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. 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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 16 17 RESPUESTA DE FRECUENCIA TRANSFORMADA DE LAPLACE SERIE Y TRANSFORMADA DE FOURIER CIRCUITOS DE FILTRO REDES DE DOS Y TRES PUERTOS 6 EL AMPLIFICADOR OPERACIONAL 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 q¼ Z t 1 i dt ¼ Z 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 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 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 K Cantidad de sustancia mol mol Intensidad luminosa candela 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 kg m/s N 2 Pa Hz 2 Presión o tensión pascal N/m Densidad kilogramo por metro cúbico kg/m3 Energía de trabajo joule (julio) Nm J Potencia watt (vatio) J/s W Carga eléctrica culombio As 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 Flujo magnético weber Vs 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 PREFIJO SÍMBOLO tera T MULTIPLE 109 PREFIX giga GSYMBOL 6 12 1010 103 109 1022 106 10323 10 mega tera kilo giga centi mega mili kilo micro centi M 26 �2 1010 29 �3 1010 k c m m T G M k c n 10�6212 10 10�9215 10 nano milli pico micro femto nano f 10�12 pico p 10�15 femto f p m m n 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–– b + 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 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 �ivbavab 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 60 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 p1 ðtÞ dt 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 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 Genera un plan para obtener una solución del problema. Acción Actúa sobre el plan. Verificación Incorrecta Correcta Solución Circuitos Eléctricos - Dorf M01_DORF_1571_8ED_SE_001-019.indd 11 Verifica que la solución propuesta sea la correcta. 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 �t/60 Alfaomega M01_DORF_1571_8ED_SE_001-019.indd 14 �� � p�1 ð�t�Þ� ¼ iv1 �¼ � De � 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 0 Z 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 3V – 2A + 3V – 2A + 3A 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) 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 pdt 0p ¼ v � i 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 and w ¼ 19 Z 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. �1 �1 25 �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 + 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 the voltage is a constant 6 V. constante Find the energy supplied over the time interval 0 to T. to T. voltaje es de 6the V.supplied Obtenga la energía alimentada de 0 a T. constant 6 V. energy supplied the time durante interval constant 6una V. Find theFind energy over theover time interval 0 to T. el0 intervalo Uncircuit circuito tiene un representado por laFigure figura 2.3-1b donde la is corriente de 5the A yis el Avhas circuit haselemento an element represented by2.3-1b 2.3-1b where the current isesa una constant A voltage and voltage is a v A an element represented by Figure where the current a constant 5 Aconstante and 5the a i – – Solution 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 the negative terminal, thelaenergy supplied by the element is given byde Dado quethe la corriente entra por lacurrent terminal negativa, energía alimentada por elbyelemento resulta node the current the negative terminal, thethe energy the element entry nodethe ofenters the i terminal, is the negative of voltage supplied v.by The current flows the is BecauseBecause thenode current enters negative the node energy supplied the element isfrom given bygiven by Solución Z Z T Z T w ¼ Twð6¼Þð5Þdtð6¼Þð30T 5ÞdtJJ¼ 30T J w ¼ 0 ð6Þð5Þdt ¼ 30T 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 (a) (b) entry node to the exit node. 