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Solutions manual for fundamentals of electric circuits 5th
edition by alexander 2019 0723 25597 16grxc5
Power Electronics (Bangalore University)
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FiFt h Edition
Fundamentals of
Electric Circuits
INSTRUCTOR
SOLUTIONS
MANUAL
Charles K. Alexander | Matthew n. o. Sadiku
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Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC
(b) q = 1. 24x1018 x [-1.602x10-19 C] = –198.65 mC
(c) q = 2.46x1019 x [-1.602x10-19 C] = –3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = –26.08 C
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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 = 1200 cos 120 t pA
i =dq/dt =  e 4t (80 cos 50 t  1000 sin 50 t )  A
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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   / 6   q(0)  (2sin(10t   / 6)  1)  C
10e -30t
q(t)   10e sin 40t  q(0) 
( 30 sin 40t - 40 cos t)
(d)
900  1600
  e - 30t (0.16cos40 t  0.12 sin 40t) C
-30t
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Chapter 1, Solution 4
q = it = 7.4 x 20 = 148 C
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Chapter 1, Solution 5
10
1
t 2 10
q   idt   tdt 
 25 C
2
4 0
0
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Chapter 1, Solution 6
(a) At t = 1ms, i 
dq 30

 15 A
dt
2
(b) At t = 6ms, i 
dq
 0A
dt
(c) At t = 10ms, i 
dq  30

 –7.5 A
dt
4
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Chapter 1, Solution 7
25A,
dq 
i
 - 25A,
dt 
 25A,
0t2
2t6
6t8
which is sketched below:
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Chapter 1, Solution 8
q   idt 
10  1
 10  1  15 μC
2
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Chapter 1, Solution 9
1
(a) q   idt   10 dt  10 C
0
3
5 1

q   idt  10  1  10 
  5 1
0
(b)
2 

 15  7.5  5  22.5C
5
(c) q   idt  10  10  10  30 C
0
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Chapter 1, Solution 10
q = it = 10x103x15x10-6 = 150 mC
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Chapter 1, Solution 11
q= it = 90 x10-3 x 12 x 60 x 60 = 3.888 kC
E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ
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Chapter 1, Solution 12
For 0 < t < 6s, assuming q(0) = 0,
t
t


0
0
q (t )  idt  q (0 )  3tdt  0  1.5t 2
At t=6, q(6) = 1.5(6)2 = 54
For 6 < t < 10s,
t
t


6
6
q (t )  idt  q (6 )  18 dt  54  18 t  54
At t=10, q(10) = 180 – 54 = 126
For 10<t<15s,
t
t


10
10
q (t )  idt  q (10 )  ( 12)dt  126  12t  246
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.
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140
120
100
q(t)
80
60
40
20
0
0
5
10
t
15
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20
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Chapter 1, Solution 13
(a) i = [dq/dt] = 20πcos(4πt) mA
p = vi = 60πcos2(4πt) mW
At t=0.3s,
p = vi = 60πcos2(4π0.3) mW = 123.37 mW
(b) W =
W = 30π[0.6+(1/(8π))[sin(8π0.6)–sin(0)]] = 58.76 mJ
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Chapter 1, Solution 14
1



(a) q   idt   0.02 1 - e -0.5t dt  0.02 t  2e -0.5t
0
(b)

1
0


 0.02 1  2e -0.5  2 = 4.261 mC
p(t) = v(t)i(t)
p(1) = 10cos(2)x0.02(1–e–0.5) = (–4.161)(0.007869)
= –32.74 mW
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Chapter 1, Solution 15
 0.006 2t
e
q   idt   0.006e dt 
0
(a)
2
2

2
- 2t

0
 0.003 e  1 
2.945 mC
(b)
-4
10di
 0.012e - 2t (10)  0.12e - 2t V this leads to p(t) = v(t)i(t) =
dt
(-0.12e-2t)(0.006e-2t) = –720e–4t µW
v
3
(c) w   pdt  -0.72 e
0
3
- 4t
 720 -4t 6
dt 
e 10 = –180 µJ
4
0
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Chapter 1, Solution 16
(a)
 30t mA, 0 < t <2
i (t )  
120-30t mA, 2 < t<4
5 V, 0 < t <2
v(t )  
 -5 V, 2 < t<4
 150t mW, 0 < t <2
p (t )  
-600+150t mW, 2 < t<4
which is sketched below.
p(mW)
300
1
2
4
t (s)
-300
(b) From the graph of p,
4
W   pdt  0 J
0
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Chapter 1, Solution 17
 p=0
 -205 + 60 + 45 + 30 + p 3 = 0
p 3 = 205 – 135 = 70 W
Thus element 3 receives 70 W.
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Solutions Manual for Fundamentals Of Electric Circuits 5th Edition by Alexander
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Chapter 1, Solution 18
p 1 = 30(-10) = -300 W
p 2 = 10(10) = 100 W
p 3 = 20(14) = 280 W
p 4 = 8(-4) = -32 W
p 5 = 12(-4) = -48 W
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