Solution Manual For Introductory Circuit Analysis, 12th Edition

Name: Solution Manual For Introductory Circuit Analysis, 12th Edition
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Chapter 1 1
Chapter 1
1. 
2. 
3. 
4.
 = t
d = 

















h1
min60
min1
s60
ft5,280
mi1
s10
ft000,20 = 1363.64 mph
5. 4 min 





min60
h1 = 0.067 h
 = h1.067
mi31

t
d = 29.05 mph
6. 100 yds 











ft5,280
mi1
yd1
ft3 = 0.0568 mi
60 mi 1 h 1 min
h 60 min 60 s
   
   
    = 0.0167 mi/s
t = 0.0568 mi
0.0167 mi/s
d
  = 3.40 s
7. a. 

















s60
min1
min60
h1
mi
ft280,5
h
mi95 = 139.33 ft/s
b. t = ft/s139.33
ft60


d = 0.431 s
c.
 = 

















 ft5,280
mi1
h1
min60
min1
s60
s1
ft60
t
d = 40.91 mph
8. 
9. 
10. 
11. MKS, CGS, C = )36(
9
5
)3268(
9
5
)32F(
9
5  = 20
SI: K = 273.15 + C = 273.15 + 20 = 293.15
12. 





J1
lb-ft0.7378
J1000 = 737.8 ft-lbs

Chapter 12
13. 0.5 yd 

















in.1
cm54.2
ft1
in.12
yd1
ft3 = 45.72 cm
14. a. °F = 2(°C) + 30° = 40° + 30° = 70°
b. °F = 9 9
( C) 32 (20 ) 32
5 5
       = 68°
c. very close
d. 30°C  90°F vs. 86°
5°C  40°F vs 41°
15. a. 14.6 b. 56.0
c. 1,046.1 d. 1
16 = 0.0625 = 0.1
e. 3.14159 = 3.1
16. a. 14.60 b. 56.04
c. 1,046.06 d. 1
16 = 0.0625 = 0.06
e. 3.14159 = 3.14
17. a. 14.603 b. 56.042
c. 1,046.060 d. 1
16 = 0.0625 = 0.063
e. 3.14159 = 3.142
18. a. 10 4 b. 10 6 c. 10 3 d. 103
e. 10 0 f. 101
19. a. 15  10 3 b. 5  103 c. 2.4  10 6 d. 60  10 3
e. 4.02  104 f. 2  1010
20. a. 4.2  10 3 + 48.0  10 3 = 52.2  10 3 = 5.22  10 4
b. 90  10 3 + 360  10 3 = 450  10 3 = 4.50  10 5
c. 50  105  6  105 = 44  105 = 4.4  104
d. 1.2  10 3 + 0.05  10 3  0.4  10 3 = 0.85  10 3 = 850
21. a. (10 2
)(103 ) = 10 5 = 100  103
b. (102 )(103
) = 10 1 = 10
c. (103
)(106 ) = 1  10 9
d. (10 2
)(105 ) = 1  103
e. (106 )(10  10 6
) = 10
f. (104
)(108 )(1028
) = 1  10 24