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 (OHMCM) 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 Circuit Elements 26 Circuit de circuitos R E S I S T26O 26 R S Elementos Circuit Elements Elements 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 + v Voltmeter – Ammeter i im = 0 + FIGURA 2.6-2 (a) Amperímetro ideal. (b) Voltímetro+ideal. vm = 0 vm = 0 Ammeter Element + Element (a) i + Circuitos Eléctricos - Dorf (a) vm = 0 i – (a) i – i Elemento v – (b) – 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 + =0 vc + =0 vc –= 0 vc –= 0 – – ic = 0 ic = 0 + ic = 0 + ic = 0 v+c vc – v+ –c v–c – Fuente de voltaje controlada por voltaje (VCVS) b es la ganancia de la VCVS. b tiene unidades de voltios/voltios ic = 0 ic = 0 + ic = 0 + ic = 0 v+c vc – v+ –c v–c – 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 ic ic ic ic + + vc + =0 vc + =0 vc –= 0 vc –= 0 – – ic ic ic ic + + – + – + – – id id id id v = ri d c vd = ric vd = ric vd = ric + + – + – + – – id id id id v = bv d c vd = bvc vd = bvc vd = bvc + + v+d v+d v–d v–d – – id = gvc id = gvc id = gvc id = gvc + + v+d v+d v–d v–d – – id = dic id = dic id = dic id = dic 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 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 is 40 is mJ. 40 mJ. Element Element 1 is a v1m is a aR I p Rp – – – – 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) i(t) = kT 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 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 + 12VV ++– 6 6V V +– – 12 – 88 kΩkΩ (b) (b) kΩ 44 kΩ 12kΩ kΩ 12 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 p5vi 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 –3.6 2.4 – 6.0 v, V (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 P2.4-10 2.4-10The Thevoltage voltagesource sourceshown shownininFigure FigureP P2.4-10 2.4-10is isanan P 2.4-6 2.4-6 A current source and two resistors are connected in P 2.4-10 voltagedesource shown in Figure P en 2.4-10 is an P fuente deshown corriente yFigure dos resistores están conecP 2.4-10 The La fuente voltaje que se muestra la figura aa series circuit P P2.4-6. Elements adjustable dcdc voltage source. InIn other words, thethe voltage vs viss is seriesinUna inthethe circuit showninin Figure 2.4-6. Elements adjustable voltage source. other words, voltage series eninserie the encircuit shown in Figure Pen 2.4-6. Elements adjustable dcuna voltage source. In other words,dethe voltage vs is tados el circuito que se muestra la figura P 2.4-6. P 2.4-10 es fuente de voltaje ajustable cd. En otras pa-a voltage, butbut thethe value ofof that constant can bebe adjusted. connected constant voltage, value that constant can adjusted. connectedininseries serieshave havethethesame samecurrent, current,sosoi1 i1¼¼is iand s and constant the ofconstante, that constant be adjusted. ¼ is and constant connected in series have the sametienen current, so i1 corriente, Los elementos conectados en serie la misma elvoltage, voltaje vbut un value voltaje perocan el valor de esa s es The tabulated data were collected asas follows. The voltage, vs,v , i2 i¼ is iin this circuit. Suppose that is i¼ 2525 mA, R1R¼ 4 V, and The tabulated data were collected follows. The voltage, in this circuit. Suppose that mA, 4 V, and labras, 2¼ s¼ 1¼ The tabulated data were collected as follows. The voltage, vs, s ipor is tanto, ins thisen circuit. Suppose that i ¼ 25 mA, R ¼ 4 V, and 2 ¼lo s 1 este circuito i1 5 isacross eacross i2 5each ifeach . Suponga que ifthe 5the constante se puede ajustar. La tabulación deacross los the datos capturaR2R¼ was set to some value, and the voltages across resistor, vava 8 V. Calculate the voltage resistor and was set to some value, and the voltages the resistor, ¼ 8 V. Calculate the voltage resistor and 2 R2 mA, set to some value, andalgún the voltages acrossalthe resistor, ¼ 8 V. Calculate voltage across each resistor and the 25 Rabsorbed yeach Rthe 8resistor. V. Calcule el voltaje a través de was dos es sigue. Se dio tipo de valor voltaje, yvaa 1 5 4 V, 2 5resistor. and vbv,como measured and recorded. Next, thethe value ofof vvssv,was power byby and , were measured and recorded. Next, value powerabsorbed each bwere s was and vb, wereameasured recorded. themidió valueyof vs was powerresistor absorbed each resistor. cada y laby potencia absorbida por cada resistor. 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 Hint: Hint:Notice Noticethethereference referencedirections directionsofofthetheresistor resistorvoltages. voltages. changed, and theelvoltages across the resistors were measured continuación, valor de v se cambió, y los voltajes a s Hint: Notice Observe the reference directions de of referencia the resistordevoltages. Sugerencia: las direcciones los vol- A 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 – – v1v1+ + tajes de los resistores. times. (The ofofvs vwere Determine thethe times. (Thevalues values notrecorded.) recorded.) Determine s were v1 + – v times. (The values of v were not recorded.) Determine the