
P 3.2-22 Determine el valor del voltaje vm en la figura P 3.2-22a.
Sugerencia: Aplicar la KVL a la ruta cerrada a-b-d-c-a en la
figura P 3.2-22b para determinar va.
Figura P 3.2-18
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 90
Respuesta: vm 5 24 V
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Problemas
91
P 3.2-26 Determine el valor del voltaje v5 para el circuito que
se muestra en la figura P 3.2-26.
Figura P 3.2-26
P 3.2-27 Determine el valor del voltaje v6 para el circuito que
se muestra en la figura P 3.2-27.
Figura P 3.2-22 (a) Un circuito que contiene una VCVS. (b)
El circuito después de haber etiquetado los nodos y algunas
corrientes y voltajes de los elementos.
P 3.2-23 Determine el valor del voltaje v6 para el circuito que
se muestra en la figura P 3.2-23.
Figura P 3.2-27
P 3.2-28 Determine el valor del voltaje v5 para el circuito que
se muestra en la figura P 3.2-28.
Figura P 3.2-23
P 3.2-24 Determine el valor del voltaje v6 para el circuito que
se muestra en la figura P 3.2-24.
Figura P 3.2-28
P 3.2-29 La fuente de voltaje en el circuito que se muestra
en la figura P 3.2-29 alimenta 2 W de potencia. El valor del
voltaje a través del resistor de 25-V es v2 5 4 V. Determine los
valores de la resistencia R1 y de la ganancia, G, de las VCCS.
Figura P 3.2-24
P 3.2-25 Determine el valor del voltaje v5 para el circuito que
se muestra en la figura P 3.2-25.
Figura P 3.2-29
P 3.2-30 Considere el circuito que se muestra en la figura
P 3.2-30. Determine los valores de
Figura P 3.2-25
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 91
(a) La corriente ia en el resistor de 20-V.
(b) El voltaje vb a través del resistor de 10-V.
(c) La corriente ic en la fuente de voltaje independiente.
Alfaomega
4/12/11 5:22 PM
92
92
Resistive Circuits
92
Circuits
CircuitosResistive
resistivos
Voltímetro
Figure P 3.2-30
Figure P 3.2-30
Figura P 3.2-30
Figure P 3.3-3
Figure P 3.3-3
Figura P 3.3-3
P 3.3-4 Determine the voltage v in the circuit shown in
P
3.3-4P Determine
el voltaje v the
en elvoltage
circuitov que
se muestra
P 3.3-4 Determine
in the
circuit shown in
Figure
3.3-4.
en la figuraFigure
3.3-4. P 3.3-4.
Section
3.3
Series
Resistors
and
Voltage
Division
Sección 3.3 Resistores en serie y división de voltaje
Section
3.3
Series
Resistorsthe
and
Voltage
, vDivision
P 3.3-1
3.3-1 Utilice
Use
voltage
division
determine
voltages
v1vol2,
P
la división
de to
voltaje
para determinar
los
P
3.3-1
Use
voltage
division
to
determine
the
voltages
v1, v2,
,
and
v
in
the
circuit
shown
in
Figure
P
3.3-1.
v
tajes
v1, v42, v3 y v4 en el circuito que se muestra en la figura
3
,
and
v
in
the
circuit
shown
in
Figure
P
3.3-1.
v
3
4
P 3.3-1.
Figure P 3.3-1
Figure P 3.3-1
Figura P 3.3-1
Figure P
P 3.3-4
3.3-4
Figura
Figure P 3.3-4
P 3.3-2 Consider the circuits shown in Figure P 3.3-2.
P
3.3-5 The
En lamodel
figuraofPa3.3-5
muestra
el modelo
de un to
ca-a
P 3.3-2
Consider
circuits
shown
inenFigure
P 3.3-2.
P 3.3-5
cablese
and
load resistor
connected
P(a)3.3-2
Considere
los circuitos
que
se muestran
la 3.3-2b
figura
Determine
the value
of thethe
resistance
R
in Figure
P
ble
y
resistor
de
carga
conectado
a
una
fuente.
Determine
la
P 3.3-5
The model
of a cable
and load
connected
to a
in Figure
P 3.3-5.
Determine
the resistor
appropriate
P 3.3-2.
(a) Determine
value of
the resistance
R into
Figure
3.3-2b is shown
that makes
the circuit the
in Figure
P 3.3-2b
equivalent
the Psource
resistencia
apropiada
delthat
cable,
R, de
modo
quevo,elremains
voltaje
source
is
shown
in
Figure
P
3.3-5.
Determine
the
appropriate
cable
resistance,
R,
so
the
output
voltage,
that
makes
the
circuit
in
Figure
P
3.3-2b
equivalent
to
the
circuit
in
Figure
P
3.3-2a.
(a)Determine el valor de la resistencia R en la figura P 3.3-2b de
salida, voV
, seand
mantenga
entre
Vthat
ysource
13the
V cuando
de
resistance,
R, 9so
outputlavoltage,
vo, remains
13 V when
the
voltage,
vfuente
s, varies
circuit
in Figure
P
(b) que
Determine
theelcurrent
i de
in Figure
the between 9cable
hace que
circuito
la 3.3-2a.
figuraP P3.3-2b.
3.3-2bBecause
sea equivavoltaje,
vs20
,between
tenga
una928
variación
entre
20resistance
V the
y 28source
V. can
La resistenV
and
13
V
when
voltage,
vs, varies
between
V
and
V.
The
cable
assume
the
i iininFigure
the
circuits
areDetermine
equivalent,
thecurrent
current
FigurePP3.3-2b.
3.3-2a Because
is
lente
al(b)
circuito
de la figura
P 3.3-2a.
cia
del cable
puede
valores
enteros
el rango de
between
20
V and
2820
V.
cable
resistance
can assume
integer
values
only asumir
in the
range
<The
R <sólo
100enV.
circuits
are
equivalent,
the
current
i
in
Figure
P
3.3-2a
is
equal
to
the
current
i
in
Figure
P
3.3-2b.
(b)Determine la corriente i en la figura P 3.3-2b. Dado que 20 , R , integer
100 V. values only in the range 20 < R < 100 V.
equal
toequivalentes,
thesupplied
current ila
incorriente
Figure
Pi 3.3-2b.
(c) los
Determine
theson
power
by
the
voltage
circuitos
desource.
la figura
(c)esDetermine
the poweri en
supplied
byPthe
voltage source.
P 3.3-2a
igual a la corriente
la figura
3.3-2b.
(c) Determine la potencia alimentada por la fuente de voltaje.
Figure P
P 3.3-5
3.3-5 Circuito
Circuit with
Figura
con auncable.
cable.
Figure P 3.3-5 Circuit with a cable.
Figure P 3.3-2
Figure P 3.3-2
Figura P 3.3-2
P
3.3-6 La
circuito
se muestra
figura
P 3.3-6
Theentrada
input toalthe
circuitque
shown
in FigureenPla3.3-6
is
P
3.3-6
es el
de
la input
fuente
va.shown
La
de este
Pofvoltaje
3.3-6
The
tode
circuit
Figure
The
output
ofsalida
thisincircuit
is P 3.3-6 is
the
voltage
the voltage
source,
vthe
a.voltaje,
circuito
esthe
el voltaje
medido
por elsource,
voltímetro
vb.output
Elproduces
circuito
The
of this circuit is
voltageby
of the voltmeter,
voltage
va. circuit
the voltage
measured
vb. This
produce
que
es proporcional
a lathat
entrada,
es decir,
thesalida
voltage
measured
voltmeter,
circuit produces
an outputuna
that
is proportional
toby
thethe
input,
is,vb. This
an output that isvbproportional
to the input, that is,
¼ k va
v
¼
k va
b
where k is the constant of proportionality.
donde k es la constante de proporcionalidad.
where
k
is
the
constant
of
proportionality.
(a) Determine the value of the output, v , when R ¼ 180 V and
b
(a)Determine
el valor dethe
la value
salida,ofvthe
180 R
V¼y 180 V and
output,Rvb5
, when
va ¼ 18(a)
V.Determine
b, cuando
P 3.3-3 The ideal voltmeter in the circuit shown in Figure
vDetermine
18 V.of the power supplied by the voltage
(b)
a 5 18 V. vthe
a ¼ value
3.3-3the
The
ideal
voltmeter
thesecircuit
Figure
P 3.3-3
3.3-3 measures
voltage
P
ElPvoltímetro
ideal
env.
el circuitoin
que
muestrashown
en la in (b)
D
etermine
el valor
de the
la
por la fuente
(b)
Determine
valuevaof
the
power supplied
by the voltage
¼alimentada
18 V.
source
when
R ¼ 180
V potencia
and
P
3.3-3
measures
the
voltage
v.
figura
P 3.3-3Rmide
el V.
voltaje
v.
(a) Suppose
Determine
the value of R1.
voltaje cuando
R5
V180
y vV
5
18
V.
2 ¼ 50
¼
18
V.to cause
source
when
R ¼resistance,
and
v
(c) de
Determine
the value
of 180
the
R,
required
a
a
(a) instead,
Suppose R
R12 ¼
¼ 50 V. Determine
the value
of of
R1. (c)Determine
(b) Suponga
Suppose,
Determine
el
requerida
para
que to cause
(c) Determine
value
theR,
resistance,
required
the output
to valor
be vbde
¼the
2laVresistencia,
whenofthe
input
is va ¼R,
18
V.
(a)
que R2 5 50
V.50
Determine
el valorthe
de Rvalue
1.
(b) Suppose,
instead, R ¼ 50 V. Determine
the value Determine
ofsalida sea
Ruponga,
vboutput
5 2 Vofto
cuando
la 2entrada
V.
2.
the
be resistance,
vb ¼
V when
the18input
is va ¼ 18 V.
a5
the
value
the
R, vrequired
to cause
(b)S
en cambio, que R1 5 150 V. Determine el valor (d) la
R2.
(c) de
Suppose,
that the voltage source supplies 1.2 W of (d)D
el valor
dethe
la
resistencia,
R,resistance,
requerida
para
que to cause
(d)a Determine
value
of the
R,
required
¼ 0.2v
ðthat
is, the
value
of the
constant
of proportionvbetermine
R2. instead,
2¼ 0.2v
(c)
Suppose,
instead,
that
the
voltage
source
supplies
1.2
W
of
v
5
0.2
v
(es
decir,
que
el
valor
de
la
constante
de
propower.
Determine
the
values
of
both
R
and
R
.
1
2
b
a
ðthat
is,
the
value
of
the
constant
of
proportionv
ality is k ¼b10Þ.
a
(c)Suponga, en cambio, que la fuente de voltaje alimenta
2
power.
Determine
the
values
of
both
R
and
R
.
1
2
ality
is
k
¼
Þ.
porcionalidad
sea
1.2 W de potencia. Determine los valores de R1 y R2.
10
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 92
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Problemas
93
Voltímetro
Voltímetro
Figura P 3.3-6
Problems
93
Figure P 3.3-6
P 3.3-7
Determine
the value
ofcircuito
voltage que
v in the
P 3.3-7 Determine
el valor
del voltaje
v en el
se circuit shown
in Figure
P 3.3-7.
muestra en la figura
P 3.3-7.
Figura P 3.3-9 Figure P 3.3-9
Figura P 3.3-7 Figure P 3.3-7
P 3.3-10
Determine
the value
of the
measured by the
P 3.3-10 Determine
el valor
del voltaje
medido
por voltage
el contameter
in
Figure
P
3.3-10.
dor en la figura 3.3-10.
P 3.3-8 Determine
la potencia
alimentada
por supplied
la fuente by
de- the dependent
P 3.3-8
Determine
the power
pendiente en elsource
circuitoinque
muestra
en lainfigura
P 3.3-8.
thesecircuit
shown
Figure
P 3.3-8.
10/23/2009
51
Voltímetro
Figure P 3.3-10
Figura P 3.3-10
Problems
51
Problems
51
P 3.3-11 For the circuit of Figure P 3.3-11, find the voltage v3
P 3.3-11 Para and
el circuito
de la
figura
P that
3.3-11,
encuentre
el
the current
i and
show
the power
delivered
to the three
Figura
P
3.3-8
50 mA
50 mA
100 Ω
Figure P
3.3-8
v3vyfor
la corriente
i
y
muestre
que
la
potencia
entregada
a
Determine
the value of v for each
Determine
of the following
thevoltaje
value of
cases.
each
of
the
following
cases.
resistors is equal to that supplied by the source.
los tres resistores es igual a la alimentada por la fuente.
–
+ va – P 3.3-9 Se puede
(a) The
switch
closed and(a)
RsaThe
closed and Rs51¼Answer:
¼manera
0 switch
(a short
0 (a short
Problems
utilizar
un is
potenciómetro
deis circuit).
¼ 3 V, i ¼ 1 A
v3 circuit).
P
3.3-9
A
potentiometer
can
be
used
as
a
transducer
+
+
The switch
is closed
and(b)
Rs The
and to
Rsv3¼553V,
¼de5 switch
V. cua-is closed
V. i 5 1 A
Respuesta:
transductor para(b)
convertir
la posición
de rotación
un
+
convert
rotational
position
ofilustra
dialesta
tois open
an
electrical
kv
k eléctrica.
vathe
(c) The
switch
is
and
R
(c)
The
and Rs ¼ 1 (an open circuit).
1aswitch
(an
open
circuit).
10 mA
Ω
10
5 VΩa una cantidad
5
drante
LaVopen
figura
P 3.3-9
s ¼
50
Ω
– 10 Va
Determine
the
valueillustrates
of v for each
ofsituation.
the following
cases.
quantity.
Figure
3.3-9
thiscuya
Figure
10 switch
kV.
(d) The
switch
is–Popen
(d)
Rs ¼The
is open and
Rs ¼ 10 kV.
–
situación.
La figura
3.3-9a
muestra
un and
potenciómetro
P
3.3-9a
shows
a
potentiometer
having
resistance
R
coni
i
–
Theb switch
closeddeand
Rs ¼El0 poten(a short circuit).
p
resistencia Rp está(a)
conectada
a unaisfuente
voltaje.
b
Rs
Problems Rs has three
51
+ tiene tres
nected
to aswitch
voltage
source.
(b)
The
andThe
Rs ¼
5 V.más
ciómetro
terminales,
unaisenclosed
cada
terminal
ypotentiometer
otra
one
at each
endand
and
connected
a sliding
k va conectada
(c) Thedeslizante
switch
is llamado
open
Rsone
¼1
(an
open to
circuit).
10 ΩFigure P 2.7-10
5 V a unterminals,
contacto
wiper.
Un
voltí+
+
+ +
50 mA
called
a
wiper.
A
voltmeter
measures
the
voltage
¼
10
kV.
(d)
The
switch
is
open
and
R
metro –mide el contact
voltaje
entre
el
wiper
y
una
de
las
terminales
Determine
the
value
of
v
for
each
of
the
following
cases.
s
+
+
+
+
100
Ω
100
100
Ω
Ω
100
Ω
v
v
v12of
v 12 V −
12 V −the wiper and one end
12
V the
V− potentiometer.
ib del potenciómetro.
−
between
Rs
is closed and Rs− ¼ 0 (a short circuit).
Figura
P
3.3-11
Figure
P
3.3-11
ansducersSection 2.8 Transducers (a) The switch
−
− −
Figure
P 3.3-9b
showsdespués
the circuit
afterelthe potentiomeLa figura 3.3-9b
muestra
el circuito
de que
+
(b) The switch is closed and Rs ¼ 5 V.
terreemplazado
is replaced
a model
potentiometer
potentiometer
P 2.8-1
circuit
Forofthe
Figure
potentiometer
2.8-2, the circuit
of Figureby
2.8-2,
theof thedel
potenciómetro
es
por
un modelo
potencióme- that consists of +
k va
(c) The switch is (a)
open and Rs ¼+1 (an open (a)
circuit).
P 3.3-12 Consider
the voltage
in Figure
10 Ω
5V
P
que se divider
muestra shown
en
(b)3.3-12
(b) de voltaje
resistors.
a depends
on+ the
u, of theConsidere el divisor
rent and potentiometer
current source
resistance
current
and
areque
1.1
potentiometer
mA two
resistance
areparameter
1.1
mA
tro,
consta
de dos
resistores.
El
parámetro
a depende
del angle,
+ isThe
100
Ω
100
Ω
¼
10
kV.
(d) The
switch
open
and
R
P
3.3-12
when
R
¼
8
V.
It
is
desired
that
the
output
power
v
–
v
12
V −a ¼ angle,
V − Also,la
u
ectively. Calculate
and 100 the
kV,irequired
respectively.
angle,Calculate
u,, so
the
required
sois sgiven
ángulo,
del dial.
Aquí
y u, uestán
dados
en
dial.
Here
in12
degrees.
infigura
FigurePP3.3-12 cuando R1 51 8 V. Lo deseable es que la
Figure
P 2.9-3360� , and
Figure
P grados.
2.9-3
absorbed
by
R
be
4.5
W.
Find
the
voltage
v
and
the
ransducers
b
potencia
de
salida
absorbida
por
R
sea
de
4.5
W.
Encuentre
el
−
1
o
−
1
voltage isthat
23 V.
the measured voltage
is 23
V.figura
Además,
en la
3.3-9b,
el voltímetro
sido reemplas
3.3-9b,Pthe
voltmeter
has beenha
replaced
by an R
open
circuit, and
required
source vvs.s.
voltaje
v
y
la
fuente
requerida
o
e potentiometer circuit of Figure
2.8-2,
the
zado
por khas
un¼
circuito
abierto,
a la vezby
que
ha etiquetado
el been labeled.
voltage
measured
voltmeter,
vm, has
590 sensor Phas
2.8-2
an associated
An AD590constant
sensor
anthe
associated
constant
¼these
Section
2-10
HowkCan
We
Section
Check2-10
. . . How
?
Can +We Check . . . ?
(a)
(b)
urrent and potentiometer
resistance
are 1.1
mA The
voltaje
medido
por el
voltímetro,
vm. Lais
al circuito
es
+entrada
input
to
the
circuit
the
angle
u,
and
the
output
is the
mA
1 � Kv. ¼
has a voltage
The20sensor
V; andhas
thea measured
voltage v ¼ P202.10-1
V;
and
the
measured
+ P
+ es
100
Ω the
Ω
The
circuit
shown
Pinel
2.10-1
Figure
The
2.10-1
is used
shown
to
testFigure P 2.10-1 is used to test
pectively. Calculate the required
angle, u,, yso
el ángulo
lavoltage
salida
el
voltaje
medido
por
contador,
vmcircuit
. 100
v in
v
.
measured
by
meter,
v
12
V Figure
12
V
P
2.9-3
m
−
own
in Figure
current,
2.8-3,
is 4 as
mA
shown
< i <in13Figure
mA in2.8-3,
a the
is 4−CCVS.
mA < i Your
< 13 mA
a theclaims
lab inpartner
CCVS.
that
Your
this
lab
measurement
partner
claims that this measurement
ed voltage
is
23
V. i(t),
ducers
(a)la
Show
that
the output−isa la
proportional
to the input. −
(a) Muestre
salida
proporcional
entrada.
s
. Find the range
laboratory
of measured
setting. Find
temperature.
the rangeque
of
measured
temperature.
shows
thatesthe
gain of the CCVS
shows
isthat
�20the
V/A
gain
instead
of theofCCVS
þ20 is �20
V/A instead of þ20 o
D590
sensorcircuit
has anofassociated
constant
entiometer
Figure 2.8-2,
the Rkp ¼
1 kV
and Can
vs ¼
V.
Express
Let
Section
2-10
We
Check
. the
. ? output as a
(b)
Sean
5 (b)
1 kV
y vR 5¼
24
V.How
Exprese
la24
salida
como .una
V/A. Dos pyou
agree?
Justify
V/A.
your
answer.
Do
you
agree?
Justify
your
answer.
(a)
(b)output when
and
arethe
1.1función
mA de la entrada.
haspotentiometer
a voltage v ¼resistance
20 V; and
measured
es
el valor
de laisinsalida
cuando
of
the
input.
What
the
value
the
P function
2.10-1�¿Cuál
The
circuit
shown
Figure
P of
2.10-1
is used to test
witches Section 2.9 Switches
vely.
the required
u,5
so
45°?
el
cuando
vangle
V? vm that
P 3.3-12Figure P 3.3-12
uPes
¼CCVS.
45ángulo
? What
islab
the partner
when
¼ 10this
V?Figura
hown Calculate
in Figure 2.8-3,
is 4 mAangle,
< i < 13
mA
inFigure
a¿Cuál
m 5 10
2.9-3
the
Your
claims
measurement
ne
the
2.9-1
i, atof
t Determine
¼measured
1 s and at
the
t¼
current,
4 s fori, at t ¼ 1 s and at t ¼ 4 s for
tage
iscurrent,
23
V.
g.
Find
theP
range
temperature.
shows– that
the
gain
of
the
CCVS
is
�20
V/A
instead
of
þ20
4 0 . 0
2 . 0
– 2 . 04 0 . 0
ure P 2.9-1.the circuit of FigureCircuitos
P 2.9-1. Eléctricos - Dorf
Alfaomega
sensor has an associated constant k ¼ Section
V/A.2-10
Do
you
agree?
Justify
your
answer.
How
Can
We
Check
.
.
.
?
Ammeter
Ammeter
Voltmeter
Voltmeter
t=2s
t=2s
witches
a voltage v ¼ 20 V; and the measured P 2.10-1 The circuit shown in Figure P 2.10-1 is used to test
in Figure
2.8-3, isi, 4atmA
13 mA
in a4 s for
mine
the current,
t ¼<1is<and
at t ¼
the CCVS. Your lab partner claims that this measurement
tthe
= 3range
s
t=3s
4R0 . 0
– 2 . 0 R
nd
of
measured
temperature.
gure P 2.9-1.
shows that the
gain of the CCVS is �20 V/A instead of þ20
M03_DORF_1571_8ED_SE_053-107.indd 93
+
5 kΩ
15t =V 2 s
+
10 V
CCVS
Ammeter
+
+
+V
V/A.
vo answer.
5 kΩ Do+ you agree?
10 VJustify your
Voltmeterv CCVS
o
V
+
4/12/11 5:22 PM
94
94
94
Resistive Circuits
Circuits
Resistive
Circuitos resistivos
P 3.3-13
3.3-13 Consider
Consider the
the voltage
voltage divider
divider circuit
circuit shown in
in Figure
Figure
P
P
3.3-13 Considere
el circuito
divisor de shown
voltaje que
se
P 3.3-13.
3.3-13. The
The resistor
resistor R
R represents
represents aa temperature
temperature sensor.
sensor. The
The
P
muestra en la figura P 3.3-13.El resistor R representa�un sensor
C, by
by the
the
resistance R,
R, in
in V,
V, is related
related to the
the temperature
temperature T,
T, in
in � C,
resistance
de
temperatura
La is
resistenciatoR,
en V, está relacionada
con la
equation
equation
temperatura
T, en °C mediante la ecuación
1
R¼
¼ 50
50 þ
þ1T
T
R
22
(a) Determine
Determine el
thevoltaje
meter medido,
voltage, vvvm
corresponding
to atemtem(a)
que corresponde
las
m,,, corresponding
(a)
Determine
the
meter
voltage,
to
� m
peratures 00�� C,
C,0 °C,
75�� C
C
and
100
� C.
temperaturas
75and
°C y100
100
peratures
75
C.°C.
(b) D
Determine
thevalor
temperature,
T, corresponding
corresponding
to the
the meter
meter
(b)
etermine el
de la temperatura,
T, que corresponda
(b)
Determine
the
temperature,
T,
to
voltages
8
V,
10
V,
and
15
V.
avoltages
los voltajes
medidos
8
V,
10
V
y
15
V.
8 V, 10 V, and 15 V.
Voltímetro
Figura
Figure P
P 3.3-13
3.3-13
Figure
P
3.3-13
P 3.3-14 Considere el circuito que se muestra en la figura
P 3.314.
3.3-14 Consider
Consider the
the circuit
circuit shown
shown in
in Figure
Figure P
P 3.3-14.
3.3-14.
P
P
3.3-14
(a) Determine
Determine el
thevalor
valuedeof
oflathe
the
resistanceR R
Rrequerida
required para
to cause
cause
(a)
resistencia
que
(a)
Determine
the
value
resistance
required
to
¼ 17.07
17:07 V.
V.
vvvooo 5
¼
17:07
V.
(b) Determine
Determine el
thevalor
value
ofvoltaje
the voltage
voltage
when
R=
=V.14
14 V.
V.
(b)
delof
vo cuando
R5R
14
(b)
Determine
the
value
the
vvoo when
(c)
Determine
the
power
supplied
by
the
voltage
source
when
(c)
alimentada
porvoltage
la fuente
de voltaje
(c) Determine
Determine la
thepotencia
power supplied
by the
source
when
¼ 14:22
14:22
V.14.22 V.
cuando
vo 5V.
vvoo ¼
energy
energy
storedstored
in a battery
in a b
P 2.5-3
P 2.5-3
The current
The current
sourcesource
and voltage
and voltage
sourcesource
in theincircuit
the circuit
voltage
voltage
and the
and
battery
the batc
shownshown
in Figure
in Figure
P 2.5-3
P 2.5-3
are connected
are connected
in parallel
in parallel
so thatsothey
that they
with the
withunits
the units
of Ampe
of
The current
sourcesource
and voltage
and voltage
both have
both the
have
same
the same
voltage,
voltage,
vs. The
vs.current
having
having
a capacity
a capacity
of 800o
sourcesource
are also
areconnected
also connected
in series
in series
so thatsothey
that both
they have
both the
have the
a current
a current
of 25 of
mA.
25 (a)
mA
same same
current,
current,
is. Suppose
is. Suppose
that vsthat
¼ 12
vs V
¼ and
12 Visand
¼ 3iA.
3 A. Calculate
s ¼Calculate
discharge
discharge
the
battery?
the
batte
(b
the
power
the
power
supplied
supplied
by
each
by
source.
each
source.
Figure P
P 3.4-2
3.4-2
Figure
to
the
to
load
the
during
load
during
the
tim
Figura
P
3.4-2
Answer:
Answer:
The voltage
The voltage
sourcesource
supplies
supplies
�36 W,
�36and
W,the
andcurrent
the current
P 3.4-3
3.4-3
The
ideal
voltmeter
in the
the circuit
circuit shown
shown in
in Figure
Figure
sourcesource
supplies
supplies
36
W.36 en
W.in
P
The
ideal
voltmeter
P
3.4-3
El
voltímetro
ideal
el circuito
que se muestra
en la
P
3.4-3
measures
the
voltage
v.
P 3.4-3P measures
voltagev.v.
figura
3.4-3 midethe
el voltaje
is
i
(a) Suppose
Suppose R
R2 ¼
¼ 6 V. Determine
Determine
the value of
of R
R1 sand
and of the
the
+
(a)
+
+theelvalue
1 R y of
(a)
Suponga
R22 56 6V.V.
Determine
valor
de
de la
1
–
+
+
current
i.
is
isvs
vs vs – vs –
current
i.
corriente i.
¼
6
V.
Determine
the
value
of
R
and
(b)
Suppose,
instead,
R
–
–
Determine the value of R22 and
(b) S
Suppose,
R11 ¼R6 V.
(b)
uponga, instead,
en cambio,
1 5 6 V. Determine el valor de R2
of the
the current
current i.
i.
bat
yofde
la corriente
i.
(c) Instead,
Instead, choose
choose R
R1 and
and R
R2 to
to minimize
minimize the
the power
power absorbed
absorbed
(c)
(c)
En cambio,
elija
R11 y R2 2para minimizar la potencia absorFigure
Figure
P 2.5-3
P 2.5-3
FigureFigure
P 2.5-6
P 2.5-6
by any
any
one resistor.
resistor.
by
bida
porone
algún resistor.
P 2.5-4
P 2.5-4
The current
The current
sourcesource
and voltage
and voltage
sourcesource
in theincircuit
the circuit
Section
Section
2.6 Voltmet
2.6 Volt
shownshown
in Figure
in Figure
P 2.5-4
P 2.5-4
are connected
are connected
in parallel
in parallel
so thatsothey
that they
P 2.6-1
P 2.6-1
For the
Forcircuit
the c
both have
both the
have
same
the same
voltage,
voltage,
vs. The
vs.current
The current
sourcesource
and voltage
and voltage
Voltímetro
sourcesource
are also
areconnected
also connected
in series
in series
so thatsothey
that both
they have
both the
have the
(a) What
(a) What
is theisvalue
the va
o
that vsthat
¼ 12
vs V
¼ and
12 Visand
¼ 2iA.
2 A. Calculate
same same
current,
current,
is. Suppose
is. Suppose
s ¼Calculate
(b) How
(b) How
muchmuch
powerpois
the power
the power
supplied
supplied
by each
by source.
each source.
Figure P
P 3.4-3
3.4-3
Figure
Figura P 3.4-3
+ 5 +.
+
+
+
+
vcircuit
isvs i invthe
P 3.4-4
3.4-4 Determine
Determine the
theicorriente
current
inmuestra
Figure
s – vsshown
s current
s el circuit
– sein
Determine
iiinenthe
shown
Figure
P
la
circuito
que
–
–
P
3.4-4.
P 3.4-4.
en
la figura P 3.4-4.
is
is
Voltmete
Vo
FigureFigure
P 2.5-4
P 2.5-4
P 2.5-5
P 2.5-5
(a) Find
(a) the
Findpower
the power
supplied
supplied
by thebyvoltage
the voltage
sourcesource
shownshown
in in
FigureFigure
P 2.5-5
P 2.5-5
when when
for t �
for0t we
� 0have
we have
Figura
Figure P
P 3.3-14
3.3-14
Figure
P
3.3-14
Sección 3.4 Resistores en paralelo y división
Section 3.4
3.4 Parallel Resistors and Current Division
Section
de
corriente Parallel Resistors and Current Division
i2,
P 3.4-1
3.4-1 Use
Use
current
divisiondeto
tocorriente
determine
the determinar
currents ii1,, las
current
division
determine
the
currents
P
Utilice
la división
para
1 i2,
,
and
i
in
the
circuit
shown
in
Figure
P
3.4-1.
icorrientes
3
4
in Figure
3.4-1. en la figura
i3, and i4 in
i1, the
i2, i3circuit
e i4 enshown
el circuito
que sePmuestra
P 3.4-1.
+
and and
v ¼ 2vcos
¼ 2t V
cos t V
12 V 12
– V
+
–
R
FigureFigure
P 2.6-1
P 2.6-1
Figure P
P 3.4-4
3.4-4
Figura
Figure
P
3.4-4
P 2.6-2
P 2.6-2
The current
The curre
sou
i ¼ 10i ¼
cos10t mA
cos t mA
What What
valuesvalues
do thedomet
th
(b) Determine
(b)
Determine
the
the energy
supplied
this
byvoltage
this
voltage
for for
P
3.4-5
Considere
elenergy
circuito
quesupplied
se
muestra
en
la source
figura
P 3.4-5
3.4-5
Consider
the
circuit
shown
inby
Figure
P
3.4-5
whensource
P
Consider
the
circuit
shown
in
Figure
P
3.4-5
when
period
the
0 � RR
t021�
�110
s.
P
4period
V
6t s.�
V
R2 5 the
10 V.
Seleccione
la
V�
� the
Rcuando
6V
and
¼
V.1ySelect
source
is so that
1 �
44 3.4-5
V
R
1 � 6 V and R2 ¼ 10 V. Select the source is so that
de
modo
que
v
se
mantenga
9iV y 13 V.
fuente
i
remains
between 99 V
Vo and
and 13
13 V.
V. entre
s
vvo remains
between
i
AmmetA
o
v
FigureFigure
P 2.5-5
P 2.5-5
Figure P
P 3.4-1
3.4-1
Figura
Figure
P
3.4-1
P
los circuits
circuitosshown
que seinmuestran
la figura
P 3.4-2
3.4-2 Considere
Consider the
the
Figure P
Pen
3.4-2.
P
3.4-2
Consider
circuits shown
in Figure
3.4-2.
P 3.4-2.
(a) Determine
Determine the
the value
value of
of the
the resistance
resistance R
R in
in Figure
Figure P
P 3.4-2b
3.4-2b
(a)
(a)Determine
el
valor
de la
resistencia
R de la
figura P 3.4-2b
that
makes
the
circuit
in
Figure
P
3.4-2b
equivalent
to the
the
that makes the circuit in Figure P 3.4-2b equivalent to
que
hace
el circuito
de la figura P 3.4-2b sea equivacircuit
in que
Figure
P 3.4-2a.
3.4-2a.
circuit
in
Figure
P
al circuito
de la figura
3.4-2a.
(b) lente
Determine
the voltage
voltage
in P
Figure
P 3.4-2b.
3.4-2b. Because
Because the
the
(b)
Determine
the
vv in
Figure
P
(b)D
etermine
elequivalent,
voltaje v de
lavoltage
figura Pv 3.4-2b.
Dado
que los
circuits
are
the
in
Figure
P
3.4-2a
is
circuits are equivalent, the voltage v in Figure P 3.4-2a is
circuitos
son voltage
equivalentes,
el voltaje
v de la figura P 3.2equal
to
the
v
in
Figure
P
3.4-2b.
equal to the voltage v in Figure P 3.4-2b.
es igual al
de la figura
3.4-2b.
(c) 4a
Determine
thevoltaje
powervsupplied
supplied
by P
the
current source.
source.
(c)
Determine
the
power
by
the
current
(c) Determine la potencia alimentada por la fuente de corriente.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 94
–
+
+
–
v
4Ω 4Ω
i
P 2.5-6
2.5-6
FigureFigure
P 2.5.6
P 2.5.6
showsshows
a battery
a battery
connected
connected
to a load.
to a load.+ 12+V 12 V
Figura
PP3.4-5
–
–
Figure
Pload
3.4-5
Figure
3.4-5
The P
The
in
load
Figure
in Figure
P 2.5.6
P 2.5.6
mightmight
represent
represent
automobile
automobile
head-head-
i
P 3.4-6
La
entrada
al circuito
que
se
en
figura
lights,lights,
a digital
a digital
camera,
camera,
or a cell
or aphone.
cellmuestra
phone.
The energy
Thelaenergy
supplied
supplied
P 3.4-6
3.4-6esThe
Thecorriente
input to
to the
the
circuit
shown
in Figure
Figurei .P
P 3.4-6
3.4-6
is the
the
P
detola
fuente
de
salida
de
3.4-6
input
circuit
in
is
a La
by thebyla
battery
the
battery
to load
is
load
given
isshown
given
bycorriente,
by
current
of the
theescurrent
current
source,
The
output
of this circuiti .isEste
the
Z por
Zel tamperímetro,
este
circuito
la corriente
medida
current
of
source,
iiaa.. The
t2output
2 of this circuitb is the
FigureFigure
P 2.6-2
P 2.6-2
.viThis
This
circuit
produces
an
current measured
measured
by the
the
ammeter,
bproporcional
circuito
produce una
salida
a
la
entrada,
wque
¼ es
w ii¼
dt
vi
dt
.
circuit
produces
an
current
by
ammeter,
b
output
that
is
proportional
to
the
input,
that
is,
t
t
1
1
es
decir,that is proportional to the input, that is,
output
P 2.6-3
P 2.6-3
An ideal
An ideal
voltm
ib ¼
¼ kk iiais constant
WhenWhen
the battery
the battery
voltage
and the
and
load
theresistance
load resistance
is
is
ivoltage
b is constant
a
more
realistic
more
realistic
model
mode
of a
where
isthen
theconstante
constant
ofcurrent
proportionality.
donde
kkk es
la
proporcionalidad.
fixed,
fixed,
the
then
battery
the de
battery
current
will be
will
constant
be constant
and and
where
is
the
constant
of
proportionality.
2.6-3a2.6-3a
showsshows
a circuit
a circu
wit
,
when
R
¼
24
V
and
(a)
Determine
the
value
of
the
output,
i
V and
(a) Determine the value ofwthe
output,
Þ when
¼salida,
viwðt¼
vii ðit,b1tb2,cuando
� t1 Þ RR¼52424
2 �
v
v
.
In
Figure
.
In
Figure
P
2.6-3b,
P
2.6th
(a)Determine
el
valor
de
la
V
e
m
m
b
ia ¼ 2.1
2.1 A.
A.
an ideal
an voltmeter,
ideal voltmeter,
an ope
iiaa ¼
5capacity
2.1 A.
The
The
capacity
of a battery
of a battery
is the isproduct
the product
of theofbattery
the battery
current
current
the 100-V
the 100-V
resistor,
resistor,
and th
and time
and required
time required
to discharge
to discharge
the battery.
the battery.
Consequently,
Consequently,
the the
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
battery
Figure P 2.5-3
load
Figure P 2.5-6
P 2.5-4 The current source and voltage source in the circuit
Section 2.6 Voltmeters and Ammeters
shown in Figure P 2.5-4 are connected in parallel so that they
For the circuit of Figure P 2.6-1:
both have the same voltage, vs. The current source and voltage P 2.6-1
Problems
95
Problemas
95
source are also connected in series so that they both have the (a)
What is the value of the resistance R?
same current, is. Suppose that vs ¼ 12 V and is ¼ 2 A. Calculate (b) How much power is delivered by the volta
(b)
Determine
the value
the resistance,
R, required
cause
P 3.4-9
Determine
the value
of the voltage
in Figure
P 3.4-9.
theR,power
supplied
eachPsource.
(b)
Determine
el valor
de laofresistencia,
requerida
paratoby
que
3.4-9
Determine
el valor
del voltaje
v en lavfigura
P 3.4-9.
1.5 A when
thesea
input
¼ 2 A.
the output
la salida
sea ibto5be
1.5ibA¼cuando
la salida
ia 5is 2iaA.
– . 5
+ 5 . 0
(c)
Determine
the value
the resistance,
R, required
cause
(c)
Determine
el valor
de laofresistencia,
R, requerida
paratoque
+
ib 0.4
¼ 0.4
ia ðthat
is,elthe
value
constant
proportionAmme
+
Voltmeter
ib 5
ia (es
decir,
valor
de of
la the
constante
deof
proporciovs
is
vs
–
4
nalidad
ality es
is k ¼ 10
Þ.
–
is
Figure P 2.5-4
Amperímetro
P 2.5-5
Figure
P 3.4-9
Figura
P 3.4-9
12 V
R
+
–
(a) Find the power supplied by the voltage source shown in
Figure P 2.5-5 when for
tP�3.4-10
0 weUnhave
A
solarsolar
photovoltaic
panelsemay
be represented
by the
P 3.4-10
panel
fotovoltaico
puede
representar por
P 3.4-7 Figura P 3.4-7 muestra un amplificador de transistor. v el¼ modelo
the load
circuit
model
shownque
in Figure
P 3.4-10,
RLP is3.4-10,
de circuito
se muestra
en lawhere
figura
Figure
P
2.6-1
2
cos
t
V
3.4-7seleccionar
Figure P 3.4-7
shows adetransistor
The values
HayP que
los valores
R1 y R2amplifier.
. Las resistencias
resistor.
of the resistances
R1 and
donde
RL esDetermine
el resistorthe
de values
carga. Determine
los valores
de R
las
L.
and
of
R
and
R
are
to
be
selected.
Resistances
R
and
R
are para
used to
1
2
1
2
R1 y R2 se utilizan para polarizar el transistor, es decir,
resistencias R1 y RL.
P 2.6-2 The current source in Figure P 2.6-2 su
bias thelas
transistor,
that is,
to create useful
conditions.
i ¼ 10 cos t mA
constituir
condiciones
operativas
útiles. operating
En este problema
What values do the meters in Figure P 2.6-2 re
and
R
so
that
v
¼ 5energy
V. Wesupplied by this voltage source for
In
this
problem,
we
want
to
select
R
b
the
vb 5Determine
52V. Esperamos
que
queremos seleccionar R1 y R2 para que1(b)
expect
value
of ib to be approximately
10 mA.0iWhen
theCuando
period
t �bi1, 1�s.
el valor
dethe
ib sea
aproximadamente
de 10 µA.
1 � 10i
ib as negligible,
is,suponer
to assume
10ib, it isescustomary
sea insignificante,
es that
decir,
lo habitual
tratar queto
ib treat
i
Ammeter
that
R2 comprise
a voltage
divider.
1 and
queibi ¼50.0.InEn
esecase,
caso,RR
incluye
un divisor
de voltaje.
Figura P 3.4-6
Figure P 3.4-6
1
–
+
b
(a)
Select values
for para
R1 and
somodo
that que
vb ¼vb55V,
(a)
Seleccione
los valores
R1 y R
R22 de
5 V,and
y lathe v
4Ω
R2 sea
is no
more
than
5 mW. Figure P 3.4-10
total power
absorbedpor
byRR11yand
potencia
total absorbida
R2Figure
no
de
más
de
5
mW.
P 2.5-5
(b)
inferiorinferior
transistor
could
to be
than Figura P 3.4-10
(b)
UnAn
transistor
podría
hacercause
que ibibfuera
máslarger
grande
P 3.4-11 Determine the power supplied by i the dependent
Using
the values
ofPR2.5-6
R
R2 desde
la(a),
de expected.
lo esperado.
Utilizando
los valores
de RFigure
1 and
1 2y from
+
P part
2.5.6
showssource
a battery
connected
to a load.
V la fuente dein
Figure
Pla3.4-11.
P
3.4-11
Determine
potencia
alimentada
– 12por
determine
the valueelofvalor
vb that
result
ib ¼P15de
mA. might represent
parte
(a), determine
dewould
vThe
podría
resultar
b que
load
infrom
Figure
2.5.6
automobile
headpendiente
en
la
figura
P
3.4-11.
ib 5 15 µA.
lights, a digital camera, or a cell phone. The energy supplied
by the battery to load is given by
Z t2
w¼
vi dt
V
+
2A
Figure P 2.6-2
t1
P 2.6-3 An ideal voltmeter is modeled as an o
more realistic model of a voltmeter is a large resist
2.6-3a shows a circuit with a voltmeter that measu
w ¼ viðt2 � t1 Þ
vm. In Figure P 2.6-3b, the voltmeter is replaced by
P 3.4-11
The capacity of a battery is theFigure
product
of the battery current an ideal voltmeter, an open circuit. Ideally, there i
and time required to discharge
the Pbattery.
Figura
3.4-11 Consequently, the the 100-V resistor, and the voltmeter measures vm
Figure P 3.4-7
P 3.4-12 The voltmeter in Figure P 3.4-12 measures the value
Figura P 3.4-7
of the voltage
vm. en la figura P 3.4-12 mide el valor del
P 3.4-12
El voltímetro
P 3.4-8
Determine
the value
of the current
in the circuit
P 3.4-8
Determine
el valor
de la corriente
i en el icircuito
que voltaje v .
m
(a) Determine
the value of the resistance R.
shown inenFigure
P 3.4-8.
se muestra
la figura
P 3.4-8.
Determine
the value
the powerR.supplied by the current
(a)(b)
Determine
el valor
de laofresistencia
source.
(b)Determine el valor de la potencia alimentada por la fuente
When the battery voltage is constant and the load resistance is
fixed, then the battery current will be constant and
de corriente.
Voltímetro
Figure P 3.4-12
Figura P 3.4-12
Figura
P 3.4-8
Figure
P 3.4-8
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 95
P 3.4-13
Determine
valuesdeoflas
theresistencias
resistances RR11 yand
P 3.4-13
Determine
losthe
valores
R2 R2
forelthe
circuit
shown
in Figure
3.4-13.
para
circuito
que
se muestra
en laPfigura
P 3.4-13.
Alfaomega
4/12/11 5:22 PM
the units
the units
of Am
o
The current
source
source
and voltage
and voltagewith with
bothboth
havehave
the same
the same
voltage,
voltage,
vs. The
vs. current
is is
having
a capacity
a capacity
of 8
source
source
are also
are also
connected
connected
in series
in series
so that
so they
that they
bothboth
havehave
the thehaving
a current
of 25ofmA.
25 m
(
same
same
is. Suppose
is. Suppose
that vthat
12
V12
and
Viand
isA.
¼ 3Calculate
A. Calculatea current
Figure
Figure
Pcurrent,
2.5-4
Pcurrent,
2.5-4
s ¼v
s¼
s¼3
discharge
discharge
the
battery?
the
bat
the
power
the
power
supplied
supplied
by
each
by
each
source.
source.
2.12 S U M M A R Y
Rthe
the load
during
durin
P A2.5-5
P 2.5-5
dependent
provides
asupplies
current
(or a�36
voltage)
that
is currentto thetoload
The engineer uses models, called circuit elements, to repreAnswer:
Answer:
Thesource
The
voltage
voltage
source
source
supplies
�36
W,
and
W, the
and
current
the
+
+
12 V12–V –
dependent
on
another
variable
elsewhere
in
the
circuit.
The
sent the devices that make up a circuit. In this book, we
supplies
supplies
36
W.
36
W.by the
(a)source
Find
(a) source
Find
the
power
the
power
supplied
supplied
by voltage
the voltage
source
source
shown
shown
in in
constitutive
equations
offor
dependent
sources
are summarized
Problems
Problems
47 47
consider
only
linear
elements
or
linear
models
of
devices.
A
Figure
Figure
P
2.5-5
P
2.5-5
when
when
t
for
�
0
t
�
we
0
have
we
have
96
Circuitos resistivos
in Table 2.7-1.
device is linear if it satisfies the properties of both superpoFigure
Figure
P 2.6-1
P 2.6-1
v¼
v2 ¼
cos
2 tcos
V t V is is
+ circuit
+
The short circuit and open
are special cases of
sition
and
homogeneity.
+
+
energy
energy
stored
stored
in
a
battery
in
a
battery
is
equal
is
equal
to
the
to
product
the
product
of
the
of
battery
the
battery
P 2.5-3
P 2.5-3
The The
current
current
source
source
and and
voltage
voltage
source
source
in the
in circuit
the circuit
P 3.4-17
Considere
la
combinación
de
los
resistores
que
se
vs
is vscircuit
vs isvsan–ideal
and and sources. iA
s short
independent
– voltage source
Theshown
relationship
directions
of
P 2.6-2
The The
current
curren
sou
voltage
voltage
the
andbattery
the
battery
capacity.
capacity.
The
capacity
is usually
is usually
givengivenP 2.6-2
shown
in Figure
in Figure
P between
2.5-3
P 2.5-3
are the
connected
arereference
connected
in parallel
in parallel
so that
so the
that
theythey
– The
muestran
enand
la figura
P 3.4-17.
R–pindenotará
lacapacity
resistencia
equihaving
v(t)
¼
0.
The
current
a
short
circuit
is
determined
by
i
¼
10
i
¼
cos
10
t
cos
mA
t
mA
current
and
of voltage,
a circuit
issource
important.
The
values
values
do the
do met
the
withwith
the units
the units
of Ampere-hours
of Ampere-hours
(Ah).(Ah).
A new
A new
12-V12-V
battery
batteryWhatWhat
both
both
have
have
thevoltage
same
the same
voltage,
vs. The
velement
current
current
source
and voltage
and
voltage
valente.
s. The
theDetermine
restDetermine
of the circuit.
open
circuit
an
current
source
(b)
(b)
the energy
theAn
energy
supplied
supplied
byisthis
by ideal
this
voltage
voltage
source
source
for for
voltage
polarity
marks
oneinterminal
and
the other
�.
The
having
a capacity
a capacity
of 800
of mAh
800 mAh
is connected
is connected
to a load
to a load
that draws
that draws
source
source
are
also
are also
connected
connected
series
in series
soþthat
so that
they
they
bothboth
have
have
the thehaving
having
i(t)
¼
0.
an open circuit
is determined
the Figure
period
the
period
0The
� 0Rtvoltage
� 400
1t �
s.across
1 s.Determine
(a)
Suponga
que
40
V.
el
rango
corresFigure
P
2.5-3
P
2.5-3
Figure
P 2.5-6
P 2.5-6
element
voltage
and
current
adhere
to
the
passive
convena current
of 25ofmA.
25 mA.
(a) How
(a) How
longlong
will will
it take
it take
for the
for load
the load
to to Figure
samesame
current,
current,
is. Suppose
is. Suppose
that vthat
V12
and
V iand
is A.
¼ 3Calculate
A. Calculatea current
s ¼ v12
s¼
s¼3
by the restde
of the
circuit.
pondiente
valores
deOpen
Rp. circuits and short circuits can also
tion
if
the
current
is
directed
from
the
terminal
marked
þ
to
discharge
discharge
the battery?
the battery?
(b) How
(b) How
much
energy
will will
be supplied
be supplied
the power
the power
supplied
supplied
by each
by each
source.
source.
imuch
i energy
Ammet
Am
described
asThe
special
cases
of
resistors.
A Determine
resistor
P 2.5-4
P 2.5-4
current
current
source
source
and
voltage
and voltage
source
source
in theinwith
circuit
the circuit
(b)Sbe
uponga,
enThe
cambio,
R5
0 (un
cortocircuito).
the terminal marked �.
– required
– to discharge
to
the
to
load
the
load
during
during
the
time
the
time
required
to
discharge
the
battery?
the
battery?
Section
Section
2.6
2.6
Voltm
Vo
Answer:
The
voltage
voltage
source
supplies
supplies
�36
�36
W, and
W,When
the
and current
the
(G
¼P1)
is connected
aare
short
circuit.
A
with
shown
P 2.5-4
2.5-4
are
connected
in parallel
in resistor
parallel
so that
so they
that they
valorshown
deinR
RFigure
.in 0Figure
Resistors
areThe
widely
usedsource
as
circuit
elements.
thecurrentelresistance
p¼
Figura
PAnswer:
3.4-13
v
v
source
source
supplies
supplies
36
36
W. adhere to the passive conven- (c)Suponga,
ivis
ivsopen
P 2.6-1
For the
For circ
the
conductance
G same
¼the0 same
(R
1)(un
bothboth
have
the
voltage,
The
. current
Thecircuit.
current
source
source
and voltage
and voltageP 2.6-1
enhave
cambio,
R¼
5 voltage,
circuito
abierto).
Deters. an
resistor
voltage
andW.
current
4Ω4Ω
E1C02_1 P10/23/2009
44
Figure
Figure
Pvalor
2.5-5
P also
2.5-5
An
ideal
ammeter
current
flowing
itshave
mine
elsource
de
source
are
areRalso
connected
connected
inthe
series
in
series
so that
so they
that through
they
bothboth
have
the the(a) What
3.4-14
Determine
los Ohm’s
valores law;
de lastheresistencias
R1 y Rthe
p.measures
tion, resistors
obey
voltage across
2
(a) What
is theis valu
the
is is
and
has
zero
voltage
its
terminals.
An2Calculate
ideal
uponga,
en
cambio,
que
la resistencia
equivalente
es
that
vacross
that
12
V12
and
Viand
isA.
¼
A. Calculate(b) How
samesame
current,
current,
is. Suppose
is+. Suppose
paraterminals
el circuitoofque
muestraisenrelated
la figura
3.4-13.
s ¼v
s¼
s¼2
theseresistor
to Pthe
current into the (d)Sterminals
i much
i much
(b) How
powep
+
+ +
+
PRvoltmeter
2.5-6
P 80
2.5-6
Figure
Figure
P supplied
2.5.6
Pthe
2.5.6
shows
shows
asource.
battery
aRsource.
battery
connected
connected
toand
a to
load.
a load. + 12
vvalor
veach
R
measures
voltage
across
its
terminals
has
–
–
5
V.
Determine
el
de
R.
the
power
the
power
supplied
by
each
by
+
+
p
positive terminal ias
vis ¼
delivered to a
vs power
vs
–
– V12 V
vs Ri.
vs The
s
–
–
The
The
loadload
incurrent
Figure
in Figure
P 2.5.6
Pto2.5.6
might
might
represent
represent
automobile
automobile
terminal
equal
zero.
Ideal
voltmeters
act like headopenheadresistance is p ¼ i2R ¼ v–2=R –watts.
lights,
lights,
a digital
a digital
camera,
camera,
or a or
cell
a cell
phone.
phone.
The The
energy
energy
supplied
supplied
+ 5+
circuits,
and
ideal
ammeters
act
like
short
circuits.
An independent source provides a current or a voltage
battery
battery
load load
+ given
+
byTransducers
the
by battery
the battery
to
load
to load
is given
isthat
by
byv physical
are
devices
convert
quantities,
+
+
Voltm
independent of other circuit variables. The voltage of an
is
is v
vZs
s –vs –
t2
Figure
Figure
P 2.5-3
P 2.5-3
Figure
Figure
P 2.5-6
P 2.5-6 position,–Zsto t2an
such
as
rotational
quantity such as Figure
– electrical
Figure
P 2.6-2
P 2.6-2
independent voltage source is specified, but the current is
¼wdescribe
¼ vi dtvi
dt transducers: potenvoltage. In this chapter,wwe
two
is
is
44
Circuit Elements
not. Conversely,
the current of an independent current
t1
t1
P 2.5-4
P 2.5-4
The The
current
current
source
source
and and
voltage
voltage
source
source
in the
in circuit
the circuit tiometers and temperature sensors.
P 2.6-3
P 2.6-3
An ideal
An ideal
voltm
v
source is specified whereas the voltage is not. The voltages
Section
Section
2.6
2.6
Voltmeters
Voltmeters
and
and
Ammeters
Ammeters
When
the
battery
battery
voltage
voltage
is constant
is constant
and to
the
andconnect
load
the load
resistance
resistance
is is
shown
shown
in Figure
in Figure
P 2.5-4
P 2.5-4
are connected
are connected
in parallel
in parallel
so that
so that
theytheyWhen
Figure
Figure
Pthe
2.5-4
Pwidely
2.5-4
Switches
are
used
in
circuits
and
dismore
more
realistic
realistic
model
model
of
a
Figura
P
3.4-14
of independent voltage sources and currents of independent
fixed,
fixed,
then
then
thethe
battery
the
current
current
be2.6-1:
constant
constant
2.6-1
P
2.6-1
For
For
circuit
thebattery
circuit
of
Figure
ofwill
Figure
Pwill
Pbe
2.6-1:
bothboth
have
the
the
voltage,
vs. The
vs. The
current
current
source
source
and voltage
and voltageP
connect
elements
and
circuits.
They
can
alsoand
beand
used to 2.6-3a
2.12
S sources
Uhave
Msame
Maresame
Avoltage,
RY
2.6-3a
shows
shows
a circuit
a circui
wit
current
frequently
used
as the inputs to electric
R
source
source
are
also
are
also
connected
connected
in
series
in
series
so
that
so
that
they
they
both
both
have
have
the
the
P
2.5-5
P
2.5-5
A
dependent
source
provides
a
current
(or
a
voltage)
that is v . In
ð
ð
Þ
Þ
w
¼
w
vi
¼
t
vi
t
�
t
�
t
2
2
1
1
create
discontinuous
voltages
or
currents.
(a)
What
(a)
What
is
the
is
value
the
value
of
the
of
resistance
the
resistance
R?
R?
The
engineer
uses el
models,
circuit elements,
repre. In Figure
P+ 2.6-3b,
P+2.6-3
th
P 3.4-15
Determine
valor called
de la corriente
medida topor
el
m vmFigure
circuits.
12 V 12–V –
Suppose
that up
vthat
V12
and
VIn
iand
2is A.
¼book,
2Calculate
A. Calculate
same
current,
current,
is. Suppose
is.3.4-15.
dependent
on
another
variable
elsewhere
in
the
circuit.
The
s¼
s¼
s¼
(b)
How
(b)
How
much
much
power
power
is
delivered
is
delivered
by
the
by
voltage
the
voltage
source?
source?
contador
en la
figura
P
sentsame
the
devices
that
make
av12
circuit.
this
we
an
ideal
an
ideal
voltmeter,
voltmeter,
an
ope
an
The(a)
The
capacity
capacity
of power
athe
of
battery
apower
battery
is supplied
theis product
the
product
of
of battery
the
battery
current
current
Find
(a)
Find
the
supplied
by the
by voltage
the voltage
source
source
shown
shown
in in
the
power
the power
supplied
supplied
by
each
by each
source.
source.
constitutive
equations
dependent
are
summarized
consider
only
linear
elements
or linear
models of devices. A
the 100-V
resistor,
resistor,
and th
a
and
and
time
time
required
required
to
discharge
toofdischarge
the
battery.
Consequently,
Consequently,
the thethe 100-V
Figure
Figure
P 2.5-5
P 2.5-5
when
when
forthe
t for
�battery.
0tsources
�
we
0have
we
have
Figura
3.4-172.7-1.
in PTable
device is linear if it satisfies the properties of both superpoFigure
Figure
P 2.6-1
P 2.6-1
2v cos
¼ 2tcos
V– t .V– 5 . 0 5 0
+ 5 +and
. 50 open
. 0v ¼circuit
The short circuit
are
special cases of
sition and homogeneity. + +
Amperímetro
P
3.4-18
Considere
la
combinación
de
los
resistores
que
se
Ammeter
Ammeter
and
and
+
+
Voltmeter
Voltmeter
independent sources. A short circuit is an ideal voltage source
vs vs directions of the
is is vs the
vs reference
The relationship between
–
–
P 2.6-2
P 2.6-2
The The
current
cur
muestran
en v(t)
la figura
P 3.4-18.
la resistencia
equi- by
having
¼ 0. The
currentR
indenotará
a10
is determined
– element is important. The
i p¼
i short
¼cos
10circuit
tcos
mAt mA
current and voltage of a–circuit
WhatWhat
values
values
do the
do m
valente.
is B
is other
the
of the
circuit.
circuit
isby
anthis
ideal
current
source
(b)rest
Determine
(b)
Determine
the An
energy
theopen
energy
supplied
supplied
by voltage
this
voltage
source
source
for for
O
EMS
voltage polarity marks one terminal P
þRand
theL
�. The
having
i(t)
¼
0.
The
voltage
across
an
open
circuit
is
determined
the
period
the
period
0
�
0
t
�
1
t
�
s.
1
s.
(a)
Suponga
que
50
V
R
800
V.
Determine
el
rango
coelement voltage and current adhere to the passive convenFigure
Figure
P 2.5-4
P 2.5-4
by the rest of the
and short circuits can also
rrespondiente
de circuit.
valoresOpen
de Rcircuits
tion if the current is directed from the terminal marked þ to
p.
i
Amm
that
the
model
isaslinear.
(b)
Use
the
model
toi predict
the value
described
special
cases
of
resistors.
A Determine
resistor
with
Section
2.2
Engineering
and Linear Models
(b)
Sbe
uponga,
en cambio,
R
5
0
(un
cortocircuito).
the
terminal
marked
�.
–
–
R R
P
2.5-5
P
2.5-5
A
dependent
source
provides
a
current
(or
a
voltage)
that
is
uit elements, to reprevvalor
corresponding
current
40 mA.
(c)A
Use
(Ga ¼
1) isofai ¼
short
circuit.
resistor
with
1 the1 model
de+ R
R¼
.+ 0 to
areon
widely
used
as circuit
When
the
PResistors
2.2-1 An
element
has
voltage
v andelements.
current
shown
in ofelresistance
12 V12–V p –
v
v
dependent
another
variable
elsewhere
insource
thei as
circuit.
The
2 A 2 A
cuit. In this book, we
(a)
Find
(a)
Find
the
power
the
power
supplied
supplied
by
the
by
voltage
the
voltage
source
shown
shown
in
in
to
predict
the
value
of
i
corresponding
to
a
voltage
of
v
¼
3 V.
conductance
G
¼
0
(R
¼
1)
is
an
open
circuit.
(c)
Suponga,
en
cambio,
R
5
(un
circuito
abierto).
Deterresistor
and
current
adhere
to
the
passive
convenFigure
P voltage
2.2-1a.
Values
the
current
i and
corresponding
4Ω 4Ω
constitutive
equations
ofof
dependent
sources
are
summarized
r models of devices. A Figura
Figure
Figure
Pammeter
2.5-5
P 2.5-5
Figure
Figure
P 2.5-5
Pobey
2.5-5
when
when
for
tfor
� 0t �
we
0 have
we
have
An
ideal
measures
the
current
flowing
through
.
mine
el
valor
de
R
p
tion,
resistors
Ohm’s
law;
the
voltage
across
the
P
3.4-15
voltage
v have
been tabulated as shown in Figure P 2.2-1b. Hint: Plot the data. We expect the data points to lie onitsa
in
Table
2.7-1.
erties of both superpoand
has zero
across
its terminals.
