Chapter 1
Chapter 1
Quantities and Units
Section 1-2 Scientific Notation
1.
(a)
3000 = 3  103
2.
(a)
1
= 0.002 = 2  103
500
(b)
1
= 0.0005 = 5  104
2000
(c)
1
= 0.0000002 = 2  107
5,000,000
3.
(a)
8400 = 8.4  103
(b)
99,000 = 9.9  104
4.
(a)
0.0002 = 2  104
(b)
0.6 = 6  101
(c)
7.8  102 (already in scientific notation)
(a)
32  103 = 3.2  104
(b)
6800  106 = 6.8  103
(c)
870  108 = 8.7  1010
(a)
2  105 = 200,000
(b)
5.4  109 = 0.0000000054
(c)
1.0  101 = 10
7.
(a)
2.5  106 = 0.0000025
8.
(a)
4.5  106 = 0.0000045
(b)
8  109 = 0.000000008
(c)
4.0  1012 = 0.0000000000040
5.
6.
(b)
75,000 = 7.5  104
(b)
5.0  102 = 500
2
(c)
2,000,000 = 2  106
(c)
0.2  106 = 2  105
(c)
3.9  101 = 0.39
Chapter 1
9.
10.
11.
12.
(a)
9.2  106 + 3.4  107 = 9.2  106 + 34  106 = 4.32  107
(b)
5  103 + 8.5  101 = 5  103 + 0.00085  103 = 5.00085  103
(c)
5.6  108 + 4.6  109 = 56  109 + 4.6  109 = 6.06  108
(a)
3.2  1012  1.1  1012 = 2.1  1012
(b)
2.6  108  1.3  107 = 26  107  1.3  107 = 24.7  107
(c)
1.5  1012  8  1013 = 15  1013  8  1013 = 7  1013
(a)
(5  103)(4  105) = 5  4  103 + 5 = 20  108 = 2.0  109
(b)
(1.2  1012)(3  102) = 1.2  3  1012 + 2 = 3.6  1014
(c)
(2.2  109)(7  106) = 2.2  7  10 9  6 = 15.4  1015 = 1.54  1014
(a)
1.0  103
= 0.4  103  2 = 0.4  101 = 4
2.5  102
(b)
2.5  106
= 0.05  106  (8) = 0.05  102 = 5
50  108
(c)
4.2  108
= 2.1  108  (5) = 2.1  1013
2  105
13.
(a)
(b)
(c)
8 10  4    2 10  80 10  4    2 10  84 10
 3 10  5 10   9 10  15 10   9 10  6 10
 2.2 10 1.1  5.5 10   1110
4

2
3
7
5
12
12
2

4

2
12
12
4
Section 1-3 Engineering Notation and Metric Prefixes
14.
The powers of ten used in engineering notation are multiples of 3:
10-12, 10-9, 10-6, 10-3, 103, 106, 109, 1012
15.
(a) 89000 = 89  103
(b) 450,000 = 450  103
(c) 12,040,000,000,000 = 12.04  1012
3
3
 2  102  4.20 102
Chapter 1
16.
(a) 2.35  105 = 235  103
(b) 7.32  107 = 73.2  106
(c) 1.333  109 (already in engineering notation)
17.
(a) 0.000345 = 345  106
(b) 0.025 = 25  103
(c) 0.00000000129 = 1.29  109
18.
(a) 9.81  103 = 9.81  103
(b) 4.82  104 = 482  106
(c) 4.38  107 = 438  109
19.
(a) 2.5  103 + 4.6  103 = (2.5 + 4.6)  103 = 7.1  103
(b) 68  106 + 33  106 = (68 + 33)  106 = 101  106
(c) 1.25  106 + 250  103 = 1.25  106 + 0.25  106 = (1.25 + 0.25)  106 = 1.50  106
20.
(a) (32  103)(56  103) = 1792  10(3 + 3) = 1792  100 = 1.792  103
(b) (1.2  106)(1.2  106) = 1.44  10(6  6) = 1.44  1012
(c) (100)(55  103) = 5500  103 = 5.5
21.
22.
(a)
50
= 22.7  103
2.2  103
(b)
5  103
= 0.2  10(3  (6)) = 0.2  109 = 200  106
25  10 6
(c)
560  103
= 0.848  10(3  3) = 0.848  100 = 848  103
660  103
(a)
89,000 = 89  103 = 89 k
(b)
450,000 = 450  103 = 450 k
(c)
12,040,000,000,000 = 12.04  1012 = 12.04 T
4
Chapter 1
(a)
0.000345 A = 345  106 A = 345 A
(b)
0.025 A = 25  103 A = 25 mA
(c)
0.00000000129 A = 1.29  109 A = 1.29 nA
24.
(a)
31  103 A = 31 mA
(b)
5.5  103 V = 5.5 kV
(c)
20  1012 F = 20 pF
25.
(a)
3  106 F = 3 F
(b)
3.3  106  = 3.3 M
(c)
350  109 A = 350 nA
26.
(a)
2.5  1012 A = 2.5 pA
(b)
8  109 Hz = 8 GHz
(c)
4.7  103  = 4.7 k
(a)
7.5 pA = 7.5  1012 A
(b)
3.3 GHz = 3.3  109 Hz
(c)
280 nW = 2.8  107 W
(a)
5 A = 5  106 A
(b)
43 mV = 43  103 V
(c)
275 k = 275  103 
(d)
10 MW = 10  106 W
23.
27.
28.
Section 1-4 Metric Unit Conversions
29.
30.
(a)
(5 mA) (1  103 A/mA) = 5  103 A = 5000 A
(b)
(3200 W)(1  103 W/W) = 3.2 mW
(c)
(5000 kV)(1  103) MV/kV = 5 MV
(d)
(10 MW)(1  103 kW/MW) = 10  103 kW = 10,000 kW
(a)
1 mA 1  103 A

= 1  103 = 1000
1 A 1  10 6 A
(b)
0.05 kV 0.05  103 V

= 0.05  106 = 50,000
3
1 mV
1  10 V
(c)
0.02 k 0.02  103 

= 0.02  103 = 2  105
1 M
1  106 
(d)
155 mW 155  103 W

= 155  106 = 1.55  104
3
1 kW
1  10 W
5
Chapter 1
31.
32.
(a)
50 mA + 680 A = 50 mA + 0.68 mA = 50.68 mA
(b)
120 k + 2.2 M = 0.12 M + 2.2 M = 2.32 M
(c)
0.02 F + 3300 pF = 0.02 F + 0.0033 F = 0.0233 F
(a)
10 k
10 k

= 0.8197
2.2 k  10 k 12.2 k
(b)
250 mV 250  103

= 5000
50 V
50  10 6
(c)
1 MW 1  106

= 500
2 kW 2  103
Section 1-5 Measured Numbers
33.
34.
The significant digits are shown in bold face.
(a) Three: 1.00 x 103
(b) Two: 0.0057
(c) Five: 1502.0
(d) Two: 0.000036
(e) Three: 0.105
(f) Two: 2.6 x 102
(a) 50,505 rounds to 5.05 x 104
(b) 220.45 rounds to 220
(c) 4646 rounds to 4.65 x 103
(d) 10.99 rounds to 11.0
(e) 1.005 rounds to 1.00
6
Chapter 2
Voltage, Current, and Resistance
Note: Solutions show conventional current direction.
Section 2-2 Electrical Charge
1.
29 e  1.6  1019 C/e = 4.64  1018 C
2.
17 e  1.6  1019 C/e = 2.72  1018 C
3.
Q = (charge per electron)(number of electrons) = (1.6  1019 C/e)(50  1031e) = 80  1012 C
4.
(6.25  1018 e/C)(80  106 C) = 5  1014 electrons
Section 2-3 Voltage
5.
(a)
V 
W 10 J

= 10 V
Q 1C
(c)
V 
W 100 J

=4V
Q 25 C
(b)
V 
W
5J

= 2.5 V
Q 2C
6.
V 
W 500 J

=5V
Q 100 C
7.
V 
W 800 J

= 20 V
Q 40 C
8.
W = VQ = (12 V)(2.5 C) = 30 J
9.
I=
10.
Four common sources of voltage are dc power supply, solar cell, generator, and battery.
11.
The operation of electrical generators is based on the principle of electromagnetic induction.
12.
A power supply converts electricity in one form (ac) to another form (dc). The other sources
convert other forms of energy into electrical energy.
Q
t
Q = It = (2 A)(15 s) = 30 C
W 1000 J

V=
= 33.3 V
Q
30 C
7
Chapter 2
Section 2-4 Current
13.
14.
15.
16.
A current source provides a constant current of 100 mA regardless of the load value.
Q 75 C
= 75 A

t
1s
Q 10 C
(b)
I 
= 20 A
t
0.5 s
Q 5C
(c)
I 
= 2.5 A
t
2s
Q 0.6 C
= 0.2 A
I 
t
3s
I
(a)
Q
t
Q 10 C
=2s

t=
I
5A
I
17.
Q = It = (1.5 A)(0.1 s) = 0.15 C
18.
I=
Q
t
574  1015 electrons
= 9.18  102 C
6.25  1018 electrons/C
9.18  102 C
= 367 mA
I=
250  103 s
Q=
Section 2-5 Resistance
19.
(a)
(b)
(c)
20.
(a)
(b)
(c)
1
1

= 0.2 S = 200 mS
R 5
1
1

G=
= 0.04 S = 40 mS
R 25 
1
1

G=
= 0.01 S = 10 mS
R 100 
G=
1
1

= 10 
G 0.1 S
1
1
R=
=2

G 0.5 S
1
1

R=
= 50 
G 0.02 S
R=
8
Chapter 2
21.
(a)
(b)
Red, violet, orange, gold: 27 k  5%
Brown, gray, red, silver: 1.8 k  10%
22.
(a)
Rmin = 27 k  0.05(27 k) = 27 k  1350  = 25.65 k
Rmax = 27 k + 0.05(27 k) = 27 k + 1350  = 28.35 k
(b)
Rmin = 1.8 k  0.1(1.8 k) = 1.8 k  180  = 1.62 k
Rmax = 1.8 k + 0.1(1.8 k) = 1.8 k + 180  = 1.98 k
23.
330 : orange, orange, brown. gold
2.2 k: red, red, red, gold
56 k: green, blue, orange, gold
100 k: brown, black, yellow, gold
39 k: orange, white, orange, gold
24.
(a)
brown, black, black, gold: 10   5%
(b)
green, brown, green, silver: 5.1 M  10%
(c)
blue, gray, black, gold: 68   5%
25.
26.
(a) red, violet, orange, silver : 27 k + 10%
(b) brown, black, brown, silver: 100  + 10%
(c) green, blue, green , gold: 5.6 M + 5%
(d) blue, gray, red, silver: 6.8 k + 10%
(e) orange, orange, black, silver: 33  + 10%
(f) yellow, violet, orange, gold: 47 k + 5%
330  : (b) orange, orange, brown; 2..2 k: (d) red, red, red; 56 k: (l) green, blue, orange;
100 k: (f) brown, black, yellow; 39 k: (a) orange, white, orange
27.
28.
29.
(a)
0.47 : yellow, violet, silver, gold
(b)
270 k: red, violet, yellow, gold
(c)
5.1 M: green, brown, green, gold
(a)
red, gray, violet, red, brown: 28.7 k  1%
(b)
blue, black, yellow, gold, brown: 60.4  1%
(c)
white, orange, brown, brown, brown: 9.31 k  1%
(a)
(b)
(c)
14.7 k  1%: brown, yellow, violet, red, brown
39.2   1%: orange, white, red, gold, brown
9.76 k  1%: white, violet, blue, brown, brown
9
Chapter 2
30.
500 , There is equal resistance on each side of the contact.
31.
4K7 = 4.7 k
32.
(a)
(b)
(c)
4R7J = 4.7   5%
5602M = 56 k  20%
1501F = 1500   1%
Section 2-6 The Electric Circuit
33.
See Figure 2-1.
Figure 2-1
34.
See Figure 2-2.
Figure 2-2
35.
Circuit (b) in Figure 2-68 can have both lamps on at the same time.
36.
There is always current through R5.
37.
See Figure 2-3.
Figure 2-3
10
Chapter 2
38.
See Figure 2-4.
Figure 2-4
Section 2-7 Basic Circuit Measurements
39.
See Figure 2-5.
Figure 2-5
40.
See Figure 2-6.
41.
Position 1: V1 = 0 V, V2 = VS
Position 2: V1 = VS, V2= 0 V
42.
Figure 2-6
See Figure 2-7.
Figure 2-7
11
Chapter 2
43.
See Figure 2-8.
44.
See Figure 2-8.
Figure 2-8
45.
On the 600 V scale (middle AC/DC scale): 250 V
46.
R = 10  10  = 100 
47.
(a)
(b)
(c)
48.
0.9999 + 0.0001 = 1.0000
Resolution = 0.00001 V
2  10  = 20 
15  100 k = 1.50 M
45  100  = 4.5 k
12
Chapter 2
49.
See Figure 2-9.
Figure 2-9
13
Chapter 3
Ohm’s Law
Note: Solutions show conventional current direction.
Section 3-1 The Relationship of Current, Voltage, and Resistance
1.
(a)
(b)
(c)
(d)
(e)
(f)
2.
I=
3.
V = IR
4.
R=
5.
See Figure 3-1.
I=
I=
I=
I=
I=
I=
I=
I=
I=
I=
I=
When voltage triples, current triples.
When voltage is reduced 75%, current is reduced 75%.
When resistance is doubled, current is halved.
When resistance is reduced 35%, current increases 54%.
When voltage is doubled and resistance is halved, current quadruples.
When voltage and resistance are both doubled, current is unchanged.
V
R
V
I
0V
100 
10 V
100 
20 V
100 
30 V
100 
40 V
100 
50 V
100 
60 V
100 
70 V
100 
80 V
100 
90 V
100 
100 V
100 
=0A
= 100 mA
= 200 mA
= 300 mA
= 400 mA
= 500 mA
= 600 mA
= 700 mA
Figure 3-1
= 800 mA
= 900 mA
The graph is a straight line indicating a linear relationship
between V and I.
=1A
14
Chapter 3
6.
R=
(a)
(b)
(c)
(d)
(e)
1V
= 200 
15 mA
1.5 V
I=
= 7.5 mA
200 
2V
I=
= 10 mA
200 
3V
I=
= 15 mA
200 
4V
I=
= 20 mA
200 
10 V
I=
= 50 mA
200 
7.
Pick a voltage value and find the corresponding value of current by projecting a line up from the
voltage value on the horizontal axis to the resistance line and then across to the vertical axis.
V
1V
R1 =

= 500 m
I
2A
V
1V
R2 =
=1

I 1A
V
1V

R3 =
=2
I 0.5 A
8.
See Figure 3-2.
Figure 3-2
15
Chapter 3
I=
I=
I=
I=
I=
9.
2V
8.2 k
4V
8.2 k
6V
8.2 k
8V
8.2 k
10 V
8.2 k
= 0.244 mA
= 0.488 mA
= 0.732 mA
= 0.976 mA
= 1.22 mA
See Figure 3-3.
Figure 3-3
I=
I=
I=
I=
I=
2V
1.58 k
4V
1.58 k
6V
1.58 k
8V
1.58 k
10 V
1.58 k
= 1.27 mA
= 2.53 mA
= 3.80 mA
= 5.06 mA
= 6.33 mA
16
Chapter 3
10.
(a)
(b)
(c)
50 V
= 15.2 mA
3.3 k
75 V
= 19.2 mA
I=
3.9 k
100 V
I=
= 21.3 mA
4.7 k
I=
Circuit (c) has the most current and circuit (a) has the least current.
11.
VS
10 V

= 0.2 k = 200 
30 mA 50 mA
VS = (200 )(30 mA) = 6 V (new value)
The battery voltage decreased by 4 V (from 10 V to 6 V).
R=
12.
The current increase is 50%, so the voltage increase must also be 50%.
VINC = (0.5)(20 V) = 10 V
V2 = 20 V + VINC = 20 V + 10 V = 30 V (new value)
13.
See Figure 3-4.
10 V
(a)
I=
= 10 A
1
20 V
= 20 A
I=
1
30 V
= 30 A
I=
1
40 V
= 40 A
I=
1
50 V
I=
= 50 A
1
60 V
= 60 A
I=
1
70 V
= 70 A
I=
1
80 V
I=
= 80 A
1
90 V
= 90 A
I=
1
100 V
= 100 A
I=
1
(b)
10 V
=2A
5
20 V
I=
=4A
5
30 V
=6A
I=
5
40 V
=8A
I=
5
50 V
I=
= 10 A
5
60 V
= 12 A
I=
5
70 V
= 14 A
I=
5
80 V
I=
= 16 A
5
90 V
= 18 A
I=
5
100 V
= 20 A
I=
5
I=
17
(c)
10 V
= 0.5 A
20 
20 V
I=
=1A
20 
30 V
I=
= 1.5 A
20 
40 V
=2A
I=
20 
50 V
I=
= 2.5 A
20 
60 V
=3A
I=
20 
70 V
= 3.5 A
I=
20 
80 V
I=
=4A
20 
90 V
= 4.5 A
I=
20 
100 V
=5A
I=
20 
I=
Chapter 3
(d)
I=
I=
I=
I=
I=
I=
I=
I=
I=
I=
14.
10 V
100 
20 V
100 
30 V
100 
40 V
100 
50 V
100 
60 V
100 
70 V
100 
80 V
100 
90 V
100 
100 V
100 
= 0.1 A
= 0.2 A
= 0.3 A
= 0.4 A
= 0.5 A
= 0.6 A
= 0.7 A
= 0.8 A
= 0.9 A
=1A
Figure 3-4
Yes, the lines on the IV graph are straight lines.
Section 3-2 Current Calculations
15.
(a)
I=
(b)
I=
(c)
I=
(d)
I=
(e)
I=
V
R
V
R
V
R
V
R
V
R
5V
=5A
1
15 V

= 1.5 A
10 
50 V

= 500 mA
100 
30 V

= 2 mA
15 k
250 V

= 44.6 A
5 . 6 M

18
Chapter 3
16.
(a)
I=
(b)
I=
(c)
I=
(d)
I=
(e)
I=
V
R
V
R
V
R
V
R
V
R
9V
= 3.33 mA
2.7 k
5.5 V

= 550 A
10 k
40 V

= 588 A
68 k
1 kV

= 455 mA
2 .2 k 
66 kV

= 6.6 mA
10 M

V 12 V

= 1.2 A
R 10 
17.
I=
18.
R = 3300   5%
Rmax = 3300  + (0.5)(3300 ) = 3465 
Rmin = 3300   (0.5)(3300 ) = 3135 
V
12 V
Imax = s 
= 3.83 mA
Rmin 3135 
V
12 V
Imin = s 
= 3.46 mA
Rmax 3465 
19.
R = 47 k  10%
Rmin = 47 k  0.1(4.7 k) = 42.3 k
Rmax = 47 k + 0.1(4.7 k) = 51.7 k
V
25 V

Imin =
= 484 A
Rmax 51.7 k
V
25 V

Imax =
= 591 A
Rmin 42.3 k
V
25 V
Inom =

= 532 A
R
47 k
20.
R = 37.4 
V
12 V
I= 
= 0.321 A
R 37.4 
21.
I = 0.642 A
Yes, the current exceeds the 0.5 A rating of the fuse.
22.
VR(max) = 120 V  100 V = 20 V
VR ( max ) 20 V
Imax =
= 2.5 A

Rmin
8
A fuse with a rating of less than 2.5 A must be used. A 2-A fuse is suggested.
19
Chapter 3
23.
I
V
3V

 9.1mA
R 330
Section 3-3 Voltage Calculations
24.
V  IR  180A  27k   4.86V
25.
(a)
(b)
(c)
(d)
(e)
V = IR = (2 A)(18 ) = 36 V
V = IR = (5 A)(56 ) = 280 V
V = IR = (2.5 A)(680 ) = 1.7 kV
V = IR = (0.6 A)(47 ) = 28.2 V
V = IR = (0.1 A)(560 ) = 56 V
26.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
V = IR = (1 mA)(10 ) = 10 mV
V = IR = (50 mA)(33 ) = 1.65 V
V = IR = (3 A)(5.6 k) = 16.8 kV
V = IR = (1.6 mA)(2.2 k) = 3.52 V
V = IR = (250 A)(1 k) = 250 mV
V = IR = (500 mA)(1.5 M) = 750 kV
V = IR = (850 A)(10 M) = 8.5 kV
V = IR = (75 A)(47 ) = 3.53 mV
27.
VS = IR = (3 A)(27 ) = 81 V
28.
(a)
(b)
(c)
29.
Wire resistance = RW =
(a)
(b)
(c)
V = IR = (3 mA)(27 k) = 81 V
V = IR = (5 A)(100 M) = 500 V
V = IR = (2.5 A)(47 ) = 117.5 V
(10.4 CM   / ft)(24 ft)
= 0.154 
1624.3 CM
V
6V
I=
= 59.9 mA

R  RW 100.154 
VR = (59.9 mA)(100 ) = 5.99 V
R 
VRW = I  W  = (59.9 mA)(0.154 /2) = 4.61 mV
 2 
Section 3-4 Resistance Calculations
30.
(a)
(b)
(c)
V 10 V

=5
I
2A
V 90 V
R=
=2

I 45 A
V 50 V
R=
= 10 

I
5A
R=
20
Chapter 3
(d)
(e)
31.
V 5.5 V
= 550 m

I
10 A
V 150 V

R=
= 300 
I
0.5 A
R=
(a)
R=
(b)
R=
(c)
R=
(d)
R=
(e)
R=
V
I
V
I
V
I
V
I
V
I
10 kV
= 2 k
5A
7V
= 3.5 k

2 mA
500 V

= 2 k
250 mA
50 V

= 100 k
500 A
1 kV

= 1 M
1 mA

6V
V

= 3 k
I 2 mA
32.
R=
33.
(a)
34.
Measure the current with an ammeter connected as shown in Figure 3-5, then calculate the
unknown resistance as R = 12 V/I.
35.
36.
RFIL: =
V 120 V
= 150 

I
0.8 A
100 V
V

= 133 
I 750 mA
Figure 3-5
V 100 V
= 100 

R=
I
1A
The source can be shorted if the rheostat is set to 0 .
R=
Rmin + 15  =
120 V
= 60 . Thus Rmin = 60   15  = 45 
2A
The rheostat must actually be set to slightly greater than 45  so that the current is limited to
slightly less than 2 A.
37.
110 V
= 110 
1A
Rmin = 110   15  = 95 
Rmin + 15  =
21
Chapter 3
Section 3-5 Introduction to Troubleshooting
38.
The 4th bulb from the left is open.
39.
It should take five (maximum) resistance measurements.
Multisim Troubleshooting and Analysis
40.
RB is open.
41.
RA = 560 k, RB = 2.2 M, RC = 1.8 k, RD = 33 
42.
No fault. I = 1.915 mA, V = 9.00 V
43.
V = 18 V, I = 5.455 mA, R = 3.3 k
44.
R is leaky.
22
Chapter 4
Energy and Power
Section 4-1 Energy and Power
1.
volt = joule/coulomb
ampere = coulomb/s
VI = (joule/coulomb)(coulomb/s) = joule/s
2.
1 kWh = (1000 joules/s)(3600 s) = 3.6  106 joules
3.
1 watt = 1 joule/s
P = 350 J/s = 350 W
4.
P=
5.
P=
6.
(a)
(b)
(c)
(d)
1000 W = 1  103 W = 1 kW
3750 W = 3.75  103 W = 3.75 kW
160 W = 0.160  103 W = 0.160 kW
50,000 W = 50  103 W = 50 kW
7.
(a)
(b)
(c)
(d)
1,000,000 W = 1  106 W = 1 MW
3  106 W = 3 MW
15  107 W = 150  106 = 150 MW
8700 kW = 8700  103 W = 8.7  106 W = 8.7 MW
8.
(a)
(b)
(c)
(d)
1 W = 1000  103 W = 1000 mW
0.4 W = 400  103 W = 400 mW
0.002 W = 2  103 = 2 mW
0.0125 W = 12.5  103 W = 12.5 mW
9.
(a)
(b)
(c)
(d)
2 W = 2,000,000 W
0.0005 W = 500 W
0.25 mW = 250 W
0.00667 mW = 6.67 W
10.
(a)
1.5 kW = 1.5  103 W = 1500 W
(b) 0.5 MW = 0.5  106 W = 500,000 W
(c)
350 mW = 350  103 W = 0.350 W
(d) 9000 W = 9000  106 W = 0.009 W
Energy = W = Pt = (100 mW)(24 h)(3600 s/h) = 8.64  103 J
11.
W 7500 J

t
5h
7500 J
7500 J

= 417 mW
(5 h)(3600 s/h) 18000 s
1000 J
= 20 kW
50 ms
23
Chapter 4
12.
300 W = 0.3 kW
(30 days)(24 h/day) = 720 h
(0.3 kW)(720 h) = 216 kWh
13.
1500 kWh/31 days = 48.39 kWh/day
(48.39 kWh/day)/24 h) = 2.02 kW/day
14.
5  106 watt-minutes = 5  103 kWminutes
(5  103 kWmin)(1 h/60 min) = 83.3 kWh
15.
6700 Ws
= 0.00186 kWh
(1000 W/kW)(3600 s/h)
16.
W = Pt
P = I2R = (5 A)2(47 ) = 1175 W
W
25 J

t=
= 0.0213 s = 21.3 ms
P 1175 W
Section 4-2 Power in an Electric Circuit
17.
RL =
V 75 V
= 37.5 

I
2A
18.
P = VI = (5.5 V)(3 mA) = 16.5 mW
19.
P = VI = (120 V)(3 A) = 360 W
20.
P = I2R = (500 mA)2(4.7 k) = 1.175 kW
21.
P = I2R = (100 A)2(10 k) = 100 W
22.
P=
V 2 (60 V)2

= 5.29 W
R
680 
23.
P=
V 2 (1.5 V)2

= 40.2 mW
R
56 
24.
P = I2R
P 100 W
= 25 
R= 2 
I
(2 A)2
25.
(a)
P=
(b)
If the resistor is disconnected after 1 minute, the power during the first minute is equal to
the power during the two minute interval. Only energy changes with time.
V 2 (12 V) 2
= 14.4 W

R
10 
W = Pt = (14.4 W)(2 min)(1/60 h/min) = 0.48 Wh
24
Chapter 4
Section 4-3 Resistor Power Ratings
26.


From Activity 1:
VR1 = 3.25 V and R1 = 18 VR2 = 6.5 V and R2 = 39 VR3 = 10 V and R3 = 68 
The power rating for each resistor is determined as follows:
V 2  3.25V 
 0.59WChoose next highest standard value of 1 W.
PR1  R1 
18
R1
2
V 2  6.5V 
 1.1WChoose next highest standard value of 2 W.
PR 2  R 2 
39
R2
2
V 2 10V 
 1.5WChoose next highest standard value of 2 W.
PR 3  R 3 
68
R3
2
27.
A 2 W resistor should be used to provide a margin of safety. A resistor rating greater
than the actual maximum power should always be used.
28.
P = I2R = (10 mA)2(6.8 k) = 0.68 W
Use at least the next highest standard rating of 1 W.
29.
Use the 12 W resistor to allow a minimum safety margin of greater than 20%. If the 8 W
resistor is used, it will be operating in a marginal condition and its useful life will be reduced.
Section 4-4 Energy Conversion and Voltage Drop in Resistance
30.
See Figure 4-1.
Figure 4-1
Section 4-5 Power Supplies and Batteries
31.
VOUT =
32.
PAVG =
PL RL  (1 W)(50 ) = 7.07 V
V 2 (1.25) 2 V

= 156 mW
R
10 
25
Chapter 4
33.
W = Pt = (0.156 W)(90 h) = (0.156 W)(324,000 s) = 50,544 J
34.
Ampere-hour rating = (1.5 A)(24 h) = 36 Ah
35.
I=
80 Ah
=8A
10 h
36.
I=
650 mAh
= 13.5 mA
48 h
37.
PLost = PIN  POUT = 500 mW  400 mW = 100 mW
P 
 400 mW 
% efficiency =  OUT 100%  
100% = 80%
 500 mW 
 PIN 
38.
POUT = (efficiency)PIN = (0.85)(5 W) = 4.25 W
39.
Assume that the total consumption of the power supply is the input power plus the power lost.
POUT = 2 W
P 
% efficiency =  OUT 100%
 PIN 


POUT
2W
PIN = 
100%  
100% = 3.33 W
 60% 
 % efficiency 
Energy = W = Pt = (3.33 W)(24 h) = 79.9 Wh  0.08 kWh
Multisim Troubleshooting and Analysis
40.
V = 24 V, I = 0.035 A, R = 680 
41.
V = 5 V, I = 5 mA, R = 1 k
42.
I = 833.3 mA
26
Chapter 5
Series Circuits
Note: Solutions show conventional current direction.
Section 5-1 Resistors in Series
1.
See Figure 5-1.
Figure 5-1
2.
R1, R2, R3, R4, and R9 are in series (pin 5 to 6).
R7, R13, R14 and R16 are in series (pin 1 to 8).
R6, R8, and R12 are in series (pin 2 to 3).
R5, R10, R11, and R15 are in series (pin 4 to 7).
See Figure 5-2.
3.
R1-8 = R13 + R7 + R14 + R16
= 68 k + 33 k + 47 k + 22 k
= 170 k
4.
R2-3 = R12 + R8 + R6 = 10  + 18  + 22 
= 50 
5.
R1, R7, R8, and R10 are in series.
R2, R4, R6, and R11 are in series.
R3, R5, R9, and R12 are in series.
Figure 5-2
27
Chapter 5
Section 5-2 Total Series Resistance
6.
RT = 1  + 2.2  + 5.6  + 12  + 22  = 42.8 
7.
(a)
(b)
(c)
(d)
RT = 560  + 1000  = 1560 
RT = 47  + 56  = 103 
RT = 1.5 k + 2.2 k + 10 k = 13.7 k
RT = 1 M + 470 k + 1 k + 2.2 M = 3.671 M
8.
(a)
(b)
(c)
RT = 1 k + 5.6 k + 2.2 k = 8.8 k
RT = 4.7  + 10  + 12  + 1  = 27.7 
RT = 1 M + 560 k + 5.6 M + 680 k + 10 M = 17.84 M
9.
RT = 12(5.6 k) = 67.2 k
10.
RT = 6(56 ) + 8(100 ) + 2(22 ) = 336  + 800  + 44  = 1180 
11.
RT = R1 + R2 + R3 + R4 + R5
R5 = RT  (R1 + R2 + R3 + R4)
= 17.4 k  (5.6 k + 1 k + 2.2 k + 4.7 k) = 17.4 k  13.5 k = 3.9 k
12.
RT = 3(5.6 k) + 1 k + 2(100 ) = 16.8 k + 1 k + 200  = 18 k
Three 5.6 k resistors, one 1 k resistor, and two 100  resistors.
Other combinations are possible.
13.
RT = 1 k + 5.6 k + 2.2 k + 4.7  + 10  + 12  + 1 
+ 1 M + 560 k + 5.6 M + 680 k + 10 M
= 17.848827.7 M  17.8 M
14.
Position 1:
RT = R1 + R3 + R5 = 510  + 820  + 680  = 2.01 k
Position 2:
RT = R1 + R2 + R3 + R4 + R5 = 510  + 910  + 820  + 750  + 680  = 3.67 k
Section 5-3 Current in a Series Circuit
V
12 V

= 100 mA
RT 120 
15.
I=
16.
I = 5 mA at all points in the series circuit.
17.
See Figure 5-3. The current through R2, R3, R4, and R9 is also measured by this set-up.
28
Chapter 5
Figure 5-3
18.
See Figure 5-4.
Figure 5-4
Section 5-4 Application of Ohm’s Law
19.
(a)
(b)
20.
RT = R1 + R2 + R3 = 2.2 k + 5.6 k + 1 k = 8.8 k
V
5.5 V

= 625 A
I=
RT 8.8 k
RT = R1 + R2 + R3 = 1 M + 2.2 M + 560 k = 3.76 M
V
16 V
I=

= 4.26 A
RT 3.76 M
(a)
I = 625 A
V1 = IR1 = (625 A)(2.2 k) = 1.375 V
V2 = IR2 = (625 A)(5.6 k) = 3.5 V
V3 = IR3 = (625 A)(1 k) = 0.625 V
(b)
I = 4.26 A
V1 = IR1 = (4.26 A)(1 M) = 4.26 V
V2 = IR2 = (4.26 A)(2.2 M) = 9.36 V
V3 = IR3 = (4.26 A)(560 k) = 2.38 V
29
Chapter 5
21.
22.
23.
RT = 3(470 ) = 1.41 k
V
48 V

(a)
I=
= 34 mA
RT 1.41 k
48 V
= 16 V
3
(b)
VR =
(c)
P = (34 mA)2(470 ) = 0.543 W
V
5V
= 2.24 k

I 2.23 mA
R
2.24 k
Reach = T 
= 560 
4
4
V
21.7 V
R1 = 1 
= 330 
I
65.8 mA
V
6.58 V
R1 = 3 
= 100 
I
65.8 mA
RT =
V2
14.5 V

= 220 
I
65.8 mA
V
30.9 V
R4 = 4 
= 470 
I
65.8 mA
R2 =
24.
V1 = IR1 = (12.3 mA)(82 ) = 1.01 V
V 12 V  2.21 V  1.01 V
= 714 
R2 = 2 
I
12.3 mA
V
2.21 V
R3 = 3 
= 180 
I 12.3 mA
25.
(a)
(b)
(c)
26.
R T = R 1 + R 2 + R3 + R 4
12 V
12 V
 (R1 + R2 + R3) =
 1200  = 1531   1200  = 331 
R4 =
7.84 mA
7.84 mA
12 V
12 V

= 9.15 mA
Position B: I =
R2  R3  R4 1311 
12 V
12 V

Position C: I =
= 14.3 mA
R3  R4 841 
12 V 12 V

= 36.3 mA
Position D: I =
R4
331 
No
Position A:
RT = R1 = 1 k
V
9V
I=

= 9 mA
RT 1 k
Position B:
RT = R1 + R2 + R5 = 1 k + 33 k + 22 k = 56 k
V
9V

I=
= 161 A
RT 56 k
Position C:
30
Chapter 5
RT = R1 + R2 + R3 + R4 + R5 = 1 k + 33 k + 68 k + 27 k + 22 k = 151 k
V
9V

I=
= 59.6 A
RT 151 k
Section 5-5 Voltage Sources in Series
27.
VT = 5 V + 9 V = 14 V
28.
VT = 12 V  3 V = 9 V
29.
(a)
VT = 10 V + 8 V + 5 V = 23 V
(b)
VT = 50 V + 10 V + 25 V = 85 V
Section 5-6 Kirchhoff’s Voltage Law
30.
VS = 5.5 V + 8.2 V + 12.3 V = 26 V
31.
VS = V1 + V2 + V3 + V4 + V5
20 V = 1.5 V + 5.5 V + 3 V + 6 V + V5
V5 = 20 V  (1.5 V + 5.5 V + 3 V + 6 V) = 20 V  16 V = 4 V
32.
(a)
(b)
By Kirchhoff’s voltage law:
15 V = 2 V + V2 + 3.2 V + 1 V + 1.5 V + 0.5 V
V2 = 15 V  (2 V + 3.2 V + 1 V + 1.5 V + 0.5 V) = 15 V  8.2 V = 6.8 V
VR = 8 V, V2R = 2(8 V) = 16 V, V3R = 3(8 V) = 24 V, V4R = 4(8 V) = 32 V
VS = VR + VR + V2R + V3R + V4R = 11(VR) = 88 V
11.2 V
= 200 mA
56 
4.4 V
R4 =
= 22 
200 mA
33.
I=
34.
R1 =
35.
Position A:
RT = R1 + R2 + R3 + R4 = 1.8 k + 1 k + 820  + 560  = 4.18 k
Voltage drop across R1 through R4:
V = IRT = (3.35 mA)(4.18 k) = 14 V
V5 = 18 V  14 V = 4 V
V1
5.6 V

= 560 
I 10 mA
22 mW
P
R2 = 22 
= 220 
(10 mA)2
I
9V
RT =
= 900 
10 mA
R3 = RT  R1  R2 = 900   560   200  = 120 
31
Chapter 5
36.
Position B:
RT = R1 + R2 + R3 = 1.8 k + 1 k + 820  = 3.62 k
Voltage drop across R1 through R3:
V = IRT = (3.73 mA)(3.62 k) = 13.5 V
V5 = 18 V  13.5 V = 4.5 V
Position C:
RT = R1 + R2 = 1.8 k + 1 k = 2.8 k
Voltage drop across R1 and R2:
V = IRT = (4.5 mA)(2.8 k) = 12.6 V
V5 = 18 V  12.6 V = 5.4 V
Position D:
RT = R1 = 1.8 k
Voltage drop across R1:
V = IRT = (6 mA)(1.8 k) = 10.8 V
V5 = 18 V  10.8 V = 7.2 V
Position A:
V1 = (3.35 mA)(1.8 k) = 6.03 V
V2 = (3.35 mA)(1 k) = 3.35 V
V3 = (3.35 mA)(820 ) = 2.75 V
V4 = (3.35 mA)(560 ) = 1.88 V
V5 = 4.0 V
Position B:
V1 = (3.73 mA)(1.8 k) = 6.71 V
V2 = (3.73 mA)(1 k) = 3.73 V
V3 = (3.73 mA)(820 ) = 3.06 V
V5 = 4.5 V
Position C:
V1 = (4.5 mA)(1.8 k) = 8.1 V
V2 = (4.5 mA)(1 k) = 4.5 V
V5 = 5.4 V
Position D:
V1 = (6 mA)(1.8 k) = 10.8 V
V5 = 7.2 V
Section 5-7 Voltage Dividers
37.
V27  27  

100 = 4.82%
VT  560  
38.
(a)
(b)
39.
 56  
12 V = 4.31 V
VAB = 
 156  
 5.5 k 
8 V = 6.77 V
VAB = 
 6.5 k 
VA = VS = 15 V
32
Chapter 5
 R2  R3 
 13.3 k 
VB = 
 15 V = 10.6 V
VS  
 18.9 k 
 R1  R2  R3 


R3
 3.3 k 
VC = 
15 V = 2.62 V
VS  
R
R
R


 18.9 k 
2
3 
 1
40.