procedimiento se repitió varias veces. (No se registraron los s – 1 + value of the resistance, R. value of the resistance, R. i1 i1 R1R1 i2 i2 + + value of the resistance, R. valores de v .) Determine el valor de la resistencia, R. s i is is is if i i11 R R11 i22 R2R R 2 R22 + v+2v v2 2 v–2 – – – + + 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 4040 ΩΩ 40 Ω RR R ++ + vbvb vb – – – 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 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 isconconFigureP P2.4-10 2.4-10 source and absorbs W.VSubsequently, thispotencia heater isabsorcon- Figure conecta 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 Figure P 2.4-10 2.4-10 Figura P nected to a constant 220-V source. What power does it absorb be de la fuente desource? 220 V?What ¿Cuál lathe resistencia del calentador? from thethe 220-V resistance from 220-V source? Whatisesisthe 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. Section 2.5 Sources Section 2.5Independent Independent Sources Sección 2.5 Fuentes independientes Hint: Model the electric heater as a resistor. Section 2.5 Independent Sources P 2.4-8 El equipo portátil de alumbrado para una mina se loPP 2.4-8 The portable lighting equipment for a mine is located 2.4-8 The portable lighting equipment for a mine is located P P 2.5-1 AA current source and a voltage source areare connected 2.5-1Una current source and ay voltage source connected P 2.4-8 Themetros portable lighting equipment fordeacd. mine isluces located P 2.5-1 2.5-1 fuentesource de corriente una fuente voltaje están caliza ameters 100 fuente de alimento Las dea a P A current and a voltage sourcedeare connected 100 from source. The lights use 100meters fromitsde itsdcladcsupply supply source. Themine mine lights use in parallel with a resistor as shown in Figure P 2.5-1. All ofof thethe in parallel with a resistor as shown in Figure P 2.5-1. All 100 meters from its de dc 5supply source. The mine lights use a in conectados en paralelo con un resistor, como se muestra enthe la la mina usan un total kW y funcionan a 120 V cd. Determiparallel with a resistor as shown in Figure P 2.5-1. All of total ofof 5 kW and operate at at 120 VV dc.dc. Determine thethe required total 5 kW and operate 120 Determine required elements connected inin parallel have thethe same voltage, vs vins in this elements connected parallel have same voltage, this total of 5 kW and operate at 120 V dc. Determine the required figura P 2.5-1. En este circuito todos los elementos conectados ne el área seccional requerida de cables de cobre que se usaron elements connected in parallel have the same voltage, v in this s cross-sectional cross-sectionalarea areaofofthethecopper copperwires wiresused usedtotoconnect connectthethe circuit. 1515V,V,is i¼ 5 5V.V. Suppose that vs v¼ ¼3 3A,A,and andRvR¼ ¼ circuit. Suppose that s ¼ sSuponga cross-sectional area ofdethe copper wires used to requiere connect que the circuit. en paralelo tienen el mismo voltaje, v . que = 15 V, para conectar la fuente las luces de la mina si se Suppose that vs ¼ 15 V, is s ¼ 3 A, and R s¼ 5 V. source toto thethe mine lights if if wewe require that thethe power lost inin thethe (a)(a)Calculate source mine lights require that power lost the current i iininthe thethepower Calculate current theresistor resistorand and power source to the mine lights if we require that the 3 A, y R 5the 5 the V.current (a) Calcule corriente el resistor y la la potencia perdida en el cableado sea menor a opower igual alost 5%indethe la if 5 Calculate i in lathe resistori en and the power copper copperwires wiresbebeless lessthan thanororequal equaltoto5 5percent percentofofthethepower power (a) absorbed by the resistor. (b) Change the current source current absorbed by the resistor. (b) Change the current source current copper wires be less than or equal tomina. 5 percent of the power absorbed potencia absorbida por el resistor. (b) Cambie la corriente de la potencia requerida por las luces de la by the resistor. (b) Change the current source current required requiredbybythethemine minelights. lights. 