An
ideal
(d)
Sterminals
uponga,
en
cambio,
lavoltage
resistencia
Rprepre5
Figure
Figure
Pline.
2.6-1
PObtain
2.6-1
vis ¼related
v2 ¼
cos2linear.
tto
cos
V the
tV
terminals
of
the resistor
i
i
a linear
model
ofequivalente
the
elementesby
Determine
whether
the
The short
circuit
andelement
open is
circuit
are current
special into
casestheof straight
P V.
2.5-6
PDetermine
2.5-6
Figure
Figure
P 2.5.6
P voltage
2.5.6
shows
shows
a battery
a its
battery
connected
connected
to ahas
to
load.
a load. + 12+ V 12 V
voltmeter
measures
the
across
terminals
and
150
el
valor
de
R.
positive
terminal
as
v
¼
Ri.
The
power
delivered
to
a
P 3.4-16
Considere
la
combinación
de
los
resistores
que
se
senting
that
straight
line
by
an
equation.
–
–
and
and
independent sources.
A short
circuit
The
The
load
load
inThe
Figure
incurrent
Figure
P to
2.5.6
P in
2.5.6
might
represent
represent
automobile
automobile
headheadnce directions of the muestran
v, V is i,anAideal voltage source
current
equal
zero.
Ideal
voltmeters
actsupplies
like40
open
2.6-2
P 2.6-2
The
current
source
source
Figure
inmight
Figure
P 2.6-2
P 2.6-2
supplies
W.
40 W.
i v2=R
resistance
¼The
R current
¼
watts.
en v(t)
laisfigura
Pi23.4-16.
Rpindenotará
la resistencia
equi- by P terminal
having
¼p 0.
short
circuit
is
determined
i ¼ 10
i ¼acos
10
t
cos
mA
t
mA
lights,
lights,
a
digital
a
digital
camera,
camera,
or
a
or
cell
a
phone.
cell
phone.
The
The
energy
energy
supplied
supplied
ment is important. The valente.
circuits,
anddoideal
ammeters
act
circuits.
–3
–3 a current
What
What
values
values
the
do meters
the
in Figure
in like
Figure
Pshort
2.6-2
P 2.6-2
read?read?
i meters
An
independent
provides
orcurrent
a voltage
+ source
theDetermine
restDetermine
of the circuit.
open
circuit
an
source
(b)
(b)
the
energy
theAn
energy
supplied
supplied
by ideal
this
voltage
voltage
source
source
for for Transducers
by the
by battery
the battery
load
to load
is that
given
is convert
given
by byphysical quantities,
þ and the other �. The
–2
–4 byisthis
aretodevices
independent
of
other
circuit
variables.
The
voltage
of
an
+
Z t2 Z t2
having
i(t)
¼
0.
is determined
the period
the
period
0vThe
� 0Rtvoltage
� 320
1t �
s.across
1 s.Determine
Suponga
que
20
V.
el rango
corres0an open0 circuit
o the passive conven- (a)independent
such as rotational position, to an
electrical
quantity such as
Figure
Figure
P 2.6-2
P 2.6-2
voltage
source
iscircuits
specified,
but the
current
is
v,¼
V vi dt
i, vi
A dt
by
the
rest
of
the
circuit.
Open
and
short
circuits
can
also
w
¼
w
v
2
12
.
pondiente
de
valores
de
R
terminal marked þ to
p
voltage.
In
this
chapter,
we
describe
two
transducers:
poten–
i
i
not.
Conversely,
the
current
of
an
independent
current
Ammeter
Ammeter –3.6
t
t
–30
1
1
described
as special
of
resistors.
A Determine
resistor with
4
32
(b)Sbe
uponga,
en cambio,
R 5 cases
0 (un
cortocircuito).
–
tiometers and temperature
sensors.
P 2.6-3
P 2.6-3
An ideal
An idea
vo
– voltage
–
source
is specified
whereas the
is6 not. A
The
voltages
20
2.4
When
When
theare
battery
thewidely
battery
voltage
voltage
isinconstant
is constant
and
andload
the load
resistance
short circuit.
resistor
with
elresistance
valor de R
Rp¼
. 0 (G ¼ 1) is a60
elements. When the
Switches
used
circuits
to the
connect
and resistance
dis- is ismoremore
realistic
realistic
model
moo
v andv currents of independent
ofconductance
independentGvoltage
sources
Voltmeter
Voltmeter
50
6.0
fixed,fixed,
then then
the battery
the
battery
current
current
will
becan
constant
bealso
constant
andused
andto
¼ 0 (R
is an
open circuit.
Suponga, en cambio,
R¼
5 1)(un
circuito
abierto). Detero the passive conven- (c)current
(b)
connect
and
circuits.
Theywill
be
2.6-3a
2.6-3a
shows
shows
a circuit
a ci
4 Ω 4elements
Ω
sources
are(a)
frequently
used
as theflowing
inputs to
electricits
+ v+ –v –
Figure
Figure
P
2.5-5
P
2.5-5
An
ideal
ammeter
measures
the
current
through
mine
el
valor
de
R
.
ð
ð
Þ
Þ
w
¼
w
vi
¼
t
vi
t
�
t
�
t
(a)
(b)
p
e voltage across the
21
1
create discontinuous voltages or 2currents.
v
v
.
In
Figure
.
In
Figure
P
2.6-3b
P
2.
m
m
circuits.
Figure
P 2.2-1
andcambio,
has zeroque
voltage
across its terminals.
An es
ideal
uponga,
en
la resistencia
equivalente
o the current into the (d)Sterminals
icapacity
i of a of
an
ideal
an
ideal
voltmeter,
voltmete
an
The
The
capacity
battery
a
battery
is
the
is
product
the
product
of
the
of
battery
the
battery
current
current
Figure
+
+P 2.2-2
PRvoltmeter
2.5-6
P 2.5-6
Figure
Figure
P 2.5.6
Pthe
2.5.6
shows
adebattery
a battery
connected
connected
toand
a to
load.
a load.
voltage
across
its
terminals
has
Determine
elshows
valor
R.
2
A battery.
2the
A battery.
12
V12time
V required
p 5 40 V.measures
power delivered to a
–and
–time
the 100-V
resistor,
resisto
an
and
required
to discharge
to discharge
the
Consequently,
Consequently,
the thethe 100-V
PThe
2.2-2
linear
element
voltage
vvoltmeters
and current
as shown
The
loadA
load
incurrent
Figure
in Figure
P 2.5.6
Ptohas
2.5.6
might
might
represent
represent
automobile
automobile
headterminal
equal
zero.
Ideal
acti like
openheadFigura P 3.4-18
P 2.2-3 A linear element has voltage v and current i as shown
in
Figure
Pand
Values
the
current
i andThe
corresponding
lights,
lights,
a digital
a2.2-2a.
digital
camera,
camera,
or of
a or
cell
a cell
phone.
The
energy
energy
supplied
supplied
circuits,
ideal
ammeters
act
likephone.
short
circuits.
current or a voltage
in Figure
2.2-3a. al
Values
of the
i and en
corresponding
voltage
v have
been
as
shown
in Figurequantities,
P 2.2-2b. P 3.4-19
byTransducers
the
by battery
the
battery
to
load
totabulated
load
is given
isthat
given
by
by physical
LaP entrada
circuito
quecurrent
se muestra
la figura
are
devices
convert
es. The voltage of an
Z t2 that expresses v as a P 3.4-19
voltage
vlahave
beendetabulated
asis.shown
in Figure
P 2.2-3b.
Represent
the element
by anZtoequation
t2an
es
corriente
la
fuente,
La
salida
es
la
corriente
such
as
rotational
position,
electrical
quantity
such
as
Figure
Figure
P 2.6-2
P 2.6-2
fied, but the current is
w
¼
w
¼
vi
dt
vi
dt
Represent
the
element
by
an
equation
that
expresses
function
of
i.
This
equation
is
a
model
of
the
element.
(a)
Verify
medida por el contador, io. Un divisor de corriente conectav laas a
voltage. In this chapter, we describePtwo
independent current
R Otransducers:
B L E M S potent1
t1
fuente
alP medidor.
Dadas
las
observaciones
siguientes:
tiometers and temperature sensors.
P 2.6-3
2.6-3
An ideal
An
ideal
voltmeter
voltmeter
is modeled
is modeled
as anasopen
an open
circuit.
circuit.
A A
ge is not. The voltages
When
When
the battery
the
battery
voltage
voltage
is constant
is constant
and to
the
andconnect
load
the load
resistance
resistance
is is
Switches
are
widely
used
in
circuits
and
dismore
more
realistic
realistic
model
model
of
a
voltmeter
of
a
voltmeter
is
a
large
is
a
large
resistance.
resistance.
Figure
Figure
P P
urrents of independent
(a) La entrada is 5 5 A hace que la salida sea io 5 2 A.
fixed,
fixed,
thenelements
then
the battery
the battery
current
willThey
will
be constant
be
constant
connect
andcurrent
circuits.
can
alsoand
beand
used to (b)that
2.6-3a
2.6-3a
shows
circuit
a la
circuit
with
aUse
voltmeter
athe
voltmeter
thatW.
measures
that
measures
the
the
voltage
the
model
linear.
(b)with
model
to predict
thevoltage
value
Section
2.2 Engineering
and Linear
Models
s the inputs to electric
Cuando
is shows
5a2is
A,
fuente
alimenta
48
ðt2or
¼wvi¼
vi�ðtcurrents.
t21 Þ� t1 Þ
create discontinuous w
voltages
vmv. In
vmFigure
. In Figure
P 2.6-3b,
Pto2.6-3b,
voltmeter
the of
voltmeter
is replaced
the
by
model
the
model
of of
of
corresponding
a the
current
i ¼is40replaced
mA.
(c)by
Use
the
model
P 2.2-1 An element has voltage v and current i as shown in Determine
los
valores
de
las
resistencias
R
R2. there
Figura
3.4-16
1 aythere
anpredict
ideal
an ideal
voltmeter,
voltmeter,
an
open
an
open
circuit.
circuit.
Ideally,
Ideally,
is no
iscurrent
no
current
in in
The PThe
capacity
capacity
of a of
battery
a battery
is theis product
the product
of the
of battery
the battery
current
current
to
the
value
of
i
corresponding
to
voltage
of
v
¼
3
V.
Figure P 2.2-1a. Values of the current i and corresponding
12=V,12the
V, the
the 100-V
the 100-V
resistor,
resistor,
and the
and voltmeter
the voltmeter
measures
measures
vmi =vmi
and and
timetime
required
required
to discharge
to discharge
the battery.
the battery.
Consequently,
Consequently,
the the
the dataEléctricos
points to -lie
on a
voltage v have been tabulated as shown in Figure P 2.2-1b. Hint: Plot the data. We expect
Alfaomega
Circuitos
Dorf
straight line. Obtain a linear model of the element by repreDetermine whether the element is linear.
senting that straight line by an equation.
44
Circuit Elements
+
+
v
96
+
+
+
ROBLEM
S
M03_DORF_1571_8ED_SE_053-107.indd
+
+
i
v, V
i, A
–3
–4
0
–3
–2
0
i
+
4/12/11 5:22 PM
Problemas
Amperímetro
Figura P 3.4-19
97
(a) Determine el valor de la resistencia R en la figura P 3.6-1b
que hace que el circuito en la figura P 3.6-1b sea equivalente al circuito de la figura P.3.6-1a.
(b) Encuentre la corriente i y el voltaje v que se muestran en la
figura P 3.6-1b. Por la equivalencia, la corriente i y el voltaje
v que se muestran en la figura P 3.6-1a son iguales a la corriente i y al voltaje v que se muestran en la figura P 3.6-1b.
(c) Encuentre la corriente i2 que se muestra en la figura P 3.6-a,
utilizando la división de corrientes.
Sección 3.5 Fuentes de voltaje en serie y fuentes
de voltaje en paralelo
P 3.5-1 Determine la potencia alimentada por cada fuente en
el circuito que se muestra en la figura P 3.5-1.
Figura P 3.5-1
P 3.5-2 Determine la potencia alimentada por cada fuente en
el circuito que se muestra en la figura P 3.5-2.
Figura P 3.5-2
P 3.5-3 Determine la potencia recibida por cada resistor en el
circuito que se muestra en la figura P 3.5-3.
Figura P 3.6-1
P 3.6-2 El circuito que se muestra en la figura P 3.6-2a ha
sido dividido en tres partes. En la figura P 3.6-2b, la parte extrema derecha ha sido reemplazada con un circuito equivalente. El resto del circuito no se ha modificado. El circuito se ha
modificado más en la figura 3.6-2c. Ahora las partes intermedia y extrema derecha han sido reemplazadas por una resistencia equivalente única. La parte extrema izquierda permanece
sin sufrir modificaciones.
(a) Determine el valor de la resistencia R1 en la figura P 3.6-2b
que hace que el circuito en la figura p 3.6-2b sea equivalente al circuito de la figura P 3.6-2a.
(b) Determine el valor de la resistencia R2 en la figura P 3.6-2c
que hace que el circuito de la figura P 3.6-2c sea equivalente al circuito en la figura P 3.6-2b.
(c) Encuentre la corriente i1 y el voltaje v1 que se muestran en
la figura P 3.6-2c. Por la equivalencia, la corriente i1 y el
voltaje v1 que se muestran en la figura P 3.62b son iguales
a la corriente i1 y al voltaje v1 que se muestran en la figura
P 3.6-2c.
Sugerencia: 24 ⫽ 6(i1 ⫺2) ⫹ i1R2
Figura P 3.5-3
Sección 3.6 Análisis de circuitos
P 3.6-1 El circuito que se muestra en la figura P 3.6-1a ha
sido dividido en dos partes. En la figura P 3.6-1b, la parte del
lado derecho ha sido reemplazada con un circuito equivalente.
La parte izquierda del circuito no ha sido modificada.
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 97
(d) Encuentre la corriente i2 y el voltaje v2 que se muestran en
la figura P 3.6-2b. Por la equivalencia, la corriente i2 y el
voltaje v que se muestran en la figura P 3.6-2a son iguales
a la corriente i2 y el voltaje v2 que se muestran en la figura
P 3.6-2b.
Sugerencia: Utilice la división de corrientes para calcular
i2 a partir de i1.
(e) Determine la potencia absorbida por la resistencia de 3-⍀
que se muestra a la derecha de la figura P 3.6-2a.
Alfaomega
6/9/11 11:27 AM
98
Circuitos resistivos
Figura P 3.6-4
P 3.6-5 El voltímetro en el circuito que se muestra en la figura P 3.6-5 muestra que el voltaje a través del resistor de 30-V
es de 6 voltios. Determine el valor de la resistencia R1.
Sugerencia: Utilice la división de voltaje dos veces.
Respuesta: R1 5 40 V
Voltímetro
Figura P 3.6-2
P 3.6-3 Encuentre i, utilizando las reducciones de circuito
apropiadas y el principio del divisor de corriente para el circuito de la figura P 3.6-3.
Figura P 3.6-5
P 3.6-6 Determine los voltajes va y vc y las corrientes ib e id
para el circuito que se muestra en la figura P 3.6-6.
Respuesta: va 5 22 V, vc 5 6 V, ib 5 216 mA e id 5 2 mA
Figura P 3.6-3
P 3.6-4
(a)Determine los valores de R1 y R2 en la figura P 3.6-4b que
hacen que el circuito en la figura P 3.6-4b sea equivalente
al circuito de la figura 3.6-4a.
(b)Analice el circuito en la figura P 3.6-4b para determinar
los valores de las corrientes ia e ib.
(c)Puesto que los circuitos son equivalentes, las corrientes ia
e ib que se muestran en la figura P 3.6-4b son iguales a las
corrientes ia e ib que se muestran en la figura p 3.6-4a. Con
base en este hecho, determine los valores del voltaje v1 y
la corriente i2 que se muestran en la figura P 3.6-4a.
Figura P 3.6-6
P 3.6.-7 Determine el valor de la resistencia R en la figura
P 3.6-7.
Respuesta: R 5 28 kV
Figura P 3.6-7
P 3.6-8 La mayoría hemos experimentado los efectos de un
suave choque eléctrico. Pero los efectos de un choque eléctrico fuerte pueden ser devastadores e incluso fatales. El choque
es el resultado del paso de la corriente a través del cuerpo.
Una persona puede ser modelada como una red de resistencias. Considere el circuito modelo que se muestra en la figura
P 3.6-8. Determine el voltaje desarrollado a través del corazón
y la corriente que fluye a través del corazón de la persona que
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 98
Circuitos Eléctricos - Dorf
4/12/11 5:22 PM
Problems
Problemas
Problems
99
99 99
the voltage developed across the heart and the current flowing P 3.6-12 The ohmmeter in Figure P 3.6-12 measures the
sostiene
con firmeza
terminal
dewhen
una fuente
de firmly
voltajegrasps
y
ohmímetro
laoffigura
PP3.6-12
mide
la resisthe resistor
circuit.
The
value
through
heartuna
of across
the
person
he the
or she
equivalent
resistance,
Rde
eq,in
the
voltagethe
developed
the heart
and
current
flowing PP3.6-12
3.6-12ElThe
ohmmeter
Figure
3.6-12
measures
theof
la otra
terminal
está
conectada
al
suelo.
El
corazón
está
retencia
equivalente,
R
,
del
circuito
del
resistor.
El
valor
dethe
eq R , ofRthe
the equivalent
resistance,
one endthe
of aheart
voltage
source
whose
other
is connected
to the equivalent
,
depends
on
the
value
of
eq
resistor
circuit.
The
value
of
through
of the
person
when
he end
or she
firmly grasps
resistance,
eq
presentado
por Rh. El
suelo tiene una
alhas
flujo
de la tola resistencia
equivalente, Req, depende del valor de la resisfloor.
is represented
by other
Rresistencia
resistance
R.
resistance
h. The
equivalent
resistance, Req, depends on the value of the
one
endThe
of aheart
voltage
source whose
endfloor
is connected
to the the
corriente
igual a R
estáperson
de pie,isdescalza
sobre
el ontencia R.
f, y la
current
topersona
Rf, and by
the
standing
barefoot
R.
floor.
Theflow
heartequal
is represented
Rh. The floor
has resistance
to resistance
(a) Determine
the value of the equivalent resistance, Req,
suelo.
Este
tipo
de
accidente
podría
ocurrir
en
una
alberca
o
the floor.
type
at a swimming
current
flowThis
equal
toof
Rfaccident
, and the might
personoccur
is standing
barefoot pool
on (a)
Determine
elthe
valor
de ofla the
resistencia
equivalente,
when
R
¼
9
V.value
(a)
Determine
equivalent
resistance, RReqeq, ,
en unordesembarcadero.
La resistencia
Ru de laRparte
andsuperior
lower-body
boat This
dock.type
Theofupper-body
resistance
the floor.
accident might
occur at auswimming
pool (b)
cuando
R¼
599V.
V. value of the resistance R required to cause
Determine
the
when
R
del cuerpo
y la R
resistencia
RL person
de la parte
inferior varían de una
varyupper-body
from
to person.
LThe
orresistance
boat dock.
resistance
Ru and lower-body (b)
etermine
elthe
valor
de la
the equivalent
resistance
be RReqrequerida
¼R12
V. paratohacer
(b)DDetermine
value
ofresistencia
the to
resistance
required
cause
persona a otra.
resistance RL vary from person to person.
que
la
resistencia
equivalente
sea
R
5
12
V.
eq
the equivalent resistance to be Req ¼ 12 V.
Ohmímetro
L
Figure P 3.6-8
Figura P 3.6-8
Figure P 3.6-8
Figura
P 3.6-12
Figure
P 3.6-12
P 3.6-9 Determine the value of the current i in Figure 3.6-9.
Figure P 3.6-12
P 3.6-9 Determine el valor de la corriente i en la figura 3.6-9.
P
3.6-13
Encuentre
las terminales
en laPfigura
PAnswer:
3.6-9 Determine
the
value
of
the
current
i
in
Figure
3.6-9.
P 3.6-13
Find thelaRReqeq atenterminals
a-b ina-b
Figure
3.6-13.
i ¼ 0.5 mA
P
3.6-13.
También
determine
i,
i
e
i
.
Respuesta: i 5 0.5 mA
1
2
Also
determine
i,
i
,
and
i
.
1
2
P 3.6-13 Find the Req at terminals a-b in Figure P 3.6-13.
Answer: i ¼ 0.5 mA
Also
determine
. i1 ¼ 5=3 A, i2 ¼ 5=2 A
1, and
Answer:
Req ¼i,8 iV,
i ¼ 5i2A,
Respuesta:
Answer: Req ¼ 8 V, i ¼ 5 A, i1 ¼ 5=3 A, i2 ¼ 5=2 A
Figure P 3.6-9
Figura P 3.6-9
Figure P 3.6-9
P 3.6-10 Determine the values of ia, ib, and vc in Figure
P 3.6-10
Determine los valores de ia, ib y vc en la figura
P3.6-10
3.6-10.
P
Determine the values of ia, ib, and vc in Figure
P 3.6-10.
P 3.6-10.
Figura
P 3.6-13
Figure
P 3.6-13
Figure P 3.6-10
Figura
P 3.6-10
Figure
P 3.6-10
Figure P 3.6-13
P 3.6-14 Todas las resistencias en el circuito que se muestra
P 3.6-14 All of the resistances in the circuit shown in Figure
en la figura P 3.6-14 son múltiplos de R. Determine el valor
3.6-14 All
are of
multiples
of R. Determine
theshown
value of
R.
P 3.6-11 Find i and Req a-b if vab ¼ 40 V in the circuit of dePP
3.6-14
the resistances
in the circuit
in Figure
R.
P 3.6-11
Encuentre
i
y
R
si
v
5
40
V
en
el
circuito
de
Figure
P
3.6-11.
P 3.6-14 are multiples of R. Determine the value of R.
P 3.6-11 Find i and Req
eq a-b
a-b if vab
ab ¼ 40 V in the circuit of
la figura
PP3.6-11.
Figure
3.6-11.
Answer:
Req a�b ¼ 8 V, i ¼ 5=6 A
Answer: Req a�b ¼ 8 V, i ¼ 5=6 A
Respuesta:
Figure
P 3.6-14
Figura
P 3.6-14
Figure P 3.6-14
Figure P 3.6-11
Figura
P 3.6-11
Figure
P 3.6-11
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 99
P 3.6-15
circuit
shown
in FigureenP la
3.6-15
contains
seven
P 3.6-15
El The
circuito
que
se muestra
figura
P 3.6-15
each
havingshown
resistance
R.con
ThePuna
input
to contains
this circuit
is the
contiene
siete
resistores,
cada inuno
resistencia
R.seven
La
Presistors,
3.6-15
The
circuit
Figure
3.6-15
voltage
source
voltage,
v
.
The
circuit
has
two
outputs,
v
and
entrada
a
este
circuito
es
el
voltaje
de
la
fuente
de
voltaje,
v
s
a
s. vb.
resistors, each having resistance R. The input to this circuit is the
Express
each voltage,
output
asvsa. The
function
the two
input.
vb. of
Exprese
cada
salida
Elvoltage
circuito
tiene
dos salidas,
va ycircuit
source
has
outputs,
va como
and vb.
una
función
deoutput
la entrada.
Express
each
as a function of the input.
Alfaomega
4/12/11 5:23 PM
10 0
Circuitos resistivos
Figura P 3.6-15
P 3.6-16 El circuito que se muestra en la figura P 3.6-16 contiene tres resistores 10-V, 1>4 W. (Los resistores de 1>4 de watt
pueden disipar 1>4 de watt con toda seguridad.) Determine el
rango de los voltajes de la fuente de voltaje, vs, de modo que
ninguno de los resistores absorba más de 1>4 w de potencia.
Figura P 3.6-16
P 3.6-17 Los cuatro resistores que se muestran en la figura
P 3.6-17 representan indicadores de tensión. Los indicadores
de tensión son transductores que miden la tensión resultante
cuando un resistor es estirado o comprimido. Los indicadores
de tensión se emplean para medir fuerzas, desplazamientos
o presiones. Los cuatro indicadores de tensión en la figura
P 3.6-17 tienen cada uno una resistencia nominal (sin tensiones) de 200 V y cada uno absorbe sin peligro 0.5 mW. Determine el rango de los voltajes de la fuente de voltaje, vs,
de modo que ningún indicador de tensión absorba más de
0.5 mW de potencia.
200 Ω
+
200 Ω
–
vs
200 Ω
+
Figura P 3.6-18
P 3.6-19 Determine los valores de v1, v2, i3 v4, v5 e i6 en la
figura P 3.6-19.
vo
−
200 Ω
Figura P 3.6-17
P 3.6-18 El circuito que se muestra en la figura P 3.6-18b se
ha obtenido del circuito que se muestra en la figura P 3.6-18a
por el reemplazo de las combinaciones de resistencias en serie
y en paralelo por resistencias equivalentes.
(a)Determine los valores de las resistencias R1, R2 y R3 en
la figura P 3.6-18b de modo que el circuito en la figura
P 3.6-18b sea equivalente al circuito que se muestra en
la figura P 3.6-18a.
(b) Determine los valores de v1, v2 e i en la figura P 3.6-18b.
(c)Como los circuitos son equivalentes, los valores de v1, v2
e i en la figura P 3.6-18a son iguales a los valores de v1, v2 e
i en la figura P 3.6-18b. Determine los valores de v4, i5, i6
y v7 en la figura P 3.6-18a.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 100
Figura P 3.6-19
P 3.6-20 Determine los valores de i, v y Req para el circuito
que se muestra en la figura 3.6-20, dado que vab 5 18 V.
Circuitos Eléctricos - Dorf
4/12/11 5:23 PM
Problemas
101
por el contador, vo. Muestre que la salida de este circuito es
proporcional a la entrada. Determine el valor de la constante
de proporcionalidad.
s
Voltímetro
o
Figura P 3.6-20
P 3.6-21 Determine el valor de la resistencia R en el circuito
que se muestra en la figura P 3.6-21, dado que Req 5 9 V.
Respuesta: R 5 15 V
Figura P 3.6-24
P 3.6-25 La entrada al circuito en la figura P 3.6-25 es el voltaje de la fuente de voltaje, vf. La salida es la corriente medida
por el contador, io. Muestre que la salida de este circuito es
proporcional a la entrada. Determine el valor de la constante
de proporcionalidad.
s
Amperímetro
Figura P 3.6-21
o
P 3.6-22 Determine el valor de la resistencia R en el circuito
que se muestra en la figura P 3.6-22, dado que Req 5 40 V.
Figura P 3.6-25
Figura P 3.6-22
P 3.6-26 Determine el voltaje medido por el voltímetro en el
circuito que se muestra en la figura P 3.6-26.
P 3.6-23 Determine los valores de r, la ganancia de la CCVS,
y de g, la ganancia de la VCCS, para el circuito que se muestra
en la figura P 3.6-23.
Voltímetro
Figura P 3.6-23
P 3.6-24 La entrada al circuito en la figura P 3.6-24 es el voltaje de la fuente de voltaje, vs. La salida es el voltaje medido
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 101
Figura P 3.6-26
Alfaomega
4/12/11 5:23 PM
102
Circuitos resistivos
P 3.6-27 Determine la corriente medida por el amperímetro
en el circuito que se muestra en la figura P 3.6-27.
Amperímetro
Amperímetro
Figura P 3.6-29
P 3.6-30 El ohmímetro de la figura P 3.6-30 mide la resistencia equivalente del circuito de resistores conectado a los
probadores del medidor.
Figura P 3.6-27
P 3.6-28 Determine el valor de la resistencia R que hace que
el voltaje medido por el voltímetro en el circuito que se muestra en la figura P 3.6-28 sea 6 V.
(a)Determine el valor de la resistencia R requerido para que
la resistencia equivalente sea Req 5 12 V.
(b)Determine el valor de la resistencia equivalente cuando
R 5 14 V.
Ohmímetro
Figura P 3.6-30
P 3.6-31 El voltímetro en la figura P 3.6-31 mide el voltaje a
través de la fuente de corriente.
Voltímetro
(a)Determine el valor del voltaje medido por el contador.
(b)Determine la potencia alimentada por cada elemento del
circuito.
Voltímetro
Figura P 3.6-28
P 3.6-29 La entrada al circuito que se muestra en la figura
P 3.6-29 es el voltaje de la fuente de voltaje, vf. La salida es la
corriente medida por el contador, im.
(a)Suponga que vs 5 15 V. Determine el valor de la resistencia R que hace que el valor de la corriente medida por el
contador sea im 5 12 A.
(b)Suponga que vs 5 15 V y R 5 80 V. Determine la corriente medida por el amperímetro.
(c)Suponga que R 5 24 V. Determine el valor del voltaje de
entrada, vs, que hace que el valor de la corriente medida
por el contador sea im 5 3 A.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 102
Figura P 3.6-31
P 3.6-32 Determine la resistencia medida por el ohmímetro
en la figura P 3.6-32.
Ohmímetro
Figura P 3.6-32
Circuitos Eléctricos - Dorf
4/12/11 5:23 PM
Problems
P 3.6-33 Determine the resistance measured by the ohmmeter in Figure P 3.6-33.
103
P 3.6-36 Consider the circuit shown in Figure P 3.6-36.
Given
Problems
Problems
103 103
2
1
3Problems
Problems103
103
Problems
103
103103
Problems
103 103
Problems
103
vProblems
v2 ;Problems
v2 ¼ vs ; i3 ¼ i1 ; andProblemas
Problems
103
4 ¼Problems
103
3
5
8 Problems103
P 3.6-33
P 3.6-33
Determine
Determine
thethe
resistance
resistance
measured
measured
by by
thethe
ohmmeohmmeP 3.6-36
P
3.6-36
Consider
Consider
thethe
circuit
shown
shown
in
Figure
Figure
P 3.6-36.
P 3.6-36.
determine
the
values
ofcircuit
Rthe
, Rcircuit
and
R
. in
P3.6-33
P3.6-33
Determine
Determine
thethe
resistance
resistance
measured
measured
by
by
thethe
ohmmeohmmeP 3.6-36
P3.6-36
Consider
Consider
shown
in
Figure
P3.6-36.
1the
2,circuit
4shown
PP
3.6-33
3.6-33
Determine
Determine
the
theresistencia
resistance
resistance
measured
measured
by
by
the
ohmmeohmmePP
3.6-36
3.6-36
Consider
Consider
the
thecircuito
circuit
circuit
shown
shown
in
in
Figure
Figure
3.6-36.
P3.6-33
3.6-33
Determine
the
resistance
the
resistance
measured
measured
by
the
by
ohmmethe
ohmmeP3.6-36
3.6-36
Consider
Consider
the
the
circuit
circuit
shown
shown
ininFigure
Figure
in
Figure
PP3.6-36.
3.6-36.
P3.6-36.
3.6-36.
3.6-33
Determine
la
medida
por
elthe
ohmímetro
3.6-36
Considere
el
que
se
muestra
enPFigure
la3.6-36.
PPPter
3.6-33
Determine
the
resistance
measured
by
the
ohmmePPPGiven
3.6-36
Consider
the
circuit
shown
in
Figure
PPP
3.6-36.
PPFigure
3.6-33
Determine
the
resistance
measured
by
the
ohmmePP 3.6-36
Consider
the
circuit
shown
in
Pfigura
3.6-33
Determine
the
resistance
measured
by
the
ohmme3.6-36
Consider
the
circuit
shown
in
Figure
3.6-36.
ter
in
in
Figure
PFigure
3.6-33.
PDetermine
3.6-33.
Given
P
3.6-33
Determine
the resistance
measured
by the ohmmeP 3.6-36
Consider
the circuit
shown
in Figure
P 3.6-36.
ter
ter
in
in
Figure
P
P
3.6-33.
3.6-33.
Given
Given
ter
ter
in
in
Figure
Figure
P
P
3.6-33.
3.6-33.
Given
Given
en
la
figura
3.6-33.
P
3.6-36.
Dado
ter
in
ter
Figure
in
Figure
P
3.6-33.
P
3.6-33.
Given
Given
ter
in
Figure
P
3.6-33.
Given
ter
in
Figure
P
3.6-33.
Given
ter in Figure
P 3.6-33.
Given
Hint: Given
Interpret 2v22¼ 232vs2; 1i3 1¼ 15 1i11; and 3v4 3¼ 38 v323 as current
ter in Figure
P 3.6-33.
iv3 s2v¼
;i¼
; and
v¼
¼
vi2; ¼
iand
v 3; v23v; 2 ;
v v¼2 ¼
22vv222sv;¼
33v342v;¼
11
;11i1i1;1i35;¼
¼
and
; 4;and
i1v1;v14;iand
22s3¼i¼
31
4;32;¼¼
and voltagev2vv2division.
¼
v8v¼
v222¼
v3v¼
;¼
¼
v4v43¼
¼
and
v4v1i41¼
¼¼
sv;vs;;iiv
3i3
2vv
2; v8v
;s32¼
i5isv3¼
and
v8and
;3; v2 ;
i5and
v3v22¼
si1i;1;i;3and
33v
¼
¼
and
s3¼
1iy1;5
2 28
3333vvss23;23¼
8888vv2248;48¼
and
3v355s55; i1i135; 5¼
5iv144; ¼
8v2 ;
3
5
8
determine
determine
thethe
values
values
ofvalues
of
R1,RR
,R
and
,, 1and
R
.R
1,2of
21R
42and
4. R4R
determine
determine
the
the
values
of
R
R
,
R
,
,
and
.
.
Ohmímetro
2
4
determine
los
de
R
R22,14,.and
determine
determine
the
thevalores
values
values
of
RR
,,,,RR
,2,y,and
and
RR
determine
determine
thevalues
values
theof
values
R
and
R4.4. R4.
determine
the
values
of
RR2122of
and
11
determine
the
of
R
1of
Figure P 3.6-33
determine
the
values
of
RR
RR4424.4.,..Rand
11, R
determine
the values
of21, Rand
1, R2, and R4.
2 2
1 1
3 3
2
2
1
1
3 current
Hint:
Hint:
Interpret
Interpret
v
v
¼
¼
v
;
v
i
;
¼
i
¼
i
;
i
and
;
and
v
v¼
v248vv¼
as3as
2 2 322v2s23v¼
4yand
Hint:
Hint:
Interpret
Interpret
v33;¼
;1is1v1i11;i35;¼
i;¼
i51;v41;iand
;¼
vcurrent
asas
current
current
Sugerencia:
Interprete
como
8¼
1¼
3vcurrent
2v2s;3i2¼
s52v
31i1;and
1viand
4323¼
2v
3v
5i¼
8v
82current
2v¼
Hint:
Hint:
Interpret
Interpret
v
v
¼
¼
v
;
i
¼
i
and
¼
as
Hint:
Hint:
Interpret
Interpret
v
¼
¼
;
v43838v4338v¼
¼
as
current
as
current
Hint:
Interpret
v
¼
v
;
i
¼
i
;
and
v
¼
vv2v¼
2
2
s
s
3
3
1
1
4
4
Hint:
Interpret
v
v
;
i
¼
i
;
and
v
vas
as
current
2
2
s
3
s
3
1
1
P 3.6-34 Consider the circuit shown in Figure P 3.6-34. and
2
s
3
1
4
222v84as
Hint:
Interpret
v
¼
and
23333 vs ;3 3i3s 2¼
1 1 5v4 ¼and
2 23current
3v35555;i1i ;5 5¼
8as
8current
2
and
voltage
voltage
division.
division.
8
Hint:
Interpret
v
¼
i
;
and
v
¼
v2 2 as current
divisiones
de
corriente
y
voltaje.
and
and
voltage
voltage
division.
division.
2
s 3
1
4
3
5
8
and
andand
voltage
voltage
division.
division.
and
and
voltage
voltage
division.
division.
and
voltage
division.
voltage
division.
Given the values of the following currents and voltages:
and
voltage
division.
and voltage division.
Figure
Figure
PFigure
3.6-33
P i3.6-33
Figure
P
P3.6-33
3.6-33
¼3.6-33
0:625
A; v2 ¼ �25 V; i3 ¼ �1:25 A;
Figura
3.6-33
Figure
Figure
PP
3.6-33
3.6-33
Figure
P 3.6-33
Figure
PPPFigure
3.6-33
Figure
P1P3.6-33
Figure
3.6-33
Figure
P 3.6-33
and v4 ¼ �18:75 V;
PP3.6-34
P3.6-34
3.6-34
Consider
Consider
the
circuit
circuit
shown
shown
inshown
in
Figure
Figure
PFigure
3.6-34.
P figura
3.6-34.
Considere
elthe
circuito
que
seshown
muestra
en
la
P 3.6-34
P3.6-34
3.6-34
Consider
Consider
the
circuit
in
Figure
P 3.6-34.
P3.6-34.
Figure P 3.6-36s
Rcircuit
R
andin
RFigure
.ininFigure
theConsider
values
ofthe
Rthe
PPPPdetermine
3.6-34
3.6-34
Consider
Consider
the
the
circuit
circuit
shown
in
Figure
PPPP
3.6-34.
3.6-34.
Pvalues
3.6-34
Consider
Consider
the
the
circuit
circuit
shown
shown
Figure
in
Figure
P3.6-34.
3.6-34.
3.6-34
Consider
the
circuit
shown
in
Figure
3.6-34.
1,circuit
2,shown
3,shown
4
PPthe
3.6-34
in
PP 3.6-34.
3.6-34
Consider
the
circuit
shown
in
Figure
3.6-34.
Given
the
values
of
of
the
the
following
following
currents
currents
and
and
voltages:
voltages:
P Given
3.6-23.
Dados
los
valores
de
las
corrientes
y
voltajes
siP
3.6-34
Consider
the
circuit
shown
in
Figure
P 3.6-34.
Given
Given
the
the
values
values
of
of
the
the
following
following
currents
currents
and
and
voltages:
voltages:
Given
Given
the
the
values
values
of
the
the
following
currents
currents
and
andand
voltages:
voltages:
Given
Given
thevalues
values
theof
values
the
offollowing
following
the following
currents
currents
and
and
voltages:
voltages:
Given
the
values
of
the
following
currents
and
voltages:
Given
the
ofoffollowing
the
currents
voltages:
Given
the
values
of
the
following
currents
and
voltages:
guientes:
Given
the values
of the following
currents
and voltages: P 3.6-37 Consider the circuit shown in Figure P 3.6-37. Given
¼i10:625
A;
A;
v
v
¼
�25
¼
�25
V;
V;
i
¼
i
�1:25
¼
�1:25
A;
A;
i1 ¼
i1 0:625
2
2
3
3
¼
¼
0:625
0:625
A;
A;
v
v
¼
¼
�25
�25
V;
V;
i
i
¼
¼
�1:25
�1:25
A;
A;
i
2¼2v V;
3¼3i �1:25
¼
0:625
A;
A;0:625
vvvA;
¼
�25
V;
ii3i�25
¼
�1:25
A;
A;
ii1i1i11¼
0:625
¼
A;
�25
¼
i�1:25
V;
�1:25
A; A;
i0:625
¼
A;
¼¼
�25
V;
¼
A;
2v2A;
3¼
¼¼1i0:625
vv2�25
¼
�25
V;
i3�1:25
¼
�1:25
i10:625
10:625
3�1:25
3 ¼
3i3V;
2
2
4
¼
V;
¼
A;
22 ¼
and
v4�25
v¼
�18:75
¼
�18:75
V;
V;A;
4and
0:625
A;2�25
v2 2and
¼
V;
i¼
¼
�1:25
A;
i1 1¼A;
and
v�18:75
¼
�18:75
�18:75
V;
V;
4v
4v3 �18:75
Figura P 3.6-36 i2 ¼ is ; v3 ¼ v1 ; and i4 ¼ i2 ;
and
and
v
v
¼
¼
�18:75
V;
V;
and
and
v
¼
¼
�18:75
and
v
¼
�18:75
V;
4
4
and
v
¼
�18:75
V;
4
4
4
y
and
v
¼
4 �18:75 V; V; V;
4
5
3
5
Figure
Figure
P
3.6-36
P
3.6-36
and
v
¼
�18:75
V;
,RR21R,,31,R
and
,, 2and
.R4.4 R4R. 4.
determine
determine
thethe
values
values
ofvalues
of
R1,RR
Figure
P P3.6-36
3.6-36
1,2of
3R
43and
R, R
,and
determine
determine
thethe
values
3R
Figure
Figure
PFigure
3.6-36
3.6-36
Figure
3.6-36
P
3.6-36of R , R , and R .
Figure
PPPFigure
3.6-36
Figure
PPthe
3.6-36
,1,of
,RRRR212R
,12,,,RRRR323R
,,231,,,2and
and
R
R4,434.4.,.and
.Rand
determine
determine
the
thevalues
values
of
of
RRRR
R
R44. . R4.
determine
determine
the
values
the
values
of
Figure
3.6-36
and
RR
determine
the
values
of
11of
R
determine
the
values
of
2,and
determine
values
determine
the
values
of
,
and
3,3
1
2
4
1
2
3
Figure
P
3.6-36
P 3.6-37 Considere el circuito que se muestra en la figura
determine
the de
values
R , R , R , and R4.
determine
los valores
R1, Rof
2, R31 y R24. 3
PP3.6-37
P3.6-37.
3.6-37
Consider
Consider
the
the
shown
inand
Figure
in iFigure
P4 Figure
3.6-37.
P as
3.6-37.
Given
Given
2circuit
2shown
P3.6-37
P
3.6-37
Consider
the
the
circuit
circuit
shown
in
in
Figure
P
P
3.6-37.
3.6-37.
Given
Given
Dado
¼
icircuit
;circuit
vcircuit
¼
vshown
;shown
¼Figure
i3.6-37.
current
and
Interpret
i2Consider
scircuit
3circuit
1shown
2Figure
PPPPHint:
3.6-37
3.6-37
Consider
Consider
the
the
shown
shown
in
inshown
Figure
Figure
3.6-37.
Given
P3.6-37
3.6-37
Consider
Consider
the
the
in
3.6-37.
PGiven
3.6-37.
Given
Given
3.6-37
Consider
the
shown
in
Figure
PPP
3.6-37.
Given
5 circuit
3circuit
5P
PP3.6-37
Consider
the
in4inFigure
PP3.6-37.
Given
3.6-37
Consider
the
circuit
shown
in
Figure
3.6-37.
Given
P 3.6-37
Consider
the
circuit
shown
in
Figure
P
3.6-37.
Given
2 2 2 22 2 2 2
4 4 44
i2v2s32;¼v¼i3s2;i¼
v32and
i
i2 ¼
;
and
i
¼
i
¼
;
;
i
i2 ¼
e
22i2i2s2;i¼
2s2vv2;231v;¼
4
4
voltage division.
2
2
4
4
424 i¼
1 v14v; 1and
42 i2 ;i2 ;
;4;¼
and
¼
¼
;1;;¼
and
and
i5is¼
i;s;i;vvvi3v53¼
v3;v¼
i5i2¼
i;24;i;¼
ii2i2i22¼
4; i2 ;
;¼
¼
i4i¼
v113;2;iand
¼
i4iand
¼
1vv
v3v3i3s¼
i54and
vand
i5¼
i52i2¼
4iv441¼
52¼
3and
¼
¼and
s3i3;s¼
2i2;5
5555isis2;525¼
3s33;3vv1133;33¼
55455i2i245;45¼
;
i
v
5
3
5i2 ;
1
R
,R
and
,, 1and
R
.
R
.
determine
thethe
values
ofvalues
R1,R
determine
values
of
1,2of
21R
4
4
5
3
R
,
R
,
and
,
and
R
R
.
.
determine
determine
the
the
values
of
R
2
2R. ,. and
4R
4 . 5
determine
los
valores
de
R
,
R
y
R
.
,
,
R
,
and
and
R
determine
determine
the
the
values
values
of
of
R
R
1
2
4
,
R
,
and
R
.
determine
determine
the
values
the
values
of
R
of
,
R
,
and
R
.
determine
the
values
of
R
1
1
2
2
4
4
R
,
and
R
.
determine
the
values
of
1 R424. 44
4
, Rand
determine
the values
of R11, 2Rof
12212
22
2 2values
4 R44. 4 4
, and
determine
1,;2R
v ; 2¼
vi32s ;32i¼
;v¼
and
and
¼
i;4 and
current
as 2current
and
and and
Hint:
Hint:
Interpret
Interpret
i2 ¼
i2the
;133;v¼
v232i1;42iv;and
ii¼
as
current
current
Hint:
Hint:
Interpret
Interpret
2i22is22;5ii¼
2sv
4i44i2445as
5¼
5¼
2¼
4¼ icurrent
4ias
2s;i2v32¼
3v21;and
1ivand
42i4as
2i current
2iv
5i¼
5;¼
3and
5as
22sv¼
Hint:
Hint:
Interpret
Interpret
i
i
¼
¼
i
i
;
v
¼
v
;
¼
¼
i
i
and
Sugerencia:
Interprete
e
di-and
¼
v
v
¼
;
and
i
¼
i¼
current
andand
and
Hint:
Hint:
Interpret
Interpret
i
¼
i
;
v
¼
v
;
and
i
¼
i
as
current
and
Hint:
Interpret
i
2
2
s
s
3
3
1
1
4
4
2
2
¼
i
;
v
v
;
and
i
¼
Hint:
Interpret
i
2
2
s
3
3
1
1
4
45i52current
2as
2 asand
2
s
3
1
4
2
and
as
current
and
Hint: Interpret
i2 ¼25555 isi;55v¼s3 2¼5i333;3 vv13; 3¼
1 2 3i4 ¼45555 i2as
4como
5i current
5
v
;
and
i
¼
as
current
and
Hint:
Interpret
2
s
3
1
4
2
5
3
5
voltage
voltage
division.
division.
visiones
de corriente
y de voltaje.
voltage
voltage
division.
division.
voltage
voltage
division.
division.
voltage
voltage
division.
division.
voltage
division.
voltage
division.
voltage
division.
voltage
division.
Figure P 3.6-34
Figure P 3.6-37
P 3.6-35 Consider the circuits shown in Figure P 3.6-35. The
s
equivalent circuit is obtained from the original circuit by replac- P 3.6-38 Consider
the circuit shown in Figure P 3.6-38.
ing series and parallel combinations of resistors with equivalent (a) Suppose i ¼ 1 i . What is the value of the resistance R?
Figure
Figure
PFigure
3.6-34
P 3.6-34
3
Figure
P value
P3.6-34
3.6-34
3 1
Figure
Figure
PPPPFigure
3.6-34
3.6-34
resistors.
The
of the current in the equivalent circuit is is ¼
Figure
3.6-34
P 3.6-34
Figure
3.6-34
Figure
PP3.6-34
Figure
3.6-34
(b) Figure
Suppose,
instead,
v2 ¼ 4.8 V. What is the value of the
Figure
Figure
PP
3.6-37
P3.6-37
3.6-37
Figura
P 3.6-34
Figura
PFigure
3.6-34
Figure
P
P
3.6-37
3.6-37
0.83.6-35
A. Determine
thethe
values
of
R
, Figure
R5Figure
, v2, Pand
i3. TheThe Figure
Figure
PPPPFigure
3.6-37
3.6-37
Figure
3.6-37
Presistance
3.6-37 of the parallel resistors?
Figure
3.6-37
1, R
Figure
PP3.6-37
Figure
3.6-37
P 3.6-35
P
Consider
Consider
the
circuits
circuits
shown
shown
in2shown
in
3.6-35.
P 3.6-35.
equivalent
P
P
3.6-35
3.6-35
Consider
Consider
the
the
circuits
circuits
shown
in
in
Figure
Figure
P
P
3.6-35.
3.6-35.
The
The
Figure
P
3.6-37
PP
3.6-35
3.6-35
Consider
the
the
circuits
circuits
shown
shown
in
inshown
Figure
Figure
PPPPFigure
3.6-35.
3.6-35.
The
3.6-35
P Consider
3.6-35
Consider
thefrom
circuits
the
circuits
shown
Figure
in
3.6-35.
PThe
3.6-35.
The
The
PPPequivalent
3.6-35
Consider
the
circuits
shown
in
Figure
3.6-35.
The
PP3.6-35
Consider
the
circuits
shown
inincircuit
Figure
PPby
3.6-35.
The
3.6-35
Consider
the
circuits
shown
in
Figure
3.6-35.
The
3-6-35
Considere
los
circuitos
que
se
muestran
en
la
figura
3.6-38
Considere
que
seshown
muestra
en
laPfigura
equivalent
circuit
is
obtained
isConsider
obtained
from
the
the
original
original
circuit
by
replacreplacPPThe
3.6-38
P(c)
3.6-38
Consider
Consider
theel
the
circuit
circuit
shown
shown
in
Figure
in
Figure
PFigure
3.6-38.
P 3.6-38.
Suppose,
instead,
Rcircuito
¼thecircuit
20circuit
V. shown
What
is in
the
value
of3.6-38.
the
P circuit
3.6-35
Consider
the
circuits
shown
in
Figure
P
3.6-35.
equivalent
equivalent
circuit
circuit
isobtained
is
obtained
obtained
from
from
the
the
original
original
circuit
circuit
by
by
replacreplacP3.6-38
Consider
Consider
the
Figure
P3.6-38.
i
40
V
equivalent
equivalent
circuit
circuit
is
is
obtained
obtained
from
from
the
the
original
original
circuit
circuit
by
by
replacreplacP
P
3.6-38
3.6-38
Consider
Consider
the
thecircuit
circuit
shown
in
in
Figure
Figure
PFigure
3.6-38.
3.6-38.
equivalent
equivalent
circuit
circuit
is
is
obtained
from
from
the
original
the
original
circuit
circuit
by
replacby
replacPP
3.6-38
P3.6-38
3.6-38
Consider
Consider
the
circuit
theshown
circuit
shown
shown
ininFigure
Figure
inPP
P3.6-38.
3.6-38.
P 3.6-38.
equivalent
circuit
is
obtained
from
the
original
circuit
by
replacs
P
3.6-38
Consider
the
circuit
shown
in
Figure
3.6-38.
equivalent
circuit
is
obtained
from
the
original
circuit
by
replacP
3.6-38
Consider
the
circuit
shown
in
P
circuit
is
obtained
from
the
original
circuit
by
replacP
3.6-38
Consider
the
circuit
shown
in
Figure
P
3.6-38.
Pequivalent
3.6-35.
El
circuito
equivalente
se
obtuvo
a
partir
del
circuito
P
3.6-38.
a
c
ing
ing
series
series
and
and
parallel
parallel
combinations
combinations
of
resistors
of
resistors
with
with
equivalent
equivalent
current
in
the
40-V
resistor?
1
1
equivalent
circuit
is obtained
from of
the
original
circuit
by replacP 3.6-38
Consider
the
circuit
shown
in
Figure
P
3.6-38.
ing
ing
series
series
and
and
parallel
parallel
combinations
combinations
of
resistors
resistors
with
with
equivalent
equivalent
1
1
(a)
(a)
Suppose
Suppose
i
i
¼
¼
i
.
i
What
.
What
is
the
is
the
value
value
of
of
the
the
resistance
resistance
R?
R? R?R?
3 3 311i1133i¼
1 1¼ i1 1.i1What
ing
ingseries
series
and
and
parallel
combinations
combinations
ofof
ofresistors
resistors
with
equivalent
equivalent
+ combinaciones
–las
(a)(a)
Suppose
Suppose
. is
is
isthe
the
value
value
of
of
the
the
resistance
resistance
ingseries
series
ing
series
and
and
parallel
parallel
combinations
combinations
ofresistors
resistors
of with
resistors
with
with
equivalent
equivalent
ing
series
and
parallel
combinations
resistors
with
equivalent
ing
and
parallel
combinations
of
with
equivalent
ing
series
and
parallel
combinations
resistors
with
equivalent
original
por
elparallel
reemplazo
de
de
resistores
3i¼
(a)
(a)
Suppose
Suppose
ii3i3i33¼
¼
i1.1.3.i1.What
the
the
value
value
of
of
the
the
resistance
resistance
R?
R? R?
resistors.
resistors.
The
The
value
value
of
the
of
the
current
current
incurrent
the
inof
the
equivalent
equivalent
circuit
circuit
is
icircuit
iequivalent
¼is
(a)
(a)
Suppose
Suppose
What
iis
.What
What
the
isWhat
value
the
value
of
the
of
resistance
the
resistance
R? R?
(a)
Suppose
¼
i¼
the
value
of
the
resistance
R?
1i1¼
(a)
Suppose
ii3i33133i¼
iWhat
isisV.
the
value
of
the
resistance
13What
1. 31.What
1is
sis¼
s is
ing
series
and
parallel
combinations
of
resistors
with
(a)
Suppose
¼
What
is
the
value
of
the
resistance
R?
1v¼
resistors.
resistors.
The
The
value
value
of
of
the
the
current
in
in
the
the
equivalent
equivalent
circuit
i
i
¼
¼
3
3
s
s
3
(b)
(b)
Suppose,
Suppose,
instead,
instead,
v
¼
4.8
4.8
V.
What
is
the
is
the
value
value
of
of
the
the
(a)
Suponga
que
.
¿Cuál
es
el
valor
de
la
resistencia
R?
Hint:
Interpret
i
as
current
division.
2
2
(a)
Suppose
i
¼
i
.
What
is
the
value
of
the
resistance
R?
3
1
resistors.
resistors.
The
The
value
value
of
ofthe
the
current
in
in
the
the
equivalent
circuit
circuit
isis
isicircuit
¼
¼isis¼
(b)(b)
Suppose,
Suppose,
instead,
instead,
4.8
V.V.
What
What
isvalue
isthe
the
value
value
ofof
the
resistors.
resistors.
The
The
value
value
ofcurrent
the
ofcurrent
current
the
current
inequivalent
the
invequivalent
equivalent
the
equivalent
circuit
is
is¼(b)
i(b)
33
1v2v2¼¼4.8
resistors.
The
value
of
the
current
in
the
equivalent
circuit
isisissis
¼
resistors.
The
value
of
the
in
the
circuit
s ¼
3¼
resistors.
The
value
of
the
current
in
the
equivalent
circuit
is
¼
en
serie
yDetermine
en
paralelo
con
resistores
equivalentes.
El
valor
de
Suppose,
Suppose,
instead,
instead,
v
v
¼
¼
4.8
4.8
V.
V.
What
What
is
is
the
the
value
of
of
the
the
0.8
0.8
A.
A.
Determine
the
the
values
values
of
of
R
R
,
R
,
,
R
R
,
,
R
,
,
v
and
,
and
i
.
i
.
(b)
(b)
Suppose,
Suppose,
instead,
instead,
v
v
¼
4.8
¼
4.8
V.
What
V.
What
is
the
is
value
the
value
of
the
ofthe
the
(b)
Suppose,
instead,
v
4.8
V.
What
is
the
value
of
the
2
2
(b)
Suppose,
instead,
v
¼
4.8
V.
What
is
the
value
of
the
1
1
2
2
5
5
2
2
3
3
resistors.
The
value
of
the
current
in
the
equivalent
circuit
is
i
¼
2
2
2
(b)
Suppose,
instead,
v
¼
4.8
V.
What
is
the
value
of
the
0.8
0.8
A.
A.
Determine
Determine
the
the
values
values
of
of
R
R
,
R
,
R
,
R
,
R
,
v
,
v
,
and
,
and
i
.
i
.
2
s
2ofv2of
1R
1
2
2, ,and
5and
532
2i, and
2v
3 3i .
R
equivalent
equivalent
resistance
resistance
the
the
parallel
parallel
resistors?
resistors?
(b)
S
uponga,
en
cambio,
5
4.8
V.
¿Cuál
es
el
valor
de
la
Ω
18
Ω
32
Ω
(b)
Suppose,
instead,
v
¼
4.8
V.
What
is
the
value
of
the
1
0.8
0.8
A.
A.
Determine
Determine
the
the
values
values
of
of
R
R
,
,
R
R
,
,
R
,
,
v
v
i
.
.
equivalent
equivalent
resistance
resistance
of
of
the
the
parallel
parallel
resistors?
resistors?
0.8
A.
0.8
Determine
A.
Determine
the
values
the
values
of
of
R
,
R
,
v
R
,
,
and
i
.
0.8
A.
Determine
the
values
of
R
,
R
,
R
,
v
,
and
i
.
2
0.8
A. Determine
values
,25and
1111, R
21
21, R
55 R
2 2,52 v
333.2 i3.3
3
22es
la
en Determine
el circuito
equivalente
0.8corriente
A. Determine
the the
values
Rof
equivalent
equivalent
resistance
resistance
of
ofthe
the
parallel
parallel
resistors?
resistors?
equivalent
equivalent
resistance
resistance
of
the
of
parallel
the
parallel
resistors?
resistors?
equivalent
resistance
of
the
parallel
resistors?
equivalent
resistance
of
the
parallel
resistors?
0.8 A.
the of
values
of
Ri251s52,1,5
Rv25220.8
, and
R25A.
, vi3Determine
equivalent
resistance
of
the
parallel
resistors?
2, and i3.
resistencia
equivalente
de
los
resistores
en the
paralelo?
(c)(c)
Suppose,
Suppose,
instead,
instead,
R
R
¼¼
20
20
V.¼
V.
What
What
isWhat
the
is
value
of
of
thethe
equivalent
resistance
of
the
parallel
resistors?
(c)(c)
Suppose,
Suppose,
instead,
instead,
R20
R
20
20
V.V.
What
isvalue
isvalue
the
value
value
of
thethe
los valores de R1, R2, R5, 40
v2 e40
. 4040
is10
isΩ i i
V i3V
(c)
(c)Suppose,
Suppose,
instead,
instead,
Rinstead,
R¼
¼
V.
V.
What
What
isis
is
the
the
value
of
of
the
the
(c)
Suppose,
(c)
Suppose,
instead,
¼¼
R20
20
¼
V.
20
What
V.
What
the
isthe
the
value
value
ofofthe
the
of
the
(c)
Suppose,
instead,
RRresistor?
¼
20
V.
What
the
value
of
the
V
(c)
Suppose,
instead,
RR20
¼
20
V.
What
isisel
the
value
of
(c)
Suppose,
instead,
¼
20
V.
What
is
the
value
of
the
a a 40
(c)
Suponga,
en
cambio,
R
5
V.
¿Cuál
es
valor
de
la
current
current
incurrent
the
in
the
40-V
40-V
resistor?
iisisisVVisics s csis c c
40
V40VV40
40
(c)
Suppose,
instead,
R
¼
20
V.
What
is
the
value
of
the
a VaVV40
current
in
in
the
the
40-V
40-V
resistor?
resistor?
40
s
aaaab a–40
ccccids cc c
a+–V+a – 40
current
current
in
in
the
the
40-V
40-V
resistor?
resistor?
current
current
in
the
in
40-V
the
40-V
resistor?
resistor?
V
current
in
the
40-V
resistor?
current
in
the
40-V
resistor?
i
+
+
–
current current
inenthe
40-V
ic3
Ra+2––+++ –s v+2 – c
corriente
de 40-V?
1in 1theresistor?
––––++
40-V
resistor?
a
1 1current
+
Hint:
Hint:
Interpret
Interpret
i3 el
i3 resistor
¼
i¼
as
current
division.
division.
1 1¼
Hint:
Hint:
Interpret
Interpret
i3current
current
division.
division.
1i11i1133ias
3¼
– +
1i1asas
3as
1as
1as
– +
3i¼
Hint:
Hint:
Interpret
Interpret
i
i
¼
¼
i
i
as
current
division.
division.
Hint:
Hint:
Interpret
Interpret
i
i
¼
i
current
ascurrent
current
division.
division.
Hint:
Interpret
i
¼
i
current
division.
3
3
1
1
Hint:
Interpret
i
¼
i
as
current
division.
3
3
1
1
3
1
Hint:
Interpret
i
¼
i
as
current
division.