Chapter 1 3
22. a. (50  10 3
)(2  103 ) = 100  10 0 = 100
b. (2.2  10 3 )(2  103 ) = 4.4  10 0 = 4.4
c. (82  106
)(1.2  106 ) = 98.4
d. (30  104 )(4  103 )(7  108
) = 840  10 1 = 8.40  103
23. a. 10 2
/10 4 = 102 = 10  103
b. 102 /103 = 105 = 10  106
c. 10 4
/103 = 10 7 = 10  106
d. 107 /102 = 1.0  109
e. 10 38 /104 = 1.0  10 42
f. 2
10/100  = 10 1
/102 = 1  10 3
24. a. (2  10 3
)/(8  105
) = 0.25  10 8 = 2.50  107
b. (4  103
)/(4  10 6
) = 4/4  109 = 1  109
c. (22  105
)/(5  105
) = 22/5  10 0 = 4.4
d. (78  10 18
)/(4  106
) = 1.95  1025
25. a. (10 2
) 3 = 1.0  10 6 b. (104
) 1/2 = 10.0  103
c. (104
) 8 = 100.0  10 30 d. (107
) 9 = 1.0  1063
26. a. (2  10 2
) 2 = 4  10 4
b. (5  103
) 3 = 125  109
c. (4  103
)(3  103
) 2 = (4  103
)(9  10 4
) = 36  10 1 = 360
d. ((2  103
)(0.8  10 4
)(0.003  10 5
)) 3 = (4.8  10 3
) 3 = (4.8) 3  (10 3
) 3
= 110.6  10 9 = 1.11  10 11
27. a. (103
) 2 = 1.0  106
b. 3
42
10
)10)(10( 
= 102
/10 3 = 1.0  105
c. 4
4
4
26
4
223
10
10
10
)10)(10(
10
)10()10( 
 = 1.0  108
d. 4
43
10
)10)(10(
 = 10 7
/104 = 1.0  10 11
e. (1  104
) 3
(102
)/106 = (1012
)(102
)/10 6 = 1010
/10 6 = 1.0  1016
f.  
    10
1
)10)(10(
1
10)10(
)10)(10(
34
322
322
 


= 1.0  101
28. a. 4
222
103
)10()103(

 = (9  10 4
)(102
)/(3  10 4
) = (9  10 6
)/(3  10 4
) = 3  10 2 = 300
b. 3
8
3
24
108
1016
)20(
)104(



 = 2  10 5
c. 4
8
22
24
104
1036
)102(
)106(
 



 = 9.0  1012

Chapter 14
d. 5
2
5
3/16
102
103
102
)1027(




 
= 1.5  107 = 150.0  109
e. 4
8
4
26
4
223
102
1048
102
)103)(1016(
102
)103()104(
 






 = 24.0  10 12
f. (16  106
) 1/2
(105
) 5
(2  102
) = (4  103
)(1025
)(2  102
) = 8  10 20 = 800.0  10 18
g.
2 1/ 23 3 2 2 4
5 2
(3 10 ) 1.60 10 (2 10 )(8 10 )
(7 10 )
 

             

9 4 2 1/ 2
10
5 1 6
10 10
(27 10 )(2.56 10 )(16 10 )
49 10
(69.12 10 )(4 10 ) 276.48 10
49 10 49 10
 

  
 
  
 
  
 
 
= 5.64  10 4 = 56.4  10 3
29. Scientific: a. 2.05 × 10 1
b. 5.04 × 10 4
c. 6.74 × 104
d. 4.60 × 102
Engineering: a. 20.46 × 10 0
b. 50.42 × 10 3
c. 674.00 × 106
d. 46.00 × 103
30. Scientific a. 5.0 × 102
b. 4.5 × 10 1
c. 1/32 = 0.03125 = 3.125 × 102
d. 3.14159 = 3.142 × 10 0
Engineering: a. 50.0 × 103
b. 0.045 × 10 3
c. 31.25 × 103
d. 3.142 × 10 0

Chapter 1 5
+2
31. a. 6  10 4 = 0.06  10 +6 = 0.06 × 10+6
3
3
b. 0.4  103 = 400  106 = 400 × 106
+3
2 +3 +3
c. 50  10 5 = 5000  10 3 = 5  10 6 = 0.005  10 9 = 0.005 × 10 9
+2 3 3
+4 3 3
d. 12  107 = 0.0012  103 = 1.2  106 = 1200  109 = 1200 × 10 9
4 +3 +3
3
32. a. 0.05  10 0 s = 50  103 s = 50 ms
+3
+3
b. 2000  106 s = 2  103 s = 2 ms
3
3
c. 0.04  103 s = 40  106 s = 40
s
+3
+6
d. 8400  1012 s  0.0084  106 s = 0.0084
s
6

Loading page 6...