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, required by the mine el lights. is ¼s de 5 Acorriente and recalculate i, innuevo the resistor and thei, Sugerencia: Modele equipo de iluminación y el cableado to power absorbed by power absorbed bythetheresistor. resistor. Hint: Model both thethe lighting equipment and thethe wire asas resistors. Hint: Model both lighting equipment and wire resistors. en el resistor y laby potencia absorbida por el resistor. absorbed the resistor. como resistencias. Hint: Model both the lighting equipment and the wire as resistors. power Answer: Answer:i ¼ i ¼3 3A Aand andthetheresistor resistorabsorbs absorbs4545WWboth bothwhen when P 2.4-9 The resistance of a practical resistor depends on P 2.4-9 The resistance of a practical resistor depends on 3A Respuesta: y el the resistor absorbe 45 W is 5 P unaresistor útil resistor dependedepends de la resisi ¼i 5 3 3AAand resistor absorbs 45cuando W both when P 2.4-9 2.4-9 La Theresistencia resistancedeof practical on Answer: is i¼ 3 3A Aand 5 5A.A. andwhen whenis i¼ the nominal resistance and the resistance tolerance as follows: s ¼ s ¼ the nominal resistance and the resistance tolerance as follows: 5 5 when A. is ¼ 5 A. is and tencia nominal y de la tolerancia de la resistencia como sigue: icuando s ¼ 3 A the nominal resistance the resistance as follows: and tolerance i i t t t t i �� Rnom RR �� Rnom Rnom1 � Rnom1 þ 1 �t 1 þt Rnom 1 � 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 vs v + + is i RR is the nominal resistance andPor t isejemplo, the resistance where Rnomexpresada resistencia porcentaje. retolerance For vs s+– – toleranceexpressed expressedasen asa percentage. aunpercentage. Forexample, example,a 100-V, a un 100-V, is s R – tolerance expressed as a percentage. For example, a 100-V, sistor de 2% de resistencia de 100 V, tendrá una resistencia 2 2percent percentresistor resistorwill willhave havea aresistance resistancegiven givenbyby 2 percent dada por resistor will have a resistance given by 9898 VV �� RR �� 102 VV 102 Figure 98 V � R � 102 V FigureP P2.5-1 2.5-1 Figura P Figure P 2.5-1 2.5-1 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, , and one El se muestra figura 2.4-9input, tienevuna entraThecircuito circuit que shown in FigureenPla2.4-9 hasP one , and one output, vov. oThe circuit is given byby s PP 2.5-2 AA current and a voltage source connected output, .salida, Thegain gainof ofthis this circuit given 2.5-2Una current source and ay voltage source are connected fuentesource de corriente una fuente deare voltaje están da, vs, yvuna . La ganancia deis este circuito la da output, gainvoof this circuit is given by P 2.5-2 A current source and a voltage source are connected o. The in series with a resistor as shown in Figure P 2.5-2. All oflaof the in series with a resistor as shown in Figure P 2.5-2. All conectados en serie con un resistor, como se muestra en fi-the vovo R2R2 in series with a resistor as shown in Figure P 2.5-2. All of the gain ¼¼ vo ¼¼ R2 ganancia gain elements connected inin series have thethe same current, itienen this elements connected series have same current, , in this gura P 2.5-2. Todos los elementos conectados en serie la s,isin R1Rþ R2R gain ¼ vs v¼ elements connected in series have the same current, is, in this 1þ is i¼ 3 3A, circuit. Suppose vs v¼ vs s R1 þ R2 2 misma corriente, isthat , en este Suponga que vand 5V. ¼1010V,V, A,and R¼V, ¼5is5V. circuit. Suppose that scircuito. s ¼ s 5R10 Suppose that vs ¼ 10 V, is ¼ 3 A, and R ¼ 5 V. Determine the range of possible values ofof thethe gain R1Ris is circuit. Determine elthe rango deof los valores posibles degain lawhen ganancia Determine range possible values when 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 Determine the range of possible values of the gain when R1 is1 (a) Calculate the voltage v across the resistor and the power thetheresistance 2 2percent resistor isV cuando R1 es laofresistencia de 2% de un resistorand de R100 ythe la resistance ofa a100-V, 100-V, percent resistor and isthe 2R2 absorbed byby thethe resistor. (b)(b) Change the voltage source voltage potencia absorbida por el resistor. (b) Cambie la corriente absorbed resistor. Change the voltage source voltage the resistance of a 100-V, 2 percent resistor and R2 is the absorbed by the resistor. (b) Change the voltage source voltage 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 de 2resistance to vlas vfuente ¼¼5 5V de recalculate resistor voltaje a vs 5 the 5 the Vvoltage, y voltage, calculev,de nuevo elthe voltaje, Vand and recalculate v,across acrossthe resistor resistance ofen a 400-V, 5 percent resistor. Express in to to v ¼s 5 V and recalculate the voltage, v, across the resistor la ganancia términos deand una nominalthe y gain de una terms ofofa anominal