3
1
1
3
3
3
33 1i 3 ¼ i como
3
R1 R1 R R
original
circuit
32
32
Ω
Ω
18
18
Ω
Ω
32
32
Ω
Ω
división
de corriente.
Sugerencia:
Interprete
Hint:
Interpret
as
current
division.
3
1
32
32
Ω
Ω
18
18
Ω
Ω
32
32
Ω
Ω
1 1
3
RR
R
R111 RR11 RR1
R11
1
Figure P 3.6-35
32Ω
Ω32ΩΩ32 Ω
18
18Ω
Ω18ΩΩ1832
32Ω
Ω32ΩΩ3232
Ω
Ω
32
Ω
18
Ω
32
Ω
32
18
32
32
Ω
18
Ω
32
Ω
Ω
32 Ω32
10
Ω
is Ω
Ω 32 Ω
18 Ω 4018
32
Ω
V 10
10
10
Ω
Ω
a
c Ω
10
10Ω
Ω10ΩΩ10
10
Ω
10
10
Ω
– + 10
d–10dΩ i3d id3 i i
b b Rb R
vR2 Ωv–2+ +
b2+ R+
2
v
–i i d i 3 3i
2v2– d
2R
2
bbbb bbRRR bRR
dd
v2+
––v–v2d+
i33 i 3 3
2d––vd
2 i3–
v2+
2 vv
R222b2 +2+++
22 –+
v
b original
d
R
– 3d 3 i3
v
6
Ω
R
R
2
+
–
2
original
circuit
circuit
2 circuit
5circuit i3
2 original
original
original
original
circuit
circuit
original
original
circuit
circuit
original
circuit
original
circuit
original
circuit
circuito
original
circuit
Ωi
i40
40 40
V 28
Voriginal
s 40
sV V is is
a a 40
c
V40iiisVisiVs40isciV
40
a40aVV
s c is c c
40
V40
s Vc
aaaab a–a40
+–Va
+ –s40
+
– + c
a
cc dis cc cc
+
––––+a
+
+
–
–
+
+
–
– ++
– +
6 equivalent
Ω6 Ω 6 6
Rcircuit
5 R5 R5R5
ΩΩ
6666Ω
Ω66ΩΩ 6 R
R
Ω
R
5555 RR
Ω
R
55 RR5
6Ω
Ω
R
5
6
Ω
28 28
Ω Ω2828
5
ΩΩ
28
28
Ω
Ω
28
Ω
28
Ω
28
Ω
28 Ω
28 Ω
Ω
28
28 Ω
b b bb
d d dd
bbbb bb b
dddd dd d
b
d
b
equivalent
equivalent
circuit
circuit
equivalent
equivalent
circuit
circuit d
equivalent
equivalent
circuit
circuit
equivalent
equivalent
circuit
circuit
equivalent
circuit
equivalent
circuit
circuito
equivalente
equivalent
circuit
equivalent circuit
Figure
Figure
PFigure
3.6-35
P 3.6-35
Figure
P P3.6-35
3.6-35
Figure
Figure
PP
3.6-35
3.6-35
Figure
3.6-35
P 3.6-35
Figure
PPPFigure
3.6-35
Figure
PP3.6-35
Figure
3.6-35
Figura
3.6-35
Figure
P 3.6-35
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 103
Figure P 3.6-38
P 3.6-39 Consider the circuit shown in Figure P 3.6-39.
(a) Suppose v3 ¼ 1 v1 . What is the value of the resistance R?
4 3.6-38
Figure
Figure
PFigure
3.6-38
P 3.6-38
Figure
P P3.6-38
Figura
PFigure
3.6-38
Figure
PPPPFigure
3.6-38
3.6-38
Figure
3.6-38
P 3.6-38
Figure
3.6-38
(b)Figure
Suppose
iP
1.2
A. What is the value of the resistance R?
3.6-38
2P¼
Figure
3.6-38
Figure
P 3.6-38
(c) Suppose R ¼ 70 V. What is the voltage across the
P 3.6-39 Considere el circuito que se muestra en la figura
P 3.6-39
P 3.6-39
Consider
Consider
thethe
circuit
circuit
shown
shown
inshown
Figure
in Figure
PFigure
3.6-39.
P 3.6-39.
20-V
resistor?
P
P3.6-39
Consider
Consider
thethe
circuit
circuit
shown
inFigure
inFigure
P3.6-39.
P3.6-39.
PP
3.6-39
3.6-39
Consider
Consider
the
thecircuit
circuit
shown
shown
in
in
Figure
Figure
3.6-39.
3.6-39.
3.6-39
P3.6-39
3.6-39
Consider
Consider
the
circuit
the
circuit
shown
shown
inPP
P3.6-39.
3.6-39.
PPP3.6-39.
3.6-39
Consider
the
circuit
shown
in
Figure
3.6-39.
PP3.6-39
Consider
the
circuit
shown
ininFigure
PP3.6-39.
3.6-39
Consider
shown
in
Figure
PPFigure
3.6-39.
1Consider
1the 1circuit
P
3.6-39
the
circuit
shown
in
Figure
PR?
3.6-39.
1
.
v
What
.
What
is
the
is
the
value
value
of
the
of
the
resistance
resistance
R? R?R?
(a)(a)
Suppose
Suppose
v3 ¼
v3 4¼
1
1
¼
¼
v
v
.
What
.
What
is
is
the
the
value
value
of
of
the
the
resistance
resistance
(a)
(a)
Suppose
Suppose
v
v
1v
1
4
1What
3v1.What
1.1What
1visis
11v3¼
4
4
¼
¼
v
.
the
the
value
value
of
of
the
the
resistance
resistance
R?
R?
(a)
(a)
Suppose
Suppose
v
v
v
¼
.
What
is
the
is
value
the
value
of
the
of
resistance
the
resistance
R? R?
(a)
(a)
Suppose
Suppose
v
¼
v
.
What
is
the
value
of
the
resistance
R?
(a)
Suppose
v
(a)
Suponga
v3434345
¿Cuál
el
valor
la
resistencia
¼
vWhat
What
isthe
the
value
of
the
resistance
(a)
Suppose
v1.2
1is the
3i33 ¼
vi1A.
the
value
ofde
resistance
R?
(a)
Suppose
v3¼
1.1.1What
4¼
4v is
114.3¼
(b)
(b)
Suppose
Suppose
i2que
What
ises
value
value
ofvalue
the
ofthe
the
resistance
resistance
R?R?
R?R?
421.2
2 ¼
.
What
is
the
value
of
the
resistance
(a)Suppose
v12¼
(b)
(b)
Suppose
i
1.2
1.2
A.
A.
What
What
is
is
the
the
value
of
of
the
the
resistance
R?R?
3 A.
1
4A.
(b)
(b)
Suppose
Suppose
i
i
¼
¼
1.2
1.2
A.
A.
What
What
is
is
the
the
value
value
of
of
the
the
resistance
resistance
R?
R?
(b)
(b)
Suppose
Suppose
i
i
¼
1.2
¼
1.2
What
A.
What
is
the
is
value
the
value
of
the
of
resistance
the
resistance
R?
(b)
Suppose
i
¼
1.2
A.
What
is
the
value
of
the
resistance
R?
(b)
Suppose
i
¼
1.2
A.
What
is
the
value
of
the
resistance
R?
2
2
2
(b)
SSuppose
uponga
5
1.2
A.
¿Cuál
es
el
valor
de
la resistencia
R?
(b)
i22 R¼
¼i222¼
1.2
A.
What
isV.
the
value
of
the
resistance
R?
(c)
(c)
Suppose
Suppose
Rque
70
70
V.
V.
What
What
is
is
the
voltage
voltage
across
the
the R?
(b)
iR
¼
1.2
A.
What
isthe
the
value
of across
the resistance
R?
(c)
(c)
Suppose
Suppose
R
¼
70
70
V.
What
What
is
is
the
the
voltage
voltage
across
across
the
the
2¼
(c)
(c)Suppose
Suppose
Rque
R¼
¼
70
70
V.
V.
What
What
isis
ises
the
the
voltage
voltage
across
across
the
the
(c)
Suppose
(c)
Suppose
R
¼
R
70
¼
V.
70
What
V.
What
is
the
is
the
voltage
voltage
across
across
the
the
(c)
Suppose
R
¼
70
V.
What
the
voltage
across
the
(c)
Suppose
R
¼
70
V.
What
is
the
voltage
across
the
(c)
Suponga
R
5
70
V.
¿Cuál
el
voltaje
a
través
del
(c)
Suppose
R
¼
70
V.
What
is
the
voltage
across
the
20-V
20-V
resistor?
resistor?
(c)
Suppose
R ¼ 70 V. What is the voltage across the
20-V
20-V
resistor?
resistor?
20-V
20-V
resistor?
resistor?
20-V
20-V
resistor?
resistor?
20-V
resistor?
20-V
resistor?
resistor
de
20-V?
20-V
resistor?
20-V
resistor?
Alfaomega
4/12/11 5:23 PM
3
32 Ω
32 Ω
3 1
10 Ω
+ v2 – d
i3
l circuit
V
is
c
Circuitos resistivos
104
Figure P 3.6-38
R5
8Ω
nt circuit
d
(d)Suponga que R 5 30 V. ¿Cuál es el valor de la corriente Sugerencia: Utilice los lineamientos dados en la sección 3.7
para etiquetar el diagrama del circuito. Utilice MATLAB para
en este
30-V? the circuit shown in Figure P 3.6-39.
P resistor
3.6-39 de
Consider
resolver las ecuaciones que representan el circuito.
Sugerencia:
de corriente.
¼ 14 v1 .como
Whatdivisión
is the value
of the resistance R?
(a) Interprete
Suppose vv3 5
(b) Suppose i2 ¼ 1.2 A. What is the value of the resistance R?
(c) Suppose R ¼ 70 V. What is the voltage across the
20-V resistor?
Figura P 3.6-39
Figura P 3.7-1
P 3.6-40 Considere el circuito que se muestra en la figura
P 3.6-40. Dado que el voltaje de la fuente de voltaje dependiente es va 5 8 V, determine los valores de R1 y vo.
P 3.7-2 Determine la potencia alimentada por cada fuente,
independiente y dependiente, en el circuito que se muestra en
la figura P 3.7-2.
Sugerencia: Utilice los lineamientos dados en la sección 3.7
para etiquetar el diagrama de circuito. Utilice MATLAB para
resolver las ecuaciones que representan el circuito.
o
Figura P 3.6-40
P 3.6-41 Considere el circuito que se muestra en la figura
P 3.6-41. Dado que la corriente de la fuente de corriente dependiente es ia 5 2 V, determine los valores de R1 e io.
Figura P 3.7-2
Sección 3.8 ¿Cómo lo podemos comprobar . . . ?
o
Figura P 3.6-41
P 3.6-42 Determine los valores de ia, ib, i2 y v1 en el circuito
que se muestra en la figura P 3.6-42.
P 3.8-1 Un programa de análisis por computadora, utilizado para el circuito de la figura P 3.8-1, proporciona los siguientes voltajes y corrientes de la derivación: i1 5 20.833 A,
i2 5 20.333 A, i3 5 21.167 A y v 5 220 V. ¿Estas respuestas
son correctas?
Sugerencia: Verifique que se satisfaga la KCL en el nodo central, y que se satisfaga la KVL en torno al circuito cerrado exterior que consta de dos resistores de 6-V y la fuente de voltaje.
5Ω
8Ω
2Ω
+ v1 –
+
6V –
20 Ω
ia
4ia
i2
12 Ω
24 Ω
ib
Figura P 3.6-42
Sección 3.7 Análisis de circuitos resistivos
utilizando MATLAB
P 3.7-1 Determine la potencia alimentada por cada una de las
fuentes, independientes y dependientes, en el circuito que se
muestra en la figura P 3.7-1.
Alfaomega
M03_DORF_1571_8ED_SE_053-107.indd 104
Figura P 3.8-1
P 3.8-2 El circuito de la figura P 3.8-2 se asignó como un
problema de tarea. La respuesta en la parte posterior del libro
dice que la corriente, i, es 1.25 A. Verifique esta respuesta,
utilizando la división de corrientes.
Circuitos Eléctricos - Dorf
4/12/11 5:23 PM
Problemas
105
P 3.8-6 El análisis por computadora del circuito en la figura
P 3.8-6 muestra que ia 5 20.5 mA e ib 5 4.5 mA. ¿Estuvo
correcto el análisis hecho por computadora?
Figura P 3.8-2
P 3.8-3 El circuito de la figura P 3.8-3 se construyó en el
laboratorio, y se midió que vo fuera de 6.25 V. Verifique esta
medición, utilizando el principio del divisor de voltaje.
Sugerencia: Primero, verifique que se cumpla con las ecuaciones de la KVL para los cinco nodos cuando ia 5 0.5 mA e ib 5
4.5 mA. Luego, verifique que se cumpla con la ecuación de
la KVL para el enlace inferior izquierdo (a-e-d-a). (Las ecuaciones de la KVL para el resto de enlaces no son útiles porque
cada una implica un voltaje desconocido.)
o
Figura P 3.8-3
P 3.8-4 El circuito de la figura P 3.8-4 representa un sistema
eléctrico de un automóvil. Un reporte establece que iH 5 9 A,
iB 5 29 A e iA 5 19.1 A. Verifique que este resultado sea el
correcto.
Figura P 3.8-6
Sugerencia: Compruebe que en cada nodo la KCL se haya
satisfecho y que en torno a cada circuito cerrado se satisfaga
la KVL.
P 3.8-7 Verifique que las corrientes y voltajes del elemento
que se muestran en la figura P 3.8-7 cumplen con las leyes de
Kirchhoff:
Luces
(a)Compruebe que las corrientes dadas satisfacen las ecuaciones de la KCL que corresponden a los nodos a, b y c.
(b)Compruebe que los voltajes dados satisfacen las ecuaciones de la KVL que corresponden a los circuitos cerrados
a-b-d-c-a y a-b-c-d-a.
Batería
Alternador
Figura P 3.8-4 Modelo de circuito eléctrico del sistema eléctrico
de un automóvil.
P 3.8-5 El análisis por computadora del circuito en la figura
P 3.8-5 muestra que ia 5 20.5 mA e ib 5 22 mA. ¿Estuvo
correcto el análisis hecho por computadora?
Sugerencia: Verifique que se satisfagan las ecuaciones de la
KVL para los tres enlaces cuando ia 5 20.5 mA e ib 5 22 mA.
Figura P 3.8-7
*P 3.8-8 La figura P 3.8-8 muestra un circuito y algunos datos que se corresponden. Los datos tabulados proporcionan
valores de la corriente, i, y el voltaje, v, que corresponden a
diversos valores de la resistencia R2.
Figura P 3.8-5
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 105
(a)Utilice los datos de las filas 1 y 2 de la tabla para encontrar
los valores de vs y R1.
(b)Utilice los resultados de la parte (a) para verificar que los
datos tabulados son consistentes.
(c)Llene las entradas faltantes en la tabla.
Alfaomega
4/12/11 5:23 PM
106
Resistive Circuits
106
Circuitos resistivos
current, i, and voltage, v, corresponding to several values of
thelaresistance
de
corriente,R2i,. el voltaje, v, que corresponden a diversos
valores
resistencia
(a) Usede
theladata
in rowsR12.and 2 of the table to find the values
of Utilice
is and R
1. datos de las filas 1 y 2 de la tabla para encon(a)
los
(b) Use
thelosresults
of de
partis (a)
verify that the tabulated data
trar
valores
y Rto
1.
are
consistent.
(b)Utilice los resultados de la parte (a) para verificar que
(c) Filllosindatos
the missing
entries
in the table.
tabulados
son consistentes.
(c)Llene las entradas faltantes en la tabla.
s
/2009
10/23/200947
47
s
Figure P 3.8-8
Problems
Problems
47
47
Figura P 3.8-8
*P 3.8-9 Figure P 3.8-9 shows a circuit and some corre3.8-9
energy
energy
stored
stored
in a battery
inFigure
a battery
is P
equal
is equal
to thetoproduct
the product
of theofbattery
the battery
P
2.5-3
P 2.5-3
The current
The current
source
source
and
voltage
andThe
voltage
source
source
in the
inun
circuit
the
circuit
sponding
data.
tabulated
data
provide
ofdatos
the
*P
3.8-9
La
figura
3.8-9
muestra
circuito
y values
algunos
106
voltage
voltage
and
the
and
battery
the
battery
capacity.
capacity.
The capacity
The capacity
is usually
is usually
givengiven
shownshown
in Figure
in Figure
P 2.5-3
P
2.5-3
are
connected
are
connected
in
parallel
in
parallel
so
that
so
they
that
they
correspondientes. Los datos tabulados proporcionan valores Figura P 3.8-9
with with
the units
the units
of Ampere-hours
of Ampere-hours
(Ah).(Ah).
A new
A 12-V
new 12-V
battery
battery
The current
source
source
and voltage
and voltage
both have
both have
the same
the same
voltage,
voltage,
vs. The
vs.current
having
having
a capacity
a capacity
of 800ofmAh
800 mAh
is connected
is connected
to a load
to athat
loaddraws
that draws
source
source
are also
are connected
also connected
in series
in series
so that
sothey
that both
they have
both have
the the
a current
a current
of 25ofmA.
25 (a)
mA.How
(a) How
long will
long itwill
takeit for
takethe
forload
the to
load to
same same
current,
current,
is. Suppose
is. Suppose
that vthat
vs ¼
V and
12 Visand
¼ 3isA.
¼Calculate
3 A. Calculate
s ¼ 12
discharge
discharge
the battery?
the battery?
(b) How
(b) How
muchmuch
energy
energy
will be
will
supplied
be supplied
the power
the power
supplied
supplied
by each
by each
source.
source.
to thetoload
the during
load during
the time
the required
time required
to discharge
to discharge
the battery?
the battery?
Answer:
Answer:
The voltage
The voltage
source
source
supplies
supplies
�36 W,
�36and
W,the
andcurrent
the current
its
source
source
supplies
supplies
36 W.36 W.
Resistive
Circuits
i
Design Problems
Problemascurrent,
de idiseño
i, and
i voltage, v, corresponding to several values of
i
s in sFigure DP 3-1 uses a potentiomto be 200 V � 5 percent. That is, 190 V � RL � 210 V. The
DP 3-1 The circuit shown
+
+ to several values of
+
+ the resistance R2. current, i, and voltage, v, corresponding
vsource
v is aR12
capaz
5 W.
estesource
circuito
utilizando
resisto– de
–alimentar
voltage
VR�Diseñe
1 percent
capable
of supplying
eter
to
produce
a
variable
voltage.
The
voltage
v
varies
as
a
+v +
m 3-1 utiliza
is 3-1
is vEl
vs vs que
PD
en resistance
la figura PD
s se– muestrathe
.
R
s circuito
–
2
res
de
1/8
de
watt
de
5%
para
R
y
R
,
de
modo
que
el
voltaje
(a)
Use
the
data
in
rows
1
and
2
of
the
table
to
find
the
values
5
W.
Design
this
circuit,
using
5
percent,
1=8-watt
resistors
for
knob
connected
to
the
wiper
of
the
potentiometer
is
turned.
1
2
–
–
un potenciómetro
para producir una voltaje variable. El voltaje
a of
través
de
RsoLtosea
Use the data
in rows
1 and 2R
the R
table
findthe
thevoltage
values across RL is
of iR
R1R. 2 (a)
and
,
that
and
so
that
following
three
Specify
the
resistances
s 1and
1
2
batterybattery load load
vm varía al girar un botón conectado al contacto deslizante del
of i (a)
andtoRverify
.
(b)
Use the results of part
that the tabulated data
requirements areEspecifique
satisfied:
vo ¼ 4 V � 10%
potenciómetro.
las resistenciass R1 y 1R2 de modo
Figure
Figure
P 2.5-3
P 2.5-3
Figure
Figure
P 2.5-6
2.5-6
(b) Use the results
of
partP (a)
to verify that the tabulated data
are
consistent.
que
cumpla
con
tres
requerimientos
The
voltage
vmlos
varies
from
8 V to 12 Vsiguientes:
as the wiper moves (A 5 percent, 1/8-watt 100-V resistor has a resistance between
1. se
consistent.
(c) Fill in the missingare
entries
in the table.
(Un resistor de 100 V de 1/8 de watt de 5%, tiene una resistenfrom source
one
of voltage
thesource
potentiometer
to the other end of the 95
and 105 V and can safely dissipate 1/8-W continuously.)
P 2.5-4
P 2.5-4
The current
The1.current
source
andend
and
source
in thealincircuit
the
El voltaje
vvoltage
moverse
el
control
desliFill circuit
in the
missing
entries
in
the
cia
de table.
entre
95Ammeters
y 105
V y puede disipar con toda seguridad 1/8
m varía de 8 a 12 V (c)
Section
Section
2.6
Voltmeters
2.6
Voltmeters
and
and
Ammeters
potentiometer.
shownshown
in Figure
in Figure
P 2.5-4
P 2.5-4
are de
connected
are
in lado
parallel
in del
parallel
so
that
sothey
that they
zante
unoconnected
a otro
potenciómetro.
W de manera continua.)
P 2.6-1
2.6-1
For the
Forcircuit
the circuit
of Figure
of Figure
P 2.6-1:
P 2.6-1:
both have
both have
the same
the2.same
voltage,
vs. The
vssource
.current
The current
source
source
and
and
Thevoltage,
voltage
supplies
lessvoltage
thanvoltage
0.5 W
ofPpower.
2. connected
La fuente
deseries
voltaje
de 0.5the
W de potencia.
source
source
are also
are connected
also
in series
in
so that
soalimenta
they
that both
theymenos
have
both have
the
(a) What
(a) What
is theisvalue
the value
of theofresistance
the resistance
R? R?
Each
ofvthat
R1, 12
, and
Risand
than 0.25 W.
3.is. Suppose
P¼dissipates
thatresistencia
vRs2¼
V
and
12, V
2isA.
¼Calculate
2 A.less
Calculate
same same
current,
current,
is. Suppose
s¼
(b)0.25
How
(b)W.How
muchmuch
powerpower
is delivered
is delivered
by thebyvoltage
the voltage
source?
source?
3.
Cada
R
1 R2 y Rp disipa menos de
the power
the power
supplied
supplied
by each
by each
source.
source.
o
3.8-8
is
+
isvs
–
+
vs
–
+ 5 +. 50 . 0
– . –5 0
. 5 0
Voltmeter
Voltmeter
Ammeter
Ammeter
Figure DP 3-2
vs + vs +
–
–
shows a circuit and some correis
Figure
3.8-9
9ated
Figure
3.8-9 shows
circuit
someiPs corredata Pprovide
valuesa of
the and
data.
The
Figure
Figure
P tabulated
2.5-4
P 2.5-4 data provide values of the
Figura PD 3-2
Voltímetro
Figure P 3.8-9
P 2.5-5
P 2.5-5
+
12 V 12
– V
(a) Find
(a) Find
the power
the power
supplied
supplied
by thebyvoltage
the voltage
source
source
shownshown
in in
Figure
Figure
P 2.5-5
P 2.5-5
whenwhen
for t for
� 0t we
� 0have
we have
v¼
2vcos
¼ 2t V
cos t V
Figura
3-1
Figure PD
DP
3-1
+
–
L
DP 3-3 A phonograph pickup, stereo amplifier, and speaker are
PD 3-3inEn
la figura
PD 3-3a
se muestran
fonocaptor
shown
Figure
DP 3-3a
and redrawn
as aun
circuit
model de
as
fonógrafo,
un amplificador
de estéreothe
y altoparlantes,
vueltos
shown
in Figure
DP 3-3b. Determine
resistance R soy that
the
a trazar vcomo
un the
modelo
de circuito
como
el que se
voltage
across
speaker
is 16 V.
Determine
themuestra
power
en la figura
Determine la resistencia R de modo que
delivered
to PD
the 3-3b.
speaker.
R
elRvoltaje
v a través del altoparlante
1
1sea de 16 V. Determine la
2 A
2 A
potencia transmitida al altoparlante.
Fonógrafo
Figure
Figure
P 2.6-1
P 2.6-1
Amplificador
Altopralante
+
+
and and
PD 3-2 La resistencia RL en la figura PD 3-2 es
la resistencia
Problems
2.6-2
P 2.6-2
The current
The current
source
source
in Figure
in Figure
P 2.6-2
P 2.6-2
supplies
supplies
40 W.40 W.
DP 3-2 The resistance RL in Figure DP 3-2 isPthe
equivalent
i ¼ 10
i¼
cos
10be
ttransductor
mA
cos
to
200t mA
V � 5depercent.
ThatSeis,
V �values
Rdo
210
V.meters
The
Figure DP 3-1 uses aequivalente
potentiomde
un
presiones.
ha190
especificaL �
What
What
values
the
do
meters
the
in
Figure
in
Figure
P
2.6-2
P
2.6-2
read?
read?
resistance of a pressure transducer. This resistance is specified
(b)The
Determine
(b)
Determine
energy
the
supplied
supplied
this
by
voltage
this
source
source
for
be
200
V for
� 5decir,
percent.
That
190 V � RL � 210 V. The
voltage
source
a to
12
V�
15%.
percent
source
capable
of supplying
etage.
circuit
shown
in Figure
DPenergy
3-1
aby
potentiomvoltage
vthe
varies
as
a uses
que
esta
resistencia
sea
deisvoltage
200
V
Es
190
V is,
mdo
the
period
thevoltage.
period
0 �R
t0The
�1210
tvoltage
s.� V.
1 s.La
voltage
source
� 11%
percent
sourcefor
capable of supplying
5vm
W.
Design
this
5 is
percent,
resistors
uce
a variable
varies
a circuit,
fuente
deasvoltaje
es unausing
fuente
dea 12 V 1=8-watt
of the
potentiometer
is
turned.
L�
W.voltage
Design across
this circuit,
RL isusing 5 percent, 1=8-watt resistors for
of the potentiometer
is R
turned.
dcted
R2 to
so the
that wiper
the following
three R1 iand
2, so that5the
i
Ammeter
Ammeter
R24
, so
that10%
the voltage across RL is
following
three R1 and
resistances R1 and RAlfaomega
2 so that the
Circuitos Eléctricos - Dorf
vo ¼
V�
–
–
ts 8are
4 V � 10%
m
V satisfied:
to 12 V as the wiper moves v (A v5 percent, 1/8-watt 100-V resistor hasvoa ¼
Voltmeter
Voltmeter
resistance
between
4 Ω1/8-W
4 Ω continuously.)
ltage
vm varies
from
8 Vend
to 12
moves
ntiometer
toP the
other
of V
theas the
5 percent,
1/8-watt
100-V
resistor has a resistance between
95wiper
and 105
V and(A
can
safely dissipate
+ v+ – v –
Figure
Figure
2.5-5
P 2.5-5
ne end of the potentiometer to the other end of the 95 and 105 V and can safely dissipate 1/8-W continuously.)
i
i
M03_DORF_1571_8ED_SE_053-107.indd 106
ometer.
2.5-6
P 2.5-6
Figure
Figure
Pof2.5.6
P 2.5.6
showsshows
a battery
a battery
connected
connected
to a load.
to a load. + 12+V 12 V
ies P
less
than
0.5 W
power.
2A 2A
–
–
4/12/11 5:23 PM
E1C03_1
11/25/2009
107
E1C03_1 11/25/2009
E1C03_1 11/25/2009
107
107
Problemas
de diseño
Design Problems
Fonocaptor
Amplificador
107107
La The
salida
del divisor
de corriente
es proportional
proporcional
entrada.
output
of the current
divider is
to athelainput.
Design
Problems
107
La The
constante
de
proporcionalidad,
g,
se
denomina
la
ganancia
constant of proportionality,
g, is called the gain
Design Problems
107 of the
del current
divisordivider
de corriente
y la daby
and is given
Altoparlante
The output of the current dividerRis proportional to the input.
1
The output
the current
divider
is proportional
to the
the gain
input.of the
g¼
The of
constant
of proportionality,
g, is called
R
þ
R
The constant
of
proportionality,
g,
is
called
the
gain
of
the
1
2
current divider and is given by
current The
divider
andsupplied
is given by
by the current source is
power
La potencia alimentada
� � por
�R1 de corriente es
g ¼la�fuente
R1 RR21 R1 þ R2 R1 R2 2
g
¼
p ¼ vs is ¼ is R1 þ R2 is ¼
i ¼ Rin is 2
R
R2Rcurrent
R
Ris2R2 s2
R
R
The
power
supplied
by
the
source
1þ
1þ
1
2
1
Figure DP 3-3 A phonograph stereo system.
 vs is by ithe
is  Rent is 2
is is
s current source
The powerp supplied
R1  R2
� �R1  R2 ��
where
Figura PD 3-3 Un sistema de fonógrafo estereofónico.
� �
�1�R2
R
R1 R2 2
DP 3-4 A Christmas tree light set is required that will operate
is11R
p ¼ vs is ¼ R1isR2 Rin ¼ RR
¼
is ¼ Rin is 2
R22 2
R
þ
R
R
R2Rin is 2
i
p
¼
v
i
¼
i
¼
1
s s
s
s 2 R1 þ R2 1isþ ¼
from a DP
6-V3-3
battery
on a treestereo
in a city
park. The heavy-duty donde
Figure
A
phonograph
system.
R 1 þ R2
R1 þ R2
PDFigure
3-4 Se
que un conjunto
de luces para arbolitos de
DPrequiere
3-3 can
A phonograph
system.
where
battery
provide 9 stereo
A for the
four-hour period of operation
is called the input resistance of the current divider.
where
Navidad funcione
con
una
batería
de
6
V
en
un
árbol
de
un
parque
R1 R2
each3-4
night.
Design a parallel
setset
of is
lights
(select
thewill
maximum
DP
A Christmas
tree light
required
that
operate
R
R2to have a gain, g ¼ 0.65.
RRent
(a) Design a current divider
in1¼
DP 3-4La
Abatería
Christmas
tree when
light
isresistance
required
that
willThe
citadino.
trabajo
puede
9operate
Aheavy-duty
durante
number
of de
lights)
the
of
each
bulb
is 12 V.
R
þ R2a gain, g ¼ 0.65, and an
from
a 6-V
battery
onpesado
aset
tree
in
a cityproveer
park.
R
¼
in divider to 1have
(b) Design a current
R1 þ Rof2 the current
a battery
6-V
on a tree
in
a city
park.
heavy-duty
un from
periodo
de battery
cuatro
horas
de
operación
cadaThe
noche.
Diseñe
un
can
provide
9
A
for
the
four-hour
period
of
operation
is
called
the
input
resistance
DP
3-5
The9input
thefour-hour
circuit shown
in Figure
DP 3-5 is the
Rof
10000 V. divider.divider.
in ¼
batteryde
canluces
provide
A foratoparallel
the
period
of operation
is called
theinput
inputresistance,
the
conjunto
enDesign
paralelo
(seleccione
el número
máximo
de
se denomina
laresistance
resistencia
de current
entrada del divisor de corriente.
each
night.
set of
lights
(select
the maximum
voltage
source
voltage,
v
.
The
output
is
the
voltage
v
.
The
s
o
(a)
Design
a
current
divider
to have a gain, g ¼ 0.65.
eachcuando
night.
Design
a parallel
set
of
lights
(selectsea
theeach
maximum
luces)
la
resistencia
de
cada
bombillo
de
12
V.
number
of
lights)
when
the
resistance
of
bulb
is
12
V.
(a)
Design
a
current
divider
to
have
gain,
0.65.g ¼ 0.65, and an
is related
input byof each bulb is 12 V.
(b)
Design
a
current
divider
toa que
have
ag ¼
gain,
numberoutput
of lights)
when to
thethe
resistance
Diseñe
un divisor
de corriente
tenga
ganancia,
(b) (a)
Design
a current
divider
to ¼have
a gain,
g ¼una
0.65,
and an g 5 0.65.
DP 3-5 The input to the circuit shown in Figure DP 3-5 is the
input
resistance,
R
10000
V.
in
(b)
Diseñe
un divisor
de
corriente
que tenga una ganancia, g 5 0.65
DP 3-5voltage
The input
to
the
circuit
shown
in
Figure
DP
3-5
is
the
R
input
resistance,
R
¼
10000
V.
2
in
source
voltage,
vs. se
The
output¼en
isgvthe
voltage
. The
al circuito
que
muestra
las figura
PDvo3-5
PD 3-5 La entrada
o ¼
s voltage
y una resistencia de entrada , Rent 5 10 000 V.
voltageoutput
sourceisvoltage,
vtos.vthe
The
output
is2 vthe
v . The
R1 vþby
relatedde
input
es el voltaje
de la fuente
voltaje,
. RLa
salida es el ovoltaje v .
output is related to the input by
s
o
La salida se relaciona con la entradaR por
2
The output of the voltage
divider
isv proportional
to the input.
voR
s ¼ gvs
22
R¼
R1vvssþ
vvooof
¼proportionality,
¼R2gv
The constant
g, ssis called the gain of the

gv
R
þgiven
R
R11 is
R22 by
voltage divider and
o
Figure DP 3-6
The output of the voltage dividerRis2 proportional to the input.
s
s
La The
salida
del divisor
de voltaje
proporcional
a lainput.
entrada.
output
the voltage
divider
is
to the
g es
¼proportional
The of
constant
of proportionality,
g, is called
the gain of the
Rse
R2the gainganancia
La The
constante
deofproporcionalidad,
denomina
1þ
constant
proportionality,
g,g,isby
called
of the del
voltage
divider
and is given
power
voltage
divider
and
isdagivenby
bythe voltage source is
DP 3-7DP
Design
Figure
3-6 the circuit shown in Figure DP 3-7 to have an
divisor
deThe
voltaje
y lasupplied
R�2
Figure output
DP 3-6v ¼ 8.5 V when the input is v ¼ 12 V. The circuit should
�gR¼
o
s
2
2
2v
vs
vs
s
require
Figura
PDno
3-6more than 1 mW from the voltage source.
p ¼ vs is g¼¼vsR1 þ RR2 1 þ R¼2
¼
R1 þ
R2 source
R1 þisR2 Rin
The power supplied by the
voltage
DP 3-7 Design the circuit shown in Figure DP 3-7 to have an
The power supplied by the voltage source is
DP 3-7output
Design
the8.5
circuit
shown
in Figure
DP
to have
an should
�
�
when
theque
input
vs ¼
123-7
V. la
The
circuit
vo ¼
PD 3-7
Diseñe
elVcircuito
se is
muestra
en
figura
PD 3-7 que
La potencia
alimentada �
por la fuente
de
voltaje
es
2
2
where
�vs
V when
the
input
is
v
¼
12
V.
The
circuit
should
output vrequire
vs 2
vs
o ¼ 8.5 no
s
2
morevothan
1 mW
from thelavoltage
source.
tenga una salida
5 8.5
V cuando
entrada
sea vs 5 12 V.
p ¼ vs is ¼ vvss
vs¼
vs ¼
no more than 1 mW from the voltage source.
RRin1 ¼
þ¼R12 þs 2R2R1¼þvRs 22 Rin require
p ¼ vs is ¼ vs
El circuito podría requerir no más de 1 mW desde la fuente de
R1 vþs R2  R1vþ
R2  Rin
p

v
i

v
s
s
s
is called the input
voltage Rdivider.
R1resistance
 R2 ofRthe
voltaje.
1  R2
ent
where
where (a) Design a voltage divider to have a gain, g ¼ 0.65.
Rin ¼ R1 þ R2
(b) Design a voltage
to2 have a gain, g ¼ 0.65, and an
Rin ¼divider
R1 þ R
Figure DP 3-7
is called
the
input
resistance
of
divider.
¼
2500
V. voltage
input
resistance,
R
R
5
R
1
R2 the
in
ent
1
is called the input resistance of the voltage
divider.
(a) Design a voltage divider to have a gain, g ¼ 0.65.
DP 3-8 Design the circuit shown in Figure DP 3-8 to have an
(a) Design
a resistencia
voltage
divider
to havetoadel
gain,
¼ 0.65.
se denomina
de divider
entrada
divisor
deg ¼
voltaje.
(b) la
Design
a voltage
have
a ggain,
0.65, and an
s
o
output ioDP
¼ 1.8
Figure
3-7mA when the input is is ¼ 5 mA. The circuit should
(b) Design input
a voltage
dividerRto ¼
have
a gain,
g ¼ 0.65, and an
2500
V.
resistance,
in
Figure require
DP 3-7 no more than 1 mW from the current source.
¼ 2500
V.tenga una ganancia, g 5 0.65.
inputun
resistance,
(a)Diseñe
divisor deRinvoltaje
que
3-8 Design the circuit shown in Figure DP 3-8 to have an
(b)Diseñe un divisor de voltaje que tenga una ganancia, g 5 0.65,DP 3-8DPDesign
the
shown
Figure
DP5 mA.
3-8 to
have
an should
output
io ¼
1.8circuit
mA when
the in
input
is is ¼
The
circuit
Figura
PD
3-7
y una resistencia de entrada, Rent 5 2 500 V.
output iorequire
¼ 1.8 mA
whenthan
the input
is from
is ¼ 5the
mA.
The circuit
should
no
more
1
mW
current
source.
Figure DP 3-5
require no more than 1 mW from the current source.
donde
PD 3-8 Diseñe el circuito que se muestra en la figura PD 3-8 que
DP 3-6 The input to the circuit shown in Figure DP 3-6 is the tenga una salida io 5 1.8 mA cuando la entrada sea is 5 5 mA.
Figure
3-5 current, is. The output is the current io. The El circuito debería requerir no más de 1 mW desde la fuente de
current DP
source
s
o
Figure DP 3-5
output is related to the input by
corriente.
DP 3-6 The input to the circuit
shown in Figure DP 3-6 is the
Figure DP 3-8
R
1 in Figure DP 3-6 is the
DP 3-6current
The input
to the
circuit
io ¼ishown
is ¼isgithe
source
current,
current io. The
s
o
s. The output
Figura
PD
3-5
R
þ
R
current output
sourceiscurrent,
output
1 byis2 the current io. The
s. The
related ito
the input
output is related to the input by
Figure DP 3-8
R1
ioR¼
is ¼ gien
PD 3-6 La entrada al circuito
s la figura PDFigure DP 3-8
1 que se muestra
R1isþ¼Rgi
io ¼
2 s
3-6 es la corriente de la fuente
R1 þ Rde
2 corriente, is. La salida es la
corriente io. La salida se relaciona con la entrada por
R1
io 
is  gis
R1  R2
Figura PD 3-8
Circuitos Eléctricos - Dorf
M03_DORF_1571_8ED_SE_053-107.indd 107
s
s
Alfaomega
4/12/11 5:23 PM
Métodos
de análisis de
circuitos resistivos
CAPÍTULO
E N E STE CAPÍTULO
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Introducción
Análisis de voltajes de nodos de circuitos
con fuentes de corriente
Análisis de voltajes de nodos de circuitos
con fuentes de corriente y de voltaje
Análisis de voltajes de nodos con fuentes
dependientes
Análisis de corrientes de enlaces con
fuentes de voltaje independientes
Análisis de corrientes de enlaces con
fuentes de corriente y de voltaje
Análisis de corrientes de enlaces con
fuentes dependientes
4.1
4.8
4.9
4.10
4.11
4.12
4.13
omparación entre el método de voltajes de
C
nodos y el método de corrientes de enlaces
Análisis de corrientes de enlaces utilizando
MATLAB
Uso de PSpice para determinar los voltajes
de nodos y las corrientes de enlaces
¿Cómo lo podemos comprobar . . . ?
EJEMPLO DE DISEÑO — Despliegue
angular del potenciómetro
Resumen
Problemas
Problemas de PSpice
Problemas de diseño
INTRODUCCIÓN
Para analizar un circuito eléctrico se requiere escribir y despejar un conjunto de ecuaciones. Aplicamos las
leyes de la corriente y el voltaje de Kirchhoff para obtener algunas ecuaciones. Las ecuaciones constitutivas de los elementos del circuito, como la ley de Ohm, proporcionan las ecuaciones restantes. Las variables desconocidas son las corrientes y voltajes de los elementos. El despeje de las ecuaciones proporciona
los valores de las corrientes y los voltajes de los elementos.
Este método funciona bien con circuitos pequeños, pero el conjunto de ecuaciones puede ser muy
extenso, incluso en circuitos de dimensiones moderadas. Un circuito con sólo 6 elementos tiene 6 corrientes del elemento y 6 voltajes del elemento. Podríamos tener 12 ecuaciones en 12 incógnitas. En este
capítulo consideramos dos métodos para escribir un conjunto más pequeño de ecuaciones simultáneas:
• El método de los voltajes de nodos
• El método de las corrientes de enlaces
108
El método de los voltaje de nodos utiliza un nuevo tipo de variable llamado voltaje de nodos. Las “ecuaciones de voltaje de nodos” o, más sencillo, las “ecuaciones nodales” son un conjunto de ecuaciones
simultáneas que representan un circuito eléctrico dado. Las variables desconocidas de las ecuaciones de
voltajes de nodos son los voltajes del nodo. Después de resolver las ecuaciones de los voltajes de nodos,
determinamos los valores de las corrientes y los voltajes de los elementos para los valores de los voltajes
de nodos.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 108
Circuitos Eléctricos - Dorf
4/12/11 5:25 PM
Análisis de voltajes de nodos de circuitos con fuentes de corriente
109
Es más fácil escribir ecuaciones de voltajes de nodos para algunos tipos de circuitos que para
otros. Empezando con el caso más fácil, aprenderemos cómo escribir ecuaciones de voltaje de nodos
para circuitos que constan de:
• Resistores y fuentes de corriente independientes
• Resistores y fuentes de corriente y de voltaje independientes
• Resistores y fuentes de corriente y de voltaje independientes y dependientes
El método de corrientes de enlaces utiliza un nuevo tipo de variable denominado corriente de enlace.
Las “ecuaciones de corrientes de enlaces” o, más sencillo, las “ecuaciones de enlaces” son un conjunto de ecuaciones simultáneas que representan un circuito eléctrico dado. Las variables desconocidas
de las ecuaciones de corrientes de enlaces son las corrientes de enlaces. Después de despejar las ecuaciones de corrientes de enlaces, determinamos los valores de las corrientes y voltajes de los elementos
a partir de los valores de las corrientes de enlaces.
Es más fácil escribir ecuaciones de corrientes de enlaces para algunos tipos de circuitos que
para otros. Empezando con el caso más fácil, aprenderemos cómo escribir ecuaciones de corrientes de
enlaces para circuitos que constan de:
• Resistores y fuentes de voltaje independientes
• Resistores y fuentes de corriente y de voltaje independientes
• Resistores y fuentes de corriente y de voltaje independientes y dependientes
4.2
N Á L I S I S D E V O LTA J E S D E N O D O S D E C I R C U I T O S
A
CON FUENTES DE CORRIENTE
Considere el circuito que se muestra en la figura 4.2-1a. El circuito contiene cuatro elementos: tres
resistores y una fuente de corriente. Los nodos de un circuito son los lugares en que los elementos están
conectados entre sí. El circuito que se muestra en la figura 4.2-1a tiene tres nodos. Lo común es trazar
los elementos horizontal o verticalmente y conectar estos elementos con líneas horizontales y verticales
que representan los cables. En otras palabras, los nodos se dibujan como puntos o bien utilizando líneas
horizontales o verticales. La figura 4.21b muestra el mismo circuito, pero trazado de nuevo de modo
que los tres nodos están dibujados como puntos en vez de líneas. En la figura 4.2-1b, los nodos están
etiquetados como nodo a, nodo b, y nodo c.
R1
R1
b
a
is
R2
R2
R3
is
R3
c
(a)
(b)
R1
Voltímetro
+
va
R2
is
R3
c
(c)
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 109
b
a
–
Voltímetro
+
vb
–
FIGURA 4.2-1 (a) Circuito con tres
nodos. (b) El circuito después de que
se han etiquetado los tres nodos y se
ha seleccionado y marcado un nodo de
referencia. (c) Utilizando voltímetros
para medir los voltajes de los nodos.
Alfaomega
4/12/11 5:25 PM
110
Métodos de análisis de circuitos resistivos
La KCL no se aplica si se llama al nodo de referencia. Cualquier nodo del circuito puede ser seleccionado como nodo de referencia. Con frecuencia elegiremos el nodo al final del circuito para que sea el
nodo de referencia. (Cuando el circuito contiene un alimentador de energía aterrizado, el nodo de tierra
del alimentador de energía es el que se suele seleccionar como nodo de referencia.) En la figura 4.21b, el
nodo c es el seccionado como nodo de referencia y marcado con el símbolo que lo identifica como tal.
El voltaje en cualquier nodo del circuito, respecto del nodo de referencia, se denomina voltaje de
nodos. En la figura 4.2-1b hay dos nodos de voltaje: el voltaje en el nodo a respecto del nodo de referencia, el nodo c, y el voltaje en el nodo b, de nuevo con respecto al nodo de referencia, el nodo c. En la
figura 4.2-1c, se han agregado voltímetros para medir los voltajes en los nodos. Para medir el voltaje de
nodos en el nodo a, conecte el probador rojo del voltímetro en el nodo a y conecte el probador negro en
el nodo c, que es el de referencia. Para medir el voltaje de nodos en el nodo b, conecte el probador rojo
del voltímetro en el nodo b, y el probador negro en el nodo c, que es el nodo de referencia.
Los voltajes de nodos en la figura 4.2-1c se pueden representar como vac y vbc, pero es convencional excluir el subíndice c y referirse a ellos como va y vb. Observe que el voltaje de nodos en el
nodo de referencia es vcc 5 vc 5 0 V porque un voltímetro que mide el voltaje de nodos en el nodo de
referencia podría tener ambos probadores conectados en el mismo punto.
Uno de los métodos comunes de analizar un circuito eléctrico es escribir y despejar un conjunto
de ecuaciones simultáneas denominado ecuaciones nodales. Las variables desconocidas en las ecuaciones de nodos son los voltajes de nodos del circuito. Determinamos los valores de los voltajes del
nodo despejando las ecuaciones nodales.
Para escribir un conjunto de ecuaciones nodales, se hacen dos cosas:
1. Expresar las corrientes del elemento como funciones de los voltajes de nodos.
2. A
plicar la ley de la corriente de Kirchhofff (KCL) en cada uno de los nodos del circuito,
excepto en el nodo de referencia.
Considere el problema de expresar corrientes de elementos como funciones de voltajes de
nodos. Aun cuando nuestro objetivo es expresar corrientes de elementos como funciones, empezaremos por expresar voltajes de elementos como funciones de los voltajes de nodos. La figura 4.2-2
muestra cómo se hace. Los voltímetros en la figura 4.2-2 miden los voltajes de nodos, v1 y v2, en los
nodos del elemento de circuito. El voltaje del elemento se ha etiquetado como va. Aplicando la ley del
voltaje de Kirchhoff al circuito cerrado que se muestra en la figura 4.2-2 resulta
va 5 v1 2 v2
Esta ecuación expresa el voltaje del elemento, va, como una función de los voltajes de nodos, v1 y v2.
(Hay una manera fácil de recordar esta ecuación. Observe la polaridad de referencia del voltaje del
elemento, va. El voltaje del elemento es igual al voltaje del nodo en el nodo cercano a la polaridad de
referencia 1 menos el voltaje de nodos en el nodo cercano a la polaridad de referencia 2.)
Ahora considere la figura 4.2-3. En la figura 4-2-3a, aplicamos lo que hemos aprendido a usar
para expresar el voltaje de un elemento de circuito como una función de voltajes de nodos. El elemento
+ va –
v1
Voltímetro
v2
+
+
v1
v2
–
–
Voltímetro
FIGURA 4.2-2 Voltajes de nodos, v1 y v2, y voltaje del elemento, va, de un elemento del circuito.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 110
Circuitos Eléctricos - Dorf
4/12/11 5:25 PM
E1C04_1
E1C04_1
11/25/2009
11/25/2009
111
111
Análisis de Node
voltajes
de nodos
de circuitos
conwith
fuentes
de corriente
Voltage
Analysis
of
Circuits
Current
Sources
Node
Voltage
Analysis
of Circuits
Circuits
with
Current
Sources
Node
Voltage
Analysis
of
with
Current
Sources
Node Voltage Analysis of Circuits with Current Sources
v1 – v2
vv11 –– vv22
ii =
2
= Rv1 –– v
i = v1R
R v2
v1
R i = v2 R
Rvv
vv
R
R
v111
v222
R
v1
v2
R
+
v1 – v2
–
4.3-3 Los voltajes
+
vv –– vv
–– FIGURA
FIGURE
4.2-3
Node
voltages,
v
+
FIGURE
4.2-3
Node
voltages,
+
v111 – v222
– de nodos,
v1 y4.2-3
v2, y los
voltajes
de vv11
FIGURE
Node
voltages,
+
v1 – v2
–
FIGURE
Node
voltages, v11
and
v2, and4.2-3
element
voltage,
111
111
111
111
111
i=
v1
vv
v111
v1
+
+
+
+
+
v1 – v2
vv –– vv
v111 – v222
v1 – v2
v2
vv
v222
v2
–
––
–
–
Vf
v1
V
Vss
vv11
+
–V
Vss
v1
+
v1
+–
–
+
v1 –+v2–
+–
+
vv11 –– vv22
+
+
v1 – v2
+
v1 – v2
v2
vv
v222
v2
–
––
–
–
and vv2,, vand
and
element
voltage,
elemento,
v2, de (a)
un elemento
and
voltage,
1 2element
vvv22,22 ,,and
voltage,
vand
of
aaelement
(a)
generic
circuit
11 �
v
�
of
(a)
generic
circuit
de circuito
(b) fuente
de
v1 � v2 ,genérico,
of a (a) generic
circuit
velement,
a
(a)
generic
circuit
element,
(b)
voltage
source,
and
1 � v2 , of
(b)
voltage
source,
and
voltaje,
y
(c)
resistor.
(a)
(b)
(c)
element, (b) voltage source, and
element,
(b)
voltage
source,
and
(c)
resistor.
(a)
(b)
(c)
(c)
resistor.
(a)
(b)
(c)
(c)
resistor.
(a) en la figura 4.2-3a podría
(b) ser cualquier cosa: un
(c)resistor, una (c)
resistor.
de circuito
fuente
de corriente, una fuen(a)
(b)
(c)
anything:
a resistor,
a current
source,
dependent
voltage
source,
and
so
In
4.2-3b
and
c,
anything:
resistor,
current source,
source,
dependent
voltage
source,
and
so on.
on. In
In Figures
Figures
4.2-3b and
andde
c,
te de corrientes
dependiente,
etcétera.aaa En
las figuras
4.2-3b
y c, and
consideramos
tipos específicos
anything:
aaa resistor,
aaa current
voltage
source,
so
4.2-3b
anything:
resistor,
current
source,
a dependent
dependent
voltage
source,
and
so on.
on.
In Figures
Figures
4.2-3b
and c,
c,
we
consider
specific
types
of
circuit
element.
In
Figure
4.2-3b,
the
circuit
element
is
aa voltage
source.
we
consider
specific
types
of
circuit
element.
In
Figure
4.2-3b,
the
circuit
element
is
voltage
source.
elementos
de
circuito.
En
la
figura
4.2-3b,
el
elemento
de
circuito
es
una
fuente
de
voltaje.
El
voltaje
we
consider
specific types
of
element.
In Figure
4.2-3b,
the
element
is
voltage source.
we
consider
types
of circuit
circuit
element.
Figure
4.2-3b,
thelacircuit
circuit
element
is aaV
source.
, and
once
as
The
element
voltage
has
been
represented
twice,
once
as
the
voltage
source
voltage,
ssvoltage
and
once
as aaa
The
elementspecific
voltage
has been
been
represented
twice,
onceelas
as
the voltage
voltage
source
voltage,
VV
de elemento
se
ha representado
dos veces,
unaIncomo
voltaje
de
fuente
de
voltaje,
, y otra
como
,,,sand
once
as
The
element
voltage
has
represented
twice,
once
the
source
voltage,
V
s
and
once
as
The
element
voltage
has
been
represented
twice,
once
as
the
voltage
source
voltage,
V
vv22v..1Noticing
that
the
reference
polarities
for
V
and
�
v
are
the
function
of
the
node
voltages,
vv11 �
sv
1
2
s
una
función
de
los
voltajes
de
nodos,
2
v
.
Es
interesante
observar
que
las
polaridades
de
referencia
�
Noticing
that
the
reference
polarities
for
V
and
v
�
v
are
thea
function
of
the
node
voltages,
1
2
s
2
vv2 .. Noticing
that
the
reference
polarities
for
V
vv1 �
vv2 are
the
function
of
the
node
voltages,
vv1 �
s and
�
Noticing
that
the
reference
polarities
for
V
and
�
are
the
function
of
the
node
voltages,
same
þ
the
left),
we
2
1
2
s
para V(both
2 on
v2 son
mismas
(ambas
1 a la izquierda), escribimos
same
(both
on
the las
left),
we 1write
write
s y v1 þ
same
same (both
(both þ
þ on
on the
the left),
left), we
we write
write
V
5
v
2
v
¼
v
�
v
V
s
1
2
� vv22
V s ¼ vv11 �
V
¼atengamos
vvoltage
V ss ¼
Éste es
un important
resultado importante.
Cuantas
veces
1 � v2 una fuente de voltaje conectada entre dos
This
is
an
result.
Whenever
we
have
source
connected
between
two
nodes
of
This
isde
anun
important
result.
Whenever
weelhave
have
voltage
sourcedeconnected
connected
between
two función
nodes of
ofdeaaa
This
an
important
Whenever
we
aaa voltage
source
two
nodes
nodosis
circuito,result.
podemos
expresar
voltaje
de la fuente
voltaje, Vbetween
una
s, como
This
is
an
important
result.
Whenever
we
have
voltage
source
connected
between
two
nodes
of
,
as
a
function
of
the
node
voltages,
v
and
circuit,
we
can
express
the
voltage
source
voltage,
V
s
1
as aa function
function of
of the
the node
node voltages,
voltages, vv11 and
and vvv222...a
circuit,
we can
can
expressv the
the
voltage
source voltage,
voltage, V
Vss,, as
circuit,
we
express
source
los voltajes
de nodos,
y vvoltage
1the
2.
,
as
a
function
of
the
node
voltages,
v
and
v 2.
circuit,
we
can
express
voltage
source
voltage,
V
Frequently,
we
know
the
value
of
the
voltage
source
voltage.
For
example,
suppose
that
s
1
Frequently,
weconocemos
know the
the value
value
ofdel
the
voltage
source
voltage.
ForPor
example,
suppose
that
Frequently,
we
know
the
voltage
source
For
example,
Con frecuencia
el
valorof
voltaje
de la
fuentevoltage.
de voltaje.
ejemplo,suppose
suponga that
que
Frequently,
we
know
the
value
of
the
voltage
source
voltage.
For
example,
suppose
that
¼
12
V.
Then
V
s
¼ 12
12 V.
V.Entonces
Then
V
Vs 5
V
¼ 12
12 V.
V. Then
Then
V sss ¼
12 5
v 2
v
12
¼
�
12 ¼
¼ vvv1111 �
� vvv2222
12
12
¼
v
�
v
Esta
ecuación
se
relaciona
con
los
valores
de
los
dos
voltajes
de nodos.
1
2
This
equation
relates
the
values
of
two
of
the
node
voltages.
This equation
equation
relates the
the
valueslaof
offigura
two of
of
the node
node
voltages.
This
relates
values
two
the
voltages.
A
continuación,
considere
4.2-3c.
En
ella
el
elemento
de circuito es un resistor. AplicaThis equation
relates Figure
the values
of two
of the 4.2-3c,
node voltages.
Next,
consider
4.2-3c.
In
Figure
the
circuit
element
is
resistor.
We
will
use
Next,
consider
Figure
4.2-3c.
Incorriente
Figure
4.2-3c,
the circuit
circuit
element
is aaa resistor.
resistor.
Wedewill
will
use
consider
Figure
4.2-3c.
In
Figure
4.2-3c,
the
element
is
We
use
remosNext,
la
ley
de
Ohm
para
expresar
la
del
resistor,
i,
como
una
función
de voltajes
nodos.
Next,
consider
Figure
4.2-3c.
In
Figure
4.2-3c,
the
circuit
element
is
a
resistor.
We
will
use
Ohm’s
law
to
express
the
resistor
current,
i,
as
a
function
of
the
node
voltages.
First,
we
express
the
Ohm’s
law
to
express
the
resistor
current,
i,
as
a
function
of
the
node
voltages.
First,
we
express
the
Ohm’s
law
to
express
the
resistor
current,
as
aa function
of
the
node
voltages.
First,
we
express
En primer
lugar,
expresamos
el voltaje
deli,
resistor
como una
función
de voltajes
de nodos,
v1 2 the
v.
Ohm’s
law
to
express
the
resistor
current,
i,
as
function
of
the
node
voltages.
First,
we
express
�
v
.
Noticing
that
the
resistor
voltage,
v
�
vv222,,
resistor
voltage
as
a
function
of
the
node
voltages,
v
11 � v22 . Noticing that the resistor voltage, v11 � the
resistor
voltage
as
a
function
of
the
node
voltages,
v
Como elvoltage
voltajeas
dela function
resistor, vof
2
v2node
, y lavoltages,
corriente,v1i,�
sevapegan
a la that
convención
pasiva,
aplicamos
la
.
Noticing
the
resistor
voltage,
v
�
v
resistor
the
2
1
2,
1
. Noticing
resistor
voltage
asi,
aadhere
function
node voltages,
v1 �we
and
the
current,
to
the
passive
convention,
use
Ohm’s
law
to
write
and
theOhm
current,
i,escribir
adhere
to of
thethe
passive
convention,
wev2use
use
Ohm’sthat
lawthe
to resistor
write voltage, v1 � v2 ,
ley de
para i,
and
the
current,
adhere
to
the
passive
convention,
we
Ohm’s
law
to
write
and the current, i, adhere to the passive convention,
vv11 �
22 use Ohm’s law to write
� vvvwe
i¼
2
¼ vv11 �
�
v
iii ¼
R
R 2
¼ R
R
En ocasiones,
el valor
de resistance.
la resistencia.
ejemplo,
cuando
5 this
8 V,equation
esta ecuación
se
Frequently,
we
know
of
For
example,
when
R
becomes
Frequently,
weconocemos
know the
the value
value
of the
the
resistance.
For Por
example,
when
R¼
¼ 88RV,
V,
this
equation
becomes
Frequently,
example,
when
convierte we
Frequently,
we know
know the
the value
value of
of the
the resistance.
resistance. vvFor
For
example,
when R
R¼
¼ 88 V,
V, this
this equation
equation becomes
becomes
vv22
11 �
�
i¼
¼ vv11 �
� vv22
iii ¼
¼ 888
and
This
equation
expresses
the
resistor
current,
i,
as
a8function
of
the
node
voltages,
and vvv2v22...1 y v2.
This
equation expresa
expresses
the
resistor
current,
i,i,as
ascomo
function
of the
thedenode
node
voltages,devvv111nodos,
Esta ecuación
la the
corriente
delcurrent,
resistor,i,
una función
los voltajes
and
This
expresses
resistor
aaa function
of
voltages,
and
v2. to
This equation
equation
expresses
the
resistor
current,
i,
as
function
of
the
node
voltages,
v
Next,
let’s
write
node
equations
to
represent
the
circuit
shown
in
Figure
4.2-4a.
The
input
this
1
A continuación,
escribamos
ecuaciones
nodales
para representar
el circuito
seinput
muestra
en
Next,
let’s write
write node
node
equations
to represent
represent
the circuit
circuit
shown in
in Figure
Figure
4.2-4a.que
The
input
to this
this
Next,
let’s
equations
to
the
shown
4.2-4a.
The
to
Next,
let’s
write
node
equations
to
represent
the
circuit
shown
in
Figure
4.2-4a.
The
input
to
this
.
To
write
node
equations,
we
will
first
express
the
resistor
currents
as
circuit
is
the
current
source
current,
i
ss.circuito
la
figura
4.2-4a.
