Bài giải phần giải mạch P11

Chapter 11, Solution 1.v( t ) = 160 cos(50t ) i( t ) = -20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°) i( t ) = 20 cos(50t + 60°) p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°) p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W p( t ) = 800 + 1600 cos(100t + 60°) WP= 1 1 Vm I m cos(θ v − θi ) = (160)(20) cos(60°) 2 2P = 800 WChapter 11, Solution 2.First, transform the...
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Bài giải phần giải mạch P11Chapter 11, Solution 1. v( t ) = 160 cos(50t ) i( t ) = -20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°) i( t ) = 20 cos(50t + 60°) p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°) p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W p( t ) = 800 + 1600 cos(100t + 60°) W 1 1 P= Vm I m cos(θ v − θi ) = (160)(20) cos(60°) 2 2 P = 800 WChapter 11, Solution 2. First, transform the circuit to the frequency domain. 30 cos(500t )  → 30 ∠0° , ω = 500 0.3 H  → jωL = j150 1 -j 20µF  → = = - j100 jωC (500)(20)(10 -6 ) I I2 -j100 Ω I1 + 30∠0° V j150 Ω 200 Ω − 30∠0° I1 = = 0.2∠ − 90° = - j0.2 j150 i1 ( t ) = 0.2 cos(500 t − 90°) = 0.2 sin(500 t ) 30∠0° 0.3 I2 = = = 0.1342∠26.56° = 0.12 + j0.06 200 − j100 2 − j i 2 ( t ) = 0.1342 cos(500 t + 25.56°) I = I 1 + I 2 = 0.12 − j0.14 = 0.1844 ∠ - 49.4° i( t ) = 0.1844 cos(500t − 35°)For the voltage source, p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.1844 cos(500t − 35°) ]At t = 2 s , p = 5.532 cos(1000) cos(1000 − 35°) p = (5.532)(0.5624)(0.935) p = 2.91 WFor the inductor, p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.2 sin(500t ) ]At t = 2 s , p = 6 cos(1000) sin(1000) p = (6)(0.5624)(0.8269) p = 2.79 WFor the capacitor, Vc = I 2 (- j100) = 13.42∠ - 63.44° p( t ) = v( t ) i( t ) = [13.42 cos(500 − 63.44°) ] × [ 0.1342 cos(500t + 25.56°)At t = 2 s , p = 18 cos(1000 − 63.44°) cos(1000 + 26.56°) p = (18)(0.991)(0.1329) p = 2.37 WFor the resistor, VR = 200 I 2 = 26.84 ∠25.56° p( t ) = v( t ) i( t ) = [ 26.84 cos(500t + 26.56°) ] × [ 0.1342 cos(500t + 26.56°) ]At t = 2 s , p = 3.602 cos 2 (1000 + 25.56°) p = (3.602)(0.1329 2 p = 0.0636 WChapter 11, Solution 3. 10 cos(2t + 30°)  → 10∠30° , ω= 2 1H  → jωL = j2 1 0.25 F  → = -j2 jωC I 4Ω I1 2Ω I2 + 10∠30° V j2 Ω -j2 Ω − ( j2)(2 − j2) j2 || (2 − j2) = = 2 + j2 2 10 ∠30° I= = 1.581∠11.565° 4 + 2 + j2 j2 I1 = I = j I = 1.581∠101.565° 2 2 − j2 I2 = I = 2.236 ∠56.565° 2 For the source, 1 S = V I* = (10∠30°)(1.581∠ - 11.565°) 2 S = 7.905∠18.43° = 7.5 + j2.5 The average power supplied by the source = 7.5 W For the 4-Ω resistor, the average power absorbed is 1 2 1 P = I R = (1.581) 2 (4) = 5 W 2 2 For the inductor, 1 2 1 S = I 2 Z L = (2.236) 2 ( j2) = j5 2 2 The average power absorbed by the inductor = 0 W For the 2-Ω resistor, the average power absorbed is 1 2 1 P = I 1 R = (1.581) 2 (2) = 2.5 W 2 2 For the capacitor, 1 2 1 S= I 1 Z c = (1.581) 2 (- j2) = - j2.5 2 2 The average power absorbed by the ...