R3
 680  
VOUT(min) = 
12 V = 3.80 V
VS  
 2150  
 R1  R2  R3 
 R2  R3 
 1680  
VOUT(max) = 
12 V = 9.38 V
VS  
 2150  
 R1  R2  R3 
41.
RT = 15R
 R 
VR = 
90 V = 6 V
 15 R 
 2R 
V2R = 
90 V = 12 V
 15 R 
 3R 
V3R = 
90 V = 18 V
 15 R 
 4R 
V4R = 
90 V = 24 V
 15 R 
 5R 
V5R = 
90 V = 30 V
 15 R 
42.
VAF = 100 V
R 
 86.6 k 
VBF =  BF VAF  
100 V = 79.7 V
 108.6 k 
 RAF 
R 
 76.6 k 
VCF =  CF VAF  
100 V = 70.5 V
 108.6 k 
 RAF 
R 
 20.6 k 
VDF =  DF VAF  
100 V = 19.0 V
 108.6 k 
 RAF 
R
VEF =  EF
 RAF
43.

 5.6 k 
VAF  
100 V = 5.16 V
 108.6 k 

V1
10 V

= 1.79 mA
R1 5.6 k
V2 = IR2 = (1.79 mA)(1 k) = 1.79 V
V3 = IR3 = (1.79 mA)(560 ) = 1.0 V
V4 = IR4 = (1.79 mA)(10 k) = 17.9 V
I=
33
Chapter 5
44.
See Figure 5-5 for one possible solution:
RT = 18 k + 33 k + 22 k + 27 k = 100 k
30 V
IT =
= 300 A
100 k
 82 k 
VA = 
30 V = 24.6 V
 100 k 
 49 k 
VB = 
30 V = 14.7 V
 100 k 
 27 k 
VC = 
30 V = 8.1 V
 100 k 
Figure 5-5
P1 = I T2 R1 = (300 A)2 27 k = 2.43 mW
P2 = I T2 R2 = (300 A)2 22 k = 1.98 mW
P3 = I T2 R3 = (300 A)2 33 k = 2.97 mW
P4 = I T2 R4 = (300 A)2 18 k = 1.62 mW
All resistors can be 1/8 W.
45.
See Figure 5-6 for one possible solution.
RT = 12.1 k
 10.1 k 
VOUT(max) = 
120 V = 100.2 V
 12.1 k 
 1 k 
VOUT(min) = 
120 V = 9.92 V
 12.1 k 
These values are within 1% of the specified values.
120 V 120 V

IMAX =
= 9.9 mA
RT
12.1 k
Figure 5-6
Section 5-8 Power in Series Circuits
46.
PT = 5(50 mW) = 250 mW
47.
VT = V1 + V2 + V3 + V4 = 10 V + 1.79 V + 1 V + 17.9 V = 30.69 V
PT = VTI = (30.69 V)(1.79 mA) = 54. 9 mW
48.
Since P = I2R and since each resistor has the same current, the 5.6 k resistor is the limiting
element in terms of power dissipation.
0.25 W
Pmax
= 6.68 mA

5.6 k
5.6 k
V5.6 k = (6.68 mA)(5.6 k) = 37.4 V
V1.2 k = (6.68 mA)(1.2 k) = 8.02 V
V2.2 k = (6.68 mA)(2.2 k) = 14.7 V
V3.9 k = (6.68 mA)(3.9 k) = 26.1 V
VT(max) = 37.4 V + 8.02 V + 14.7 V + 26.1 V = 86.2 V
Imax =
34
Chapter 5
49.
50.
12 V
V1
= 2.14 A

R1 5.6 M
V
4.8 V
R2 = 2 
= 2.2 M
I
2.14 A
P3 = I2R3
P
21.5 W
= 4.7 M
R3 = 32 
I
 214 A)2 
RT = R1 + R2 + R3 = 5.6 M + 2.2 M + 4.7 M = 12.5 M
I=
(a)
P = I2R
P
R= 2
I
R1 + R2 + R3 = 2400 
1  1  1 
 W  W  W
 8    4    2  = 2400 
I2
I2
I2
7 
 W
 8  = 2400 
I2
7 
 W
8 
2
= 0.0003646 A2
I = 
2400 
I=
0.0003646 A 2 = 19.1 mA
(b)
VT = IRT = (19.1 mA)(2400 ) = 45.8 V
(c)
R1 =
0.125 W
P1
= 343 

2
I
(19.1 mA)2
0.25 W
P
R2 = 22 
= 686 
(19.1 mA)2
I
0.5 W
P
R3 = 32 
= 1.37 k
I
(19.1 mA)2
Section 5-9 Voltage Measurements
51.
VAG = 100 V (voltage from point A to ground)
Resistance between A and C:
RAC = 5.6 k + 5.6 k = 11.2 k
Resistance between C and ground:
RCG = 1 k + 1 k = 2 k
35
Chapter 5
 2 k 
VCG = 
100 V = 15.2 V
 13.2 k 
 1 k 
 1 k 
VDG = 
VCG  
15.2 V = 7.58 V
 2 k 
 2 k 
 11.2 k 
VAC = 
100 V = 84.9 V
 13.2 k 
 5.6 k 
 5.6 k 
VBC = 
VAC  
84.9 V = 42.5 V
 11.2 k 
 11.2 k 
VBG = VCG + VBC = 15.2 V + 42.5 V = 57.7 V
52.
Measure the voltage at point A with respect to ground and the voltage at point B with respect to
ground. The difference is VR2.
VR2 = VB  VA
53.
RT = R1 + R2 + R3 + R4 + R5 = 56 k + 560 k + 100 k + 1 M + 100 k = 1.816 M
VT = 15 V  9 V = 6 V
V
6V
I= T 
= 3.3 A
RT
1.816 M
V1 = IR1 = (3.3 A)(56 k) = 185 mV
VA = 15 V  V1 = 15 V  185 mV = 14.82 V
V2 = IR2 = (3.3 A)(560 k) = 1.85 V
VB = VA  V2 = 14.82 V  1.85 V = 12.97 V
V3 = IR3 = (3.3 A)(100 k) = 330 mV
VC = VB  V3 = 12.97 V  330 mV = 12.64 V
V4 = IR4 = (3.3 A)(1 M) = 3.3 V
VD = VC  V4 = 12.64 V  3.3 V = 9.34 V
54.
VAC = VA – VC = 14.82 V – 12.64 V = 2.18 V
55.
VCA = VC – VA = 12.64 V – 14.82 V = -2.18 V
Section 5-10 Troubleshooting
56.
There is no current through the resistors which have zero volts across them; thus, there is an
open in the circuit. Since R2 has voltage across it, it is the open resistor.
12 V will be measured across R2.
57.
(a)
(b)
Zero current indicates an open. R4 is open since all the voltage is dropped across it.
VS
10 V

= 33.3 mA
R1  R2  R3 300 
R4 and R5 have no effect on the current.
There is a short from A to B, shorting out R4 and R5.
36
Chapter 5
58.
R2 = 0 
RT = R1 + R3 + R4 + R5 = 400 
V
10 V
IT = S 
= 25 mA
RT 400 
59.
The results in Table 5-1 are correct.
60.
If 15 k is measured between pins 5 and 6, R3 and R5 are shorted as indicated in Figure 5-7.
61.
In this case, there is a short between the points indicated in Figure 5-7.
Figure 5-7
62.
(a)
(b)
(c)
R11 has burned out because it has the highest resistance value (P = I2R).
Replace R11 (10 k).
RT = 47.73 k
0.5 W
P11

Imax =
= 7.07 mA
R11
10 k
Vmax = ImaxRT = (7.07 mA)(10 k) = 70.7 V
Multisim Troubleshooting and Analysis
63.
7.481 k
66.
6V
64.
R2 is open.
67.
R1 is shorted
65.
R3 = 22 
37
Chapter 6
Parallel Circuits
Note: Solutions show conventional current direction.
Section 6-1 Resistors in Parallel
1.
See Figure 6-1.
Figure 6-1
2.
R1, R2 and R5 are not individually in parallel with the other resistors. The series combination of
R1, R2, and R5 is in parallel with the other resistors.
3.
R1, R2, R5, R9, R10 and R12 are in parallel.
R4, R6, R7, and R8 are in parallel.
R3 and R11 are in parallel.
Section 6-2 Voltage in a Parallel Circuit
4.
5.
V1 = V2 = V3 = V4 = 12 V
V
12 V
IT = T 
= 21.8 mA
RT 550 
The total current divides equally among the four equal parallel resistors.
21.8 mA
= 5.45 mA
I1 = I2 = I3 = I4 =
4
The resistors are all in parallel across the source. The voltmeters each measure the voltage
across a resistor, so each meter indicates 100 V.
6.
Position A: RT = R1  R4 = (1.0 k)  (2.7 k) = 730 
Position B: RT = R1  R3 = (1.0 k)  (2.2 k) = 688 
Position C: RT = R1  R2 = (1.0 k)  (1.8 k) = 643 
7.
Position A:
V1 = 15 V, V2 = 0 V, V3 = 0 V, V4 = 15 V
Position B:
V1 = 15 V, V2 = 0 V, V3 = 15 V, V4 = 0 V
Position C:
V1 = 15 V, V2 = 15 V, V3 = 0 V, V4 = 0 V
38
Chapter 6
8.
15V
= 20.6 mA
730 
15V
Position B: IT =
= 21.8 mA
688 
15V
= 23.3 mA
Position C: IT =
643 
Position A: IT =
Section 6-3 Kirchhoff’s Current Law
9.
IT = 250 mA + 300 mA + 800 mA = 1350 mA = 1.35 A
10.
IT = I1 + I2 + I3 + I4 + I5
I5 = IT  (I1 + I2 + I3 + I4)
= 500 mA  (50 mA + 150 mA + 25 mA + 100 mA) = 500 mA  325 mA = 175 mA
11.
VS = I1R1 = (1 mA)(47 ) = 47 mV
V
47 mV
R2 = S 
= 22 
I 2 2.14 mA
V
47 mV
= 100 
R3 = S 
I 3 0.47 mA
I4 = IT  (I1 + I2 + I3) = 5.03 mA  3.61 mA = 1.42 mA
V
47 mV
R4 = S 
= 33 
I 4 1.42 mA
12.
IT = 1.25 A + 0.833 A + 0.833 A + 10 A = 12.92 A
I4 = 15 A  12.92 A = 2.08 A
See Figure 6-2.
Figure 6-2
13.
VT = ITRT = (100 mA)(25 ) = 2500 mV = 2.5 V
VT
2.5 V
I220  =

= 11.4 mA
220  220 
14.
IT = 4IRUN + 2ITAIL
= 4(0.5A) + 2(1.2 A) = 4.4 A
39
Chapter 6
15.
(a) IT = 4IRUN + 2ITAIL + 2IBRAKE = 4(0.5A) + 2(1.2 A) + 2(1 A) = 6.4 A
(b) IGND = IT = 6.4 A
Section 6-4 Total Parallel Resistance
16.
RT =
17.
(a)
(b)
(c)
(d)
18.
(a)
(b)
(c)
1
= 568 k
1
1
1
1
1




1 M 2.2 M 5.6 M 12 M 22 M
(560 )(1 k)
= 359 
560   1 k
(47 )(56 )
RT =
= 25.6 
47   56 
1
RT =
= 819 
1
1
1


1.5 k 2.2 k 10 k
1
RT =
= 997 
1
1
1
1



1 M 470 k 1 k 2.7 M
RT =
(560 )(220 )
= 158 
560   220 
(27 k)(56 k)
RT =
= 18.2 k
27 k  56 k
(1.5 k)(2.2 k)
= 892 
RT =
1.5 k  2.2 k
RT =
6.8 k
= 0.567 k = 567 
12
19.
RT =
20.
Five 470  resistors in parallel:
470 
R1 =
= 94 
5
Ten 1000  resistors in parallel:
1000 
= 100 
R2 =
10
Two 100  resistors in parallel:
100 
R3 =
= 50 
2
21.
RT =
1
= 24.6 
1
1
1


94  100  50 
40
Chapter 6
R1R2
R1  R2
RT(R1 + R2) = R1R2
RTR1 + RTR2 = R1R2
RTR1 = R1R2  RTR2
RTR1 = R2(R1  RT)
RT R1
(389.2 )(680 )

R2 =
= 910 
R1  RT 680   389.2 
22.
RT =
23.
(a)
RT = R1 = 510 k
(b)
RT = R1  R2 =
(c)
1
= 245 k
1
1

510 k 470 k
RT = R1 = 510 k
1
RT = R1  R2  R3 =
= 193 k
1
1
1


510 k 470 k 910 k
Section 6-5 Application of Ohm’s Law
24.
(a)
(b)
R 240 

= 80 
3
3
120 V
IT =
= 1.5 A
80 
25.
RT =
26.
RT =
27.
1
= 10.2 
1
1
1


33  33  27 
V
10 V
IT =

= 980 mA
RT 10.2 
1
= 334 
RT =
1
1
1


1 k 4.7 k 560 
V
25 V

IT =
= 74.9 mA
RT 334 
RT =
VS
5V

= 4.5 k
I T 1.11 mA
Reach = 4RT = 4(4.5 k) = 18 k
VS
110 V
= 50 mA
Rfilament 2.2 k
When one bulb burns out, the others remain on.
I=

41
Chapter 6
28.
(a)
I2 = IT  I1 = 150 mA  100 mA = 50 mA
10 V
= 100 
R1 =
100 mA
10 V
R2 =
= 200 
50 mA
(b)
I3 =
100 V
= 100 mA
1 k
100 V
= 147 mA
I2 =
680 
I1 = IT  I2  I3 = 500 mA  247 mA = 253 mA
100 V
= 395 
R1 =
253 mA
29.
Imax = 0.5 A
15 V 15 V

RT(min) =
= 30 
I max 0.5 A
(68 ) Rx
= RT(min)
68   Rx
(68 )Rx = (30 )(68  + Rx)
68Rx = 2040 + 30Rx
68Rx  30Rx = 2040
38Rx = 2040
Rx = 53.7 
30.
Position A:
24 V
= 42.9 A
I1 =
560 k
24 V
= 109 A
I2 =
220 k
24 V
I3 =
= 88.9 A
270 k
IT = 42.9 A + 109 A + 88.9 A = 241 A
Position B:
I1 = 42.9 A
I2 = 109 A
I3 = 88.9 A
24 V
= 24 A
I4 =
1 M
24 V
= 29.3 A
I5 =
820 k
24 V
I6 =
= 10.9 A
2.2 M
IT = 42.9 A + 109 A + 88.9 A + 24 A + 29.3 A + 10.9 A = 305 A
42
Chapter 6
Position C:
I4 = 24 A
I5 = 29.3 A
I6 = 10.9 A
IT = 24 A + 29.3 A + 10.9 A = 64.2 A
31.
100 V
= 83.3 mA
1.2 k
I2 = 250 mA  83.3 mA = 166.7 mA
IT = 250 mA + 50 mA = 300 mA
100 V
R1 =
= 2 k
50 mA
100 V
= 600 
R2 =
166.7 mA
I3 =
Section 6-6 Current Sources in Parallel
32.
(a)
(b)
(c)
IL = 1 mA + 2 mA = 3 mA
IL = 50 A  40 A = 10 A
IL = 1 A  2.5 A + 2 A = 0.5 A
33.
Position A: IR = 2.25 mA
Position B: IR = 4.75 mA
Position C: IR = 4.75 mA + 2.25 mA = 7 mA
Section 6-7 Current Dividers
34.
35.
 R2 
 2.7 k 
I1 = 
 IT  
 3 A = 2.19 A
 3.7 k 
 R1  R2 
 R1 
 1 k 
I2 = 
 IT  
 3 A = 0.811 A
 3.7 k 
 R1  R2 
(a)
(b)
 R2 
 2.2 M 
I1 = 
 IT  
10 A = 6.88 A
 3.2 M 
 R1  R2 
I2 = IT  I1 = 10 A  6.88 A = 3.12 A
R 
Ix =  T  I T
 Rx 
RT = 525 
 525  
10 mA = 5.25 mA
I1 = 
 1000  
 525  
10 mA = 2.39 mA
I2 = 
 2.2 k 
43
Chapter 6
 525  
10 mA = 1.59 mA
I3 = 
 3 .3 k  
 525  
10 mA = 0.772 mA
I4 = 
 6.8 k 
36.
37.
 1 1 1
= R / 1     = 0.48R
1
1
1
1
 2 3 4



R 2 R 3R 4 R
 RT 
 RT 
 0.48 R 
 0.48 R 
IR =  10 mA  
10 mA = 4.8 mA; I2R = 
10 mA = 2.4 mA;
10 mA  
 R 
 2R 
 R 
 2R 
R 
 RT 
 0.48R 
 0.48 R 
I3R =  T 10 mA  
10 mA = 1.59 mA; I4R = 
10 mA = 1.2 mA
10 mA  
 3R 
 4R 
 3R 
 4R 
1
RT =
RT = 773 
I3 = IT  I1  I2  I3 = 15.53 mA  3.64 mA  6.67 mA  3.08 mA = 2.14 mA
R 
I1 =  T  I T
 R1 
R
R1 =  T
 I1
R
R2 =  T
 I2
R
R3 =  T
 I3
R
R4 =  T
 I4
38.

 IT


 IT


 IT


 IT

 773  

15.53 mA = 3.3 k
 3.64 mA 
 773  

15.53 mA = 1.8 k
 6.67 mA 
 773  

15.53 mA = 5.6 k
 2.14 mA 
 773  

15.53 mA = 3.9 k
 3.08 mA 
(a)
IT = 10 mA, IM = 1 mA
VM = IMRM = (1 mA)(50 ) = 50 mV
ISH1 = 9 mA
V
50 mV
RSH1 = M 
= 5.56 
I SH1 9 mA
(b)
IT = 100 mA, IM = 1 mA
VM = IMRM = (1 mA)(50 ) = 50 mV
ISH2 = 99 mA
V
50 mV
RSH2 = M 
= 0.505 
I SH2 99 mA
44
Chapter 6
39.
(a)
(b)
50 mV
= 1 m
50 A
50 mV
ISH =
= 50 A
1 m
50 mV
Imeter =
= 5 A
10 k
RSH =
Section 6-8 Power in Parallel Circuits
40.
PT = 5(250 mW) = 1.25 W
41.
(a)
RT =
(b)
RT =
42.
(1 M)(2.2 M)
= 687.5 k
1 M  2.2 M
PT = I 2RT = (10 A)2(687.5 k) = 68.8 W
1
1

= 525 
1
1
1
1
1
1
1
1






R1 R2 R3 R4 1 k 2.2 k 3.3 k 6.8 k
PT = I 2RT = (10 mA)2(525 ) = 52.5 mW
P = VI
Ieach =
P 75 W

= 625 mA
V 120 V
IT = 6(625 mA) = 3.75 A
43.
P1 = PT  P2 = 2 W  0.75 W = 1.25 W
P
2W
VS = T 
= 10 V
I T 200 mA
P 0.75 W
I2 = 2 
= 75 mA
10 V
VS
V
10 V
R2 = S 
= 133 
I 2 75 mA
I1 = IT  I2 = 200 mA  75 mA = 125 mA
V
10 V
R1 = S 
= 80 
I1 125 mA
44.
(a)
(b)
PT = I T2 RT = (50 mA)2 1 k = 2.5 W
P
2.5 W
Number of resistors = n = T 
= 10
Peach 0.25 W
R
RT =
n
R = nRT = 10(1 k) = 10 k
45
Chapter 6
I T 50 mA

= 5 mA
10
n
(c)
I=
(d)
VS = ITRT = (50 mA)(1 k) = 50 V
Section 6-10 Troubleshooting
45.
Ieach =
P 75 W
= 625 mA

V 120 V
IT = 5(625 mA) = 3.13 A
46.
1
= 47.5 
1
1
1
1
1




220  100  1 k 560  270 
10 V
= 210.5 mA
IT =
47.5 
The measured current is 200.4 mA, which is 10.1 mA less than it should be. Therefore, one of
the resistors is open.
RT =
V
10 V
= 990   1 k

I 10.1 mA
The 1 k resistor (R3) is open.
1
RT =
= 2.3 k
1
1
1


4.7 k 10 k 8.2 k
25 V
IT =
= 10.87 mA
2.3 k
The meter indicates 7.82 mA. Therefore, a resistor must be open.
25 V
I3 =
= 3.05 mA
8.2 k
I = IT  IM = 10.87 mA  7.82 = 3.05 mA
This shows that I3 is missing from the total current as read on the meter. Therefore,
R3 (8.2 k) is open.
R? =
47.
48.
25 V
= 5.32 mA
4.7 k
25 V
I2 =
= 2.5 mA
10 k
25 V
I3 =
= 3.05 mA
8.2 k
R1 is open producing a total current of
IT = I2 + I3 = 2.5 mA + 3.05 mA = 5.55 mA
I1 =
46
Chapter 6
49.
Connect ohmmeter between the following pins:
Pins 1-2
Correct reading: R = 1 k  3.3 k = 767 
R1 open: R = 3.3 k
R2 open: R = 1 k
Pins 3-4
Correct reading: R = 270   390  = 159.5 
R3 open: R = 390 
R4 open: R = 270 
Pins 5-6
Correct reading: R = 1 M  1.8 M  680 k  510 k = 201 k
R5 open: R = 1.8 M  680 k  510 k = 251 k
R6 open: R = 1 M  680 k  510 k = 226 k
R7 open: R = 1 M  1.8 M  510 k = 284 k
R8 open: R = 1 M  1.8 M  680 k = 330 k
50.
Short between pins 2 and 4:
(a)
R1-2 = R1  R2  R3  R4  R11  R12 + R5  R6  R7  R8  R9  R10
= 10 k  2.2 k  2.2 k  3.3 k  18 k  1 k + 4.7 k 4.7 k  6.8 k 
5.6 k  1 k  5.6 k = 940 
(b)
R2-3 = R5  R6  R7  R8  R9  R10 = 4.7 k  4.7 k  6.8 k  5.6 k  1 k 5.6 k
= 518 
(c)
R3-4 = R5  R6  R7  R8  R9  R10 = 4.7 k  4.7 k  6.8 k  5.6 k  1 k 5.6 k
= 518 
R1-4 = R1  R2  R3  R4  R11  R12 = 10 k  2.2 k  2.2 k  3.3 k  18 k 1 k
= 422 
(d)
51.
Short between pins 3 and 4:
(a)
R1-2 = (R1  R2  R3  R4  R11  R12) + (R5  R6  R7  R8  R9  R10) = 940 
(b) R2-3 = R5  R6  R7  R8  R9  R10 = 518 
(c)
R2-4 = R5  R6  R7  R8  R9  R10 = 518 
(d)
R1-4 = R1  R2  R3  R4  R11  R12 = 422 
Multisim Troubleshooting and Analysis
52.
RT = 547.97 
53.
R2 is open.
54.
R1 = 890 
55.
VS = 3.3 V
56.
R1 is open.
47
Chapter 7
Series-Parallel Circuits
Note: Solutions show conventional current direction.
Section 7-1 Identifying Series-Parallel Relationships
1.
See Figure 7-1.
Figure 7-1
2.
See Figure 7-2.
Figure 7-2
3.
(a)
(b)
(c)
R1 and R4 are in series with the parallel combination of R2 and R3.
R1 is in series with the parallel combination of R2, R3, and R4.
The parallel combination of R2 and R3 is in series with the parallel combination of R4 and
R5. This is all in parallel with R1.
4.
(a)
R2 is in series with the parallel combination of R3 and R4.
This series-parallel combination is in parallel with R1.
All of the resistors are in parallel.
R1 and R2 are in series with the parallel combination of R3 and R4.
R5 and R8 are in series with the parallel combination of R6 and R7.
These two series-parallel combinations are in parallel with each other.
(b)
(c)
48
Chapter 7
5.
See Figure 7-3.
Figure 7-3
6.
See Figure 7-4.
Figure 7-4
7.
See Figure 7-5.
Figure 7-5
49
Chapter 7
Section 7-2 Analysis of Series-Parallel Resistive Circuits
8.
9.
R1R2
R1  R2
R1 RT
(1 k)(667 )

R2 =
= 2.0 k
R1  RT 1 k  667 
RT =
(c)
R2
100 
= 133 
= 56  + 27  +
2
2
1
1
= 680  + 99.4  = 779 
RT = R1 
= 680  +
1
1
1
1
1
1




680  330  180 
R2 R3 R4
RT = R1  (R2  R3 + R4  R5) = R1  (2.154 k + 3.59 k) = 852 
(a)
RT = R1  (R2 + R3  R4) = 1 k  (1 k + 2.2 k  3.3 k) = 699 
(b)
RT =
(c)
RA = R1 + R2 +
(a)
IT =
(a)
(b)
10.
11.
RT = R1 + R4 +
1
= 406 k
1
1
1
1



1 M 1 M 3.3 M 6.2 M
R3 R4
(10 k)(4.7 k)
= 1 k + 1 k +
= 5.2 k
R3  R4
10 k  4.7 k
R6 R7
6 .8 k 
RB = R5 + R8 +
= 8.5 k
= 3.3 k + 1.8 k +
R6  R7
2
1
1

RT =
= 3.23 k
1
1
1
1


RA RB
5.2 k 8.5 k
1.5 V
= 11.3 mA
133 
I1 = I4 = 11.3 mA
11.3 mA
I2 = I3 =
= 5.64 mA
2
V1 = (11.3 mA)(56 ) = 633 mV
V4 = (11.3 mA)(27 ) = 305 mV
V2 = V3 = (5.64 mA)(100 ) = 564 mV
50
Chapter 7
12.
3V
= 3.85 mA
779 
V1 = (3.85 mA)(680 ) = 2.62 V
V2 = V3 = V4 = VS  ITR1 = 3 V  (3.85 mA)(680 ) = 383 mV
I1 = IT = 3.85 mA
V
383 mV
I2 = 2 
= 563 A
R2
680 
V
383 mV
I3 = 3 
= 1.16 mA
R3
330 
V
383 mV
I4 = 4 
= 2.13 mA
R4
180 
(b)
IT =
(c)
I1 =
(a)
IT =
5V
5V
Iright =
= 5 mA
= 871 A
1 k
5.74 k
 3.3 k 
I2 = 
871 A = 303 A
 9.5 k 
 6.2 k 
I3 = 
871 A = 568 A
 9.5 k 
 5.6 k 
I4 = 
871 A = 313 A
 15.6 k 
 10 k 
I5 = 
871 A = 558 A
 15.6 k 
V1 = VS = 5 V
V2 = V3 = (303 A)(6.2 k) = 1.88 V
V4 = V5 = (313 A)(10 k) = 3.13 V
1V
= 1.43 mA
699 
 2.32 k 
I1 = 
1.43 mA = 1 mA
 3.32 k 
V1 = (1 mA)(1 k) = 1 V
 1 k 
I2 = 
1.43 mA = 431 A
 3.32 k 
V2 = (431 A)(1 k) = 431 mV
 3.3 k 
I3 = 
431 A = 259 A
 5.5 k 
V3 = (259 A)(2.2 k) = 570 mV
V4 = V3 = 570 mV
570 mV
I4 =
= 173 A
3.3 k
51
Chapter 7
(b)
V1 = V2 = V3 = V4 = 2 V
2V
I1 =
= 2 A
1 M
2V
I2 =
= 606 nA
3.3 M
2V
I3 =
= 323 nA
6.2 M
2V
I4 =
= 2 A
1 M
(c)
IT =
5V
= 1.55 mA
3.23 k
 5.2 k 
I5= 
1.55 mA = 588 A
 13.7 k 
V5 = (588 A)(3.3 k) = 1.94 V
I
588 A
I6 = I7 = 5 
= 294 A

2
V6 = V7 = (294 A)(6.8 k) = 2 V
I8 = I5 = 588 A
V8 = (588 A)(1.8 k) = 1.06 V
 8.5 k 
I1 = I2 = 
1.55 mA = 962 A
 13.7 k 
V1 = V2 = (962 A)(1 k) = 962 mV
 4.7 k 
I3 = 
962 A = 308 A
 14.7 k 
V3 = V4 = (308 A)(10 k) = 3.08 V
 10 k 
I4 = 
962 A = 654 A
 14.7 k 
13.
SW1 closed, SW2 open:
RT = R2 = 220 
SW1 closed, SW2 closed:
RT = R2  R3 = 220   2.2 k = 200 
SW1 open, SW2 open:
RT = R1 + R2 = 100  + 220  = 320 
SW1 open, SW2 closed:
RT = R1 + R2  R3 = 100  + 200  = 300 
14.
RAB = (10 k + 5.6 k)  4.7 k = 15.6 k  4.7 k = 3.61 k
The 1.8 k and the two 1 ks are shorted).
52
Chapter 7
15.
16.
VAG = 100 V
RAC = (4.7 k + 5.6 k)  10 k = 5.07 k
RCG = 2 k  1.8 k = 947 
 5.07 k 
VAC = 
100 V = 84.2 V
 6.02 k 
 947  
VCG = 
100 V = 15.7 V
 6.02 k 
 1 k 
 1 k 
VDG = 
VCG  
15.7 V = 7.87 V
 2 k 
 2 k 
 5.6 k 
 5 .6 k  
VBC = 
VAC  
84.2 V = 45.8 V
 10.3 k 
 10.3 k 
VBG = VCG + VBC = 15.7 V + 45.8 V = 61.5 V
 56 k 
VA = 
50 V = 3.91 V
 716 k 
VC = 50 V
 616 k 
VB = 
50 V = 43.0 V
 716 k 
 100 k 
VD = 
50 V = 4.55 V
 1.1 M 
17.
Measure the voltage at point A with respect to ground and the voltage at point B with respect to
ground. The difference is VR2.
VR2 = VB  VA
18.
RT = (10 k  (4.7 k + 5.6 k)) + (1.8 k  (1 k + 1 k))
= 10 k  10.3 k + 1.8 k  2 k
= 5.07 k + 947 k = 6.02 k
19.
RT = (R1 + R2 + R3)  R4  (R5 + R6)
= (100 k + 560 k + 56 k)  1.0 M  (1.0 M + 100 k)
= 716 k  1.0 M  1.1 M = 303 k
20.
Resistance of the right branch:
RR = R2 + R5  R6 + R7 + R8 = 330  + 600  + 680  + 100  = 1710 
Resistance of the left branch:
RL = R3 + R4 = 470  + 560  = 1030 
Total resistance:
RT = R1 + RL  RR = 1 k + 643  = 1.64 k
100 V
IT =
= 60.9 mA
1.64 k
Current in the right branch:
 RL 
 1030  
IR = 
 IT  
 60.9 mA = 22.9 mA
 2740  
 RL  RR 
Current in the left branch:
 RR 
 1710  
IL = 
 IT  
 60.9 mA = 38.0 mA
 2740  
 RL  RR 
53
Chapter 7
With respect to the negative source terminal:
VA = ILR4 = (38.0 mA)(560 ) = 21.3 V
VB = IR(R7 + R8) = (22.9 mA)(780 ) = 17.9 V
VAB = VA  VB = 21.3 V  17.9 V = 3.4 V
21.
(a)
 R1 
I2 = 
 IT
 R1  R2 
 47 k 
1 mA = 
 IT
 47 k  R2 
47 k + R2 = (47 k)IT
Also,
V
220
IT =

RT 33 k  (47 k) R2
(47 k)  R2
Substituting the expression for IT into 47 k + R2 = (47 k)IT.