La
entrada
a
este
en
la
corriente
de
la
fuente
de
corriente,
i
.
Para
escribir
To
write
node
equations,
we
will
first
express
the
resistor
currents
as
circuit
is
the
current
source
current,
i
s
write
node
equations,
we
will
first
express
the
resistor
currents
as
circuit
is
the
current
source
current,
iis.. To
To
write
node
equations,
we
will
first
express
the
resistor
currents
as
circuit
is
the
current
source
current,
functions
of
the
node
voltages
and
then
apply
Kirchhoff’s
current
law
at
nodes
a
and
b.
The
resistor
voltages
s
ecuaciones
nodales,
primero
expresaremos
las
corrientes
del
resistor
como
funciones
de
voltajes
de
functions
of
the
node
voltages
and
then
apply
Kirchhoff’s
current
law
at
nodes
a
and
b.
The
resistor
voltages
functions
of
node voltages
and
then
apply
Kirchhoff’s
current
law
at
aa and
The resistor
voltages
functions
of the
the
voltages
and
then
apply
Kirchhoff’s
current
lawthen
at nodes
nodes
andab.
b.currents
resistor
voltages
are
expressed
as
functions
of
the
node
voltages
in
Figure
4.2-4b,
and
the
resistor
are
expressed
nodos
y después
aplicaremos
lanode
ley voltages
de
la corriente
de4.2-4b,
Kirchhoff
en
los
yThe
b. Los
voltajes
del
are
expressed
asnode
functions
of the
the
node
voltages
in Figure
Figure
4.2-4b,
and
then
thenodos
resistor
currents
are
expressed
are
expressed
as
functions
of
in
and
then
the
resistor
currents
are
expressed
are
expressed
as
functions
of
the
node
voltages
in
Figure
4.2-4b,
and
then
the
resistor
currents
are
expressed
as
functions
of
the
node
voltages
in
Figure
4.2-4c.
resistor
se
expresan
como
funciones
de
los
voltajes
de
nodos
en
la
figura
4.2-4b,
y
luego
las
corrientes
as
functions
of
the
node
voltages
in
Figure
4.2-4c.
as
functions of
the
node
voltages
in
4.2-4c.
as
of expresan
the
node como
voltages
in Figure
Figure
4.2-4c.
The
node
equations
representing
the
in
Figure
4.2-4
are
applying
Kirchhoff’s
delfunctions
resistor
se
funciones
decircuit
los voltajes
de nodos
en obtained
la figura by
4.2-4c.
The node
node
equations
representing
the
circuit
in Figure
Figure
4.2-4 are
are
obtained
by
applying Kirchhoff’s
Kirchhoff’s
The
equations
representing
the
circuit
in
4.2-4
obtained
by
applying
The
node
equations
representing
the
circuit
in
Figure
4.2-4
are
obtained
by
applying
Kirchhoff’s
Las
ecuaciones
nodales
que
representan
el
circuito
en
la
figura
4.2-4
se
obtienen
aplicando
la
current
law
at
nodes
a
and
b.
Using
KCL
at
node
a
gives
current
law
at
nodes
a
and
b.
Using
KCL
at
node
a
gives
current
law at
a and
KCL at
aa gives
current
at nodes
nodes
and b.
b. Using
Using
atanode
node
ley de lalaw
corriente
de aKirchhoff
en losKCL
nodos
la
KCL
en
el
nodo
a
nos
da
vvyaa b. Usar
vvaagives
�
v
� vbb
vva þ
�
ð4:2-1Þ
i ¼
¼R
þ vvaa R
ð4:2-1Þ
�1 vvbb
a2 þ
ð4:2-1Þ
iiisss ¼
R
R
2
1
¼
þ
ð4:2-1Þ
(4.2-1)
s
R
R
2
1
R2
R1
Similarly,
the
KCL
equation
at
node
b
is
Similarly, the
the KCL
KCL equation at
at node bb is
is
Similarly,
Del mismothe
modo,
ecuaciónatdenode
KCLbenis elvnodo
b es
Similarly,
KCLlaequation
equation
node
�
� vvvbbb ¼ vvvbbb
vvvaaa �
ð4:2-2Þ
ð4:2-2Þ
�v ¼R
v
aR
ð4:2-2Þ
(4.2-2)
R111 b ¼
Rb333
¼R
ð4:2-2Þ
R
R1
R3
If
¼
R
¼
0:5
V,
A,
and
Eqs.
4.2-1
and
4.2-2
may
be
rewritten
as
If
R
¼
V;
R222 5
¼R
R333 5
¼ 0.5
0:5 V
V,eand
and
¼
A, ecuaciones
and Eqs.
Eqs. 4.2-1
4.2-1
and
4.2-2semay
may
be rewritten
rewritten
as
Si R
R111 ¼
5 111 V;
V, R
R
is 5iiss4¼
A,44las
4.2-1and
y 4.2-2
pueden
reescribiras
así
If
4.2-2
be
If R
R11 ¼
¼ 11 V;
V; R
R22 ¼
¼R
R33 ¼
¼ 0:5
0:5 V,
V, and
and iiss ¼
¼ 44 A,
A, vvand
and
Eqs.
vvaa and 4.2-2 may be rewritten as
vvbb 4.2-1
aa �
�
v
v
�
v
ð4:2-3Þ
4¼
(4.2-3)
þ 0:5
ð4:2-3Þ
¼ vaa � vbb þ
vaa
ð4:2-3Þ
444 ¼
0:5
þ 0:5
ð4:2-3Þ
¼ 111 þ
v �
� vvv1bbb ¼ vvvbbb0:5
vvvaaa �
ð4:2-4Þ
(4.2-4)
¼ 0:5
ð4:2-4Þ
vb
a�
ð4:2-4Þ
1 vb ¼
0:5
ð4:2-4Þ
111 ¼ 0:5
0:5
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 111
Alfaomega
4/12/11 5:25 PM
112
112
112
112
112
112
112
Methods
of
Analysis
of
Resistive
Circuits
Methods
of
Analysis
of
Resistive
Circuits
Methods
Methods
ofof
Analysis
ofAnalysis
Analysis
ofof
Resistive
ofResistive
Resistive
Circuits
Circuits
Métodos
de
análisis
de
circuitos
resistivos
Methods
Circuits
R
R1R111
R
b
RR
bb
1R
b
11
b bb
v
+
–
v
1
+
–
v1v11– –––
++v1vv
+++
11 – ++
++
+
+ +
R
R3R333
vbvbbb RRR
R2R222
vv
R
v
R
v
RR
b v–bb 3R
2R22
33
–
–
–
– ––
a
a aaaaa
i
isisiiisssis
s
+
++
+
+++
v
v
vavaaa
vavv
–aa
– –––––
(a)
(a)
(a)
(a)
(a)
(a)
a
a aaaaa
i
isisiiisssis
s
+
++
+
+++
v
v
vavaaa
vavv
–aa
– –––––
R
R1R111
R
RR
1R
11
b
bb
b
b bb
(v
–
v
)
+
–
(v(v
vbvbbb) ))–)––
a–
++
(v(v
+
a–aa–
v––v)v
++(v
– –+
a(v
+
aa–b vbb)+–++
++
+
R
R3R333
vbvbbb RRR
R2R222
vv
R
v
R
v
RR
b v–bb 3R
2R22
33
–
–
–
– ––
(b)
(b)
(b)
(b)
(b)
(b)
i
isisiiisssis
s
v
va–vaaa–v––––vv
vbvbbb
v
v
vavv
–b v
aaR
bb
R
R
1
R
R1R111
R
1
1
R
R
11 RRR
a
b
R
1
a
bb
b
a aaaa 1 1R11
b bb
v
vbvbbb
v
v
v
aa
vbvv
v
v
+
va vavaa
bb
vbvbbb) )))–)–––
(v(v
a–
++
(v(v
+(v
–––v)v
v
(v
+
a
R
a–aa–
v
+
–
–
R3R333
+
+
R
–
v
)
(v
a
a
b
b
+
–
R
++ a
++ RR
R2R222 +++
b +++
R
3R
RR
33
+
+
2R
2
2
v
R
R3R333
vavaaa
vbvbbb RRR
R2R222 v vv
R
v
v
R
vbvv
RR
3
av
a
b 3R
2R
2
3
a
b
–
–
2
– –––––
– –––––
(c)
(c)
(c)
(c)
(c)
(c)
FIGURA 4.2-4
FIGURE
4.2-4
(a)
Circuito
con
FIGURE
4.2-4
FIGURE
FIGURE
4.2-4
4.2-4
FIGURE
4.2-4
(a)
A
circuit
with
tres
resistores.
(b) three
(a)
A
circuit
with
three
(a)(a)
(a)
AA
circuit
Acircuit
circuit
with
with
three
three
with
resistors.
(b)
The
Los
voltajes
de
losthree
resistors.
(b)
The
resistors.
resistors.
(b)
(b)
The
The
resistors.
(b)
The
resistores
expresados
resistor
voltages
resistor
voltages
resistor
resistor
voltages
voltages
resistor
voltages
como
funciones
de
expressed
as
functions
expressed
as
functions
expressed
expressed
as as
functions
asfunctions
functions
expressed
los
voltajes
de
nodos.
of
the
node
voltages.
of
the
node
voltages.
of(c)
the
ofLas
the
node
node
voltages.
voltages.
of
the
node
voltages.
corrientes
de
(c)
The
resistor
currents
(c)
The
resistor
currents
(c)
(c)
The
The
resistor
resistor
currents
currents
(c)
The
resistor
currents
los
resistores
expresadas
expressed
as
functions
expressed
as
functions
expressed
expressed
as as
functions
asfunctions
functions
expressed
como
funciones
de los
of
the
node
voltages.
of
the
node
voltages.
ofvoltajes
the
ofthe
the
node
node
voltages.
voltages.
of
node
voltages.
de nodos.
Solving
Eq.
for
vbb gives
El
despeje
de4.2-4
la ecuación
4.2-4
Solving
Eq.
4.2-4
for
b gives
Solving
Solving
Eq.
Eq.
4.2-4
4.2-4
forfor
for
vbvvvgives
gives para vb resulta
Solving
Eq.
4.2-4
bb gives
v
vavvvaaaaa
vbb ¼
¼
bb ¼
¼
vbvvv¼
b 33
333
Substituting
Eq.
4.2-5
into
Eq.
4.2-3
gives
Al
sustituir
la
ecuación
4.2-5
en
la
ecuación
4.2-3
da
Substituting
Eq.
4.2-5
into
Eq.
4.2-3
gives
Substituting
Substituting
Eq.
Eq.
4.2-5
4.2-5
into
into
Eq.
Eq.
4.2-3
4.2-3
gives
gives
Substituting
Eq.
4.2-5
into
Eq.
4.2-3
gives
v
8
vavvvaaaaaþ
8 888vvaa
2v
44 ¼
vaa �
aa ¼
þ
2v
¼
�
¼
aa ¼
2v
þ2v
4 4¼4¼
¼
vavv�
vaaa�
� 33þþ
¼
¼
v
3a vaa
a2v
a
3 33
3 333va
Solving
Eq.
4.2-6
for
v
gives
aa4.2-6
Al
despejar
la4.2-6
ecuación
para va da
Solving
Eq.
4.2-6
for
a gives
Solving
Solving
Eq.
Eq.
4.2-6
4.2-6
forfor
for
vavvvgives
gives
Solving
Eq.
aa gives
3
3 333 V
vaa ¼
¼
VV
aa ¼
¼
vavvv¼
2V V
a
2 222
Finally,
Eq.
4.2-5
gives
Finally,
Eq.
4.2-5
gives
Finalmente,
la
ecuación
Finally,
Finally,
Eq.
Eq.
4.2-5
4.2-5
gives
gives4.2-5 da
Finally,
Eq.
4.2-5
gives
1
1 111 V
vbb ¼
¼
VV
bb ¼
¼
VV
vbvvv¼
b 22
222
Thus,
the
node
voltages
of
this
circuit
are
Thus,
the
node
voltages
of
this
circuit
are
Thus,
Thus,
thethe
the
node
node
voltages
voltages
ofof
of
this
this
circuit
circuit
are
Entonces,
los
voltajes
de
nodos
de
esteare
circuito
son
Thus,
node
voltages
this
circuit
are
33
1
3
1 111 V
3
V
and
vvbb ¼
vvaa ¼
3V VV
¼
and
¼
VV
aa ¼
bb ¼
v
and
v
¼
and
¼
¼
vavv¼
y
2
2V V
b
V
and
v
a
b
2 222
2 222
ð4:2-5Þ
(4.2-5)
ð4:2-5Þ
ð4:2-5Þ
ð4:2-5Þ
ð4:2-5Þ
(4.2-6)
ð4:2-6Þ
ð4:2-6Þ
ð4:2-6Þ
ð4:2-6Þ
ð4:2-6Þ
X
A
P
Node
Equations
XXM
AAM
M
PPEL
LLE
EE 4
E
X
A
M
P
L
E
442....-2
2
11 Node
Node
Equations
EE
X
PM
-211----1
Node
Equations
Equations
jEeA
m
pM
lLP
oL
4 .4
Ecuaciones
nodales
E
X
A
E
2
1
Node
Equations
Determine
the
value
of
the
resistance
R
in
the
circuit
shown
in
Figure
4.2-5a.
Determine
the
value
of
the
resistance
Rin
in
the
circuit
shown
in
Figure
4.2-5a.
Determine
Determine
the
the
value
value
ofof
of
thethe
the
resistance
resistance
Rel
in
the
the
circuit
circuit
shown
shown
inin
in
Figure
Figure
4.2-5a.
Determine
the
value
resistance
R
in
the
circuit
shown
Figure
4.2-5a.
Determine
el
valor
de
la
resistencia
RRen
circuito
que
se
muestra
en la4.2-5a.
figura
4.2-5a.
Solution
Solution
Solución
Solution
Solution
Solution
Let
vaa el
denote
the
node
voltage
aa and
vvbb denote
node
voltage
at
node
b.
voltmeter
in
Figure
4.2-5
Let
denote
the
node
voltage
at
node
and
the
node
voltage
at
node
b.
The
voltmeter
in
Figure
4.2-5
Sea
voltaje
del
nodo
a, y at
vbat
elnode
voltaje
devbnodos
enthe
elthe
nodo
b.
El
voltímetro
en
laThe
figura
4.2-5 in
mide
el valor
del
b denote
Let
Let
vavvvdenote
vaaaadenote
denote
thethe
the
node
node
voltage
voltage
node
at
node
a and
and
denote
the
node
node
voltage
voltage
at at
node
atnode
node
b.b.
The
b.
The
voltmeter
voltmeter
Figure
inFigure
Figure
4.2-5
4.2-5
Let
node
voltage
at
node
aaand
vvdenote
the
node
voltage
The
voltmeter
in
4.2-5
bbdenote
.
In
Figure
4.2-5b,
the
resistor
currents
are
expressed
as
measures
the
value
of
the
node
voltage
at
node
b,
v
b
.
In
Figure
4.2-5b,
the
resistor
currents
are
expressed
as
measures
the
value
of
the
node
voltage
at
node
b,
v
b
voltaje
de
nodos
enofelof
nodo
b,
vvoltage
En
la at
figura
4.2-5b,
corrientes
de losthe
resistores
se
expresan
como
funciones
In
In
Figure
Figure
4.2-5b,
4.2-5b,
the
resistor
resistor
currents
currents
areare
are
expressed
expressed
asas
as
measures
measures
thethe
the
value
value
of
the
the
node
node
voltage
at
node
node
b,b,b,
vbv.vbblas
b. voltage
b. . In
Figure
4.2-5b,
the
resistor
currents
expressed
measures
value
the
node
at
node
functions of the node voltages. Apply KCL at node a to obtain
functions
of
the
node
voltages.
Apply
KCL
atat
node
to
obtain
de
los voltajes
de
nodos.
Aplique
la KCL
enatelat
nodo
obtener
functions
functions
ofof
of
thethe
the
node
node
voltages.
voltages.
Apply
Apply
KCL
KCL
node
node
aaato
apara
to
obtain
obtain
functions
node
voltages.
Apply
KCL
node
obtain
vvaa a vvto
vvbb
aa �
�
a
a
v
v
v
�
v
v
�
vbbb ¼
þ
0
44 þ
a
a
a
a
bv
v
v
�
a
a
þ
¼
þ
þ
þ
¼
0 000
4þ
4
þ
þ 55 ¼¼
4 þ 10
10
1010
10 5 55
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 112
Circuitos Eléctricos - Dorf
4/12/11 5:25 PM
Node
Node Voltage
Voltage Analysis
Analysis of
of Circuits
Circuits with
with Current
Current Sources
Sources
Voltage
Analysis
of
Circuits
with
Current
Sources
Análisis de Node
voltajes
de
nodos
de
circuitos
con
fuentes
de corriente
Node Voltage Analysis of Circuits with Current
Sources
113
113
113
113
113
55 . . 00 00
5
.. 0
5Voltmeter
0 0
0
Voltmeter
Voltmeter
Voltímetro
Voltmeter
aa
aa
44AA
4A
4A
55ΩΩ
5
5Ω
Ω
Ω
10
10ΩΩ RR
10 Ω
R
Ω
R
10 Ω
bb
b
b
44AA
4A
4A
44AA
4A
4A
vva
a
vva
10
v10
aa
10
10
vva ––vvb
a
b
vvaa 5–– vvbb
aa a 5 b
bb
5
5
aa
bb
55ΩΩ
vvb
5
5Ω
Ω
b
Ω
RR
10
vvRbb 44AA
10ΩΩ
R
b 4A
10 Ω
R
Ω
R
10 Ω
R
R 4A
(a)
(a)
(a)
(a)
(b)
(b)
(b)
(b)
Using
Using vvbb ¼
¼55VV gives
gives
Using
v
¼
b
Using vbvb¼5555V
VVgives
gives
Utilizar
da
FIGURE
FIGURE 4.2-5
4.2-5 (a)
(a) The
The
FIGURA
4.2-5
(a)
circuit
for
Example
4.2-1.
FIGURE
4.2-5
(a)
The
circuit
for
Example
4.2-1.
FIGURE
4.2-5
(a)
The
Circuito
para
elafter
ejemplo
(b)
circuit
the
circuit
for
Example
(b)The
The
circuit
after4.2-1.
the
circuit
for
4.2-1.
4.2-1.
(b)
ElExample
circuito
(b)
The
circuit
after
the
resistor
currents
are
resistor
currents
are
(b)
The
circuit
after
the
después de las corrientes
expressed
as
functions
resistor
currents
are
expressed
as
of
resistor
currents
are of
del
resistor
se functions
expresan
the
node
voltages.
expressed
as
functions
of
the node
voltages.
expressed
as functions
como
funciones
de los of
the
the node
nodedevoltages.
voltages.
voltajes
nodos.
vvaa vvaa�
�55¼ 0
44þ
þvva 5�
þ10
vvaa þ
5 ¼0
a�5
10þ
44 þ
þ 55 ¼
¼ 00
þ 10
Solving
10
5
Solving for
for vvaa,, we
we get
get
Solving
for
we
Al
despejar
Solving
for vvvaaa,,, obtenemos
we get
get
vvaa ¼
¼�10
�10VV
210
vvvaaa 5
¼
�10
V
¼
�10
V
Next,
apply
KCL
at
node
b
to
obtain
Next, apply KCL at node b to obtain
�
�
A
continuación,
aplicar
la
KCL
en
el
nodo
b
para
obtener
�
�
Next,
apply
KCL
at
node
b
to
obtain
v
�
v
v
Next, apply KCL at node b to obtain
vaa � vbb� þ vbb� 4 ¼ 0
�
þvvRb � 4 ¼ 0
��
�vva �
55 vvbb � þ
a�
b
�
þ RR �
� 44 ¼
¼ 00
�
5
Using
v
¼
�10
V
and
v
¼
5
V
gives
5
R
a
b
Using va ¼ �10 V and vb ¼ 5 V gives
Utilizando
V yvvbv¼
55 5VVgives
resulta ���10 � 5�� 5
Using
V
a 5 210
Using vvaa ¼
¼v�10
�10
V and
and
b b¼ 5 V gives
� �10 � 5�
5 4¼0
�
þR
�4¼0
���10
�
55� þ
55 �
5
�10
�
5
R�
þ
44 ¼
�
þ
�
¼ 00
�
55
R
R
Finally,
solving
for
R
gives
Finally, solving for R gives
Finalmente,
despejar
da
Finally,
for
Finally, solving
solving
for R
RR,gives
gives
R
¼
¼555V
V
RR5
V
R
R¼
¼ 55 V
V
AAM
EEjEeXXm
pMlPP
oLLE4
.42..2
-22--22
E4
E
PLE 4.2-2
E XX AA M
MPLE 4.2-2
Node
Ecuaciones
nodales
Node Equations
Equations
Node
Node Equations
Equations
Obtain
equations
for
circuit
in
Obtener
lasnode
ecuaciones
nodales
para
el circuito
en4.2-6.
la figura 4.2-6.
Obtainthe
the
node
equations
forthe
the
circuit
inFigure
Figure
4.2-6.
Obtain
the
node
equations
for
the
circuit
in
Figure
Obtain the node equations for the circuit in Figure 4.2-6.
4.2-6.
Solución
Solution
Solution
Solution
Sea
eldenote
voltaje
nodos
en el nodo
a, va,a,
Let
the
voltage
atat node
vvbbvoltaje
denote
b el
Solution
Let vvvaaa denote
thedenode
node
voltage
node
denotede
aa
aa a
i2i
2
ii2i2
2
RR
55
R
R
R55 5
bb
b
b
b
Let
vvnode
denote
the
node
voltage
a,
vvthe
denote
nodos
el
nodo
vc el
voltaje
nodos
el
nodo
denote
node
the
voltage
at
b,b, and
vvcnode
aa en
b
Let
denote
theb,
node
voltage
atde
node
a, en
denote
node c.
the node
voltage
atynode
node
andat
bthe
c denote
denote
the
node
the
node
voltage
at
node
b,
and
v
Aplique
la
KCL
al
nodo
a
para
obtener
voltage
at
node
c.
Apply
KCL
at
node
a
to
obtain
c
RR1
RR2
RR4
i1i
denote
the
node
the
node
voltage
at
node
b,
and
v
voltage
at �node c. Apply
KCL
at cnode
to obtain
1
2
4
1
��
�� KCL
�� at
��aa to
��
RR3
voltage
at
node
c.
Apply
node
obtain
�
R
R
R
i
R
R
R
1
2
i
1
3
1
voltage
at
node
c.
Apply
KCL
at
node
a
to
obtain
1
2
vvaa�
vvcc�
vvaa�
vvcc�
vvaa�
vvbb�
R1
R2
R44 4
i1
�
�
�
R
R
�
�
�
�
R33 3
þii11�
þii22�
¼00
��vva R�
� �va �
� �va �
vc � þ
vc � þ
vb � ¼
�
�
11vc þ i � va R
22vc þ i � va R
55vb ¼ 0
a�
R
R
R
�
1
2
þ i1 �
þ i2 �
¼0
�
cc
R
R
R
RR6
R1los
R22 equation
R55
6
1 the
Separate
terms
vvaa va
Separe
términos
dethis
esta
ecuaciónthat
queinvolve
impliquen
cc
Separate
the
terms of
of
this
equation
that
involve
R
R
c
R66 6
Separate
the
terms
of
this
equation
that
involve
v
from
the
terms
that
involve
v
and
the
terms
that
a
de
los
términos
que
impliquen
v
y
los
que
incluyan
v
b
Separate
the
terms
of
this
equation
that
involve
v
b
c
from the terms that involve vb and the terms that
a
FIGURE
4.2-6
The circuit
for
Example
4.2-2
FIGURE
4.2-6
forpara
Example
4.2-24.2-2.
from
the
that
involve
vvb and
that
involve
to
FIGURA
4.2-6The
Elcircuit
circuito
el ejemplo
para
obtener
from
thevvccterms
terms
that �
involve
and the
the terms
terms
that FIGURE
involve
to obtain.
obtain.
b�
4.2-6
The
circuit
for
Example
��
�
�
�
FIGURE 4.2-6 The circuit for Example 4.2-2
4.2-2
�
�
�
�
�
to
obtain.
involve
v
1
11vcc to11obtain.
1
1
11�
involve
�
�
�
�
� 11 þ
�vvaa�
� 11�
�vvbb�
� 11 þ
þ
þ
þ
�
�
þ
¼ii11þ
þii22
1
1
112 �vvcc ¼
RR
111 þ RR
122 þ RR
155 v � RR
155 v � RR
111 þ RR
ii1 þ
ii2
2 v
aa �
bb �
cc ¼
v
v
v
þ
þ
þ
¼
þ
1
2
R
R
R
R11 R
R22 R
R55
R55
R11 R
R22
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 113
i3i
3
ii3i3
3
Alfaomega
4/12/11 5:26 PM
E1C04_1
11/25/2009
114
114
114
114
114
Methods of Analysis of Resistive Circuits
Métodos of
deAnalysis
análisis of
deResistive
circuitosCircuits
resistivos
Methods
Methods of Analysis of Resistive Circuits
There is a pattern in the node equations of circuits that contain only resistors and current sources. In the node
the
sum of the
reciprocals
ofsólo
the
resistances
all resistors
connected
to
equation
nodeena,las
thethe
coefficient
of va is of
There
aatpattern
in
node equations
that
contain
only
resistors
and current
sources.
In the En
node
Hay
unispatrón
ecuaciones
nodales
decircuits
circuitos
que
contienen
resistores
yof
fuentes
de corriente.
la
There
aat
pattern
node
of de
circuits
that
contain
only
resistors
andlascurrent
sources.
Inconnected
thelos
node
minus
sum
ofthe
theresistances
resistances
ofallall
resistors
node a.is The
coefficient
of
vbelisequations
the
sum
of
reciprocals
of
of
resistors
to
equation
node
a, in
thethe
coefficient
of va isthe
ecuación
nodal
en
el
nodo
a,
coeficiente
va of
es the
la suma
de los
recíprocos
de
resistencia
de connected
todos
rethevsum
sum
of the
the
reciprocals
of
the
resistances
ofof
allallresistors
to
equation
at node
a, the
coefficient
of va isthe
between
node
b and
node
is minus
sumdeoflathesuma
of the
resistances
ofconnected
all
resistors
minus
reciprocals
ofreciprocals
thede
resistances
connected
node
a. conectados
The
coefficient
ofa. vThe
sistores
al
nodo
a.
Elcoefficient
coeficiente
vof
lathe
resta
los recíprocos
deresistors
las resistencias
de
cde
b is
b es
is
minus
the
sum
of
the
reciprocals
of
the
resistances
of
all
resistors
connected
node
a.
The
coefficient
of
v
b node
connected
between
a.los
The
right-hand
of this
equation
algebraic
sum de
of current
source
the
sum
of the
reciprocals
of
the
resistances
of
resistors
between
b and node
node ca.and
The
coefficient
vc is minus
todos
losnode
resistores
conectados
entre
nodos
a y b. side
El coeficiente
vc es is
la the
resta
de la
suma
losall
recíprocos
between
node
b and
node
The
coefficient
vc is minus
the sum
of the
the resistances
of
all resistors
currents
directed
into
node
a. node
connected
between
ca.and
a. The
right-hand
side
of this
equation
is the
algebraic
sum
of current
source
de
las resistencias
denode
todos
los
resistores
conectados
entre
los
nodos
creciprocals
y a. El
ladoof
derecho
de esta
ecuación
es la
connected
between
node
c and
node
a. Thede
right-hand
of thisalequation
Apply
KCL
atcorrientes
node
ba.tode
obtain
currents
directed
into
node
suma
algebraica
de
la fuente
corriente side
dirigidas
nodo a. is the algebraic sum of current source
�
� �
� � �
currents
directed
into
node
a.to obtain
Apply
KCL
at node
Aplique
la KCL
en elbnodo
b paravobtener
v � vc
v
a � vb
� �� b
� � � b � þ i3 ¼ 0
Apply KCL at node b to�i
obtain
2 þ�
� vb � �vb R� vc � �R
vb �
�va R
�i2 þ va �5 vb � vb �3 vc � vb4 þ i3 ¼ 0
R
R
R
5
3
4 þvbi3and
�
�
þ
¼ 0the terms that involve vc to obtain
�i
Separate the terms of this equation that
2 involve va from the terms that involve
R�5
R3�
� �que impliquen
� R4�que incluyan v y los términos que impliSepare
los
términos
de
esta
ecuación
v
de
los
términos
a terms
b
Separate the terms of this equation that
that involve
vb and the terms
that involve vc to obtain
1 involve v1a from
1 the
1
1
� vbthat
�vc v¼
va þ � va from
þ the
þ terms
� �involve
i3 �the
i2 terms that involve vc to obtain
�� that�involve
Separate
the terms
of this equation
quen
vc para
obtener
b and
R1
R1 �
�R1 �
�R1
�R1 �
� 15 va þ 13 þ 14 þ 15 vb � 13 vc ¼ i3 � i2
R
R
R
R
R3 vcof¼circuits
5
3
4
5
v
v
þ
þ
þ
�
i3 � i2that contain only resistors and
�
As expected, this node equation adheresa to the pattern for nodeb equations
R5
R3 R 4 R5
R3
sum
of thethat
reciprocals
of theresistors
resistances
current
sources.
the node
equation
at node
the coefficient
vb is theof
As
expected,
thisInnode
equation
adheres
to theb,pattern
for node of
equations
circuits
contain only
and
Como
se
esperaba,
esta
ecuación
apega
para
nodales
decontain
circuitos
que
contienen
As
expected,
this
equation
adheres
tosethe
for
node
equations
of
circuits
resistors
the
sum
the
of theonly
resistances
ofand
all
of all
resistors
connected
to node
b.nodal
The
coefficient
of
vpatrón
theof
sum
ofreciprocals
thethat
reciprocals
of
the
resistances
current
sources.
Innode
the
node
equation
at node
b,pattern
the al
coefficient
of
vecuaciones
a is minus
b is
solamente
resistores
y node
fuentes
deb.corriente.
En
laThe
ecuación
nodal
en
nodo
b,of
el
coeficiente
dereciprocals
vbthe
es resistances
la suma
de
the
sum
the
reciprocals
of
current
sources.
In the
equation
at coefficient
node
b,b.the
coefficient
of vof
b is
minus
the
sum
ofofthe
of
resistors
connected
between
node
aThe
and
node
vel
sum
the
reciprocals
the
resistances
ofthe
all
of
all resistors
connected
to node
of
vcoefficient
c is of
a is minus the
los
recíprocos
de
las
resistencias
de
los
resistores
conectados
al
nodo
b.
El
coeficiente
de
v
es
la
resta
de
la
suma
is
minus
the
sum
of
the
reciprocals
of
the
resistances
of
all
of
all
resistors
connected
to
node
b.
The
coefficient
of
v
a
a
resistances
of all resistors
between
c and nodeofb.vThe
right-hand
sideofofthe
this
equation of
is the
the sum
reciprocals
resistors
connected
betweenconnected
node a and
node b.node
The coefficient
c is minus
de
los recíprocos
las resistencias
todos
los
resistores
conectados
entre
los the
nodos
a of
yofb.
Elreciprocals
coeficiente
vc
minus
sum
the
the
resistors
connected
between
nodecurrents
a de
and
node
b.node
The
c is
algebraic
sum
of de
current
source
directed
into
node
b. ofb.vThe
resistances
of all
resistors
connected
between
ccoefficient
and
node
right-hand
side
this
equation of
isdethe
es
la
resta
de
la
suma
de
los
recíprocos
de
las
resistencias
de
todos
los
resistores
conectados
entre
los
nodos
c
y
b.
resistances
of
all
resistors
connected
between
node
c
and
node
b.
The
right-hand
side
of
this
equation
is
the
Finally,
the pattern
for currents
the node directed
equationsinto
of circuits
algebraic
sum use
of current
source
node b. that contain only resistors and current sources to
El
lado
derecho
de
esta
ecuación
es
la
suma
algebraica
de
corrientes
de
la
fuente
de
corriente
dirigidas
al
nodo
b.
algebraic
sum
of
current
source
currents
directed
into
node
b.
obtainFinally,
the node
at node
c: node equations of circuits that contain only resistors and current sources to
useequation
the pattern
for the
�for the
� equations
� nodales
� of circuits
�de circuitos
� resistors
Finally,
useequation
the
pattern
that contain
only
andresistores
current sources
to
utilice
elatpatrón
de node
ecuaciones
que contiene
solamente
y fuentes
obtainFinalmente,
the node
node
1
1
1
1
1
1 c:1
�nodeþc: nodal
�va �en
� el nodo
� vb þc:� þ þ þ � vc ¼ i1
obtain
the node
atecuación
de
corriente
paraequation
obtener la�
R1 �
R1
R1
R1 �
�R1
�R1 �
�R1
� 11 þ 12 va � 13 vb þ 11 þ 12 þ 13 þ 16 vc ¼ i1
� R1 þ R2 va � R3 vb þ R1 þ R2 þ R3 þ R6 vc ¼ i1
R3
R 1 R2 R3 R6
R1 R 2
EEj eXmA pMlPoL E4 .42. -23- 3
EXAMPLE 4.2-3
EXAMPLE 4.2-3
Node Equations
Ecuaciones
nodales
Node Equations
Node Equations
Determine los
the voltajes
node voltages
for para
the circuit
in Figure
4.2-6 when
2 A;
i3 2¼A,
3 A;
R ¼
1 ¼ 1 A;
2 ¼
Determine
de nodos
el circuito
en la figura
4.2-6 icuando
i1 i5
1 A,
i2 5
i3 5R13 ¼
A, 5RV;
1 5 52 V,
¼R
10
V;
R
¼
4
V;
R
¼
5
V,
and
R
¼
2
V.
2 2V;
Determine
the
node
voltages
for
the
circuit
in
Figure
4.2-6
when
i
¼
1
A;
i
¼
2
A;
i
¼
3
A;
R
¼
5
V;
R
4
5
6
1
2
3
1
2 ¼
R
5 R23V,
5
10
V,
R
5
4
V
y
R
5
2
V.
3
4
the
node
the 6circuit in Figure 4.2-6 when i1 ¼ 1 A; i2 ¼ 2 A; i3 ¼ 3 A; R1 ¼ 5 V; R2 ¼
V;
R4 voltages
¼ 4 V; Rfor
2Determine
V; R3 ¼ 10
5 ¼ 5 V, and R6 ¼ 2 V.
2Solution
V; R3 ¼ 10 V; R4 ¼ 4 V; R5 ¼ 5 V, and R6 ¼ 2 V.
Solución
The ecuaciones
node equations
are son
Las
nodales
Solution
Solution
The
node equations are
�
�
� �
�
�
1 1 1
1
1 1
� þ þ � va � � � vb � � þ � vc ¼ 1 þ 2
�15 12 15�
�15�
�15 12�
þ�1 þ 1 �va � 1 vb�� 1 �
þ 1 vc ¼ 1 þ 2
51�þ
2
5
5
þ v1a � 1 v1b � 5 þ 12 �vc ¼ 1 þ 2
1
�5� �2va þ5 � þ 5þ � vb �5� 2 � vc ¼ �2 þ 3
1 15 14�
1�
�15�
�10
�10
�
�� 1 v�
a þ �1 þ�
1 þ 1 vb � 1 �vc ¼ �2 þ 3
1 1� 5 va þ1 10 þ 5 1þ 4 1 vb �1 101 vc ¼ �2 þ 3
v þ�
�� þ � v5a � � �10
5 1 þ4 1 þ 1 þ101�vc ¼ �1
1� b �
5 2 10 2�
�15 12�
�10
� 1 þ 1 va � 1 vb þ 1 þ 1 þ 1 þ 1 vc ¼ �1
� 5 þ 2 va � 10 vb þ 5 þ 2 þ 10 þ 2 vc ¼ �1
0:9v
5 2
10 a � 0:2v5b � 0:7v
2 c10¼ 3 2
0:9v
�
0:2v
�
0:7v
a
b
c c¼¼3 1
�0:2va þ 0:55vb � 0:1v
0:9va � 0:2vb � 0:7vc ¼ 3
þ 0:1v
0:55v
� 0:1v ¼ 1
�0:2vaa �
�0:7v
b bþ 1:3vc c¼ �1
�0:2va þ 0:55vb � 0:1vc ¼ 1
�0:7v
� 0:1vb þ
1:3vc ¼ como
�1
Las
nodales
utilizando
The ecuaciones
node equations
cansebepueden
writtenescribir
using amatrices
asmatrices
�0:7v
a � 0:1vb þ 1:3vc ¼ �1
The node equations can be written using matrices
AAvv 5
¼as
bb
The node equations can be written using matrices
as
Av ¼ b
Av ¼ b
The node equations are
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 114
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Node
Voltage
Analysis
of
Circuits
with
Current
and
Voltage
Sources
Node
NodeVoltage
VoltageAnalysis
Analysisof
ofCircuits
Circuitswith
withCurrent
Currentand
andVoltage
VoltageSources
Sources
Análisis de voltajes de nodos de circuitos con fuentes de corriente y de voltaje
115
115
115
115
where
where
where
donde
222
0:9
�0:2
0:9
0:9 �0:2
�0:2
4
4
4
A
¼
�0:2
0:55
AA¼¼ �0:2
�0:2 0:55
0:55
�0:7
0:1
�0:7
�0:7 0:1
0:1
222 333
333
222 333
�0:7
�0:7
333
�0:7
vvavaa
4
5
4
5
4
4
5
4
5
5
4
5
�0:1
;
b
¼
1
and;
v
¼
and;
�0:1
�0:1 ; ;bb¼¼ 11 and;
y, v v¼¼ vvbvbb555
1:3
�1
vvcvcc
1:3
�1
1:3
�1
Estamatrix
ecuación
de matriz
se despeja
utilizando
en
la fiThis
matrix
equation
solved
using
MATLAB
ininFigure
Figure
4.2-7.
This
This
matrix
equation
equation
isisissolved
solved
using
usingMATLAB
MATLABMATLAB
in
Figure4.2-7.
4.2-7.
gura 4.2-7.
222v 333 2227:1579333
vavaa
7:1579
7:1579
¼¼444vvbvbb555¼
¼¼4445:0526
5:0526
vvv¼
5:0526555
3:4737
3:4737
3:4737
vvcvcc
En consecuencia,¼va7:1579
5 7.1579
V,¼
vb5:0526
5 5.0526
V y vvc 5
3.4737
V.
V;
V,
and
¼¼3:4737
3:4737
VV
Consequently,
¼7:1579
7:1579V;
V;vvbvbb¼
¼5:0526
5:0526V,
V,and
andvcvcc¼
3:4737V
Consequently,
Consequently,vvavaa¼
FIGURE
4.2-7
Using
MATLAB
totosolve
solve
the
FIGURE
FIGURE 4.2-7
4.2-7 Using
UsingMATLAB
MATLABto
solvethe
the
FIGURA 4.2-7 Uso de MATLAB para despejar
node
equation
ininExample
Example
4.2-3.
node
nodeequation
equationin
Example4.2-3.
4.2-3.
la ecuación nodal del ejemplo 4.2-3.
EJERCICIO
4.2-1Determine
Determine
los
voltajes
de nodos
v y for
vb, para
el circuito
de la figura
E 4.2-1.
EXERCISE
4.2-1
Determine
the
node
voltages,
and
the
circuit
of
Figure
4.2-1.
EXERCISE
EXERCISE
4.2-1
4.2-1
Determinethe
thenode
nodevoltages,
voltages,vvavaaand
andvvabvb,b, ,for
forthe
thecircuit
circuitof
ofFigure
FigureEEE4.2-1.
4.2-1.
Respuesta: va 5 3 V y vb 5 11 V
¼¼333V
VVand
and
¼¼11
11
VV
Answer:
andvvbvbb¼
11V
Answer:
Answer:vvavaa¼
EJERCICIO 4.2-2 Determine los voltajes de nodos va y vb, para el circuito de la figura E 4.2-2.
EXERCISE
4.2-2
Determine
the
node
voltages,
and
for
the
circuit
of
Figure
4.2-2.
EXERCISE
EXERCISE
4.2-2
4.2-2 Determine
Determinethe
thenode
nodevoltages,
voltages,vvavaaand
andvvbvb,b, ,for
forthe
thecircuit
circuitof
ofFigure
FigureEEE4.2-2.
4.2-2.
Respuesta: v 5 24>3 V y v 5 24V
a
b
Answer:
¼¼�4=3
�4=3
VVand
and
¼¼444V
VV
Answer:
Answer:vvavaa¼
�4=3V
andvvbvbb¼
3A
2Ω
333AAA
a
aaa
ΩΩ
222Ω
3Ω
1A
2Ω
a
b
aaa
bbb
ΩΩ
222Ω
4Ω
ΩΩ
444Ω
ΩΩ 111AAA
333Ω
FIGURA E 4.2-1 FIGURE
4.2-1
FIGURE
FIGUREEEE4.2-1
4.2-1
b
bbb
3Ω
3A
333AAA
ΩΩ
333Ω
4A
444AAA
FIGURA E 4.2-2
FIGURE
4.2-2
FIGURE
FIGUREEEE4.2-2
4.2-2
4.3
N Á L I S I S D E V O LTA J E S D E N O D O S D E C I R C U I T O S
A
C O N F U E N T E S D E C O R R I E N T E Y D E V O LTA J E
4.3
N
O
DDEEEV
VVO
O
AAG
G
AAN
N
AALLLYYYSSSIIISSSO
O
U
4.3
4.3 N
NO
OD
OLLLTTTA
GEEEA
NA
OFFFCCCIIIRRRCCCU
UIIITTTSSS
En la sección anterior determinamos los voltajes de nodos de circuitos que únicamente incluían fuenW
HHCC
U
N
AAN
N
DDsección
VVO
O
TTA
AAG
G
O
U
WIIITTTH
CU
URRRRRREEEN
NTTTEn
A
ND
V
OLLLT
GEEESSSO
Ocircuitos
U
URRRCCCEEE
SSS fuentes tanto de corriente
tes deW
corriente
independientes.
esta
consideramos
con
como de voltaje independientes.
In
the
preceding
section,
we
determined
the
node
voltages
of
circuits
with
independent
current
sources
In
Inthe
thepreceding
preceding
section,
section,
wedetermined
determinedun
the
the
node
nodevoltages
voltages
of
offuente
circuits
circuits
with
with
independent
independent
current
current
sources
sources
En primer
lugar we
consideremos
circuito
con una
de
voltaje
entre tierra
y uno
de los
only.
In
this
section,
we
consider
circuits
with
both
independent
current
and
voltage
sources.
only.
only.
In
In
this
this
section,
section,
we
we
consider
consider
circuits
circuits
with
with
both
both
independent
independent
current
current
and
and
voltage
voltage
sources.
sources.
nodos restantes. Puesto que hay libertad de seleccionar el nodo de referencia, esta disposición en parFirst
we
consider
the
circuit
with
voltage
source
between
ground
and
one
of
the
other
nodes.
First
First
weconsider
consider
the
thecircuit
circuitwith
withaaavoltage
voltagesource
sourcebetween
betweenground
groundand
andone
oneof
ofthe
theother
othernodes.
nodes.
ticular
se we
logra
fácilmente.
Because
we
are
free
totoselect
select
the
reference
node,
this
particular
arrangement
easily
achieved.
Because
Becausewe
weare
arefree
freeto
selectthe
thereference
referencenode,
node,this
thisparticular
particulararrangement
arrangementisisiseasily
easilyachieved.
achieved.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 115
Alfaomega
4/12/11 5:26 PM
116
116
116
116
116
116
Methods of Analysis of Resistive Circuits
Methods
Methods of
of Analysis
Analysis of
of Resistive
Resistive Circuits
Circuits
Methods of Analysis of Resistive Circuits
Métodos of
deAnalysis
análisis of
deResistive
circuitosCircuits
resistivos
Methods
vs
vvsss
v
s
vs
vs
+
–+
+
+
–
–
+
–
–+
–
a
aa
a
a
a
R2
R
R2
R222
R2
R2
R1
R11
R
1
R
1
R1
R1
Supernode
Supernode
Supernode
b
b
b
b
b
b
vs
Supernode
Supernodo
va
Supernode
vvs
va
+vss–s
v
vs+v –
vaa
R3
R33
R
3
R
3
R3
R3
is
is
iisss
is
is
FIGURE 4.3-1 Circuit with an independent
FIGURE 4.3-1
4.3-1 Circuit
Circuit with
with an
an independent
independent
FIGURE
voltage source
an independent
current source.
FIGURE
4.3-1and
Circuit
with an
independent
voltage
source
and
an
independent
currentdesource.
source.
FIGURE
4.3-1
Circuit
with
an
independent
FIGURA
4.3-1and
Circuito
con una fuente
voltaje
voltage
source
and
an independent
independent
current
voltage
source
an
current
source.
voltage
source yand
independent
current
source.
independiente
unaanfuente
de corriente
independiente
Alfaomega
va va
a
R1
R1
R
1
1
R
1
R1 R1
++ s––
+–
+ –+ –
vb
vvbbb
v
vvbb
b
R2
R
R2
R222
RR22
is
is
iisss
is is
FIGURE 4.3-2 Circuit with a supernode
FIGURE 4.3-2
4.3-2 Circuit
Circuit with
with a supernode
supernode
FIGURE
that incorporates
va and with
vb. aa supernode
FIGURE
4.3-2 Circuit
that
incorporates
v
and
v
.
FIGURE
4.3-2
Circuit
with
a
supernode
a
b
that incorporates
incorporates
and con
FIGURA
4.3-2 Circuito
that
vvaaa and
vvbbb.. un supernodo
thatincorpora
incorporates
que
va y vvba. and vb.
Such a circuit is shown in Figure 4.3-1. We immediately note that the source is connected between
Such aa circuit
circuit is
is shown
shown in
in Figure
Figure 4.3-1.
4.3-1. We
We immediately
immediately note
note that
that the
the source
source is
is connected
connected between
between
Such
terminal
a and
and,
therefore,
Such
a circuit
isground
shown
inmuestra
Figure
4.3-1.
We immediately
note that
theobservamos
source is
connected
between
Un
circuito
como
ese
se
en
la
figura
4.3-1.
Inmediatamente
que
la
fuente
está
Such
a
circuit
is
shown
in
Figure
4.3-1.
We
immediately
note
that
the
source
is
connected
between
terminal
a
and
ground
and,
therefore,
terminal aa and
and ground
ground and,
and, therefore,
therefore,
terminal
¼
v
v
a
s
conectada
entre
la
terminal
a
y
la
tierra,
por
lo
que
terminal a and ground and, therefore,
¼v
v ¼
¼ vvvssss the KCL equation at node b to obtain
vvvaaaa 5
Thus, va is known and only vb is unknown. We
write
¼
vs the KCL equation at node b to obtain
v
Thus, vvaa is
is known
known
and only
only
v is
is
unknown.
WeaEscribimos
write
Thus,
and
unknown.
We
write
the
KCL
equation
atla node
node
toelobtain
obtain
Entonces,
es conocida
y sólo
vbunknown.
es incógnita.
la ecuación
deat
KCL bbento
nodo b para
Thus,
va isvaknown
and
only
vvbbb is
We
write
the
KCL
equation
v
v
va KCL
b�
Thus,
the
equation at node b to obtain
vbbb write
�
vaa
þ vvbb �
is ¼We
obtenerva is known and only vb is unknown.
v
v
v þ
¼R
þ vb R�2 va
i ¼
vbb33 þ
vb R
� va
iisss ¼
R
R
R
3
i
¼
þ
R3
R222
s
However, va ¼ vs . Therefore,
R
R
3
2
However, vvaa ¼
¼ vss.. Therefore,
Therefore,
However,
However,
va ¼v vv5
Therefore,
s. v
v
v
�
v
b
b
Sin
embargo,
,
por
lo
que,
b
However, va ¼avs . Therefore,
� vs
is ¼ vvbb þ vvbb �
vb3 þ
vb R�2 vvsss
iiss ¼
¼R
þ
vb3 þ vb R
� vs
i ¼R
R
R
iss ¼ vR
R222
33 þ
Then, solving for the unknown node voltage
b3, we get
R
R
2
Then, solving
solving
for the
the unknown
unknown
node
voltagede
v ,, we
we get
get
Then,
for
node
voltage
Entonces,
al despejar
incógnitanode
del voltaje
vRb, obtenemos
Then,
solving
for
the la
unknown
voltage
vvbb2bnodos
,Rwe
get
R
i
þ
v
3
s
Then, solving for the unknown node voltage
vb22,R
getR
R
Rwe
þ
R333vvsss
vb ¼ R
33iiss þ
R
R
i
þ
R
v
v
¼
R
þ
R
2
3
s
3
b
¼ R2 RR32isþþRR3 3 vss
vvbb ¼
R222 þ
þR
R333
v
¼
R
b
Next, let us consider the circuit of Figure 4.3-2, which
includes a voltage source between two nodes.
R
þ
R
2
3 el cual
Next,
let
us
consider
the
circuit
of
Figure
4.3-2,
which
includes
a voltage
voltage
source
between
two nodes.
nodes.
A continuación,
consideremos
de
la figura
4.3-2,
incluyesource
una fuente
de voltaje
entre
Next,
let the
us consider
consider
the circuit
circuit
ofcircuito
Figure
4.3-2,
which
includes
source
between
two
Because
source
voltage
is el
known,
use
KVL
to obtain
Next,
let
us
the
of
Figure
4.3-2,
which
includes
aa voltage
between
two
nodes.
Next,
let
us
consider
the
circuit
of
Figure
4.3-2,
which
includes
a
voltage
source
between
two
nodes.
Because
the
source
voltage
is
known,
use
KVL
to
obtain
dos
nodos.
Dado
que
se
conoce
el
voltaje
de
la
fuente,
usamos
la
KVL
para
obtener
Because
the
source
voltage
is
known,
use
KVL
to
obtain
Because the source voltage is known, use KVL
to ¼
obtain
v
va � vto
Because the source voltage is known, use KVL
obtain
� vvvbbbb 5
¼ vvvssss
vvvaaa 2
�
¼
vsb
vvaa �
or
� vvbs ¼
¼
vvvvaaa �
oorbien
2
� vvvvbss 5
¼ vvvvbsbb
or
�
or
vaa �
vs ¼
b
To
account
forcuenta
the fact
that
the source
voltage
is
we consideramos
consider both tanto
node aeland
node
b as
Para
tener
en
que
el
voltaje
de
la
fuente
esknown,
conocido,
nodo
a como
or
v
�
v
¼
vknown,
a
s
b
To account
account for
for the
the fact
fact that
that the
the source
source voltage
voltage is
is
we consider
consider both
both node
node aa and
and
node
b as
as
To
known,
we
node
b
part
of
one
larger
node
represented
by
the
shaded
ellipse
shown
in
Figure
4.3-2.
We
require
a
larger
To
account
for
the
fact
that
the
source
voltage
is
known,
we
consider
both
node
a
and
node
b
as
el b como
parte de
un
nodo
más
grande
representado
por la elipse
sombreada
que
sea muestra
en
la
To
account
for
the
fact
that
the
source
voltage
is
known,
we
consider
both
node
and
node
b
as
part
of
one
larger
node
represented
by
the
shaded
ellipse
shown
in
Figure
4.3-2.
We
require
a
larger
part
ofbecause
one larger
larger
node
represented
by grande
the
shaded
ellipse
shown
in Figure
Figure
4.3-2.
Wenodo
require
larger
vb represented
are
Thisshaded
larger
node
called
a supernode
or
a generalized
node
va and
part
of
one
node
by
the
ellipse
in
4.3-2.
We
require
larger
figura
4.3-2.
Requerimos
undependent.
nodo más
porque
va yisshown
voften
son dependientes.
Este
másaagrande
b
part
one larger
node
by the
Figure
4.3-2. Weor
require
a larger
and
v represented
are dependent.
dependent.
Thisshaded
largerellipse
node is
isshown
often in
called
a supernode
supernode
or
a generalized
generalized
nodeofbecause
because
v and
are
This
larger
node
often
called
node
node.
KCL says
the
algebraic
sum of
the larger
currents
entering
adice
supernode
is zero.
That
we
andsupernodo
vvbbb are
dependent.
This
node
is often
called
aa supernode
or means
aa generalized
node
vvaaathat
suele because
denominarse
o nodo
generalizado.
La
KCL
que la
suma
algebraica
de that
las coand
v
are
dependent.
This
larger
node
is
often
called
a
supernode
or
a
generalized
node
because
v
node.
KCL
says
that
the
algebraic
sum
of
the
currents
entering
a
supernode
is
zero.
That
means
that we
we
a
b
node.
KCL
says
that
the
algebraic
sum
of
the
currents
entering
a
supernode
is
zero.
That
means
that
apply
KCL
to
a
supernode
in
the
same
way
that
we
apply
KCL
to
a
node.
node.
KCL
says
that
the
algebraic
sum
of
the
currents
entering
a
supernode
is
zero.
That
means
that
we
rrientes que entran a un supernodo es cero. Eso significa que aplicamos la KCL a un supernodo de la
node.
KCL
says
that
the
algebraic
sum
of
the
currents
entering
a
supernode
is
zero.
That
means
that
we
apply
KCL
to
a
supernode
in
the
same
way
that
we
apply
KCL
to
a
node.
apply
KCL
to aaque
supernode
inla the
the
same
way
that we
we apply
apply KCL
KCL to
to aa node.
node.
apply
to
supernode
same
that
mismaKCL
manera
se aplicain
KCL
a unway
nodo.
apply KCL to a supernode in the same way that we apply KCL to a node.
A supernode consists of two nodes connected by an independent or a dependent voltage source.
A supernodo
supernodeconsta
consists
of
two
nodes
connected
byuna
an fuente
independent
or aaindependiente
dependent voltage
voltage
source.
A
supernode
consists
of
two
nodes
connected
by
an
independent
or
dependent
source.
Un
deof
dos
nodos
conectados
por
de voltaje
o dependiente.
A
supernode
consists
two
nodes
connected
by
an
independent
or
a dependent
voltage
source.
A supernode consists of two nodes connected by an independent or a dependent voltage source.
We then can write the KCL equation at the supernode as
We then
then can
can
write escribir
the KCL
KCL equation
equation
at
the
supernode
as
Entonces
podemos
ecuaciónat
dethe
lavKCL
el supernodo
como
We
write
the
at
the
supernode
as
ven
We
then can
write
the
KCLlaequation
as
asupernode
b
v
þ
¼
i
We then can write the KCL equation at thevvasupernode
as
a
b
s
v
b
v þ
v ¼
þR
¼ iiss
R
vaa111 þ
vbb222 ¼
i
R
R
R
R
þ
¼
iss
R
R
1
2
However, because va ¼ vs þ vb , we have R
However,
because
¼
þ
we
have 1 R2
Sin
embargo,
comovvvaaa¼
5vv þ
1vvvbbb,,, we
tenemos
However,
because
have
However,
because
v ¼ vvsssþ
v , we have
v þ vb vb
However, because vaa ¼ vss þ vbb , we have vvsss þ
þ v þ vvbb ¼ is
vs R
þ1 vvbbb þ
v ¼
þR
¼ iiss
vs R
þ
vbb22 ¼
þR
is
R11 vb þ
R
2
¼
R
R
Then, solving for the unknown node voltage
vb, weR2get is
R11de
2
Entonces,
despejando
la
incógnita
del
voltaje
nodos
v
,
Then,
solving
for
the
unknown
node
voltage
v
,
we
get
Then, solving
solving for
for the
the unknown
unknown node
node voltage
voltage vvbbb,, we
we get
getb tenemos
Then,
Then, solving for the unknown node voltage R
vb1,Rwe
get
i � R 2 vs
R11R
R222iisss �
�R
R22vvss
vb ¼ R
R
R
i
�RR
¼
v
R
1
2
2 vs
b
1
¼ R1 RR2 issþ
�
R
vvbb ¼
þR
R2 2 vs
R111 þ
þ
vb ¼de ambos
R
R222
Podemos ahora compilar un resumen
métodos
tratar con fuentes de voltaje indeR1 þ Rof2 dealingdewith
We can now compile a summary of both methods
independent voltage sources in
pendientes
en
un
circuito,
y
deseamos
despejar
por
medio
de
los
métodos
de voltajevoltage
de nodos,
como
We
can
now
compile
a
summary
of
both
methods
of
dealing
with
independent
voltage
sources
in
Wewe
canwish
now to
compile
summary
of both
bothmethods,
methodsas
of recorded
dealing with
with
independent
sources
in
a circuit
solve aabysummary
node voltage
in Table
4.3-1. voltage sources
We
can
now
compile
of
methods
of
dealing
independent
in
se
registra
en
la
tabla
4.3-1.
We
can
now
compile
a
summary
of
both
methods
of
dealing
with
independent
voltage
sources
in
a
circuit
we
wish
to
solve
by
node
voltage
methods,
as
recorded
in
Table
4.3-1.
circuit we
we wish
wish to
to solve
solve by
by node
node voltage
voltage methods,
methods, as
as recorded
recorded in
in Table
Table 4.3-1.
4.3-1.
aa circuit
a circuit we wish to solve by node voltage methods, as recorded in Table 4.3-1.
M04_DORF_1571_8ED_SE_108-161.indd 116
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Node Voltage Analysis of Circuits with Current and Voltage Sources
Node Voltage
Analysis
of
with Current
and
Sources
Voltage
Analysis
of Circuits
Circuits
Current
and Voltage
Voltage
Análisis de Node
voltajes
de nodos
de circuitos
conwith
fuentes
de corriente
y deSources
voltaje
117
117
117
117
Table
4.3-14.3-1
Node
Voltage Analysis
Method
withwith
a Voltage
Source
Table
Node
Analysis
Method
a
Source
Table
Node Voltage
Voltage
Analysis
with
a Voltage
Voltage
Source
Tabla4.3-1
4.3-1 Método
de análisis
del Method
voltaje de
nodos
con una
fuente de voltaje
CASE
METHOD
CASE
METHOD
CASO
MÉTODO
CASE
METHOD
equal
to the source
voltage
accounting
for theteniendo
polarities
1. The
voltage
source
connects
node
q
and
Set
v
q
1. The
La fuente
desource
voltajeconnects
conecta node
el nodo
q y Set
al voltage
voltaje
de
la fuente,
enand
cuenta
vq igual
to
source
accounting
for
and
1.
voltage
qq and
vvEstablecer
q equal
equal the
to the
the
source
accounting
for the
the polarities
polarities
andlas
1.
The
voltage
source
connects
node
and
Set to
q write
proceed
KCL
at thevoltage
remaining
nodes.
the
reference
node
(ground).
el
nodo
de
referencia
(tierra).
polaridades
y
procesos
para
escribir
la
KCL
en
los
nodos
restantes.
proceed
to
write
the
KCL
at
the
remaining
nodes.
the
reference
node
(ground).
proceed to write the KCL at the remaining nodes.
the reference node (ground).
2. The voltage source lies between two
Create a supernode that incorporates a and b and equate the sum of all the
2.
voltage
between
two
Create
a supernode
that
aa and
equate
the
of
2. The
La fuente
desource
voltajelies
está
entre dos
Crear
un supernodo
que incorpore
a y bbb eand
igualar
a cero
la suma
2.
The
voltage
Create
supernode
thattoincorporates
incorporates
and
and
equate
the sum
sum
of all
alldethe
the
nodes,
a and
b. source lies between two currents
intoa the
supernode
zero.
nodes,
a
and
b.
currents
into
the
supernode
to
zero.
nodos,
a
y
b.
todas
las
corrientes
en
el
supernodo.
nodes, a and b.
currents into the supernode to zero.
EEjXeAmMp Pl LoE 44..33--11 Ecuaciones
nodales
un circuito
que contiene
Node
Equations
forpara
a Circuit
Containing
E
Node
for
E XX AA M
MP
PL
LE
E 4
4 .. 3
3 -- 1
1fuentes
NodedeEquations
Equations
for aa Circuit
Circuit Containing
Containing
voltaje
Voltage
Sources
Voltage
Voltage Sources
Sources
Determine los voltajes de nodos para el circuito que se muestra en
Determine
the the
nodenode
voltages for the
circuit
shown in Figure
4.3-3.
