tailieunhanh - Bài giải phần giải mạch P1

Chapter 1, Solution 1 (a) q = x [ C] = C (b) q = 1. 24x1018 x [ C] = C (c) q = x [ C] = C (d) q = x [ C] = C Chapter 1, Solution 2 (a) (b) (c) (d) (e) i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ. | Chapter 1 Solution 1 a q x C C b q 1. 24x1018 x C C c q x C C d q x C C Chapter 1 Solution 2 a i dq dt 3 mA b i dq dt 16t 4 A c i dq dt -3e-t 10e-2t nA d i dq dt 1200n cos 120nt pA e i dq dt - e 4 80 cos50t 1000 sin50t fA Chapter 1 Solution 3 a q t J i t dt q 0 3t 1 C b q t J 2t s dt q v t2 5t mC c q t J 20 cos 10t n 6 q 0 2sin 10t n 6 1 fC _ q t J 10e-30t sin40t q 0 l0e 3 -30sin40t-40 cos t d J 900 1600 - e-30t t sin 40t C Chapter 1 Solution 4 10 0 q J idt J . . .- 5 5sin 6 n t dt cos6n t 6n 1 - mC 6n Chapter 1 Solution 5 q e-2tdt mC -1e-2t 2 2 0 2 1 - e4 mC 490 pC Chapter 1 Solution 6 dq 80 a At t 1ms i 40 mA dt 2 ------- dq b At t 6ms i 0 mA dt c At t 10ms i d - 20 mA Chapter 1 Solution 7 i dq dt 25A -25A 25A 0 t 2 2 t 6 6 t 8 which is sketched below Chapter 1 Solution 8 q J idt 10 x 1 15 liC Chapter 1 Solution 9 a q J idt J10 dt 10 C q idt 10 x 1 10 - x 5 x 1 b 105 10 - 25 2 J c q J idt 10 10 10 30 C Chapter 1 Solution 10 q ixt 8x103x15x10 6 120 fiC Chapter 1 Solution 11 q it 85 x10-3 x 12 x 60 x 60 3 672 C E pt ivt qv 3672 J Chapter 1 Solution 12 For 0 t 6s assuming q 0 0 t t q t J idt q 0 J 3tdt 0 At t 6 q 6 6 2 54 For 6 t 10s