220

47 k + R2 = 47 k 
 33 k  (47 k) R2 

47 k  R2 


(47 k) R2 
(47 k + R2)  33 k 
 = 47 k(220)
47 k  R2 

(80 k)R2 = 47 k(220)  (47 k)(33 k)
47 k(220  33 k)
R2 =
= 109.9 k  110 k
80 k
(b)
P2 = I 22 R2 = (1 mA)2 110 k = 0.11 W = 110 mW
22.
RAB = R1  (R2 + R7 + R8) = 1 k  (2.2 k + 3.3 k + 4.7 k) = 1 k  10.2 k = 911 
RAG = R8  (R1 + R2 + R7) = 4.7 k  (1 k + 2.2 k + 3.3 k) = 4.7 k  6.5 k = 2.73 k
RAC = (R1 + R2)  (R7 + R8) = (1 k + 2.2 k)  (3.3 k + 4.7 k) = 3.2 k  8 k = 2.29 k
RAD = RAC + R3  (R4 + R5 + R6) = 2.29 k + 1 k  10.2 k = 3.20 k
RAE = RAC + (R3 + R4)  (R5 + R6) = 2.29 k + 3.2 k  8 k = 4.58 k
RAF = RAC + R6  (R3 + R4 + R5) = 2.29 k + 4.7 k  6.5 k = 5.02 k
23.
RAB = (R1 + R2)  R4  R3 = 6.6 k  3.3 k  3.3 k = 1.32 k
Note: R5 and R6 is shorted out (ACD) and is not a factor in the total resistance.
RBC = R4  (R1 + R2)  R3 = 1.32 k
RCD = 0 
54
Chapter 7
24.
V2 = V5  V6 = 5 V  1 V = 4 V
2W
= 0.5 A
I2 = I6 =
4V
I5 = I8  I6 = 1 A  0.5 A = 0.5 A
I1 = I2 + I5 + I4 = 0.5 A + 0.5 A + 1 A = 2 A
I3 = IT  I1 = 4 A  2 A = 2 A
V7 = VS  V3 = 40 V  20 V = 20 V
20 W
= 10 V
V1 =
2A
V4 = V3  V1 = 10 V
V8 = V4  V5 = 5 V
R1 =
R2 =
R3 =
R4 =
R5 =
R6 =
10 V
=5
2A
4V
=8
0.5 A
20 V
= 10 
2A
10 V
= 10 
1A
5V
= 10 
0.5 A
1V
=2
0.5 A
20 V
=5
4A
5V
R8 =
=5
1A
R7 =
Section 7-3 Voltage Dividers with Resistive Loads
25.
26.
 56 k 
VOUT(unloaded) = 
15 V = 7.5 V
 112 k 
56 k in parallel with a 1 M load is
(56 k)(1 M)
= 53 k
Req =
56 k  1 M
 56 k 
VOUT(loaded) = 
15 V = 7.29 V
 109 k 
See Figure 7-6.
 6.6 k 
VA = 
12 V = 8 V
 9.9 k 
 3.3 k 
VB = 
12 V = 4 V
 9.9 k 
With a 10 k resistor connected from tap A to ground:
(6.6 k)(10 k)
RAB =
= 3.98 k
6.6 k  10 k
 3.98 k 
VA(loaded) = 
12 V = 6.56 V
 7.28  
Figure 7-6
27.
The 47 k will result in a smaller decrease in output voltage because it has less effect on the
circuit resistance than does the smaller resistance.
55
Chapter 7
28.
RT = 10 k + 5.6 k + 2.7 k = 18.3 k
 R2  R3 
 8.3 k 
VOUT(NL) = 
 22 V = 9.98 V
VS = 
 18.3 k 
 R1  R2  R3 
With a 100 k load:
( R2  R3 ) RL
(8.3 k)(100 k)
RT = R1 +
= 17.7 k
= 10 k +
108.3 k
R2  R3  RL
 7.7 k 
VOUT = 
 22 V = 9.57 V
 17.7 k 
(8.3 k)(33 k)
= 6.63 k
8.3 k  33 k
6.63 k


VAB = 
 22 V = 8.77 V



10
k
6.63
k


29.
RAB =
30.
RT = 10 k + 5.6 k + 2.7 k = 18.3 k
22 V
= 1.2 mA
I=
18.3 k
(8.3 k)(33 k)
RT = 10 k +
= 16.6 k
8.3 k  33 k
22 V
= 1.33 mA
I=
16.6 k
31.
See Figure 7-7.
10 V
= 2 k
RT =
5 mA
R1 = R2 + R3
R2 = R3
R1 = 2R2
R1 + 2R2 = 2 k
2R2 + 2R2 = 2 k
4R2 = 2 k
R2 = R3 = 500 
R1 = R2 + R3 = 1000 
Figure 7-7
With a 1 k load on the lower tap:
1 k  500  = 333 
10 V
= 5.46 mA
IT =
1 k  500   333 
Vlower tap = (333 )(5.46 mA) = 1.82 V
Vupper tap = (500  + 333 )(5.46 mA) = 4.55 V
56
With a 1 k load on the upper tap:
10 V
IT =
= 6.67 mA
1 k  1 k / 2
Vupper tap = (500 )(6.67 mA) = 3.33 V
3.33 V
= 1.67 V
Vlower tap =
2
Chapter 7
32.
Position 1:
RT = 10 k + 30 k  68 k = 10 k + 20.82 k = 30.8 k
 20.8 k 
V1 = 
120 V = 81.0 V
 30.8 k 
 20 k 
V2 = 
81 V = 54.0 V
 30 k 
 10 k 
V3 = 
81 V = 27.0 V
 30 k 
Position 2:
RT = 20 k + 20 k  68 k = 20 k + 15.5 k = 35.5 k
 10 k  15.5 k 
V1 = 
120 V = 86.2 V
35.5 k



15.5
k


V2 = 
81 V = 52.4 V

35
.
5
k


10
k



V3 = 
52.4 V = 26.2 V
20
k



Position 3:
RT = 30 k + 10 k  68 k = 30 k + 8.72 k = 38.7 k
 20 k  8.72 k 
V1 = 
120 V = 89.0 V
38.7 k


 10 k  8.72 k 
V2 = 
81 V = 58.0 V
38.7 k


 8.72 k 
V3 = 
81 V = 27.0 V
 38.7 k 
33.
(a)
 R2 
 270 k 
VG = 
VDD  
16 V = 1.75 V
 2.47 M 
 R1  R2 
VS = VG + 1.5 V = 1.75 V + 1.5 V = 3.25 V
VDD  VG 16 V  1.75 V

= 6.48 A
R1
2.2 M
V
1.75 V
I2 = I1 = G 
= 6.48 A
R2 270 k
V
3.25 V
IS = S 
= 2.17 mA
RS 1.5 k
ID = IS = 2.17 mA
(b)
I1 =
(c)
VD = VDD  IDRD = 16 V  (2.17 mA)(4.7 k) = 16 V  10.2 V = 5.8 V
VDS = VD  VS = 5.8 V  3.25 V = 2.55 V
VDG = VD  VG = 5.8 V  1.75 V = 4.05 V
57
Chapter 7
34.
Imax = 100 mA
24 V
= 240 
RT =
100 mA
 R2 
  24 V = 6 V
 RT 
24R2 = 6RT
6(240 )
R2 =
= 60 
24
R1 = 240   60  = 180 
With load:
R2  RL = 60   1000  = 56.6 


56.6 
24 V = 5.74 V
VOUT = 
 180   56.6  
Section 7-4 Loading Effect of a Voltmeter
35.
The voltmeter presents the least load when set on the 1000 V range.
For example, assuming 20,000 /V:
Rinternal = (20,000 /V)(1 V) = 20 k on the 1 V range
Rinternal = (20,000 /V)(1000 V) = 20 M on the 1000 V range
36.
(a)
(b)
(c)
(d)
(e)
(f)
37.
Rinternal = (20,000 /V)(0.5 V) = 10 k
Rinternal = (20,000 /V)(1 V) = 20 k
Rinternal = (20,000 /V)(5 V) = 100 k
Rinternal = (20,000 /V)(50 V) = 1 M
Rinternal = (20,000 /V)(100 V) = 2 M
Rinternal = (20,000 /V)(1000 V) = 20 M




R4
1.5 V   27  1.5 V = 0.305 V actual
VR4  
 133  
R R R R 


2
3
4 
 1
(a)
Use the 0.5 V range to measure 0.305 V.
(b)
Rinternal = (20,000 /V)(0.5 V) = 10 k
27   10 k = 26.93 
 26.93  
1.5 V  0.304 V with meter connected
VR4  
 132.93  
0.305 V  0.304 V = 0.001 V less with meter
58
Chapter 7
38.
 R2 R3 R4 


3 V   99.4  3 V = 0.383 V actual
VR4  
 779.4  
R R R R 


1
 2 3 4
(a)
Use the 0.5 V range to measure 0.383 V.
(b)
Rinternal = (20,000 /V)(0.5 V) = 10 k
99.4   10 k = 98.4 
 98.4  
3 V = 0.379 V with meter connected
VR4  
 778.4  
0.383 V  0.379 V = 0.004 V less with meter
39.

RMETER = 10 V(10,000/V) = 100 k
R2
RMETER 
R2 RMETER
(100k)(100k)

 50k
R2  RMETER
200k

 R2 RMETER
VR2  
 R1  R2 RMETER
40.
R2
RMETER 

50k


 VS  
 VS  0.333VS
100

k


50

k




R2 RMETER
(100k)(10M)

 99k
R2  RMETER
10.1M
 R2 RMETER
VR2  
 R1  R2 RMETER

99k


 VS  
 VS  0.498V
 100k  99k 

Section 7-5 Ladder Networks
41.
The circuit in Figure 7-76 in the text is redrawn here in Figure 7-8 to make the analysis
simpler.
(a)
RT = 560   524.5  = 271 
(b)
IT =
(c)
 271  
221 mA = 114 mA
I2 = 
 524.5  
60 V
= 221 mA
271 
59
Chapter 7
 468.5  
114 mA = 58.7 mA
I910 = 
 910  
(d)
The voltage across the 437.5  parallel combination of the 560  and the two series
1 k resistors is determined as follows:
 468.5  
114 mA = 55 mA
I4 = 
 965.5  
V437.5  = I4(437.5 ) = (55 mA)(437.5 ) = 24.06 V
 1 k 
VAB = 
24.06 V = 12 V
 2 k 
Figure 7-8
42.
The total resistance is determined in the steps shown in Figure 7-9.
RT = 6.66 k
 1.06 k 
VA = 
18 V = 2.86 V
 6.66 k 
 1.05 k 
VB = 
2.86 V = 1.47 V
 2.05 k 
 1 k 
VC = 
1.47 V = 735 mV
 2 k 
Figure 7-9
60
Chapter 7
43.
The circuit is simplified in Figure 7-10 to determine RT.
RT = 621 
From Figure 7-10(e):
IT = I9 = IT = 16.1 mA
From Figure 7-10(c):
 420.8  
16.1 mA = 8.27 mA
I2 = 
 820  
From Figure 7-10(b):
 420.8  
16.1 mA = 7.84 mA
I3 = I8 = 
 864.5  
From Figure 7-10(a):
 424.5  
7.84 mA = 4.06 mA
I4 = 
 820  
From the original circuit:
I5 = I6 = I7 = I3  I4 = 7.84 mA  4.06 mA = 3.78 mA
Figure 7-10
44.
The currents were found in Problem 43.
V1 = ITR1 = (16.1 mA)(100 ) = 1.61 V
V2 = I2R2 = (8.27 mA)(820 ) = 6.78 V
V3 = I3R3 = (7.84 mA)(220 ) = 1.73 V
V4 = I4R4 = (4.06 mA)(820 ) = 3.33 V
V5 = I5R5 = (3.78 mA)(100 ) = 0.378 V
V6 = I6R6 = (3.78 mA)(680 ) = 2.57 V
V7 = I7R7 = (3.78 mA)(100 ) = 0.378 V
V8 = I8R8 = (7.84 mA)(220 ) = 1.73 V
V9 = I9R9 = (16.1 mA)(100 ) = 1.61 V
61
Chapter 7
45.
The two parallel ladder networks are identical; so, the voltage to ground from each output
terminal is the same; thus,
VOUT = 0 V.
Working from the right end, RT and then IT are determined as follows:
(12  + 12 )  18  = 10.3 
(22  + 10.3 )  27  = 14.7 
RT1 = 47  + 14.7  = 61.7 
R
RT(both) = T1  61.7  = 30.9 
2
2
30 V
= 971 mA
IT =
30.9 
46.
(a)
(b)
47.
(a)
(b)
(c)
V 12 V

= 1.5 V
8
8
V 12 V

VOUT =
= 0.75 V
16
16
VOUT =
V V 12 V 12 V
 

=3V+6V=9V
4 2
4
2
V V 12 V 12 V
VOUT = 


= 3 V + 0.75 V = 3.75 V
4 16
4
16
V V V V 12 V 12 V 12 V 12 V




VOUT =   
2 4 8 16
2
4
8
16
= 6 V + 3 V + 1.5 V + 0.75 V = 11.25 V
VOUT =
Section 7-6 The Wheatstone Bridge
48.
49.
50.
R 
Rx = RV  2  = (18 k)(0.02) = 360 
 R4 
SG3
119.94 




VLEFT = 
VS  
12 V = 5.997 V
 SG1 + SG3 
 120.06   119.94  
SG4
120.06 




VRIGHT = 
VS  
12 V = 6.003 V
 SG2 + SG4 
 119.94   120.06  
VOUT = VRIGHT  VLEFT = 6.003 V  5.997 V = 6 mV
(Right side positive with respect to left side)
At 60 C, RTHERM = 5 k
 R3 
 27 k 
VLEFT = 
VS  
 9 V = 7.59 V
 32 k 
 R1  R3 
 R4 
 27 k 
VRIGHT = 
VS  
 9 V = 4.50 V
 54 k 
 R2  R4 
VOUT = VLEFT  VRIGHT = 7.59 V  4.50 V = 3.09 V
62
Chapter 7
Section 7-7 Troubleshooting
(680 )(4.7 k)
= 594 
680   4.7 k
RT = 560  + 470  + 594  = 1624 
The voltmeter reading should be
 594  
12 V = 4.39 V
V? = 
 1624  
The voltmeter reading of 6.2 V is incorrect.
51.
Req =
52.
The circuit is redrawn in figure 7-11 and points are labeled.
(10 k  47 k)(100 k)
RBG =
= 36.3 k
10 k  47 k  100 k
RAG = 33 k + RBG = 33 k + 36.3 k = 69.3 k
RT = 27 k + RAG = 27 k + 69.3 k = 96.3 k
R 
 69.3 k 
VAG =  AG 18 V  
18 V = 12.95 V
 96.3 k 
 RT 
 47 k 
 47 k 
VCG = 
VBG  
6.79 V = 5.60 V
57
k



 57 k 
VAC = VAG  VCG = 12.95 V  5.60 V = 7.35 V
Both meters are correct.
Figure 7-11
53.
The 2.5 V reading indicated on one of the meters shows that the series-parallel branch
containing the other meter is open. The 0 V reading on the other meter shows that there is no
current in that branch. Therefore, if only one resistor is open, it must be the 2.2 k.
63
Chapter 7
54.
The circuit is redrawn in Figure 7-12.


12 k 12 k
150 V   6 k 150 V = 56.25 V
VA = 
 12 k 12 k  10 k 
 16 k 


The meter reading of 81.8 V is incorrect.
The most likely fault is an open 12 k resistor. This will cause the voltage at point A to be
higher than it should be. To verify, calculate VA assuming an open 12 k resistor.
 12 k 
VA = 
150 V = 81.8 V
 22 k 
 2.2 k 
VB = 
150 V = 42.3 V
 7 .8 k  
The meter is correct.
Figure 7-12
55.
56.
 1.62 k 
V3.3 k = 
(10 V) = 6.18 V
 2.62 k 
The 7.62 V reading is incorrect.
 2.2 k 
V2.2  = 
(6.18 V) = 4.25
 3.2 k 
The 5.24 V reading is incorrect.
The 3.3 k resistor must be open. If it is, then
 3.2 k 
V3.3 k = 
( 10 V) = 7.62 V
 4.2 k 
 2.2 k 
V2.2 k = 
(7.62 V) = 5.24 V
 3.2 k 
If R2 opens, VA = 15 V, VB = 0 V, and VC = 0 V
Multisim Troubleshooting and Analysis
57.
RT = 296.744 
58.
R4 is open.
59.
R3 = 560 k
60.
No fault.
61.
R5 is shorted.
62.
RX = 550 
64
Chapter 8
Circuit Theorems and Conversions
Note: Solutions show conventional current direction.
Section 8-3 Source Conversions
VS 300 V

=6A
RS 50 
RS = 50 
See Figure 8-1.
1.
IS =
2.
(a)
(b)
5 kV
= 50 A
100 
12 V
IS =
= 5.45 A
2.2 
Figure 8-1
IS =
1.6 V
= 0.2 
8.0 A
3.
RS =
4.
See Figure 8-2.
Figure 8-2
5.
VS = ISRS = (600 mA)(1.2 k) = 720 V
RS = 1.2 k
See Figure 8-3.
6.
(a)
(b)
VS = (10 mA)(4.7 k) = 47 V
VS = (0.01 A)(2.7 k) = 27 V
Figure 8-3
Section 8-4 The Superposition Theorem
7.
First, zero the 3 V source by replacing it with a short as in Figure 8-4(a).
RT = 1.955 k
2V
IT =
= 1.02 mA
1.955 k
 2.2 k 
I3 = 
1.02 mA = 577 A
 3.89 k 
65
Chapter 8
 1 k 
I5 = 
5.77 A = 180 A
 3.2 k 
Next, zero the 2 V source by replacing it with a short as in Figure 8-4(b).
RT = 1.955 k
3V
IT =
= 1.53 mA
1.955 k
 1.69 k 
I5 = 
1.53 mA = 655 A
 3.89 k 
Since both components of I5 are in the same direction, the total I5 is
I5(total) = 180 A + 665 A = 845 A
Figure 8-4
8.
From Problem 7:
RT = 1.955 k and IT = 1.02 mA
Current in R2 due to the 2 V source acting alone. See Figure 8-5(a):
 1.69 k 
I2 = 
1.02 mA = 443 A (downward)
 3.89 k 
From Problem 7:
RT = 1.955 k and IT = 1.53 mA
Current in R2 due to the 3 V source acting alone. See Figure 8-5(b):
 2.2 k 
ILeft = 
1.53 mA = 865 A
 3.89 k 
 1 k 
I2 = 
865 A = 270 A (downward)
 3.2 k 
The total current through R2 is
I2 = 443 A + 270 A = 713 A
Figure 8-5
66
Chapter 8
9.
From Problem 7:
From the 2 V source:
I4 = I3 – I5 = 577 A – 180 A = 397 A downward through R4
From the 3 V source:
I4 = IT = 1.53 mA upward through R4
I4(TOT) = 1.53 mA – 397 A = 1.13 mA upward
10.
First, zero the voltage source by replacing it with a short as shown in Figure 8-6(a):
 680  
100 mA = 79.8 mA
 852.6  
I1 = 
 220  
 79.8 mA = 17.2 mA
 1020  
I3 = 
I2 = I1 –I3 = 79.9 mA – 17.2 mA = 62.7 mA downward
Next, zero the current source by replacing it with an open as shown in Figure 8-6(b):
RT = 587.6 
I2 = IT =
20 V
= 34.0 mA downward
587.6 
I2(TOT) = 62.7 mA + 34.0 mA = 96.7 mA
11.
First, zero the voltage source by replacing it with a short as shown in Figure 8-6(a):
 680  
100 mA = 79.8 mA
I1 = 
 852.6  
 220  
79.8 mA = 17.2 mA
I3 = 
 1020  
Next, zero the current source by replacing it with an open as shown in Figure 8-6(b):
RT = 587.6 
20 V
IT =
= 34.0 mA
587.6 
 680  
34.0 mA = 15.6 mA
I3 = 
 1480  
The total I3 is the difference of the two component currents found in the above steps because
they are in opposite directions.
I3(total) = 17.2 mA  15.6 mA = 1.6 mA
67
Chapter 8
Figure 8-6
12.
(a)
Current through RL due to the 1 A source. See Figure 8-7(a):
 2.2 k 
IL = 
1 A = 361 mA (down)
 6.1 k 
Current through RL due to the 2 A source is zero because of infinite resistance (open) of
the 1 A source. See Figure 8-7(b):
IL = 0 A
Total current through RL:
IL(total) = 361 mA + 0 A = 361 mA
Figure 8-7
(b)
Current through RL due to the 40 V source is zero because of zero resistance (short) of
the 60 V source. See Figure 8-8(a):
IL = 0 A
Current through RL due to the 0.5 A source is zero because of zero resistance of the 60 V
source. See Figure 8-8(b):
IL = 0 A
Current through RL due to the 60 V source. See Figure 8-8(c):
 1.5 k 
VL = 
60 V = 43.7 V
 2.06 k 
43.7 V
V
IL = L 
= 29.1 mA
RL 1.5 k
Total current through RL:
IL = 0 A + 0 A + 29.1 mA = 29.1 mA
68
Chapter 8
Figure 8-8
13.
 R2  R3 
 7.8 k 
VRef(max)  
 30 V  15 V = 3.72 V
 30 V  15 V  
 12.5 k 
 R1  R2  R3 


R3
 6.8 k 
VRef(min)  
 30 V  15 V = 1.32 V
 30 V  15 V  
 12.5 k 
 R1  R2  R3 
14.
 R2  R3 
 16.8 k 
VRef(max)  
 30 V  15 V = 8.44 V
 30 V  15 V  
 21.5 k 
 R1  R2  R3 


R3
 6.8 k 
VRef(min)  
 30 V  15 V = 5.51 V
 30 V  15 V  
 21.5 k 
 R1  R2  R3 
15.
75 V source. See Figure 8-9(a):
Req = R2  R3  (R4 + R5) = 17.2 k
 Req 
 17.2 k 
VA = 
75 V  

 75 V = 13 V
R R 
 99.2 k 
1 
 eq
 R5 
91 k 
VA  
VB = 
13 V = 11.7 V

 101 k 
 R4  R5 
50 V source. See Figure 8-9(b):
Req = R1  R2  (R4 + R5) = 25 k
 Req 
 25 k 
VA =  
50 V   
 50 V = 21.6 V
 R  R 
 58 k 
3 
 eq
69
Chapter 8
 R5 
 91 k 
VB = 
 VA  
 (  21.6 V) = 19.5 V
 101 k 
 R4  R5 
100 V source. See Figure 8-9(c):
Req = R1  R2  R3 = 16.6 k
RT = 10 k + 91 k + 16.6 k = 117.6 k
100 V
IT =
= 850 A
117.6 k
VA = (850 A)(16.6 k) = 14.1 V
VB = (850 A)(91 k) = 77.4 V
Superimposing voltages at each point:
VA = 13 V  21.6 V + 14.1 V = 5.5 V
VB = 11.7 V  19.5 V  77.4 V = 85.2 V
VAB = 5.5 V  (85.2 V) = 90.7 V
Figure 8-9
70
Chapter 8
16.
SW1 closed. See Figure 8-10(a):
12 V
12 V
IL =

= 508 A
5.6 k  18 k 23.6 k
SW1 and SW2 closed. See Figure 8-10(b):
Current from the 12 V source (6 V source zeroed)
RT = R1 + R2  RL = 5.6 k + 8.2 k  18 k = 11.2 k
12 V
= 1.07 mA
IT =
11.2 k
 8.2 k 
IL = 
1.07 mA = 335 A
 26.2 k 
Current from the 6 V source (12 V source zeroed):
RT = R2 + R1  RL = 8.2 k + 5.6 k  18 k = 12.47 k
6V
IT =
= 481 A
12.47 k
 5.6 k 
IL = 
481 A = 114 A
 23.6 k 
IL(total) = 335 A + 114 A = 449 A
SW1, SW2, and SW3 closed. See Figure 8-10(c).
Current from the 12 V source (6 V and 9 V sources zeroed):
RT = R1 + R2  R3  RL = 5.6 k + 8.2 k  12 k  18 k = 9.43 k
12 V
IT =
= 1.27 mA
9.43 k
R R R 
 3.83 k 
IL =  2 3 L  I T  
1.27 mA = 270 A
RL
 18 k 


Current from the 6 V source (9 V and 12 V sources zeroed):
RT = R2 + R1  R3  RL = 8.2 k + 5.6 k  12 k  18 k = 11.35 k
6V
IT =
= 529 A
11.35 k
R R R 
 3.15 k 
IL =  1 3 L  I T  
 529 A = 93 A
RL
 18 k 


Current from the 9 V source (6 V and 12 V sources zeroed):
RT = R3 + R1  R2  RL = 12 k + 5.6 k  8.2 k  18 k = 14.8 k
9V
IT =
= 608 A
14.85 k
R R R 
 2.81 k 
IL =  1 2 L  I T  
 608 A = 95 A
RL
 18 k 


IL(total) = 270 A + 93 A + 95 A = 458 A
71
Chapter 8
Figure 8-10
17.
VS1 “sees” a total resistance of
RT = 10 k + (5.6 k  (10 k + (5.6 k  ((10 k + 5.6 k)
+ (10 k  (5.6 k + (10 k  5.6 k)))))))
= 10 k + (5.6 k  (10 k + (5.6 k  (15.6 k + (10 k  (5.6 k + 3.59 k))))))
= 10 k + (5.6 k  (10 k + (5.6 k  (15.6 k + (10 k  9.19 k)))))
= 10 k + (5.6 k  (10 k + (5.6 k  (15.6 k + 4.79 k))))
= 10 k + (5.6 k  (10 k + (5.6 k  20.4 k)))
= 10 k + (5.6 k  (10 k + 4.39 k))
= 10 k + (5.6 k  (14.4 k) = 10 k + 4.03 k = 14.0 k
32 V
IT(S1) =
= 2.28 mA
14.0 k
VS2 “sees” a total resistance of
RT = 5.6 k + (10 k  (5.6 k + (10 k  ((10 k + 5.6 k)
+ (5.6 k  (10 k + (5.6 k  10 k)))))))
= 5.6 k + (10 k  (5.6 k + (10 k  (15.6 k + (5.6 k  (10 k + 3.59 k))))))
= 5.6 k + (10 k  (5.6 k + (10 k  (15.6 k + (5.6 k  13.6 k)))))
= 5.6 k + (10 k  (5.6 k + (10 k  (15.6 k + (3.97 k))))
= 5.6 k + (10 k  (5.6 k + (10 k  (19.6 k)))
= 5.6 k + (10 k  (5.6 k + 6.62 k))
= 5.6 k + (10 k  12.2 k) = 5.6 k + 550 k = 11.1 k
IT(S2) =
15 V
= 1.35 mA
11.1 k
72
Chapter 8
Section 8-5 Thevenin’s Theorem
18.
19.
(a)
RTH = 27  + 75   147  = 76.7 
 75  
25 V = 8.45 V
VTH = 
 222  
(b)
RTH = 100   270  = 73 
 100  
3 V = 811 mV
VTH = 
 370  
(c)
RTH = 56 k  100 k = 35.9 k
 56 k 
(15 V  10 V) = 1.79 V
VTH = 
 156 k 
(b)
RTH = 2.2 k  (1 k + 2.2 k = 1.3 k
 2.2 k 
IAB = 
0.1 A = 40.7 mV
 5.4 k 
VTH = IAB(2.2 k)
= (40.7 mA)(2.2 k) = 89.5 V
First, convert the circuit to its Thevenin equivalent as shown in the steps of Figure 8-11.
RTH = 13.97 k
 4.12 k 
VA = 
32 V = 9.34 V
 14.12 k 
 5.6 k 
 5.6 k 
VTH = 
VA  
9.34 V = 3.35 V
 15.6 k 
 15.6 k 
VTH
3.35 V
IL =

= 116 A
RTH  RL 28.97 k
Figure 8-11
73
Chapter 8
20.
First, zero (open) the current source, remove R4, and redraw the circuit as shown in
Figure 8-12(a).
RTH = R3  (R1 + R2  R5) = 5.6 k  (1 k + 1.65 k) = 5.6 k  2.65 k = 1.8 k
2.65 k


 2.65 k 
VTH = 
50 V  
50 V = 16.1 V
 5.6 k  2.65 k 
 8.25 k 
Determine V4 due to the 50 V source using the Thevenin circuit in Figure 8-12(b).
 R4

 10 k 
V4 = 
VTH  
16.1 V = 13.6 V
 11.8 k 
 RTH  R4 
Next, zero (short) the voltage source, remove R4, and redraw the circuit as shown in
Figure 8-12(c).
RTH = R3  (R1 + R2  R5) = 5.6 k  (1 k + 1.65 k) = 5.6 k  2.65 k = 1.8 k
 2.65 k 
I3 = 
10 mA = 3.2 mA
 8.25 k 
VTH = V3 = I3R3 = (3.2 mA)(5.6 k) = 17.9 V
Determine V4 due to the current source using the Thevenin circuit in Figure 8-12(d).
 R4

 10 k 
V4 = 
VTH  
17.9 V = 15.2 V
 11.8 k 
 RTH  R4 
Use superposition to combine the V4 voltages to get the total voltage across R4:
V4 = 13.6 V + 15.2 V = 28.8 V
Figure 8-12
74
Chapter 8
21.
Looking back from the amplifier input:
RTH = R1  R2  R3 = 100   2.2 k  1.2 k = 88.6 
1 V source (Figure 8-13(a)):
 776  
1 V = 886 mV
VA = 
 876  
5 V source (Figure 8-13(b)):
 92.3  
VA = 
= 200 mV
 2292  5 V


VTH = 886 mV + 200 mV = 1.09 V
Figure 8-13
22.
Consider R6  (R7 + R8) to be the load. Thevenize to the left of point A as shown in
Figure 8-14(a).
RTH = R5 + R4  (R3 + (R1  R2)) = 1 k 4.7 k  (10 k + 6.8 k  9.1 k)
= 1 k + 4.7 k  13.89 k = 4.51 k
See Figure 8-14(b) to determine VTH:
RT = (R3 + R4)  R2 + R1 = (10 k + 4.7 k)  6.8 k + 9.1 k) = 4.65 k + 9.1 k = 13.8 k
48 V
IT =
= 3.48 mA
13.8 k


