Chapter 1, Solution 1 (a) q = 6.482x1017x [-1.602x10-19C] = -0.10384 C (b) q = 1. 24x1018x [-1.602x10-19C] = -0.19865 C (c) q = 2.46x1019x [-1.602x10-19C] = -3.941 C (d) q = 1.628x1020x [-1.602x10-19C] = -26.08 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 = 1200πcos120πt pA (e) i =dq/dt = −e−4t(80cos50t+1000sin50t) µA Chapter 1, Solution 3 (a) q(t) = ∫i(t)dt+q(0) =(3t +1) C (b) q(t) = ∫(2t+s) dt+q(v) = (t2 +5t) mC (c) q(t)= ∫20 cos (10t+π/6)+q(0)=(2sin(10t+π/6)+1)µC 10e-30t q(t) = ∫10e-30t sin40t+q(0) = (−30sin40t-40 cos t) (d) 900+1600 = −e-30t(0.16cos40t +0.12 sin 40t) C Chapter 1, Solution 4 10 −5 q = ∫idt = ∫ 5sin 6 π t dt = cos6π t 6π 0 5 = (1−cos0.06π)= 4.698 mC 6π Chapter 1, Solution 5 2 1 q = ∫idt = ∫ e-2tdt mC= - e-2t 2 0 1 = (1−e4) mC= 490 µC 2 Chapter 1, Solution 6 dq 80 (a) At t = 1ms, i = = = 40 mA dt 2 dq (b) At t = 6ms, i = = 0 mA dt dq 80 (c) At t = 10ms, i = = =-20 mA dt 4 Chapter 1, Solution 7 25A, 0< t < 2 dq  i = = -25A, 2< t < 6 dt  25A, 6< t <8  which is sketched below: Chapter 1, Solution 8 10×1 q = ∫idt = +10×1=15 µC 2 Chapter 1, Solution 9 1 (a) q = ∫idt = ∫10 dt =10 C 0 3  5×1 q = ∫ idt =10×1+10− +5×1 (b) 0  2  =15+10−25=22.5 C 5 (c) q = ∫ idt =10+10+10= 30 C 0 Chapter 1, Solution 10 q = ixt = 8x103x15x10−6 = 120 µC Chapter 1, Solution 11 q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, t t q(t)=∫idt+q(0)=∫3tdt+0=1.5t2 0 0 At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s, t t q(t)=∫idt+q(6)=∫18dt+54=18t−54 6 6 At t=10, q(10) = 180 – 54 = 126 For 10

Solution of Fundamentals of Electric Circuits

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