Determine
la
figura 4.3-3.
Determine
the node voltages
voltages for
for the
the circuit
circuit shown
shown in
in Figure
Figure 4.3-3.
4.3-3. a
Solution
Solución
–
Solution
Solution
TheEn
methods
summarized
in Table 4.3-1
are exemplified
esta solución
se ejemplifican
los métodos
resumidosinenthis
la tabla +
8V
a 6 Ω6 Ω
b 8V
c
b
+8
8– V
Vc
6
Ω
aa
b
cc
6Ω
b+ – + –
–
+
+–
4V
–
–
12 Ω
12 Ω
12
12 Ω
Ω
12 Ω
12 Ω
2A
4+V
2A
12
2
The
summarized
in
Table
are
exemplified
in
+ 4
12 Ω
Ω
4V
V
2A
A
The methods
methods
summarized
in connected
Table 4.3-1
4.3-1
are
exemplified
in this
this
solution.
The
4-V
voltage
source
to
node
a
exemplifies
4.3-1.
La
fuente
de
voltaje
de
4
V
conectada
al
nodo
ejemplifica
el
mésolution.
The
4-V
voltage
source
connected
to
node
a
exemplifies
solution.
The
4-V
voltage
source
connected
to
node
a
exemplifies
method
1.1.The
8-V
sourcebetweennodesband
todo
La
fuente
de 8 V entre los nodos bcexemplifiesmethod2.
y c cexemplifiesmethod2.
ejemplifica el método 2.
method
1.
The
method
1.method
The 8-V
8-V1sourcebetweennodesband
sourcebetweennodesband
cexemplifiesmethod2.
Using
for
the
4-V
source,
we
note
that
Utilizando
el
método
1
para
la
fuente
de
4 V,
observamos que
Using
Using method
method 11 for
for the
the 4-V
4-V source,
source, we
we note
note that
that
v
5
24
V
FIGURA 4.3-3 Un circuito que contiene dos
va ¼
a �4 V
FIGURE
4.3-34.3-3
A circuit
containing
two voltage
�4
V
vvaa ¼
¼
�4
V
FIGURE
A
containing
two
voltage
fuentes
de voltaje,
las cuales
sólo una
está
FIGURE
4.3-3
Adecircuit
circuit
containing
Utilizando
elfor
método
2 source,
para la we
fuente
dea supernode
8 V, tenemos
unsources,
only
one
of
which
is
connected
totwo
thetovoltage
Using
method
2
the
8-V
have
at
sources,
only
one
of
which
is
connected
the
Using
method
2
for
the
8-V
source,
we
have
a
supernode
at
conectada
al
nodo
de
referencia.
sources,
only
one
of
which
is
connected
to
the
Using
method
2
for
the
8-V
source,
we
have
a
supernode
at
supernodo
losnode
nodos
b y c. Los
voltajes
de cnodos
en los by
nodos breference node.
nodes
b and
c. en
The
voltages
at nodes
b and
arec related
reference
node.
nodes
b
and
c.
The
node
voltages
at
nodes
b
and
are
related
by
reference
node.
nodes
b
and
c.
The
node
voltages
at
nodes
b
and
c
are
related
by
y c se relacionan por
8 þ 88
vb ¼vvbv5
cþ
¼ vvvccc 1
þ8
vbb ¼
Writing
a KCL
equation
forKCL
the
supenode,
we have
Writing
a
KCL
equation
for
the
supenode,
we
have
Al
escribir
una
ecuación
par
el
supernodo
tenemos
Writing a KCL equation for the supenode,
have
vb �we
vb vc
vvbbva�
�þvvaa þþvvbb þ¼vvcc2¼ 2
6 6 12þ 1212þ 12 ¼ 2
6
12 12
or oorbien
3 v3b 3vþv 1
vcþv¼v 5
24
þ2421þ
va
b
c
b
c
or
3 vb þ vc ¼
¼24
24 þ222vvvaaa
Using
va ¼ �4 V
and and
vb ¼v vc¼þv 8 þto8eliminate
va and
vb, we
have
Using
eliminate
vvaa vand
we
Utilizando
24and
V yvbbvb¼5vccvþ
eliminar
Using vvaa ¼
¼v�4
�4
V
8 8to
topara
eliminate
and
vbbb,,, tenemos
we have
have
a 5V
c1
a y vv
3ðvc3þ
8
Þ
þ
v
¼
24
þ
�4Þ Þ
sþ v ¼ 242ðþ
v
ð
þ
8
Þ
3ðvcc þ 8Þ þ vss ¼ 24 þ 22ðð�4
�4Þ
Solving
this this
equation
for vfor
get
c, we
Solving
equation
v
,
we
get
Solving this esta
equation
for vvccc,, tenemos
we get
Despejando
ecuación
vc ¼v �2
V
¼
�2
vvccc 5
¼ 22
�2 V
V
NowAhora
we
calculate
v
to
be
calculamos
que
v
sea
b
b
Now
Now we
we calculate
calculate vvbb to
to be
be
vvb v5
vc81¼88�2
5 22
vb ¼
þ 81
¼865
V6 V
cþ
vbb ¼
¼ vvcc þ
þ8¼
¼ �2
�2 þ
þ 88 ¼
¼ 66 V
V
EM
j ePm
p l4o. 34-. 2
3 - 2Supernodes
Supernodos
E XE
A
LE
E XX AA M
MP
PL
LE
E 4
4 .. 3
3 -- 2
2 Supernodes
Supernodes
aa
12 12
V V
12
12 V
V b bb
b
– –
+
–
++
Solución
1.51.5
A A
1.5
1.5 A
A
Solution
Podemos
escribir
la
primera
ecuación
nodal
considerando
la
Solution
Solution
We fuente
can write
the firstElnode
equation
by considering
voltage
de voltaje.
voltaje
de la fuente
de voltajethe
está
relacio-
a a
+
Determine
valores
denode
los voltajes
devnodos,
y vbthe
para el
Determine
the los
values
of the
voltages,
vvb,a for
a and
Determine
the
values
of
the
node
voltages,
vvaa and
vvbb,, for
the
Determine
the
values
of
the
node
voltages,
and
for
the
circuito
que
se
muestra
en
la
figura
4.3-4.
circuit
shown
in Figure
4.3-4.
circuit
shown
in
Figure
4.3-4.
circuit shown in Figure 4.3-4.
–
+ +
+
+
va va 6 Ω6 Ω 3.53.5
A A
v
6
3.5
vaa
6Ω
Ω
3.5 A
A
– –
––
+ +
+
+
vb vb 3 Ω3 Ω
v
3
vbb
3Ω
Ω
– –
––
We
can
write
first
node
by
considering
the
WeThe
can voltage
write the
the
firstvoltage
node equation
equation
by
considering
the voltage
voltage
source.
source
related
to the
nodenode
voltages
by
nado
con losvoltage
voltajes
de nodosispor
source.
source. The
The voltage source
source voltage
voltage is
is related
related to
to the
the node voltages
voltages by
by
FIGURA
4.3-4
delExample
ejemplo 4.3-2.
4.3-2.
vb �v va�¼v 12
)
v
¼
v
þ
12
FIGURE
4.3-4
TheCircuito
circuit
for
b
a
FIGURE
vbb � vaa ¼
¼ 12
12 )
) vvbb ¼
¼ vvaa þ
þ 12
12
FIGURE 4.3-4
4.3-4 The
The circuit
circuit for
for Example
Example 4.3-2.
4.3-2.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 117
Alfaomega
4/12/11 5:26 PM
E1C04_1 11/25/2009
E1C04_1 11/25/2009
11/25/2009
E1C04_1
E1C04_1
11/25/2009
Métodos de análisis de circuitos resistivos
Methods of Analysis of Resistive Circuits
Methods
ofAnalysis
Analysisof
ofResistive
ResistiveCircuits
Circuits
Methods of
of
Methods
Analysis of
Resistive Circuits
33ΩΩ
33Ω
ΩΩ
3
1.5 A
1.5 A
1.5 AA
1.5
1.5
A
–
––
––
–
––
FIGURA4.3-5
4.3-5Method
Método1 1For
para
el ejemplo
4.3-2.
FIGURE
Example
4.3-2.
FIGURE4.3-5
4.3-5Method
Method11For
ForExample
Example4.3-2.
4.3-2.
FIGURE
FIGURE 4.3-5 Method 1 For Example 4.3-2.
–
b
vv
–v–bbb
–
––
+++
++
+b+
vv+
12 V
a
b
12 V
a
b
12 VV
12
+ aa
12 V
bb
a
b
+
v
6Ω
3.5 A
++ a
va+
6Ω
3.5 A
–
vv
3.5 AA
66Ω
ΩΩ
3.5
–vaaa
6
3.5
A
–
–
–
+
3.5 AA
3.5
3.5 AA
3.5
3.5
A
bb
bb
b
+
12 VV
12
12 VV
12
12
V
++ +++
1.5 AA
1.5
1.5 AA
1.5
1.5
A
ii
aa
i
aa
ii
++ a
+a+
vva+
66ΩΩ
vaa
v
66Ω
ΩΩ
–
–va
6
–
––
–
118
118
118
118
118
118
118
118
118
+
+
v
3Ω
++ b
vb+
3Ω
–
vv
33Ω
ΩΩ
–vbbb
3
–
––
FIGURA 4.3-6 Método 2 para el ejemplo 4.3-2.
FIGURE 4.3-6 Method 2 for Example 4.3-2.
FIGURE 4.3-6
4.3-6Method
Method22for
forExample
Example4.3-2.
4.3-2.
FIGURE
FIGURE
4.3-6 Method
2 for Example
4.3-2.
escribir
la segunda
ecuación we
nodal,
debemos
decidir
quéabout
hacerthe
respecto
la corriente
de(Notice
la fuente
dethere
voltaToPara
write
the second
node equation,
must
decide what
to do
voltagedesource
current.
that
je.
(Observe
que
no
es
fácil
expresar
la
corriente
de
la
fuente
de
corriente
en
términos
de
voltajes
de
nodos.)
En
To
write
thesecond
second
nodethe
equation,
we
mustcurrent
decidewhat
what
todo
do
about
thevoltage
voltagesource
source
current.
(Notice
thatthere
there
To
write
the
node
equation,
we
must
decide
to
about
the
current.
(Notice
that
isTo
nowrite
easy the
way
to express
voltagewe
source
in
terms
ofabout
the node
voltages.)
In this
example,
we illustrate
second
node
equation,
must
decide
what
to
do
the
voltage
source
current.
(Notice
that
there
este
ejemplo,
ilustramos
dos
métodos
de
escritura
de
la
segunda
ecuación
nodal.
no
easyway
way
to
express
the
voltage
source
currentin
interms
termsof
ofthe
thenode
nodevoltages.)
voltages.)In
Inthis
thisexample,
example,we
weillustrate
illustrate
isisno
no
easy
to
express
the
voltage
source
current
two
methods
of to
writing
thethe
second
node
equation.
is
easy
way
express
voltage
source
current
in terms
of the
node voltages.)
In this
example, we
illustrate
Métodoof
Asignar
unsecond
nombre
avoltage
laequation.
corriente
decurrent.
fuente Apply
de voltaje.
Aplicar
la
KCL
en ambos
nodos
de la
two methods
methods
of1:writing
writing
the
second
node
equation.
two
the
node
Method
1:
Assign
a
name
to
the
source
KCL
at
both
of
the
voltage
source
nodes.
two methods of writing the second node equation.
fuente
de
voltaje.
Eliminar
la
corriente
de
la
fuente
de
voltaje
de
las
ecuaciones
KCL.
Method
1:Assign
Assign
aname
name
tothe
the
voltage
source
current.Apply
ApplyKCL
KCLatatboth
bothof
ofthe
thevoltage
voltagesource
sourcenodes.
nodes.
Method
1:
to
voltage
source
current.
Eliminate
the voltage
source
current
from
the KCL
equations.
Method
1:
Assign
aa name
to
the
voltage
source
current.
Apply KCL
at both of
the voltage
source nodes.
La figura
4.3-5
muestra
el circuito
después
de
etiquetar
la
corriente
de laThe
fuente
de equation
voltaje. La
ecuación
Eliminate
the
voltage
source
current
from
the KCL
KCL
equations.
Eliminate
the
voltage
source
current
from
the
equations.
Figure
4.3-5
shows
the
circuit
after
labeling
the
voltage
source
current.
KCL
at
node
a isde
Eliminate the voltage source current from the KCL equations.
la KCL
en
el
nodo
a
es
Figure 4.3-5
4.3-5 shows
shows the
the circuit
circuit after
after labeling
labeling the
the voltage
voltage
source current.
current. The
The KCL
KCL equation
equation atat node
node aa isis
Figure
va source
Figure
4.3-5 shows
the circuit
after labeling
1:5the
þ ivoltage
¼ vva source current. The KCL equation at node a is
a
1:5þ
þ i¼
¼6va
1:5
1:5
þ ii ¼
6
The KCL equation at node b is
66
The
KCL
equation
at
node
b
is
La ecuación
KCL en
nodobb bis
The
KCL equation
equation
at elnode
node
ises
vb
The
KCL
at
i þ 3:5 þ vvb¼ 0
¼00
iþ
þ3:5
3:5þ
þ3vbb ¼
ii þ
3:5 þ
3¼ 0
Combining these two equations gives
33
Combining
thesede
two
equations
gives� da v � v
Combining
these
two
equations
gives
La combinación
estas
dos ecuaciones
v
v
Combining
these
two
equations
gives
� þ bv ��¼ av ) �2:0 ¼ avþ bv
1:5 � �
�3:5
vaa 3vvbb
vbb� 6vvaa
3
6
v
v
1:5�
� 3:5
3:5þ
þ b ¼
¼ a )
) �2:0
�2:0¼
¼ aþ
þ b
1:5
� 3:5
þ 33 ¼
1:5
6 ) �2:0 ¼ 66 þ 33
3
66
6
3
Method 2: Apply KCL to the supernode corresponding to the voltage source. Shown in Figure 4.3-6, this
Método
2:
Aplicar
la KCL
al supernodo
que
corresponda
la the
fuente
desource.
voltaje.
En
la figura
4.3-6
se
muesMethod
2:Apply
Apply
KCL
to
the
supernode
corresponding
toathe
the
voltage
source.
Shown
in
Figure
4.3-6,
this
Method
2:
KCL
to
the
supernode
corresponding
to
voltage
Shown
in
Figure
4.3-6,
this
supernode
separates
the voltage
source
and its nodes
from the rest
of
circuit.
(In this
small
circuit,
the
rest
of
Method
2:
Apply
KCL
to
the
supernode
corresponding
to
the
voltage
source.
Shown
in
Figure
4.3-6,
this
tra
esta
fuente
que
separa
la
fuente
de
voltaje
y
sus
nodos
del
resto
del
circuito.
(En
este
pequeño
circuito,
el
resto
supernode
separates
thevoltage
voltagenode.)
sourceand
andits
itsnodes
nodesfrom
fromthe
therest
restof
ofthe
thecircuit.
circuit.(In
(Inthis
thissmall
smallcircuit,
circuit,the
therest
rest
of
supernode
separates
the
source
of
the
circuit
is
just
the
reference
supernode
separates
thereferencia
voltage source
and its nodes from the rest of the circuit. (In this small circuit, the rest of
del circuit
circuito
esjust
sólo
la
deltonodo.)
the
circuit
is
just
the
reference
node.)
the
is
the
reference
node.)
Apply
KCL
to
the
supernode
get
the circuit
is just
the reference
node.) obtener
Aplicar
la KCL
al supernodo
Apply
KCL
to the
the
supernodepara
to get
Apply
KCL
to
supernode
to
vget
vb
va vb
a
Apply
KCL
to
the supernode
3:5 þ vvb ) �2:0 ¼ vvþ
1:5 to
¼ get
vvvaþ
vvb
a
a 3v
b
6
3
þ3:5
3:5þ
þvb
) �2:0
�2:0¼
¼6vaa þ
þ bb
1:5¼
¼ aþ
)
1:5
3:5 þ
)
�2:0
¼
þ
1:5
¼ 66 þ
3
6 33 the supernode is a shortcut for
This is the same equation that was obtained
1. Applying KCL
6 using method
33
66 to
3
This
is
the
same
equation
that
was
obtained
using
method
1.
Applying
KCL
to the
the supernode
supernode
shortcutpara
for
This
is
the
same
equation
that
was
obtained
using
method
1.
Applying
KCL
to
isis aaa shortcut
shortcut
for
doings
things:
Ésta isesthree
la misma
ecuaciónthat
quewas
se obtenía
conusing
el método
1. Aplicar
la KCL
al supernodo
es un cortocircuito
This
the
same
equation
obtained
method
1. Applying
KCL
to
the supernode
is
for
doings three
three things:
things:
doings
doings
threetres
things:
hacer estas
cosas:
1. Labeling the voltage source current as i
1. Labeling
Etiquetarthe
la corriente
de la fuente
deas
voltaje
como i
Labeling
the
voltage source
source
current
as
i
1.1.
voltage
current
Labeling KCL
the voltage
as ii
2.1. Applying
at bothsource
nodes current
of the voltage
source
2. Applying
Aplicar laKCL
KCLatat
a los
dos
nodos
lavoltage
fuente de
voltaje
Applying
KCL
both
nodes
ofde
the
voltage
source
2.2.
both
nodes
of
the
source
Applying KCL
at both
nodes
of
the
voltage source
3.2. Eliminating
i from
the KCL
equations
3. Eliminating
Eliminar i dei from
las
ecuaciones
la KCL
Eliminating
from
the KCL
KCLde
equations
3.3.
the
equations
3.
Eliminating ii from
the
KCL equations
InEn
summary,
equations
are son
resumen,the
lasnode
ecuaciones
nodales
In summary,
summary, the
the node
node equations
equations are
are
In
vb � va ¼ 12
In
summary, the
node equations
are
�vvvbaa ¼
¼12
12
�
vvvabbb�
va ¼
¼ 12
and
y
vvbb �2:0
vvvaþ
a 3v
6
a þ
þ b¼
¼�2:0
�2:0
and
and
and
66 þ 333 ¼ �2:0
6
Despejando
las ecuaciones
nodales queda
Solving
the node
equations gives
Solving the
the node
node equations
equations gives
gives
Solving
212
y vvbb 5
va v¼a 5
�12
V;V,and
¼ 00 V
V
Solving the node equations gives
v
¼
�12
V;
and
v
¼00VVvalores son correctos si los sustituimos en
v
¼
�12
V;
and
v
¼
a
b
a
b
(Debía
sorprendernos
que
v
sea
0
V,
pero
es
fácil
verificar
que
estos
�12toV;check
and that
vb ¼these
0 V values are correct by substituting them
(We might be surprised that vbb is 0 V, but itvais¼easy
las the
ecuaciones
nodales.)that
(We
might
be
surprised
that vvbb isis 00 V,
V, but
but itit isis easy
easy to
to check
check that
that these
these values
values are
are correct
correct by
by substituting
substituting them
them
(We
might
be
surprised
into
node
equations.)
(We might be surprised that vb is 0 V, but it is easy to check that these values
are correct
by substituting
them
into
the
node
equations.)
into the
the node
node equations.)
equations.)
into
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 118
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
E1C04_1
E1C04_1
11/25/2009
11/25/2009
119
119
Node
Voltage
Analysis
of Circuits
with
Current
and Voltage
Sources
Análisis de Node
voltajes
de nodos
de circuitos
conwith
fuentes
de corriente
y deSources
voltaje
Voltage
Analysis
of Circuits
Current
and Voltage
Node
Node Voltage
Voltage Analysis
Analysis of
of Circuits
Circuits with
with Current
Current and
and Voltage
Voltage Sources
Sources
119
119
119
119
119
EmXXpAAlM
M
P4L .E34
3 3Ecuaciones
Node Equations
Equations
for aaun
Circuit
Containing
E j eE
oP
-3
nodales para
circuito
que contiene
L EL 4
.. 3
-- 3
Node
for
Circuit
Containing
E
Node
for
aa Circuit
Containing
E XX AA M
MP
PLE
E 4
4 .. 3
3 -- 3
3 Voltage
Node Equations
Equations
for
Circuit
Containing
Sources
fuentes
de
voltaje
Voltage
Sources
Voltage
Voltage Sources
Sources
1010
V V
10 10
V V
+ –10
+ –V
Determine the
the
node
voltages
for the
the
circuit
shownque
in Figure
Figure
4.3-7.en la
Determine
losthe
voltajes
de nodos
para
el circuito
seFigure
muestra
Determine
node
voltages
for
circuit
shown
in
4.3-7.
Determine
node
voltages
for
the
circuit
shown
in
4.3-7.
Determine
the
node
voltages
for
the
circuit
shown
in
Figure
4.3-7.
figura 4.3-7.
+–
+
+ ––
Solution
Solution
Solution
WeSolution
will calculate
calculate the
the node
node voltages
voltages of
of this
this circuit
circuit by
by writing
writing aa KCL
KCL
Solución
We
will
1010
Ω Ω
Ω Ω
b b 4040
40 40
Ω
10 10
Ω
c
We
calculate
the
voltages
of
this
circuit
by
writing
aa ecuaKCL
40 Ω
Ω
10 Ω
Ω b bb
We will
will
calculate
the node
node
voltages
of circuito
thisthe
circuit
by
writing
KCL aa aa
c cc
equation
for
the
supernode
corresponding
to
10-V
voltage
source.
Calcularemos
los
voltajes
de
nodos
en
este
escribiendo
una
equation
for
the
supernode
corresponding
to
the
10-V
voltage
source.
c
a
equation
for
supernode
10-V
source.
equation
for
the
supernode
corresponding
to
the
10-V
voltagede
source.
First
notice
that
ción
KCL
para
elthe
supernodo
quecorresponding
corresponda ato
la the
fuente
de voltage
voltaje
10 V.
– –
First
notice
that
5
A
2
A2 A
–
12
V
+
5
A
12
V
First
that
––12 V
5 A5 A
2 A2
First notice
notice
that que
+ +
Primero
observamos
12 V
+
5A
2A
A
+ 12 V
¼ �12
�12 V
V
vvbb v¼
¼
�12
V
�12VV
vbvbb5¼212
and
that
and
that
y que
and
and that
that
FIGURE
4.3-74.3-7
The circuit
circuit
for del
Example
4.3-3.
FIGURA
Circuito
ejemplo
4.3-3.
FIGURE
4.3-7
The
for
Example
4.3-3.
10
FIGURE
4.3-7
The
circuit
for
Example
4.3-3.
vvva a¼
¼5vvvc cþ
þ110
10
FIGURE
4.3-7
The
circuit
for
Example
4.3-3.
a va ¼c vc þ 10
va ¼ vc þ 10
Al
escribir
una ecuación
para
el supernodo,
tenemos
Writing
KCL
equationKCL
for the
the
supernode,
we have
have
Writing
aa KCL
equation
for
supernode,
we
Writing
Writing aa KCL
KCL equation
equation for
for the
the supernode,
supernode,
we
have vc � vb
vv a �
�we
vv b have
vb v¼
a va �b vþ
þ vc vv�cc �
v10
� v¼bb 55 5
þbb þ
22 þ
a�v
40
2
þ
¼5
1010 þ 2 þ 4040 ¼
10
40
oorbien
or
or
or
444 vvvaa 1
5 120
þ vvvc 2
� 55 vvvbb ¼
¼
120
�
44a vvþaa þ
þc vvcc �
� 55b vvbb ¼
¼ 120
120
Using vva ¼
¼ vvc5þ
þ 10
101and
and
¼5�12
�12
topara
eliminate
va vand
and
vb,,, tenemos
we have
have
Using
to
eliminate
v
v
we
Utilizando
10
yvvbbvv¼
212
eliminar
y
v
a va v¼
c vcvþ
a
b
a
c
b
a
b
Using
10
and
¼
�12
to
eliminate
v
and
v
,
we
have
Using va ¼ vc þ 10 and vbb ¼ �12 to eliminate vaa and vbb, we have
þ 10
10ÞÞ þ
þ vvc �
� 55ðð�12
�12ÞÞ ¼
¼ 120
120
44ððvv4ccðvþ
þ 10
10ÞÞ þ
þc vvcc �
� 55ðð�12
�12ÞÞ ¼
¼ 120
120
4ðvcc þ
,
we
get
Solving
this
equation
for
v
c
Despejando
esta
ecuación
para
v
,
tenemos
Solving
this equation
for vc,vwe
c get
Solving
we get
get
Solving this
this equation
equation for
for vcc,, we
¼ 444 V
V
vvvccc v5
¼
V
vcc ¼
¼ 44 V
V
EJERCICIO
4.3-1
los voltajes
para
circuito
de la figura 4.3-1.
EXERCISE 4.3-1
4.3-1
FindEncuentre
the node
node voltages
voltages
forde
thenodos
circuit
of el
Figure
E 4.3-1.
4.3-1.
EXERCISE
the
for
the
circuit
of
Figure
E
EXERCISE
Find
EXERCISE 4.3-1
4.3-1Find
Find the
the node
node voltages
voltages for
for the
the circuit
circuit of
of Figure
Figure E
E 4.3-1.
4.3-1.
Sugerencia:
una ecuación
de supernode
KCL para corresponding
el
supernodo que
corresponda
a la fuente
Hint: Write
Write aaEscriba
KCL equation
equation
for the
the
supernode
corresponding
to the
the 10-V
10-V voltage
voltage
source.de voltaje
Hint:
KCL
for
to
source.
Hint:
Write aa KCL
KCL equation
equation for
for the
the supernode
supernode corresponding
corresponding to
to the
the 10-V
10-V voltage
voltage source.
source.
deHint:
10 V. Write
þ 10
10 vvb
vvbb vþ
b
Answer:
2
þ
¼
5
)
v
þ
¼
30
V
and
v
¼
40
V
þ
10
v
b
a
bb þ 10
bb 5 ) vb ¼ 30 V and va ¼ 40 V
Answer:
2 þ v20
þ 30 v¼
Answer:
Answer: 22 þ
¼ 55 )
þ 2020 þ
) vvbb ¼
þ 30 ¼
¼ 30
30 V
V and
and
¼ 40
40 V
V
Respuesta:
y vvaa ¼
30
30
20
EXERCISE 4.3-2
4.3-2 Find
Find the
the voltages
voltages vva and
and vvb for
for the
the circuit
circuit of
of Figure
Figure E
E 4.3-2.
4.3-2.
EXERCISE
a v and
EXERCISE
the
the
of
E
EJERCICIO
4.3-2 Find
Encuentre
los voltajes
vb para
el circuito
de la figura
E 4.3-2.
EXERCISE 4.3-2
4.3-2
Find
the voltages
voltages
vaa andbvvavbby for
for
the circuit
circuit
of Figure
Figure
E 4.3-2.
4.3-2.
ðv þ 88ÞÞ �
� ðð�12
�12ÞÞ vvb
Answer: ðvbbððvvþb þ
¼ 33 )
) vvb ¼
¼ 88 V
V and
and vva ¼
¼ 16
16 V
V
þÞÞ b vv¼
88ÞÞ �
ðð�12
Answer:
þ
�
�12þ
b
Answer:
40 bb ¼
10
þ
Respuesta:
y a vvaa ¼
Answer:
¼ 33 )
) b vvbb ¼
¼ 88 V
V and
and
¼ 16
16 V
V
þ
40
1010
40
40
10
10 V
10 VV V b
a
aa a +10–10
10 V bb b
a ++––+ –
b
+–
2A
22A2
A
2A
A
20 Ω
20
ΩΩ
2020
20 Ω
Ω
FIGURE E
E 4.3-1
4.3-1
FIGURA
FIGURE
E
4.3-1
FIGURE
FIGURE E
E 4.3-1
4.3-1
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 119
30 Ω
30
ΩΩ
3030
30 Ω
Ω
10 Ω
a
10Ω
Ω
10
10
10 Ω
Ω aa aa
5A
55A5
A
5A
A
–
––
+ –12 V
–12 V
++ 12
+
+ V12
12 VV
8V
88VV8 V
+ –8 V
++––+ –
+–
3A
33AA3 A
3A
b
bb b
b
40 Ω
40Ω
Ω
40
40
40 Ω
Ω
FIGURE E
E 4.3-2
4.3-2
FIGURE
FIGURA
E 4.3-2
FIGURE
FIGURE E
E 4.3-2
4.3-2
Alfaomega
4/12/11 5:26 PM
E1C04_1
E1C04_1 11/25/2009
11/25/2009
E1C04_1 11/25/2009
120120
120
120
120
120
120
120
120
Methods
of Analysis
of Resistive
Circuits
Métodos
deAnalysis
análisis
deResistive
circuitos
resistivos
Methods
of
of
Circuits
Methods
ofof
Analysis
ofof
Resistive
Circuits
Methods
Analysis
Resistive
Circuits
Methods of Analysis of Resistive Circuits
A D
NEÁ V
L IOSLI S
E ISS D E N O D O S
4.44.4 N O
T ADGEE VAONLTA
A L YJ S
4.4 NN
O
D
V
T
A
G
A
N
A
I SN T E S
4.4
OO
DN
E EV
OO
LL
T
A
G
E EA
N
A
LL
YY
SS
IIS
C
F
U
E
N
T
E
S
D
E
P
E
N
E
4.4 WNI TOHD D
E EVPOELNTDAEGNETASNOAULRYD
CSEI S
S
WWI TI THHDDE EP PE ENNDDE ENNT TSSOOUURRCCE ESS
WITH DEPENDENT SOURCES
When a circuit contains a dependent source the controlling current or voltage of that
Whena aun
circuit
contains
dependent
sourcethe
thecontrolling
controlling
current
voltage
thatde
When
circuit
contains
a adependent
source
ororvoltage
ofofthat
Cuando
circuito
contiene
una
fuente
dependiente,
la corrientecurrent
o el voltaje
controladores
dependent
be expressed
as a source
functionthe
of controlling
the node voltages.
When a source
circuit must
contains
a dependent
current or voltage of that
dependent
sourcemust
mustse
expressed
afunction
function
ofthe
thenode
node
voltages.
dependent
bebe
expressed
asasa como
voltages.
esa
fuente source
dependiente
deben
expresar
unaof
función
de los
voltajes de nodos.
dependent source must be expressed as a function of the node voltages.
It is then a simple matter to express the controlled current or voltage as a function of the node voltages.
Entonces
es
sencillo
expresar
la the
corriente
o voltajes
controlados
como
una función
de
los
voltajes
It It
is isthen
simple
matter
totoexpress
controlled
current
ororvoltage
ofof
the
node
voltages.
thenaequations
simple
matter
express
the
controlled
current
voltageasas
asainafunction
afunction
function
the
node
voltages.
The
node
are then
obtained
using
the techniques
the
previous
two
sections.
It is
then
aasimple
matter
to express
the
controlled
current
ordescribed
voltage
of the
node
voltages.
de node
nodos,
y las ecuaciones
nodales
se
obtienen
mediantedescribed
las
técnicas
descritas
en las
dos
sec­cio­
The
equations
are
then
obtained
using
the
techniques
ininthe
previous
two
sections.
The
node
equations
are
then
obtained
using
the
techniques
described
the
previous
two
sections.
The node equations are then obtained using the techniques described in the previous two sections.
nes anteriores.
E XEAjMe PmLpEl o4 . 44.-41- 1 Node
Equations
for a Circuit
Ecuaciones
nodales
un Containing
circuito
que contiene
Node
Equations
forapara
aCircuit
Circuit
Containing
E EX XXA AA
MM
P LP EL E4 4
. 4. 4
- 1- 1 Node
Equations
for
Containing
E X A MMPPLLEE 4 . 4 - 1 a Dependent
Node
Equations
for
a
Circuit
Containing
Source
una
fuente
dependiente
a aDependent
DependentSource
Source
a Dependent
Source
i
Determine
thelos
node
voltages
for the
circuit
shownque
in Figure
4.4-1.
x
a xi ii 6 Ωa6 iΩ
Determine
voltajes
de nodos
para
el
circuito
se
muestra
en la figura
Determinethe
thenode
nodevoltages
voltages
thecircuit
circuit
showninin
Figure
4.4-1.
Determine
forforthe
shown
Figure
4.4-1.
6Ω
a
aaxixx 6 6Ω
Determine
the
node
voltages
for
the
circuit
shown
in
Figure
4.4-1.
Ω
4.4-1.
a
Solution
Solution
Solution
The
controlling current of the dependent source is ix. Our first task
Solución
Solution
The controlling current of the dependent source is
i . Our first task
+
–
+8 V
+
8V
6Ω
b
b bb
b
2A
3Ω
3Ω
Ω
3bΩ
3
3Ω
c3 Ω
c cc
c
c
+
+
3ix2 A + +3i
+ + 8 V–
3ix– + x
2A
A 3ix3i
–
2A
8V
2
The
controlling
current
dependent
source
is isix.ixOur
first
The
controlling
currentof
ofthe
the
dependent
source
Our
firsttask
task
– +– 8 V
3ixx – ––
2A
is The
to
thiscontroladora
current
asof
ade
function
of dependiente
thesource
node is
voltages:
Laexpress
corriente
ladependent
fuente
ix. Nuestra
primera
controlling
current
the
ixes
.. Our
first
task
– 8V
–
is istotoexpress
this
current
asasa afunction
ofofthe
node
voltages:
express
this
current
function
the
node
voltages:
es expresar
corriente
como
los voltajes de nodos:
istarea
to express
this esta
current
as a function
the nodedevoltages:
va �una
vofb función
vbvb
ix ¼ vav�
a�
ix i¼
x ¼ v6a � vb
ixx ¼
66
6
FIGURE 4.4-1
A circuit
CCVS.con una
FIGURA
4.4-1with
Un acircuito
The value of the node voltage at node a is set by the 8-V voltage
FIGURE
4.4-1
A
circuit
with
CCVS.
FIGURE
4.4-1
AA
circuit
with
a CCVS.
FIGURE
4.4-1
circuit
with
aa CCVS.
The
value
of
the
node
voltage
at
node
a
is
set
by
the
8-V
voltage
CCVS.
The
value
of
the
node
voltage
at
node
a
is
set
by
the
8-V
voltage
FIGURE
4.4-1
A
circuit
with
a CCVS.
El valor
de voltage
nodos enat elnode
nodoa ais losetestablece
la fuente
de voltaje
source
to bedel
The
value
of voltaje
the node
by the 8-V
voltage
source
source
tobebe
beque sea
de 8 Vtoto
para
source
va ¼ 8 V
8 8VV
vav¼
a¼
�8 vVb
v a 8¼
So
ix ¼a 8 8��vbvbb
86 � vb
¼
So
i
ix ¼
SoPor lo tanto,
x
So
ixx ¼ 6 6
6
The node voltage at node c is equal to the voltage of the dependent source, so
The
node
voltage
atatnode
is isequal
totothe
of
dependent
source,
sosolo tanto,
El voltaje
de
nodos
ennode
el cnodo
cequal
es igual
alvoltage
voltaje
deofthe
lathe
fuente
dependiente,
The
node
voltage
c
the
voltage
dependent
source,por
� of
The node voltage at node c is equal to the voltage
the�dependent
source,
so
�8�� vb ��
vb
�
�
8
�
v
v
vc ¼ 3ix ¼ 3 8 � vb bb¼ 4 � vb bb
ð4:4-1Þ
ð4:4-1Þ
vcv¼
ð4:4-1Þ
3i3i
3 3 86 � vb ¼¼4 4��
2 vb
c ¼
(4.4-1)
x ¼
xx ¼
vcc ¼ 3i
¼ 4 �2 2
ð4:4-1Þ
66
x ¼ 3
6
2
Next, apply KCL at node b to get
A continuación,
lab bKCL
al nodo b para obtener
Next,
apply
atatnode
totoget
Next,
applyKCL
KCLaplique
node
get
Next,
apply
KCL
at node
b to
get
8 � vb
vb � vc
8 8��vbþ
vbv�
vcv c
ð4:4-2Þ
b�
bb 2þ¼
2
ð4:4-2Þ
86 � vvbþ
(4.4-2)
2 ¼¼ v3bb � vcc
ð4:4-2Þ
ð4:4-2Þ
66 þ2 ¼ 33
6
3
Using Eq. 4.4-1 to eliminate vc from Eq. 4.4-2 gives
Utilizando
la ecuación
4.4-1vpara
eliminar
vc degives
la
ecuación 4.4-2 resulta
Using
Eq.
4.4-1
totoeliminate
Eq.
4.4-2
c from
Using
Eq.
4.4-1
eliminate
Eq.
4.4-2
gives
cvfrom
c
�
Using Eq. 4.4-1 to eliminate vc from Eq. 4.4-2 gives�
�4 �� vvb �b�
v
�
�
b
8 � vb
�4 �
4 2�bvvbb� vb v 4 4
bb �
8�
vbb 2 ¼ vbvvb�
8�
vbþ
b 4
4
�2 2¼¼ vbv��
86 � vbþþ2 2¼¼
2 bb 3�4
3
¼
66 þ2 ¼
3 3 2 ¼2 2 �3 3
6
2 3
3
Solving
for vb gives
Despejando
resulta
b
gives
Solving
Solving
forforvbvvbgives
b
Solving for vb gives
vb ¼ 7 V
7V
7V
vbv¼
b ¼
vbb ¼ v7 V 1
b
vbb 1V1
Then,
vc ¼ 4 � vb¼
Entonces,
vb¼
¼
4
�
Then,
4
�
Then,
vcv¼
2
2¼ 1VV
c
c
Then,
vc ¼ 4 �2 2 ¼2 2 V
2
2
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 120
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Node Voltage Analysis with Dependent Sources
Node
Voltage
Analysis
withwith
Dependent
Sources
Node
Voltage
Analysis
with
Dependent
Sources
Node
Voltage
Analysis
Dependent
Sources
Node
Voltage
Analysis
with
Dependent
Sources
Node
Voltage
Analysis
with
Dependent
Sources
Análisis Node
de voltajes
deAnalysis
nodos con
fuentes
dependientes
Voltage
with
Dependent
Sources
EXAMPLE 4.4-2
E XE
A
PM
LMEPP L4
. 444-..244 -- 22
EM
XA
A
LE
X
X jA
AM
MmP
P LE
E 4
EE
X
4.. 44. 4-- 2-22
E
X
A eM
PpLLlEEo4
121
121 121
121
121
121
121
121
4vx
Determine the node voltages for the circuit shown in Figure 4.4-2.
4v
a 4vx 4v
b
4vxxx
4v
Determine
the the
node
voltages
the
circuit
in que
Figure
4.4-2.
Determine
the node
node
voltages
for the
the
circuit
shown
in Figure
Figure
4.4-2.
+4v
–x4vbx
Determine
los
voltajes
defor
nodos
paracircuit
elshown
circuito
se
muestra
en la
a
a
b
Determine
voltages
for
shown
in
4.4-2.
x
a
b
a
b
Determine
the
node
voltages
for
the
circuit
shown
in
Figure
4.4-2.
+
+
–
Determine
the node voltages for the circuit shown in Figure 4.4-2.
aa a +++ ––––
bb b
Solution
figura 4.4.-2.
–
+ –+ –
Solution
Solution
Solution
The
controlling voltage of the dependent source is vx. Our first task
Solution
4Ω
10 Ω
3 A – v––––x
Solution
Theis
controlling
voltage
of
the
dependent
source
is
v
.
Our
first
task
The
controlling
voltage
of
the
dependent
source
is
v
.
Our
first
task
Solución
xis voltages:
The
controlling
voltage
ofasthe
the
dependent
source
vxxx.. Our
Our first
first task
task 3 A 33 AA vx –vvx4 –Ω 44 ΩΩ
10 Ω10
10 Ω
Ω
to controlling
express thisvoltage
voltageof
a function
of source
the node
The
dependent
is
v
+
v
4
Ω
10 Ω
Ω
x
x. Our first task
3A
A
controlling
voltage
ofaas
the
dependent
is vvvoltages:
Ω4 Ω
10
3
xx
is toThe
this
voltage
asde
of
the
node
voltages:
isexpress
to
express
this
voltage
as
a function
function
ofsource
the node
node
voltages:
El
voltaje
controlador
lafunction
fuente
dependiente
es
.
Nuestra
primera
10 Ω
3 A+ vv+xxx vx 44 Ω
10
Ω
3
A
is
to
express
this
voltage
a
of
the
¼ �va
vxvoltages:
+
is
to express
express
this este
voltage
as aacomo
function
of the
the node
node
+
is
to
this
voltage
as
function
of
voltages:
tarea
es expresar
voltaje
una función
los
voltajes
vde
¼a�v
�vaa de nodos:
x ¼
++ +
¼
vvaxxx�v
The difference between the node voltages at nodes
and
b
is
set
a
¼
�v
v
a
vx 5
2vat
xxa¼and
a nodes
Theby
difference
the the
node
voltages
a vand
b�v
is bbabset
The
difference
between
the node
node
voltages
at nodes
nodes
is set
set
The
difference
between
voltages
at
and
is
voltage
ofbetween
the
dependent
source:
The
difference
between
the
node
voltages
at
nodes
and
is
set
The
difference
between
the
node
voltages
at
nodes
aaaestablece
and
b is
is
set
La
diferencia
entre
los
voltajes
de
los
nodos
a
y
b
la
el
voltaje
The
difference
between
the
node
voltages
at
nodes
a
and
b
set
by voltage
of
the
dependent
source:
by
voltage
of
the
dependent
source:
FIGURE 4.4-2 A circuit with a VCVS.
by
voltage
of
the
dependent
source:
by fuente
voltagedependiente:
of the
the dependent
dependent
source:
4 vx ¼ 4ð�va Þ ¼ �4 va
va � vb ¼ source:
by
voltage
of
source:
FIGURE
4.4-24.4-2
A circuit
withwith
a VCVS.
FIGURE
4.4-2
A circuit
circuit
with
a VCVS.
VCVS.
de
by
voltage
of
thev dependent
FIGURE
A
FIGURE
4.4-2
A circuit
circuit
with aaa VCVS.
VCVS.
�
v
¼
4
v
¼
4
ð
�v
Þ
¼
�4
v
�
v
¼
4
v
¼
4
ð
�v
Þ
¼
�4
v
v
FIGURE
4.4-2
A
with
a v aa b� v bb ¼x4 v xx ¼ 4ða�v aaÞ ¼ �4
a v aa
FIGURA
4.4-2
Un circuito
con
una VCVS.
FIGURE
4.4-2
A
circuit
with
a
VCVS.
a
b
x
a
a
�
v
¼
4
v
¼
4
ð
�v
Þ
¼
�4
v
v
Simplifying this equation
gives
vaa � vbb ¼ 4 vxx ¼ 4ð�vaaÞ ¼ �4 vaa
Simplifying
this
equation
gives
Simplifying
this
equation
gives
Simplifying this
this equation
equation gives
gives
ð4:4-3Þ
vb ¼ 5 va
Simplifying
Simplificando
ecuación
resulta
Simplifying
thisesta
equation
gives
va 55 vvaa
ð4:4-3Þ
vb ¼vvbb5¼
¼
ð4:4-3Þ
ð4:4-3Þ
Applying KCL to the supernode corresponding to vthe
voltage source gives
¼dependent
55 vvvaaaa
ð4:4-3Þ
vvbbbb ¼
5
(4.4-3)
5
ð4:4-3Þ
Applying
KCLKCL
to the
supernode
corresponding
to the
dependent
voltage
source
gives
Applying
KCL
to the
the supernode
supernode
corresponding
to the
thevadependent
dependent
voltage
source
gives
Applying
to
corresponding
to
voltage
source
gives
v
b
Applying
KCL
to
the
supernode
corresponding
to
the
dependent
voltage
source
gives
Al aplicarKCL
la KCL
al supernodo
fuente
de voltaje
dependiente
da
Applying
to the
supernodecorrespondiente
corresponding ato3la
voltage
source gives
ð4:4-4Þ
¼
vthe
vaþ
a vvdependent
b vvbb
ð4:4-4Þ
3 ¼33 ¼
vbbb
þ10
ð4:4-4Þ
¼þv4vaaaa þ
ð4:4-4Þ
v
(4.4-4)
þ 10
ð4:4-4Þ
¼ 4410
10
þ
ð4:4-4Þ
334¼
Using Eq. 4.4-3 to eliminate vb from Eq. 4.4-4 gives
4 10
10
4
from
Eq.
4.4-4
gives
Using
Eq.
4.4-3
to
eliminate
v
from
Eq.
4.4-4
gives
Using
Eq.
4.4-3
to
eliminate
v
from
Eq. 4.4-4
4.4-4
gives
Using
Eq. 4.4-3
4.4-3
to eliminate
eliminate
vbb from
Utilizando
la ecuación
4.4-3b para
eliminar
vb de la
vecuación
5va 4.4-4
3 da
Eq.
gives
Using
Eq.
to
a
Eq.
4.4-4
gives
Using
Eq.
4.4-3
to
eliminate
vvbbb from
3v¼
5va3¼ 33va
a vva5v
aþa 5v
10
va433 vvaa
¼aaaa ¼
3 ¼33 ¼
5v
þ5v
¼
¼þv4vaaa þ
þ 10
¼ 4410
104¼ 4vaa
334¼
Solving for va, we get
4 þ 10
10 ¼ 444 va
4
Solving
for
v
,
we
get
Solving
for
v
,
we
get
a v
Solving
for
we get
get
Despejando
vaaa,, obtenemos
va ¼ 4 V
Solving
for
we
Solving
for
vvaaa,, we
get
va ¼vvaa4¼
V 4V
¼
V
Finally,
vvvbaa¼
¼5444vV
V
a ¼ 20 V
¼
a
Finalmente,
va 5¼va20
Finally,
vb ¼vvbb5¼
¼
¼V20
20 V
V
Finally,
¼
Finally,
¼ 555 vvvaaaa ¼
¼
20 V
V
Finally,
20
Finally,
vvbbb ¼
E XE Aj eMmP pL lEo4 4
. 4. 4- 3- 3
E XE
A
PM
LMEPP L4
. 444-..344 -- 33
EM
XA
A
L EE
X
X
A
M
P
L
E
E XX AA M
M PP LL EE 4
4 .. 44 -- 33
E
10 Ω
20 Ω
Determine los voltajes de nodos correspondientes a los nodos a y b para el
a
b
10 Ω
20 Ω
Determine
the
node
voltages
corresponding
to nodes a and b for the circuit
a
b
circuito
que
se
muestra
en
la
figura
4.4-3.
10 Ω10
20 Ω20
10 Ω
Ω a
20 Ω
Ωb
Determine
the
node
voltages
corresponding
to
nodes
a
and
b
for
the
circuit
Determine
the
node
voltages
corresponding
to
nodes
a
and
b
for
the
circuit
a
10
Ω
20
Ω
Determine
the
node
voltages
corresponding
to
nodes
a
and
b
for
the
circuit
a
bbb
shown
in
Figure
4.4-3.
a
10 Ω
Ω
20 Ω
Ω
Determine the
the node
node voltages
voltages corresponding
corresponding to
to nodes
nodes aa and
and bb for
for the
the circuit
circuit
10
20
Determine
aa
bb
shown
in
Figure
4.4-3.
shown
in
Figure
4.4-3.
ia
shown
in Figure
Figure 4.4-3.
4.4-3.
Solución
+
shown
in
5ia
6V
shown
in
Figure
4.4-3.
Solution
La corriente
controladora
de la fuente dependiente es ia. Nuestra primera + – +–+ 6 V i iai
5ia
Solution
a
Solution
a
i
5ia 5i
5i
Solution
The
controlling
current
of the dependent
source
is iade
. Our
first
task is
tonodos.
express– 6+++–+–V 666 VVV
iiaaa
tarea
es
expresar
esta
corriente
como
una
función
los
voltajes
de
Solution
5iaaa
i
5i
Solution
a
a
Thethis
controlling
current
of the
dependent
source
is Apply
iais
.isOur
first first
task
is tois
The
controlling
current
ofof
the
dependent
source
iaa.. Our
Our
first
task
isexpress
toget
express ––– 66 VV
5i
a
The
controlling
current
of
the
dependent
source
i
task
to
express
current
as
a
function
the
node
voltages.
KCL
at
node
a
to
Aplique
la
KCL
al
nodo
a
para
obtener
The controlling
controlling current
current of
of the
the dependent
dependent source
source is
is iiaaa.. Our
Our first
first task
task is
is to
to express
express
thisThe
current
as
a
function
of
the
node
voltages.
Apply
KCL
at
node
a
to
get
this
current
as
a
function
of
the
node
voltages.
Apply
KCL
at
node
a
to
get
this current
current as
as aa function
function of
of6the
the
node
voltages.
Apply
KCL at
at node
node aa to
to get
get
� node
vnode
va � Apply
vb
this
voltages.
Apply
KCL
a
this
current as
a function6of
�6the
6va�
� vvaa¼ ivoltages.
� vvbb KCL at node a to get FIGURA 4.4-3 Circuito con una CCCS.
avþ
a �vvava b�
þiiaa þ
6 10
�¼vvaaaia¼
� vvbbb
¼
þvvaaa20
FIGURE 4.4-3 A circuit with a CCCS.
106al�
20 �
¼de
iaaa þ
þ
10
20
10
20
¼
i
El nodo
está conectado
nodo
referencia
un cortocircuito,
Node
a is aconnected
to the reference
node
by
a shortpor
circuit,
so va ¼ 0 V. deFIGURE
4.4-34.4-3
A circuit
withwith
a CCCS.
FIGURE
4.4-3
A circuit
circuit
with
a CCCS.
CCCS.
10
20
FIGURE
A
10
20
FIGURE 4.4-3
4.4-3 A
A circuit
circuit with
with aaa CCCS.
CCCS.
FIGURE
Node
a
is
connected
to
the
reference
node
by
a
short
circuit,
so
v
¼
0
V.
Node
a
is
connected
to
the
reference
node
by
a
short
circuit,
so
v
¼
0
V.
modo aque
vthis
0 V.to
Althe
estenode
valorby
dea short
va enand
la ecuación
a vaa anterior
Node
is connected
connected
reference
circuit,
so
¼gives
0 V.
V. y FIGURE
4.4-3 A
circuit with
a CCCS.
a 5value
into
the preceding
equation
simplifying
Substituting
of
vasustituir
a¼
Node
a
is
to
the
reference
node
by
a
short
circuit,
so
v
0
a
Node
a isthis
connected
the
reference
nodeequation
byequation
a short
circuit,
¼ gives
0gives
V.
into
the
preceding
and and
simplifying
Substituting
value
of vtoaof
into
the preceding
preceding
equation
and
simplifying
Substituting
thisresulta
value
of
va into
simplificando,
12simplifying
þ so
vb vagives
the
simplifying
Substituting
this
value
into the
the
preceding equation
equation
and
gives
Substituting
this
value
of vvvaaaainto
ð4:4-5Þ
ia 12
¼
preceding
and
simplifying
gives
Substituting
this
value
of
þ12
vbþ
12
þ vvbb
(4.4-5)
ð4:4-5Þ
ia ¼iiaa ¼
1220
þ vvbbb
ð4:4-5Þ
¼12
ð4:4-5Þ
þ
ð4:4-5Þ
iiaaa ¼
¼20
20
20
ð4:4-5Þ
Next, apply KCL at node b to get
20
20
Next,
apply
KCLKCL
ataplique
node
b latoKCL
Next,
apply
KCL
at node
node
bget
to get
get
A
continuación,
al nodo b para obtener
Next,
apply
at
to
0 � vb
Next,
apply
KCL at
at
node bbb to
to
get
Next,
apply
KCL
node
get
¼
ð4:4-6Þ
0 �00vb�
� vb 5 ia
20
(4.4-6)
ia 55 iiaa
ð4:4-6Þ
0�
�¼vvvbbbb5¼
¼
ð4:4-6Þ
ð4:4-6Þ
0
20 20
¼ 55 iiaaa
ð4:4-6Þ
20 ¼
ð4:4-6Þ
Using Eq. 4.4-5 to eliminate ia from Eq. 4.4-6 gives
20
20
� 4.4-6,�resulta
Using
Eq.
4.4-5
to
eliminate
ia from
Eq. Eq.
4.4-6
Using
Eq.
4.4-5
to eliminate
eliminate
from
Eq.
gives
Al
utilizar
la
ecuación
4.4-5
para
eliminar
i4.4-6
de lagives
ecuación
a gives
Using
Eq.
4.4-5
to
iiaaa from
from
4.4-6
0
�
v
Using
Eq.
4.4-5
to
eliminate
i
Eq.
4.4-6
gives
� �
�12 þ�vb �
�
b
a
Using Eq. 4.4-5 to eliminate ia from Eq.04.4-6
gives
5�
�þ12
�
�00vb�
� vvbb¼ 12
12vbþ
þ vvbb�
20
20
b
b
¼ 5 12
0 �¼vvbb5¼
12 þ
þ vb
200 �
¼ 555 20
20 ¼
20vb
20
20
20
20
Solving for vb gives
20
20
Despejando
Solving
for vfor
Solving
for
gives
b gives
vb ¼ �10 V
Solving
vvvbbbb resulta
gives
Solving
for
v
gives
b
v
5
210
V
Solving for vb gives
V V
vb ¼vvvbbb�10
¼ �10
�10
V
¼
¼
�10
V
¼
�10
V
�10
V
vvbbb ¼
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 121
Alfaomega
4/12/11 5:26 PM
122122
122
Methods
Analysis
Resistive
Circuits
Métodos
deof
deof
resistivos
Methods
ofanálisis
Analysis
ofcircuitos
Resistive
Circuits
88
ΩΩ
8Ω
+ +
VV
– –+ 6 6
– 6V
aa
a
ia ia
ia
1212
Ω
12 Ω
Ω
20 Ω
20
20 Ω
Ω a
+
bb
b
+
+
+
4i4i
a4ia – –
a
–
+
–
+
6
V
–+ V 6
– 6V
va va
va
–
FIGURA E 4.4-1 Circuito con una CCVS.
FIGURE
FIGURE E
E 4.4-1
4.4-1 A
A circuit
circuit with
with aa CCVS.
CCVS.
15
aa 15 Ω
15 Ω
Ω b
++
+
4va 4va
4v–a
bb
+
+
–
–
––
FIGURA
E 4.42 4.4-2
Circuitocircuit
con una VCVS.
FIGURE
FIGURE E
E 4.4-2 A
A circuit with
with aa VCVS.
VCVS.
EJERCICIO 4.4-1 Encuentre el voltaje de nodos vb para el circuito que se muestra en la
EXERCISE
figura
E 4.4-1. 4.4-1
EXERCISE
4.4-1 Find
Find the
the node
node voltage
voltage vvbb for
for the
the circuit
circuit shown
shown in
in Figure
Figure E
E 4.4-1.
4.4-1.
Sugerencia:
Aplique
KCLaaalto
a para
expresar
ia como
de Substitute
los voltajesthe
deresult
nodos.
Hint:
KCL
express
iia as
a function
of the una
nodefunción
voltages.
into
Hint: Apply
Apply
KCL at
atlanode
node
tonodo
express
a as a function of the node voltages. Substitute the result into
Sustituya
el
resultado
en
v
=
4i
y
despeje
v
.
4i
and
solve
for
v
.
vvb ¼
b
a
b
a
b
b ¼ 4ia and solve for vb.
66 vvb vvb
b
b
Answer:
Respuesta:
Answer: �
� ¼
¼ 00 )
) vvbb ¼
�8 þ
þ �
¼ 4:5
4:5 V
V
12
8 44 12
EJERCICIO 4.4-2 Encuentre los voltajes de nodos para el circuito que se muestra en la
EXERCISE
EXERCISE
4.4-2 Find
Find the
the node
node voltages
voltages for
for the
the circuit
circuit shown
shown in
in Figure
Figure E
E 4.4-2.
4.4-2.
figura
E 4.4-2. 4.4-2
Hint:
voltage
source
is
expressed
as
Sugerencia:
El voltaje controlador
de dependent
la fuente dependiente
es unvoltage,
voltaje so
de ititnodos,
de modo
que ya
Hint: The
The controlling
controlling
voltage of
of the
the
dependent
source is
is aa node
node
voltage,
so
is already
already
expressed
as aa
function
of
the
node
voltages.
Apply
KCL
at
node
a.
está
expresado
como
unavoltages.
función Apply
de los voltajes
nodos.
function
of the
node
KCL at de
node
a. Aplique una KCL en el nodo a.
vva �
6 va � 4va
a � 6 þ va � 4va ¼ 0 ) va ¼ �2 V
Answer:
Answer: 20 þ 15
Respuesta:
¼ 0 ) va ¼ �2 V
20
15
4.5 A
NÁLISIS DE CORRIENTES DE ENLACES CON
4.5
M
S
H
R
R
T A
WITH
4.5 F U
MEE
EN
ST
HEC
CSU
UD
RE
RE
EVN
N
AN
NJA
AELL IY
YNS
SDII S
S
HI E N T E S
OTLTA
E PWEINT D
II N
D
E
P
E
N
D
E
N
T
V
O
L
T
A
G
E
S
O
U
R
C
N D E P E N D E N T V O L T A G E S O U R CE
ES
S
En ésta y en las siguientes secciones consideramos el análisis de circuitos aplicando la ley del voltaje
this
succeeding
sections,
of
using
Kirchhoff’s
law
deIn
Kirchhoff
en torno
a una we
rutaconsider
cerrada. the
Unaanalysis
ruta cerrada
(también
conocida
comovoltage
loop) se
In
this and
and (KVL)
succeeding
sections,
we
consider
the
analysis
of circuits
circuits
using
Kirchhoff’s
voltage
law
(KVL)
around
path.
A
closed
aa loop
is
by
starting
at
and
path
traza
empezando
un nodo
y se
poror
una
ruta de
tal modo
se vuelva
al nodo
original
(KVL)
around aaenclosed
closed
path.
Acontinúa
closed path
path
or
loop
is drawn
drawn
by que
starting
at aa node
node
and tracing
tracing aasin
path
such
that
to
original
node
pasar
unareturn
vez por
un nodo
intermedio.
suchmás
thatdewe
we
return
to the
the
original
node without
without passing
passing an
an intermediate
intermediate node
node more
more than
than once.
once.
mesh
is
aa special
case
aa loop.
UnA
caso especial
un
circuito cerrado.
Aenlace
mesh es
is un
special
case of
ofen
loop.
Un A
enlace
es una circuito
cerradonot
que no contiene
ningún otrowithin
circuito cerrado dentro de sí.
A mesh
mesh is
is a loop
loop that
that does
does not contain
contain any
any other
other loops
loops within it.
it.
El análisis de corriente de enlaces es aplicable sólo a redes planares. Un circuito planar es aquel
Mesh
current
analysis is
only to
A
planar
circuit
is
can be
Meshtrazar
current
is applicable
applicable
to planar
planardenetworks.
networks.
A no
planar
circuit
is one
one that
that
que se puede
enanalysis
un plano,
sin
cruces. only
Un ejemplo
un circuito
planar
se muestra
en lacan
fi- be
drawn
on
a
plane,
without
crossovers.
An
example
of
a
nonplanar
circuit
is
shown
in
Figure
4.5-1,
in
drawn
on
a
plane,
without
crossovers.
An
example
of
a
nonplanar
circuit
is
shown
in
Figure
4.5-1,
gura 4.5-1, en el cual el cruce está identificado y no se puede eliminar volviendo a dibujar el circuito. in
which
the
crossover
is
identified
and
cannot
be
removed
by
redrawing
the
circuit.
For
planar
networks,
which
theplanares,
crossover
identified
cannot
becomo
removed
by redrawing
the circuit.
planar
networks,
Para
redes
losisenlaces
en and
la red
se ven
ventanas.
Hay cuatro
enlacesFor
en el
circuito
que
the
meshes
in
the
network
look
like
windows.
There
are
four
meshes
in
the
circuit
shown
in
Figure
4.5-2.
the
meshes
in
the
network
look
like
windows.
There
are
four
meshes
in
the
circuit
shown
in
Figure
4.5-2.
se muestra en la figura 4.5-2.