R2
 6.8 k 
I4 = 
 IT  
 3.48 mA = 1.1 mA
R
R
R


 21.5 k 
3
4 
 2
V4 = I4R4 = (1.1 mA)(4.7 k) = 5.17 V
VX = 48 V  V4 = 48 V  5.17 V = 42.8 V
VTH = VA = VX = 42.8 V
75
Chapter 8
The Thevenin circuit is shown in Figure 8-14(c). The current into point A is determined for
each value of R8.
When R8 = 1 k:
RL = 12 k  (8.2 k + 1 k) = 5.21 k
VTH
42.8 V
IA =

= 4.41 mA
RTH  RL 9.72 k
When R8 = 5 k:
RL = 12 k  (8.2 k + 5 k) = 6.29 k
VTH
42.8 V

IA =
= 3.97 mA
RTH  RL 10.8 k
When R8 = 10 k:
RL = 12 k  (8.2 k + 10 k) = 7.23 k
VTH
42.8 V
IA =

= 3.66 mA
RTH  RL 11.7 k
Figure 8-14
76
Chapter 8
23.
See Figure 8-15.
 1.2 k 
 2.2 k 
VTH = VA  VB = 
12 V = 8.25 V  7.13 V = 1.12 V
12 V  
 2.02 k 
 3.2 k 
RTH = 1 k  2.2 k + 820   1.2 k = 688  + 487  = 1175 
VTH
1.12 V
IL =

= 100 A
RTH  RL 11,175 
Figure 8-15
24.
See Figure 8-16.
VR3 = (0.2 mA)(15 k) = 3 V
V  VR 3 10 V  3 V

= 35 k
R4 = S
I4
0.2 mA
 R2 
 12 k 
VA = 
VS  
10 V = 5.46 V
 22 k 
 R1  R2 
 R4 
 35 k 
VB = 
VS  
10 V = 7V
 50 k 
 R3  R4 
VTH = VBA = VB  VA = 7 V  5.46 V = 1.54 V
RTH = R1  R2 + R3  R4 = 5.46 k + 10.5 k = 15.96 k
Figure 8-16
77
Chapter 8
Section 8-6 Norton’s Theorem
25.
(a)
See Figure 8-17(a).
RN = 76.7 
(b)
RT = 166.9 
See Figure 8-17(b).
RN = 73 
3V
IN =
= 11.1 mA
270 
25 V
= 150 mA
166.9 
 75  
 75  
IN = 
 IT  
150 mA = 110 mA
 102  
 102  
IT =
(c)
See Figure 8-17(c).
(56 k)(100 k)
RN =
= 35.9 k
156 k
5V
IN =
= 50 A
100 k
(d)
Figure 8-17
78
See Figure 8-17(d).
(3.2 k)(2.2 k)
RN =
= 1.3 k
5.4 k
 2.2 k 
IN = 
0.1 A = 68.8 mA
 3.2 k 
Chapter 8
26.
First, RN is found by circuit simplification as shown in Figure 8-18(a).
RN = 14.0 k
The current IN through the shorted AB terminals is found as shown in Figure 8-18 (b).
RT = 14.0 k as viewed from the source
32 V
IT =
= 2.29 mA
14.0 k
 5.6 k 
I1 = 
2.29 mA = 668 A
 19.2 k 
 5.6 k 
IN = 
668 A = 240 A
 15.6 k 
Finally, the current through RL is determined by connecting RL to the Norton equivalent circuit
as shown in Figure 8-18(c).
 14.0 k 
IL = 
240 A = 116 A
 29.0 k 
Figure 8-18
79
Chapter 8
27.
The 50 V source acting alone. Short AB to get IN. See Figure 8-19(a):
RT = R3 + R1  R4 = 5.6 k + 1 k  10 k = 6.51 k
50 V
IT =
= 7.68 mA
6.51 k
 R4 
10 k 
 IT  
IN = 
7.68 mA = 6.98 mA
R
R

 11 k 
4 
 1
See Figure 8-19(b):
RN = R2  (R1 + R3 R4) = 3.3 k  (1 k + 5.6 k 10 k) = 3.3 k  4.59 k = 1.92 k
See Figure 8-19(c):
 RN 
 1.92 k 
IR5 = 
 IN  
 6.98 mA = 2.57 mA (from B to A)
 5.22 k 
 RN  R5 
The 10 mA source acting alone. Short AB to get IN. See Figure 8-19(d):
 R3 R4 


5.6 k 10 k
10 mA  
10 mA   3.59 k 10 mA = 7.82 mA
IN = 
R R R 
 1 k  5.6 k 10 k 
 4.59 k 
3
4 
 1


RN = 1.92 k
See Figure 8-19(e):
 1.9 k 
IR5 = 
7.82 mA = 2.85 mA (from B to A)
 5.22 k 
V5 = I5R5 = (5.42 mA)(3.3 k) = 17.9 V
Figure 8-19
80
Chapter 8
28.
See Figure 8-20(a):
RN = R2  (R3 + R4  (R5 + R6  (R7 + R8 )))
= 6.8 k  (10 k + 4.7 k  (1 k + 6.89 k))
= 6.8 k  (10 k + 2.95 k) = 4.46 k
See Figure 8-20(b):
RT = R2  (R4 + R3  (R5 + R6  (R7 + R8)))
= 6.8 k  (4.7 k + 10 k  (1 k + 6.89 k))
= 6.8 k  (4.7 k + 4.41 k) = 3.89 k
48 V
= 12.3 mA
IT =
3.89 k
9.11 k
9.11 k




I2 = 
 IT  
12.3 mA = 7.07 mA
 6.8 k  9.11 k 
 6.8 k  9.11 k 
 6.8 k 
I4 = 
12.3 mA = 5.27 mA
 15.9 k 
 7.89 k 
I3 = 
5.27 mA = 2.62 mA
 15.9 k 
IN = I2 + I3 = 7.07 mA + 2.62 mA = 9.69 mA
See Figure 8-20(c):
 4.46 k 
I1 = 
9.69 mA = 3.18 mA
 13.6 k 
Figure 8-20
81
Chapter 8
29.
Using the results of Problem 23:
V
1.12 V
= 953 A
IN = TH 
RTH 1175 
RN = RTH = 1175 
See Figure 8-21.
Figure 8-21
30.
See Figure 8-22(a):
RN = 10 k  (15 k + 8.2 k  22 k) = 6.77 k
See Figure 8-22(b):
RT = 8.2 k  15 k + 22 k = 27.3 k
12 V
= 440 A
IT =
27.3 k
 8 .2 k  
IN1 = 
440 A = 156 A down
 23.3 k 
See Figure 8-22(c):


15 k
10 mA   15 k 10 mA = 7.15 mA down
IN2 = 
 15 k  22 k 8.2 k 
 20.97 k 


See Figure 8-22(d):
IN = IN1 + IN2 = 156 A + 7.15 mA = 7.31 mA
Figure 8-22
82
Chapter 8
31.
RN = 220   100   330  = 56.9 
Find IN1 due to the 3 V source, as shown in Figure 8-23(a).
3V
= 9.1 mA (down)
IN1 =
330 
Find IN2 due to the 8 V source, as shown in Figure 8-23(b).
8V
= 80 mA (up)
IN2 =
100 
Find IN3 due to the 5 V source, as shown in Figure 8-23(c).
5V
IN1 =
= 22.7 mA (down)
220 
The Norton equivalent is shown in Figure 8-23(d).
IN(tot) = IN1 + IN2 + IN3 = 9.1 mA  80 mA + 22.7 mA = 48.2 mA
56.9 
Figure 8-23
83
Chapter 8
Section 8-7 Maximum Power Transfer Theorem
RL = RS = 12 
RL = RS = 8.2 k
RL = RS = 4.7  + 1   2  = 6.37 
RL = RS = 47  + 680  = 727 
32.
(a)
(b)
(c)
(d)
33.
See Figure 8-24.
As seen by RL:
RS = 8.2  + 2.94  = 11.1 
For maximum power transfer:
RL = RS = 11.1 
Figure 8-24
34.
Refer to Problem 33 and Figure 8-24.
RL+ = RL + 0.1RL = 11.1  + 1.11  = 12.21 
RTH = RS = 11.1 
IL due to the 1.5 V source:


15  16.4 
1.5 V   7.79  1.5 V = 936 mV
VTH = 
 4.7   15  16.4  
 12.49  


VTH
936 mV
IL =

= 40 mA
23.4 
RTH  RL 
IL due to the 1 mA source:


4.7  16.4 
1 mA   3.65  1 mA = 196 A
I15 = 
 15   4.7  16.4  
 18.65  


VTH = I15(15 ) = (196 A)(15 ) = 2.94 mV
VTH
2.94 mV
IL =

= 126 mA
23.4 
RTH  RL 
IL(total) = 40 mA + 126 A = 40.126 mA
PL = I L2 RL  = (40.126 mA)212.21  = 19.7 mW
84
Chapter 8
35.
For maximum power transfer, RTH = RLADDER
The voltage across RTH = 24 V (one half of VTH)
24 V
= 48 
RTH =
0.5 A
RLADDER = 48 
RLADDER = ((R4  (R5 + R6) + R3)  R2) + R1

 69 R4

 10 47

 69  R4
= 26
69 R4
47 
 10
69  R4

69 R4
26  69 R4

 10 
 57 
69  R4
47  69  R4

69 R4  26   26 
1     57  10  21.53
69  R4  47   47 
69R4 = 69(48.17) + 48.17R4
R4(69  48.17) = 69(48.17)
69(48.17)
= 160 
R4 =
69  48.17
Section 8-8 Delta-Wye (Y) and Wye-Delta (Y-) Conversions
36.
RA RC
(560 k)(1 M)

= 183 k
RA  RB  RC
3.06 M
RB RC
(1.5 M)(1 M)
R2 =

= 490 k
RA  RB  RC
3.06 M
RA RB
(560 k)(1.5 M)
R3 =

= 275 k
RA  RB  RC
3.06 M
(a)
R1 =
(b)
R1 =
RA RC
(1 )(2.2 )

= 373 m
RA  RB  RC
5.9 
RB RC
(2.2 )(2.7 )

R2 =
= 1.01 
RA  RB  RC
5 .9 
RA RB
(1 )(2.7 )
R3 =

= 4.58 m
RA  RB  RC
5 .9 
85
Chapter 8
37.
38.
R1R2  R1R3  R2 R3 (12 )(22 )  (12 )(18 )  (22 )(18 ) 876


= 39.8 
R2
22 
22
R R  R1R3  R2 R3 (12 )(22 )  (12 )(18 )  (22 )(18 ) 876
RB = 1 2


= 73 
R1
12 
12
R R  R1R3  R2 R3 (12 )(22 )  (12 )(18 )  (22 )(18 ) 876


RC = 1 2
= 48.7 
R3
18 
18
(a)
RA =
(b)
RA =
R1R2  R1R3  R2 R3 (6.8 k)(3.3 k)  (6.8 k)(4.7 k)  (3.3 k)(4.7 k)

= 21.2 k
R2
3.3 k
R R  R1R3  R2 R3 (6.8 k)(3.3 k)  (6.8 k)(4.7 k)  (3.3 k)(4.7 k)
RB = 1 2

= 10.3 k
R1
6.8 k
R R  R1R3  R2 R3 (6.8 k)(3.3 k)  (6.8 k)(4.7 k)  (3.3 k)(4.7 k)
RC = 1 2

= 14.9 k
R3
4 .7 k 
Convert the delta formed by R3, R4, and R5 to a Wye configuration. See Figure 8-25:
R3 R4
(22 k)(12 k)

RY1 =
= 6.13 k
43.1 k
R3  R4  R5
(22 k)(9.1 k)
R3 R5

RY2 =
= 4.65 k
43.1 k
R3  R4  R5
R4 R5
(12 k)(9.1 k)

RY3 =
= 2.53 k
43.1 k
R3  R4  R5
RT = (R1 + RY1)  (R2 + RY2) + RY3
= (10 k + 6.13 k)  (39 k + 4.65 k) + 2.53 k = 11.78 k + 2.53 k = 14.3 k
136 V 136 V

IT =
= 9.5 mA
14.3 k
RT


R2  RY 2
 43.65 k 
 IT  
IR1 = IRY1 = 
9.5 mA = 6.94 mA
 59.78 k 
 R1  RY 1  R2  RY 2 
IR2 = IRY2 = IT  IR1 = 9.5 mA  6.94 mA = 2.56 mA
VB = VA  IR1R1 = 136 V  (6.94 mA)(10 k) = 66.6 V
VC = VA  IR2R2 = 136 V  (2.56 mA)(39 k) = 36.16 V
In the original circuit:
V
66.6 V
IR4 = B 
= 5.55 mA
R4 12 k
V
36.16 V
IR5 = C 
= 3.97 mA
R5
9.1 k
V V
66.6 V  36.16 V
IR3 = B C 
= 1.38 mA
R3
22 k
Figure 8-25
86
Chapter 8
Multisim Troubleshooting and Analysis
39.
R1 is leaky.
40.
VTH = 17.478 V; RTH = 247.279 
41.
IN = 0.383 mA; RN = 9.674 k
42.
R3 is shorted.
43.
IAB = 1.206 mA; VAB = 3.432 V
87
Chapter 9
Branch, Loop, and Node Analysis
Note: Solutions show conventional current direction.
Section 9-1 Simultaneous Equations in Circuit Analysis
1.
100I1 + 50I2 = 30
75I1 + 90I2 = 15
30  50 I 2
I1 =
100
 30  50 I 2 
75
 + 90I2 = 15
 100 
22.5  37.5I2 + 90I2 = 15
52.5I2 = 7.5
I2 = 143 mA
100I1 + 50(0.143) = 30
I1 = 371 mA
2.
(a)
4 6
= 12  12 = 0
2 3
(b)
9 1
= 45  0 = 45
0 5
(c)
12
2
(d)
100
50
= 2000  1500 = 3500
30  20
15
= 12  (30) = 18
1
1 4
4 2
3.
(a)
4.
(a)
I1 =
6 3
12  12

=0A
1 2
 3  14
7 3
(b)
I2 =
7 6
 6  28

=2A
1 2
 3  14
7 3
1 0 2 1 0
5 4
15 4
2 10
0 2 10
= (1)(4)(0) + (0)(1)(2) + (2)(5)(10)  [(2)(4)(2) + (10)(1)(1) + (0)(5)(0)]
= (0 + 0  100)  (16 + 10 + 0) = 100 + 6 = 94
(b)
0 .5
1
 0.8 0.5
1
0.1
1.2
1.5 0.1
1 .2
 0.1  0.3
5  0 .1  0 .3
= (0.5)(1.2)(5) + (1)(1.5)(0.1) + (0.8)(0.1)(0.3)
 [(0.8)(1.2)(0.1) + (0.3)(1.5)(0.5) + (5)(0.1)(1)]
= (3  0.15 + 0.024)  (0.096  0.255 + 0.5) = 2.874  0.371 = 2.50
88
Chapter 9
5.
(a)
0 20 25 0
25
10 12
5 10 12
 8 30 16 8 30
= 25(12)(16) + (0)(5)(8) + (20)(10)(30)
 [(8)(12)(20) + (30)(5)(25) + (16)(10)(0)]
= 10800  5670 = 16,470
(b)
1.08 1.75
0.55 1.08 1.75
2.12 0.98 0
3.49 1.05 1
0
1
2.12
3.49
= (1.08)(2.12)(1.05) + (1.75)(0.98)(1) + (0.55)(0)(3.49)
 [(1)(2.12)(0.55) + (3.49)(0.98)(1.08) + (1.05)(0)(1.75)]
= 4.119 + 2.528 = 1.591
6.
The characteristic determinant was evaluated as 2.35 in Example 9-4. The determinant for I3 is
as follows:
2 0.5
0
2 0.5
0.75
0 1.5 0.75 0 = (0 + 2.25 + 0)  (0 + 0.6  0.375) = 2.25  0.225 = 2.025
3 0.2  1
3 0 .2
I3 =
7.
2.025
= 862 mA
2.35
The characteristic determinant is:
2 6 10 2 6
3 7 8 3 7
10 5 12 10 5
= (2)(7)(12) + (6)(8)(10) + (10)(3)(5)
 [(10)(7)(10) + (5)(8)(2) + (12)(3)(6)]
= 462  836 = 374
I1 =
9 6 10 9 6
3 7 8 3 7
0 5 12 0 5
374
(9)(7)(12)  (6)(8)(0)  (10)(3)(5)  [(0)(7)(10)  (5)(8)(9)  (12)(3)(6)]
=
374
606  144 462

= 1.24 A
=
374
374
89
Chapter 9
2 9
3 3
I2 =
10 2 9
8 3 3
10 0 12 10 0
374
(2)(3)(12)  (9)( 8)(10)  (10)(3)(0)  [(10)(3)(10)  (0)( 8)(2)  ( 12)(3)(9)]
=
374
792  24 768
=

= 2.05 A
374
374
2 6 9 2 6
3 7 3 3 7
I3 =
10
5 0 10
5
374
(2)(7)(0)  (6)(3)(10)  (9)(3)(5)  [(10)(7)(9)  (5)(3)(2)  (0)(3)(6)]
=
374
45  660 705

=
= 1.89 A
374
374
8.
The calculator results are:
V1 = 1.61301369863
V2 = 1.69092465753
V3 = 2.52397260274
V4 = 4.69691780822
9.
X1 = .371428571429 (I1 = 371 mA)
X2 = .142857142857 (I2 = 143 mA)
10.
X1 = 1.23529411765 (I1 = 1.24 A)
X2 = 2.05347593583 (I2 = 2.05 A)
X3 = 1.88502673797 (I3 = 1.89 A)
Section 9-2 Branch Current Method
11.
The sum of the currents at the node is zero. Currents into the node are assumed positive and
currents out of the node are assumed negative.
I1  I2  I3 = 0
12.
I1  I2  I3 = 0
8.2I1 + 10I2 =12
10I2 + 5.6I3 = 6
Solving by substitution:
I1 = I2 + I3
8.2(I2 + I3) + 10I2 = 12
8.2I2 + 8.2I3 = 10I2 = 12
90
Chapter 9
18.2I2 + 8.2I3 = 12
12  8.2 I 3
I2 =
18.2
 12  8.2 I 3 
 10
 + 5.6I3 = 6
 18.2 
120  82 I 3
+ 5.6I3 = 6
18.2
10.11I3 = 0.59
I3 = 58.4 mA
10I2 + 5.6(0.058) = 6
10I2 + 0.325 = 6
I2 = 633 mA
I1 = I2 + I3 = 633 mA + 58.4 mA = 691 mA
13.
The branch currents were found in Problem 12.
I1 = 691 mA
I2 = 633 mA
I3 = 58.4 mA
V1 = I1R1 = (691 mA)(8.2 ) = 5.66 V (+ on left)
V2 = I2R2 = (633 mA)(10 ) = 6.33 V (+ at top)
V3 =I3R3 = (58.4 mA)(5.6 ) = 325 mV (+ on left)
14.
I1  I2 = 100 mA
12  VA VA

= 0.1
47
100
100(12  VA)  47VA = 470
1200  100VA  47VA = 470
147VA = 730
VA = 4.97
12 V  4.97 V 7.03 V
I1 =

= 150 mA
47 
47 
4.97 V
I2 =
= 49.7 mA
100 
I3 = 100 mA (current source)
15.
Current source zeroed (open). See Figure 9-1(a).
 R2 
 100  
VAB = V2 = 
VS  
12 V = 8.16 V
 147  
 R1  R2 
Voltage source zeroed (shorted). See Figure 9-1(b).
VAB = V3 = I3R3 = (100 mA)(68 ) = 6.8 V
 R1 
 47  
I2 = 
 IS  
100 mA = 31.97 mA
 147  
 R1  R2 
VAG = V2 = (31.97 mA)(100 ) = 3.197 V
VAB = VAG  VBG = 3.197  6.8 V = 9.997 V
Superimposing:
VAB = 8.16 V + (9.997 V) = 1.84 V
91
Figure 9-1
Chapter 9
Section 9-3 Loop Current Method
16.
The characteristic determinant is:
0.045 0.130 0.066 0.045 0.130
0.177 0.042 0.109 0.177 0.042
0.078 0.196 0.290 0.078 0.196
= (0.045)(0.042)(0.290) + (0.130)(0.109)(0.078) + (0.066)(0.177)(0.196)
 [(0.078(0.042)(0.066) + (0.196)(0.109)(0.045) + (0.290)(0.177)(0.130)]
= 0.00394  0.00785 = 0.00391
17.
1560I1  560I2 = 6
560I1 + 1380I2 = 2
 6  560
I1 =
 2 1380
 8280  1120
 9400


= 5.11 mA
1560  560
2,152,800  313,600 1,839,200
 560 1380
1560  6
 560  2
 3180  3360
I2 =

= 3.52 mA
1,839,200
1,839,200
18.
Using the loop currents from Problem 17:
I1 k = I1 = 5.11 mA
I820  = I2 = 352 mA
I560  = I1  I2 = 5.11 mA + 3.52 mA = 1.59 mA
19.
Using the branch currents from Problem 18:
V1 k = I1 k(1 k) = (5.11 mA)(1 k) = 5.11 V (+ on right)
V560  = I560 (560 ) = (1.59 mA)(560 ) = 890 mV (+ on bottom)
V820  = I820 (820 ) = (3.52 mA)(820 ) = 2.89 V (+ on right)
20.
57I1 10I2 = 1.5
10I1 + 41.7I2  4.7I3 = 3
4.7I2 + 19.7I3 = 1.5
92
Chapter 9
21.
The equations were developed in Problem 20. The characteristic determinant is as follows
with the k units omitted for simplicity:
57 10
0 57 10
 10 41.7 4.7 10 41.7
0 4.7 19.7 0 4.7
= (57)(41.7)(19.7) + (10)(4.7)(0) + (0)(10)(4.7)
 [(0)(41.7)(0) + (4.7)(4.7)(57) + (19.7)(10)(10)]
= 46,824.93  3,229.13 = 43,595.8
1.5 10
0 1.5 10
43,595.8I1 =  3 41.7 4.7 3 41.7
1.5 4.7 19.7 1.5 4.7
= (1.5)(41.7)(19.7) + (10)(4.7)(1.5) + (0)(3)(4.7)
 [(1.5)(41.7)(0) + (4.7)(4.7)(1.5) + (19.7)(3)(10)]
1302.735  624.135
678.6

= 15.6 mA
I1 =
43,595.8
43,595.8
57 1.5
0 57 1.5
43,595.8I2 =  10 3 4.7 10 3
0 1.5 19.7 0 1.5
= (57)(3)(19.7) + (1.5)(4.7)(0) + (0)(10)(1.5)
 [(0)(3)(0) + (1.5)(4.7)(57) + (19.7)(10)(1.5)]
3368.7  697.35 2671.35
= 61.3 mA
I2 =

43,595.8
43,595.8
Substituting into the third equation to get I3:
19.7I3 = 1.5 + 4.7I2
1.5  4.7(0.0613 A)
= 61.5 mA
I3 =
19.7
93
Chapter 9
22.
Use the loop currents from Problem 21:
I47 = I1 = 15.6 mA
I27 = I2 = 61.3 mA
I15 = I3 = 61.5 mA
I10 = I1  I2 = 15.6 mA  (61.3 mA) = 76.9 mA
I4.7 = I2  I3 = 61.3 mA  61.5 mA = 123 mA
23.
See Figure 9-2.
The loop equations are:
(10 + 4.7 + 2.2)I1  (4.7 + 2.2)I2 = 8 V
(2.2 + 4.7 + 8.2 + 3.9)I2  (2.2 + 4.7)I1 = 0 V
16.9I1  6.9I2 = 8
6.9I1 + 19I2 = 0
8  6.9
0
19
(8)(19)
152
152
= 555 mA



I1 =
16.9  6.9
(16.9)(19)  (6.9)(6.9) 321.1  47.61 273.49
 6 .9
19
16.9 8
I2 =
 6 .9 0
 (8)(6.9)
55.2
55.2
= 202 mA



16.9  6.9
(16.9)(19)  (6.9)(6.9) 321.1  47.61 273.49
 6 .9
19
VA = (I1  I2)2.2  = (555 mA  202 mA) 2.2  = (353 mA)2.2  = 776.6 mV
VB = I2(3.9 ) = (202 mA)(3.9 ) = 787.8 mV
VAB = VA  VB = 776.6 mV  787.8 mV = 11.2 mV
Figure 9-2
94
Chapter 9
24.
See Figure 9-3.
The loop equations are:
(10 + 4.7 + 2.2)I1  4.7I2  2.2I3 = 8 V
(4.7 + 8.2 + 10)I2  4.7I1  10I3 = 0
(2.2 + 10 + 3.9)I3  2.2I1  10I2 = 0
16.9I1  4.7I2  2.2I3 = 8 V
4.7I1 + 22.9I2  10I3 = 0
2.2I1  10I2 + 16.1I3 = 0
Figure 9-3
The characteristic determinant is:
4.7 2.2 16.9
4.7
 4.7 22.9 10 4.7
 2.2 10 16.1 2.2
22.9
10
16.9
= (16.9)(22.9)(16.1) + (4.7)(10)(2.2) + (2.2)(4.7)(10)
 [(2.2)(22.9)(2.2) + (10)(10)(16.9) + (16.1)(4.7)(4.7)]
= 6024.061  2156.485 = 3867.576
16.9 8 2.2 16.9 8
3867.576I2 =  4.7 0 10 4.7 0
 2.2 0 16.1 2.2 0
= (16.9)(0)(16.1) + (8)(10)(2.2) + (2.2)(4.7)(0)
 [(2.2)(0)(2.2) + (0)(10)(16.9) + (16.1)(4.7)(8)]
I2 =
176  605.36
781.36

= 202 mA
3867.576
3867.576
16.9 4.7 8 16.9 4.7
3867.576I2 =  4.7 22.9 0 4.7 22.9
 2.2 10 0 2.2 10
= (16.9)(22.9)(0) + (4.7)(0)(2.2) + (8)(4.7)(10)
 [(2.2)(22.9)(8) + (10)(0)(16.9) + (0)(4.7)(4.7)]
I3 =
376  403.04
779.04

= 201 mA
3867.576
3867.576
IBA = I2  I3 = 202 mA  201 mA = 1 mA
95
Chapter 9
25.
See Figure 9-4.
(R1 + R2 + R3)IA  R2IB  R3IC = 0
R2IA + (R2 + R4)IB  R4IC = VS
R3IA  R4IB + (R3 + R4 + RL)IC = 0
5.48IA  3.3IB  1.5IC = 0
3.3IA + 4.12IB  0.82IC = 15
1.5IA  0.82IB + 4.52IC = 0
Coefficients are in k.
Figure 9-4
26. Using a calculator to solve for the loop currents:
IA = 7.63 mA, IB = 10.6 mA, IC = 4.46 mA
IRL = IC = 4.46 mA
27. IR3 = IA – IC = 7.63 mA – 4.46 mA = 3.17 mA
VR3 = IR3R3 = (3.17 mA)(1.5 k) = 4.76 V
Section 9-4 Node Voltage Method
28.
See Figure 9-5.
The current equation at node A is:
I1  I2  I3 = 0
Using Ohm’s law substitutions for the currents:
30  VA VA  40 VA
=0


82
68
147
30 VA VA 40 VA




0
82 82 68 60 147
Multiply each term in the last equation by (82)(68)(147) = 819,672 to eliminate the denominators.
9996(30)  9996VA  12,054VA + 12,054  5576VA = 0
782,040  27,626VA = 0
782,040
= 28.3 V
VAB = VA =
27,626
Figure 9-5
96
Chapter 9
29.
Use VAB = 28.3 V from Problem 28.
30 V  VAB 30 V  28.3 V

I1 =
= 20.6 mA
82 
82 
V  40 V 28.3 V  40 V

I2 = AB
= 172 mA
68 
68 
28.3 V
VAB

I3 =
= 193 mA
147  147 
30.
See Figure 9-6.
I1  I2  I3 = 0
I3 + I4  I5 = 0
Figure 9-6
Substituting into the first equation and simplifying:
1.5  VA VA VA  VB


=0
47
10
27
1.5 VA VA VA VB




=0
47 47 10 27 27
 27VA  126.9VA  47VA VB  1.5


126.9
27
47
200.9VA VB 1.5


126.9
27 47
1.58VA  0.037VB = 0.0319
Substituting into the second equation and simplifying:
VA  VB 3  VB VB  1.5


=0
27
4 .7
15
3
VA VB
V
V 1.5


 B  B
=0
27 27 4.7 4.7 15 5
0.037VA  0.037VB  0.213VB  0.067VB + 0.738
0.037VA  0.317VA = 0.738
97
Chapter 9
31.
See Figure 9-7.
Node A: I1  I2  I3 = 0
Node B: I3  I4  I5 = 0
I1 =
9 V  VA
R1
VA
R2
V  VB
I3 = A
R3
V  4.5 V
I4 = B
Figure 9-7
R4
V  1.5 V
I5 = B
R5
9  VA VA VA  VB
Node A:


=0
56
27
91
9 VA VA VA VB




=0
56 56 27 91 91
 2457VA  5096VA  1512VA VB 9


0
137,592
91 56
I2 =
9
 9065 
1
 
=0
VA   VB 
56
 91 
 137,592 
0.0659VA + 0.0109VB = 0.1607
Node B:
VA  VB VB  4.5 VB  15


=0
91
33
82
VA VB VB 4.5 VB 15





0
91 91 33 33 82 82
VA  2706VA  7462VA  3003VA  (32)(4.5)  (33)(15)

=0

91
246,246
2706
VA 131,171VB 864


=0
91 246,246 2706
0.0109VA  0.0535VB = 0.3193
The characteristic determinant is:
 0.0659
0.0109
= 0.0035  0.0001 = 0.0034
0.0109  0.0535
0.0034VA =
VA =
 0.1607
0.0109
= 0.0086  0.0035 = 0.0051
0.3193  0.0535
0.0051
= 1.5 V
0.0034
98
Chapter 9
0.0034VB =
VB =
32.
 0.0659
0.1607
0.0109  0.3193
= 0.0210  0.0018 = 0.0192
 0.0192
= 5.65 V
0.0034
See Figure 9-8.
Node A: I1  I2 + I3 + I4 = 0
Node B: I2 + I5  I6 = 0
Node C: I3 + I7 + I8 = 0
24 V  VA
1 k
VA  VB
I2 =
1 k
V  VA
I3 = C
1 k
VA
I4 =
1 k
I1 =
24 V  VB
1 k
VB  18 V
I6 =
1 k
10 V  VC
I7 =
1 k
18 V  VC
I8 =
1 k
I5 =
Figure 9-8
The k and V units are omitted for simplicity and the denominators are all 1.
Node A: (24  VA)  (VA  VB) + (VC  VA)  VA = 0
4VA + VB + VC = 24
Node B: (VA  VB) + (24  VB) +(VB  18) = 0
VA  3VB = 42
Node C: (VC  VA) + (10  VC) + (18  VC) = 0
VA  3VC = 28
The characteristic determinant is:
4
1
1
1 3
0 = (4)(3)(3)  (1)(3)(1)  (1)(1)(3) = 36 + 3 + 3 = 30
1
0 3
 24
1
1
30VA =  42  3
0 = (24)(3)(3)  (28)(3)(1)  (1)(42)(3) = 2166  84  126
 28
0 3
= 426
VA =
 426
= 14.2 V
 30
99
Chapter 9
 4  24
1
1  42
0 = (4)(42)(3) + (1)(28)(1)  (1)(42)(1)  (42)(1)(3)
30VB =
1  28  3
= 504  28 + 42  72 = 562
VB =
 562
= 18.7 V
 30
4
30VC =
1  24
1 3
42 = (4)(3)(28) + (1)(42)(1)  (1)(3)(24)  (1)(1)(28)
1
0  28
= 336  42  72  28 = 422
VC =
33.
 422
= 14.1 V
 30
See Figure 9-9.
4.32 V
= 2.16 mA
I7 =
2 k
VC = +4.32 V  20 V = 15.7 V
 5.25 V  ( 15.7 V) 10.43 V

I6 =
= 522 A
20 k
20 k
5.25 V
= 328 A
I4 =
16 k
I1 = I6  I4 = 522 A  328 A = 193 A
VA = 5.25 V + (193 A)(8 k) = 5.25 V + 1.55 V = 3.70 V
3.70 V
I2 =
= 370 A
10 k
I5 = I7  I4  I2 = 2.16 mA  328 A  370 A = 1.46 mA
VB = (1.46 mA)(4 k) = 5.85 V
V  VB  3.70 V  (5.85 V) 2.14 V
I3 = A


= 179 A
12 k
12 k
12 k
I8 = I3 + I5 = 179 A + 1.46 mA = 1.64 mA
100
Figure 9-9
Chapter 9
Multisim Troubleshooting and Analysis
34.
No fault.
35.
No fault.
36.
VA = 0.928 V; VB = 5.190 V
37.
R4 is open.
38.
V1 = 4.939 V; V2 = 2.878 V
39.
Lower fuse is open.
40.
R3 is open.
41.
R4 is open.
101
Chapter 10
Magnetism and Electromagnetism
Note: Solutions show conventional current direction.
Section 10-1 The Magnetic Field
1.
Since B =
2.
B=
3.
B=
4.

