tailieunhanh - Bài giải phần giải mạch P6

Chapter 6, Solution 1. i=C dv = 5 2e −3t − 6 + e −3 t = 10(1 - 3t)e-3t A dt ( ) p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W Chapter 6, Solution 2. 1 2 1 Cv1 = (40)(120) 2 2 2 1 2 1 w2 = Cv1 = (40)(80) 2 2 2 w1 = ∆w = w 1 − w 2 = 20(120 2 − 80 2 ) = 160 kW Chapter 6, Solution 3. i=C 280 − 160 dv = 40x10 −3 = 480 mA dt 5 Chapter 6, Solution 4. v= = 1 t idt + v(0) C ∫o 1 6 sin. | Chapter 6 Solution 1. i C 5 2e-3t - 6 e-3t 10 1 - 3t e-3t A dt V s------------ p vi 10 1-3t e-3t 2t e-3t 20t 1 - 3t e-6t W Chapter 6 Solution 2. 1 9 1 z _ W1 -Cv2 40 120 2 11 W2 Cv2 40 80 2 Aw w1 - w2 20 1202 - 802 160 kW Chapter 6 Solution 3. i C-d 40x10-3 280-160 480 mA Chapter 6 Solution 4. v 1 ftidt v 0 CJo 1 I 6sin 4tdt 1 2 1 - cos 4t Chapter 6 Solution 5. 1t v 77 f idt v 0 Co For 0 t 1 i 4t v ----1 - ft 4t dt 0 100t2 kV 20x10-6 Jo v 1 100 kV For 1 t 2 i 8 - 4t v 1 20x10-6 J1 8 - 4t dt v 1 100 4t - t2 - 3 100 kV Thus v t Chapter 6 Solution 6. 100t2kV 0 t 1 100 4t -12 - 2 kV 1 t 2 i C 30x10 6 x slope of the waveform. dt For example for 0 t 2 dv 10 dt 2x10-3 i C 30x10-6x 10 150mA dt 2x10-3 Thus the current i is sketched below. Chapter 6 Solution 7. v iidt v t ----- - i 4tx10-3 dt 10 C 50x10-3 o 2t2 ----- 10 10 V 50 --------------- Chapter 6 Solution 8. a i C -100ACe 100t - 600BCe-600t 1 dt i 0 2 -100AC - 600BC ------- 5 -A - 6B 2 v 0 v 0- --- 50 A B 3 Solving 2 and 3 leads to A 61 B -11 1 2 1 3 b Energy - Cv2 0 - x4x10 3x2500 5 J c From 1 i -100x61x4x10-3 e 100t - 600x11x4x10-3 e 600t - 100t - A Chapter 6 Solution 9. v t 12 W - e-t dt 0 12 t e-t v v 2 12 2 e-2 V p iv 12 t e-t 6 1-e-t 72 t-e-2t p 2 72 2-e-4 W Chapter 6 Solution 10 dv dv i C 2 x10 3 dt dt v 5 16 0 t 1 s 1 t 3 is 16t 64-16t 3 t 4