Cruce
Crossover
Crossover
iisis
s
FIGURE
4.5-1
Nonplanar
circuit
with
crossover.
FIGURA
4.5-1
Circuito
no planar
unaacruce.
FIGURE
4.5-1
Nonplanar
circuitcon
with
crossover.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 122
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Análisis de corrientes de enlaces con fuentes de voltaje independientes
123
R2
R1
M4
M3
vs
+
–
M1
R4
R3
R6
M2
FIGURA 4.5-2 Circuito con cuatro enlaces. Cada enlace
está identificado por líneas punteadas.
R5
Están identificados como Mi. El enlace 2 contiene los elementos R3, R4 y R5. Observe que el resistor
R3 es común para el enlace 1 y el enlace 2.
Definimos una corriente de enlaces como la corriente que fluye a través de los elementos que
constituyen el enlace. La figura 4.5-3a muestra un circuito que tiene dos enlaces con las corrientes de
enlaces etiquetadas como i1 e i2. Usaremos la convención de una corriente de enlaces que fluye en el
sentido de las manecillas del reloj, como se muestra en la figura 4.5-3a. En la figura 4.5-3b se han
insertado amperímetros en los enlaces para medir las corrientes de enlaces.
Uno de los métodos normales para analizar un circuito eléctrico es escribir y despejar un conjunto de ecuaciones simultáneas denominadas ecuaciones de enlaces. Las variables desconocidas en
las ecuaciones de enlaces son las corrientes de enlaces del circuito. Determinamos los valores de las
corrientes de enlaces despejando las ecuaciones de enlaces.
Para escribir un conjunto de ecuaciones de enlaces, se hacen dos cosas:
1. Expresar voltajes de elementos como funciones de las corrientes de enlaces
2. Aplicar la ley de los voltajes de Kirchhoff a cada uno de los enlaces del circuito
Considere el problema de expresar voltajes de elementos como funciones de corrientes de enlaces. Aun cuando nuestro objetivo es expresar voltajes de elementos como funciones de las corrientes
de enlaces, empezaremos por expresar corrientes de elementos como funciones de las corrientes de
enlaces. La figura 4.5-3b muestra cómo se hace esto. Los amperímetros en la figura 4.5-3b miden las
corrientes de enlaces, i1 e i2. Los elementos C y E están en el enlace derecho pero no en el izquierdo.
Aplique la ley de la corriente de Kirchhoff en el nodo c y luego en el nodo f para ver que las corrientes
en los elementos C y E son iguales a la corriente de enlaces del enlace de la derecha, i2, como se
i1
i2
Amperímetro
Amperímetro
b
a
i1
i2
A
i1
d
(a)
i1
c
i2
B
ib
D
i1
e
i2
C
i2
E
f
(b)
FIGURA 4.5-3 (a) Un circuito con dos enlaces. (b) Inserción de amperímetros para medir las corrientes de enlaces.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 123
Alfaomega
4/12/11 5:26 PM
124
124
124
124
Methods of
of Analysis
Analysis of
of Resistive
Resistive Circuits
Circuits
Methods
Methods de
of Analysis
of Resistive
Circuits
Métodos
análisis de
circuitos resistivos
ii11i
ii22i
1
2
ii11i
1
3A
A
3
3A
ii11i –––ii22i
1
(a)
(a)
(a)
ii22i
2
ii11i –––ii22i
2
1
(b)
(b)
(b)
ii11i
1
+
+
+
vvv
–––
R
R
R
2
ii22i
2
= ii1 –– ii2
ii i=
= 1i – 2i
1
2
(c)
(c)
(c)
FIGURA
corrientes
i1 e element
i2, y la corriente
i1 2i
(a) elemento
FIGURE 4.5-4
4.5-4 Las
Mesh
currents,deii11 enlaces,
and ii22,, and
and
current, iide
i2 , of aa (a)
(a)
generic
circuit
element, de
(b) circuito
current
2, de un
1 �elemento,
FIGURE
4.5-4
Mesh
currents,
and
FIGURE(b)
4.5-4
Mesh
currents,y i(c)
i2, and element
element current,
current, 1i1��i2i2, , of
of a (a) generic
generic circuit
circuit element,
element, (b)
(b) current
current
1 and
genérico,
fuente
de
corriente
resistor.
source,
and
(c)
resistor.
source,
source,and
and(c)
(c)resistor.
resistor.
E are
are equal
to
the
mesh
current
of mismo
the right
rightmodo,
mesh,los
i , as
shown in
in Figure
4.5-3b.
Similarly,
elements
A and
and
E
current
of
mesh,
4.5-3b.
Similarly,
elements
A
muestra
en to
la
figura
4.5-3b.
Del
elementos
y D sólo
están
en el enlace
izquierdo.
E areequal
equal
tothe
themesh
mesh
current
ofthe
the right
mesh,i2i22,,as
asshown
shown inAFigure
Figure
4.5-3b.
Similarly,
elements
A and
D are
are
only in
in the
the
left
mesh. The
The A
currents
iniguales
elements
Acorriente
and D
D are
aredeequal
equal
to the
the
mesh
current
of the
the left
left
D
only
left
mesh.
currents
in
elements
A
and
to
mesh
current
of
Las
corrientes
en
los
elementos
y
D
son
a
la
enlaces
del
enlace
de
la
izquierda,
D are only in the left mesh. The currents in elements A and D are equal to the mesh current of the left
,
as
shown
in
Figure
4.5-3b.
mesh,
i
1
imesh,
,
como
se
muestra
en
la
figura
4.5-3b.
,
as
shown
in
Figure
4.5-3b.
i
1mesh, 1
i1, as shown in Figure 4.5-3b.
. Applying
Element
B is
is in
in both
both
meshes.
The current
current
of element
element
B has
has been
labeled
as iibbcomo
El
elemento
en ambos
enlaces.
La corriente
del elemento
se halabeled
etiquetado
ib. Al
Element
B
meshes.
The
of
B
as
Applying
Element
BB
is está
in both
meshes.
The current
of element
B hasBbeen
been
labeled
as i b.. Applying
Kirchhoff’s
current
law at
at node
node
b in
in Figure
Figure
4.5-3b
gives
aplicar
la leycurrent
de la corriente
de Kirchhoff
en el4.5-3b
nodo gives
b en la figura 4.5-3b resulta
Kirchhoff’s
law
b
Kirchhoff’s current law at node b in Figure 4.5-3b gives
i ¼
5 ii1 �
2i
i11 � ii222
iiibbb ¼
b ¼ i1 �iib2, como una función de las corrientes de enlaEsta ecuación expresa la corriente del elemento,
This
equation expresses
expresses the element
element current, iibb,, as
as a function of
of the mesh
mesh currents, ii11 and
and i .
ces i1 eThis
i2. equation
This
equation expressesthe
the elementcurrent,
current, ib, asaafunction
function ofthe
the meshcurrents,
currents, i1 andii222..
Figure
4.5-4a
shows
a
circuit
element
that
is
in
two
meshes.
The
current
of
the
circuit
element
is
La
figura
4.5-4a
muestra
un elemento
circuito
quemeshes.
está
en dos
enlaces.
del
elemenFigure
4.5-4a
shows
aacircuit
element
that
isisin
The
current
of
the
element
is
Figure
4.5-4a
shows
circuit
elementde
that
intwo
two
meshes.
The
currentLa
ofcorriente
thecircuit
circuit
element
is
expressed
as
a
function
of
the
mesh
currents
of
the
two
meshes.
The
circuit
element
in
Figure
4.5-4a
to
de
circuito
se
expresa
como
una
función
de
las
corrientes
de
enlaces
de
los
dos
enlaces.
El
elemento
expressed
expressedas
asaafunction
functionof
ofthe
themesh
meshcurrents
currentsof
ofthe
thetwo
twomeshes.
meshes.The
Thecircuit
circuitelement
elementin
inFigure
Figure4.5-4a
4.5-4a
could
be anything:
anything:
a resistor,
currentser
source,
a dependent
and so
so
In Figures
Figures
de
circuito
en la figura
4.5-4aapodría
cualquier
cosa: unvoltage
resistor,source,
una fuente
deon.
corriente,
una4.5-4b
fuencould
couldbe
be anything:aaresistor,
resistor,aacurrent
currentsource,
source,aadependent
dependentvoltage
voltagesource,
source,and
and soon.
on.In
In Figures4.5-4b
4.5-4b
te
de
voltaje
dependiente,
etcétera.
En
las
figuras
4.5-4b
y
c,
consideramos
tipos
específicos
de
eleand
c,
we
consider
specific
types
of
circuit
element.
In
Figure
4.5-4b,
the
circuit
element
is
a
current
and
andc,c,we
weconsider
considerspecific
specifictypes
typesof
ofcircuit
circuitelement.
element.In
InFigure
Figure4.5-4b,
4.5-4b,the
thecircuit
circuitelement
elementisisaacurrent
current
mentos
de
circuito.
En
la
figura
4.5-4b,
el
elemento
de
circuito
es
una
fuente
de
corriente.
La
corriente
source.
The
element
current
has
been
represented
twice,
once
as
the
current
source
current,
3
A,
and
source.
The
current
been
represented
twice,
once
the
source
current,
33A,
and
source.
Theelement
element
currenthas
hasdos
been
represented
twice,
onceas
asde
thelacurrent
current
source
current,
A,
and
del
elemento
se
ha
representado
veces,
una
como
la
corriente
fuente
de
corriente,
3
A,
y
otra
�
i
.
Noticing
that
the
reference
directions
for
3
A
and
once
as
a
function
of
the
mesh
currents,
i
1
2
that
the
reference
directions
for
33AA and
once
as
aafunction
of
the
mesh
currents,
i1i ��i2i . .Noticing
Noticing
that
the
reference
directions
for
and
once
as
function
of
the
mesh
currents,
1i 2i2 . Observando que las direcciones de referencia para
como
corrientes
1 points
2
� ii22una
arefunción
differentde(one
(one
pointsde
up,enlaces,
the other
other
points
down), we
we write
write
are
different
points
up,
the
down),
ii11i �
�
i
are
different
(one
points
up,
the
other
points
down),
we
write
3 1A e i21 2i2 son diferentes (una apunta hacia arriba, la otra hacia abajo), escribimos
�3
¼
i �
i
�3
23
5
2i
�3¼
¼iii1111�
�i2i222
Esta
valores
dos
corrientes
de enlaces.
This ecuación
equation relaciona
relates the
thelosvalues
values
of de
two
ofde
thelasmesh
mesh
currents.
This
relates
of
currents.
This equation
equation
relatesconsideramos
the values of
of two
two
of the
the
meshen
currents.
A
continuación
la
figura
4.5-4c,
la cualelement
el elemento
de circuito
es un
Next
consider
Figure
4.5-4c.
In
Figure
4.5-4c,
the
circuit
is aa resistor.
resistor.
We will
will
useresistor.
Ohm’s
Next
Figure
4.5-4c.
In
4.5-4c,
the
circuit
is
We
use
Nextlaconsider
consider
Figure
4.5-4c.
InFigure
Figure
4.5-4c,del
theresistor,
circuitelement
element
a resistor.de
We
useOhm’s
Ohm’s
Usaremos
leythe
deresistor
Ohm
para
expresar
los
voltajes
v,
comoisfunciones
laswill
corrientes
de
law
to
express
voltage,
v,
as
functions
of
the
mesh
currents.
First,
we
express
the
resistor
law
the
resistor
v,v,as
of
mesh
currents.
First,
we
express
the
resistor
lawto
toexpress
express
theexpresamos
resistorvoltage,
voltage,
asfunctions
functions
ofthe
the
mesh
currents.
First,
we
expressde
theenlaces,
resistor
enlaces.
Primero,
la
corriente
del
resistor
como
una
función
de
las
corrientes
� i .. Noticing
Noticing that
that the
the resistor
resistor current,
current, ii11 �
� i , and the
the
current as
as aa function
function of
of the mesh
mesh currents,
currents, ii11 �
thatv,the
i1 �ii222,, and
and the
as a function
the
mesh currents,
i1 �i1ii22–i
icurrent
2i2. Observando
queoflathe
corriente
del resistor,
, y el voltaje,
se resistor
apegan acurrent,
la convención
pasiva,
2 .2Noticing
1current
voltage,
v,
adhere
to
the
passive
convention,
we
use
Ohm’s
law
to
write
voltage,
adhere
to
passive
convention,
usamos
de Ohm
para
escribir
voltage,lav,v,ley
adhere
to the
the
passive
convention, we
we use
use Ohm’s
Ohm’s law
law to
to write
write
vv ¼
¼R
Rððii11 �
� ii22 ÞÞ
v ¼ Rði1 � i2 Þ
Con
frecuencia,
conocemos
el valor
la resistencia.
Por ejemplo,
cuando
8 V,equation
esta ecuación
se
Frequently,
we know
know
the value
value
of the
thederesistance.
resistance.
For example,
example,
when
R¼
¼ 88RV,
V,5this
this
equation
becomes
Frequently,
we
the
of
For
when
R
becomes
Frequently,
convierte
en we know the value of the resistance. For example, when R ¼ 8 V, this equation becomes
¼ 88ððii11 �
�i Þ
vvv¼
¼ 8ði1 �ii222ÞÞ
This ecuación
equation expresa
expresses
the resistor
resistor
voltage,v,v,
v,como
as aa una
function
of the
the
mesh
currents,
and ii22..i1 e i2.
Esta
el voltaje
del resistor,
función
de las
corrientes
de ienlaces,
This
This equation
equation expresses
expresses the
the resistor voltage,
voltage, v, as
as a function
function of
of the mesh
mesh currents,
currents, ii111 and
and i2.
A
continuación
escribamos
las
ecuaciones
de
enlaces
para
representar
el
circuito
que
se muestra
Next,
let’s
write
mesh
equations
to
represent
the
circuit
shown
in
Figure
4.5-5a.
The
input
to this
this
Next,
let’s
write
mesh
equations
to
represent
the
circuit
shown
in
Figure
4.5-5a.
The
input
to
Next,
let’s
write
mesh
equations
to
represent
the
circuit
shown
in
Figure
4.5-5a.
The
input
to this
en
la
figura
4.5-5a.
La
entrada
a
este
circuito
es
el
voltaje
de
la
fuente
de
voltaje,
v
.
Para
escribir
.
To
write
mesh
equations,
we
will
first
express
the
resistor
circuit
is
the
voltage
source
voltage,
v
s
write mesh
we will
first
the
resistor
circuit
source
voltage,
vvss.. To
mesh equations,
equations,
willfunciones
first express
express
thecorrienresistor
circuit isis the
the voltage
voltage
sourcedebemos
voltage, expresar
s To write
ecuaciones
enlaces,of
los voltajes
del resistorwe
como
las
voltages as
as de
functions
ofantes
the mesh
mesh currents
currents
and then
then
apply Kirchhoff’s
Kirchhoff’s
voltage
law to
to the
thedemeshes.
meshes.
The
voltages
functions
the
and
apply
voltage
law
The
voltages
as
functions
of
the
mesh
currents
and
then
apply
Kirchhoff’s
voltage
law
to
the
meshes.
The
tes
de
enlaces
y
luego
aplicar
la
ley
del
voltaje
de
Kirchhoff
a
los
enlaces.
Las
corrientes
del
resistor
resistor
currents
are
expressed
as
functions
of
the
mesh
currents
in
Figure
4.5-5b,
and
then
the
resistor
resistor
currents
are
expressed
as
functions
of
the
mesh
currents
in
Figure
4.5-5b,
and
then
the
resistor
resistor
currents
are
expressed
as
functions
of
the
mesh
currents
in
Figure
4.5-5b,
and
then
the
resistor
se
expresan
funciones
de las corrientes
de enlaces
figura4.5-5c.
4.5-5b, y luego los voltajes del
voltages
are como
expressed
as functions
functions
of the
the mesh
mesh
currentsenin
inlaFigure
Figure
voltages
expressed
as
of
4.5-5c.
voltagesseare
are
expressed
asfunciones
functionsde
of las
thecorrientes
mesh currents
currents
in Figure
resistor
expresan
como
de mesh.
enlaces
en will
la 4.5-5c.
figura
4.5-5c.
We
may
use
Kirchhoff’s
voltage
law
around
each
We
use the
the
following convention
convention
We
voltage
law
around
mesh.
will
use
following
Wemay
mayuse
useKirchhoff’s
Kirchhoff’s
voltage
law
aroundeach
each
mesh.aWe
We
will
use the
following la
convention
Podemos
aplicar
la
ley
del
voltaje
de
Kirchhoff
en
torno
cada
enlace.
Utilizaremos
siguienfor
obtaining
the
algebraic
sum
of
voltages
around
a
mesh.
We
will
move
around
the
mesh
in the
the
for
obtaining
of
aa mesh.
move
around
mesh
forconvención
obtaining the
the
algebraic
of voltages
voltages around
around
mesh.
We will
will
move
around
the
mesh in
in en
the
te
paraalgebraic
obtener sum
lasum
suma
algebraica
de voltajes
en We
torno
a un
enlace.
Nosthe
moveremos
clockwise
direction.
If
we
encounter
the
þ
sign
of
the
voltage
reference
polarity
of
an
element
voltage
clockwise
direction.
we
the
voltage
reference
of
voltage
torno
al enlace
en elIfIf
sentido
de las manecillas
del
reloj.
Si encontramos
el signo
1
de
la polaridad
clockwise
direction.
weencounter
encounter
theþþsign
signof
ofthe
the
voltage
referencepolarity
polarity
ofan
anelement
element
voltage
before the
the �
� sign,
sign, we
we add
add that
that voltage.
voltage. Conversely,
Conversely, if
if we
we encounter
encounter the
the –– of
of the voltage
voltage reference
reference
before
de
referencia
voltaje
de un
del signo
el voltaje.
el contrario,
si
before
the �del
sign,
we add
thatelemento
voltage. antes
Conversely,
if 2,
we agregamos
encounter the
– of the
thePor
voltage
reference
polarity
of
an
element
voltage
before
the
þ
sign,
we
subtract
that
voltage.
Thus,
for
the
circuit
of
polarity
before
sign,
subtract
that
Thus,
for
of
encontramos
signo 2voltage
de la polaridad
deþþ
referencia
voltaje
de voltage.
un
elemento
antes
del
signo
polarity of
of an
anelelement
element
voltage
before the
the
sign, we
wedel
subtract
that
voltage.
Thus,
for the
the circuit
circuit1,
of
Figure 4.5-5c,
4.5-5c,
we have
have
Figure
we
restamos
ese voltaje.
Por lo tanto, para el circuito de la figura 4.5-5c, tenemos
Figure 4.5-5c,
we have
þ R i þ R ði � i Þ ¼ 0
ð4:5-1Þ
mesh 1:
1: �vss þ
ð4:5-1Þ
mesh
enlace
2v
(4.5-1)
ð4:5-1Þ
mesh 1: �v
�v s þRR111ii111þþRR333ððii111��ii222ÞÞ¼¼00
ð
i
�
i
Þ
þ
R
i
¼
0
ð4:5-2Þ
mesh
2:
�R
3
1
2
2
2
ð
i
�
i
Þ
þ
R
i
¼
0
ð4:5-2Þ
mesh
2:
�R
3
1
2
2
2
enlace
(4.5-2)
ð4:5-2Þ
mesh 2: �R3 ði1 � i2 Þ þ R2 i2 ¼ 0
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 124
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Mesh Current Analysis with Independent Voltage Sources
Mesh Current Analysis with Independent Voltage Sources
Mesh Current
Analysis
Independent
Sources
Análisis de corrientes
de enlaces
con with
fuentes
de voltaje Voltage
independientes
R1
R1
R11
vs
vs
vss
+
–+
–
+
–
i1
i1
i11
R2
R2
R22
R3
R3
R33
vs
vs
vss
i2
i2
i22
(a)
(a)
(a)
vs +
vs –+
vss +–
–
+ R1i1 –
+ R1i1 –
+ R11i11 –
R1
i1
R1
i1
R11 R3
i11
i1
R3
i1
R33
i11i – i
1
2
i1 – i2
i11 – i22
+
–+
–
+
–
i1
i1
i11
+ R2i2 –
+ R2i2 –
+ R22i22 –
i2
2
+ R
i2
+ R2
R
+3(i1R–22i2) i22
R3(i1 – i2)
R–33(i11 –i i22)
2
–
i2
–
i
R1
R1
R11
R2
R2
R22
R3
i1
R3
i1
3
i11i – R
3
1 i2
i1 – i2
i11 – i22
i2
i2
i22
125
125
125
125
i2
i2
i22
(b)
(b)
(b)
2
2
(c)
(c)
(c)
FIGURE 4.5-5 (a) A circuit. (b) The resistor currents expressed as functions of the mesh currents. (c) The resistor
FIGURE 4.5-5 (a) A circuit. (b) The resistor currents expressed as functions of the mesh currents. (c) The resistor
voltages
functions
of Las
the
currents.
FIGURA
4.4-5
Circuito.
corrientes
del resistor
expresadas
como funciones
de las
corrientes
de enlaces.
FIGUREexpressed
4.5-5 (a)
(a)as
circuit.(b)
(b)
Themesh
resistor
currents
expressed
as functions
of the mesh
currents.
(c) The
resistor
voltages
expressed
asAfunctions
of the
mesh
currents.
(c)
Los voltajes
del as
resistor
expresados
comocurrents.
funciones de las corrientes de enlaces.
voltages
expressed
functions
of the mesh
Note that the voltage across R3 in mesh 1 is determined from Ohm’s law, where
Note that the voltage across R3 in mesh 1 is determined from Ohm’s law, where
Note thatque
theelvoltage
1 is determined
from Ohm’s
where
Observe
voltajeacross
a travésR3deinRmesh
porlaw,
la ley
de Ohm, donde
¼ enlace
R i ¼1Restá
ði determinado
�i Þ
3 env el
v ¼ R33 iaa ¼ R33 ði11 � i22 Þ
v ¼ R3 ia ¼ R3 ði1 � i2 Þ
where ia is the actual element current flowing downward through R3.
where
is la
the actual element
current
flowing
downward
through
R3. de R .
donde
iiiaaa es
del4.5-2
elemento
real
que
hacia abajo
a través
4.5-1 and
will enable
usfluye
todownward
determine
the
two
mesh
i1 and i2. Rewriting
3
whereEquations
is thecorriente
actual
element
current
flowing
through
R3. currents,
i2. Rewriting
Equations
4.5-1
and
4.5-2
will
enable
us
to
determine
the
two
mesh
currents,
i1 and
Las
ecuaciones
y 4.5-2
nos
permitirán
determinar
la
corrientes
de los dos
enlaces,
i1 e i2.
the two
equations,
we4.5-1
have
i2. Rewriting
Equations
4.5-1
and
4.5-2
will
enable
us
to
determine
the
two
mesh
currents,
i1 and
the
two
equations,
we
have
Reescribiendo
las dos
tenemos
the two equations,
weecuaciones
have
i1 ðR1 þ R3 Þ � i2 R3 ¼ vs
i1 ðR1 þ R3 Þ � i2 R3 ¼ vs
i1 ðR1 þ R3 Þ � i2 R3 ¼ vs
and
y
and
and
�i1 R3 þ i2 ðR3 þ R2 Þ ¼ 0
�i1 R3 þ i2 ðR3 þ R2 Þ ¼ 0
�i1 R3 þ i2 ðR3 þ R2 Þ ¼ 0
R
R
¼ 11 V,
we have
If
RR11 ¼
22 ¼
33 5
Si
5
R
5
R
V,
tenemos
If R1 ¼ R2 ¼ R3 ¼ 1 V, we have
2i1 � i2 ¼ vs
If R1 ¼ R2 ¼ R3 ¼ 1 V, we have
� i2 ¼ vvss
2i11 2
2i
¼ vs
2i1 � ii22 5
and
and
yand
�i1 þ 2i2 ¼ 0
�i1 1
þ 2i
¼ 00
2i222 5
11 þ 2i
�i
¼
0
Add twice the first equation to the second 2i
equation,
obtaining
3i ¼ 2v . Then we have
Add twice the first equation to the second equation, obtaining 3i11 ¼ 2vss . Then we have
Sume
dos
veces
la
primera
ecuación
a
la
segunda,
obteniendo
2v2v
tenemos
2vs
vs 3i13i5
have
Add twice the first equation to the second equation,
obtaining
s. Entonces
1 ¼
s . Then we
i1 ¼ 2vs and i2 ¼ vs
i1 ¼ 2v
3 ss and i2 ¼vs v3s
2v
ii11 ¼
y i2 i
 3 and
2 ¼ 3
Thus, we have obtained two independent
that are readily solved for the
3mesh current
3 3equations
Thus, we have obtained two independent3mesh current
equations that are readily solved for the
two
unknowns.
If
we
have
N
meshes
and
write
N
mesh
equations
in
terms
of are
N mesh
currents,
can
Thus,
have
obtained
twoecuaciones
independent
equations
that
readily
solved
for
the
Por lo
tanto,we
hemos
dos
deNmesh
corriente
de enlaces
independiente
que
ya
estánwe
totaltwo
unknowns.
If
weobtenido
have N meshes
and write
meshcurrent
equations
in terms
of N mesh
currents,
we
can
obtain
N
independent
mesh
equations.
This
set
of
N
equations
is
independent
and
thus
guarantees
a
two
unknowns.
If
we
have
N
meshes
and
write
N
mesh
equations
in
terms
of
N
mesh
currents,
we
can
mente
despejadas
para
las
dos
incógnitas.
Si
tenemos
N
enlaces
y
escribimos
N
ecuaciones
en
términos
obtain N independent mesh equations. This set of N equations is independent and thus guarantees a
solution
for
the
N
mesh
currents.
obtain
N for
independent
mesh
equations.
This Nsetecuaciones
of N equations
is independent
and thus
de
N corrientes
podemos
obtener
de enlaces
independientes.
Esteguarantees
conjunto dea
solution
thedeNenlaces,
mesh
currents.
A circuit
that
contains
only
independent
voltage
sources
and
resistors
results in adespecific
format
solution
for
the
N
mesh
currents.
N
ecuaciones
es
independiente
y
por
eso
garantiza
una
solución
para
las
N
corrientes
enlaces.
A circuit that contains only independent voltage sources and resistors results in a specific format
of equations
that
can
readily
be
obtained.
Consider
a
circuit
with
three
meshes,
as
shown
in
Figure
A circuit
that
contains
only
independent
and resistors
resultsas
in
acomo
specific
Un
circuito
que
contiene
fuentesConsider
devoltage
voltaje
independiente
y resistores
dashown
resultado
of equations
that
can
readily
besólo
obtained.
a sources
circuit
with
three
meshes,
in format
Figure
4.5-6.
Assign
the
clockwise
direction
to
all
of
the
mesh
currents.
Using
KVL,
we
obtain
the
three
mesh
of equations
that
can readily
be obtained.
Consider
a currents.
circuit
with
three
meshes,
as
incon
Figure
un
formato
específico
de ecuaciones
que
pueden
obtener
fácilmente.
Piense
un shown
circuito
tres
4.5-6.
Assign
the clockwise
direction
to allseof
the mesh
Using
KVL,
weenobtain
the three
mesh
4.5-6. Assign
all of the
meshlacurrents.
Using
we de
obtain
the three mesh
enlaces,
comothe
se clockwise
muestra endirection
la figurato4.5-6.
Asigne
dirección
en elKVL,
sentido
las manecillas
del
R1
R1
R1 R1
vs +
vs –+
vs vs+ –+
– –
i1
i1
i1 i1
R2
R2
R2 R2
R3
R3
R3 R3
R4
i2
R4
i
R4 R4 i i22
2
R5
i3
R5
i
R5 R5 i i33
3
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 125
+ v
–+ vg
+v vg
+ –
– – g g
FIGURE 4.5-6 Circuit with three
FIGURE 4.5-6 Circuit with three
mesh
currents
and
two voltage
sources.
FIGURE
4.5-6
Circuit
with
mesh
currents
and
twocon
voltage
sources.
FIGURA
4.5-6
Circuito
tresthree
mesh currents
and ytwo
corrientes
de enlaces
dosvoltage
fuentessources.
de voltaje.
Alfaomega
4/12/11 5:26 PM
E1C04_1
11/25/2009
126
126
126
126
126
126
Methods of
of Analysis
Analysis
of Resistive
Resistive
Circuits
Métodos
de
análisis of
de
circuitos Circuits
resistivos
Methods
Methods
of
Analysis
of
Resistive
Circuits
reloj
a todas la corrientes de enlaces. Utilizando la KVL, obtenemos las tres ecuaciones de enlaces
equations
equations
equations
R
R i þ
R
R 1i � ii22�
enlace
1: �vvs þ
¼ 00�0
mesh 1:
1:
RR1111iii1111 þ
RR4444ðððiii1111 �
ii222ÞÞÞ ¼
mesh
�v
sss þ
mesh
1:
�v
s2 þ
15 1
1iþ
4i32
1�
21i2¼ 0i12� 0
i

R
R
�
enlace
2:
R
2
2
4
þ RR55ððiii222 �
� iii333ÞÞÞ þ
þ RR
R44ðððiii222 �
� ii11ÞÞ ¼
¼ 00
mesh 2:
2: RR
R222iii222 þ
mesh
þ
mesh
2:
2�
20� i11 Þ ¼ 0
RR3 i3i3 
vvg44 
enlace
3: R
R25ð12ii3þ�Rii552ðÞ2
þ
þ
¼
0
mesh
3:
ð
i
�
i
Þ
þ
R
i
þ
v
¼
0
mesh
3:
R
5
3
2
3
3
g
mesh 3: R555 ði333 � i222 Þ þ R333 i333 þ vggg ¼ 0
Estas
ecuaciones
de enlaces
se rewritten
pueden reescribir
al conjuntar
los for
coeficientes
para
cadaas
corriente
Thesetres
three
mesh equations
equations
can be
be
rewritten
by collecting
collecting
coefficients
for each
each mesh
mesh
current
as
These
three
mesh
can
by
coefficients
current
These
three
mesh
equations
can
be
rewritten
by
collecting
coefficients
for
each
mesh
current
as
de enlaces como
mesh 1:
1: ðððRR
R þ
þ RR
R Þi
Þi111 �
� RR
R44iii22 ¼
¼ vvvss
mesh
mesh
1:
þ
Þi
�
¼
enlace
1:
1R11111
R444442i
11  R444i222  vsss
mesh
2:
�R
i
þ
R
þ
ð
R
þ
R222 þ
þ RR
R55Þi
Þi22 �
� RR
R55iii33 ¼
¼ 000
mesh
2:
�R
i
þ
R
þ
ð
R
þ
4
1
5
4
mesh 2:
2: R
�R4444i1i111
þRR5555 
þ 1R
ðR4444 
þ RR
R
þ
Þi
�
¼
enlace
22  R5552i222  R555i333  0
i
þ
ð
R
þ
R
Þi
¼
�v
mesh
3:
�R
i
þ
ð
R
þ
R
Þi
¼
�v
mesh
3:
�R
5
2
3
5
3
g
þ1R
ðR3333 
þR
R55552i
Þi3333 
¼ v
�vgggg
mesh 3:
3: R
�R5555i2i222
enlace
Hence,
we
note
that
the
coefficient
of
the
mesh
current
for the
the first
first
mesh
is the
theenlace
sum es
of
Hence,
we
note
that
the
coefficient
of
the
mesh
current
iii111enlaces
for
is
sum
of
Hence,
we note
that theque
coefficient
of thedemesh
currentde
mesh
is
the
sum
of
Por
lo tanto,
observamos
el coeficiente
la corriente
paramesh
el primer
1 for thei1first
resistances
in
mesh
1,
and
the
coefficient
of
the
second
mesh
current
is
the
negative
of
the
resistance
resistances
in
mesh
1,
and
the
coefficient
of
the
second
mesh
current
is
the
negative
of
the
resistance
resistances
in mesh 1, and
coefficient
of the second
current
is the negative
of es
thelaresistance
la
suma de resistencias
en elthe
enlace
1, y el coeficiente
de mesh
la segunda
corriente
de enlaces
negativa
, the
the equation
equation
forde
theenlaces
nth mesh
mesh
common
to meshes
meshes
and
2. In
In
general,
we
state
that for
forplanteamos
mesh current
current
for
the
nth
common
to
111 and
2.
general,
that
mesh
iiinnn,para
,
the
equation
for
the
nth
mesh
common
to
meshes
and
2.
In
general,
we
state
that
for
mesh
current
de
la resistencia
común
a los
enlaces
1 ywe
2. state
En
general,
que
la
corriente
in,
n
with
independent
voltage
sources
only
is
obtained
as
follows:
with
independent
voltage
sources
only
is
obtained
as
follows:
with
independent
onlyfuentes
is obtained
as follows:
la
ecuación
para elvoltage
enésimosources
enlace con
de voltaje
independientes sólo se obtiene como sigue:
Q
P
N
QQ
PP
NN
X
X
X
X
X
X
Q
P
N
X
X
X
R
i
þ
R
i
¼
�
ð4:5-3Þ
�
R
i
þ
R
i
¼
�
vvvsnsn
ð4:5-3Þ
�
(4.5-3)
k
q
j
n
sn
Rkkk iqqq þ
Rjjj innn ¼ �
ð4:5-3Þ
�
sn
q¼1
q¼1
q¼1
q¼1
j¼1
j¼1
j¼1
j¼1
n¼1
n¼1
n¼1
n¼1
Es
decir,
para
el enlace
n multiplicamos
porof
la all
suma
de todas RR
las
resistencias
Rj enThen
tornowe
al add
enlace.
by the
theinsum
sum
of
all resistances
resistances
around
the mesh.
mesh.
Then
we
add the
the
That
is, for
for
mesh
we multiply
multiply
the
That
is,
mesh
nnn we
iiinnn by
jj around
That
is,
for
mesh
we
multiply
n by the sum of all resistances Rjj around the mesh. Then we add the
Entonces
agregamos
los
términos
respecto
de
las
resistencias
en
común
con
otro
enlace
como
la neterms
due
to
the
resistances
in
common
with
another
mesh
as
the
negative
of
the
connecting
resistance
terms
due
to
the
resistances
in
common
with
another
mesh
as
the
negative
of
the
connecting
resistance
terms due to the resistances in common with another mesh as the negative of the connecting resistance
gativa
de la resistencia
de conexión
la corriente
el enlaceFinally,
adyacente
k, multiplicada
multiplied
by the
the mesh
mesh
currentRin
in
the adjacent
adjacent por
mesh
for all
allde
Q enlaces
adjacentenmeshes.
meshes.
Finally,
the
Rkk,,, multiplied
by
current
the
mesh
iiq for
Q
adjacent
the
R
Q
adjacent
meshes.
Finally,
the
R
kk multiplied by the mesh current in the adjacent mesh iqq
q fordeall
iindependent
para
todos
los
Q
enlaces
adyacentes.
Finalmente,
las
fuentes
voltaje
independientes
en
torno of
al
q
independent
voltage
sources
around
the
loop
appear
on
the
right
side
of
the
equation
as
the
negative
of
voltage
sources
around
the
loop
appear
on
the
right
side
of
the
equation
as
the
negative
independent voltage sources around the loop appear on the right side of the equation as the negative of
circuito
cerrado
aparecen
a
la
derecha
de
la
ecuación
como
la
negativa
de
las
fuentes
de
voltaje
enthe
voltage
sources
encountered
as
we
traverse
the
loop
in
the
direction
of
the
mesh
current.
the
voltage
sources
encountered
as
we
traverse
the
loop
in
the
direction
of
the
mesh
current.
the voltage
sourcescruzamos
encounteredcircuito
as wecerrado
traverse the
loop in
direction
of the mesh
current.
contradas
dirección
lathe
corriente
deflow
enlaces.
Recuerde
que
Rememberconforme
that the
the preceding
precedingel
result is
is obtained
obtained en
assuming
allamesh
mesh
currents
flow
clockwise.
Remember
that
result
assuming
all
currents
clockwise.
Remember
that
the
preceding
result
is
obtained
assuming
all
mesh
currents
flow
clockwise.
el resultado
anterior
se
obtuvo
suponiendo
que
todas
las
corrientes
de
enlaces
fluyen
en
el
sentido
de
The general
general matrix
matrix equation
equation for
for the
the mesh
mesh current
current analysis
analysis for
for independent
independent voltage
voltage sources
sources
The
The
general
matrix
equation
for
the
mesh
current
analysis
for
independent
voltage
sources
las
manecillas
del
reloj.
present in
in aaa circuit
circuit is
is
present
present
circuit
is
Lainecuación
matriz
general para el análisis de la corriente de enlaces para fuentes de voltaje
ð4:5-4Þ
R iii ¼
¼ vvvsss
ð4:5-4Þ
R
independiente en un circuito es
ð4:5-4Þ
R
¼
s
R
i
5
v
(4.5-4)
s
where R
R is
is aaa symmetric
symmetric matrix
matrix with
with aaa diagonal
diagonal consisting
consisting
of the
the sum
sum of
of resistances
resistances in
in each
each mesh
mesh and
and
where
of
where
R
is
symmetric
matrix
with
diagonal
consisting
of
the
of
resistances
in
each
mesh
and
donde
R
es
una
matriz
simétrica
con
una
diagonal
queofconsta
desum
la suma
de las resistencias
en cada
the
off-diagonal
elements
are
the
negative
of
the
sum
the
resistances
common
to
two
meshes.
The
the
off-diagonal
elements
are
the
negative
of
the
sum
of
the
resistances
common
to
two
meshes.
The
the off-diagonal
elements
are thelanegative
of the sum of the resistances common to two meshes. The
enlace
los
elementos
diagonal
matrix yiii consists
consists
of the
thefuera
meshdecurrent
current
as son la negativa de la suma de las resistencias comunes a
matrix
of
mesh
as
matrix
consists
of
the
mesh
current
as
dos enlaces. La matriz i consta de la corriente de enlaces
2 33
3 como
22
iii111
6 ii221 77
7
66
6 i22 77
7
66
6
7
66
___ 7
7
6
7
6
7
iii ¼
¼
7
¼ 66
6
7
___ 7
7
6
7
66
7
6
7
4
5
44 ___ 55
iiiNNN
N
is
For
N
mesh
currents,
the
source
matrix
v
is
For
N
mesh
currents,
the
source
matrix
v
s
For Nlasmesh
currents,dethe
sourcelamatrix
Para
N corrientes
enlaces,
matrizvsfuente
v22s es 33
ss is
2v 3
vvs1s1
s1 7
6 vvs2s1
66
7
s2 7
6 vs2
7
66
6 __s2 7
7
66
77
7
_
6
7
6
7
¼
vvvsss ¼
6
7
¼
6
7
6
7
_
s
6
7
_
_
6
7
66
7
4 __ 7
5
44
5
_ 5
vvvsN
sN
sN
sN
is
the
algebraic
sum
of
the
voltages
of
the voltage
voltage
sources
injotaésimo
the jth
jth mesh
mesh
with
the
where
v
is
the
algebraic
sum
of
the
voltages
of
the
sources
in
the
with
where
sjes
donde
la
suma
algebraica
de
los
voltajes
de
las
fuentes
de voltaje
en el
enlace
conthe
el
is
the
algebraic
sum
of
the
voltages
of
the
voltage
sources
in
the
jth
mesh
with
the
where vvvsjsjsj
sj
appropriate
sign
assigned
to
each
voltage.
appropriate
sign
assigned
each
voltage.
signo
apropiado
cada
voltaje.
appropriate
signasignado
assignedato
to
each
voltage.
For the
the
circuit of
of
Figure
4.5-6
and
the
matrix Eq.
Eq. 4.5-4,
4.5-4,
wetenemos
have
For
circuit
4.5-6
and
matrix
we
have
Para
el circuito
de Figure
la figura
4.5-6
y lathe
ecuación
4.5-4
matriz,
For
the
circuit
of
Figure
4.5-6
and
the
matrix
Eq.
4.5-4,
we
have
2
3
22
33
þ RR
R444ÞÞÞ
�R444
R111 þ
ðððRR
�R
000
þ
�R
1
4
4
4 �R
5
�R444
R¼
¼ 44
R222 þ
þ RR44 þ
þ RR55ÞÞ
�R555 55
ðððRR
�R
R
�R
�R
R
¼
4
2 þ R44 þ R55 Þ
5
�R555
R333 þ
þ RR55ÞÞ
000
�R
ðððRR
�R
5
3 þ R55 Þ
Note that
thatque
R is
is aes
symmetric
matrix,
as we
we expected.
expected.
Observe
R
una matrizmatrix,
simétrica,
como
se esperaba.
Note
R
as
Note
that
R
is
aa symmetric
symmetric
matrix,
as
we
expected.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 126
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Análisis de corrientes de enlaces con fuentes de corriente y de voltaje
127
EJERCICIO 4.5-1 Determine el valor del voltaje medido por el voltímetro en la figura E 4.5-1.
6Ω
3Ω
+
–
12 V
–
+
8V
Voltímetro
6Ω
FIGURA E 4.5-1
Respuesta: 21 V
4.6
NÁLISIS DE CORRIENTES DE ENLACES
A
C O N F U E N T E S D E C O R R I E N T E Y D E V O LTA J E
Hasta aquí solamente hemos considerado circuitos con fuentes de voltaje independientes para análisis por el método de corrientes de enlaces. Si el circuito tiene una fuente de corriente independiente,
como se muestra en la figura 4.6-1, aceptamos que la segunda corriente de enlaces es igual a la negativa de la corriente de la fuente de corriente. Entonces podemos escribir
i2 5 2is
y solamente necesitamos determinar la primera corriente de enlaces i1. Escribiendo KVL para el primer enlace, obtenemos
1R1 1 R22i1 2 R2i2 5 vs
Dado que i2 5 2is, tenemos
vs R 2 is
(4.6-1)
i1 
R1  R2
donde is y vs son fuentes de magnitud conocida.
Si nos encontramos con un circuito como el que se muestra en la figura 4.6-2, tenemos una fuente de corriente is que tiene un voltaje desconocido vab a través de sus terminales. Fácilmente podemos
observar que
i2 2 i1 5 is
(4.6-2)
al escribir la KCL en el nodo a. Las dos ecuaciones de enlaces son
R1
vs
+
–
i1
enlace 1: R1i1 1 vab 5 vs
(4.6-3)
enlace 2: 1R2 1 R32i2 2 vab 5 0
(4.6-4)
R3
R2
i2
is
vs
+
–
R1
R2
a
is
i1
i2
R3
b
FIGURA 4.6-1 Circuito con una fuente de voltaje
independiente y una fuente de corriente independiente.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 127
FIGURA 4.6-2 Circuito con una fuente de corriente
independiente común a ambos enlaces.
Alfaomega
4/12/11 5:26 PM
128
128
128
128
128
Methods of Analysis of Resistive Circuits
Métodos
de
análisis
de
circuitos
resistivos
Methods
Methods
ofof
Analysis
ofAnalysis
Analysis
ofof
Resistive
ofResistive
Resistive
Circuits
Circuits
Methods
Circuits
Observamos
si agregamos
las ecuaciones
y 4.6-4 eliminamos
vab, obteniendo
We
note thatque
if we
add Eqs. 4.6-3
and 4.6-4, 4.6-3
we eliminate
vab, obtaining
We
We
note
note
that
that
if ifwe
if we
we
add
add
Eqs.
Eqs.
4.6-3
4.6-3
and
and
4.6-4,
4.6-4,
wewe
we
eliminate
eliminate
vabvv,ababobtaining
obtaining
We
note
that
add
Eqs.
4.6-3
and
4.6-4,
eliminate
,, obtaining
R
i

1
R

R
2�
i

v
R11i11 þ ðR22 þ R33Þi22 ¼ vss
R1RiR111iþ
i11þ
ðþRð2ðRRþ
R
þ3RÞi
R332ÞiÞi¼
v¼s vvss
22þ
22 ¼
Sin
embargo,
dado
que
i
5
i
1
i
,
obtenemos
However, because i2 ¼ i2s þ is1 , we1 obtain
However,
However,
because
because
i2 i¼
i22 ¼
i¼s iþ
issþ
iþ
we
obtain
obtain
However,
because
1 ,i1iwe
1, , we
Robtain
i  1R  R 2 1i  i 2�  v
R11i11 þ ðR22 þ R33Þðiss þ i11Þ ¼ vss
R
R
i
þ
i11þ
ðþRð2ðRRþ
R
þ3RÞR3ð3iÞsÞððiþ
issþ
iþ
R
i
ÞÞ¼v¼s vvss
1
1
1
1 Þi1i1¼
1
22þ
o bien
or
vs 1R2  R32is
oror
or
i1  v � ðR þ R Þi
(4.6-5)
s 1 R
22  R
33 s
ð4:6-5Þ
i1 ¼
ð�Rð2ðRRþ
Þiss
R
þ3RÞi
R33sÞi
vs v�
vsRs�
22þ
ð4:6-5Þ
ð4:6-5Þ
i1 i¼
i11 ¼¼ R1 þ R2 þ R3
ð4:6-5Þ
R1RRþ
R
þ
þ
R
þ3RR33 independientes al registrar la relación
Por lo tanto, nos inclinamos por las fuentes
de
2RR
2þ
11þ
2corriente
Thus,
we account
for independent
current
sources
byderecording
the
relationship
between
the
mesh
entreThus,
las
corrientes
defor
enlaces
y la corriente
de
lasources
fuente
corriente.
Si
la fuente de
corriente
influye
Thus,
wewe
we
account
account
for
independent
independent
current
current
sources
byby
by
recording
recording
thethe
the
relationship
relationship
between
between
thethe
the
mesh
mesh
Thus,
account
for
independent
current
sources
recording
relationship
between
mesh
currents
and
the
current
source
current.
If
the
current
source
influences
only
one
mesh
current,
we
write
en
sóloand
una
corriente
desource
enlaces,
escribimos
la ecuación
que
relaciona
esa
corriente
de enlaces
con
currents
currents
and
thethe
the
current
current
source
source
current.
current.
If If
the
Ifthe
the
current
current
source
source
influences
influences
only
only
one
one
mesh
mesh
current,
current,
wewe
we
write
write
currents
and
current
current.
current
source
influences
only
one
mesh
current,
write
the
equationde
that
relates
that
mesh current
to the current
source current
and
write
the KVLrestantes.
equationsSifor
la
corriente
la
fuente
de
corriente
y
escribimos
las
ecuaciones
KVL
para
los
enlaces
la
thethe
the
equation
equation
that
that
relates
relates
that
that
mesh
mesh
current
current
toto
the
tothe
the
current
current
source
source
current
current
and
and
write
write
thethe
the
KVL
KVL
equations
equations
forfor
for
equation
that
relates
that
mesh
current
current
source
current
and
write
KVL
equations
the
remaining
meshes.
If theen
current
source influences
two escribimos
mesh currents,
we writeKVL
the KVL
equation
for
fuente
de
corriente
influye
dos
corrientes
de
enlaces,
la
ecuación
para
los
dos
enthethe
the
remaining
remaining
meshes.
meshes.
If If
the
Ifathe
the
current
current
source
source
influences
influences
two
two
mesh
mesh
currents,
currents,
we
we
write
write
thethe
the
KVL
KVL
equation
equation
for
remaining
meshes.
current
source
influences
two
mesh
currents,
we
write
KVL
equation
for
the
terminals
of
the
current
source.
Then,
adding
thesefor
two
both
meshes,
assuming
voltage
ab across
laces,
suponiendo
un avoltaje
vabvab
avvvtravés
de
las
terminales
dethe
la
fuente
de
corriente.
Luego,
agregando
across
across
thethe
the
terminals
terminals
ofof
of
the
the
current
current
source.
source.
Then,
Then,
adding
adding
these
these
two
two
both
both
meshes,
meshes,
assuming
assuming
voltage
voltage
across
terminals
current
source.
Then,
adding
these
two
both
meshes,
assuming
aavoltage
ab
ab
mesh
equations,
we obtain
an equation
independent
of vab.independiente de v .
estas
dos
ecuaciones
de enlaces,
obtenemos
una ecuación
ab
mesh
mesh
equations,
equations,
wewe
we
obtain
obtain
anan
an
equation
equation
independent
independent
ofof
of
vabvv.abab..
mesh
equations,
obtain
equation
independent
E
X A M P L E 4 . 6 - 1 Mesh Equations
EEj E
eEm
p
oLPPEL4
.E64
-..16
de enlaces
X
A
PM
LE4
.46
-61--11Ecuaciones
Mesh
Mesh
Equations
Equations
XXM
AAlM
Mesh
Equations
Consider the circuit of Figure 4.6-3 where R1 ¼ R2 ¼ 1 V and
Considere
elcircuit
circuito
deFigure
la figura
4.6-3,
donde
5and
1Vy
Consider
Consider
thethe
the
circuit
ofof
of
Figure
Figure
4.6-3
4.6-3
where
where
R RR¼¼R
¼RRR¼5¼1¼RV
V
and
and
Consider
circuit
4.6-3
where
11V
R3 ¼ 2 V. Find the three mesh currents. 1 11 21 22 2
R
5
2
V.
Encuentre
las
corrientes
de
estos
tres
enlaces.
2¼V.
Find
Find
thethe
the
three
three
mesh
mesh
currents.
currents.
R3RR¼
22 V.
Find
three
mesh
currents.
333 ¼
Solution
Solución
Solution
Solution
Solution
Because the 4-A source is in mesh 1 only, we note that
4A 4A
4 A44AA
i1 i1
R1 R1i1 ia1i1 a R2 R2
R2RR22
R1RR11
a aa
Como
la
4-A
sólo
está
en
el1enlace
1,
observamos
Because
Because
thefuente
the
4-A
4-A
source
source
is is
in
is in
in
mesh
mesh
1 only,
only,
wewe
we
note
note
that
thatque
Because
the
4-A
source
mesh
1only,
note
that
+
+
i2
i
R3 R3
10 V10
5 A 5 iA3 i3
–V
+ +
+ – i i2i
R3RR33
10
10
V
V
10
V
2 22
5 A55AAi3 i3i3
¼
4
i
– ––
i1 5
1 4
4¼44
i1 i¼
i11 ¼
Para
la fuente
5-A tenemos
b b
For the
5-A source,
we have
b bb
For
For
thethe
the
5-A
5-A
source,
source,
wewe
we
have
have
For
5-A
source,
have
i 2i 55
(4.6-6)FIGURE
FIGURA
Circuito
fuentes de
4.6-34.6-3
Circuit
with con
twodos
independent
ð4.6-6Þ
i22 � i33 ¼ 5
FIGURE
4.6-3
4.6-3
Circuit
Circuit
with
with
two
two
independent
independent
FIGURE
4.6-3
Circuit
with
two
independent
corriente
independientes.
i�
i33 ¼
5¼55
ð4.6-6Þ
ð4.6-6Þ FIGURE
i2 i�
i22�
ð4.6-6Þ
3 i¼
current sources.
Si escribimos la KVL para los enlaces 2 y 3, obtenemos
current
current
sources.
sources.
current
sources.
Writing KVL for mesh 2 and mesh 3, we obtain
Writing
Writing
KVL
KVL
forfor
for
mesh
mesh
2 2and
2 and
and
mesh
mesh
3,enlace
wewe
we
obtain
obtain
Writing
KVL
mesh
mesh
3,3,
obtain
2: RR11ði1i
2 i ) 1 v 5 10
(4.6-7)
ð4.6-7Þ
mesh 2:
2 2� i1 Þ1 þ vabab¼ 10
i22�
i�
vþabvvab¼
10
¼10
10
ð4.6-7Þ
ð4.6-7Þ
mesh
mesh
2:2:
R
2:1RðR1i12ðði�
ð4.6-7Þ
mesh
1 Þii1þ
1ÞÞþ
ab ¼
enlace
3: RR22ð1i
2i1i1Þ2þ1RR3 i33i3�2vab
vab¼500
(4.6-8)
mesh 3:
i33�
ð4.6-8Þ
mesh
mesh
3: 3:
R
3:2RðR2i23ðði�
i33�
i�
R
þ3RiR333i�
i33�
v�abvvab¼
0¼00
ð4.6-8Þ
ð4.6-8Þ
mesh
ð4.6-8Þ
1 Þi1iþ
1ÞÞþ
ab ¼
Se sustituye i1 5 4 y se agregan las ecuaciones 4.6-7 y 4.6-8 para obtener
We substitute i1 ¼ 4 and add Eqs. 4.6-7 and 4.6-8 to obtain
4 and
and
add
add
Eqs.
Eqs.
4.6-7
4.6-7
and
and
4.6-8
4.6-8
toto
to
obtain
obtain
We
We
substitute
substitute
i1 i¼
i11 ¼4¼4and
add
Eqs.
4.6-7
and
4.6-8
obtain
We
substitute
(4.6-9)
R1 ði2 � 4Þ þ R2 ði3 � 4Þ þ R3 i3 ¼ 10
ð4.6-9Þ
R1RðR1i12ðði�
i22�
4�Þ44þ
R
þ2RðR2i23ðði�
i33�
4�Þ44þ
R
þ3RiR333i¼
i33 ¼10
¼10
10
ð4.6-9Þ
ð4.6-9Þ
ÞÞþ
ÞÞþ
ð4.6-9Þ
i53 ,1substituting
into laEq.
4.6-9, we
have
From
Eq. 4.6-6,4.6-6,
i2 ¼ i5 þ
De
la
ecuación
5
i
,
sustituimos
ecuación
4.6-9
y
obtenemos
3
iþ
substituting
into
into
Eq.
Eq.
4.6-9,
4.6-9,
wewe
we
have
have
From
From
Eq.
Eq.
4.6-6,
4.6-6,
i2 i¼
i22 ¼5¼þ
525þ
into
Eq.
4.6-9,
have
From
Eq.
4.6-6,
3 ,ii3substituting
3, , substituting
R1 ð5 þ i3 � 4Þ þ R2 ði3 � 4Þ þ R3 i3 ¼ 10
iþ
i33�
4�Þ44þ
R
þ2RðR2i23ðði�
i33�
4�Þ44þ
R
þ3RiR333i¼
i33 ¼10
¼10
10
R1RðR151ððþ
55þ
ÞÞþ
ÞÞþ
3 i�
Using the values for the resistors, we obtain
Using
Using
thethe
the
values
values
forfor
for
thethe
the
resistors,
resistors,
wewe
we
obtain
obtain
Using
values
resistors,
obtain
Utilizando
los
valores
para
los resistores,
obtenemos
13
33
i3 ¼
1313
13A and i2 ¼ 5 þ i3 ¼
3333
33A
eand
4A AA
and
i22 ¼5¼þ
i33 ¼¼ 4A AA
i2 i¼
i3 i¼
iþ
¼
i
¼
and
55þ
i
¼
3 33
4 44
4 44
Another
technique
for
the mesh
analysis
method
when a una
current
source
is common
to twoa
Otra
técnica
para for
elfor
método
de análisis
de
enlaces
cuando
fuente
de corriente
es to
común
Another
Another
technique
technique
for
thethe
the
mesh
mesh
analysis
analysis
method
method
when
when
a acurrent
a current
current
source
source
is isis
common
common
to
two
two
Another
technique
mesh
analysis
method
when
source
common
to
two
meshes
involves
the
concept
of
a
supermesh.
A
supermesh
is
one
mesh
created
from
two
meshes
that
dos
enlaces
implica
elconcept
concepto
un superenlace.
Un superenlace
es created
uncreated
enlace
creado
ameshes
partir
de
dos
meshes
meshes
involves
involves
thethe
the
concept
concept
ofof
of
a supermesh.
asupermesh.
supermesh.
AA
supermesh
Asupermesh
supermesh
is is
one
isone
one
mesh
mesh
created
from
from
two
two
meshes
that
that
meshes
involves
ade
mesh
from
two
meshes
that
have
a current
source
in
common,
as shown
in Figure
4.6-4.
enlaces
que
tienen
una
fuente
de corriente
enincomún,
como
se muestra en la figura 4.6-4.
have
have
a acurrent
a current
current
source
source
inin
in
common,
common,
asas
as
shown
shown
in
Figure
Figure
4.6-4.
4.6-4.
have
source
common,
shown
in
Figure
4.6-4.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 128
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Análisis de corrientes de enlaces con fuentes de corriente y de voltaje
i3
17
129
27
37
+
10 V –
i1
5A
i2
27
17
FIGURA 4.6-4 Circuito con un superenlace que
incorpora los enlaces 1 y 2. El superenlace está
indicado por la línea punteada.
Superenlace
Un superenlace es un enlace más grande, creado a partir de dos enlaces que tienen en común
una fuente de corriente independiente o dependiente.
Por ejemplo, considere el circuito de la figura 4.6-4. La fuente de corriente 5-A es común a los
enlaces 1 y 2. El superenlace consta de los enlaces internos 1 y 2. Al escribir la KVL en torno a la
periferia del superenlace mostrado por la línea punteada, obtenemos
10 þ 1ði1 i3 Þ þ 3ði2 i3 Þ þ 2i2 ¼ 0
Para el enlace 3 tenemos
1ði3 i1 Þ þ 2i3 þ 3ði3 i2 Þ ¼ 0
Finalmente, la ecuación que relaciona la corriente de la fuente de corriente con las corrientes de enlaces es
i1 ⫺ i2 ⫽ 5
Entonces las tres ecuaciones se pueden reducir a
superenlace:
1i1 þ 5i2 4i3 ¼ 10
enlace 3:
1i1 3i2 þ 6i3 ¼ 0
fuente de corriente: 1i1 1i2
¼5
Por consiguiente, al despejar simultáneamente las tres ecuaciones encontramos que i2 ⫽ 2.5 A,
i1 ⫽ 7.5 A e i3 ⫽ 2.5 A.
El método de análisis de las corrientes de enlaces utilizado cuando hay una fuente de corriente
se resume en la tabla 4.6-1.
Tabla 4.6-1 Métodos de análisis de corrientes de enlaces con una fuente de corriente
CASO
MÉTODO
1.
Una fuente de corriente aparece en
la periferia de sólo un enlace, n.
Igualar la corriente de enlaces in con la corriente de la fuente de corriente,
teniendo en cuenta la dirección de la fuente de corriente.
2.
Una fuente de corriente es común a
dos enlaces.
A. Suponga un voltaje vab a través de las terminales de la fuente de
corriente, escriba las ecuaciones de la KVL para los dos enlaces, y
agréguelas para eliminar vba, o bien,
B. Cree un superenlace en la periferia de los dos enlaces y escriba una
ecuación de la KVL en torno a la periferia del superenlace. Además,
escriba la ecuación restrictiva para las dos corrientes de enlaces en
términos de la fuente de corriente.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 129
Alfaomega
6/24/11 4:20 PM
E1C04_1
E1C04_1
11/25/2009
11/25/2009
130
130
130
130
130
130
Methods of Analysis of Resistive Circuits
Métodos de análisis de circuitos resistivos
Methods of
of Analysis
Analysis of
of Resistive
Resistive Circuits
Circuits
Methods
E X A M P L E 4 . 6 - 2 Supermeshes
mP
pL
lo
Superenlaces
EEXXjAAe M
M
E 4
44...6
66---2
22 Supermeshes
Supermeshes
E
PL
E
Determine the values of the mesh currents, i1 and i2, for the circuit shown in Figure 4.6-5.
Determine the
los valores
dethe
las mesh
corrientes
de enlaces,
i2, para
el circuito
queinseFigure
muestra
en la figura 4.6-5.
Determine
values of
of
currents,
and iii12,,efor
for
the circuit
circuit
shown
4.6-5.
Determine
the values
the mesh
currents,
ii11 and
the
shown
in Figure 4.6-5.
2
9Ω
99Ω
ΩΩ
9
12 V
12 VV
12
12
V
+
–+
+
+
––
–
FIGURE 4.6-5
FIGURA
FIGURE 4.6-5
4.6-5
FIGURE
4.6-5
i1
i1
ii1
1
9Ω
9Ω
Ω
9
9
Ω
3Ω
33Ω
ΩΩ
3
1.5 A
1.5 AA
1.5
1.5
A
6Ω
66Ω
ΩΩ
6
i2
i2
ii2
2
12
12
12
12
V
V
V
V
+
–+
+
+
––
–
i1
i1
ii1
1
1.5
1.5
1.5
1.5
A
A
A
A
+
+
+
+
v
vv
–v
–
––
3Ω
3Ω
Ω
3
3
Ω
i2
i2
ii2
2
6Ω
6Ω
Ω
6
6
Ω
FIGURE 4.6-6 Method 1 of Example 4.6-2.