A


A
, when A increases, B (flux density) decreases.
1500 Wb
= 3000 Wb/m2 = 3000 T
0.5 m 2

A
There are 100 cm per meter:
1m
1 m2

100 cm 10,000 cm 2
Converting 150 cm 2 to m2:
 1 m2

 = 0.015 m2
A = 150 cm 2 
2 
10
,
000
cm


3
 = BA = (2.5  10 T)(0.015 m2) = 37.5 Wb
1 T = 104 gauss
 1T 
B = (0.6 gauss)  4
 = 60 T
 10 gauss 
5.
 104 gauss 
B = (100,000 T) 
 = 1000 gauss
 1T

Section 10-2 Electromagnetism
6.
The compass needle turns 180.
7.
r =
8.
Reluctance =

0
0 = 4  107 Wb/At  m
750  106 Wb/At  m
= 597
r =
4  10-7 Wb/At  m
1
0.28 m

= 233,333 At/Wb
-7
A (150  10 Wb/At  m)(0.08 m 2 )
102
9.
Fm = NI = (50 t)(3 A) = 150 At
Section 10-3 Electromagnetic Devices
10.
The plunger is retracted when the solenoid is activated.
11.
(a)
(b)
12.
When SW1 is closed, there is current through the relay coil, moving the armature from contact
1 to contact 2. This action causes current through lamp 1 to stop and current to begin through
lamp 2.
13.
When there is current through the coil of a d’Arsonval meter movement, it creates a magnetic
field around the coil. This reinforces the permanent field on one side of the coil and weakens it
on the other, causing the coil to move because of the differential field strength.
The electromagnetic field causes the plunger to move when the solenoid is activated.
The spring force returns the plunger to its inactive position.
Section 10-4 Magnetic Hysteresis
14.
Fm = 150 At
F
150 At
H= m 
= 750 At/m
l
0.2 m
15.
The flux density can be changed without altering the core characteristics by changing the
current or changing the number of turns.
16.
(a)
(b)
(c)
17.
Fm NI (500 t)(0.25A)


= 417 At/m
l
l
0.3 m
Fm
NI
=

reluctance l / A
 = ro = (250)(4  107) = 3142  107 Wb/At m
A = (2 cm)(2 cm) = (0.02 m)(0.02 m) = 4  104 m2
(500 t)(0.25 A)
125 At
=

= 5.23 Wb
6

2.39
10
At/Wb


0.3 m


  3142  107 Wb/At  m  4  104 m 2 


 5.23 Wb
= 0.13 T
B= 
A 4  104 m 2
H=
Material A has the most retentivity.
Section 10-5 Electromagnetic Induction
18.
The induced voltage doubles when the rate of change of magnetic flux doubles.
19.
The strength of the magnetic field, the length of the conductor exposed to the field, and the
velocity of the conductor relative to the field.
103
Chapter 10
20.
 d 
Vind = N   = 50(3500  103 Wb/s = 175 V
 dt 
21.
Lenz’s law defines the polarity of the induced voltage.
22.
The magnetic field is not changing, therefore, there is no induced voltage.
Section 10-6 The DC Generator
23.
The commutator and brush assembly electrically connect the loop to the external circuit.
24.
60 rps  2 peaks/rev = 120 peaks/s
25.
See Figure 10-1.
Figure 10-1
26.
I A  I L  I F  12 A + 1 A = 13 A
27.
(a) PL = IL VL = (12 A)(14 V) = 168 W
(b) PF = IF VL = (1 A)(14 V) = 14 W
Section 10-7 The DC Motor
28.
(a) P = 0.105Ts = (0.105)(3.0 N-m)(1200 rpm) = 378 W
(b) 378 W/746 W/hp = 0.51 hp
29.
PT = Pint + PL = 12 W + 50 W = 62 W
PL = 50 W
Efficiency = PL/PT = 50 W/62 W = 81%
104
Chapter 11
Introduction to Alternating
Current and Voltage
Section 11-1 The Sinusoidal Waveform
1.
(a)
(b)
(c)
2.
(a)
(b)
(c)
3.
T=
4.
T=
5.
T
1
1
= 1 Hz

T 1s
1
1
f= 
= 5 Hz
T 0.2 ms
1
1
f= 
= 20 Hz
T 50 ms
(d)
f=
(e)
(f)
1
1

= 1s
f 1 Hz
1
1
T=

= 16.7 ms
f
60 Hz
1
1

= 2 ms
T=
f 500 Hz
T=
(d)
(e)
(f)
1
1
= 1 kHz

T 1 ms
1
1
= 2 kHz
f= 
T 500 s
1
1
f= 
= 100 kHz
T 10 s
f=
1
1

= 1 ms
f 1 kHz
1
1
T=

= 5 s
f
200 kHz
1
1

T=
= 200 ns
f 5 MHz
T=
10 s
= 2 s
5 cycles
1
1

= 20 s
f 50 kHz
10 ms
= 500 cycles
0.02 ms
1
1

 0.1ms
f 10kHz
Time for 100 cycles = 100(0.1 ms) = 10 ms
Section 11-2 Sinusoidal Voltage and Current Values
6.
(a)
(b)
(c)
Vrms = 0.707Vp = 0.707(12 V) = 8.48 V
Vpp = 2Vp = 2(12 V) = 24 V
Vavg = 0 V over a full cycle.
Vavg = 0.637(12 V) = 7.64 over a half cycle.
105
Chapter 11
7.
(a)
(b)
(c)
8.
Ip = 1.414Irms = 1.414(5 mA) = 7.07 mA
Iavg = 0 A over a full cycle
Iavg = 0.637Ip = 0.637(7.07 mA) = 4.5 mA over a half cycle
Ipp = 2Ip = 2(7.07 mA) = 14.14 mA
Vp = 25 V
Vpp = 2Vp = 50 V
Vrms = 0.707Vp = 17.7 V
Vavg = 0.637Vp = 15.9 V
Section 11-3 Angular Measurement of a Sine Wave
9.
(a)
(b)
(c)
10.
(a)
(b)
(c)
  rad 

30 
 180 
  rad 
45 

 180 
  rad 

78 
 180 

rad
6

rad
4
39
rad
90

 57.3 
 rad 
 = 22.5
8
 rad 

 57.3 
 rad 
 = 60
3
 rad 

 57.3 
 rad 
 = 90
2
 rad 
(d)
(e)
(f)
(d)
(e)
(f)
3
  rad 
rad

135 
4
 180 
10
  rad 
rad
200 

9
 180 
5
  rad 
rad

300 
3
 180 
 3
 57.3 
 rad 
 = 108
 5
 rad 
 6
 57.3 
rad 

 = 216
 5
 rad 
 57.3 
(1.8 rad)
 = 324
 rad 
11.
 = 45  30 = 15 A leading B
12.
With respect to 0: sine wave with a peak at 75 is shifted 15 to left. Sine wave with a peak
at 100 is shifted 10 to right.
Phase difference =  = 100  75 = 25
13.
See Figure 11-1.
Figure 11-1
106
Chapter 11
Section 11-4 The Sine Wave Formula
14.
Vp = 1.414(20 V) = 28.28 V
(a)
v = Vpsin  = (28.28 V)sin15 = 7.32 V
(b) v = Vpsin  = (28.28 V)sin 33 = 15.4 V
(c)
v = Vpsin  = (28.28 V)sin 50 = 21.7 V
(d) v = Vpsin  = (28.28 V)sin 110 = 26.6 V
(e)
v = Vpsin  = (28.28 V)sin 70 = 26.6 V
(f)
v = Vpsin  = (28.28 V)sin 145 = 16.2 V
(g) v = Vpsin  = (28.28 V)sin 250 = 26.6 V
(h) v = Vpsin  = (28.28 V)sin 325 = 16.2 V
15.
(a)
(b)
(c)
(d)
(e)
(f)
16.
Vp = 1.414Vrms = 1.414(6.37 V) = 9 V

(a)
= 22.5
8
v = (9 V)sin 22.5 = 3.44 V

(b)
= 45
4
v = (9 V)sin 45 = 6.36 V

= 90
(c)
2
v = (9 V)sin 90 = 9 V
3
= 135
(d)
4
v = (9 V)sin 135 = 6.36 V
i = Ipsin  = (100 mA)sin 35 = 57.4 mA
i = Ipsin  = (100 mA)sin 95 = 99.6 mA
i = Ipsin  = (100 mA)sin 190 = 17.4 mA
i = Ipsin  = (100 mA)sin 215 = 57.4 mA
i = Ipsin  = (100 mA)sin 275 = 99.6 mA
i = Ipsin  = (100 mA)sin 360 = 0 mA
17.
vB = (15 V)sin (30 + 30) = 13.0 V
vB = (15 V)sin (30 + 45) = 14.5 V
vB = (15 V)sin (30 + 90) = 13.0 V
vB = (15 V)sin (30 + 180) = 7.5 V
vB = (15 V)sin (30 + 200) = 11.5 V
vB = (15 V)sin (30 + 300) = 7.5 V
18.
(a)
(b)
(c)
(d)
(e)
(f)
(e)
(f)
(g)
 = 180
v = (9 V)sin 180 = 0 V
3
= 270
2
v = (9 V)sin 270 = 9 V
2 = 360
v = (9 V)sin 360 = 0 V
vB = (15 V)sin (  30) = (15 V)sin(30  30) = (15 V)sin(0) = 0 V
vB = (15 V)sin (  30) = (15 V)sin(45  30) = (15 V)sin(15) = 3.88 V
vB = (15 V)sin (  30) = (15 V)sin(90  30) = (15 V)sin(60) = 13.0 V
vB = (15 V)sin (  30) = (15 V)sin(180  30) = (15 V)sin(150) = 7.5 V
vB = (15 V)sin (  30) = (15 V)sin(200  30) = (15 V)sin(170) = 2.60 V
vB = (15 V)sin (  30) = (15 V)sin(300  30) = (15 V)sin(270) = 15 V
107
Chapter 11
19.
1
1

= 4.55 s
f
2.2 kHz
At t = 0.12 ms = 120 s:
 120 s 
= 
 360 = 94.9
 455 s 
25 V
Vp =
= 35.4 V
0.707
v = (35.4 V)sin 94.9 = 35.3 V
At t = 0.2 ms = 200 s:
 200 s 
= 
 360 = 158
 455 s 
v = (35.4 V)sin 158 = 13.3 V
v = 35.4 V  13.3 V = 22.1 V
T=
Section 11-5 Introduction to Phasors
20.
See Figure 11-2.
Figure 11-2
21.
See Figure 11-3.
Figure 11-3
22.
 = 2f
(a)
(b)
(c)

60
= 9.55 Hz
2  2
 360
f=
= 57.3 Hz

2  2
2


f=
= 0.318 Hz
2  2
f=

108
Chapter 11
23.


1256
= 200 Hz
2
(d)
f=
(a)
v = Vpsin(t  /4) = (1 V)sin[2(5 kHz)(30 s)  /4]
= (1 V)sin(0.3  0.25) = (1 V)sin(0.05) = (1 V)(0.156) = 156 mV
(b)
v = Vpsin(t  /4) = (1 V)sin[2(5 kHz)(75 s)  /4]
= (1 V)sin(0.75  0.25) = (1 V)sin(0.5) = (1 V)(1) = 1 V
(c)
v = Vpsin(t  /4) = (1 V)sin[2(5 kHz)(125 s)  /4]
= (1 V)sin(1.25  0.25) = (1 V)sin() = (1 V)(0) = 0 V
2
Section 11-6 Analysis of AC Circuits
24.
(a)
(b)
(c)
(d)
(e)
 Vp 
 10 V 
 = 7.07 mA
Irms = 0.707   0.707
 R
 1 k 
2
Iavg =  10 mA = 6.37 mA
 
10 V
Ip =
= 10 mA
1 k
Ipp = 2(10 mA) = 20 mA
i = Ip = 10 mA
25.
V2(rms) = V4  V3 = 65 V  30 V = 35 V
V2(p) = 1.414(35 V) = 49.5 V
V2(AVG) = 0.637(49.5 V) = 31.5 V
V1(rms) = Vs  V4 = 120 V  65 V = 55 V
V1(p) = 1.414(55 V) = 77.8 V
V1(AVG) = 0.637(77.8 V) = 49.6 V
26.
Ipp =
16 V 16 V

= 16 mA
R1
1 k
 I pp 
 16 mA 
  0.707
Irms = 0.707 
 = 5.66 mA
 2 
 2 
VR4 = IrmsR4 = (5.66 mA)(560 ) = 3.17 V rms
Applying Kirchhoff’s voltage law:
VR1 + VR2 + VR3 + VR4 = Vs
0.707(8 V) + 5 V + VR3 + 3.17 V = 0.707(30 V)
VR3 = 21.21 V  5.66 V  5 V  3.17 V = 7.38 V
27.
Vp = (1.414)(10.6 V) = 15 V
Vmax = 24 V + Vp = 39 V
Vmin = 24 V  Vp = 9 V
109
Chapter 11
28.
Vp = (1.414)(3 V) = 4.242 V
VDC = Vp = 4.24 V
29.
Vmin = VDC  Vp = 5 V  6 V = 1 V
30.
Vrms = 0.707Vp = 0.707(150 V) = 106.1 V
V2
V 2 (106.1 V) 2 (200 V) 2
= 112.6 W + 400 W = 513 W
P = Pac + PDC = rms  S 

RL
RL
100 
100 
Section 11-7 The Alternator (AC Generator)
31.
f = (number of pole pairs)(rps) = (1)(250 rps) = 250 Hz
32.
f = (number of pole pairs)(rps)
3600 rpm
rps =
= 60 rps
60 s/m
f = (2 pole pairs)(60 rps) = 120 Hz
f
400 Hz
rps =

= 200 rps
pole pairs
2
33.
34.
f = (number of pole pairs)rps
#pole pairs 
f
400Hz
400Hz


8
rps  3000rpm  50rps


 60s/m 
# poles = 2(# pole pairs) = 2 X 8 = 16
Section 11-8 The AC Motor
35.
A one-phase motor requires a starting winding or other means to produce torque for starting the
motor, whereas a three-phase motor is self-starting.
36.
The field is set up by current in the stator windings. As the current reaches a peak in one
winding, the other windings have less current and hence less effect on the field. The result is a
rotating field.
Section 11-9 Nonsinusoidal Waveforms
37.
The approximate values determined from the graph are:
tr  3.5 ms  0.5 ms = 3.0 ms
tf  16 ms  13 ms = 3.0 ms
tW  14.5 ms  2.5 ms = 12.0 ms
Amplitude = 5 V
110
Chapter 11
38.
1
= 0.5 ms = 500 s
2 kHz
 1 s 
t 
100% = 0.2%
% duty cycle =  W 100%  
T 
 500 s 
T=
39.
Vavg = baseline + (duty cycle)(amplitude)
t
1 s
= 0.167
duty cycle = W 
T
6 s
Vavg = 5 V + (0.167)(5 V) = 5.84 V
40.
(a)
(b)
41.
42.
Vavg = baseline + (duty cycle)(amplitude)
Vavg = 1 V + (0.25)(2.5 V) = 0.375 V
(b)
Vavg = baseline + (duty cycle)(amplitude)
Vavg = 1 V + (0.67)(3 V) = 3.01 V
(a)
(a)
(b)
44.
45.
 1 s 

100% = 25%
100%  

 4 s 

 20 ms 
100%  
100% = 66.7%

 30 ms 
(a)
(b)
43.
t
% duty cycle =  W
T
t
% duty cycle =  W
T
1
1

= 250 kHz
T 4 s
1
1
f= 
= 33.3 Hz
T 30 ms
f=
1
1

= 50 kHz
T 20 s
1
1
f= 
= 10 Hz
T 100 ms
f=
area under curve (0 V  1 V  2 V  3 V  4 V  5 V  6 V)(1 ms)

period
7 ms
21V  ms
=3V
=
7 ms
Average value =
1
1

= 25 kHz (fundamental)
T 40 s
3rd harmonic = 75 kHz
5th harmonic = 125 kHz
7th harmonic = 175 kHz
9th harmonic = 225 kHz
11th harmonic = 275 kHz
13th harmonic = 325 kHz
f=
111
Chapter 11
46.
f=
1
1

= 25 kHz
T 40 s
Section 11-10 The Oscilloscope
47.
Vp = (3 div)(0.2 V/div) = 600 mV
T = (10 div)(50 ms/div) = 500 ms
48.
Vp(in) = (1 div)(5 V/div) = 5 V
Tin = (2 div)(0.1 ms/div) = 200 s
1
fin =
= 5 kHz
200 s
Rtot = 560  + (470   (560  + 470 )) = 560  + 323  = 883 
 470   323  

 323  
470 

5 V = 835 mV

V p (in )  
Vp(out) = 
 1030   883  
 470   560   883  
fout = fin = 5 kHz
49.
Vp(out) = (3 div)(0.2 V/div) = 0.6 V
Tout = (10 div)(50 ms/div) = 500 ms
1
fout =
= 2 Hz
500 ms
Rtot = 1 k + 1 k  3.2 k = 1762 
 1762  
 1762  
(3.2 V)(0.6 V) = 4.44 V
(3.2 V)V p ( out )  
Vp(in) = 
 762  
 762  
fout = fin = 2 Hz
Multisim Troubleshooting and Analysis
50.
VR1 = 199.411 Vpp = 70.509 Vrms; VR2 = 111.685 Vpp = 39.487 Vrms
51.
VR1 = 16.717 Vpp = 5.911 Vrms; VR2 = 36.766 Vpp = 13.005 Vrms; VR3 = 14.378 Vpp = 5.084 Vrms
52.
R2 open.
55.
VMIN = 2.000 Vp; VMAX = 22.000 Vp
56.
VMIN = 4.000 Vp; VMAX = 16.000 Vp
53.
No fault.
54.
112
R1 is open.
Chapter 12
Capacitors
Section 12-1 The Basic Capacitor
1.
(a)
(b)
(c)
Q 50 C
= 5 F

V
10 V
Q = CV = (0.001 F)(1 kV) = 1 C
Q
2 mC
V=

= 10 V
C 200 F
C=
2.
(a)
(b)
(c)
(0.1 F)(106 pF/F) = 100,000 pF
(0.0025 F)(106 pF/F) = 2500 pF
(4.7 F)(106 pF/F) = 4,700,000 pF
3.
(a)
(b)
(c)
(1000 pF)(106 F/pF) = 0.001 F
(3500 pF)(106 F/pF) = 0.0035 F
(250 pF)(106 F/pF) = 0.00025 F
4.
(a)
(b)
(c)
(0.0000001 F)(106 F/F) = 0.1 F
(0.0022 F)(106 F/F) = 2200 F
(0.0000000015 F)(106 F/F) = 0.0015 F
5.
W=
6.
1
1
CV 2  (1  1000 F)(500 V) = 125 J
2
2
1
W =  CV 2
2
2W 2(10 mJ)
C= 2 
= 2 F
(100 V) 2
V
7.
(a)
(b)
(c)
(d)
8.
C=
9.
C=
Air:  = r0 = 1(8.85  1012 F/m) = 8.85  1012 F/m
Oil:  = r0 = 4.0(8.85  1012 F/m) = 35.4  1012 F/m
Glass:  = r0 = 7.5(8.85  1012 F/m) = 66.4  1012 F/m
Teflon:  = r0 = 2.0(8.85  1012 F/m) = 17.7  1012 F/m
A r (8.85  1012 F/m) (1.44  103 )(5)(8.85  1012 F/m)

= 0.001 F
d
6.35  105 m
A r (8.85  1012 F/m)
d
2
(0.05 m )(1.0)(8.85  1012 F/m)
=
= 983 pF
4.5  104 m
113
Chapter 12
10.
A r 8.85  1012
d
Cd
(1)(8  105 )
= 3.6  106 m2
A=

 r 8.85  1012 (2.5)(8.85  1012 )
C=
l = A = 1.9  103 m (almost 1.2 miles on a side!)
The capacitor is too large to be practical and will not fit in the Astrodome!
A r 8.85  1012 (0.09)(2.5)(8.85  1012 )

= 24.9 nF = 0.0249 F
d
(8.0  105 )
11.
C=
12.
T = 50 C
(200 ppm/C)50 C = 10,000 ppm
1  103
C =
(10  103 ppm) = 10 pF
6
1  10
C75 = 1000 pF  10 pF = 990 pF
13.
T = 25 C
(500 ppm/C)25 C = 12,500 ppm
(1  106 pF/F)(0.001 F) = 1000 pF
1000
(12.5  103 ppm) = +12.5 pF
C =
6
1  10
Section 12-2 Types of Capacitors
14.
The plate area is increased by increasing the number of layers of plate material and dielectric.
15.
Ceramic has the highest dielectric constant (r = 1200).
16.
See Figure 12-1.
Figure 12-1
17.
Aluminum, tantalum; electrolytics are polarized, others are not.
18.
(a)
(b)
(c)
(d)
Encapsulation
Dielectric (ceramic disk)
Plate (metal disk)
Conductive leads
19.
(a)
(b)
(c)
(d)
0.022 F
0.047 F
0.001 F
220 pF
114
Chapter 12
Section 12-3 Series Capacitors
20.
CT =
21.
(a)
(b)
(c)
22.
(a)
(b)
(c)
1000 pF
= 200 pF
5
1
= 0.688 F
1
1

1 F 2.2 F
1
= 69.7 pF
CT =
1
1
1


100 pF 560 pF 390 pF
1
= 2.64 F
CT =
1
1
1
1



10 F 4.7 F 47 F 22 F
CT =
CT = 0.688 F
 C 
 0.688 F 
V1F =  T 10 V  
10 V = 6.88 V
 1 F 
 1 F 
 CT 
 0.688 F 
V2.2F = 
10 V  
10 V = 3.13 V
 2.2 F 
 2.2 F 
CT = 69.7 pF
 CT 
 69.7 pF 
V100pF = 
100 V  
100 V = 69.7 V
 100 pF 
 100 pF 
 CT 
 69.7 pF 
V560pF = 
100 V  
100 V = 12.4 V
 560 pF 
 560 pF 
 CT 
 69.7 pF 
V390pF = 
100 V  
100 V = 17.9 V
 390 pF 
 390 pF 
CT = 2.64 F
 C 
 2.64 F 
V10F =  T  30 V  
 30 V = 7.92 V
 10 F 
 10 F 
 CT 
 2.64 F 
V4.7F = 
 30 V  
 30 V = 16.9 V
 4.7 F 
 4.7 F 
 C 
 2.64 F 
V47F =  T  30 V  
 30 V = 1.69 V
 47 F 
 47 F 
 C 
 2.64 F 
V22F =  T  30 V  
 30 V = 3.60 V
 22 F 
 22 F 
115
Chapter 12
23.
C 
Vx =  T VS
 Cx 
CV
(1 F)(8 V)
= 0.667 F
CT = x x 
12 V
VS
C
Cx =  T
V
24.

 0.667 F 
12 V = 2 F
VS  
 4V 

QT = Q1 = Q2 = Q3 = Q4 = 10 C
10 C
Q
= 2.13 V
V1 = 1 
C1 4.7 F
Q
10 C
= 10 V
V2 = 2 
C2
1 F
10 C
Q
V3 = 3 
= 4.55 V
C3 2.2 F
Q
10 C
=1V
V4 = 4 
C4 10 F
Section 12-4 Parallel Capacitors
25.
(a)
(b)
CT = 47 pF + 10 pF + 1000 pF = 1057 pF
CT = 0.1 F + 0.01 F + 0.001 F + 0.01 F = 0.121 F
26.
(a)
Q = CV
Q47pF = (47 pF)(10 V) = 470  1012 C = 470 pC
Q10pF = (10 pF)(10 V) = 100  1012 C = 100 pC
Q1000pF = (1000 pF)(10 V) = 10,000  1012 C = 0.01 C
(b)
Q = CV
Q0.1F = (0.1 F)(5 V) = 0.5 C
Q0.01F = (0.01 F)(5 V) = 0.05 C
Q0.001F = (0.001 F)(5 V) = 0.005 C
Q10000pF = (10000 pF)(5 V) = 0.05 C
(a)
= 2.62 F
1
1
1


10 F 10 F 2.2 F  3.3 F
1
1
1
CT =


1
1
1
1
1
1



100 pF 100 pF 1000 pF 470 pF 0.001 F 470 pF
= 50 pF + 319.7 pF + 319.7 pF = 689 pF
27.
(b)
CT =
1
116
Chapter 12
(c)
1
CT =
1
1
1
1

1 F 1 F
28.
(a)
(b)
 1 F = 1.6 F
 1 F
 1 F
CT = 2.62 F
 2.62 F 
5 V = 2.62 V
VAB = 
 5 F 
470 pF
See Figure 12-2(a):
For this part of the circuit:
1
CT =
= 319.7 pF
1
1

0.001 F 470 pF
 319.7 pF 
10 V = 3.20 V
VAB = 
 0.001 F 
Figure 12-2
(c)
See Figure 12-2(b).
CAB = 1.5 F
For this part of the circuit:
1
CT =
= 0.6 F
1
1

1 F 1.5 F
 0.6 F 
10 V = 4 V
VAB = 
 1.5 F 
29.
(a)
(b)
CT = C1,2 + C3,4 = 0.00872 F + 0.0256 F = 0.0343 F
QT = CTVT = (0.0343 F)(12 V) = 0.411 C
 C2 


0.068 F
V1 = 
VT  
12 V = 10.47 V
 0.01 F  0.068 F 
 C1  C2 
 C1 


0.01 F
V2 = 
VT  
12 V = 1.54 V
 0.01 F  0.068 F 
 C1  C2 
 C4 


0.056 F
V3 = 
VT  
12 V = 6.52 V
 0.047 F  0.056 F 
 C3  C4 
 C3 


0.047 F
V4 = 
VT  
12 V = 5.48 V
 0.047 F  0.056 F 
 C3  C4 
117
Chapter 12
Section 12-5 Capacitors in DC Circuits
30.
(a)
(b)
(c)
(d)




31.
(a)
(b)
(c)
(d)
5
5
5
5
32.
 = RC = (10 k)(0.001 F) = 10 s
vC = VF(1  et/RC) = 15 V(1  e10s/10s) = 15 V(1  e1) = 9.48 V
(a)
(b) vC = VF(1  et/RC) = 15 V(1  e20s/10s) = 15 V(1  e2) = 13.0 V
vC = VF(1  et/RC) = 15 V(1  e30s/10s) = 15 V(1  e3) = 14.3 V
(c)
(d) vC = VF(1  et/RC) = 15 V(1  e40s/10s) = 15 V(1  e4) = 14.7 V
vC = VF(1  et/RC) = 15 V(1  e50s/10s) = 15 V(1  e5) = 14.9 V
(e)
33.
 = RC = (1 k)(1.5 F) = 1.5 ms
vC = Viet/RC = (25 V)e1.5ms/1.5ms = (25 V)e1 = 9.20 V
(a)
(b) vC = Viet/RC = (25 V)e4.5ms/1.5ms = (25 V)e3 = 1.24 V
(c)
vC = Viet/RC = (25 V)e6ms/1.5ms = (25 V)e4 = 0.458 V
(d) vC = Viet/RC = (25 V)e7.5ms/1.5ms = (25 V)e5 = 0.168 V
34.
(a)
(b)
(c)
vC = VF(1  et/RC) = 15 V(1  e2s/10s) = 15 V(1  e0.2) = 2.72 V
vC = VF(1  et/RC) = 15 V(1  e5s/10s) = 15 V(1  e0.5) = 5.90 V
vC = VF(1  et/RC) = 15 V(1  e15s/10s) = 15 V(1  e1.5) = 11.7 V
35.
(a)
(b)
(c)
vC = Viet/RC = (25 V)e0.5ms/1.5ms = (25 V)e0.333 = 17.9 V
vC = Viet/RC = (25 V)e1ms/1.5ms = (25 V)e0.667 = 12.8 V
vC = Viet/RC = (25 V)e2ms/1.5ms = (25 V)e1.333 = 6.59 V
36.
vC = VF(1  et/RC) = VF  VFet/RC
VFet/RC = VF  vC
V  vC
et/RC = F
VF
= RC = (100 )(1 F) = 100 s
= RC = (10 M)(47 pF) = 470 s
= RC = (4.7 k)(0.0047 F) = 22.0 s
= RC = (1.5 M)(0.01 F) = 15 ms
= 5RC = 5(56 )(47 F) = 13.2 ms
= 5RC = 5(3300 )(0.015 F) = 247.5 s
= 5RC = 5(22 k)(100 pF) = 11 s
= 5RC = 5(5.6 M)(10 pF) = 280 s
 V  vC 

ln et/RC = ln F
 VF 
 V  vC 
t


= ln F
RC
 VF 
 v
t = RC ln1  C
 VF



6V 

t = (2.2 k)(0.01 F) ln 1 
 = 15.2 s
 12 V 
118
Chapter 12
37.
 v
t = RC ln1  C
 VF
38.
v
t = RC ln C
 Vi
39.
40.
41.

