tailieunhanh - 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°) W P= 1 1 Vm I m cos(θ v − θi ) = (160)(20) cos(60°) 2 2 P = 800 W Chapter 11, Solution 2. First, transform the. | Chapter 11 Solution 1. v t 160cos 50t i t -20 sin 50t - 30 2 cos 50t - 30 180 - 90 i t 20cos 50t 60 p t v t i t 160 20 cos 50t cos 50t 60 p t 1600 cos 100t 60 cos 60 W p t 800 1600cos 100t 60 W 1 1 P 2VmIm cos 6v -6i 160 20 cos 60 P 800 W Chapter 11 Solution 2. First transform the circuit to the frequency domain. 30cos 500t ------- 30Z0 o 500 H ----- joL j150 1 - j 20pF ------ ----- ----- -j100 H joC 500 20 10-6 J I i2 -j100 Q v I1 30Z0 V 200 Q 30Z0 I1 ------- - 90 1 j150 i1 t 500t - 90 500t 30Z0 I -------- ------ 2 200 - j100 2 - j J i2 t cos 500t I I1 12 - i t 500t - 35 For the voltage source p t v t i t 30cos 500t x cos 500t - 35 At t 2 s p 1000 cos 1000 - 35 p p W For the inductor p t v t i t 30cos 500t x 500t At t 2s p 6 cos 1000 sin 1000 p 6 p W For the capacitor Vc 12 -j100 p t v t i t cos 500 - x cos 500t At t 2 s p 18cos 1000- cos 1000 p 18 p W For the resistor VR 20012 p t v t i t 500t x 500t At t 2 s p 1000 p p W Chapter 11 Solution 3. 10cos 2t 30 ---- 10Z30 o 2 1 H ------ joL j2 F 1 ----- -j2 joC j I 4 Q i1 2 Q ----r- WV w I2 10Z30 V g j2 Q -j2 Q j2 2 - j2 j2 2- j2 2 j2 10Z30 I ---------- 4 2 j2 j2 T 2-1 j i 1 2 2 - j2 12 --- 1 2 2 For the source S VI 2 10Z30 S The average power supplied by the source W For the 4-Q resistor the average power absorbed is 12 1 P 21 R 2 2 4 5W For the inductor 12 1 S - IJ Z L 2 2 j2 j5 The average power absorbed by the inductor 0