Chapter 16
+3
increase by 3
10 0
e. 100  10 3  103 m = 0.1  10 3 m = 0.1 km
3
33. a. 1.5 min 





min1
s60 = 90 s
b. 2 × 102 h 











min1
s60
h1
min60 = 72 s
c. 0.05 s 





 s10
s1
6
 = 0.05  106
s = 50  10 3
s
d. 0.16 m 




  m10
mm1
3 = 0.16  10 3 mm = 160 mm
e. 1.2  107 s 




  s10
ns1
9 = 1.2  10 2 ns = 120 ns
f. 4  108 s 

















h24
day1
min60
h1
s60
min1 = 4629.63 days
34. a. 80  103 m 





m1
cm100 = 8000  103 cm = 8 cm
b. 60 cm 











m1000
km1
cm100
m1 = 60  105 km
c. 12 × 103 m 6
1 m
10 m


 
   = 12  103 × 10 +6
m = 12 × 10 3
m
d. 60 cm2 1 m 1 m
100 cm 100 cm
   
   
    = 60  104 m 2
35. a. 100 in. 





in.39.37
m1 = 2.54 m
b. 4 ft 











in.39.37
m1
ft1
in.12 = 1.22 m

Loading page 7...

Chapter 1 7
c. 6 lb 





lb1
N45.4 = 26.7 N
d. 60  10 3 dynes 











N4.45
lb1
dynes10
N1
5 = 0.13 lb
e. 150,000 cm 











in.12
ft1
cm2.54
in.1 = 4921.26 ft
f. 0.002 mi 

















in.39.37
m1
ft1
in.12
mi1
ft5280 = 3.22 m
36. 5280 ft, 5280 ft 





ft3
yd1 = 1760 yds
5280 ft 











in.39.37
m1
ft1
in.12 = 1609.35 m, 1.61 km
37. 60 mi 5280 ft 12 in. 1 m 1 h 1 min
h 1 mi 1 ft 39.37 in. 60 min 60 s = 26.82 m/s
38. 10 km 























ft5280
mi1
in.12
ft1
m1
in.39.37
km1
m1000 = 6.214 mi
 = min6.5
mi1 , t =
min6.5
mi1
mi6.214


d = 40.39 min
39. 100 yds 











ft1
in.12
yd1
ft3 = 3600 in  3600 quarters
40. 60 mph: t = 500 mi
60 mi/h
d
  = 8.33 h = 8 h:19.8 min
75 mph: t = 500 mi
75 mi/h
d
  = 6.67 h = 6h:40.2 min
difference = 1h:28.6 minutes
41. d =
t =   























cm100
m1
min1
s60
h1
min60
h016.0
s
cm
600 = 345.6 m

Loading page 8...

Chapter 18
42. d = 86 stories


















ft
12
9
step1
story
ft14 = 1605 steps
 = 





 seconds60
minute1
seconds5.802
second
steps2
steps1605

d
t
t
d = 13.38 minutes
43. d = (86 stories) 











ft5,280
mile1
ft1204
story
ft14 = 0.228 miles
min 10.22 min
mile 0.228 miles
 = 44.82 min/mile
44. minute
ft1056
1mile
ft5,280
min5
mile1
mile
min5 





 , distance = 86 stories 





story
ft14 = 1204 ft =
min
ft
1056
ft1204


d
t
t
d = 1.14 minutes
45. a. 





J1054.35
Btu1
J5 = 4.74  103 Btu
b. 24 ounces 











gallons172.264
m1
ounces128
gallon1 3
= 7.1  104 m 3
c. 1.4 days 





day1
s400,86 = 1.21  10 5 s
d. 1 m3












gallon1
pints8
m1
gallons264.172
3 = 2113.38 pints
46. 6(4 × 2 + 8) = 96
47. (42 + 6/5)/3 = 14.4
48.
2
2 2
5 3
 = 5.044
49. MODE = DEGREES: cos 21.87 = 0.928
50. MODE = DEGREES: tan1
(3/4) = 36.87
51.  2
400 /(6 10 / 5) = 7.071

Loading page 9...

Loading page 10...