8V 

 = (10 k)(0.001 F) ln1 
 = 7.62 s
 15 V 


3V 
 = (1 k)(1.5 F) ln
 = 3.18 ms

 25 V 

Looking from the capacitor, the Thevenin resistance is:
RTH = R3 + R1  R2  R4 = 1 k + 1 k  2.2 k  1.5 k = 1.47 k
 = RTHC = (1.47 k)0.0022 F) = 3.00 s
 v 
t = RC ln1  C 
 VF 
t
 10 s

R=
= 7.86 k
 vC 
 7 .2 
 (1000 pF)ln1 

C ln1 
10 

 VF 
See Figure 12-3(a).
1 = (R1 + R2)C = (57 k)(1 F) = 57 ms
2 = (R2 + R3)C = (43 k)(1 F) = 43 ms
vC = 20 V(1  e10 ms/57 ms) = 3.22 V
See Figure 12-3(b).
vC = (3.22 V)e5 ms/43 ms = 2.85 V
Figure 12-3
119
Chapter 12
Section 12-6 Capacitors in AC Circuits
42.
(a)
(b)
(c)
1
1

= 3.39 k
2fC 2(1 kHz)(0.047 F)
CT = 10 F + 15 F = 25 F
1
1
XC =

= 6.37 k
2fCT 2(1 Hz)(25 F)
1
CT =
= 0.5 F
1
1

1 F 1 F
1
1
XC =

= 5.31 k
2fCT 2(60 Hz)(0.5 F)
XC =
43.
CT for each circuit was found in Problem 27.
1
1

= 30.4 
(a)
XC =
2fCT 2(2 kHz)(2.62 F)
1
1
(b) XC =

= 116 k
2fCT 2(2 kHz)(689 pF)
1
1

= 49.7 
(c)
XC =
2fCT 2(2 kHz)(1.6 F)
44.
(a)
For XC = 100 :
1
1
= 33.9 kHz
f=

2X C C 2(100 )(0.047 F)
For XC = 1 k:
1
1

= 3.39 kHz
f=
2X C C 2(1 k)(0.047 F)
(b)
For XC = 100 :
1
1
= 63.7 Hz
f=

2X C C 2(100 )(25 F)
For XC = 1 k:
1
1

= 6.37 Hz
f=
2X C C 2(1 k)(25 F)
(c)
For XC = 100 :
1
1
= 3.18 kHz
f=

2X C C 2(100 )(0.5 F)
For XC = 1 k:
1
1

f=
= 318 Hz
2X C C 2(1 k)(0.5 F)
120
Chapter 12
Vrms
20 V

= 200 
I rms 100 mA
45.
XC =
46.
Vrms = IrmsXC
1
= 3.39 k
2(10 kHz)(0.0047 F)
Vrms = (1 mA)(3.39 k) = 3.39 V
XC =
47.
48.
1
= 3.39 k
2fC
Ptrue = 0 W
2
Pr = I rms
X C =(1 mA)2(3.39 k) = 3.39 mVAR
XC =
C5-6 = 0.006 F, C4-5-6 = 0.053 F, C3-4-5-6 = 0.012 F, C2-3-4-5-6 = 0.034 F
CT = 0.008 F, XCT = 66.3 k
V
10 V
IC1 = s 
= 151 A
X CT 66.3 k
C 
 0.008 F 
10 V = 8.00 V
VC1 =  T Vs  
 0.01 F 
 C1 
VC2 = Vs  VC1 = 10 V  8.00 V = 2.00 V
XC2 = 24.1 k
V
2.00 V
IC2 = C 2 
= 83.0 A
X C 2 24.1 k
C

 0.012 F 
2.00 V = 1.6 V
VC3 =  3 4 5  6 VC 2  
 0.015 F 
 C3 
XC3 = 35.4 k
V
1.6 V
IC3 = C 3 
= 45.2 A
X C 3 35.4 k
VC4 = VC2  VC3 = 2.00 V  1.6 V = 400 mV
XC4 = 11.3 k
V
400 mV
IC4 = C 4 
= 35.4 A
X C 4 11.3 k
C 
 0.006 F 
400 mV = 240 mV
VC5 =  5  6 VC 4  
 0.01 F 
 C5 
XC5 = 53.1 k
V
240 mV
IC5 = IC6 = C 5 
= 4.52 A
X C 5 53.1 k
VC6 = VC4  VC5 = 400 mV  240 mV = 160 mV
121
Chapter 12
49.
VC2 = VC3 = (4 mA)XC3 = (4 mA)(750 ) = 3 V
1
1

= 141.5 kHz
f=
2X C 3C3 2(750 )(0.0015 F)
1
1
XC2 =

= 511.3 
2fC2 2(141.5 kHz)(0.0022 F)
V
3V
= 5.87 mA
IC2 = C 2 
X C 2 511.3 
IC1 = ICT = IC2 + IC3 = 5.87 mA + 4 mA = 9.87 mA
VC1 = 5 V  3 V = 2 V
V
2V
XC1 = C1 
= 203 
I C1 9.87 mA
1
1

= 0.00541 F
C1 =
2fX C1 2(141.5 kHz)(203 )
50.
CT(3,5,6)
1
1
= 0.0043 F

1
1
1
1
1
1




C3 C5 C6
0.015 F 0.01 F 0.015 F
CT(2,3,5,6) = 0.022 F + 0.0043 F = 0.0263 F
1
1
CT =

= 0.00725 F
1
1
1
1


C1 C2,3,5,6
0.01 F 0.0263 F
C
VC1 =  T
 C1
C
VC2 =  T
 C2

 0.00725 F 
10 V  
10 V = 7.25 V
 0.01 F 


 0.00725 F 
10 V  
10 V = 3.30 V
 0.022 F 

 CT(3,5,6)
VC3 = 
 C3
 CT(3,5,6)
VC5 = 
 C5

 0.0043 F 
VC 2  
 3.30 V = 945 mV
 0.015 F 


 0.0043 F 
VC 2  
 3.30 V = 1.42 V
 0.01 F 

VC6 = 945 mV
Section 12-7 Capacitor Applications
51.
The ripple voltage is reduced when the capacitance is increased.
52.
The reactance of the bypass capacitor should ideally be 0  in order to provide a short to
ground for ac.
122
Chapter 12
Section 12-8 Switched-Capacitor Circuits
53.
54.
T
10 s

= 4.55 k
C 2200 pF
1
1
T= 
= 125 s
f 8 kHz
T 125 s
= 1.25 M
R= 
C 100 pF
R=
Multisim Troubleshooting and Analysis
55.
VC = 3.103 V; VC2 = 6.828 V; VC3 = 2.069 V
56.
VC1 = 48.837 V; VC2 = 51.163 V; VC3 = 51.163 V; VC4 = 51.163 V
57.
IC(1kHz) = 1.383 mA; IC(500Hz) = 0.691 mA; IC(2kHz) = 2.768 mA
58.
C4 is open.
59.
C4 is shorted.
123
Chapter 13
Inductors
Section 13-1 The Basic Inductor
1.
(a)
(b)
(c)
(d)
1 H  1000 mH/H = 1000 mH
250 H  0.001 mH/H = 0.25 mH
10 H  0.001 mH/H = 0.01 mH
0.0005 H  1000 mH/H = 0.5 mH
2.
(a)
(b)
(c)
(d)
(300 mH)(103) = 300,000 H
(0.08 H)(106) = 80,000 H
(5 mH)(103) = 5000 H
(0.00045)(103) = 0.45 H
3.
4.
 10 mA 
 di 
 = 50 mV
vind = L   5 H
 dt 
 s 
 di 
v = L 
 dt 
di v 50 mV
 
= 2000 A/s
dt L 25 mH
6.
 di 
 200 mA 
vind = L   100 mH
 = 20 mV
 dt 
 1s 
N 2A
L=
l
(30 mH)(0.05 m)
Ll
N=
= 3536 turns

A
(1.2  10  6 )(10  10 5 m 2 )
7.
W=
1 2 1
LI  (4.7 mH)(20 mA) 2 = 0.94 J
2
2
8.
L=
N 2 A
; Inductor 2 has 4 times the inductance of inductor 1.
l
5.
124
Chapter 13
9.
N 2A
l
1 = r0 = 200 0
2 = r0 = 150 0
L=
2
150 3
=

200 4
1
Therefore, coil 2 has 3/4 the inductance of coil 1.
3
L2 = L1
4
10.
A =  r2 = (0.0035 m)2 = 38.5  106 m2
N 2A 1002 (4  106 H/m)(38.5  106 m 2 )
L=
=
= 138 H
l
0.035 m
Section 13-3 Series and Parallel Inductors
11.
LT = 5 H + 10 H + 20 H + 40 H + 80 H = 155 H
12.
Lx = 50 mH  10 mH  22 mH = 18 mH
13.
LT = L1 + L2 + L3 = 50 mH + 500 H + 0.01 mH = 50.5 mH
14.
Position 1:
LT = 330 H + 680 H = 1010 H
Position 2:
LT = 680 H + 800 H = 1480 H
Position 3:
LT = 800 H
Position 4:
LT = 1.5 mH + 800 H = 2300 H
15.
LT =
16.
1
= 7.14 H
1
1
1
1



75 H 50 H 25 H 15 H
L1 (12 mH)
L1  12 mH
( 8 mH)L1 + (8 mH)(12 mH) = (12 mH)L1
(4 mH)L1 = 96 mH2
96 mH 2
L1 =
= 24 mH
4 mH
8 mH =
125
Chapter 13
17.
(a)
(b)
(c)
18.
(a)
(b)
(c)
(10 H)(5 H)
= 4.33 H
10 H  5 H
100 mH
LT =
= 50 mH
2
1
LT =
= 57.1 H
1
1
1


100 H 200 H 400 H
LT = 1 H +
(100 mH)(50 mH) (60 mH)(40 mH)

= 33.3 mH + 24 mH = 57.3 mH
150 mH
100 mH
(12 mH)(6 mH)
LT =
= 4 mH
18 mH
(2 mH)(4 mH)
LT = 4 mH +
= 5.33 mH
6 mH
LT =
Section 13-4 Inductors in DC Circuits
19.
(a)
(b)
(c)
20.
21.
L 100 H
= 1 s

R 100 
L 10 mH
= 
= 2.13 s
R 4.7 k
3H
L
= 
= 2 s
R 1.5 M
=
(a)
 50 H 
L
 = 4.46 s
5 = 5   5
R
 56  
(b)
 15 mH 
L
 = 22.7 s
5 = 5   5
R
 3300  
(c)
 100 mH 
L
 = 22.7 s
5 = 5   5
R
 22 k 
=
(a)
(b)
(c)
(d)
(e)
L 10 mH

= 10 s
R 1 .0 k 
vL = (15 V)e10s/10s = (15 V)e1 = 5.52 V
vL = (15 V)e20s/10s = (15 V)e2 = 2.03 V
vL = (15 V)e30s/10s = (15 V)e3 = 747 mV
vL = (15 V)e40s/10s = (15 V)e4 = 275 mV
vL = (15 V)e50s/10s = (15 V)e5 = 101 mV
126
Chapter 13
22.
IF =
(a)
(b)
(c)
(d)
(e)
V
15V
= 15 mA

R 1.0 k
iL = 15 mA(1 - e10s/10s ) = 15 mA(1 - e1 )= 9.48 mA
iL = 15 mA(1 - e20s/10s )= 15 mA(1 - e2 )= 13.0 mA
iL = 15 mA(1 - e30s/10s )= 15 mA(1 - e3 )= 14.3 mA
iL = 15 mA(1 - e40s/10s )= 15 mA(1 - e4 )= 14.7 mA
iL = 15 mA(1 - e50s/10s )= 15 mA(1 - e5 )= 14.9 mA
L 75mH

 9.15s
R 8.2kΩ
23.

24.
The time constant is  
L 75 mH

= 9.15 s. For the increasing exponential, the final
R 8.2 k
V
10 V
= 1.22 mA
current is IF = S 
R 8.2 k
(a)
(b)
(c)
At 10 s, i = 1.22 mA(1  e10s/9.15s) = 0.81 mA
At 20 s, i = 1.22 mA(1  e20s/9.15s) = 1.08 mA
At 30 s, i = 1.22 mA(1  e30s/9.15s) = 1.17 mA
25.
vL = (15 V)et/(L/R) = (15 V)e2s/10s = (15 V)e0.2 = 12.3 V
vL = (15 V)et/(L/R) = (15 V)e5s/10s = (15 V)e0.5 = 9.10 V
vL = (15 V)et/(L/R) = (15 V)e15s/10s = (15 V)e1.5 = 3.35 V
26.
For the decreasing exponential, the initial current is 1.22 mA and the final current is 0. The
current is solved by subtracting 50 s from the given times to account for the time when the
falling square wave occurs.
(a)
(b)
(c)
At 65 s, i = 1.22 mA(e15s/9.15s) = 0.237 mA
At 75 s, i = 1.22 mA(e25s/9.15s) = 0.079 mA
At 85 s, i = 1.22 mA(e35s/9.15s) = 0.027 mA
27.
VL = (15 V)et/10 s
5V
et/10 s =
15 V
5
t = (10 s) ln 
 15 
t = 11.0 s
28.
(a)
(b)
29.
The polarity is positive at the top of the inductor.
24 V
24 V

= 1.22 mA
IF =
RW
8.2 k
The time constant is found by first
thevenizing the bridge. Figure 13-1
shows the Thevenin circuit.
127
Chapter 13
=
L 3.3 mH

= 0.722 s
R 4.57 k
Figure 13-1
The current at 1.0 s is I = 0.569 mA(1  e1.0s/0.722s) = 0.426 mA
The current after 5 is 0.569 mA.
30.
(a)
(b)
31.
When the switch is open, the circuit
appears as in Figure 13-2.
RT = (R1 + R3)  (R2 + R4)
= 8 k  11.5 k = 4.72 k
L 3.3 mH

= 0.699 s
R 4.72 k
i = 0.569 mA(e1.0s/0.699s) = 136 A
=
Figure 13-2
Section 13-5 Inductors in AC Circuits
32.
The total inductance for each circuit was found in Problem 17.
(a)
XL = 2fLT = 2(5 kHz)(4.33 H) = 136 k
(b) XL = 2fLT = 2(5 kHz)(50 mH) = 1.57 k
(c)
XL = 2fLT = 2(5 kHz)(57.1 H) = 1.79 
33.
The total inductance for each circuit was found in Problem 18.
(a)
XL = 2fLT = 2(400 Hz)(57.3 mH) = 144 
(b) XL = 2fLT = 2(400 Hz)(4 mH) = 10.1 
(c)
XL = 2fLT = 2(400 Hz)(5.33 mH) = 13.4 
34.
L2 L3
(20 H)(40 H)
 50 H 
= 63.3 mH
L2  L3
60 H
XL(T) = 2fLT = 2(2.5 kHz)(63.3 H) = 995 m
XL2 = 2fL2 = 2(2.5 kHz)(20 H) = 314 m
XL3 = 2fL2 = 2(2.5 kHz)(40 H) = 628 m
V
10 V
= 10.1 A
IT = rms 
X LT 995 m
LT = L 1 +




X L3
628 m
IL2 = 
 IT  
10.1 A = 6.73 A
 314 m  628 m 
 X L 2  X L3 




X L2
314 m
IL3 = 
 IT  
10.1 A = 3.37 A
 314 m  628 m 
 X L 2  X L3 
35.
(a)
LT = 57.3 mH
10 V
V
= 20 

XL =
I 500 mA
128
Chapter 13
(b)
(c)
XL = 2fLT
XL
20 

= 55.5 Hz
f=
2LT 2(57.3 mH)
LT = 4 mH
XL = 20 
XL
20 
= 796 Hz
f=

2LT 2(4 mH)
LT = 5.33 mH
XL = 20 
20 
XL

f=
= 597 Hz
2LT 2(5.33 mH)
36.
XLT = 995 m
2
Pr = I rms
X LT = (10.1 mA)2(995 m) = 101 VAR
37.
XL1 = 2(3 kHz)(5 mH) = 94.2 
XL3 = 2(3 kHz)(3 mH) = 56.5 
VL3 = IL3XL3 = (50 mA)(56.5 ) = 2.83 V
VL1 = 10 V  2.83 V = 7.17 V
V
7.17 V
IL1 = L1 
= 76.1 mA
X L1 94.2 
IL2 = IL1  IL3 = 76.1 mA  50 mA = 26.1 mA
Multisim Troubleshooting and Analysis
38.
VL1 = 1.158 V; VL2 = 3.579 V; VL3 = 5.263 V
39.
VL1 = 12.953 V; VL2 = 11.047 V; VL3 = 5.948 V
VL4 = 5.099 V; VL5 = 5.099 V
40.
IL(10kHz) = 0.016 A; IL(20kHz) = 7.855 mA; IL(5kHz) = 0.032 mA
41.
L3 is open.
42.
L2 is shorted.
129
Chapter 14
Transformers
Section 14-1 Mutual Inductance
1.
LM = k L p Ls  0.75 (1 H )(4 H) = 1.5 H
2.
LM = k L1L2
k=
LM

L1L2
1 H
= 0.25
(8 H)(2 H)
Section 14-2 The Basic Transformer
3.
n=
N s 1000

=4
Np
250
n=
N s 100

= 0.25
N p 400
4.
Ns = 2Np = 2(25) = 50 turns
5.
See Figure 14-1.
N 
(a)
Vs =  s V p = 10(10 V) = 100 V rms
 Np 


N 
(c)
Vs =  s V p = 0.2(100 V) = 20 V rms
 Np 


(b)
Figure 14-1
130
N 
Vs =  s V p = 2(50 V) = 100 V rms
 Np 


Chapter 14
Section 14-3 Step-Up and Step-Down Transformers
N s 720 V

=3
N p 240 V
6.
n=
7.
N 
Vs =  s V p = 5(120 V) = 600 V
 Np 


8.
N s Vs

N p Vp
 Np 
1
Vs   60 V = 6 V
Vp = 

 10 
 Ns 
9.
Vs
N
30 V
 s 
= 0.25
V p N p 120 V
10.
Vs = (0.2)(1200 V) = 240 V
11.
N s Vs

N p Vp
 Np 
10
Vs    6 V = 60 V
Vp = 

1
 Ns 
12.
(a)
(b)
(c)
13.
(a)
(b)
N 
 1 
VRL =  s V p   120 V = 6 V
 Np 
 20 


VRL = 0 V (transformers do not couple dc)
N 
VRL =  s V p = 4(10 V) = 40 V
 Np 


VL = (0.1)Vs = (0.1)(100 V) = 10 V
Vp = 20VL = 20(12 V) = 240 V
Section 14-4 Loading the Secondary
14.
Is N p

I p Ns
 Np 
1
1
 I p    I p   100 mA = 33.3 mA
Is = 

3
3
 Ns 
VL = 3(20 V) = 60 V
V
60 V
RL = L 
= 1.8 k
I s 33.3 mA
131
Chapter 14
15.
Ns
= 0.5
Np
(a)
2
 Np 
1 
 RL  
Rreflect = 
 300  = 1200 

 0 .5 
 Ns 
30 V
Ip =
= 25 mA
1200 
2
 Np 
 I p = 2(25 mA) = 50 mA
Is = 

 Ns 
N 
Vs =  s V p = 0.5(30 V) = 15 V
 Np 


PL = VsIs = (15 V)(50 mA) = 750 mW
(b)
(c)
(d)
Section 14-5 Reflected Load
16.
17.
 Np
Rp = 
 Ns
2

1
 RL    680  = 27.2 
5

2
Rp = 300 , RL = 1 k
Ns
RL
1 k
= 1.83


Np
Rp
300 
Section 14-6 Impedance Matching
2
18.
 Np 
 RL
Rp = 

 Ns 
2
Rp
 Np 

 N   R
L
 s
Np
Ns
n=

Rp
RL

16 
 4 =2
4
Ns 1
 = 0.5
Np 2
132
Chapter 14
2
19.
 Np 
 RL = 16 
Rp = 

 Ns 
2
N 
R
4
n = s  L 
 Np 
R p 16 


4
N
 0.25 = 0.5
n= s 
16 
Np
2
Ip =
25 V
= 781 mA
16   16 
 1 
1
Is =   I p  
781 mA = 1562 mA
 0 .5 
n
Pspeaker = I s2 RL = (1562 mA)24  = 9.76 W
20.
Position 1:
RL = 560  + 220  + 1 k = 1780 
Ns
RL
1780 


= 13.34
Np
Rp
10 
Ns = Ns1 + Ns2 + Ns3 = 13.34Np = 13.34(1000) = 13,340 turns (total secondary turns)
Position 2:
RL = 220  + 1 k = 1220 
Ns
RL
1220 


= 11.05
Np
Rp
10 
Ns2 + Ns3 = 11.05Np = 11.05(1000) = 11,050 turns
Position 3:
RL = 1 k
Ns
RL
1000 


= 10
Thus, Ns2 = 11,050  10,000 = 1,050 turns
Np
Rp
10 
Ns3 = 10Np = 10(1000) = 10,000 turns
Ns1 = 13,340  11,050 = 2,290 turns
Section 14-7 Transformer Ratings and Characteristics
21.
PL = Pp  Plost = 100 W  5.5 W = 94.5 W
22.
P
% efficiency =  out
 Pin
23.
Coefficient of coupling = 1  0.02 = 0.98

94.5 W 
100%  
100% = 94.5 %

 100 W 

133
Chapter 14
24.
(a)
(b)
(c)
Pa 1 kVA

= 1.67 A
Vs 600 V
Vs
600 V
RL(max) =

= 359 
I L (max) 1.67 A
IL(max) =
Vs
= 359 
IL
1
Cmax =
= 7.39 F
2(60 Hz)(359 )
XC =
25.
kVA = (2.5 kV)(10 A) = 25 kVA
26.
(a)
(b)
(c)
Vp = 2400 V
N s Vs
120 V


= 0.05
N p V p 2400 V
Pa 5 kVA

= 41.7 A
Vs 120 V
5 kVa
Ip =
= 2.08 A
2400 V
Is =
Section 14-8 Tapped and Multiple-Winding Transformers
27.
 50 
120 V = 12.0 V
 500 
 100 
V2 = 
120 V = 24.0 V
 500 
 100 
V3 = 
120 V = 24.0 V
 500 
V1 = 
V4 = V2 + V3 = 48.0 V
28.
N s1 Vs 24 V


=2
N p V p 12 V
Ns2
6V

= 0.5
N p 12 V
N s3
3V

= 0.25
N p 12 V
29.
N 
 200 
Vs =  s V p  
120 V = 48 V
 Np 
 500 


N 
 250 
Vs =  s V p  
5 V = 25 V
 Np 
50




134
Chapter 14
30.
(a)
(b)
31.
(a)
See Figure 14-2.
 100 
100 T: Vs = 
240 V = 12 V
 2000 
 200 
200 T: Vs = 
240 V = 24 V
 2000 
 500 
500 T: Vs = 
240 V = 60 V
 2000 
 1000 
1000 T: Vs = 
240 V = 120 V
 2000 
Figure 14-2
Ns = 400 T + 300 T = 700 turns
N 
 700 
VRL =  s V p  
 60 V = 35 V
N 
1200


p


V
35 V
= 2.92 A
IRL = RL 
RL 12 
 300 
VC = 
 60 V = 15 V
 1200 
V
15 V
= 1.5 A
IC = C 
X C 10 
(b)
1
1
1
1
1




2
2
(2.94)(12 ) (16)(10 )
Rp  N p 
 Np 

 RL 
 X CL
 N 700 
 N 300 
1
1
=
= 28.3 mS + 6.25 mS = 29.0 mS

35.3  160 
1
= 34.5 
Rp =
29.0 mS
Section 14-9 Troubleshooting
32.
Open primary winding. Replace the transformer.
33.
If the primary shorts, excessive current is drawn which potentially can burn out the source
and/or the transformer unless the primary is fused.
34.
Some, but not all, of the secondary windings are shorted.
Multisim Troubleshooting and Analysis
35.
Turns ratio = 0.5
36.
Secondary winding is open.
37.
R2 is open.
135
Chapter 15
RC Circuits
Section 15-1 The Complex Number System
1.
A complex number indicates both magnitude and angle of quantity.
2.
See Figure 15-1.
Figure 15-1
3.
See Figure 15-2.
Figure 15-2
136
Chapter 15
4.
(a)
(b)
(c)
3, j5
+7, j1
+10, +j10
5.
(a)
(b)
(c)
5, +j3 and +5, j3
1, j7 and 1, +j7
10, +j10 and +10, j10
6.
(a)
(b)
(c)
3 + j5
2 + j1.5
10  j14
7.
C = 102  152 = 18.0
8.
(a)
(b)
(c)
(d)
  40 
402  402 tan 1 
 = 56.645
 40 
  200 
50  j200 = 502  2002 tan 1 
 = 20676
 50 
  20 
35  j20 = 352  202 tan 1 
 = 40.329.7
 35 
 45 
98 + j45 = 982  452 tan 1   = 10824.7
 98 
40  j40 =
9.
(a)
(b)
(c)
(d)
100050 = 643  j766
15160 = 14.1 + j5.13
25135 = 17.7  j17.7
3180 = 3 + j0
10.
(a)
(b)
(c)
10120 = 10240
3285 = 32275
5310 = 550
11.
(a)
(b)
(c)
(d)
40  j40 is in the fourth quadrant.
50  j200 is in the fourth quadrant.
35  j20 is in the fourth quadrant.
98 + j45 is in the first quadrant.
12.
(a)
(b)
(c)
10120 is in the second quadrant.
3285 is in the first quadrant.
5310 is in the fourth quadrant.
13.
(a)
(b)
(c)
(d)
12(180  65) = 12115
20(180 + 50) = 20230
100(360  170) = 100190
50(360  200) = 50160
137
Chapter 15
14.
15.
16.
17.
(a)
(9 + j3) + (5 + j8) = 14 + j11
(b)
(3.5  j4) + (2.2 + j6) = 5.7 + j2
(c)
(18 + j23) + (30  j15) = 12 + j8
(d)
1245 = 8.49 + j8.49
2032 = 17.0 + j10.6
(8.49 + j8.49) + 17.0 + j10.6) = 25.5 + j19.1
(e)
3.875 = 0.984 + j3.67
(0.984 + j3.67) + (1 + j1.8) = 1.98 + j5.47
(f)
6030 = 52  j30
(52  j30) + (50  j39) = 102  j69
(a)
(2.5 + j1.2)  (1.4 + j0.5) = 1.1 + j0.7
(b)
(45  j23)  (36 + j12) = 81  j35
(c)
(8  j4)  325 = (8  j4)  (2.72 + j1.27) = 5.28  j5.27
(d)
48135  3360 = (33.9 + j33.9)  (16.5  j28.6) = 50.4 + j62.5
(a)
(4.548)(3.290) = 14.4138
(b)
(120220)(95200) = 11,40020
(c)
4  j3 = 536.9
(3150)(536.9) = 15113
(d)
67 + j84 = 107.551.4
(107.551.4)(10240) = 10,96591.4
(e)
15  j10 = 1833.7
25  j30 = 39.1129.8
(1833.7)(39.1129.8) = 704164
(f)
0.8 + j0.5 = 0.9432
1.2  j1.5 = 1.9251.3
(0.9432)(1.9251.3) = 1.8119.3
(a)
(b)
(c)
(d)
850
= 3.2(50  39) = 3.211
2.539
63  91
= 7(91 10) = 7101
910
2830
2830

= 1.52(30  (40.6)) = 1.5270.6
14  j12 18.4  40.6
40  j30 50  36.9

= 2.79(36.9  26.6) = 2.7963.5
16  j8
17.926.6
138
Chapter 15
18.
(a)
(b)
(c)
(d)
2.565  1.8  23
(1.06  j2.27)  (1.66  j0.70)
=
1.237
1.237
 0.6  j2.97 3.03101.4

= 2.5364.4
=
1.237
1.237
(10015)(85  j150) (10015)(172.460.46)

= 335106
25  j45
51.561.0
(25090  17575)(50  j100) ( j250  45.29  j169.04)(111.8  63.43)
=
(125  j90)(3550)
(154.0435.75)(3550)
(421.4883.83)(111.8  63.43)
=
= 8.7465.4
(5391.0585.75)
(1.5) 2 (3.8)  8
4
4
4
 j  j   7.77  j  j  = 7.77 + 2 + j2 = 9.77 + j2 = 9.9711.6
1.1
4
2
2
2



Part 1: Series Circuits
Section 15-2 Sinusoidal Response of Series RC Circuits
19.
fVR = 8 kHz, fVC = 8 kHz
20.
The current is sinusoidal because the voltage is sinusoidal.
Section 15-3 Impedance of Series RC Circuits
21.
(a)
(b)
Z = R  jXC = 270   j100  = 28820.3 
Z = R  jXC = 680   j1000  = 1.2155.8 k
22.
(a)
RT = R1 + R2 = 100 k + 47 k = 147 k
1
1
CT =
= 0.00688 F

1
1
1
1


0.01 F 0.022 F
C1 C2
1
1

= 231 k
XCT =
2fCT 2(100 Hz)(0.00688 F)
Z = RT  jXCT = 147 k  j231 k = 27357.5 k
Z = 273 ,  = 57.5
(b)
CT = C1 + C2 = 470 pF + 470 pF = 940 pF
1
1

XCT =
= 8.47 k
2fCT 2(20 kHz)(940 pF)
Z = R  jXCT = 10 k  j8.47 k = 13.140.3 k
Z = 13.1 k,  = 40.3
139
Chapter 15
(c)
23.
(a)
(b)
(c)
(d)
24.
(a)
RT = R1 + R2  R3 = 680  + 720  = 1400 
1
1
1


= 762 pF
CT =
1
1
C1 C2  C3

1000 pF 0.0032 F
1
1

= 2089 
XCT =
2fCT 2(100 kHz)(762 pF)
Z = RT  jXCT = 1400   j2089  = 2.5256.2 k
Z = 2.52 k,  = 56.2
1
1
= 723 k

2fC 2(100 Hz)(0.0022 F)
Z = 56 k  j723 k
XC =
1
1

= 145 k
2fC 2(500 Hz)(0.0022 F)
Z = 56 k  j145 k
XC =
1
1

= 72.3 k
2fC 2(1 kHz)(0.0022 F)
Z = 56 k  j72.3 k
XC =
1
1

= 28.9 k
2fC 2(2.5 kHz)(0.0022 F)
Z = 56 k  j28.9 k
XC =
1
1

= 339 k
2fC 2(100 Hz)(0.0047 F)
Z = 56 k  j339 k
XC =
1
1
= 67.7 k

2fC 2(500 Hz)(0.0047 F)
Z = 56 k  j67.7 k
(b)
XC =
(c)
XC =
(d)
XC =
1
1

= 33.9 k
2fC 2(1 kHz)(0.0047 F)
Z = 56 k  j33.9 k
1
1
= 13.5 k

2fC 2(2.5 kHz)(0.0047 F)
Z = 56 k  j13.5 k
140
Chapter 15
25.
(a)
R = 33 , XC = 50 
(b)
Z = 30025  = 272   j127 
R = 272 , XC = 127 
(c)
Z = 1.867.2 k = 698   j1.66 k
R = 698 , XC = 1.66 k
(d)
Z = 78945  = 558   j558 
R = 558 , XC = 558 
Section 15-4 Analysis of Series RC Circuits
26.
(a)
From Problem 21(a): Z = 28820.3 
100 V
= 34.720.3 mA
I=
288  20.3 
From Problem 21(b): Z = 1.2155.8 k
50 V
I=
= 4.1355.8 mA
1.21  55.8 k
Start with the current in polar form from Problem 26:
(b)
27.
(a)
I=
100 V
= 34.720.3 mA = (34.7 mA)cos 20.3o + j(34.7 mA)sin 20.3o
288  20.3 
= 32.5 mA + j12.0 mA
(b)
I=
50 V
= 4.1355.8 mA = (4.13 mA)cos 55.8o + j(4.13 mA)sin 55.8o
1.21  55.8 k
= 2.32 mA + j3.42mA
28.
(a)
From Problem 22(a): Z = 147 k  j231 k = 27357.5 
500 V
IT =
= 18357.5 A
273  57.5 
(b)
From Problem 22(b): Z = 10 k  j8.47 k = 13.140.3 k
80 V
IT =
= 61140.3 A
13.1  40.3 k
(c)
From Problem 22(c): Z = 1400   j2089  = 2.5256.2 k
520 V
IT =
= 1.9876.2 mA
2.52  56.2 k
141
Chapter 15
29.
Start with the current in polar form from Problem 28
(a)
IT =
500 V
= 18357.5 A = (183 A)cos 57.5o + j(183 A)sin 57.5o
273  57.5 
= 98.3 A + j154 A
(b)
IT =
80 V
= 61140.3 A = (611 A)cos 40.3o + j(611 A)sin 40.3o
13.1  40.3 k
= 466 A + j395 A
(c)
IT =
520 V
= 1.9876.2 mA = (1.98 mA)cos 76.2o + j(1.98 mA)sin 76.2o
2.52  56.2 k
= 0.472 mA + j1.92 mA
30.
31.
32.
Using the results of Problem 22:
X 
 231 k 
 =  tan 1  C    tan 1 
(a)
 = 57.5
 R 
 147 k 
X 
 8.47 k 
(b)  =  tan 1  C    tan 1 
 = 40.3
 R 
 10 k 
X 
 2089  
(c)
 =  tan 1  C    tan 1 
 = 56.2
 R 
 1400  
1
 14.5 k
2fC
X 
 14.5 k 
 =  tan 1  C    tan 1 
 = 14.5
 R 
 56 k 
See Figure 15-3.
1
= 0.069 F
CT =
1
1

0.1 F 0.22 F
1
XC =
= 154 
2(15 kHz)(0.069 F)
ZT = 50   j154  = 16272.0 
20 V
V
IT = s 
= 12.372.0 mA
Z T 162  72.0 
1
= 106 
XC1 =
2(15 kHz)(0.1 F)
1
XC2 =
= 48.2 
2(15 kHz)(0.22 F)
VC1 = ITXC1 = (12.372.0 mA)(10690 ) = 1.3018.0 V
VC2= ITXC2= (12.372.0 mA)(48.290 ) = 59318.0 mV
VR1 = VR2 = ITRT = (12.372.0 mA)(500 ) = 61572.0 mV
XC =
142
Chapter 15
Figure 15-3
33.
34.
(a)
1
= 79.6 
2(20 Hz)(100 F)
Z = 56   j79.6  = 97.354.9 
XC =
100 
= 10354.9 mA
97.3  54.9 
(b)
IT =
(c)


560 
100 V = 5.7654.9 V
VR = 
 97.3  54.9  
(d)
 79.6  90  
100 V = 8.1835.1 V
VC = 
 97.3  54.9  
Vs
10 V
= 1 k

I 10 mA
1
XC =
= 589.5 
2(10 kHz)(0.027 F)
Z=
R 2  X C2 = 1 k
R2 + (589.5 )2 = (1 k)2
R=
(1 k) 2  (589.5 ) 2 = 808 
 589.5  
 = 36.1
 808  
 = tan1 
35.
100 V
 20 
5A
Ptrue = I2RT
P
400 W
RT = true
= 16 

2
I
(5 A)2
RX = RT  R1 = 16   4  = 12 
ZT =
143
Chapter 15
ZT2  RT2  X C2
XC =
C=
36.
(a)
ZT2  RT2  (20 ) 2  (16 ) 2  144 = 12 
1
= 13.3 F
2(1 kHz)(12 )
XC =
1
= 4.08 M
2(1 Hz)(0.039 F)
 XC 
1  4.08 M 
  90  tan 
 = 0.055
 R 
 3.9 k 
 = 90 + tan1 
(b)
XC =
1
= 40.8 k
2(100 Hz)(0.039 F)
 XC 
1  40.8 k 
  90  tan 
 = 5.46
 R 
 3.9 k 
 = 90 + tan1 
(c)
XC =
1
= 4.08 k
2(1 kHz)(0.039 F)
 XC
 R
 = 90 + tan1 
(d)
XC =

1  4.08 k 
  90  tan 
 = 43.7

 3.9 k 
1
= 408 
2(10 kHz)(0.039 F)
 XC 
1  408  
 = 84.0
  90  tan 
 3.9 k 
 R 
 = 90 + tan1 
37.
X
Use the formula, Vout =  C
 ZT
Frequency
(kHz)
XC
0
1
4.08 k
2
2.04 k
3
1.36 k
4
1.02 k
5
816 
6
680 
7
583 
8
510 
9
453 
10
408 

1 V. See Figure 15-4.

ZT
5.64 k
4.40 k
4.13 k
4.03 k
3.98 k
3.96 k
3.94 k
3.93 k
3.93 k
3.92 k
Vout
1V
723 mV
464 mV
329 mV
253 mV
205 mV
172 mV
148 mV
130 mV
115 mV
104 mV
144
Figure 15-4
Chapter 15
38.
(a)
XC =
1
1

= 15.9 k
2fC 2(1 Hz)(10 F)
 XC 
1  15.9 k 
 = 90.0
  tan 
 10  
 R 
1
XC =
= 159 
2(100 Hz)(10 F)
 = tan1 
(b)
 XC 
1  159  
 = 86.4
  tan 
 10  
 R 
1
XC =
= 15.9 
2(1 kHz)(10 F)
 = tan1 
(c)
 XC 
1  15.9  
 = 57.9
  tan 
 10  
 R 
1
XC =
= 1.59 
2(10 kHz)(10 F)
 = tan1 
(d)
 XC 
1  1.59  
 = 9.04
  tan 
 10  
 R 
 = tan1 
39.
40.
 R
Use the formula, Vout = 
 ZT
Frequency
XC
(kHz)
0
1
15.9 
2
7.96 
3
5.31 
4
3.98 
5
3.18 
6
2.65 
7
2.27 
8
1.99 
9
1.77 
10
1.59 