gain a gain terms nominal gainand a ganancia gaintolerance. tolerance. v, a stravés del resistor y labypotencia absorbida por el resistor. and thethepower absorbed thetheresistor. and power absorbed by resistor. terms of ade nominal gain and a gain tolerance. and the power absorbed by the resistor. tolerancia 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 –– – – + ++ + i i ii ++ + + R2R2 vovo R2 vs R v 2 –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 m im im 1 k 1 kΩ 1 kΩ R RmRm 1 kΩ Rm m (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 thedata 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) El interruptor está cerrado y Rs ⫽ 0 (un cortocircuito). (b) El interruptor está cerrado y Rs ⫽ 5 ⍀. (c) El interruptor está cerrado y Rs ⫽ 1 (un cortocircuito) (d) 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. 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 FIGURE 3.2-3 Gustav 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 FIGURE 3.2-3 Gustav Robert Kirchhoff (1824any es instant is todo zero.instante. nodo cero en FIGURA 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. 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 Determine the the value value of of the the voltage, voltage, in in volts, volts, measured 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 vvm¼¼ ðð�12 �36 ¼3i3i 3ia a¼¼ ¼333ð�12 �12Þ ÞÞ¼¼ ¼�36 �36VV V m a 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 s 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 s todos los elementos transportan la node mismathat corriente. Para identificar un par de to elementos en two elements connected toto aaasingle has no other elements connected it.it. Notice, two elements connected single node that has no other elements connected to Notice, two elements connected to single node that has no other elements connected to it. Notice, serie, buscamos dos elementos conectados a un nodo único to quenode no tenga otros elementos and connected bbband other for example, that resistors RR andRR2 2are are both connected to and that no other for example, that resistors 1 areboth both connected toR node node andthat thatno no other FIGURE for example, that resistors R1ejemplo, 1 and R2 3.3-1 conectados a sí. Observe, por que los resistores R y están conectados al nodo FIGURE 3.3-1 1 ¼i2i, ,2so ¼ both resistors have the circuit elements are connected to node b. Consequently, i FIGURE3.3-1 3.3-1 1 so both resistors have the circuit elements are connected to node b. Consequently, i 1 2 ¼ i , so both resistors have the FIGURA circuit elements are connected to node b. Consequently, i 1 2 Single-loop circuit with a Single-loop circuit with b y que ningún otros elemento del circuito lo está. En consecuencia, i 5 i , de modo que 1 also 2 connected Single-loop circuitcerrado with aa and R are in same current. A similar argument shows that resistors R Circuito de circuito 2 3 and R are also connected in same current. A similar argument shows that resistors R R33 are also connected in voltage same current. A tiene similar argument showsUn that resistorsparecido R22 and muestra source vsv.s. voltage source ambos resistores la misma corriente. argumento que los resistores voltage vs. de único consource una fuente isisconnected with RR all series. Noticing connectedinin series with both and ,,we we say that all three series. Noticingthat that 2 inseries series withboth both R1 11Rand andRR R3,33conectado wesay saythat that allthree three series. thatRR R2conectados 2 is connected R2 y R3Noticing también están en serie. Al observar que en serie con voltaje vs. 2 está is resistors are connected in series. The order of series resistors not important. For resistors are connected in series. The order of series resistors is not important. For resistors are connected in series. Theestán order of seriesenresistors is not de important. For R1 y R3, decimos que and los tres resistores conectados serie. El orden los resistores example, the voltages currents ofof the three resistors inin Figure 3.3-1 will not change ifif example, the voltages and currents the three resistors 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 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 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 FIGURE 3.4-4 such that Recall that the equivalent circuit, Figure 3.4-12, has an equivalent conductance G FIGURE 3.4-4 that the equivalent circuit, 3.4-12, has an equivalent conductance G FIGURE