Solution Manual for Fundamentals of Electric Circuits 7th Edition by Charles K. Alexander, Matthew N. O. Sadiku All Chapters 1-19

Chapter 1, Solution 1

(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C

(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C

(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C

(d) q = 1.628x1020 x [-1.602x10-19 C] = -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 = 1200π cos 120π t pA
(e) i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A


Chapter 1, Solution 3

(a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C

(b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC
(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C

10e -30t
q(t) = ∫ 10e -30t sin 40t + q(0) = ( −30 sin 40t - 40 cos t)
(d) 900 + 1600
= − e - 30t (0.16cos40 t + 0.12 sin 40t) C



Chapter 1, Solution 4

10
−5
q = ∫ idt = ∫ 5sin 6 π t dt = cos 6π t
6π 0

5
= (1 − cos 0.06π ) = 4.698 mC
6π

,Chapter 1, Solution 5

2
1
q = ∫ idt = ∫ e dt mC = - e -2t
-2t

2 0

1
= (1 − e 4 ) 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 = 8 x10 3 x15 x10 − 6 = 120 µ C



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 = 1.5t 2
0 0
At t=6, q(6) = 1.5(6)2 = 54
For 6 < t < 10s,

, t t

∫ ∫
q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t − 54
6 6
At t=10, q(10) = 180 – 54 = 126
For 10<t<15s,
t t

∫ ∫
q (t ) = idt + q (10 ) = ( −12)dt + 126 = −12t + 246
10 10


At t=15, q(15) = -12x15 + 246 = 66
For 15<t<20s,
t

∫
q (t ) = 0 dt + q (15) =66
15
Thus,

 1.5t 2 C, 0 < t < 6s

 18 t − 54 C, 6 < t < 10s
q (t ) = 
−12t + 246 C, 10 < t < 15s
 66 C, 15 < t < 20s


The plot of the charge is shown below.

140


120


100


80
q(t)




60


40


20


0
0 5 10 15 20
t