1 V. See Figure 15-5.

ZT
Vout
18.8 
12.8 
11.3 
10.8 
10.5 
10.4 
10.3 
10.2 
10.2 
10.1 
0V
5.32 V
7.82 V
8.83 V
9.29 V
9.53 V
9.66 V
9.76 V
9.80 V
9.84 V
9.87 V
Figure 15-5
For Figure 15-91 in the text (See Figure 15-6):
1
XC =
= 816 
2(5 kHz)(0.039 F)
Z = 3.9 k  j816  = 398411.8 
10 V
V
I= s 
= 25111.8 A
Z 3984  11.8 
VR = IR = (25111.8 A)(3.90 k) = 97911.8 mV
VC = IXC = (25111.8 A)(81690 ) = 20578.2 mV
145
Chapter 15
Figure 15-6
41.
For Figure 15-92 in the text (See Figure 15-7)
1
= 15.9 
XC =
2(1 kHz)(10 F)
Z = 10   j15.9  = 18.857.8 
V
100 V
I= s 
= 53257.8 mA
Z 18.8  57.8 
VR = IR = (53257.8 mA)(100 ) = 5.3257.8 V
VC = IXC = (53257.8 mA)(15.990 ) = 8.4632.2 V
Figure 15-7
Part 2: Parallel Circuits
Section 15-5 Impedance and Admittance of Parallel RC Circuits
( R0)( X C   90) (1.20 k)(2  90 k) (1.20 k)(2  90 k)


1.2 k  j2 k
2.33   59 k
R  jX C
= 1.0331 k
42.
Z=
43.
BC = 2fC = 2f(C1 + C2) = 2(2 kHz)(0.32 F) = 4.02 mS
146
Chapter 15
1
1

= 0.676 mS
R1  R2  R3 1480 
Y = G + jBC = 0.676 mS + j4.02 mS = 4.0880.5 mS
1
1
Z=
= 24580.5 

Y 4.0880.5 mS
Z = 245 ,  = 80.5
G=
44.
(a)
1
= 332 
2(1.5 kHz)(0.32 F)
(14800 )(332  90 ) (14800 )(332  90 )
Z=

 324  77.4 
1480   j332 
1517  12.6 
Z = 324 ,  = 77.4
XC =
1
= 166 
2(3 kHz)(0.32 F)
(14800 )(166  90 ) (14800 )(166  90 )
Z=

 165  83.6 
1480   j166 
1489  6.40 
Z = 165 ,  = 83.6
(b)
XC =
(c)
XC =
(d)
XC =
1
= 99.5 
2(5 kHz)(0.32 F)
(14800 )(99.5  90 ) (14800 )(99.5  90 )

 99.3  86.2 
Z=
1480   j99.5 
1483  3.85 
Z = 99.3 ,  = 86.2
1
= 49.7 
2(10 kHz)(0.32 F)
(14800 )(49.7  90 ) (14800 )(49.7  90 )

 49.7  88.1 
Z=
1480   j49.7 
1481  1.92 
Z = 49.7 ,  = 88.1
Section 15-6 Analysis of Parallel RC Circuits
45.
(680)(90  90 )
= 54.337.1 
68   j90 
VC = VR = Vs = 10 0 V
ZT =
100 V
= 18437.1 mA
54.3  37.1 
100 V
IR =
= 1470 mA
680 
IT =
147
Chapter 15
IC =
46.
47.
100 V
= 11190 mA
90  90 
1
= 67.7 
2(50 kHz)(0.047 F)
1
XC2 =
= 145 
2(50 kHz)(0.022 F)
V
80 V
IC1 = s 
= 11890 mA
XC1 67.7  90 
80 V
V
IC2 = s 
= 55.290 mA
XC2 145  90 
V
80 V
IR1 = s 
= 36.40 mA
R 1 2200 
80 V
V
IR2 = s 
= 44.40 mA
R 2 1800 
IT = IR1 + IR2 + IC1 + IC2
= 36.4 mA + 44.4 mA + j118 mA + j55.2 mA = 80.8 mA + j173.2 mA = 19165.0 mA
IT = 191 mA,  = 65.0
XC1 =
(a)
XC = XC1  XC2 = 21   15  = 8.75 
1
1

BC =
= 114 mS
X C 8.75 
1
1
G=

= 100 mS
R 10 
YT = 100 mS + j114 mS = 15248.8 mS
1
1

ZT =
= 6.5948.8 
YT 15248.8 mS
(b)
IR =
(c)
ICT =
(d)
IT =
(e)
 = 48.8
1000 mV
= 100 mA
100 
1000 mV
= 11.490 mA
8.75  90 
1000 mV
= 15.248.8 mA
6.59  48.8 
148
Chapter 15
48.
(a)
CT = C1 + C2 = 0.047 F + 0.022 F = 0.069 F
1
XCT =
= 4613 
2(500 Hz)(0.069 F)
(5.60 k)(4613  90 )
= 3.5650.5 k
Z=
7255  39.5 
(b)
IR =
(c)
ICT =
1000 mV
Vs

= 21.790 A
XCT 4613  90 
(d)
IT =
1000 mV
Vs

= 28.150.5 A
Z 3560  50.5 
(e)
 = 50.5
Vs 1000 mV
= 17.90 A

R
5.60 k
49.
RT = 22 k, CT = 32 pF
1
XCT =
= 49.7 k
2(100 kHz)(32 pF)
(220 k)(49.7  90 k)
= 20.123.9 k = 18.4 k  j8.14 k
Z=
54.4  66.1 k
Req = 18.4 k, XCeq = 8.14 k
1
Ceq =
= 196 pF
2 (100 kHz)(8.14 k)
50.
XC =
1
= 15.9 k
2(1 kHz)(0.01 F)
X 
 = tan 1  C 
 RT 
XC
= tan 
RT
15.9 k
XC
= 27.6 k
RT =

tan
tan 30
R1R2
RT =
R1  R2
RT(R1 + R2) = R1R2
R1RT + R2RT = R1R2
R1(RT  R2) = R2RT
(47 k)(27.6 k)
R2 RT
R1 =
= 66.7 k

R2  RT
19.4 k
149
Chapter 15
Part 3: Series-Parallel Circuits
Section 15-7 Analysis of Series-Parallel RC Circuits
51.
See Figure 15-8.
1
1

XC1 =
= 106 
2fC1 2(15 kHz)(0.1 F)
1
1
XC2 =
= 226 

2fC2 2(15 kHz)(0.047 F)
1
1

XC3 =
= 48.2 
2fC3 2(15 kHz)(0.22 F)
ZC2R1 = R1  jXC2 = 470   j226  = 52225.7 
ZC3 = jXC3 = 48.290 
ZR2R3 = R2 + R3 = 330  + 180  = 5100 
Combining the three parallel branches:
1
1

1
1
1
1
1




ZC2R1 ZC3 Z R2R3
522  25.7  48.2  90  5100 
1
1
=

1.9225.7 mS  20.790 mS  1.960 mS 1.73 mS  j0.833 mS  j20.7 mS  1.96 mS
1
1

= 45.980.3  = 7.73   j45.2 
=
3.69 mS  j21.5 mS 21.880.3 mS
ZA =
1
ZT = XC1 + ZA = j106  + 7.73   j45.2  = 7.73   j151  = 15187.1 
X 
 106  90  
VC1 =  C1 120 V  
120 V = 8.422.9 V
 151  87.1  
 ZT 
Z 
 45.9  80.3  
VZA =  A 120 V  
120 V = 3.656.8 V
 151  87.1  
 ZT 
 X 
 226  90  
VC2 =  C2 VZA  
3.656.8 V = 1.5857.5 V
 522  25.7  
 ZC2R1 
 R1 
4700 

VZA  
VR1 = 
3.656.8 V = 3.2932.5 V

 522  25.7  
 ZC2R1 
VC3 = VZA = 3.656.8 V
 R2 
3300  
VZA  
VR2 = 
3.656.8 V = 2.366.8 V

Z
 5100  
 R2R3 
 R3 
1800 
VZA  
VR3 = 

 5100 
 Z R2R3 

3.656.8 V = 1.296.8 V

150
Chapter 15
57.5
Figure 15-8
52.
From Problem 51:
ZT = 7.73   j151 
The j term is larger, therefore the circuit is predominantly capacitive.
53.
See Figure 15-9.
Using the results of Problem 51:
120 V
120 V
IT =

= 79.587.1 mA
ZT
151  87.1 
VZA  3.656.8 V 

 = 6.9932.5 mA
Z C2R1  522  25.7  
V
 3.656.8 V 
IC3 = ZA  
 = 75.796.8 mA
ZC3  48.2  90  
IC2R1 =
IR2R3 =
VZA  3.656.8 V 

 = 7.166.8 mA
Z R2R3  5100  
Figure 15-9
151
2.9
Chapter 15
54.
RT = R1 + R2  R3 = 47  + 42.9  = 89.9 
1
XC =
= 339 
2(1 kHz)(0.47 F)
ZT = 89.9   j339 = 35175.1 
150 V
V
(a)
IT = s 
= 42.775.1 mA
Z T 351  75.1 
(b)  = 75.1
R 


470 V
150 V = 2.0175.1 V
VR1 =  1 Vs  
(c)
 351  75.1  
 ZT 
(d)
(e)
(f)
55.
R R 
 42.90 V 
150 V = 1.8375.1 V
VR2 =  2 3 Vs  




351
75
.
1
Z


T


VR3 = VR2 = 1.8375.1 V
X 
 339  90  
150 V = 14.514.9 V
VC =  C Vs  
 351  75.1  
 ZT 
For I = 0 A, VA = VB and VR1 = VR2
1
XC1 =
= 3.39 k
2(1 kHz)(0.047 F)
VR1 = VR2



2.2 k
1 k

V  
 (2.2 k) 2  (3.39 k) 2  s  (1 k) 2  X 2
C2



Cancelling the Vs terms and solving for XC2:

 
2.2 k
1 k



 (2.2 k) 2  (3.39 k) 2 
2
2

  (1 k)  X C 2

V
 s





(1 k) 2 (2.2 k) 2  (3.39 k) 2
2.2 k
Squaring both sides to eliminate the radicals and solving for XC2:
(1 k) 2 (2.2 k) 2  (3.39 k) 2
(1 k) 2  X C2 2 
( 2 .2 k ) 2
(1 k) 2  X C2 2 

XC2 =
C2 =



(1 k) 2 (2.2 k) 2  (3.39 k) 2
 (1 k) 2 = 1541 
2
(2.2 k)
1
= 0.103 F
2(1 kHz)(1541 )
152
Chapter 15
56.
1
= 2.89 k
2(2.5 kHz)(0.022 F)
ZC3R6 = R6  jXC3 = 820   j2.89 k = 374.2 k
1
XC2 =
= 1.35 k
2(2.5 kHz)(0.047 F)
XC2 R 4
(1.35  90 k)(9100 )

ZC = ZC2R4 =
= 75434  = 625   j422 
R4  jX C 2
1.63  56 k
ZR5C2R4 = R5 + ZC = 1 k + 625   j422  = 1625   j422  = 1.6814.6 k
ZB = (R5 + ZC)  ZC3R6 = 1.6814.6 k  374.2 k
(1.68  14.6 k)(3  74.2 k)
=
= 1.2632.9 k = 1.06 k  j684 
4  55.9 k
ZR3ZB = R3 + ZB = 680  + 1.06 k  j684  = 1.74 k  j684  = 1.8721.5 k
ZA = R2  ZR3ZB = 2200   1.8721.5 k
(2200 )(1.87  21.5 k)
= 1982.3  = 197.8   j7.95 
=
2.08  19.2 k
1
XC1 =
= 4.24 k
2(2.5 kHz)(0.015 F)
ZT = R1  jXC1 + ZA
= (1 k  j4.24 k) + (197.8   j7.95 ) = 1197.8   j4248  = 4.4174.3 k
XC3 =
Z 
 198  2.3  
VA =  A Vs  
100 V = 44972.0 mV
 4.41  74.3 k 
 ZT 
 ZB 
1.26  32.9 k 
VA  
VB = 
44972.0 mV = 30360.6 mV

 1.87  21.5 k 
 Z R3ZB 
 ZC 
754  34  
VB  
VC = 
30360.6 mV = 13641.2 mV

 1.68  14.6 k 
 Z R5C2R4 
 R6 
820  0  
VB  
VD = 
30360.6 mV = 83135 mV

 3  74.2 k 
 ZC3R6 
57.
Use the voltages found in Problem 56:
V
83135 mV
IC3 = IR6 = D 
= 101135 A = 71.4 A + j71.4 A
R6
8200 
V
13641.2 mV
IR4 = C 
= 14941.2 A = 112 A + j98 A
R4
9100 
V
13641.2 mV
IC2 = C 
= 101131 A = 66 A + j76 A
XC2
1.350 k
IR5 = IR4 + IC2
= (112 A + j98 A) + (66 A + j76 A)
= 46 A + j174 A = 18075.1 A
IR3 = IR5 + IC3
= (46 mA + j174 A) + (71.4 A + j71.4 A)
153
Chapter 15
= 25.4 A + j245 A = 24684.3 A
V
44972.0 mV
IR2 = A 
= 2.0472.0 mA = 0.630 mA + j1.94 mA
R2
2200 
IR1 = IC1 = IR2 + IR3
= (0.630 mA + j194 mA) + (25.4 A + j245 A)
= 605 A + j2185 A = 2.2774.5 mA
58.
See Figure 15-10.
Figure 15-10
154
Chapter 15
Part 4: Special Topics
Section 15-8 Power in RC Circuits
2
Ptrue
 Pr2  (2 W ) 2  (3.5 VAR ) 2 = 4.03 VA
59.
Pa =
60.
From Problem 33:
IT = 103 mA, XC = 79.6 
Ptrue = IT2 R = (103 mA)2(56 ) = 594 mW
Pr = IT2 X C = (103 mA)2(79.6 ) = 845 mVAR
61.
Using the results from Problem 49:
Req = 18.4 k
XCeq = 8.14 k
Zeq = Req  jXCeq = 18.4 k  j8.14 k = 20.123.9 k
 = 23.9
PF = cos  = cos (23.9) = 0.914
62.
From Problem 54:
IT = 42.7 mA, RT = 89.9 , XC = 339 , ZT = 35175.1 
Ptrue = IT2 RT = (42.7 mA)2(89.9 ) = 164 mW
Pr = IT2 X C = (42.7 mA)2(339 ) = 618 mVAR
Pa = IT2 ZT = (42.7 mA)2(351 ) = 640 mVA
PF = cos(75.1) = 0.257
63.
240 V
= 4.8 A
50 
240 V
ILB =
= 3.33 A
72 
(a)
ILA =
(b)
PFA = cos  = 0.85;  = 31.8
PFB = cos  = 0.95;  = 18.2
XCA = (50 )sin(31.8) = 26.3 
XCB = (72 )sin(18.2) = 22.5 
2
PrA = I LA
X CA = (4.8 A)226.3  = 606 VAR
2
PrB = I LB
X CB = (3.33 A)222.5  = 250 VAR
(c)
RA = (50 )cos(31.8) = 42.5 
RB = (72 )cos(18.2) = 68.4 
2
PtrueA = I LA
RA = (4.8 A)242.5  = 979 W
2
PtrueB = I LB
RB = (3.33 A)268.4  = 759 W
155
Chapter 15
(d)
(e)
PaA =
(979 W)2  (606 VAR)2 = 1151 VA
PaB =
(759 W) 2  (250 VAR)2 = 799 VA
Load A
Section 15-9 Basic Applications
1
1
=
= 9278 Hz
2 6(10 k)(0.0022 F)
2 6RC
64.
fr =
65.
Vout1

Vin1
R
R  X C2
2
= 0.707
R = 0.707 R 2  X C2
R
= 1.414R
0.707
R 2  X C2 = (1.414)2R2
R 2  X C2 
X C2 = 2R2  R2 = R2(2  1) = R2
XC = R
1
=R
2fC
1
1

C=
= 0.0796 F
2fR 2(20 Hz)(100 k)
66.
XC =
1
= 1.13 k
2(3 kHz)(0.047 F)



Rin ( B )
10 k




V
Vin(B) = 
(
)
out
A


2
2
2
 Rin ( B )  X C 2 
 10 k)  (1.13 k)


Signal loss = 50 mV  49.7 mV = 300 V

 50 mV = 49.7 mV


Section 15-10 Troubleshooting
67.
After removing C, the circuit is reduced to Thevenin’s equivalent:
(4.7 k)(5 k)
= 2.42 k
Rth =
9.7 k
 5 k 
10 V = 5.15 V
Vth = 
 9 .7 k  
Assuming no leakage in the capacitor:
1
XC =
= 1592 
2(10 Hz)(10 F)
156
Chapter 15
 XC 
 1592  90  
100 V  
100 V = 3.2171.3 V
Vout = 

 4962  18.7  
 R  jXC 
With the leakage resistance considered:


 1592  90  
XC
Vth  
5.150 V = 2.8356.7 V
Vout = 

 2897  33.3  
 R th  jXC 
68.
(a)
The leakage resistance effectively appears in parallel with R2.
Thevenizing from the capacitor:
Rth = R1  R2  Rleak = 10 k  10 k  2 k = 1.43 k
 R 2 R leak 


Vin   1.670 k 10 V = 1430 mV
Vth = 
 11.670 
R R R



2
leak 
 1
1
= 3.38 k
XC =
2(10 Hz)(4.7 F)


 3.38  90 k 
XC
Vth  
1430 mV = 13222.9 mV
Vout = 

 3.67  67.1 k 
 R th  jXC 
(b)
1
= 3.39 M
2(100 Hz)(470 pF)
Req = R1  (R2 + R3) = 2.2 k  2 k = 1.05 k
ZT = Req + XC  Rleak
(3.39  90 M)(20 k)
= 1.050 k +
2 k  j3.39 M
(3.39  90 M)(20 k)
= 1.050 k +
3.39  90 M
= 1.050 k + 20 k = 3.050 k
 R eq 
 1.050 k 
Vin  
VR1 = 
50 V = 1.720 V
 3.050 k 
 ZT 
XC =
 R3 
 10 k 
VR1  
1.720 V = 8600 mV
Vout = 

 20 k 
 R2  R3 
69.
(a)
(b)
(c)
(d)
Vout = 0 V (less than normal)
1
XC =
= 3.39 k
2(10 Hz)(4.7 F)
 XC 
 3.39  90 k 
Vin  
10 V = 320-71.3 mV (greater than normal)
Vout = 

 10.6  18.7 k 
 R  jXC 
 R2 
 100 k 
Vin  
10 V = 5000 mV (greater than normal)
Vout = 
 200 k 
 R1  R 2 
Vout = 0 V (less than normal output)
157
Chapter 15
70.
(a)
(b)
(c)
(d)
(e)
Vout = 0 V (less than normal)
 R3 
 10 k 
Vin  
50 V = 2.50 V (greater than normal)
Vout = 

 20 k 
 R2  R3 
XC = 3.39 M




10 k
R3
Vin  
50 V = 1.4790 mV
Vout = 

 3.39  90 M 
 R 2  R 3  jXC 
(greater than normal)
Vout = 0 V (less than normal)


 2.20 k 
R1
Vin  
50 V = 3.2490 mV
Vout = 






3
.
39
90
M
R
jX
1
C




(greater than normal)
Multisim Troubleshooting and Analysis
71.
No fault.
72.
C1 is leaky.
73.
R1 is open.
74.
No fault.
75.
No fault.
76.
C2 is open.
77.
fc = 48.41 Hz
78.
fc = 3.422 kHz
158
Chapter 16
RL Circuits
Part 1: Series Circuits
Section 16-1 Sinusoidal Response of RL Circuits
1.
fVR = 15 kHz, fVL = 15 kHz
2.
The current is sinusoidal because the voltage is sinusoidal.
Section 16-2 Impedance of Series RL Circuits
Z = R + jXL = 100  + j50  = 11226.6 
Z = R + jXL = 1.5 k + j1 k = 1.8033.7 k
3.
(a)
(b)
4.
See Figure 16-1.
(a)
RT = 56  + 10  = 66 
LT = 50 mH + 100 mH = 150 mH
XL = 2fLT = 2(100 Hz)(150 mH) = 94.2 
Z = RT + jXL = 66  + j94.2  = 11555.0 
Z = 115 ,  = 55.0
(b)
LT = 5 mH  8 mH = 3.08 mH
XL = 2fLT = 2(20 kHz)(3.08 mH) = 387 
Z = RT + jXL = 560  + j387  = 68134.6 k
Z = 681 ,  = 34.6
Figure 16-1
159
Chapter 16
5.
6.
7.
(a)
XL = 2fL = 2(100 Hz)(0.02 H) = 12.6 
Z = 12  + j12.6  = 17.446.4 
(b)
XL = 2fL = 2(500 Hz)(0.02 H) = 62.8 
Z = 12  + j62.8  = 64.079.2 
(c)
XL = 2fL = 2(1 kHz)(0.02 H) = 126 
Z = 12  + j126  = 12784.6 
(d)
XL = 2fL = 2(2 kHz)(0.02 H) = 251 
Z = 12  + j251  = 25187.3 
(a)
Z = 20  + j45  = R + jXL
R = 20 , XL = 45 
(b)
Z = 50035  = 410  + j287  = R + jXL
R = 410 , XL = 287 
(c)
Z = 2.572.5 k = 752  + j2.38 k = R + jXL
R = 4752 , XL = 2.38 k
(d)
Z = 99845  = 706  + j706  = R + jXL
R = 706 , XL = 706 
L1  L2 = 3.11 mH, R1  R2 = 476 
RT = R1 + R1  R2 = 330  + 476  = 806 
LT = L3 + L1  L2 = 1000 H + 3.11 mH = 4.11 mH
Section 16-3 Analysis of Series RL Circuits
8.
9.
RT = 806 
XLT = 2fLT = 2(10 kHz)(4.11 mH) = 258 

RT
VRT = 
2
 R  X2
LT
 T


806 
Vs  
2
2


 (806 )  (258 )


 5 V = 4.76 V



X LT
VLT = 
 R2  X 2
LT
 T


258 
Vs  
 (806 ) 2  (258 ) 2




 5 V = 1.52 V


XL3 = 2fL3 = 2(10 kHz)(1000 H) = 62.8 
X 
 62.8  
VL3 =  L 3 VLT  
1.52 V = 0.370 V
 258  
 X LT 
160
Chapter 16
10.
11.
(a)
From Problem 3(a): Z = 11226.6 
100 V
I=
= 89.426.6 mA
11226.6 
(b)
From Problem 3(b): Z = 1.8033.7 k
50 V
I=
= 2.7833.7 mA
1.8033.7 k
(a)
From Problem 4(a): Z = 11555.0 
50 V
IT =
= 43.555.0  mA
11555.0 
(b)
From Problem 4(b): Z = 68134.6 k
80 V
IT =
= 11.834.6 mA
68134.6 k
12.
XL = 2fL = 2(60 Hz)(0.1 H) = 37.7 
 37.7  
X 
 = 38.7
 = tan 1  L   tan 1 
 R 
 47  
13.
 = 38.7 from Problem 12.
Double L:
XL = 2fL = 2(60 Hz)(0.2 H) = 75.4 
 75.4  
X 
 = 58.1
 = tan 1  L   tan 1 
 R 
 47  
 increases by 19.4 from 38.7 to 58.1
14.
See Figure 16-2. The circuit phase angle was determined to be 38.7 in Problem 12. This is
the phase angle by which the source voltage leads the current; it is the same as the angle
between the resistor voltage and the source voltage. The inductor voltage leads the resistor
voltage by 90. Assume that 10 V is the rms value of Vs.
XL = 37.7 
 XL 
 37.790  
 37.790  
Vs  
100 V  
100 V = 6.2553.1 V
VL = 
 47   j37.7  
 60.338.7  
 R  jX L 
 R 
 470  
Vs  
100 V = 7.7938.7 V
VR = 
 60.338.7  
 R  jX L 
161
Chapter 16
Figure 16-2
15.
(a)
f = 60 Hz
XL = 2(60 Hz)(100 mH) = 37.7 
Z = R + jXL = 150  + j37.7  = 154.714.1 
 1500  
R
50 V = 4.8514.1 V
VR =  Vs  
Z
 154.714.1  
 37.790  
X 
50 V = 1.2275.9 V
VL =  L Vs  
 Z 
 154.714.1  
(b)
f = 200 Hz
XL = 2(200 Hz)(100 mH) = 125.7 
Z = R + jXL = 150  + j125.7  = 195.740.0 
 1500  
R
50 V = 3.8340.0 V
VR =  Vs  
Z
 195.740.0  
 125.790  
X 
50 V = 3.2150.0 V
VL =  L Vs  
 Z 
 195.740.0  
(c)
f = 500 Hz
XL = 2(200 Hz)(100 mH) = 314 
Z = R + jXL = 150  + j314  = 34864.5 
 1500  
R
50 V = 2.1664.5 V
VR =  Vs  
Z
 34864.5  
 31490  
X 
50 V = 4.5125.5 V
VL =  L Vs  
 Z 
 34864.5  
(d)
f = 1 kHz
XL = 2(1 kHz)(100 mH) = 628 
Z = R + jXL = 150  + j628  = 645.776.6 
 1500  
R
50 V = 1.1676.6 V
VR =  Vs  
Z
 645.776.6  
 62890  
X 
50 V = 4.8613.4 V
VL =  L Vs  
 Z 
 645.776.6  
162
Chapter 16
16.
Vs = VL1 + VL2 + VR1 + VR2
= 1590 V + 8.590 V + 6.90 V + 20 V = 8.90 V + 23.590 V
= 8.9 V + j23.5 V = 25.169.3 V
Vs = 25.1 V,  = 69.3
17.
18.
(a)
XL = 2(1 Hz)(10 mH) = 62.8 m
X 
 62.8 m 
 =  tan 1  L    tan 1 
 = 0.0923
 39  
 R 
(b)
XL = 2(100 Hz)(10 mH) = 628 m
X 
 628 m 
 =  tan 1  L    tan 1 
 = 9.15
 39  
 R 
(c)
XL = 2(1 kHz)(10 mH) = 6.28 
X 
 6.28  
 =  tan 1  L    tan 1 
 = 58.2
 39  
 R 
(d)
XL = 2(10 kHz)(10 mH) = 62.8 
X 
 62.8  
 =  tan 1  L    tan 1 
 = 86.4
 39  
 R 
(a)
 = tan1 
(b)
(c)
19.
 R
 XL

1  39  
  tan 
 = 89.9
 62.8 m 

 R 
1  39  
 = tan1 
  tan 
 = 80.9
 628 m 
 XL 
 R 
1  39  
 = tan1 
  tan 
 = 31.8
 6.28  
 XL 
 R
 XL
 = tan1 
(a)
XL = 2(1 Hz)(10 mH) = 62.8 m
Z = 39  + j62.8 m = 390o 
X
Vout   L
 Z
(b)
`

1  39  
  tan 
 = 3.60
 62.8  

(d)
 62.890o m 

o
o
V

 500 mV = 80.590 V
 in 
o

 390  
XL = 2(100 Hz)(10 mH) = 628 m
Z = 39  + j628 m = 390o 
X
Vout   L
 Z
 62890o m 

o
o
V

 500 mV  80590 V
 in 
o

 390  
163
Chapter 16
(c)
XL = 2(1 kHz)(10 mH) = 6.28 
Z = 39  + j6.28  = 39.59.14o 
X
Vout   L
 Z
(d)
 6.2890o  

o
o
V

 500 mV = 7.9580.9 mV
 in 
o

 39.59.14  
XL = 2(10 kHz)(10 mH) = 62.8 
Z = 39  + j62.8  = 73.958.2o 
X
Vout   L
 Z
 62.890o  

o
o
V

 500 mV = 42.531.8 mV
 in 
o

 73.958.2  
Part 2: Parallel Circuits
Section 16-4 Impedance and Admittance of Parallel RL Circuits
20.
XL = 2fL = 2(2 kHz)(800 H) = 10.1 
 1   1 
YT = G  jBL = 
 = 83.3 mS  j99.0 mS = 12949.9 mS
  j
 12    10.1  
1
1

= 7.7549.9 
Z=
YT 129  49.9 mS
21.
From Problem 20:
Z=
1
1
= 7.7549.9  = (7.75  cos 49.9o + j(7.75 sin49.9o

YT 129  49.9 mS
= 4.99  + j5.93 
22.
(a)
f = 1.5 kHz
XL = 2fL = 2(1.5 kHz)(800 H) = 7.54 
 1   1 
YT = G  jBL = 
 = 83.3 mS  j133 mS = 15758.0 mS
  j
 12    7.54  
1
1

= 6.3758.0 
Z=
YT 157  58.0 mS
(b)
f = 3 kHz
XL = 2fL = 2(3 kHz)(800 H) = 15.1 
 1   1 
YT = G  jBL = 
 = 83.3 mS  j66.2 mS = 10638.5 mS
  j
 12    15.1  
1
1

= 9.4338.5 
Z=
YT 106  38.5 mS
(c)
f = 5 kHz
XL = 2fL = 2(5 kHz)(800 H) = 25.1 
164
Chapter 16
 1   1 
YT = G  jBL = 
  j
 = 83.3 mS  j39.8 mS = 92.325.5 mS
 12    25.1  
1
1
Z=

= 10.825.5 
YT 92.3  25.5 mS
(d)
23.
f = 10 kHz
XL = 2fL = 2(10 kHz)(800 H) = 50.3 
 1   1 
YT = G  jBL = 
 = 83.3 mS  j19.9 mS = 85.613.4 mS
  j
 12    50.3  
1
1

= 11.713.4 
Z=
YT 85.6  13.4 mS
XL = 2fL
X
12 
f= L 
= 2.39 kHz
2L 2(800 H)
Section 16-5 Analysis of Parallel RL Circuits
100 V
= 4.550 mA
2.20 k
100 V
= 2.8690.0 mA
IL =
3.590 k
IT = IR + IL = 4.55 mA  j2.86 mA = 5.3732.2 mA
24.
IR =
25.
(a)
XL = 2fL = 2(2 kHz)(25 mH) = 314 
RX L
(560 )(314 )
= 274 
Z=

2
2
R  XL
(560  2  (314 ) 2
 XL 
1  314  
 = 60.7
 = 90  tan 
 R 
 560  
Z = 27460.7 
 = 90  tan 1 
(b)
IR =
500 mV
= 89.30 mA
5600 
(c)
IL =
500 mV
= 15990 mA
31490 
(d)
IT =
500 mV
= 18260.7 mA
27460.7 
(e)
 = 60.7 (from part a)
165
Chapter 16
26.
(a)
XL = 2fL = 2(2 kHz)(330 H) = 4.15 
1
1
Z=

 1   1   1   1 

   j
  j
 
 R   X L   56    4.15  
=
27.
1
1

= 4.1385.8 
17.9 mS  j241 mS 242  85.8 mS
(b)
IR =
500 mV
= 8930 mA
560 
(c)
IL =
500 mV
= 12.090 mA
4.1590 
(d)
IT = IR + IL = 893 mA  j12 A = 12.085.8 A
(e)
 = 85.8
ZT =
( R1  R2 ) X L
( R1  R2 )
2
X L2

(11.5 k)(5 k)
(11.5 k) 2  (5 k) 2
= 4.59 k
 XL 
1  5 k 
  90  tan 
 = 66.5
 R 
 11.5 k 
ZT = 4.5966.5  = 1.83 k + j4.21 k
1.83 k resistance in series with 4.21 k inductive reactance.
 = 90  tan 1 
28.
IT = IR1 + IR2R3 + IL1 = 50 mA + 30 mA + 8.390 mA
= 80 mA + 8.390 mA = 8 mA  j8.3 mA = 11.546.1 mA
IT = 11.5 mA,  = 46.1
Part 3: Series-Parallel Circuits
Section 16-6 Analysis of Series-Parallel RL Circuits
29.
See Figure 16-3.
XL1 = XL2 = 2(400 Hz)(50 mH) = 125.6 
ZR3L1L2 = R3 + XL1  XL2 = 33  + j62.8  = 70.962.3 
R2  ZR3L1L2 = 220   70.962.3  = 18.713.5  = 18.2  + j4.37 
ZT = R1 + R2  ZR3L1L2 = 56  + 18.2  + j4.37  = 64.33.89 
R 
 560  
250 V = 21.83.89 V
VR1 =  1 Vs  
Z
64
.
3

3
.
89




 T
 R Z R3L1L2 
 18.713.5  
Vs  
VR2 =  2
 64.33.89  250 V = 7.279.61 V

Z


T


 R3 
 330  
VR2  
7.279.61 V = 3.3853.3 V
VR3 = 




70
.
9
62
.
3
Z


 R3L1L2 
166
Chapter 16
X
XL2 
 62.890  
VR2  
VL1 = VL2 =  L1
 70.962.3  7.279.61 V = 6.4437.3 V



 Z R3L1L2 
Figure 16-3
30.
LT = L1  L2 = 25 mH
XLT = 2(400 Hz)(25 mH) = 62.8 
Combining R3, L1, and L2:
ZA = 33  + j62.8  = 70.962.3 
Combining ZA with R2 in parallel:
(220 )(70.962.3 ) 156062.3  2
ZB =
= 17.410.1  = 17.1  + j3.05 

22   33   j70.9 
89.752.2 
ZT = R1 + ZB = 56  + 17.1  + j3.05  = 73.1  + j3.05 
The circuit is predominantly resistive because the resistance is greater than the reactance in the
expression for ZT.
167
Chapter 16
31.
See Figure 16-4.
Using the results of Problem 29:
21.8  3.89 V
V
IR1 = IT = R1 
= 3893.89 mA
R1
560 
V
7.279.61 V
IR2 = R2 
= 3309.61 mA
R2
220 
3.38  53.3 V
V
IR3 = R3 
= 102-53.3 mA
R3
330 
V
6.4437.3 V
IL1 = IL2 = L1 
= 51.3-52.7 mA
XL1 125.690 
52.7
Figure 16-4
32.
3.89
IR1 = IT
51.3 mA
XL1 = 2(80 kHz)(10 mH) = 5 k
XL2 = 2(80 kHz)(8 mH) = 4 k
Z1 = 5.6 k + j4 k = 6.8835.5 k
Combining R2 in parallel with Z1:
(3.30 k)(6.8835.5 k) 22.835.5 k
Z2 =