FIGURA
ejemplo 4.6-2.
FIGURE4.6-6
4.6-6Método
Method111del
of Example
Example
4.6-2.
FIGURE
4.6-6
Method
of
4.6-2.
The circuit for Example 4.6-2.
El circuito
el ejemplo
4.6-2.
The
circuit para
for Example
Example
4.6-2.
The
circuit
for
4.6-2.
Solution
Solución
Solution
We
can write
the first
mesh equation
byde
considering
the current source.
The
source
current isde
related
to the
Solution
Podemos
escribir
la primera
ecuación
enlaces considerando
la fuente
decurrent
corriente.
La corriente
la fuente
de
We can
can
write the
the
first mesh
mesh equation
equation by
by considering
considering the
the current source. The current source current is related to the
mesh
currents
byfirst
We
write
corriente
se relaciona
con las corrientes
de enlaces por current source. The current source current is related to the
mesh
currents
by
i1 � i2 ¼ 1:5 ) i1 ¼ i2 þ 1:5
mesh currents by
� ii2 ¼
¼ 1:5
1:5 )
) i1 ¼
¼ ii2 þ
þ 1:5
1:5
ii11 �
2
2 the
To write the second mesh equation, we must
decide
what to doi1about
current source voltage. (Notice that there
Para
escribir
latosegunda
de
enlaces
debemos
decidir
que
hacer
respecto
del
voltaje
de
la fuente
de
coTo
write
the
second
mesh ecuación
equation,
we
must decide
decide
what
to do
doofabout
about
the current
current
source
voltage.
(Notice
that
there
is
no
easy
way
express
the
current
source
voltage
in terms
the mesh
currents.)
In this
example,
we that
illustrate
To
write
the
second
mesh
equation,
we
must
what
to
the
source
voltage.
(Notice
there
rriente.
(Observe
que
no
es
fácil
expresar
el
voltaje
de
la
fuente
de
corriente
en
términos
de
corrientes
de
enlaces.)
is
no
easy
way
to
express
the
current
source
voltage
in
terms
of
the
mesh
currents.)
In
this
example,
we
illustrate
two
is nomethods
easy wayoftowriting
expressthe
thesecond
currentmesh
sourceequation.
voltage in terms of the mesh currents.) In this example, we illustrate
En este
ejemplo
ilustramos
dos
métodos
deequation.
escritura
la segunda
ecuación
enlaces.
two
methods
of1:writing
writing
the
second
mesh
Method
Assign the
a name
to the
current
sourcede
voltage.
Apply
KVL tode
both
of the meshes. Eliminate the
two
methods
of
second
mesh
equation.
Método
1:
Asignar
un
nombre
al
voltaje
de
la
fuente
de
corriente.
Aplicar
la KVL
los dos enlaces.
ElimiMethod
1:
Assign
a
name
to
the
current
source
voltage.
Apply
KVL
to
both
of the
thea meshes.
meshes.
Eliminate
the
currentMethod
source 1:
voltage
the KVL
Assignfrom
a name
to the equations.
current source voltage. Apply KVL to both of
Eliminate
the
nar elFigure
voltaje
de
la
fuente
de
corriente
desde
las
ecuaciones
de
la
KVL.
current
source
voltage
from
the
KVL
equations.
shows
thethe
circuit
labeling the current source voltage. The KVL equation for mesh 1 is
current source4.6-6
voltage
from
KVLafter
equations.
La figura
4.6-6
muestra
el circuito
etiquetar
voltaje
de la fuente
de corriente.
Figure
4.6-6
shows
the circuit
circuit
afterdespués
labelingdethe
the
current el
source
voltage.
The KVL
KVL
equationLa
forecuación
mesh 11 de
is
Figure
4.6-6
shows
the
after
labeling
current
source
voltage.
The
equation
for
mesh
is
9i1 þ v � 12 ¼ 0
la KVL para el enlace 1 es
þ
�
12
¼5000
9i
9i111 þ
1 vvv�
212
12¼
9i
The KVL equation for mesh 2 is
La ecuación
de la KVL
para el2 enlace
2 es
The
KVL equation
equation
for mesh
mesh
is
The
KVL
for
2 is
3i
3i2 þ
16i
6i2 �
2vv¼500
3i22 þ
þ 6i
6i22 �
� vv ¼
¼ 00
3i
2
2
La combinación
estas
dos ecuaciones
Combining
these de
two
equations
gives da
Combining these
these two
two equations
equations gives
gives
Combining
9i1 þ ð3i2 þ 6i2 Þ � 12 ¼ 0 ) 9i1 þ 9i2 ¼ 12
9i1 þ
þ ðð3i
3i2 þ
þ 6i
6i2 ÞÞ �
� 12 ¼
¼ 00 )
) 9i
9i1 þ
þ 9i
9i2 ¼
¼ 12
12
1 superenlace
2
2 corresponding
2 de
Método 2:
2: Apply
AplicarKVL
la KVL
al
que12
corresponda
a la1current
fuente
corriente.
Este
se
Method
to9ithe
supermesh
to the
source.
Shown
insuperenlace
Figure 4.6-7,que
this
Method
2:perimeter
Apply
KVL
toperímetro
themeshes
supermesh
corresponding
tocurrent
the current
current
source.
Shown
in
Figure
4.6-7,
this
muestra
enislathe
figura
4.6-7
esthe
elto
dethat
loseach
dos enlaces,
losto
cuales
contienen
cada
unotolain
fuente
de4.6-7,
corriente.
supermesh
of
two
contain the
source.
Apply
KVL
the
supermesh
tothis
get
Method
2:
Apply
KVL
the
supermesh
corresponding
the
source.
Shown
Figure
supermesh
is
the perimeter
perimeter
of the
the para
two meshes
meshes
that each
each contain the
the current source.
source. Apply
Apply KVL
KVL to
to the
the supermesh
supermesh to
to get
get
Aplicar la is
KVL
al superenlace
obtener
supermesh
the
of
two
3i2 þthat
6i2 � 12contain
¼ 0 ) current
9i1 þ 9i2 ¼ 12
9i1 þ
þ 3i
3i2 þ
þ 6i
6i2 �
� 12
12 ¼
¼ 00 )
) 9i
9i1 þ
þ 9i
9i2 ¼
¼ 12
12
9i1 þ
9i
1obtained
2
2
1
2
This is the same equation that was
using
method
1. Applying
KVL
to the supermesh is a shortcut for
This
is
the
same
equation
that
was
obtained
using
method
1.
Applying
KVL
to
the supermesh
supermesh
isatajo
shortcut
for
doing
three
things:
Ésta
es
la
misma
ecuación
que
se
obtuvo
con
el
método
1.
Aplicar
la
KVL
al
superenlace
es unis
para hacer
This is the same equation that was obtained using method 1. Applying KVL to
the
aa shortcut
for
doing
three
things:
estas
tres
cosas:
doing three things:
1. Labeling the current source voltage as v
1. Labeling
Etiquetarthe
el voltaje
la fuente
de corriente
como v.
1.
currentdesource
source
voltage
as vv
1.
Labeling the
current
voltage
as
2.
Applying
KVL
to both meshes
that contain
the current source
2. Applying
Aplicar laKVL
KVLto
a los
don
enlaces
quecontain
contienen
fuentesource
de corriente.
2.
both
meshes
that
the la
current
2.
Applying
KVL
to
both
meshes
that
contain
the
current
source
3.
Eliminating
v
from
the
KVL
equations
3. Eliminar v a partir de las ecuaciones de la KVL.
3. Eliminating
Eliminating vv from
from the
the KVL
KVL equations
equations
3.
9Ω
9Ω
9Ω
Ω
9
12 V +– +
12 V + –
12 V
V +–
12
–
i1
i1
ii1
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 130
1
3Ω
3Ω
3Ω
Ω
3
1.5 A
1.5 A
1.5 A
A
1.5
i2
i2
ii2
2
6Ω
6Ω
6Ω
Ω
6
FIGURE 4.6-7
FIGURA
FIGURE 4.6-7
4.6-7
FIGURE
4.6-7
Method 2 of Example 4.6-2.
Método 22 of
delExample
ejemplo 4.6-2.
4.6-2.
Method
Method
2 of Example
4.6-2.
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Análisis de corrientes de enlaces con fuentes dependientes
131
En resumen, las ecuaciones de enlace son
i1 5 i2 1 1.5
y
9i1 1 9i2 5 12
Despejar las ecuaciones nodales da por resultado
i1 5 1.4167 A e i2 5 283.3 mA
EJERCICIO 4.6-1 Determine el valor del voltaje medido por el voltímetro en la figura
E 4.6-1.
9V
Voltímetro
+–
3 4
A
4Ω
2Ω
3Ω
FIGURA E 4.6-1
Sugerencia: Escriba y despeje una ecuación de enlaces única para determinar la corriente en el resistor de 3 V.
Respuesta: 24 V.
EJERCICIO 4.6-2 Determine el valor de la corriente medida por el amperímetro en la fi-
gura E 4.6-2.
15 V
+–
3A
Amperímetro
6Ω
3Ω
FIGURA E 4.6-2
Sugerencia: Escriba y despeje una ecuación de enlaces única.
Respuesta: 23.67 A
4.7
NÁLISIS DE CORRIENTES DE ENLACES
A
CON FUENTES DEPENDIENTES
Cuando un circuito contiene una fuente dependiente la corriente controladora o el voltaje de esa
fuente dependiente se debe expresar como una función de las corrientes de enlaces.
De este modo, entonces ya es sencillo expresar la corriente controlada o el voltaje como una función
de las corrientes de enlaces. Las ecuaciones de enlaces se pueden obtener entonces de la aplicación de
la ley del voltaje de Kirchhoff para los enlaces del circuito.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 131
Alfaomega
4/12/11 5:26 PM
132
132
132
132
132
Methods of
of Analysis
Analysis of
of Resistive
Resistive Circuits
Circuits
Methods
Methods
of
Analysis
of
Resistive
Circuits
Methods
Métodosof
deAnalysis
análisisof
deResistive
circuitosCircuits
resistivos
E XX AA M
MP
PL
LE
E 4 . 7 - 1 Mesh Equations and
X A MPPLLEE 44..77--11 Mesh
Mesh Equations
Equations
and
and
E jEE
eX
mAp M
l o 4 . 7- 1 Ecuaciones
de Sources
enlaces
Dependent
Dependent
Sources
Dependent
Sources
y fuentes
dependientes
N TT EE R
RA
AC
C TT II V
V EE EE X
XA
AM
MP
P LL EE
II N
E
I IENNJTTEEEM
RRA
TTI IIVV
PAC
LCO
NEETEEEXRXAAAM
CMTPPILVLEO
Consider the circuit shown in Figure 4.7-1a. Find the value of the voltage measured by the voltmeter.
Consider
the
circuit shown
shown
in
Figureen
4.7-1a.
Find
the value
value
of the
theelvoltage
voltage
measured
by the
the voltmeter.
voltmeter.
Consider
Figure
4.7-1a.
Find
the
of
measured
by
Considerethe
el circuit
circuito
que sein
muestra
la figura
4.7-1a.
Encuentre
valor del
voltaje medido
por el voltímetro.
Ω
32 Ω
Ω
32
32ΩΩ
32
+
+
24 V
V
–+ 24
–
+
24 V
–– 24 V
Ω
32 Ω
Ω
32
32ΩΩ
32
iiaa
iaia
5iaa
5i
5ia
5i
a
(b)
(b)
(b)
Voltímetro
Voltmeter
Voltmeter
Voltmeter
Voltmeter
5ia
5i
a
5ia
5i
a
iiaa
iaia
Ω
32 Ω
Ω
32
32ΩΩ
32
+
+
24 V
V
–+ 24
–
+
24 V
–– 24 V
Ω
32 Ω
Ω
32
32ΩΩ
32
(a)
(a)
(a)
––
––
v
vm
m
vm
v+
+
m
++
24 V
V
24
24VV
24
Ω
32 Ω
Ω
32
32ΩΩ
32
+
+
–+
–
+
––
1
1
11
Ω
32 Ω
Ω
32
32ΩΩ
32
iiaa
iaia
(c)
(c)
(c)
2
2
22
5iaa
5i
5ia
5i
a
––
––
v
vm
m
vm
v+
+
m
++
FIGURE 4.7-1
4.7-1 (a)
(a) The
The circuit
circuit
FIGURE
FIGURA
(a)
circuito
FIGURE4.7-1
4.7-1
(a)El
The
circuit
FIGURE
4.7-1
(a)
The
circuit
considered
in Example
Example
4.7-1.
considered
in
4.7-1.
considerado
enExample
el ejemplo
4.7-1.
considered
in
4.7-1.
considered
in
Example
4.7-1.
(b)
The
circuit
after
replacing
the
(b) El
The
circuitluego
after replacing
the
(b)
circuito
de
reemplazar
(b)
The
circuit
after
replacing
the
(b)
The circuit
after
replacing
the
voltmeter
by
an
open
circuit.
voltmeter
by
an
open
circuit.
elvoltmeter
voltímetro
por
un circuito
abierto.
by
an
open
circuit.
voltmeter
by
an
open
circuit.
(c) El
The
circuitdespués
after labeling
labeling
the
(c)
The
circuit
after
the
(c)
circuito
de haber
(c)The
The
circuitafter
afterlabeling
labeling
the
(c)
circuit
the
meshes.
meshes.
etiquetado
los enlaces.
meshes.
meshes.
Solution
Solución
Solution
Solution
Figure
4.7-1b
shows
the circuit
afterdespués
replacing
by an
open circuit
labeling
the
La figura
4.7-1b
muestra
el circuito
de the
quevoltmeter
el voltímetro
ha equivalent
sido reemplazado
por unand
circuito
abierto
Figure 4.7-1b
4.7-1b shows
shows the
the circuit
circuit after
after replacing
replacing the
the voltmeter
voltmeter by
by an
an equivalent
equivalent open
open circuit
circuit and
and labeling
labeling the
the
Figure
m, measured
by the
voltmeter.
Figurev4.7-lc
shows
theelcircuit
after numbering
the meshes.
Leteli11circuito
and i22
voltage,
vm
equivalente
ymeasured
que el voltaje
se
ha etiquetado
, medido
por
voltímetro.
La
figura 4.7-1c
muestra
m
,
by
the
voltmeter.
Figure
4.7-lc
shows
the
circuit
after
numbering
the
meshes.
Let
i
and
voltage,
v
m
1
,mesh
measured
by the
voltmeter.
Figure
4.7-lc shows
the circuit
after
numbering
meshes. Let i1 and ii22
voltage,
vm
denote
the
currents
in meshes
2, corrientes
respectively.
luego de
numerar
los enlaces.
Sean i111eand
i2 las
de enlaces
en los
enlaces
1 y 2, the
respectivamente.
denote
the
mesh currents
currents
in meshes
meshes
and
2, respectively.
respectively.
denote
the
mesh
in
1
and
2,
current in
a short
circuit. This
circuitpara
is
The
controlling
current ofdethe
source, iaai,a,isesthe
La corriente
controladora
la dependent
fuente
dependiente,
la
en
cortocircuito.
Ésteshort
es común
thecorriente
current in
in aun
a short
short
circuit. This
This
short
circuit isis
The
controlling
current of
of the
the
dependent
source, iiaa,, isis the
current
circuit.
short
circuit
The
controlling
current
dependent
source,
common
to
meshes
1
and
2.
The
short-circuit
current
can
be
expressed
in
terms
of
the
mesh
currents
as
los enlacesto1meshes
y 2. La corriente
del cortocircuito
se
puedecan
expresar
en términos
de lasofcorrientes
enlacesas
como
common
and 2.2. The
The
short-circuit current
current
be expressed
expressed
in terms
terms
the mesh
meshde
currents
common
to meshes 11 and
short-circuit
be
in
of the
currents
as
iaa ¼ ican
11 � i22
5iii11��
2iii22
¼
iiiaa¼
The dependent source is in only one mesh, mesh 2. aThe 1reference
direction of the dependent source current does
The
dependent
sourceisisestá
inonly
only
onemesh,
mesh,
meshel2.2.enlace
Thereference
reference
direction
ofreferencia
thedependent
dependent
source
current
does
La fuente
dependiente
en sólo
un
enlace,
2. La dirección
deof
de lasource
fuentecurrent
de corriente
The
dependent
in
one
The
direction
the
does
. Consequently,
not agree
with source
the reference
direction
of i22mesh
. Consequently,
Consequently,
not
agree with
with
the
reference
direction
of ii22.de
dependiente
nothe
concuerda
con
la dirección
referencia de i2. En consecuencia,
not
agree
reference
direction
of
5iaa ¼ �i22
5
2i
5iaa ¼
¼�i
�i22
5i
Solving
for i22i2,gives
Despejando
resulta
Solving for
for ii22 gives
gives
Solving
i22 ¼ �5iaa ¼ �5ði11 � i22 Þ
¼�5i
�5iaa ¼
¼�5
�5ððii11�
�ii22ÞÞ
ii22 ¼
5
Therefore;
�4i22 ¼ �5i11 ) i22 ¼ 55i11
Entonces,
Therefore;
�4i
¼
�5i
)
i
¼
Therefore;
�4i22 ¼ �5i11 ) i22 ¼ 44ii11
4
AplicarKVL
la KVL
al enlace
para obtener
Apply
to mesh
1 to 1get
Apply
KVL
to
mesh
1
to
get
Apply KVL to mesh 1 to get
3
32i11 � 24 ¼ 0 ) i11 ¼ 33A
�
24
¼
0
)
i
¼
32i
32i11� 24 ¼ 0 ) i11 ¼ 44AA
4
Consequently,
the
value
of
i
is
En consecuencia,
valorof
de22i es
� �
Consequently,
theel
value
Consequently,
the
value
of ii222isis
5 �3 � 15
i22 ¼ 55�33� ¼ 15
15A
i
¼16 AA
¼
i22 ¼ 44 44 ¼
16
4 4
16
Apply KVL to mesh 2 to get
Apply KVL
KVL to
to mesh
mesh 22 to
to get
Apply
Aplicar la KVL
al enlace 2get
para obtener
m ¼ 0 ) vm
m ¼ 32i22
32i22 � vm
32i22�
�vvmm ¼
¼00 �)
) �vvmm ¼
¼32i
32i22
32i
�
�
15
� 15�
m ¼ 32 15 ¼ 30 V
Finally;
vm
¼30
30VV
Finally;
¼32
32 16 ¼
Finalmente,
Finally;
vvmm ¼
16
16
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 132
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Análisis de voltajesMesh
de nodos
de Analysis
circuitos
con
de corriente
Mesh
Current
Analysiswith
withfuentes
Dependent
Sources
Current
Dependent
Sources
E jEE
eX
m
lMoPPLL
4EE. 4
7de enlaces
XApAM
4..2
Mesh Equations
Equations
and
77 --22Ecuaciones
Mesh
and
y fuentes
dependientes
Dependent Sources
Sources
Dependent
133
133
133
PAC
LCO
CMTPPILVLEO
VEETEEEXRXAAAM
E
I IENNJTTEEEM
RRA
TTI IIVN
Considere
el circuit
circuito
que seinin
muestra
la figura
4.7-2a.
Encuentre
valorA,
ganancia,
Consider the
the
circuit shown
shown
Figureen
4.7-2a.
Find
the value
value
of the
theelgain,
gain,
A,deof
oflathe
the
CCVS.A, de la CCVS.
Consider
Figure
4.7-2a.
Find
the
of
CCVS.
10 Ω
10ΩΩ
10
4Ω
44ΩΩ
+
+–+ 36 V
36VV
–– 36
10 Ω
10ΩΩ
10
ia
iaaia
ia Aia
Aiaaa
iaaia Ai
+
++
–7.2 V
–7.2VV
–7.2
–
––
–
–+–
++
–7.2 V
–7.2VV
–7.2
Voltímetro
Voltmeter
Voltmeter
Aia
Aiaaa
Ai
(a)
(a)
(a)
4Ω
44ΩΩ
+
+–+ 36 V
36VV
–– 36
–
–+–
++
+
36 V +–+
36VV ––
36
10 Ω
10ΩΩ
10
1
11
(b)
(b)
(b)
4Ω
44ΩΩ
ia
iaaia
2
22
Aia
Aiaaa
Ai
–
–+–
++
+
++
–7.2 V
–7.2VV
–7.2
–
––
(c)
(c)
(c)
FIGURA
circuito
considerado
ejemplo4.7-2.
4.7-2.
(b)
El
circuito
de reemplazar
el voltímetro
un circuit.
circuito
FIGURE4.7-2
4.7-2(a)
(a)El
The
circuit
considereden
Example
4.7-2.(b)
(b)The
The
circuitdespués
afterreplacing
replacing
thevoltmeter
voltmeter
byan
anpor
open
circuit.
FIGURE
4.7-2
(a)
The
circuit
considered
ininelExample
circuit
after
the
by
open
abierto.
(c)
El circuito
después
demeshes.
etiquetar los enlaces.
(c)The
Thecircuit
circuit
afterlabeling
labeling
the
meshes.
(c)
after
the
Solución
Solution
Solution
La
figura
4.7-2b
muestra
el circuito
después de
el
voltímetro
poropen
un circuito
abierto
equivalente
y de
Figure
7-2b
shows
thecircuit
circuit
afterreplacing
replacing
thereemplazar
voltmeterby
byan
anequivalent
equivalent
open
circuitand
andlabeling
labeling
thevoltage
voltage
Figure
447-2b
shows
the
after
the
voltmeter
circuit
the
etiquetar
el
voltaje
medido
por
el
voltímetro.
La
figura
4.7-2c
muestra
el
circuito
luego
de
enumerar
los
enlaces.
andi2i2denote
denotethe
the
measuredby
bythe
thevoltmeter.
voltmeter.Figure
Figure4.7-2c
4.7-2cshows
showsthe
thecircuit
circuitafter
afternumbering
numberingthe
themeshes.
meshes.Let
Leti1i1and
measured
Sean
icurrents
corrientes
enlaces
en los enlaces 1 y 2, respectivamente.
1 e i2 lasin
mesh currents
in
meshes 1de
1 and
and
respectively.
mesh
meshes
2,2, respectively.
El voltaje
a través
la dependent
fuente
dependiente
representado
de dos
maneras.
Como
1 de
a con
withAithe
the
ofsigno
reference
The
voltage
acrossde
the
dependent
source isisestá
represented
two
ways.
Ai
þþel
of
reference
The
voltage
across
the
source
represented
inin two
ways.
ItIt isis Ai
aa with
la
dirección
de
referencia
al
final,
y
como
27.2
V
con
el
signo
1
arriba.
En
consecuencia,
direction atat the
the bottom
bottom and
and �7.2
�7.2 VV with
with the
the þþ atat the
the top.
top. Consequently,
Consequently,
direction
Ai
5 2127.22
5 7:2
7.2 V
V
�7:2ÞÞ¼¼7:2
Aiaaa ¼¼��ðð�7:2
V
Ai
La corriente controladora de la fuente dependiente, ia, es la corriente en un cortocircuito. Este cortocircuito es
isthe
thecurrent
current
inaashort
short
circuit.en
This
shortcircuit
circuit
commonto
to
Thecontrolling
controlling
current
of
the
dependent
source,
circuit.
This
short
common
The
source,
aia, ,is
común
para loscurrent
enlacesof
1the
y 2.dependent
La corriente
del icortocircuito
se in
puede
expresar
términos
de lasisiscorrientes
de
meshes
1
and
2.
The
short-circuit
current
can
be
expressed
in
terms
of
the
mesh
currents
as
meshes
1
and
2.
The
short-circuit
current
can
be
expressed
in
terms
of
the
mesh
currents
as
enlaces como
Aplique
la KVL
al enlace
1get
para obtener
Apply KVL
KVL
mesh
Apply
toto mesh
11 toto get
¼ii1i11�
�i2ii22
5
2
iaiiaa¼
36¼¼00 )
) i1i1¼¼3:6
3:6AA
10i11��36
10i
Aplique
la KVL
al enlace
2get
para obtener
Apply KVL
KVL
mesh
Apply
toto mesh
22 toto get
�7:2ÞÞ¼¼00 )
) i2i2¼¼1:8
1:8AA
4i4i22þþðð�7:2
Finally;
Finally;
Finalmente,
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 133
Aiaa
Aiaa
7:2
Ai
Ai
7:2
V/A
AA¼¼
¼¼
¼¼
¼¼44V/A
3:6��1:8
1:8
iaia i1i1��i2i2 3:6
Alfaomega
4/12/11 5:26 PM
Métodos de análisis de circuitos resistivos
134
4.8
O M PA R A C I Ó N E N T R E E L M É T O D O D E
C
V O LTA J E S D E N O D O S Y E L M É T O D O
DE CORRIENTES DE ENLACES
El análisis de un circuito compuesto se suele completar tanto por el método de voltajes de nodos como
por el de corrientes de enlaces. La ventaja de utilizar estos métodos son los procedimientos sistemáticos que se proporcionan para obtener las ecuaciones simultáneas.
En algunos casos se tiene preferencia por un método sobre el otro. Por ejemplo, cuando el
circuito contiene sólo fuentes de voltaje, quizá sea más fácil utilizar el método de las corrientes de
enlaces. Pero si el circuito sólo contiene fuentes de corriente, entonces lo más sencillo sería utilizar el
método de voltajes de nodos.
Si un circuito tiene tanto fuentes de corriente como de voltaje, se puede analizar por cualquier
método. Un enfoque puede ser comparar el número de ecuaciones que se requieren para cada método.
Si el circuito contiene menos nodos que enlaces, lo sensato sería elegir el método de voltaje de nodos.
Pero si el circuito consta de menos enlaces que nodos, entonces lo prudente es seguir el método de las
corrientes de enlaces.
Otro punto a considerar al momento de seleccionar los métodos es qué información se requiere.
Si lo que necesita es conocer las diversas corrientes, lo aconsejable sería proceder directamente con
el análisis de corrientes de enlaces. Recuerde: el análisis de corrientes de enlaces sólo funciona para
redes planares.
En ocasiones es útil determinar cuál método es el más apropiado para los requerimientos del
problema y tener en consideración ambos métodos.
Ejemplo 4.8-1
Ecuaciones de enlaces
EJEMPLO INTERACTIVO
Considere el circuito que se muestra en la figura 4.8-1. Encuentre el valor de la resistencia, R.
2Ω
1A
2Ω
0.5 A
3A
Amperímetro
R
6Ω
12 Ω
FIGURA 4.8-1 El circuito considerado en el ejemplo 4.8-1.
Solución
La figura 4.8-2a muestra el circuito de la figura 4.8-1 después de reemplazar el amperímetro por un cortocircuito
equivalente y etiquetar la corriente medida por el amperímetro. Este circuito se puede analizar empleando ecuaciones de enlaces o ecuaciones nodales. Para decidir cuál podría ser más fácil, primero se cuentan los nodos y
los enlaces. Este circuito tiene cinco nodos. Seleccionar un nodo de referencia y luego aplicar la KCL a los otros
1A
2Ω
3A
1A
2Ω
0.5 A
R
6Ω
12 Ω
(a)
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 134
2Ω
3A
2
1
2Ω
R
6Ω
3
12 Ω
(b)
0.5 A
FIGURA 4.8-2 (a) El
circuito de la figura 4.8-1,
luego de reemplazar al
amperímetro por un cortocircuito. (b) El circuito,
luego de etiquetar los
enlaces.
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
E1C04_1
11/25/2009
135
Comparación entre el método
de voltajes
nodos y
el método
de corrientes
enlaces
The Node
VoltagedeMethod
and
Mesh Current
Method de
Compared
The Node Voltage Method and Mesh Current Method Compared
135
135
135
cuatro nodos
conjunto
deatcuatro
ecuaciones
nodales.
El circuito
tresnode
enlaces.
AplicarThe
la KVL
reference
nodeproducirá
and then un
applying
KCL
the other
four nodes
will produce
a settiene
of four
equations.
circuita
estos
tresmeshes.
enlaces
producirá
un conjunto
ecuaciones
dewill
enlaces.
consiguiente,
analizar
este circuito
utireference
node
andApplying
then applying
KCL
atde
thetres
other
fourwill
nodes
produce
set of
fourequations.
node
equations.
The
circuit
has
three
KVL
to
these
three
meshes
produce
a setPor
of athree
mesh
Hence,
analyzing
lizando
ecuaciones
de equations
enlaces
eninstead
vez
deofecuaciones
nodales
producirá
menor
de ecuaciones.
Además,
has three
meshes.
Applying
KVL
to
these
three
will
produce
a set
ofconjunto
threeset
mesh
equations.
Hence,notice
analyzing
this
circuit
using mesh
nodemeshes
equations
will
produce
aun
smaller
of
equations.
Further,
that
observe
que
dos mesh
de lasequations
tres corrientes
sedirectly
pueden
determinar
por
las corrientes
denotice
la fuente
this circuit
using
instead
ofenlaces
node equations
will
produce
adirectamente
smaller
set of
equations.
Further,
that
two
of the
three
currents
can
bede
determined
from
the current
source
currents.
This
makes
the
mesh
de
Esto
hace
que
ecuaciones
de circuit
enlaces
sean
másthe
fáciles
desource
despejar.
Analizaremos
estethe
circuito
twocorriente.
of theeasier
three
mesh
currents
can
be determined
directly
from
current
This makes
mesh
equations
to
solve.
Welas
will
analyze
this
by
writing
and
solving
meshcurrents.
equations.
mediante
la
escritura
y despeje
de ecuaciones
decircuit
enlaces.
equations
easier
to solve.
We
analyze
by
and
meshi3equations.
denote the mesh currents in
Figure
4.8-2b
shows
thewill
circuit
after this
numbering
thewriting
meshes.
Letsolving
i1, i2, and
La
figura
4.8-2b
muestra
el
circuito
después
de inumerar
lostoenlaces.
Sean
i1,the
i32 denote
e1-A
i3 lascurrent
corrientes
de enlaces
,
i
,
and
i
the mesh
currents
the circuit
the
meshes.
Let
i
is
equal
the
current
in
source,
so in
meshesFigure
1, 2, 4.8-2b
and 3, shows
respectively.
The after
meshnumbering
current
1
2
1
en
los
enlaces
1,
2
y
3,
respectivamente.
La
corriente
de
enlaces
i
es
igual
a
la
corriente
en
la
fuente
de
corriente
1 the current in the 1-A current source,
to
so
meshes 1, 2, and 3, respectively. The mesh currenti i1¼is1 equal
A
1
1-A, por lo que
i1 ¼current
1A
The mesh current i2 is equal to the current in the 3-A
i1 5 1 A source, so
is
equal
to
the
current
in
the
3-A
current
source, so
The
mesh
current
i
2
¼ 3 Ade corriente
La corriente de enlaces
i2 es igual a la corriente en lai2 fuente
de 3 A, de modo que
ii2 5
¼ 33circuit
A
A
that
replaced
the ammeter, so
The mesh current i3 is equal to the current in the short
1
is
equal
to
the
current
in
the
short
circuit
that
replaced
the al
ammeter,
so
The
mesh
current
i
3
La corriente de enlaces i3 es igual a la corriente en iel
cortocircuito
que reemplazó
amperímetro,
de modo que
0:5 A
3 ¼
i
¼
0:5
A
3
i
5
0.5
A
Apply KVL to mesh 3 to get
1
Aplicar
la KVL
al enlace
para obtener
Apply KVL
to mesh
3 to3get
2ði3 � i1 Þ þ 12ði3 Þ þ Rði3 � i2 Þ ¼ 0
2ði3 � i1gives
Þ þ 12ði3 Þ þ Rði3 � i2 Þ ¼ 0
Substituting the values of the mesh currents
Sustituir
los valores
de las
de
Substituting
the values
of corrientes
the
mesh
gives
2ð0:5
� 1Þcurrents
þenlaces
12ð0:5
Þda
þ Rð0:5 � 3Þ ¼ 0 ) R ¼ 2 V
2ð0:5 � 1Þ þ 12ð0:5Þ þ Rð0:5 � 3Þ ¼ 0
E j e m p l o 4 . 8 - 2 Ecuaciones nodales
E X A M P L E 4 . 8 - 2 Node Equations
E X A M P L E 4 . 8 - 2 Node Equations
)
R¼2V
EJEMPLO INTERACTIVO
INTERACTIVE EXAMPLE
INTERACTIVE EXAMPLE
Considere el circuito que se muestra en la figura 4.8-3. Encuentre el valor de la resistencia, R.
Consider the circuit
shown in Figure 4.8-3. Find the value of the resistance, R.
2Ω
Consider the circuit shown in Figure 4.8-3. Find the value of the resistance, R.
2Ω
18 V
2Ω
2A
+ –V
18
18
+ –V
+
– 16 V+ –
+
–
+
–
16 V
16 V
2Ω
2A
2A
2Ω
2Ω
Solución
16 V
R
R
R
Voltímetro
16 V
16 V
Voltmeter
Voltmeter
FIGURA 4.8-3 El circuito considerado en el ejemplo 4.8-2.
FIGURE 4.8-3 The circuit considered in Example 4.8-2.
FIGURE 4.8-3 The circuit considered in Example 4.8-2.
La figura 4.8-4a muestra el circuito de la figura 4.8-3 después de reemplazar el voltímetro por un circuito abierto
Solution
equivalente y etiquetar el voltaje medido por el voltímetro. Este circuito se puede analizar utilizando las ecuaSolution
Figure
4.8-4a
shows
theecuaciones
circuit fromnodales.
Figure 4.8-3
after replacing
thelavoltmeter
an equivalent
openlos
circuit
andy
ciones
de enlaces
o las
Para decidir
cuál será
más fácil,byprimero
se cuentan
nodos
Figure
4.8-4a
shows
the circuit
Figure
4.8-3
after
replacing
the de
voltmeter
by an
equivalent
circuit
labeling
the voltage
measured
by from
the
voltmeter.
This
circuit
can
analyzed
using mesh
oropen
node
equations.
los
enlaces.
Este
circuito
tiene
cuatro
nodos.
Seleccionar
unbenodo
referencia
y equations
luego
aplicar
la KCL
a and
los
labeling
the
voltage
measured
by
the
voltmeter.
This
circuit
can
be
analyzed
using
mesh
equations
or
node
equations.
To
decide
which
will
be
easier,
we
first
count
the
nodes
and
meshes.
This
circuit
has
four
nodes.
Selecting
a
reference
otros tres nodos producirá un conjunto de tres ecuaciones nodales. El circuito tiene tres enlaces. Aplicar la KVL
decide
which
willproducirá
beKCL
easier,
first
count
the
nodes
and
meshes.
Thisofcircuit
has four
nodes.
Selecting
a reference
node
andtres
then
applying
at we
theconjunto
other
three
nodes
will
produce
set
three
node
equations.
Theutilizando
circuit
has ecuathree
aToestos
enlaces
un
de
tres
ecuaciones
dea enlaces.
Analizar
este
circuito
node
and
then
applying
KCL
at
the
other
three
nodes
will
produce
a
set
of
three
node
equations.
The
circuit
has
meshes.
Applying
KVL
to
these
three
meshes
will
produce
a
set
of
three
mesh
equations.
Analyzing
this
circuit
using
ciones de enlaces requiere el mismo número de ecuaciones como el que se requiere para analizar el circuitothree
utimeshes.
Applying
KVL
tothe
these
three
meshes
aare
setrequired
of three
mesh
equations.
Analyzing
circuit
using
mesh
equations
requires
same
number
equations
as corrientes
analyze
circuit
using this
node
equations.
lizando
ecuaciones
nodales.
Observe
que of
unawill
de produce
las tres
deto
enlaces
sethe
puede
determinar
directamente
equations
requires
themesh
number
of equations
aredirectly
required
tovoltajes
analyze
the nodos
circuitse
using
node
equations.
Notice
that
of
the three
currents
can
be determined
from
the current
source
current,
butdeterminar
two
of the
amesh
partir
de one
la corriente
de
lasame
fuente
de corriente,
peroasdos
de
los tres
de
pueden
Noticenode
that voltages
one
of thecan
three
currents
can
befrom
determined
directly
from
the
source
current,
but
two
ofmás
the
three
be mesh
determined
directly
the voltage
source
This
makes
thesenode
equations
directamente
a partir
de
los
voltajes
de la
fuente
de voltaje.
Esto
hace
quevoltages.
las current
ecuaciones
nodales
despejen
three node
voltages
can be
determined
directly
fromy the
voltage
source
This makes the node equations
fácilmente.
Analizaremos
este
circuito
escribiendo
despejando
ecuaciones
nodales.
easier
to solve.
We will
analyze
this circuit
by writing
and
solving
nodevoltages.
equations.
easier Figure
to solve.
We
will
analyze
circuit
by writing
and solving
node
equations.
La
figura
4.8-4b
muestra
elthis
circuito
después
de
seleccionar
un
nodo
de referencia
y enumerar
los vnodos
4.8-4b
shows
the circuit
after selecting
a reference
node
and
numbering
the other
nodes. Let
1, v 2,
4.8-4b
thevoltajes
circuit
after
a nodos
reference
and
numbering
the
nodes.
v1, vbe
restantes.
Sean
v1,node
v2shows
y voltages
v3 los
de nodos
en3,
losrespectively.
1, 2node
yThe
3, respectivamente.
Elother
voltaje
de
laLet
fuente
de
denote
the
at nodes
1, selecting
2, and
voltage
of the 16-V
voltage
source
can
and
v3Figure
2,
voltaje
de 16
Vthe
senode
puede
expresar
términos
de voltajes
de nodosThe
como
voltages
at en
nodes
1,
3, respectively.
voltage of the 16-V voltage source can be
and v3 denote
expressed
in terms
of the
node voltages
as2, and
expressed in terms of the node voltages16as¼ v � 0 ) v ¼ 16 V
1
1
16 ¼ v1 � 0 ) v1 ¼ 16 V
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 135
Alfaomega
4/12/11 5:26 PM
Methods
Methods of
of Analysis
Analysis of
of Resistive
Resistive Circuits
Circuits
Métodos of
deAnalysis
análisis of
deResistive
circuitos Circuits
resistivos
Methods
136
136
136
136
2Ω
22ΩΩ
2Ω
2Ω
22ΩΩ
2Ω
18
18VVV
18
18
+ –V
+
+
16
16VVV +––+
16
16 V ––
1
11
1
2A
22AA
2A
++––
+–
2Ω
22ΩΩ
2Ω
+
++
+
16
16VVV
16
16– V
––
–
R
RR
R
18
18VVV
18
18
+ –V
+
+
16
16VVV +––+
16
16 V ––
++––
+–
2
22
2
2A
22AA
2A
2Ω
22ΩΩ
2Ω
(a)
(a)
(a)
(a)
R
RR
R
3
33
3
+
++
+
16
16VVV
16
16– V
––
–
(b)
(b)
(b)
(b)
FIGURE
FIGURE 4.8-4
4.8-4 (a)
(a) The
The
circuit
from
Figure
4.8-3
FIGURE
4.8-4
The
FIGURA
4.8-4
(a)(a)El
circuit
from
Figure
4.8-3
after
replacing
the
circuit
from
Figure
4.8-3
circuito
de
la
figura
4.8-3
after replacing the
voltmeter
an
open
after replacing
después
de by
reemplazar
voltmeter
by
anthe
open el
voltímetro
uncircuit
circuito
circuit.
(b)
The
voltmeter
by
an
open
circuit.
(b)por
The
circuit
abierto.
(b)
circuito
after
labeling
the
circuit.
(b)El
The
after
labeling
thecircuit
después
de etiquetar
nodes.
after labeling
the los
nodes.
nodos.
nodes.
El voltaje
de la fuente
de voltaje de
18 V se puede
expresar enterms
términosthe
de voltajes
de nodos como
The
The voltage
voltage of
of the
the 18-V
18-V voltage
voltage source
source can
can be
be expressed
expressed in
in terms of
of the node
node voltages
voltages as
as
The voltage of the 18-V voltage
source
can)
be expressed
in vterms
of the
node
voltages
as
v
18
¼
�
v
18
¼
16
�
)
v
¼
�2
V
1
2
2
2
18 ¼ v1 � v2 ) 18 ¼ 16 � v2 ) v2 ¼ �2 V
18 ¼ vvoltage
18 ¼
16 � v2 ) v2 ¼ �2 V
1 � v2 at)
The
voltmeter
measures
the
node
El
el voltaje
de nodos
en elatnodo
Thevoltímetro
voltmetermide
measures
the node
node
voltage
node3,3,
3,deso
somodo que
The voltmeter measures the node voltage at node v3,
so V
¼
16
V
¼ 16
16 V
vv333 5
¼
16 V
v
3
Applying
KCL
at
node
3
to
get
Aplicar
la KCL
KCL al
obtener
Applying
at nodo
node 33 para
to get
Applying KCL at node 3 to get
vv1 �
v
v
1 � v33 þ 2 ¼ v33
2 ¼ vR3
v1 �
22 v3 þ
þ2¼ R
2
R
Substituting
the
values
of
the
node
voltages
gives
Substituting
the values
of the
node voltages
gives
Sustituir
los valores
de los
nodos da
Substituting
the values
of voltajes
the nodedevoltages
gives
16
16 �
� 16
16 þ 2 ¼ 16
16 ) R ¼ 8 V
2 ¼ 16
) R ¼ 8V
16 �
22 16 þ
R
þ2¼ R
) R ¼ 8V
2
R
4.9
4.9
4.9
4.9
M
E
CU
RR
EN
M
ES
SH
H
R
NT
TA
AN
NA
A LL Y
YS
S IE
IS
SU
US
S II N
NG
GM
MA
AT
T LL A
AB
B
A N
L IC
SU
I SR
DE
M
EÁ
SH
C
U
RR
EE NCTOARNRAI EL N
YT
S I SS UDSEI N G M A T L A B
N L that
A Ccircuits
E S Uthat
T Icontain
L I Z Aresistors
N D O and
M AT
L A B or dependent sources can be analyzed
We haveEseen
independent
We have seen that circuits that contain resistors and independent or dependent sources can be analyzed
Wethe
have
seen thatway:
circuits that contain resistors and independent or dependent sources can be analyzed
in
following
in
the
following
way:
Hemos
visto queway:
los circuitos que contienen resistores y fuentes independientes o dependientes se
in
the following
pueden analizar
de lanode
siguiente
manera:
1.
1. Writing
Writing aa set
set of
of node or
or mesh
mesh equations
equations
1.
Writing
a setunofconjunto
node or de
mesh
equations
1.
Escribiendo
ecuaciones
nodales o de enlaces
2.
2. Solving
Solving those
those equations
equations simultaneously
simultaneously
2.
Solving
those
equations
simultaneously
2. Despejando simultáneamente esas ecuaciones
In
In this
this section,
section, we
we will
will use
use the
the MATLAB
MATLAB computer
computer program
program to
to solve
solve the
the equations.
equations.
Enthis
esta
sección the
usaremos
elshown
programa
de computación
MATLAB
para despejar
las ecuaciones.
In
section,
we
will use
the
MATLAB
program
solve
the
equations.
Consider
circuit
in
4.9-1a.
circuit
aa potentiometer.
In
Consider
the
circuit
shown
in Figure
Figure computer
4.9-1a. This
This
circuittocontains
contains
potentiometer.
In Figure
Figure
Considere
el circuito
quebeen
se in
muestra
en4.9-1a.
la figura
4.9-1a.
Este
circuito
contiene
un potenciómeConsider
the
circuit
shown
Figure
This
circuit
contains
a
potentiometer.
In Figure
4.9-1b,
the
potentiometer
has
replaced
by
a
model
of
a
potentiometer.
R
of
4.9-1b, the potentiometer has been replaced by a model of a potentiometer. Rpp is
is the
the resistance
resistance
of
tro. En la
4.9-1b, el has
potenciómetro
ha sido
por un modeloRde
potenciómetro.
R
4.9-1b,
thefigura
potentiometer
been replaced
by a reemplazado
model of a potentiometer.
is
the
resistance
ofp
p
R
R11
R1
+
–+
–
+
–
vv1
1
v1
R
Rpp
Rp
R
R33
R3
+
+
vv+o
o
v––o
–
(a)
(a)
(a)
R
R44 =
= aR
aRpp
R4 = aRp
R
R11
R1
R
R22
R2
vv2 ++
2 –
v2 –+
–
+
–+
–
+
–
vv1
1
v1
ii1
1
i1
R
R55 =
= (1
(1 –– a)R
a)Rpp
R5 = (1 – a)Rp
R
R33
R3
+
+
vv+o
o
v––o
–
ii2
2
i2
R
R22
R2
vv2 ++
2 –
v2 –+
–
(b)
(b)
(b)
FIGURE
FIGURE 4.9-1
4.9-1 (a)
(a) A
A circuit
circuit that
that contains
contains aa potentiometer
potentiometer and
and (b)
(b) an
an equivalent
equivalent circuit
circuit formed
formed by
by replacing
replacing the
the
FIGURA
4.9-1
(a)
Circuito
que
contiene
un
potenciómetro
y
(b)
un
circuito
equivalente
formado
por
el
reemplazo
potentiometer
with
a
model
of
a
potentiometer
ð
0
<
a
<
1Þ
.
FIGURE
4.9-1
(a)
A
circuit
that
contains
a
potentiometer
and
(b)
an
equivalent
circuit
formed
by
replacing del
the
potentiometer with a model of a potentiometer ð0 < a < 1Þ.
potenciómetrowith
con un
modelo
un potenciómetro
, 1Þ
a,
potentiometer
a model
of de
a potentiometer
ð0 <(0
a<
. 1).
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 136
Circuitos Eléctricos - Dorf
4/12/11 5:26 PM
Análisis de corrientes de enlaces utilizando MATLAB
137
es la resistencia del potenciómetro. El parámetro a varía de 0 a 1 en cuanto el contacto deslizante del
potenciómetro se mueve de un extremo al otro del potenciómetro. Las resistencias R4 y R5 se describen mediante las ecuaciones
R4 5 aRp
(4.9-1)
R5 5 11 2 a2Rp
(4.9-2)
y
Nuestro objetivo es analizar este circuito para determinar cómo cambia el voltaje de salida al cambiar
la posición del contacto deslizante del potenciómetro.
% mesh.m solves mesh equations
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% Enter values of the parameters that describe the circuit.
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% circuit parameters
R11000;
% ohms
R21000;
% ohms
R35000;
% ohms
V1 15;
% volts
V215;
% volts
Rp20e3;
% potentiometer parameters
% ohms
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% the parameter a varies from 0 to 1 in 0.05 increments.
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
a00.051;
% dimensionless
for k1length(a)
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% Here is the mesh equation, RIV:
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
R  [R1a(k)*RpR3
R3;
R3
(1a(k))*RpR2R3];
V  [ V1;
V2];
%
%
%
%
––––––
eqn.
4.9-6
––––––
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% Tell MATLAB to solve the mesh equation:
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
I  V’/R;
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% Calculate the output voltage from the mesh currents.
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Vo(k)  R3*(I(1)I(2)); % eqn. 4.9-7
end
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
% Plot Vo versus a
%–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
plot(a, Vo)
axis([0 1 15 15])
xlabel(‘a, dimensionless’)
ylabel(‘Vo, V’)
FIGURA 4.9-2 Archivo de entrada de MATLAB que se utilizó para analizar el circuito que se muestra en la figura 4.9-1.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 137
Alfaomega
4/12/11 5:26 PM
E1C04_1
E1C04_1
11/25/2009
11/25/2009
138
138
138
138
138
138
138
Methods of
of Analysis of
of Resistive Circuits
Circuits
Methods
Métodos deAnalysis
análisis deResistive
circuitos resistivos
Methods
of
Analysis
of
Resistive
Circuits
Methods of Analysis of Resistive Circuits
Voo,,VV
VVoo,,V
VV
El circuito
figura4.9-1b
4.9-1bcan
se puede
representar
ecuaciones
deas
enlaces como
The
circuiten
in laFigure
Figure
4.9-1b
can
be represented
represented
bypor
mesh
equations
as
The
circuit
in
be
by
mesh
equations
The
can
be
represented
by
mesh
equations
as
The circuit
circuit in
in Figure
Figure 4.9-1b
4.9-1b
can
be
represented
by
mesh
equations
as
R
i
þ
R
i
þ
R
ð
i
�
i
Þ
�
v
¼
0
R11i11 þ R44 i11 þ R33ði11 � i22 Þ � v11 ¼ 0
ð4:9-3Þ
(4.9-3)
ð4:9-3Þ
þR
þ½½vvRR
ððii1R
�ððiii1122Þ�
RR
i iþ
RR
i iþ
�
R
�
Þ�
�ii22vvÞÞ1��1 ¼
¼000
¼
R
1þ
1þ
1 33�
55Ri1221 i1þ
22Ri4224 i1þ
22 33�
ð4:9-3Þ
ð4:9-3Þ
RR22ias
RR33ððii11 �
RR55ii22 þ
þreacomodar
i22 þ
þ ½½vv22 �
�como
� ii22ÞÞ�� ¼
¼00
Theseecuaciones
mesh equations
equations
can be
be
rearranged
Estas
de enlaces
se rearranged
pueden
These
mesh
can
as
These
as
These mesh
mesh equations
equations can
can be
be rearranged
rearranged
as44 þ
R11 þ
þR
R
þR
R33Þi
Þi11 �
�R
R33ii22 ¼
¼ vv11
ððR
ð4:9-4Þ
ð4:9-4Þ
(4.9-4)
ððRR
þ
R
þ
R
Þi
�
R
i
¼
�R
i
þ
ð
R
þ
R
þ
R
Þi
�vv
1
4
3
1
3
2
3
1
5
2
3
þ
R
þ
R
Þi
�
R
i
�R
i
þ
ð
R
þ
R
þ
R
Þi
¼
45
321
3 11
3 3 22 ¼ �vv1221
ð4:9-4Þ
ð4:9-4Þ
�R
RR2 þ
RR33ÞiÞi2and
¼ �v
33ii11 þ
22 into Eq. 4.9-4 gives
�R
þ ððRR55 þ
þEqs.
þ
2 ¼ �v
Substituting
Eqs.
4.9-1
4.9-2
Substituting
4.9-1
and
4.9-2
into
Eq. 4.9-4
Sustituyendo
las2�ecuaciones
4.9-1
y 4.9-2
en la gives
ecuación 4.9-4
�
� into Eq.
15
Substituting
4.9-1
and
4.9-4
gives
15
15
Substituting Eqs.
Eqs.�R
4.9-1
andþ4.9-2
4.9-2
4.9-4
R
þ aR
R iinto
� REq.
¼
resulta
3 ii22 ¼
vv11 gives ð4:9-5Þ
�
11 þ aRpp þ R33��i11 � R�
3�
�
�
�
15
ð4:9-5Þ
15
R
þ
aR
þ
R
i
�
R
i
¼
v
�R
i
þ
ð
1
�
a
ÞR
þ
R
þ
R
�v
pp ppþþR3R
33ii222 ¼
�R33 i11 þ �R
� aR
aÞR
RR
3 22i11þ�
10
33�
�ð111 þ
� ¼�vv1221
10
ð4:9-5Þ
10
(4.9-5)
ð4:9-5Þ
�R
i
þ
ð
1
�
a
ÞR
þ
R
þ
R
¼
�v
i
33 i11 þ
1 �written
aÞRpp þusing
R22 þ matrices
R33 i22 ¼ �v
10
Equation �R
4.9-5
canðbe
be
written
using
matrices
as22
10
Equation
4.9-5
can
as
5
��
����as
�� como
��
��
55
Equation
4.9-5
can
written
using
matrices
La
ecuación
4.9-5
se3be
puede
escribir
usando
matrices
Equation
4.9-5
can
be
written
using
matrices
as
R
þ
aR
þ
R
�R
i
v
1
P
3
1
1
R
þ
aR
þ
R
�R
i
v
�
�
�
�
�
�
1
P
3
3
1
1
55
�
�� � ¼ �
�
RR11 þ
aR
vv1122
�R
� aaÞR
ÞR�R
þ33R
R22 þ
þR
R33 iii1221 ¼ �v
�v
Pþ
þ�R
aRP3P3 þ
þ RR33 ðð11 �
�R
0
P
00
¼
¼
�R
ðð11 �
�v
22
00
�R33
� aaÞR
ÞRPP þ
þ RR22 þ
þ RR33 ii22
�vð4:9-6Þ
ð4:9-6Þ
–5
–5
–5
ð4:9-6Þ
ð4:9-6Þ
and ii22 are
are calculated
calculated by
by using
using MATLAB
MATLAB to
to solve
solve the
the(4.9-6)
mesh
Next, ii11 and
mesh
Next,
–5
–5
and
i
are
calculated
by
using
MATLAB
to
solve
the
mesh
Next,
i
equation,
Eq.
4.9-6.
Then
the
output
voltage
is
calculated
as
–10
are
calculated
using MATLAB
to solve
Next, i11 and
equation,
Eq.
Then
theby
output
voltage
is calculated
asmesh
–10
–10
Enseguida,
i1i22e4.9-6.
i2 se
calculan
utilizando
MATLAB
para the
despejar
equation,
Eq.
4.9-6.
Then
the
output
voltage
isis calculated
as
–10
equation,
Eq.
4.9-6.
Then
the
output
voltage
calculated
as
–10
v
¼
R
ð
i
�
i
Þ
ð4:9-7Þ
la ecuación de enlaces, 4.9-6.
el voltaje deð4:9-7Þ
salida
voo ¼ Luego
R33ði11 �sei22calcula
Þ
–15
–15
–15
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
v
¼
R
ð
i
�
i
Þ
ð4:9-7Þ
00 0.1
0.1 0.2
0.2 0.3
0.3 0.4
0.4 0.5
0.5 0.6
0.6 0.7
0.7 0.8
0.8 0.9
0.9 11 como Figure 4.9-2 showsvothe
R33 ði11 � i22 Þinput file. The parameter
ð4:9-7Þ
o ¼MATLAB
–15
–15
Figure 4.9-2 shows the
MATLAB
input file. The parameter
a,sin
dimensionless
00 0.1
0.4
0.5
a,a,
dimensiones
dimensionless
v
5
R
1i
2
i
2
(4.9-7)
0.1 0.2
0.2 0.3
0.3
0.4
0.5 0.6
0.6 0.7
0.7 0.8
0.8 0.9
0.9 11
o
3
1
2
Figure
the
input
file.
The
parameter
varies
from4.9-2
toshows
in increments
increments
of 0.05.
0.05.
At
each
value
of a,
a,
Figure
4.9-2
shows
the MATLAB
MATLAB
inputAt
file.
Thevalue
parameter
aa varies
from
00 to
11 in
of
each
of
a,
a, dimensionless
dimensionless
FIGURE 4.9-3
4.9-3 Plot
Plot
ofde
versus
for the
the circuit
circuit shown
shown a
FIGURE
4.9-3
vvoovversus
aa for
from
004.9-2
to
11muestra
in
of
At
value
a,
MATLAB
solves
Eq.
4.9-6
andelthen
then
uses
Eq.entrada
4.9-7
todecalculate
calculate
the
La figura
archivo
de
MATLAB.
FIGURA
Trazoof
a varies
varies
from
to
in increments
increments
of 0.05.
0.05.
At each
each
value of
ofthe
a,
MATLAB
solves
Eq.
4.9-6
and
uses
Eq.
4.9-7
to
o versus a para el circuito que
in
Figure
4.9-1.
FIGURE
4.9-3
Plot
vvoo versus
in
Figure
FIGURE
4.9-3
Plot of
of4.9.1.
versus aa for
for the
the circuit
circuit shown
shown El
se
muestra4.9-1.
en la figura
MATLAB
solves
Eq.
then
uses
4.9-7
calculate
the
versus
output
voltage.
Finally,
MATLAB
produces
the
plotto
of
parámetro
a varía
de4.9-6
0MATLAB
a 1and
en incrementos
0.05.
valor
o
MATLAB
solves
Eq.
4.9-6
and
then
usesEq.
Eq.de
4.9-7
toEn
calculate
the
versus
aa
output
voltage.
Finally,
produces
the
plot
of
vvcada
o
in
inFigure
Figure 4.9-1.
4.9-1.
versus
output
voltage.
MATLAB
the
vvooecuación
that
is
shown
inFinally,
Figurela
4.9-3.
de
a, is
MATLAB
despeja
ecuación produces
4.9-6
y luego
utiliza
versus aa
output
voltage.in
Finally,
MATLAB
produces
the plot
plot of
ofla
that
shown
Figure
4.9-3.
that
in
4.9-3.
4.9.7
para
calcular
el voltaje
de salida. Finalmente, MATLAB produthat is
is shown
shown
in Figure
Figure
4.9-3.
ce
el
trazo
de
v
comparado
con
a que se muestra en la figura 4.9-3.
4.10
U
S
I
N
G
P
S
P
I
C
E
T
O
D
E
T
E
R
M
I
N
E
o
4.10 U S I N G P S P I C E T O D E T E R M I N E
4.10
4.10
4.10
U
C
EE E
TS
O
D
EED
T EMR
N
US
SID
IN
NE
GV
SL
CG
OA
DN
RE
M
NE
EU
N
O
D
EG
VPPO
OS
LPPT
TIIA
A
G
A
N
EM
SIIH
H
C
UR
RR
RE
EN
NT
TS
S
N
O
ETS
DT EM
S
C
N
O
D
E
V
O
L
T
A
G
E
S
A
N
D
M
E
S
H
C
U
R
R
E
N
T
S
NSOOD ED V
G EE SPA
AR
NA
D M
U
E OPLSTPAI C
D EETSEHRCMUI R
NR
AERNLTOSS V O LTA J E S
To determine
determine
the
node
voltages
of
dcCcircuit
circuit
using
PSpice,
we
To
aa dc
D E the
N Onode
D Ovoltages
S Y Lof
AS
O R Rusing
I E NPSpice,
T E S we
DE ENLACES
Alfaomega
To
To determine
determine the
the node
node voltages
voltages of
of aa dc
dc circuit
circuit using
using PSpice,
PSpice, we
we
1.
Draw
the
circuit
in
the
OrCAD
Capture
workspace
1.
Draw
the
circuit
in
the
OrCAD
Capture
workspace
Para determinar los voltajes de nodos de un circuito utilizando PSpice,
1.
circuit
in
OrCAD
1. Draw
Draw the
the
circuitPoint’
in the
thesimulation
OrCAD Capture
Capture workspace
workspace
2.
Specify
‘Bias
2.
Specify
aa ‘Bias
Point’en
simulation
1.
Dibujamos
el circuito
el área de trabajo de OrCAD Capture
2.
aa ‘Bias
2. Specify
Specify
‘Bias Point’
Point’ simulation
simulation
3.
Run
the simulation
simulation
3.
Run
the
2.
Especificamos
una simulación ‘Bias Point’
3.
3. Run
Run the
the simulation
simulation
3.
Ejecutamos
simulación
PSpice
will label
labellathe
the
nodes with
with the
the values
values of
of the
the node
node voltages.
voltages.
PSpice
will
nodes
PSpice
will
label
the
nodes
with
the
values
the
node
voltages.
An
extra
steplos
is nodos
needed
to use
use
PSpice
toof
determine
the
mesh
currents. PSpice
PSpice does
does not
not label
label the
the
PSpice
will
label
the
nodescon
with
the
valuesto
of
thevoltajes
nodethe
voltages.
An
extra
step
is
needed
to
PSpice
determine
currents.
PSpice
etiquetará
los
valores
de
los
demesh
nodos.
extra
step
isispaso
needed
to
PSpice
to
the
mesh
currents.
PSpice
does
not
the
valuesAn
ofnecesita
the
mesh
currents,
but
ituse
does
provide
the value
value
of
the
current
in
each
voltage
source.