Chapter 210
Chapter 2
1. 
2. a. F = 2
9
2
21
m)(1
C)C)(21)(109( 

r
QQ
k = 18  10 9 N
b. F = k 2
21
r
QQ =  
2
9
m)3(
C2C)1)(109(  = 2  10 9 N
c. F = 2
9
2
21
m)(10
C)C)(21)(109( 

r
QQ
k = 0.18  10 9 N
d. Exponentially, 3
1
10 m
1 m
r
r  = 10 while
9
1
9
2
18 10 N
0.18 10 N
F
F

  = 100
3. a. r = 1 mi:
1 mi 

















in.37.39
m1
ft1
in.12
mi1
ft5280 = 1609.35 m
F =
9 6 -6 3
1 2
2 2 6
(9 10 )(8 10 C)(40 10 C) 2880 10
(1609.35 m) 2.59 10
kQ Q
r
 
   
  
= 1.11
N
b. r =10 ft:
12 in. 1 m
10 ft 1 ft 39.37 in.
   
   
    = 3.05 m
F =
3 3
1 2
2 2
2880 10 2880 10
9.30(3.05 m)
kQ Q
r
 
 
  = 0.31 N
c. 





in.39.37
m1
16
in.1 = 1.59 mm
F =
3 3 31 2
2 3 2 6
2880 10 2880 10 1138.34 10
(1.59 10 m) 2.53 10
kQ Q
r
 
 
 
   
  N
= 1138.34 kN
4. 
5. Q1 = Q2 = Q;
2 22 2
2 1 1 1 1
1 22 2 2
1 2 2
;
F r F rkQ kQ k
F Q F
k kr r r
 
      
 
and 2F 
2
1
12
2
r F
r
6. F = 4
269
21
2
21
106.3
)1020)(109(




F
QkQ
r
r
QkQ = 10 mm

Loading page 11...

Chapter 2 11
7. F = 212
21
2
21
m)2(
8.1 QkQ
QkQ
r
QkQ  = 4(1.8) = 7.2
a. F = 22
21
)10(
2.7

r
QkQ = 72 mN
b. Q1/Q2 = 1/2  Q2 = 2Q1
7.2 = kQ1Q2 = (9  10 9
)(Q1)(2Q1) = 9  10 9  2
12Q
91
2
19 1018
2.7
1018
2.7


 QQ = 20
C
Q2 = 2Q1 = 2(2  105 C) = 40
C
8. V = 1.2 J
20 C
W
Q

 = 60 kV
9. W = VQ = (60 V)(8 mC) = 0.48 J
10. Q = 120 J
20 mV
W
V

 = 6 mC
18
6.242 10 electrons
1 C
 
 
  = 37.45 × 10 15 electrons
11. Q = V9
J72

V
W = 8 C
12. a. W = QV = (1 × 10 12 electrons)(40 V) = 40 × 10 12 eV
b. 40 × 1012 eV 18
1 C
6.242 10 electrons
 
   = 6.41 μJ
13. I = 12 mC
2.8 s
Q
t  = 4.29 mA
14. I = 312 C
(2)(60 s)
Q
t  = 2.60 A
15. Q = It = (40 mA)(0.8)(60 s) = 1.92 C
16. Q = It = (250 mA)(1.2)(60 s) = 18.0 C
17. t = mA2
mC6

I
Q = 3 s
18. 21.847  10 18 electrons 





 electrons106.242
C1
18 = 3.5 C
I = s12
C3.5

t
Q = 0.29 A

Loading page 12...

Chapter 212
19. Q = It = (4 mA)(90 s) = 360 mC
360 mC 




 
C1
electrons10242.6 18
= 2.25  1018 electrons
20. I = s)(1.2)(60
C86

t
Q = 1.194 A > 1 A (yes)
21. 0.84  10 16 electrons 





 electrons106.242
C1
18 = 1.346 mC
I = ms60
mC346.1

t
Q = 22.43 mA
22. a. Q = It = (2 mA)(0.01
s) = 2  1011 C
2  1011 C
18
6.242 10 electrons 1 ¢
1 C electron
   
   
  
