Solution of Fundamentals of Electric Circuits

1117 Pages·2008·17.305 MB·English

by  Charles Alexander, Matthew Sadiku

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<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)= ∫0dt+q(15)=66
15
Thus,

 1.5t2 C, 0 < t < 6s

 18t−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
)
t
(
q
60
40
20
0
0 5 10 15 20
t
Chapter 1, Solution 13

2 2
w = ∫vidt = ∫ 1200cos2 4t dt
0 0
2
=1200∫ (2cos8t-1)dt (since, cos2x = 2 cos 2x-1)
0
2
2  1 
=1200 sin8t −t =1200 sin16−2
8  4 
0
=-2.486 kJ

Chapter 1, Solution 14

1 ( ) ( )1
q = ∫idt = ∫ 101-e-0.5t dt =10 t+2e-0.5t
(a) 0 0
( )
=101+2e-0.5 −2 = 2.131 C

(b) p(t) = v(t)i(t)
p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935)
= -8.188 W

Chapter 1, Solution 15

2
2 −3
q = ∫idt = ∫ 3e-2tdt = e2t
(a) 0 2
0
( )
= −1.5e-2 −1 =1.297 C

5di
v = = −6e2t(5) = −30e-2t
(b) dt
p = vi = −90e−4tW

3
3 −90
(c) w = ∫pdt = -90∫ e-4t dt = e-4t = −22.5 J
0 −4
0
Chapter 1, Solution 16

 10t V 0< t <1
25t mA 0< t < 2 
i(t) = , v(t) = 10 V 1< t < 3
100-25t mA 2< t < 4 
 40-10t V 3< t < 4
1 2 3 4
w = ∫v(t)i(t)dt = ∫10+(25t)dt+∫10(25t)dt+∫10(100−25t)dt+∫(40−10t)(100-25t)mJ
0 1 2 3
250 1 250 2  t2 3 4
= t3 + +2504t-  + ∫ 250(4−t)2dt
 
3 2  2  3
0 1 2
4
250 250  9   t2 
= + (3)+25012− −8+2+25016t-4t2 + 
3 2  2   3 
3
= 916.7 mJ

Chapter 1, Solution 17

Σ p = 0 → -205 + 60 + 45 + 30 + p = 0
3

p = 205 – 135 = 70 W
3

Thus element 3 receives 70 W.

Chapter 1, Solution 18

p = 30(-10) = -300 W
1
p = 10(10) = 100 W
2
p = 20(14) = 280 W
3
p = 8(-4) = -32 W
4
p = 12(-4) = -48 W
5

Chapter 1, Solution 19

∑
p =0 → −4I −2x6 −13x2+5x10 =0 → I = 3 A
s s
Chapter 1, Solution 20

Since Σ p = 0

-30×6 + 6×12 + 3V + 28 + 28×2 - 3×10 = 0
0

72 + 84 + 3V = 210 or 3V = 54
0 0

V = 18 V
0

Chapter 1, Solution 21

∆ q photon 1electron
i = = 4×1011 ⋅  ⋅1.6×1019(C / electron)
∆ t  sec  8 photon 

4
= ×1011×1.6×10−19C/s = 0.8×10-8C/s = 8nA
8

Chapter 1, Solution 22

It should be noted that these are only typical answers.

(a) Light bulb 60 W, 100 W
(b) Radio set 4 W
(c) TV set 110 W
(d) Refrigerator 700 W
(e) PC 120 W
(f) PC printer 18 W
(g) Microwave oven 1000 W
(h) Blender 350 W

Chapter 1, Solution 23

p 1500
(a) i = = =12.5 W
v 120
45
(b) w = pt =1.5×103 ×45×60⋅J =1.5× kWh =1.125 kWh
60
(c) Cost = 1.125 × 10 = 11.25 cents
Chapter 1, Solution 24

p = vi = 110 x 8 = 880 W

Chapter 1, Solution 25

4
Cost = 1.2 kW × hr ×30×9 cents/kWh = 21.6 cents
6

Chapter 1, Solution 26

0.8A⋅h
(a) i = =80 mA
10h
(b) p = vi = 6 × 0.08 = 0.48 W
(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh

Chapter 1, Solution 27

(a) Let T = 4h = 4 × 36005

T
q = ∫idt = ∫3dt =3T = 3 ×4×3600 = 43.2 kC
0

T T  0.5t 
(b) W = ∫pdt = ∫vidt = ∫ (3)10+ dt
0 0  3600
4×3600
 0.25t2 
= 310t +  = 3[40×3600+0.25×16×3600]
 
 3600 
0
= 475.2 kJ

(c) W = 475.2 kWs, (J =Ws)
475.2
Cost = kWh×9 cent =1.188 cents
3600
Chapter 1, Solution 28

P 30
(a) i = = = 0.25 A
V 120

(b) W = pt =30×365×24 Wh= 262.8 kWh

Cost =$0.12 ×262.8=$31.54

Chapter 1, Solution 29

(20+40+15+45) 30
w = pt =1.2kW hr +1.8 kW hr
60 60
= 2.4+0.9 = 3.3 kWh
Cost =12 cents × 3.3= 39.6 cents

Chapter 1, Solution 30

Energy = (52.75 – 5.23)/0.11 = 432 kWh

Chapter 1, Solution 31

Total energy consumed = 365(4 +8) W
Cost = $0.12 x 365 x 12 = $526.60

Chapter 1, Solution 32

(20+40+15+45) 30
w = pt =1.2kW hr +1.8 kW hr
60 60
= 2.4+0.9 = 3.3 kWh
Cost =12 cents × 3.3= 39.6 cents
Chapter 1, Solution 33

dq
i = → q = ∫idt = 2000×3×103 = 6 C
dt

Chapter 1, Solution 34

(b) Energy = ∑pt = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
= 10,000 kWh

(c) Average power = 10,000/24 = 416.67 W

Chapter 1, Solution 35

(a) W = ∫ p(t)dt = 400×6+1000×2+200×12×1200×2+400×2

= 7200+2800=10.4 kWh

10.4 kW
(b) = 433.3 W/h
24 h

Chapter 1, Solution 36

160A⋅h
(a) i = = 4 A
40

160Ah 160,000h
(b) t = = = 6,667 days
0.001A 24h / day

Chapter 1, Solution 37

( )
q =5×1020 −1.602×10−19 = −80.1 C

W = qv = −80.1×12 = −901.2 J