Recall
An
extra
step
needed
topara
use
PSpice
to determine
determine
thethe
mesh
currents.
PSpice
does
not
label
the
values
of
the
mesh
but
it
does
provide
the
of
current
each
voltage
source.
Recall
Se
uncurrents,
extra
utilizar
PSpice
para determinar
las in
corrientes.
PSpice
nolabel
etiqueta
values
of
currents,
it
the
the
current
in
each
voltage
source.
Recall
that
0-V
voltage
source is
isbut
equivalent
to
short
circuit.
Consequently,
we can
can
insert
0-V
current
values
of the
the
mesh
currents,
but
it does
does provide
provide
the value
value of
ofel
the
current
incorriente
each
voltage
source.
Recall
that
aa 0-V
voltage
source
equivalent
to
aa short
circuit.
Consequently,
we
insert
0-V
current
los
valores
demesh
las corrientes
de
enlaces,
pero
proporciona
valor
de la
en cada
fuente
de
that
source
isis equivalent
to
avalues
circuit.
we
can
insert
0-V
current
sources
intovoltage
the circuit
circuit
without
altering
the
values
of
the mesh
mesh
currents.
We
will
insert
those
sources
that aa 0-V
0-V
voltage
source
equivalent
tode
a short
circuit.
Consequently,
wewill
can
insertthose
0-Vsources
current
sources
into
the
without
altering
the
of
the
We
insert
voltaje.
Recuerde
que
una
fuente
de voltaje
0short
V equivale
aConsequently,
un currents.
cortocircuito.
En
consecuencia,
podesources
into
the
circuit
altering
values
of
the
mesh
currents.
We
will
insert
sources
into
the
circuit
in
such
way that
that
their
currents
are
also
the mesh
mesh
currents.
Tocorrientes
determine
the
mesh
sources
into
thein
circuit
altering
the
valuesare
of
the
mesh
currents.
We
will
insert those
those
sources
into
the
circuit
such
aawithout
way
currents
the
currents.
To
determine
mesh
mos
insertar
fuentes
de without
voltaje
de
0their
V enthe
el circuito
sinalso
alterar
los
valores
de las
dethe
enlaces.
into
such
way
that
their
are
currents.
determine
the
currents
of
dcin
circuit
PSpice,
wecurrents
into the
the circuit
circuit
incircuit
such ausing
ausing
way
that
their
currents
are also
also
the
mesh
currents.
To
determine
the mesh
mesh
currents
of
aa esas
dc
we
Insertaremos
fuentes
en elPSpice,
circuito
de
tal manera
quethe
susmesh
corrientes
seanTo
también
corrientes
de
currents
of
dc
PSpice,
currents Para
of aa determinar
dc circuit
circuit using
using
PSpice, we
weenlaces de un circuito de cd usando PSpice,
enlaces.
las corrientes
de
1. Draw
Draw the
the circuit
circuit in
in the
the OrCAD
OrCAD Capture
Capture workspace.
workspace.
1.
1. Draw
Dibujamos
el circuito
enOrCAD
el área de
trabajoworkspace.
de OrCAD Capture.
1.
the
circuit
in
the
Capture
1.
Draw
the
circuit
in
the
OrCAD
Capture
2. Add
Add 0-V
0-V voltage
voltage sources
sources to
to measure
measure the
the workspace.
mesh currents.
currents.
2.
mesh
2. Add
Agregamos
fuentes
de voltaje
de 0 V para
medircurrents.
las corrientes de enlaces.
2.
0-V
voltage
sources
to
measure
the
mesh
2. Specify
Add 0-V voltage
sources
to measure the mesh currents.
3.
Bias Point
Point
simulation.
3.
Specify aa Bias
simulation.
3. Specify
Especificamos
un simulación
Bias Point.
3.
aa Bias
3. Run
Specify
Bias Point
Point simulation.
simulation.
4.
the
simulation.
4.
Run
the simulation.
4. Run
Ejecutamos
la simulación.
4.
4. Run the
the simulation.
simulation.
PSpice escribirá
will write
writelasthe
the
voltage source
source
currents
in the
the
output
file.de salida.
PSpice
corrientes
de fuente
de voltaje
en output
el archivo
PSpice
will
voltage
currents
in
file.
PSpice
PSpice will
will write
write the
the voltage
voltage source
source currents
currents in
in the
the output
output file.
file.
M04_DORF_1571_8ED_SE_108-161.indd 138
Circuitos Eléctricos - Dorf
4/12/11 5:27 PM
Uso de PSpice para determinar los voltajes de nodos y las corrientes de enlaces
139
E j e m p l o 4 . 10 - 1 Uso de PSpice para encontrar voltajes de nodos
y corrientes de enlaces
Utilice PSpice para determinar los valores de los voltajes de nodo y las corrientes de enlaces para el circuito que
se muestra en la figura 4.10-1.
5Ω
i1
v1
15 Ω
v2
0.5 A
i2
30 V
+–
20 Ω
0.2 A
10 Ω
v3
i3
i4
FIGURA 4.10-1 Circuito con voltajes de nodos v1, v2, v3
y v4 y corrientes de enlaces i1, i2, i3 e i4.
v3
25 Ω
FIGURA 4.10-2 El circuito de la figura 4.10-1 dibujado en el
área de trabajo de OrCAD. Los números blancos que se muestran
en los fondos negros son los valores de los voltajes de nodos.
Solución
La figura 4.10-2 muestra el resultado de dibujar el circuito en el área de trabajo de OrCAD (vea Apéndice A) y
ejecutar una simulación Bias Point. (En la barra de menús de OrCAD seleccione el menú PSpice y haga clic en la
opción New Simulation Profile; luego seleccione Bias Point de la lista descendente Analysis Type en el cuadro de
diálogo Simulation Settings, para especificar un punto de simulación de desvío. En la barra de menús de OrCAD
Capture, seleccione el menú PSpice, y haga clic en la opción Run Simulation Profile para ejecutar la simulación.)
PSpice etiqueta los nodos con los valores de los voltajes de nodos mediante números blancos que destacan sobre
fondos negros. Al comparar las figuras 4.10-1 y 4.10-2 vemos que los voltajes de nodos son
v1 5 26.106 V, v2 5 210.61 V, v3 5 222.34 V y v4 5 27.660 V.
La figura 4.10-3 muestra el circuito de la figura 4.10-2 después de insertar una fuente de corriente de 0 V en el
exterior de cada enlace. Las corrientes en estas fuentes de 0 V serán las corrientes de enlaces que se muestran en la
figura 4.10-1. En particular, la fuente V2 mide la corriente de enlaces i1, la fuente V3 mide la corriente de enlaces
i2, la fuente V4 mide la corriente de enlaces i3, y la fuente V5 mide la corriente de enlaces i4.
Luego ejecute una vez más la simulación (de la barra de menús de OrCAD seleccione el menú PSpice y
haga clic en la opción Run), OrCAD Capture abrirá una ventana Schematics. En la barra de menús de la ventana
Schematics seleccione el menú View y haga clic en la opción Output File. Mueva hacia abajo la barra deslizadora
FIGURA 4.10-3 Circuito de la figura 4.10-1 dibujado en el taller
de OrCAD con fuentes de voltaje de 0 V agregadas para medir las
corrientes de enlaces.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 139
Alfaomega
4/12/11 5:27 PM
E1C04_1
11/25/2009
140
140
140
140
Methods
Métodosof
deAnalysis
análisisof
deResistive
circuitosCircuits
resistivos
Methods of Analysis of Resistive Circuits
down
through
the output
file para
to find
the currents
in the voltage
sources:de voltajes:
a través
del archivo
de salida
encontrar
las corrientes
en las fuentes
down through the output file to find the currents in the voltage sources:
VOLTAGE SOURCE CURRENTS
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
NAME
CURRENT
� 6:170E � 01
V V1
V V1
� 6:170E � 01
V V2
3:106E � 01
V V2
3:106E � 01
� 3:064E � 01
V V3
V V3
� 3:064E � 01
8:106E � 01
V V4
V V4
8:106E � 01
6:106E � 01
V V5
V V5
6:106E � 01
TOTAL POWER DISSIPATION 1:85E þ 01 WATTS
TOTAL POWER DISSIPATION 1:85E þ 01 WATTS
JOB CONCLUDED
JOB CONCLUDED
PSpice
thelapassive
convention
current and
circuit
voltage
sources.
PSpice uses
utiliza
convención
pasiva for
parathe
la corriente
y elvoltage
voltajeof
deall
todos
los elements,
elementosincluding
de circuito,
incluyendo
las
PSpice uses the passive convention for the current and voltage of all circuit elements, including voltage sources.
Noticing
smallObservando
þ and � signs
on the1voltage
source
symbols
Figure
welasee
that4.10-3,
the currents
fuentes dethe
voltaje.
los signos
y 2 en los
símbolos
de lainfuente
de 4.10-3,
voltaje en
figura
vemos
Noticing the small þ and � signs on the voltage source symbols in Figure 4.10-3, we see that the currents
provided
by PSpice
are directed
form left
to rightdeinizquierda
sources VI
and V2en
and
directed
to left ina
que las corrientes
provistas
por PSpice
se dirigen
a derecha
lasare
fuentes
V1 yfrom
V2, right
y de derecha
provided by PSpice are directed form left to right in sources VI and V2 and are directed from right to left in
sources
V3,
and V5.V3,
In V4
particular,
mesh currents
are
izquierda
en V4,
las fuentes
y V5. Enthe
particular,
las corrientes
de enlaces son
sources V3, V4, and V5. In particular, the mesh currents are
i1i1¼50:3106
0:6106
0:8106
�0:3064A.
A:
0.3106A;
A,ii2i2¼
0.6106A;
A,ii3i3¼
0.8106A;
A and
e ii4i45¼
A;
¼50:6106
A;
¼50:8106
A;
and
¼20.3064
�0:3064 A:
i ¼ 0:3106
1
4.11
4.11
4.11
2
3
4
¿ C Ó M O LO P O D E M O S C O M P R O B A R . . . ?
HOW CAN WE CHECK . . . ?
HOW CAN WE CHECK . . . ?
A los ingenieros se les suele solicitar comprobar que la solución de un problema sea la correcta. Por
Engineers
aresoluciones
frequentlypropuestas
called upon
check thatdeadiseño
solution
to a problem
is indeed
correct. For
ejemplo, las
parato
se deben
comprobar
para confirmar
Engineers
are frequently
called upon
toproblemas
check that a solution
to a problem
is indeed
correct. que
For
example,
proposed
solutions
to design problems
must
be checked
to resultados
confirm that
allcomputaof the
se
ha
cumplido
con
todas
las
especificaciones.
Además,
se
deben
revisar
los
de
example, proposed solutions to design problems must be checked to confirm thatlaall
of the
specifications
have been
satisfied.
computer
output
be reviewed
to guard
against
dora para protegerse
contra
erroresIn
deaddition,
captura de
datos, así
comomust
las exigencias
de los
specifications
have been
satisfied.
In
addition,
computer
output
must
be reviewed
to comerciantes,
guard against
data-entry
errors,
and
claimsa made
by vendors must be examined critically.
las
cuales
se
deben
analizar
fondo.
data-entry errors, and claims made by vendors must be examined critically.
Engineering
students
are also
asked to secheck
the que
correctness
oflatheir
work.deFor
También a los
estudiantes
les pide
verifiquen
exactitud
susexample,
trabajos.
Engineering
students
are de
alsoingeniería
asked to check
the correctness
of their
work. For
example,
occasionally
just
a littleuntime
remains
at
the end
of an exam.
It is useful
to be able
quickly
to identify
Por
ejemplo,
tomarse
breve
lapso
antes
de
terminar
un
examen
permitiría
dar
una
vista
occasionally just a little time remains at the end of an exam. It is useful to be able quickly to rápida
identifye
those
solutions
that
need more
work. requerir un poco más de trabajo.
identificar
esas
soluciones
que
podrían
those solutions that need more work.
The
examples
techniques
useful
checkinglas
the solutions aoflos
the sort of
Los following
siguientes
ejemplos illustrate
ilustran
útiles
para for
comprobar
The
following examples
illustratetécnicas
techniques
useful
for
checking thesoluciones
solutions of thediversos
sort of
problem
discussed
in
this
chapter.
problemas
analizados
capítulo.
problem
discussed
in en
thiseste
chapter.
E j e m p l o 4 . 11- 1 ¿Cómo podemos comprobar los voltajes de nodos?
E X A M P L E 4 . 1 1 - 1 How Can We Check Node Voltages?
E X A M P L E 4 . 1 1 - 1 How Can We Check Node Voltages?
El circuito que se muestra en la figura 4.11-1a se analizó utilizando PSpice. El archivo de salida de PSpice, figura
The circuit shown in Figure 4.11-1a was analyzed using PSpice. The PSpice output file, Figure 4.11-1b, includes
4.11-1b,
incluye
de nodoswas
delanalyzed
circuito. using
¿Cómo
podemos
comprobar
que
esos
voltajes
de nodos
son
The circuit
shownlosinvoltajes
Figure 4.11-1a
PSpice.
The PSpice
output
file,
Figure
4.11-1b,
includes
the node voltages of the circuit. How can we check that these node voltages are correct?
correctos?
the node voltages of the circuit. How can we check that these node voltages are correct?
Solution
Solución
Solution
The node equation corresponding to node 2 is
La
nodal que
correspondetoalnode
nodo 22 is
es
Theecuación
node equation
corresponding
V ð 2Þ � V ð 1Þ V ð 2Þ V ð 2Þ � V ð 3Þ
V ð 2Þ � V ð 1Þ þ V ð 2Þ þ V ð 2Þ � V ð 3Þ ¼ 0
100
100
þ 200 þ
¼0
100
200
100
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 140
Circuitos Eléctricos - Dorf
4/12/11 5:27 PM
How Can
Can We
Check .. ....... ???
Check
¿Cómo lo How
podemos
comprobar
How Can We
We
Check . . . ?
11
1
1
+
12 VV +–++
12
12
12 VV –––
22
2
2
100 Ω
Ω
100
100
Ω
100
200ΩΩ
Ω
200
200
200 Ω
Ω
33
3
3
100 Ω
Ω
100
100
100 Ω
Ω
200 Ω
Ω
200
200
200 Ω
Ω
(a)
(a)
(a)
141
141
141
141
44
4
4
200 Ω
Ω
200
200
200 Ω
Ω ++
8 VV
++ 8
––
8 VV
–– 8
00
0
0
(b)
(b)
(b)
FIGURE 4.11-1
4.11-1 (a) A
A circuit and
and (b) the
the node voltages
voltages calculated using
using PSpice. The
The bottom node
node has been
been chosen as
as the
FIGURE
FIGURE 4.11-1 (a)
(a) A circuit
circuit and (b)
(b) the node
node voltages calculated
calculated using PSpice.
PSpice. The bottom
bottom node has
has been chosen
chosen as the
the
reference
node,
which
is
indicated
by
the
ground
symbol
and
the
node
number
0.
The
voltages
and
resistors
have
units
of
voltages
FIGURA
4.11-1
(a)
Circuito
y
(b)
los
voltajes
de
nodos
calculados
utilizando
PSpice.
El
nodo
inferior
se
ha
elegido
como
el
nodo
reference
reference node,
node, which
which is
is indicated
indicated by
by the
the ground
ground symbol
symbol and
and the
the node
node number
number 0.
0. The
The voltages
voltages and
and resistors
resistors have
have units
units of
of voltages
voltages
and
ohms, respectively.
respectively.
de
referencia,
el cual está indicado por el símbolo de tierra y el número de nodo 0. Los voltajes y los resistores tienen unidades de
and
and ohms,
ohms, respectively.
voltajes y ohmios, respectivamente.
where, for example, V(2) is the node voltage at node 2. When the node voltages from Figure 4.11-1b are
where, por
for ejemplo,
example,V(2)
V(2)esiselthe
nodedevoltage
at el
node
2. 2.When
thelos
node
voltages
from de
Figure
4.11-1b
are
donde,
voltaje
nodos en
nodo
Cuando
voltajes
de nodos
la figura
4.11-1b
substituted into the left-hand side of this equation, the result is
substituted
into
the
left-hand
side
of
this
equation,
the
result
is
están sustituidos en el lado izquierdo de esta ecuación, el resultado es
7:2727
7:2727 �
� 12
12 þ 7:2727
7:2727 þ 7:2727
7:2727 �
� 5:0909
5:0909 ¼ 0:011
þ 200 þ
¼ 0:011
100
100
100
200
100
The
right-hand
of
equation
should
00 instead
of
ItIt looks
like
is
Is
of
El
derechoside
de esa
ecuación
podría
ser be
0been
vez
de 0.011.
Al parecer
anda mal. ¿Una
corriente
de 0.011
Thelado
right-hand
side
of this
this
equation
should
instead
of 0.011.
0.011.
looks algo
like something
something
is wrong.
wrong.
Is aa current
current
of
only
0.011
negligible?
Probably
not
in
this
case.
If
the
node
voltages
were
correct,
then
the
currents
of
the
100-V
no
es0.011
significativa?
EnProbably
este casonot
probablemente
los voltajes
nodos
fueran
correctos,
entonces
coonly
negligible?
in this case. Ifno.
theSinode
voltages de
were
correct,
then
the currents
of the las
100-V
resistors
would
be
A
A,
current
of
not
seem
rrientes
los resistores
100 0.022
V serían
de 0.047 A yThe
0.022
A, respectivamente.
corriente
de 0.011 when
A no
resistorsde
would
be 0.047
0.047 de
A and
and
0.022
A, respectively.
respectively.
The
current
of 0.011
0.011 A
A does
doesLa
not
seem negligible
negligible
when
compared
to
currents
of
0.047
A
and
0.022
A.
parece
insignificante
le compara
las A.
corrientes de 0.047 A y 0.022 A.
compared
to currentssiofse0.047
A andcon
0.022
Is
possible
that
PSpice
would
calculate
the
incorrectly?
Probably
not,
but
PSpice
¿Es
que
PSpice
haya
calculado
incorrectamente
los voltajes
de nodos?
Quizá no,
pero
archivo
de
Is ititposible
possible
that
PSpice
would
calculate
the node
node voltages
voltages
incorrectly?
Probably
not,
butelthe
the
PSpice
input
file
could
easily
contain
errors.
In
this
case,
the
value
of
the
resistance
connected
between
nodes
entrada
decould
PSpice
podría
contener
errores.
En case,
este caso,
el valor
resistenciaconnected
conectadabetween
entre losnodes
nodos222and
y3
input file
easily
contain
errors.
In this
the value
of de
thelaresistance
and
3se
been
specified
to
changing
this
to
100
PSpice
calculates
the
especificó
erróneamente
a 200 V.
Después
deAfter
cambiar
esta resistencia
a 100 V,
los voltajes
3 has
has
been mistakenly
mistakenly
specified
to be
be 200
200 V.
V.
After
changing
this resistance
resistance
toPSpice
100 V,
V,calcula
PSpiceque
calculates
the
node
voltages
to
be
de
nodos
deben to
serbe
node
voltages
V ð1Þ ¼ 12:0; V ð2Þ ¼ 7:0; V ð3Þ ¼ 5:5; V ð4Þ ¼ 8:0
V ð1Þ ¼ 12:0; V ð2Þ ¼ 7:0; V ð3Þ ¼ 5:5; V ð4Þ ¼ 8:0
Substituting
these
voltages
the
Al
sustituir estos
eninto
las ecuaciones
nodales gives
resulta
Substituting
thesevoltajes
voltages
into
the node
node equation
equation
gives
7:0
7:0 �
� 12:0
12:0 þ 7:0
7:0 7:0
7:0 �
� 5:5
5:5 0:0
þ 200 þ
þ 100 ¼
¼ 0:0
100
100
200
100
porthese
lo que estosvoltages
voltajes desatisfy
nodos satisfacen
la ecuación nodal que corresponde
al nodo 2.
so
so these node
node voltages do
do satisfy the
the node
node equation
equation corresponding
corresponding to
to node
node 2.
2.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 141
Alfaomega
4/12/11 5:27 PM
142
Métodos de análisis de circuitos resistivos
E j e m p l o 4 . 11- 2 ¿Cómo podemos comprobar las corrientes de enlaces?
El circuito que se muestra en la figura 4.11-2a se analizó utilizando PSpice. El archivo de salida de PSpice, figura 4.11-2b, incluye las corrientes de enlaces del circuito. ¿Cómo podemos comprobar que estas corrientes de
enlaces son correctas?
1
200 Ω
3
500 Ω
5
250 Ω
7
100 Ω
i1
2
+
–
4
200 Ω
i2
250 Ω
i3
250 Ω
8V
–
+
0V
6
–
+
0V
0
(a)
(b)
FIGURA 4.11-2 (a) Un circuito y (b) las corrientes de enlaces calculadas utilizando PSpice. Los voltajes y resistencias están dados
en voltios y ohmios respectivamente.
(El archivo de salida de PSpice incluirá las corrientes a través de las fuentes de voltaje. Recuerde que PSpice
utiliza la convención pasiva, de modo que la corriente en la fuente de 8 V será –i1 en vez de i1. Las dos fuentes de 0 V
se han agregado para incluir las corrientes de enlaces i2 e i3 en el archivo de salida de PSpice.)
Solución
La ecuación de enlace correspondiente al enlace 2 es
2001i2 2 i12 1 500i2 1 2501i2 2 i32 5 0
Cuando las corrientes de enlaces de la figura 4.11-2b se sustituyen en el lado izquierdo de esta ecuación el resultado es
200120.004068 2 0.017632 1 500120.0040682 1 250120.004068 2 120.00135622 5 1.629
El lado derecho de esta ecuación debe ser 0 en vez de 1.629. Parece que algo está mal. Lo más probable es que el
archivo de entrada de PSpice contenga un error. Así es en este caso. Los nodos de las dos fuentes de 0 V de voltaje
se han instalado en el orden equivocado. Recuerde que el primer nodo debe ser el nodo positivo de la fuente de
voltaje. Una vez corregido el error, PSpice da
i1 5 0.01763, i2 5 0.004068, i3 5 0.001356
Utilizando estos valores en la ecuación de enlaces, resulta
20010.004068 2 0.017632 1 50010.0040682 1 25010.004068 2 0.0013562 5 0.0
Estas corrientes de enlaces cumplen satisfactoriamente con la ecuación de enlaces que corresponde al enlace 2.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 142
Circuitos Eléctricos - Dorf
4/12/11 5:27 PM
E1C04_1
11/25/2009
143
4.12
E J E M P LO D E D I S E Ñ O
4.12
DESIGN EXAMPLE
Ejemplo de diseño
143
Design Example
143
DESPLIEGUE ANGULAR DEL POTENCIÓMETRO
POTENTIOMETER
ANGLE
DISPLAY
Se necesita que un
circuito mida y despliegue la posición
angular
del eje de un potenciómetro.
La posición angular, u, variará de 2180° a 180°.
A circuit
is needed
to measure
andcircuito
displayque
thepudiera
angular funcionar.
position ofLos
a potentiometer
shaft.
La figura
4.12-1
ilustra un
suministradores
deThe
po�
�
angular
position,
u,
will
vary
from
�180
to
180
.
tencia de 1 15 V y 2 15 V, el potenciómetro y los resistores R1 y R2 se emplean para obtener
Figurev ,4.12-1
illustrates a circuit that could do the job. The +15-V and –15-V power
un voltaje,
i que es proporcional a u. El amplificador sirve para modificar la constante de
are used
obtain va ,voltage,
vi, that is
supplies,
the
potentiometer,
1 and R2entre
proporcionalidad para obtenerand
unaresistors
relaciónRsencilla
u y eltovoltaje,
o desplegado por el
proportional
to
u.
The
amplifier
is
used
to
change
the
constant
of
proportionality
to
obtain a
voltímetro. En este ejemplo el amplificador se utilizará para obtener la relación
simple relationship between u and the voltage, vo, displayed by the voltmeter. In this example,
the amplifier will be used to obtain the relationship voltios
(4.12-1)
vo  k u donde k  0.1
volt
grados
ð4:12-1Þ
vo ¼ k � u where k ¼ 0:1
degree
de modo que u se puede determinar multiplicando la lectura del medidor por 10. Por ejemplo,
so that
u can
determined
by multiplying
theu 5
meter
reading by 10. For example, a meter
una
lectura
debe
medidor
de 27.32
V indica que
273.2°.
reading of �7.32 V indicates that u ¼ �73:2� .
Describa la situación y los supuestos
Describe
the 4.12-2
Assumptions
El
diagramathe
de Situation
circuito en and
la figura
se obtiene por el modelado de los suministradores
Thepotencia
circuit diagram
in Figure
4.12-2ideales,
is obtained
by modeling
power supplies
de
como fuentes
de voltaje
el voltímetro
como the
un circuito
abierto, as
y elideal
povoltage
sources,
the
voltmeter
as
an
open
circuit,
and
the
potentiometer
by
two
resistors.
tenciómetro por dos resistores. El parámetro, a, en el modelo del potenciómetro varía de 0 a 1
The parameter,
a, in the
modelEso
of quiere
the potentiometer
como
u varía de 2180°
a 180°.
decir que varies from 0 to 1 as u varies from
�180� to 180� . That means
u
1
(4.12-2)
ð4:12-2Þ
a¼
�þ
2
360
+15 V
+15 V
R1
Voltímetro
R1
Rp
+
Rp
+v
i
R2
vi–
R2
Voltmeter
+ vo –
Amplificador 100 Ω
Amplifier 100 Ω
+
+–
2 MΩ
2 MΩ
–
–
+
vo
–
bvi
bvi
–15 V
–15 V
FIGURA 4.12-1 Circuito propuesto para medición y despliegue de la posición angular del eje del potenciómetro.
FIGURE 4.12-1 Proposed circuit for measuring and displaying the angular position of the potentiometer shaft.
aRp
aRp
R1
(1 – a)Rp
(1 – a)Rp
R2
100 Ω
+
R
+1
– 15 V
+
–
15 V
R2
–15 V
–15 V
+
–
+
–
+
vi
–
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 143
vi
–
+
–
2 MΩ
2 MΩ
+
–
100 Ω
bvi
bvi
+
+
vo
–
vo
–
FIGURA 4.12-2 Diagrama de circuito
con modelos de los suministradores de
FIGURE
Circuit
diagram
potencia,4.12-2
voltímetro
y potenciómetro.
containing models of the power supplies,
voltmeter, and potentiometer.
Alfaomega
4/12/11 5:27 PM
E1C04_1
11/25/2009144 144
E1C04_1
11/25/2009
E1C04_1
11/25/2009
144
144
144
144
144
144
Methods of Analysis of Resistive Circuits
Métodos de análisis de circuitos resistivos
Methods
of Analysis
of Resistive
Circuits
Methods
of Analysis
Analysis
of Resistive
Resistive
Circuits
Methods of
of
Circuits
Despejando
Solving for uu resulta
gives
Solving
for
u gives
Solving for
for uu gives
gives
Solving
�
�
1
�
� �
u ¼ � a � � �1360
2
�
1
1
¼u ¼
� a ���2360
360� ��360
uu ¼
aa �
2
ð4:12-3Þ
(4.12-3)
ð4:12-3Þ
ð4:12-3Þ
ð4:12-3Þ
Establezca
el objetivo
State the Goal
Especifique
los
valores
de R
los
resistores R1 y R2, la resistencia
delRpotenciómetro
Rp, y gain
la gaState
the
Goal
Specify
values
of
resistors
resistance
b
1 and R2, the potentiometer
P, and the amplifier
State the
the Goal
Goal
State
nancia
b values
del
amplificador
que
del medidor,
vo, esté
enand
relación
el ángulo
values
of
resistors
R1 and
Rel
the
potentiometer
resistance
REq.
thecon
amplifier
b
to
be related
to the
angle
uRby
4.12-1.
that Specify
will
cause
theresistors
meter
voltage,
v,o,the
2,voltaje
P, and
Specify
of
R11 hará
and
Rque
potentiometer
resistance
the
amplifier
gaingain
Specify
values of
resistors R
and
R
resistance
RPP,, and
the
amplifier
gain
bb
2, the potentiometer
2
uthat
porthat
la
ecuación
4.12-1.
,
to
be
related
to
the
angle
u
by
Eq.
4.12-1.
will
cause
the
meter
voltage,
v
o
to be
be related
related to
to the
the angle
angle uu by
by Eq.
Eq. 4.12-1.
4.12-1.
will cause
cause the
the meter
meter voltage,
voltage, vvoo,, to
that will
Generate a Plan
Genere
un
Generate
a Plan
Analyze
theaplan
circuit
shown in Figure 4.12-2 to determine the relationship between vi and u.
Generate
Plan
Analice
elthe
circuito
muestra
enthese
la figura
4.12-2
parathe
determinar
la relación
entre
Analyze
the
shown
Figure
4.12-2
to
therelationship
relationship
between
R2, se
and
.in
Use
values
todetermine
simplify
the
between
vivvand
Select
values
of Rcircuit
i1and
1,que
Analyze
circuit
shown
inRpFigure
4.12-2
to determine
relationship
between
vii and
u.y u.
u.
Seleccione
valores
de
R
,
R
y
R
.
Utilice
estos
valores
para
simplificar
la
relación
entre
v
,
R
,
and
R
.
Use
these
values
to
simplify
the
relationship
between
v
Select
values
of
R
,
to
be
related
to
u.
If
possible,
calculate
the
value
of
b
that
will
cause
the
meter
voltage,
v
1
2
p
2 R p. Use
p these values to simplify the relationship
o between v i and1i and
Select values of R11, R221, and
p
i
yu.
u.
Si
es
posible,
calcule
el
valor
de
b
que
hará
que
el
voltaje
del
medidor,
vand
relacione
, se
toRrelated
related
u.
If
possible,
calculate
the
value
of
b
that
will
cause
the
meter
voltage,
voo,be
,
R
and
try
the
angle
u
by
Eq.
4.12-1.
If
this
isn’t
possible,
adjust
the
values
of
R
1 v o2,, to
pbe
to to
If possible, calculate the value of b that will cause the meter voltage,
o
con
el
ángulo
u
por
la
ecuación
4.12-1.
Si
esto
no
es
posible,
ajuste
los
valores
de
R
,
R
y
R
,
R
,
and
R
and
the
angle
u
by
Eq.
4.12-1.
If
this
isn’t
possible,
adjust
the
values
of
R
again.
1
2
2 R p and
p tryp try
the angle u by Eq. 4.12-1. If this isn’t possible, adjust the values of R11, R221, and
p
yagain.
pruebe
again.de nuevo.
again.
Act on the Plan
Actúe
sobre
el Plan
plan redrawn in Figure 4.12-3. A single node equation will provide the
on the
The Act
circuit
has
been
Act
on
the
Plan
En
laThe
figura
4.12-3
sebeen
ha
dibujado
de
el 4.12-3.
circuito.
ecuación
nodal
única
proporciocircuit
has
redrawn
innuevo
AUna
single
node
equation
will
provide
u:Figure
relationship
between
between
vin
i and
The
circuit
has
been
redrawn
Figure
4.12-3.
A single
node
equation
will
provide
the the
nará
la
relación
entre
v
y
u
:
and
u:
relationship
between
between
v
1
relationship between between
vii u:
vi � ð�15Þ
� 15
vi vii and
þ
þ
¼0
vaR
15
i�
2 MV
þ
� að�15
ÞR
R
vvii vi Rvv1þ
�
15
v
�ðv1ði�15
Þ p Þ ¼0
p
2
vþ
þ
ii � 15
ii � ð�15Þ
þ
þ p R2 þ ð1 � aÞR
¼0
2 MV
1 þ aRR
MV
þRaR
aR
þ ðð11 �
� aaÞR
ÞRpp p
R11 þ
22 MV
R
R22 þ
pp
Solving for vi gives
Solving
vi gives
�
�
Despejando
ii resulta
Solving
for vvfor
i gives
2 MV Rp ð�2a � 1Þ þ R1 � R2 15 �
�� �
�
��
�
vi ¼ �
ð4:12-4Þ
MV
Rpa�
ðÞR
2a1pÞ�þ
Þ21þ
Rþ
152 þ Rp
1 2�
2 R
R¼1 �þ aRp 2 MV
R2�2�þ
þ1R
MV
R115
�
�RR
Rðpp1ð2a
�
�
�
1
2
v
ð4:12-4Þ
(4.12-4)
i
�
�
¼ ��
ð4:12-4Þ
����
��
vvii ¼
ð4:12-4Þ
� a�ÞR
RR
Rð21þ�ð1aÞR
2 �MV
RR
1 þ aRRp þ
p2 þ
1 þþ
2 þ Rp
R11 þ
þ
MV
R
þRaR
aR
þ
R
R
þ
2
MV
R
þ
ð
1
�
a
ÞR
þ
R
þ
R
p
2
p
1
2
p
p
2 Let’s put some
p
1 on2R , Rp , and R that will make
This equation is quite complicated.
restrictions
1
2
p
R
and
make
This
equation
is
quite
complicated.
Let’s
put
some
restrictions
on
R21,a,and
=
R
=
R.
Second,
require
that
and
Rposiit
possible
to
simplify
this
equation.
First,
let
R
Esta
ecuación
es
bastante
complicada.
Pongamos
algunas
restricciones
R21,, R
Rboth
quewill
ppthat
1
2
p be
2 yRRR
,
R
This equation is quite complicated. Let’s put some
restrictions
on R
1
p
1
2
p that will make
=
R
=
R.
Second,
require
that
both
R
and
it
possible
to
simplify
this
equation.
First,
let
R
much
smaller
than
2
MV
(for
example,
R
<
20
kV).
Then,
bilitarán
la
simplificación
de
esta
ecuación.
Primero,
sean
R
5
R
5
R.
Segundo,
se
requiere
1
2
require
that both R and Rpp beRp be
it possible to simplify this equation. First, let R11= R122= R.2 Second,
�
�
�
�
�
�
much
smaller
than
2
MV
(for
example,
R
<
20
kV).
Then,
que
tanto
R
como
R
sean
mucho
menores
que
2
MV
(por
ejemplo,
R
,
20 kV). Entonces,
much smaller than p2 MV
20p kV).
R þ(for
aR example,
R þ ð1 R
�<
aÞR
� 2Then,
MV 2R �þ Rp
�� � p���� ��
�� �
��
�� �
� aÞR
2 MV
Rþ
þ RaR
aRþpp aRR
Rp þ
þ Rðð11þ�
�ð1aaÞR
ÞR
�p 22�
MV
2R þ
þ2RRppþ Rp
R
�
pp side
That is, the first term in the
denominator
of the
left
ofMV
Eq. 2R
4.12-4Ris
negligible compared to
That
is,
the
first
term
in
the
denominator
of
the
left
side
of
Eq.
4.12-4
negligible
compared
the
second
term.
Equation
4.12-4
can
be
simplified
to
Es
decir,
elfirst
primer
en el denominador
lado
izquierdo
denegligible
lais ecuación
4.12-4
es
That
is, the
the
first
termtérmino
in the
the denominator
denominator
of the
the left
leftdel
side
of Eq.
Eq.
4.12-4 is
is
negligible
compared
to to
That
is,
term
in
of
side
of
4.12-4
compared
to
the
second
term.
Equation
4.12-4
can
be
simplified
to
insignificante
comparado
el segundo
La to
ecuación 4.12-4 se puede simplificar a
the second
second term.
term.
Equationcon
4.12-4
can be
betérmino.
simplified
to
Rsimplified
the
Equation
4.12-4
can
p ð2a � 1Þ15
vi ¼
R
ð
2a
�
R pð2R
2apþ
� R1pÞ151Þ15
¼vi p¼ 2R þ R
vvii ¼
2R þ
þR
Rpp p
2R
(1 – a)Rp
aRp
(1 –p a)Rp
aRp (1
aR
(1 ––– a)R
a)R
aRpppp
p
(1
a)R
aR
p
p
R1
+
–
+
+
+
–
–
–
+
io = 0
R2
R
+
R111 R1vi+
R222 R2
+ 2+ MΩ R
R
R
1
2
2 MΩ+
vvi – v2
MΩ
i
2
MΩ
–15 V –
15 V viii
2 MΩ
+
+
+
V –
15–V ––15 V–15+
15
– V
–
–15 V
V +
15
V ––
–15
15
V
–
–
+
–
+
+
+
–
–
–
iio =
=0
0io = 0
ioo =
0
+
100 Ω o
+
+
100 vΩo =
bvi Ω
100
+ bvi
+
100
Ω
100
+ Ω
v = bvi
bvi v
bv
voooo =
=–bv
bvoiiii
– iii
bv
v
=
bv
bv
i
–
––
–
FIGURE 4.12-3 The redrawn circuit showing the mode vi.
FIGURE
4.12-3
The redrawn
circuit
showing
the mode
vi . v .
FIGURA
4.12-3
El circuito,
dibujado
de nuevo,
FIGURE
4.12-3
The
redrawn
circuit
showing
themuestra
mode
vel
. modo
i
iii
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 144
Circuitos Eléctricos - Dorf
4/12/11 5:27 PM
E1C04_1
11/25/2009
145
Design
DesignExample
Example
Design
Example
Design
Example
Ejemplo
de
diseño
Design
Design Example
Example
Next,
Next,using
usingEq.
Eq.4.12-3,
4.12-3,
Next,
using
Eq.
4.12-3,
Next,
using
Eq.
4.12-3,
��
�4.12-3, ��
���
� ��
�
A
continuación,
Next,
using
4.12-3,
Next,
using Eq.
Eq. utilizando
4.12-3, la ecuación
��
��15
RR
Rp pp ��
15VV
V��
15
�
�
�
�
R
15
V
p
uuu
vvvi ii¼¼
¼
Rpp
15
15��V
V
vvi i¼¼ 2R
�� uu
2RþR
þRR
Rp pp 180
180
2R
þ
180
vi ¼ 2R
�
þþRRp p 180
� u
2R
180
2RR
þ
R¼p55kV
180
ItIt
to
pick
values
for
R
and
R
.
Let
¼
and
RR
Itisis
istime
time
to
pick
values
for
R
and
R
.
Let
R
kV
and
Rp pp¼¼
¼10
10kV;
kV;then
then
timetotopick
pickvalues
valuesfor
forRRand
andRRpp.pp.Let
LetR�R�¼¼55kV
kV
andR
10
kV;
then
ItItisistime
and
kV;
then
p p¼¼10
�
�
.. pLet
¼
555�
kV
and
R
10
kV;
then
time
pick
values
for
�7:5
�
Es
hora
deto
captar
valores
para
yppR
. Sea
R
=
kV
y
R
=
10
kV;
entonces
It is
time
to
pick los
values
for R
R and
andR R
R
Let
R
¼
kV
and
R
¼
10
kV;
then
�R
p
p
7:5VV
V�
�7:5
7:5
�
V
vvvi ii¼¼
¼�
7:5
V
�
�
7:5
vvi i¼¼ 180
�
180�V
180
vi ¼ 180
��
180
180
Referring
Referringtoto
toFigure
Figure4.12-2,
4.12-2,the
theamplifier
amplifieroutput
outputisis
isgiven
givenby
by
Referring
Figure
4.12-2,
the
amplifier
output
given
by
Referring
totoFigure
4.12-2,
the
amplifier
output
isisgiven
by
Referring
Figure
4.12-2,
the
amplifier
output
given
by
Referring
to
Figure
4.12-2,
the
amplifier
output
is
given
by
Refiriéndonos a la figura 4.12-2, la salidavdel
amplificador
está dada por
bv
v ¼¼
¼
bvi ii
bv
vvvoo ooo¼¼
bv
ii
bv
vvoo 5
¼ bv
bvii
��
��
�
�
��
7:5
7:5VV
V��
7:5
�
�
7:5
V
soso
vv ¼¼
uuu
so
¼bbb 7:5
V
7:5
por
lo tanto,
so
vvvoo ooo¼¼
b 180
�� uu
180��V
so
180
so
vo ¼ bb 180
�� u
180
Comparing
Comparingthis
thisequation
equationtoto
toEq.
Eq.4.12-1
4.12-1gives
gives 180
Comparing
this
equation
Eq.
4.12-1
gives
Comparing
this
equation
totocon
Eq.
4.12-1
gives
Comparando
esta
ecuación
la
ecuación
4.12-1
resulta
Comparing
this
equation
Eq.
4.12-1
gives
Comparing this equation to Eq. 4.12-1
��
��
� gives
�
��
��
volt
7:5
V
volt
7:5
V
volt
7:5
V
7.5VV
V�¼¼0:1
voltio
volt
7:5
bb �
0:1
volt
7:5
�V
� ¼¼
0:1

0.1
volt
7:5
0:1
bbbb 180
�
degree
degree
180
��
¼
degree
180
180
grado
¼ 0:1
0:1degree
b 180
degree
180
degree
180�
145
145
145
145
145
145
145
ð4:12-5Þ
ð4:12-5Þ
ð4:12-5Þ
ð4:12-5Þ
ð4:12-5Þ
ð4:12-5Þ
(4.12-5)
180
180
180
oroor
bb¼¼
ð0:1
orbien
¼180
0:1ÞÞÞ¼¼
¼2:4
2:4
180
ðð0:1
2:4
180
oror
bbb¼¼
Þ Þ¼¼2:4
7:5
7:5ð0:1
ðð0:1
2:4
7:5
or
b ¼7:5
0:1
Þ
¼
2:4
7:5
7:5
The
final
circuit
isis
inin
4.12-4.
Thela
final
circuit
isshown
shown
inFigure
Figure
4.12-4.
En
figura
4.12-4
se
muestra
el circuito
final.
The
final
circuit
shown
Figure
4.12-4.
The
final
circuit
is
shown
in
Figure
4.12-4.
The
The final
final circuit
circuit is
is shown
shown in
in Figure
Figure 4.12-4.
4.12-4.
Verifique
soluciónSolution
propuesta
Verify
the
Proposed
Verifythe
thela
Proposed
Solution
Verify
the
Proposed
Solution
Verify
Proposed
Solution
Verify
the
Proposed
Solution
Verify
the
Proposed
Solution
A
modo
de
verificación,
que Eq.
uEq.
= 150°.
Dewe
la see
ecuación
As
a
check,
suppose
u
¼
150
4.12-2,
As
a
check,
suppose
u
¼
150�� .��� ..From
From
Eq.
4.12-2,
we
seethat
that 4.12-2 vemos que
As a check, suppose u ¼suponga
150
From
4.12-2,
we
see
that
As
a check, suppose
u ¼ 150
. From Eq.
4.12-2,
As
Eq.
4.12-2,we
wesee
seethat
that
As aa check,
check, suppose
suppose uu ¼
¼ 150
150�.. From
From150
Eq.
111 we see that
150�� ���4.12-2,
150
1
þ
¼
0:9167
aaa¼¼
0:9167
¼150
þ
¼
150
� �� þ 1
¼0:9167
0:9167
150
aa¼¼360
2221¼¼
360
� ��þþ
0:9167
360
a ¼360
� þ22 ¼ 0:9167
360
360
2
Using
Eq.
4.12-4,
we
Using
Eq.
4.12-4,
wecalculate
calculate
Con
laEq.
ecuación
4.12-4
calculamos
Using
Eq.
4.12-4,
we
calculate
Using
4.12-4,
we
calculate
Using
Using Eq.
Eq. 4.12-4,
4.12-4, we
we calculate
calculate
22MV
ð10
ð2ð2��0:9167
MV
ð10kV
kV
0:9167��1Þ1Þ15
ÞÞ15
MV
ð10kV
kV
�0:9167
0:9167��1Þ1Þ15
ÞÞ15
222MV
ð10
ð2ðð22��
¼¼
v v¼
¼6:24
6:24
i ¼
ðþ
�
6:24
i ¼ð5
6:24
viivv¼
MV
ð10
10
kV
ð0:9167
20:9167
� 0:9167
0:9167
�
ÞÞ15
Þ15
kV
þþ
0:9167
��
10
kV
ÞðÞÞ25ððMV
kV
ð1kV
��
Þ10
kV
Þ11ÞÞÞþ
ð2ðð22��
Þ ÞÞ¼¼
ð
5
kV
þ
0:9167
�
10
kV
5
kV
þ
ð
1
�
0:9167
Þ10
kV
þ222MV
MV
�555kV
kVþþ
þ10
10kV
kV
6:24
¼
5
kV
ð
0:9167
10
kV
5
kV
þ
ð
1
Þ10
kV
þ
MV
kV
10
kV
þ
0:9167
�
10
kV
Þ
ð
5
kV
þ
ð
1
�
0:9167
Þ10
kV
Þ
þ
2
MV
ð
2
�
5
kV
þ
10
kV
Þ
¼
6:24
vii ¼ð5ð5kV
kV
þ
�
ÞÞðð55 kV
þ
ðð11 �
Þ10
kV
ÞÞ þ
22 MV
ð54.12-5
kV
þ 0:9167
0:9167
� 10
10 kV
kV
kV
þmeter
� 0:9167
0:9167
Þ10will
kV
þbe
MVðð22 �
� 55 kV
kV þ
þ 10
10 kV
kVÞÞ
Finally,
indicates
that
the
voltage
be
Finally,Eq.
Eq.
4.12-5
indicates
that
themeter
voltage
will
Finally,Eq.
Eq.4.12-5
4.12-5
indicates
that
themeter
meter
voltagewill
willmedidor
be
Finally,
indicates
that
the
be
Finalmente,
la
ecuación
4.12-5
indica
que
elvoltage
voltaje
del
será de
Finally,
meter
voltage
will
Finally, Eq.
Eq. 4.12-5
4.12-5 indicates
indicates that
thatvthe
meter
voltage
will be
be
2:4
�
6:24
¼
14:98
vvthe
�
2:4
�
6:24
¼
14:98
o oo�
�2:4
2:4
�
6:24
¼
14:98
vvvooo�3
�
6:24
¼
14:98
2.4 �� 6:24
6.24 ¼
5
14.98
� 2:4
14:98
¼
14:98
o � 2:4
This
that
the
was
Thisvoltage
voltagewill
willbe
beinterpreted
interpretedtoto
tovmean
mean
that6:24
theangle
angle
was
This
voltage
will
be
interpreted
mean
that
the
angle
was
This
voltage
will
be
interpreted
to
mean
that
the
angle
was
La
interpretación
de
este
voltaje
significa
que
el
ángulo
This
voltage
will
be
interpreted
to
mean
that
the
angle
� era
� was
This voltage will be interpreted touumean
that
the
angle
was
�
¼
149:8
¼
10
�
v
¼
149:8
¼
10
�
v
149:8� ��
¼10
10� �vvoo oo¼¼149:8
uuu¼5
5
149.8°
149:8
¼
10
�� vvvooo ¼
¼
149:8
u
¼
10
which
whichisis
iscorrect
correcttoto
tothree
threesignificant
significantdigits.
digits.
which
correct
three
significant
digits.
which
totopara
three
significant
digits.
lo
cualisis
escorrect
correcto
tres
dígitos importantes.
which
correct
significant
digits.
which
is
correct
to three
three
significant
digits.
+15
+15VV
V
+15
+15
VV
+15
+15
+15 VV
10
10kΩ
kΩ
10
kΩ
10
kΩ
10
kΩ
10
10 kΩ
kΩ
20
20kΩ
kΩ
20
kΩ
20
kΩ
20
kΩ
20
20 kΩ
kΩ
10
10kΩ
kΩ
10
kΩ
10
kΩ
10
kΩ
10
10 kΩ
kΩ
++
+
++
++
vvvi
viiv
vii
– –vii
– –––
–
Amplifier
Amplifier 100
100ΩΩ
Ω
Amplifier
100
Amplifier
100
ΩΩ
Amplificador
Amplifier
100
Amplifier 100
100 Ω
Ω
22
2MΩ
MΩ
MΩ
22
MΩ
22MΩ
MΩ
MΩ
Voltmeter
Voltmeter
Voltmeter
Voltmeter
Voltímetro
Voltmeter
Voltmeter
++
+ vovvo – ––
++
vvooo – –––
++ vov
o
++
+
++
2.4v
2.4v
i i
2.4v
–+
–+ 2.4v
2.4v
–
i i
2.4v
––
2.4viii
––
–15
–15VV
V
–15
–15
VV
–15
–15
–15 VV
FIGURE
FIGURE4.12-4
4.12-4The
Thefinal
finaldesigned
designedcircuit.
circuit.
FIGURE
4.12-4
The
final
designed
circuit.
FIGURE
4.12-4
The
designed
circuit.
FIGURA
4.12-4
Elfinal
circuito
final diseñado.
FIGURE
FIGURE 4.12-4
4.12-4 The
The final
final designed
designed circuit.
circuit.
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 145
Alfaomega
4/12/11 5:27 PM
Métodos de análisis de circuitos resistivos
146
4.13
RESUMEN
1. Etiquetar las corrientes de enlaces.
2.Expresar los voltajes de elementos como funciones de
corrientes de enlaces. La figura 4.13-1b ilustra la relación entre el voltaje a través de un resistor y las corrientes de los enlaces que incluyen al resistor.
3. Aplicar la KVL a todos los enlaces.
La solución de las ecuaciones simultáneas da por resultado el conocimiento de las corrientes de enlaces. Todos
los voltajes y corrientes en el circuito se pueden determinar cuando las corrientes de enlaces son conocidas.
Si una fuente de corriente es común a dos enlaces que se
juntan, definimos el interior de los dos enlaces como un superenlace. Entonces escribimos la ecuación de la corriente
de enlace en torno a la periferia del superenlace. Si en la
periferia de sólo un enlace aparece una fuente de corriente, podemos definir esa corriente de enlace como igual a la
corriente de la fuente, teniendo en cuenta la dirección de la
fuente de corriente.
Si el circuito contiene una fuente dependiente, primero expresamos el voltaje o corriente controladores de la fuente
dependiente como una función de las corrientes de enlaces.
A continuación expresamos el voltaje o corriente controlados como una función de las corrientes de enlaces. Finalmente, aplicamos la KVL a enlaces y superenlaces.
En general, se pueden utilizar los análisis tanto de voltajes
de nodos como de corrientes de enlaces para obtener las corrientes o los voltajes en un circuito. Sin embargo, un circuito con menos ecuaciones nodales que ecuaciones de corrientes de nodos puede requerir que seleccionemos el método de
los voltajes de nodos. Por el contrario, el análisis por medio
de corrientes de enlaces es fácilmente aplicable para un circuito con menos ecuaciones de corrientes de enlaces que las
ecuaciones de voltaje de nodos.
MATLAB reduce en gran manera la monotonía de despejar
ecuaciones de nodos o de enlaces.
El método del voltaje de nodos del análisis de circuitos identifica los nodos de un circuito donde están conectados dos
o más elementos. Cuando el circuito consta solamente de
resistores y fuentes de corriente, se toma el procedimiento
siguiente para obtener las ecuaciones nodales.
1.Elegimos un nodo para que funcione como nodo de referencia. Etiquetamos los voltajes de nodos en los demás
nodos.
2.Expresamos corrientes de elementos como funciones
de voltajes de nodos. La figura 4.13-1a ilustra la relación entre la corriente en un resistor y los voltajes en los
nodos del resistor.
3.Aplicamos la KCL en todos los nodos, excepto en el
nodo de referencia. La solución de las ecuaciones simultáneas da como resultado el conocimiento de los voltajes
de nodos. Todos los voltajes y corrientes en el circuito se
pueden determinar cuando los voltajes son conocidos.
Cuando un circuito tiene fuentes de voltaje y a la vez fuentes de corriente aún podemos aplicar el método de voltajes
de nodos utilizando el concepto de un supernodo. Un supernodo es un nodo grande que incluye dos nodos conectados
mediante una fuente de voltaje conocida. Si la fuente de voltaje está conectada directamente entre un nodo q y el nodo
de referencia, podemos establecer que vq = vs y escribir las
ecuaciones KCL en los nodos restantes.
Si el circuito contiene una fuente dependiente, primero
expresamos el voltaje controlador o corriente de la fuente
dependiente como una función de los voltajes de nodos. A
continuación expresamos el voltaje o corriente controladores como una función de los voltajes de nodos. Finalmente,
aplicamos la KCL a nodos y supernodos.
El análisis de corrientes de enlaces se complementa con la
aplicación de la KVL a los enlaces de un circuito planar.
Cuando el circuito consta solamente de resistores y fuentes de
voltaje, se aplica el procedimiento siguiente para obtener las
ecuaciones de enlaces.
va – vb
a
va
is
R2
R2
R1
R1
+ (va – vb)
+
va
–
(a)
b
–
+
vb
–
i1
vb
+ R1i1 –
R1
R3
R3
va
+
–
i1
i2
+ R2i2 –
(i1 – i2) R2
+
R3(i1 – i2)
–
R3
i2
vb +
–
(b)
FIGURA 4.13-1 Expresión de corrientes y voltajes de resistor en términos de (a) voltajes de nodos o (b) corrientes de enlaces.
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 146
Circuitos Eléctricos - Dorf
4/12/11 5:27 PM
Problemas
147
PROBLEMAS
Sección 4.2 Análisis de voltajes de nodos de
circuitos con fuentes de corriente
P 4.2-1 Los voltajes de nodos en el circuito de la figura
P 4.2-1 son v1 5 2 4 V y v2 5 2 V. Determine i, la corriente
de la fuente de corriente.
Respuesta: i 5 1.5 A
P 4.2-4 Considere el circuito que se muestra en la figura
P 4.2-4. Encuentre los valores de las resistencias R1 y R2 que
ocasionan que los voltajes v1 y v2 sean v1 5 1 V y v2 5 2 V.
500 Ω
+
v1
–
3 mA
6Ω
+
v2
–
R1
R2
5 mA
Figura P 4.2-4
i
P 4.2-5 Encuentre el voltaje v para el circuito que se muestra
en la figura P 4.2-5.
v2
v1
Respuesta: v 5 21.7 mV
4Ω
8Ω
+
v –
250 Ω
125 Ω
250 Ω
500 Ω
1 mA
500 Ω
Figura P 4.2-1
P 4.2-2 Determine los voltajes de nodos para el circuito de
la figura 4.2-2.
Figura P 4.2-5
Respuesta: v1 5 2 V, v2 5 30 V, y v3 5 24 V
1A
v1
20 Ω
10 Ω
v2
v3
P 4.2-6 Simplifique el circuito que se muestra en la figura
P 4.2-6 reemplazando resistores en serie y en paralelo con
resistores equivalentes; luego analice el circuito simplificado
escribiendo y despejando ecuaciones nodales. (a) Determine
la potencia suministrada por cada fuente de corriente. (b) Determine la potencia recibida por el resistor de 12 V.
2A
20 Ω
15 Ω
5Ω
Figura P 4.2-2
P 4.2-3 Los voltajes de nodos en el circuito de la figura
P 4.2-3 son v1 5 4 V, v2 5 15 V y v3 5 18 V. Determine i1 e i2,
las corrientes de las fuentes de corriente.
15 Ω
v2
Figura P 4.2-3
Circuitos Eléctricos - Dorf
M04_DORF_1571_8ED_SE_108-161.indd 147
120 Ω
b
a
v3
2Ω
i2
20 Ω
2 mA 60 Ω
P 4.2-7 Los voltajes de nodos en el circuito que se muestran en
la figura P 4.2-7 son va 5 7 V y vb 5 10 V. Determine los valores
de la corriente de la fuente de corriente, is, y la resistencia, R.
i1
5Ω
10 Ω
3 mA
10 Ω
Figura P 4.2-6
Respuesta: i1 5 22 A e i2 5 2 A
v1
40 Ω
12 Ω
2A
10 Ω
R
4Ω
8Ω 8Ω
is
Figura P 4.2-7
Alfaomega
4/12/11 5:27 PM
Métodos de análisis de circuitos resistivos
148
Sección 4.3 Análisis de voltajes de nodos
de circuitos con fuentes de corriente y voltaje
P 4.3-1 El voltímetro en la figura P 4.3-2 mide vc, el voltaje
de nodos en el nodo c. Determine el valor de vc.
Respuesta: vc 5 2 V
6Ω
a
6V
–
+
10 Ω
b
va 5 12 V, vb 5 9.882 V, y vc 5 5.294 V.
Determine la potencia suministrada por la fuente de voltaje.
c
+
vc
8Ω
2A
P 4.3-5 Los voltajes va, vb y vc en la figura P 4.3-5 son los
voltajes de nodos correspondientes a los nodos a, b y c.
Los valores de estos voltajes son
Voltímetro
6Ω
–
Figura P 4.3-1
Respuesta: va 5 212 V, vb 5 vc 5 4 V, vd 5 24 V, i 5 2 mA
4 kΩ
i
b
+
va
8V
c
+
–
+ 12 V
+
vb
2 mA
–
+–
vc
1 mA
–
4 kΩ
–
+
vd
–
P 4.3-3 Determine el voltaje de nodos va para el circuito de
la figura P 4.3-3.
Respuesta: va 5 7 V
+
–
va
12 V
1A
–
+
vb
2Ω
–
vc
–
Figura P 4.3-5
P 4.3-6 El voltímetro en el circuito de la figura P 4.3-6 mide
un voltaje de nodos. El valor de ese voltaje de nodos depende
del valor de la resistencia R.
(a)Determine el valor de la resistencia R que ocasionará que
el voltaje medido por el voltímetro sea 4 V.
(b)Determine el voltaje medido por el voltímetro cuando
R 5 1.2 kV 5 1200 V.
Respuesta: (a) 6 kV (b) 2V
8V
+
va
–
+
100 Ω
10 V
c
+
d
Figura P 4.3-2
+
–
3Ω
b
+
P 4.3-2 Los voltajes va, vb, vc y vd en la figura P 4.3-2 son los
voltajes de nodos correspondientes a los nodos a, b, c y d. La
corriente i es la corriente en un cortocircuito conectado entre
los nodos b y c. Determine los valores de va, vb, vc y vd y de i.
a
4Ω
a
–
100 Ω
100 Ω
Voltímetro
30 mA
Figura P 4.3-3
P 4.3-4 Determine el voltaje de nodos va para el circuito de
la figura P 4.3-4.
6 kΩ
3 kΩ
Respuesta: va 5 4 V
+
–
+
8V
Figura P 4.3-4
Alfaomega
M04_DORF_1571_8ED_SE_108-161.indd 148
R
2 mA
+
–
8V
–
250 Ω
125 Ω
500 Ω
12 V
+
–
12 V
+
va
–
500 Ω
Figura P 4.3-6
P 4.3-7 Determine los valores de los voltajes de nodos v1 y
v2, en la figura P 4.3-7. Determine los valores de las corrientes
ia e ib.
Circuitos Eléctricos - Dorf
4/12/11 5:27 PM
E1C04_1
11/25/2009
149
Problemas
Problems
1 kΩ
1 kΩ
10 V +
10 V + –
v1
–
ia
ia
5 kΩ
5 kΩ
v1
ib
3 kΩ
3 kΩ
ib
4 . 5 0
4 . 5 0
Voltímetro
Voltmeter
4 kΩ
4 kΩ
v2
149
149
v2
2 kΩ
2 kΩ
R1
R1
+
V
+ –1212
V
–
R3
R3 R2
R2
+
+ –6 V6 V
–
Figura P 4.3-7
Figure P 4.3-7
P 4.3-8 El circuito que se muestra en la figura P 4.3-8 tiene Figura P 4.3-10
P 4.3-8
The circuit
in Figure
has two inputs, v1 Figure P 4.3-10
dos entradas,
v1 yshown
v2, y una
salida,P v4.3-8
o. La salida se relaciona
and
v
,
and
one
output,
v
.
The
output
is
related
to the input by
2
o
con la entrada por la ecuación
*P 4.3-11 Determine los valores de los voltajes de nodos del
the equation
*Pcircuito
4.3-11que
Determine
theenvalues
of the
node voltages of the
se muestra
la figura
P 4.3-11.
circuit shown in Figure P 4.3-11.
o5
1 1 bv2
vo v¼
avav
1 þ bv2
3A
3A
donde a y b son constantes que dependen de R , R y R .
where a and b are constants that depend on R1, 1R2,2and 3R3.
(a)(a)