Ejemplo cálculo de Impedancia equivalente Problem 9.51 (a) 20 mH → 12.5 µF → jωL = j (103 )( 20 × 10 -3 ) = j20 1 1 = = - j80 3 j ωC j (10 )(12.5 × 10 - 6 ) Z in = 60 + j20 || ( 60 − j80) ( j20)( 60 − j80) Z in = 60 + 60 − j60 Zin = 63.33 + j23.33 = 67.494∠ 20.22° Yin = (b) 1 = ∠ ° Zin 0.0148 -20.22 S 10 mH → 20 µF → 30 || 60 = 20 j ωL = j (10 3 )(10 × 10 -3 ) = j10 1 1 = = - j50 3 j ωC j (10 )( 20 × 10 - 6 ) Z in = - j50 + 20 || ( 40 + j10) (20)( 40 + j10) Z in = - j50 + 60 + j10 Z in = 13.5 − j48.92 = 50.75∠ - 74.56° Yin = 1 = Zin 0.0197∠ 74.56° S = 5.24 + j18.99 mS Ejemplo cálculo de Impedancia equivalente Problem 9.48 → j ωL = j (10 × 10 3 )( 2 × 10 -3 ) = j20 1 1 1 µF → = = - j100 j ωC j (10 ×10 3 )(1× 10-6 ) 2 mH 50 Ω + 1∠ 0° A + V j20 Ω − + − Vin − -j100 Ω V = (1∠0°)( 50) = 50 Vin = (1∠0°)(50 + j20 − j100 ) + ( 2)(50) Vin = 50 − j80 + 100 = 150 − j80 Z in = Vin = 150 – j80 Ω 1∠0° 2V Ejemplo división de voltajes ω = 200 1 1 50 µF → = = - j100 jωC j (200)(50 × 10 -6 ) Problem 9.35 → jωL = j (200 )( 0.1) = j20 (50)(-j100 ) - j100 50 || -j100 = = = 40 − j20 50 − j100 1 - j2 0.1 H j20 j20 Vo = ( 60 ∠0°) = ( 60 ∠0°) = 17 .14 ∠90° j20 + 30 + 40 − j20 70 Thus, v o ( t ) = 17.14 sin(200t + 90° ) V v o ( t ) = 17.14 cos(200t) V or Ejemplo división de corrientes Problem 9.41 0.1 F → 1 1 = = - j5 jωC j ( 2)( 0. 1) 0.5 H → jωL = j ( 2)(0.5) = j The current I through the 2-Ω Ω resistor is I= I 1 Is = s , 1 − j5 + j + 2 3 − j4 I s = (10)( 3 − j4) = 50∠ - 53.13° Therefore, i s (t ) = 50 cos(2t – 53.13°) A where I = 10∠0°