such thatcomo Recallthat that the equivalent circuit,Figure Figure 3.4-12, has anuna equivalent conductance Gpppppppsuch FIGURE3.4-4 3.4-4 Recall 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 FIGURE 3.4-4 FIGURE 3.4-4 Recuerde que elequivalent circuito equivalente, figura3.4-12, 3.4-12,has tiene conductancia equivalente Gp, casi p such Set of NNparallel parallel N Set of parallel Set of N N such that Recall that the equivalent circuit, Figure 3.4-12, has an equivalent conductance G N FIGURE 3.4-4 X p X Setof ofN Nparallel parallel NN Set ofof NN parallel X Set parallel FIGURA 3.4-4 X N Set NN X X conductances conductances conductances G ¼ G (3.4-12) Set of N parallel G (3.4-12) N G conductances conductances Gpppppppp¼ ¼X Gnnnnnnnn (3.4-12) (3.4-12) Conjunto de N conductances conductances GG ¼¼ GG (3.4-12) (3.4-12) with a current with a current n¼1 with a current n¼1 conductances n¼1 with a current G ¼ G (3.4-12) n¼1 with a current n¼1 p n conductancias n¼1 with a current with a current n¼1 n¼1 source source source iisssii..ss.con with a current n¼1 source Therefore, paralelas Therefore, source issi.s.. source Por consiguiente, Therefore, Therefore, Therefore, source is. de una fuente G Therefore, Gnnnnnnnniiissssiisisss GG iinn ¼ ¼ (3.4-13) G ¼ (3.4-13) (3.4-13) ¼G (3.4-13) corriente is. (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 El voltaje vm y la corriente de la fuente de corriente se apegan a la convención pasiva, por lo que pR ¼ 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 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. 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. 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. in Figure 3.6-7. 2. 2. Determine 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. 2. Determine los valores de i eniiiiiila 3.6-7. 2. Determine the values of and in Figure 3.6-7. 2. Determine the values of in Figure 3.6-7. 2. Determine the values of and in Figure 3.6-7. 11 22 and iiiiin 3. circuits are equivalent, vvv11vv,,1,1,vv,v2v2v,,2,2,and 3. 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 3. 3. Because 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 3. Because the circuits are equivalent, the values of ii in Figure 3.6-6 are equal to the values of 3. Because the circuits are equivalent, the values of vv111,, vv222,, and and in Figure 3.6-6 are equal to the values of 3. 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 ¼vv22þþvv44 12 (3.7-3) 2 4 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) vv444þþvv555þþvv666 ¼¼00 (3.7-4) 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 40iiii2222;;;; vvvv4444¼ ¼80 80iiii4444;;;; vvvv5555¼ ¼48 48iiii5555;;;; vvvv6666¼ ¼32 32iiii6666 (3.7-5) (3.7-5) vv22¼¼40 2 2 4 4 5 5 6 6 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, notice that when a ¼ 1, Resistive Circuits � 14,000 þ 20,000 Resistive Circuits Circuits Resistive When a a¼5 0, v must 5 V, Resistive Circuits ¼ 12 Cuando debebe V,so por�lo tanto, Resistive Circuits When a ¼ 0,0,v vmust beser5v5V, so � 24 ¼ �5 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 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 v + FigureFigure P 2.5-5 P 2.5-5 + – 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 – vo 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. 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. donde R ¼R þR in (b) Design a voltage to2 1have a2 gain, g ¼ 0.65, and an Rin ¼divider R1 þ R Figure DP 3-7 is called the input resistance of divider. 2500 V. voltage Rent 5Rof R¼ 1 R2 the in 1the is called theinput inputresistance, resistance 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. 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 – ++ ++ ++ + + +++ + + v R R3R333 v vbvbbb RRR a R2R222 vv R vavaa RRR v vavv v b v–bb 3R 33 –––aa 2R22 – – – – – –– – –– a a aaaaa i isisiiisssis s (a) (a) (a) (a) (a) (a) a a aaaaa i isisiiisssis s 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)+–++ + ++ +++ + v R R3R333 vbvbbb RRR v a R2R222 vv R v vavaa RRR v vavv b v–bb 3R 33 –––aa 2R22 – – – – – –– – –– (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 v denote the node voltage at node a and v denote the node voltage at node b. The voltmeter in Figure 4.2-5 b Let denote the node voltage at node and the node voltage at node b. The voltmeter in Figure 4.2-5 b denote 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 Let Let vavvvdenote vaaaaaael 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 denote node voltage aaaand vvvdenote node voltage 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 at node a to obtain de los voltajes de nodos. Aplique la KCL enatelatat 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 to obtain v vaa � v � vavvvaaaaa þ vavv� vaaa� v�bvvvbbbbb ¼ 0 44 þ þ ¼ þ þ þ ¼ 0 000 4þ 4 þ þ 5 ¼¼ 4 þ 10 10 1010 10 5 555 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 Node EEjEeXXm pMlPP oLLE4 .42..2 -22--22 Ecuaciones nodales E4 Node Equations Equations E P L E 4 . 