= 1.25  10 8 ¢ = $1.25  106 = 1.25 million
b. Q = It = (100
A)(1.5 ns) = 1.5  1013 C
1.5  1013 C 










 
electron
1$
C1
electrons10242.6 18
= 0.94 million
(a) > (b)
23. Q = It = (200  103 A)(30 s) = 6 C
V = C6
J40

Q
W = 6.67 V
24. Q = It = min)5.0(
min
C420





 = 210 C
V = C210
J742

Q
W = 3.53 V
25. Q = V24
J0.4

V
W = 0.0167 C
I = 3
0.0167 C
5 10 s
Q
t 
  = 3.34 A
26. I = h40
Ah200
(hours)
ratingAh 
t = 5 A
27. Ah = (0.8 A)(75 h) = 60.0 Ah
28. t(hours) = A1.28
Ah32ratingAh 
I = 25 h

Loading page 13...

Chapter 2 13
29. 40 Ah(for 1 h): W1 = VQ = VIt = (12 V)(40 A)(1 h) 











min1
s60
h1
min60 = 1.728  10 6 J
60 Ah(for 1 h): W2 = (12 V)(60 A)(1 h) 











min1
s60
h1
min60 = 2.592  106 J
Ratio W2/W1 = 1.5 or 50% more energy available with 60 Ah rating.
For 60 s discharge: 40 Ah = It = I   











min60
h1
s60
min1
s60 = I(16.67  103 h)
and I = h1016.67
Ah40 3-
 = 2400 A
60 Ah = It = I   











min60
h1
s60
min1
s60 = I(16.67  103 h)
and I = h1016.67
Ah60 3-
 = 3600 A
I2/I1 = 1.5 or 50 % more starting current available at 60 Ah
30. 0.75(18 Ah) = 13.5 Ah   250 mA
31. (18 Ah  15.5 Ah)/18 Ah × 100% = 13.89%
32. At 100 mA, discharge time  120 H; At 25 mA, discharge time  425 h;
 300 h more at 25 mA
33. I = 3 Ah
6.0 h = 500 mA
Q = It = (500 mA)(6 h) 











min1
s60
h1
min60 = 10.80 kC
W = QV = (10.8 kC)(12 V)  129.6 kJ
34. 
35. 
36. 
37. a. 0.5 in 2.54 cm
1 in
 
   = 1.27 cm
1.27 cm 30 kV
cm
 
   = 38.1 kV
b. 1.27 cm 270 kV
cm
 
   = 342.9 kV
c. 342.9 kV:38.1 kV = 9:1

Loading page 14...

Loading page 15...

Chapter 3 15
Chapter 3
1. a. 0.5 in. = 500 mils
b. 0.02 in. 





in.1
mils1000 = 20 mils
c. 4
1 in. = 0.25 in. 





in.1
mils1000 = 250 mils
d. 10 mm = 10 × 103 m 39.37 in 1000 mils
1 m 1 in
   
      = 393.7 mils
e. 0.01 ft 











in.1
mils10
ft1
in.12 3
= 120 mils
f. 0.1 cm 











in.1
mils1000
cm2.54
in.1 = 39.37 mils
2. a. ACM = (30 mils)2 = 900 CM
b. 0.016 in. = 16 mils, ACM = (16 mils)2 = 256 CM
c. 1
8 = 0.125" = 125 mils, ACM = (125 mils)2 = 15.63  10 3 CM
d. 1 cm 











in.1
mils1000
cm2.54
in.1 = 393.7 mils, ACM = (393.7 mils)2 = 155  103 CM
e. 0.02 ft 











in.1
mils1000
ft1
in.12 = 240 mils, ACM = (240 mils)2 = 57.60  103 CM
f. 4 × 103 m 39.37 in. 1000 mils
1 m 1 in
   
       = 157.48 mils, ACM = (157.48 mils)2 = 24.8 × 10 3 CM
3. ACM = (dmils )2  dmils = CMA
a. d = CM1600 = 40 mils = 0.04 in.
b. d = 820 CM = 28.64 mils = 0.029 in.
c. d = CM40,000 = 200 mils = 0.2 in.
"

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Solution Manual For Introductory Circuit Analysis, 12th Edition

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