2 - 2 Node Equations E XX AA M M P L E 4 . 2 - 2 Node 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 Node Equations EEj eXmA pMlPoL E4 .42. -23- 3 Ecuaciones nodales E X A M P L E 4 . 2 - 3 Node Equations E X A M P L E 4 . 2 - 3 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 222 333 333 222 333 0:9 �0:2 �0:7 0:9 333 0:9 �0:2 �0:2 �0:7 �0:7 vvavaa 4 4 5 4 5 4 4 4 5 4 5 4 5 4 5 A ¼ �0:2 0:55 �0:1 ; b ¼ 1 and; v ¼ and; AA¼¼ �0:2 �0:2 0:55 0:55 �0:1 �0:1 ; ;bb¼¼ 11 and; y, v v¼¼ vvbvbb555 �0:7 0:1 1:3 �1 vvcvcc �0:7 1:3 �1 �0:7 0:1 0:1 1:3 �1 222 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 +–+ vs + v vsss +–– 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 8V a 6 Ω6 Ω b 8V c b +8 8– V Vc 6 Ω aa b cc 6Ω b+ – + – – Solution Solución – + 4V Solution Solution + TheEn methods summarized in Table 4.3-1 are exemplified esta solución se ejemplifican los métodos resumidosinenthis la tabla 4––V +– 12 Ω 12 Ω 12 12 Ω Ω 12 Ω 12 Ω 2A 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 – –– +++ 3.5 AA 66ΩΩ 3.5 6ΩΩ 3.5 AA 6 3.5 6Ω 3.5 A ++ +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 – – – + a vv –v–aaa – –– bb bb b + 1.5 AA 1.5 1.5 AA 1.5 1.5 A +a+ vv+ 12 V i i 12 V 12 VV 12 i ii 12 V ++ +++ aa aa ++ a – 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 10 Ω a 10Ω Ω 10 10 10 Ω Ω aa aa +– 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 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 6Ω b b bb b 3Ω 3Ω Ω 3bΩ 3 3Ω c3 Ω c cc c c + + Solution Solution + Solution + 8V 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 V8 V– 8 V 2A Alfaomega Circuitos Eléctricos - Dorf 3ix2 A + +3i 3ix– + x 2A A 3ix3i – 2A 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 M04_DORF_1571_8ED_SE_108-161.indd 120 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 N Á L I S I S D E C O R R I E N T E S D E E N L A C E S C O N 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 vs + vs –+ vs vs+ –+ – – R1 R1 R1 R1 R2 R2 R2 R2 R3 R3 R3 R3 i1 i1 i1 i1 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 + –+ + + –– – 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 V 12 V V 12 12 V + –+ + + –– – i1 i1 ii1 1 1.5 A 1.5 A A 1.5 1.5 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. FIGURE 4.6-5 The circuit for Example 4.6-2. FIGURA El circuito el ejemplo 4.6-2. FIGURE 4.6-5 4.6-5 The The circuit para for Example Example 4.6-2. FIGURE 4.6-5 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 Method 2 of Example 4.6-2. FIGURA Método 22 of delExample ejemplo 4.6-2. 4.6-2. FIGURE 4.6-7 4.6-7 Method Method FIGURE 4.6-7 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 ++ Ω 32 Ω Ω 32 32ΩΩ 32 + 24 V V +– 24 + 24VV +–– 24 – 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 135 135 The Node Voltage Method and Mesh Current Method Compared 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+ – 2Ω 16 V 16 V 2Ω 2Ω + – + – 2A 2A 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 R11000; % ohms R21000; % ohms R35000; % ohms V1 15; % volts V215; % volts Rp20e3; % potentiometer parameters % ohms %––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– % the parameter a varies from 0 to 1 in 0.05 increments. %––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– a00.051; % dimensionless for k1length(a) %––––––––––––––––––––––––––––––––––––––––––––––––––––––––– % Here is the mesh equation, RIV: %––––––––––––––––––––––––––––––––––––––––––––––––––––––––– R [R1a(k)*RpR3 R3; R3 (1a(k))*RpR2R3]; 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 Voltmeter + vo – Amplificador 100 Ω Rp Amplifier + Rp R2 R2 +v i 2 MΩ vi– 2 MΩ 100 Ω + +– – – + 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 aRppp p (1 a)R aR p 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 vb – – (a) b 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 v v¼ avavþ1 1 bvbv2 o 1 2 3A 3A donde a y b son constantes que dependen de R , R y R . where a and b are constants