= 2.3411.3 k = 2.29 k + j459 
8.9 k  j4 k
9.7624.2 k
Combining XL1 in series with Z2:
Z3 = 2.29 k + j5.46 k = 5.9167.5 k
Combining R1 in parallel with Z3:
(1.20 k)(5.9167.5 k) 7.0967.5 k
ZT =

= 1.109.90 k
3.46 k  j5.46 k
6.4657.6 k
(a)
(b)
(c)
(d)
(e)
VS
180 V

= 16.49.90 mA
Z T 1.109.90 k
 = 9.90 (IT = lags Vs)
VR1 = Vs = 180 V
Z 
 2.3411.3 k 
VR2 =  2 VR1  
180 V = 7.1356.2 V
 5.9167.5 k 
 Z3 
IT =
R 
 5.60 k 
VR3 =  3 VR2  
7.13  56.2 V = 5.8091.7 V
 6.8835.5 k 
 Z1 
168
Chapter 16
(f)
(g)
33.
X 
 590 k 
VL1 =  L1 VS  
180 V = 15.222.5 V
 5.9167.5 k 
 Z3 
X 
 490 k 
VL2 =  L2 VR2  
7.13  56.2 V = 4.151.70 V
 6.8835.5 k 
 Z1 
The circuit is rearranged in Figure 16-5 for easier analysis.
ZT = (R1 + XL1  R2)  (XL2 + XL3)
= (50  0  + 56.234.2 )  12090  = (50  + 46.5  + j31.6 )  (12090)
(10218.1 )(12090 )
= (10218.1 )  (12090 ) =
96.5   j152 
(10218.1 )(12090 )
= 68.050.5 
=
18057.6 
(a)
(b)
(c)
IT =
400 V
Vs
= 58850.5 mA

Z T 68.050.5 
 XL1 R 2 


Vs   56.234.2  400 V = 22.016.1 V
VL1 = 


R X

 10218.1  
L1 R 2 
 1
 XL3

 4590  
Vs  
400 V = 150 V
VA = 

 12090  
 XL2  XL3 
VB = VL1 = 22.016.1 V
VAB = VA  VB = 150 V  22.016.1 V = 15 V  21.1 V  j6.10 V
= 6.10  j6.10 = 8.63135 V
Figure 16-5
34.
See Figure 16-6.
22.016.1 V
V
IL1 = L1 
= 22073.9 mA
XL1
10090 
V
22.016.1 V
IR2 = B 
= 32416.1 mA
R2
680 
400 V
Vs
IL2 = IL3 =

= 33390 mA
X L2  X L3 12090 
IR1 = IR2 + IL1 = 32416.1 mA + 22073.9 mA
= (311 mA + j89.8 mA) + (61.0 mA  j211 mA) = 372 mA  j121 mA = 39118.0 mA
169
Chapter 16
VR1 = IR1R1 = (39118.0 mA)(500 ) = 19.618.0 V
VR2 = VL1 = 22.016.1 V
 XL2

 7590  
400 V  
400 V = 250 V
VL2 = 

 12090  
 XL2  XL3 
 XL3

 4590  
400 V  
400 V = 150 V
VL3 = 

 12090  
 XL2  XL3 
16.1
Figure 16-6
35.
R4 + R5 = 3.9 k + 6.8 k = 10.7 k
R2  (R4 + R5) = 4.7 k  10.7 k = 3.27 k
R2 + R3  (R4 + R5) = 5.6 k + 3.27 k = 8.87 k
RT = R1  (R2 + R3  (R4 + R5)) = 3.3 k  8.87 k = 2.41 k
XL = 2(10 kHz)(50 mH) = 3.14 k
X 
 3.14 k 
 = tan 1  L   tan 1 
 = 52.5 Vout lags Vin
 R 
 2.41 k 




2.41 k
RT
V  
1 V = 609 mV
VR1 = 
 R 2  X 2  in  (2.41 k) 2  (3.14 k) 2 
L 


 T
 R3 ( R4  R5 ) 
3.27 k

V  
VR3 = 
609 mV = 225 mV
 R  R ( R  R )  R1  3.27 k  5.6 k 
2
3
4
5


 R5 
6 .8 k 

VR 3  
Vout = VR5 = 
225 mV = 143 mV

 3.9 k  6.8 k 
 R4  R5 
Vout 143 mV

= 0.143
Vin
1V
36.
XL1 = 3.14 k, XL2 = 4.7 k, XL3 = 6.38 k
Z3 = R3 + jXL3 = 6.8 k + j6.28 k = 9.2642.7 k
Z2 = XL2 + R2  Z3
= 4.790 k + 4.70 k  9.2642.7 k = 4.790 k + 3.32 14.06 k
= 6.3859.7 k
Z1 = R1  Z2 = 3.30 k  6.3859.7 k = 2.4719.5 k
ZT = XL1 + Z1 = 3.1490 k + 2.4719.5 k = 4.659.6 k
170
Chapter 16
 2.4719.5 k 
10 V = 53740.1 mV
VR1 = 
 4.659.6 k 
 3.3214.1 k 
537  40.1 mV = 27985.7 mV
VR2 = 
 6.3859.7 k 
 6.80 k 
279  85.7 mV = 205128 mV
Vout = 
 9.2642.7 k 
V
205 mV
Phase shift  = 128, Attenuation = out 
= 0.205
Vin
1V
37.
12 V
= 12 
1A
2.5 kV
= 2.5 k
R2 =
1A
When the switch is thrown from position 1 to position 2, the inductance will attempt to keep
1 A flowing through R2 for a short time. This design neglects the arcing of the switch,
assuming instantaneous closure from position 1 to position 2. The value of L is arbitrary since
no time constant requirements are imposed. See Figure 16-7.
R1 =
Figure 16-7
Part 4: Special Topics
Section 16-7 Power in RL Circuits
2
Ptrue
 Pr2  (100 mW ) 2  (340 mVAR) 2 = 354 mVA
38.
Pa =
39.
XL = 2(60 Hz)(0.1 H) = 37.7 
Z = R + jXL = 47  + j37.7  = 60.338.7 
100 V
V
IT = s 
= 165.838.7 mA
Z 60.338.7 
Ptrue = IT2 R = (165.8 mA)2(47 ) = 1.29 W
Pr = IT2 X L = (165.8 mA)2(37.7 ) = 1.04 VAR
40.
 = 32.2 from Problem 22.
PF = cos  = cos (32.2) = 0.846
171
Chapter 16
41.
See Figure 16-8.
From Problem 32: ZT = 1.109.90 k = 1.08 k + j189 
IT = 16.49.90 mA
Ptrue = I R2 R = (16.5 mA)2(1.08 k) = 290 mW
Pr = IT2 X L = (16.4 mA)2(189 ) = 50.8 mVAR
Pa = IT2 ZT = (16.4 mA)2(1.10 k) = 296 mVA
PF = cos(9.90) = 0.985
Figure 16-8
42.
From Problem 33: ZT = 68.050.5  = 43.3  + j52.5 , IT = 58850.5mA
R = 43.3 .
Ptrue = IT2 R = (588 mA)2(43.3 ) = 15.0 W
Section 16-8 Basic Applications
43.
 R
Use the formula, Vout = 
 ZT
Frequency
XL
(kHz)
0
0 
1
62.8 
2
126 
3
189 
4
251 
5
314 

Vin . See Figure 16-9.

Ztot
Vout
39.0 
73.9 
132 
193 
254 
317 
1V
528 mV
296 mV
203 mV
153 mV
123 mV
Figure 16-9
44.
X 
Use the formula, Vout =  L Vin . See Figure 16-10.
 ZT 
Frequency
(kHz)
0
1
2
3
4
5
XL
ZT
Vout
0
62.8 
126 
189 
251 
314 
39.0 
73.9 
132 
193 
254 
317 
0V
42.5 mV
47.8 mV
49.0 mV
49.4 mV
49.6 mV
Figure 16-10
172
Chapter 16
45.
For Figure 16-61 in the text (See Figure 16-11(a)):
XL = 2(8 kHz)(10 mH) = 502.65 
Z = 39  + j502.65  = 504.1685.6 
390 
R


VR =   Vin  
10 V = 77.485.6 mV



Z
504.16
85.6
 


X



502.65
90




VL =  L  Vin  
10 V = 9974.44 mV
504.16
85.6
Z







For Figure 16-62 in the text (See Figure 16-11(b)):
390 
R


VR =   Vin  
 500 mV = 3.8785.6 mV
504.16
85.6
Z



 


X 
 502.6590  
VL =  L  Vin  
 500 mV = 49.94.44 mV
Z
 504.1685.6  


(a)
(b)
Figure 16-11
Section 16-9 Troubleshooting
46.
VR1 = VL1 = 18 V
VR2 = VR3 = VL2 = 0 V
47.
(a)
(b)
(c)
Vout = 0 V
Vout = 0 V
XL1 = 2(1 MHz)(8 H) = 50.26 
XL2 = 2(1 MHz)(4 H) = 25.13 
XLT = 50.26  + 25.13  = 75.39 
RT = R2 + R3 = 156 
Z = RT + jXLT = 156  + j75.39  = 173.2625.8 
50 V
= 28.925.8 mA
I=
173.2625.8 
Vout = IR3 = (28.925.8 mA)560  = 1.6225.8 V
(d)
R1  R3 = 100   56  = 35.9 
Z = 35.9  + j75.39  = 83.564.5 
50 V
= 59.964.5 mA
I=
83.564.5 
Vout = I(R1  R3) = (59.964.5 mA)35.90  = 2.1564.5 V
173
Chapter 16
Multisim Troubleshooting and Analysis
48.
No fault.
49.
L1 is leaky.
50.
No fault.
51.
L1 is open.
52.
R2 is open.
53.
No fault.
54.
fc = 16.05 MHz
55.
fc = 53.214 kHz
174
Chapter 17
RLC Circuits and Resonance
Part 1: Series Circuits
Section 17-1 Impedance of Series RLC Circuits
1.
1
1

= 677 
2fC 2(5 kHz)(0.047 F)
XL = 2fL = 2(5 kHz)(5 mH) = 157 
Z = R + jXL  jXC = 10  + j157   j677 = 10   520  = 52088.9 
Net reactance = jXL  jXC = j520 
XC =
2.
Z = R + j(XL  XC) = 47  + j45  = 65.143.8 
3.
Doubling f doubles XL and halves XC, thus increasing the net reactance and, therefore, the
impedance magnitude increases.
X2
35 
 2(80 ) 
 142.5 
2
2
ZT = 47   j142.5  = 15071.7 
XT = 2XL 
4.
Z=
R 2  ( X L  X C ) 2 = 100 
R2 + (XL  XC)2 = 1002
(XL  XC)2 = 1002  R2
XL  XC = 1002  R 2  (100 ) 2  (47 ) 2 = 88.3 
Section 17-2 Analysis of Series RLC Circuits
5.
ZT = R + jXL  jXC = 47  + j80   j35  = 47  + j45  = 65.143.8 
V
40 V
IT = s 
= 61.443.8 mA
Z T 65.143.8 
VR = ITR = (61.443.8 mA)(470 ) = 2.8943.8 V
VL = ITXL = (61.443.8 mA)(8090 ) = 4.9146.2 V
VC = ITXC = (61.443.8 mA)(3590 ) = 2.15134 V
175
Chapter 17
6.
Use the results of Problem 5.
See Figure 17-1.
Figure 17-1
7.
RT = R1  R2 = 220   390  = 141 
LT = L1 + L2 = 0.5 mH + 1.0 mH = 1.5 mH
CT = C1 + C2 = 0.01 F + 1800 pF = 0.0118 F
XLT = 236 , XCT = 540 
Ztot = RT + j(XLT  XCT) = 141   j304  = 33565.1 
V
120 V
(a)
IT = s 
= 35.865.1 mA
ZT 335  65.1 
(b)
Ptrue = IT2 RT = (35.8 mA)2(141 ) = 181 mW
(c)
Pr = IT2 X T = (35.8 mA)2(304 ) = 390 mVAR
(d)
Pa =
( Ptrue ) 2  ( Pr ) 2 = 430 mVA
Section 17-3 Series Resonance
8.
At the resonant frequency, XL = XC.
In text Figure 1759, XL = 80  and XC = 35 .
For resonance to occur, XL must decrease and XC must increase. Therefore, the resonant
frequency is lower than the frequency, producing the indicated values.
9.
VR = Vs = 12 V
10.
fr =
= 734 kHz
2 LC 2 (1 mH)(47 pF)
XL = 2frL = 2(734 kHz)(1 mH) = 4.61 k
XC = XL = 4.61 k
Ztot = R = 22 
V
12 V
= 545 mA
I= s 
Z tot 22 
11.
VC = VL = 100 V at resonance
V
10 V
= 200 
Z=R= s 
I max 50 mA
V
100 V
= 2 k
XL = XC = L 
I max 50 mA
1

1
176
Chapter 17
12.
fr =
1
2 LC

1
2 (0.008 mH)(0.015 F)
= 459 kHz
XL = 2(459 kHz)(0.008 mH) = 23.1 
X
23.1 
Q= L 
= 2.31
R
10 
f
459 kHz
BW = r 
= 199 kHz
Q
2.31
BW
199 kHz
f 1 = fr 
= 459 kHz 
= 359.5 kHz
2
2
BW
199 kHz
= 459 kHz +
= 558.5 kHz
f2 = fr +
2
2
13.
14.
Vs 7.07 V
= 707 mA at resonance

R
10 
Ihalf-power = 0.707Imax = 0.707(707 mA) = 500 mA
Imax =
At f1:
1
= 29.5 
23595 kHz)(0.015 F)
XL = 2(359.5 kHz)(0.008 mH) = 18.1 
XC  XL = 29.5   18.1  = 11.4 
 11.4  
 = 48.7 current leading
 = tan 1 
 10  
At f2:
1
XC =
= 19.0 
25585 kHz)(0.015 F)
XL = 2(588.5 kHz)(0.008 mH) = 28.1 
XL  XC = 28.1   19.0  = 9.1 
 9 .1  
 = 42.3 current lagging
 = tan 1 
 10  
XC =
r = 0
15.
Refer to Figure 17-2.
1
fr =
Choose C = 0.001 F
2 LC
(a)
fr = 500 kHz
1
f r2  2
4 LC
1
1
L=
 2
= 101 H
2 2
4 f r C 4 (500 kHz)2 (0.001 F)
Figure 17-2
(b)
fr = 1000 kHz
1
1
L=
= 25.3 H
 2
2 2
4 f r C 4 (1000 kHz)2 (0.001 F)
177
Chapter 17
(c)
fr = 1500 kHz
1
1
 2
= 11.3 H
L=
2 2
4 f r C 4 (1500 kHz)2 (0.001 F)
(d)
fr = 2000 kHz
1
1
 2
= 6.33 H
L=
2 2
4 f r C 4 ( 2000 kHz)2 (0.001 F)
Part 2: Parallel Circuits
Section 17-4 Impedance of Parallel RLC Circuits
16.
XL = 2fL = 2(12 kHz)(15 mH) = 1131 
1
1

XC =
= 603 
2fC 2(12 kHz)(0.022 F)
1
Z=
1
1
1


1000  113190  603  90 
1
= 99.74.43 
=
10 mS  j0.884 mS  j1.66 mS
17.
From Problem 16, Z = 99.74.43 
The small negative phase angle indicates a slightly capacitive circuit.
18.
The circuit was found to be capacitive in Problem 17. A decrease in frequency to a point where
XL is slightly less than XC will result in an inductive circuit.
XL < XC
1
2fC
1
f2 <
2
4 LC
1
f<
42 LC
1
f<
2 LC
2fL <
f<
1
2  mH)(0.022 F)
f < 8.76 kHz
178
Chapter 17
Section 17-5 Analysis of Parallel RLC Circuits
50 V
Vs

= 50.24.43 mA
Z 99.7  4.43 
V
50 V
IR = s 
= 500 mA
R 1000 
50 V
V
IL = s 
= 4.4290 mA
XL 113190 
V
50 V
IC = s 
= 8.2990 mA
XC 603  90 
VR = VL = VC = 50 V
19.
IT =
20.
XL = j9.42 k; XC = j72.3 k
ZT = 100   j9.42 k  j72.3 k = 58.953.9 
21.
IR =
50 V
50 V
= 500 mA; IL =
= 53190 A
1000 
9.4290 k
50 V
50 V
= 69.190 A; IT =
= 84.953.9 mA
IC =
72.3  90 k
58.9  53.9 
Section 17-6 Parallel Resonance
22.
ZT =  (infinitely high)
RW2 C
1
1
L 

fr =
= 104 kHz
2 LC
2 LC 2 (50 mH)(47 pF)
XL = 2frL = 2(104 kHz)(50 mH) = 32.7 k
32.7 k
X
Q= L 
= 1635
RW
20 
Zr = RW(Q2 + 1) = 20 (16352 + 1) = 53.5 M
1
23.
24.
Zr = 53.5 M and fr = 104 kHz from Problem 23.
6.3 V
Itot =
= 11.8 A
53.5 M
XL = 2frL = 2(104 kHz)(50 mH) = 32.7 k
6.3 V
= 164 mA
IC = IL =
2
(20 )  (32.7 k) 2
25.
Ptrue = (164 mA)2 20  = 538 mW; Pr = 0 VAR
Pa = (11.8 A)2 53.5 M = 7.45 mVA
179
Chapter 17
Part 3: Series Parallel RLC Circuits
Section 17-7 Analysis of Series-Parallel RLC Circuits
26.
27.
28.
R (  jX C )
R  jX C
(220 )( j150 )
 j33000 
= j100  +
 j100  
= j100  + 12455.7 
266  34.3
220   j150 
= j100  + 69.9   j102.4  = 69.9   j2.4  = 69.91.97 
(a)
ZT = jXL +
(b)
ZT = jXL +
R (  jX C )
R  jX C
 j120 k
= j8 k + 7.6950.2 k
= j8 k +
15.6  39.8
= j8 k + 4.92 k  j5.91 k = 4.92 k + j2.09 k = 5.3523.0 k
From Problem 26:
 = 1.97 (capacitive)
(a)
(b)
 = 23.0 (inductive)
1.5 H = 1500 mH
XL = 2(2 kHz)(1500 mH) = 18.9 k
1
XC =
= 16.9 k
2(2 k)(0.0047 F)
jX L ( R2  jX C )
ZLR2C = jXL  (R2  jXC) =
R2  jX C  jX L
(18.990 k)(27.7  37.5 k)
= 23.747.3 k = 16.1 k + j17.4 k
=
22.15.19 k
ZT = R1 + ZLR2C = 33 k + 16.1 k + j17.4 k = 49.1 k + j17.4 k = 52.1 19.5 k
R 
 330 k 
120 V = 7.6019.5 V
VR1 =  1 Vs  
 52.119.5 k 
 ZT 

Z
 23.747.3 k 
120 V = 5.4627.8 V
VL =  LR2C Vs  
 52.119.5 k 
 ZT 
 R2 


220 k
VL  
5.4627.8 V = 4.3463.3 V
VR2 = 

 27.7  37.5 k 
 R2  jX C 
 XC
VC = 
 R2  jX C

 16.9  90 k 
VL  
5.4627.8 V = 3.3324.7 V

 27.7  37.5 k 

180
Chapter 17
29.
See Figure 17-3.
XC = 16.9 k, XL = 18.9 k
Z1 = R2  jXC = 22 k  j16.9 k = 27.737.5 k
(27.7  37.5 k)(18.990 k)
Z2 = XL  Z1 =
= 23.747.3 k = 16.1 k + j17.4 k
22 k  j2 k
ZT = R1 + Z2 = 33 k + 16.1 k + j17.4 k = 49.1 k + j17.4 k = Req + jXeq
Xeq = 2fL
X eq
17.4 k
L=
= 1.38 H

2f
2( 2 kHz)
49.1 k
Figure 17-3
30.
1
= 56.4 
2(60 Hz)(47 F)
XL = 2(60 Hz)(390 mH) = 147 
ZA = R1 + jXL = 100   j147  = 17855.8 
(1000 )(17858.8 )
ZB = R2  ZA =
= 71.819.5  = 67.7  + j24.0 
200   j147 
ZT = XC + ZB = j56.4  + 67.7  + j24.0  = 67.7   j32.4  = 74.825.7 
Z 
 71.819.5  
1150 V = 11045.2 V
VR2 =  B Vs  
 74.8  25.7  
 ZT 
XC =
I2 =
VR2 11045.2 V
= 1.1045.2 A

R2
1000 
31.
From Problem 30:
I2 = 1.1045.2 A
The phase angle between I2 and the source voltage is 45.2 with I2 leading.
32.
ZA = R2  XL  XC2 =
1
 1   1   1 

  j
  j

 10 k   5 k   10 k 
1
1
=

= 7.0745 k = 5 k + j5 k
0.1 mS  j0.1 mS 0.1414  45 mS
ZB = R1  jXC1 = 3.3 k  j1 k
ZT = ZA + ZB = 8.3 k + j4 k = RT + jXT
RT = 8.3 k, XT = 4 k (inductive)
181
1.38 H
Chapter 17
33.
From Problem 32, ZT = 8.3 k + j4 k = 9.2125.7 k
V
100 V
IT = s 
= 1.0925.7 mA
Z T 9.2125.7 k
ZA = 7.0745 k from Problem 32.
Z 
 7.0745 k 
100 V = 7.6719.3 V
VZA =  A Vs  
 9.2125.7 k 
 ZT 
IR1 = IC1 = IT = 1.0925.7 mA
V
7.6719.3 V
IR2 = ZA 
= 76719.3 A
R2
100 k
7.6719.3 V
V
IC2 = ZA 
= 767109.3 A
XC2 10  90 k
V
7.6719.3 V
IL = ZA 
= 1.53-70.7 mA
XL
590 k
VR1 = ITR1 = (1.0925.7 mA)(3.30 k) = 3.6025.7 V
VR2 = VL = VC2 = VZA = 7.6719.3 V
VC1 = ITXC1 = (1.0925.7 mA)(190 k) = 1.09116 V
34.
For Vab = 0 V, Va must equal Vb.
XL1 = 226 , XL2 = 151 
 22690  
120 V = 9.3838.5 V
Va = VL1 = 
 180   j226  
It is not possible for Vab to be 0 V because the LC branch has no resistance; thus, the voltage a
to b can only have a phase angle of 0, 90, or 90 (the branch is either resonant, purely
inductive, or purely capacitive depending on the value of XL). Therefore, it is not possible for
Va to equal Vb in both magnitude and phase, which are necessary conditions.
35.
See Figure 17-4.
Figure 17-4
1
= 241 
2(3 kHz)(0.22 F)
XL1 = 2(3 kHz)(12 mH) = 226 
XL2 = 2(3 kHz)(8 mH) = 151 
XC =
182
Chapter 17
Za + Zb + Zc = 100   j226  + j151  = 100  + j377  = 39075 
(22690 )(1000 )
Za Zc

Z1 =
= 57.915 
39075 
Za  Zb  Zc
(15190 )(1000 )
Zb Zc
Z2 =
= 38.515 

Za  Zb  Zc
39075 
(22690 )(15190 )
Za Zb

Z3 =
= 86.9105 
39075 
Za  Zb  Zc
Combining R1 + Z1 in parallel with XC + Z2:
(1800  + 57.915)  (24190  + 38.515 )
= (180  + 55.9  + j14.98 )  (j241  + 37.2  + j9.96 )
= (236  + j15.0 )  (37.2   j231 )
= (2363.64 )  (23480.9 ) = 15938.9 
ZT = 15938.9  + 86.9105 
= 124   j99.8   22.5  + j83.9  = 101.5   j15.9  = 1038.9 
 159  38.9  
120 V = 18.530.0 V
VR1Z1 = VCZ2 = 
 103  8.9  
 R1 
 1800  
VR1Z1  
18.5  30.0 V = 14.133.6 V
VR1 = 
 2363.64  
 Z R1Z1 
 X 
 241  90  
18.5  30.0 V = 19.139.1 V
VC =  C VCZ2  
 234  80.9  
 ZCZ2 
Vab = VR1  VC = 14.133.6 V  19.139.1 V
= (11.7 V  j7.80 V)  (14.8 V  j12.0 V) = 3.10 V + j4.20 V = 5.22126 V
5.22126 V
Vab

I100  =
= 52.2126 mA
1000 
1000 
36.
There are two resonant frequencies. One is associated with the parallel circuit containing C
and L2. The other is associated with the series circuit consisting of C and L1.
37.
For series resonance:
1
1

fr =
= 4.11 kHz
2 L1C 2 (10 mH)(0.15 F)
XL1 = 2(4.11 kHz)(10 mH) = 258 
XC = 258 
XL2 = 2(4.11 kHz)(25 mH) = 646 
 jX C ( RW 2  jX L 2 )
 j258 (4   j646 )
Zr = RW1 + jXL1 +
= 2  + j258  +
RW 2  jX L 2  jX C
4   j646  j258
= 47590 
 Zr 
475  90 


Vs  
Vout = 
100 V
 8600  475  90 
 R  Zr 
= 4.8361.0 V
183
Chapter 17
For parallel resonance:
RW2 C
1
1
L 

fr =
= 2.60 kHz
2 LC
2 LC 2 (25 mH)(0.15 F)
1
XL = 408 
X
408 
= 102
Q = L2 
RW 2
4


Zr = RW Q 2  1  4 (1022  1) = 41.6 k
XL1 = 2FrL1 = 2(2.6 kHz)(10 mH) = 163 
Since Zr is much greater than R, RW1, or XL1 and is resistive, the output voltage is approximately:
Vout  100 V
38.
See Figure 17-5. The winding resistance is neglected because it contributes negligibly to the
outcome of the calculations.
1
1
f r2  
fr 
4 LC
2 LC
1
C 
 2
4 f r L
For fr = 8 MHz, 9 MHz, 10 MHz, and 11 MHz
1
C1 =
= 39.6 pF

4 (8 MHz)2 (10 H)
1
= 31.3 pF
C2 =

4 (9 MHz) 2 (10 H)
1
C3 =
= 25.3 pF

4 (10 MHz) 2 (10 H)
1
= 20.9 pF
C4 =

4 (11 MHz) 2 (10 H)
Figure 17-5
Part 4: Special Topics
Section 17-8 Bandwidth of Resonant Circuits
X L 2 k

= 80
R
25 
f
5 kHz
BW = r 
= 62.5 Hz
Q
80
39.
Q=
40.
BW = f2  f1 = 2800 Hz  2400 Hz = 400 Hz
f  f 2 2400 Hz  2800 Hz
fr = 1

= 2600 Hz
2
2
184
Chapter 17
41.
Pf1 = (0.5)Pr = (0.5)(2.75 W) = 1.38 W
42.
Q=
43.
BW =
fr
8 kHz

= 10
BW 800 Hz
XL(res) = QRW = 10(10 ) = 100 
XL
100 

= 1.99 mH
L=
2f r 28 kHz)
XC = XL at resonance
1
1

= 0.2 F
C=
2f r X C 28 kHz)(100 )
fr
Q
If Q is doubled, the bandwidth is halved to 200 Hz.
Multisim Troubleshooting and Analysis
44.
No fault.
45.
C1 is leaky.
46.
R1 is open.
47.
C1 is leaky.
48.
L1 is open.
49.
No fault.
50.
fc = 504.89 kHz
51.
fc = 338.698 kHz
185
Chapter 18
Passive Filters
Section 18-1 Low-Pass Filters
1.
 500  90  
 100 V = 0.491 V  j2.16 V = 2.2277.2 V rms
Vout = 
 2.2 k  j500  
2.
(a)
(d)
100 Hz is passed
3 kHz is borderline
3.
(a)
XC =
(b)
(c)
4.
(b)
(e)
1 kHz is passed
5 kHz is rejected
1
= 265 
2(60 Hz)(10 F)
 265  90  
100 V = 9.3620.7 V
Vout = 
 100   j265  
1
XC =
= 48.5 
2(400 Hz)(8.2 F)
 48.5  90  
100 V = 7.1844.1 V
Vout = 
 47   j48.5  
XL = 2(1 kHz)(5 mH) = 31.4 
 3300  
100 V = 9.965.44 V
Vout = 
 330   j31.4  
(d)
XL = 2(2 kHz)(80 H) = 1 
 100  
100 V = 9.955.74 V
Vout = 
 10   j1  
(a)
fc =
(b)
 100  90  
50 V = 3.5445 V
Vout = 
 100   j100  
1
fc =
= 413 Hz
2(47 )(8.2 F)
1
XC =
= 47.0 
2(413 Hz)(8.2 F)
1
= 159 Hz
2(100 )(10 F)
1
XC =
= 100 
2(159 Hz)(10 F)
 47  90  
50 V = 3.5445 V
Vout = 
 47   j47  
186
(c)
2 kHz is passed
Chapter 18
5.
fc =
(d)
fc =
1
= 20.0 k
2(80 H / 10 )
XL = 2(20.0 kHz)(80 H) = 10 
 100  
50 V = 3.5445 V
Vout = 
 10   j10  
1
2RC
1
C=
2Rf c
fc =
(a)
(b)
(c)
(d)
6.
1
= 10.5 kHz
2(5 mH/330 )
XL = 2(10.5 kHz)(5 mH) = 330 
 3300  
50 V = 3.5445 V
Vout = 
 330   j330  
(c)
1
= 12.1 F
2(220 )(60 Hz)
1
C=
= 1.45 F
2(220 )(500 Hz)
1
C=
= 0.723 F
2(220 )(1 kHz)
1
= 0.144 F
C=
2(220 )(5 kHz)
C=
Position 1:
1
fc =
2RCT
CT =
Position 3:
CT = 1000 pF
1
= 15.9 kHz
2(10 k)(1000 pF)
Position 4:
1
= 500 pF
CT =
1
1

0.001 F 1000 pF
1
fc =
= 31.8 kHz
2(10 k)(500 pF)
fc =
1
1
1

0.01 F 0.022 F  0.047 F
= 0.00873 F
1
= 1.82 kHz
fc =
2(10 k)(0.00873 F)
Position 2:
CT = 0.022 F + 0.047 F = 0.069 F
1
= 231 Hz
fc =
2(10 k)(0.069 F)
187
Chapter 18
7.
See Figure 18-1.
Figure 18-1
8.
(a)
(b)
(c)
(d)
9.
V
20 log out
 Vin
V
20 log out
 Vin

1V 
  20 log
 = 0 dB

1V 


3V
  20 log
 = 4.44 dB

5V

V
20 log out
 Vin
V
20 log out
 Vin

7.07 V 
  20 log
 = 3.01 dB

 10 V 


5V 
  20 log
 = 14.0 dB

 25 V 

V 
dB = 20 log out 
 Vin 
Vout
 dB 
 log 1 

Vin
 20 
 dB 
Vout = Vin log 1 

 20 
(a)
(b)
(c)
(d)
 1
Vout = (8 V) log 1   = 7.13 V
 20 
 3
Vout = (8 V) log 1 
 = 5.67 V
 20 
6
Vout = (8 V) log 1 
 = 4.01 V
 20 
  20 
Vout = (8 V) log 1 
 = 0.800 V
 20 
188
Chapter 18
10.
The output decreases at the rate of 20 dB/decade
(a)
10 kHz is 2 decades above fc: Vout = 20 dB
(b) 100 kHz is 2 decades above fc: Vout = 40 dB
(c)
1 MHz is 3 decades above fc: Vout = 60 dB
11.
The output decreases at the rate of 20 dB/decade
(a)
10 kHz is in the pass bandc: Vout = 0 dB
(b) 100 kHz is the cutoff frequency fc: Vout = 3 dB (ideally 0 dB)
(c)
1 MHz is 1 decade above fc: Vout = 20 dB
Section 18-2 High-Pass Filters
12.
13.
The output increases at the rate of 20 dB/decade
(a)
10 kHz is 1 decade below fc: Vout = -20 dB
(b) 100 kHz is the cutoff frequency fc: Vout = 3 dB (ideally 0 dB)
(c)
1 MHz is in the pass band: Vout = 0 dB
 2.20 k 
 100 V = 9.7512.8 V
Vout = 
 2.2 k  j500  
14.
(a)
(d)
1 Hz is rejected.
60 Hz is passed.
15.
(a)
XC =
(b)
(c)
(d)
(b)
(e)
20 Hz is rejected.
30 kHz is passed.
1
= 265 
2(60 Hz)(10 F)
 1000  
100 V = 3.5369.3 V
Vout = 
 100   j265  
1
XC =
= 84.7 
2(400 Hz)(4.7 F)
 470  
100 V = 4.8561.0 V
Vout = 
 47   j84.7  
XL = 2(1 kHz)(5 mH) = 31.4 
 31.490  
100 V = 94784.6 mV
Vout = 
 330   j31.4  
XL = 2(2 kHz)(80 H) = 1 
 190  
100 V = 99584.3 mV
Vout = 
 10   j1  
1
1
, fc =
2RC
2 ( L / R )
1
fc =
= 159 Hz;
2(100 )(10 F)
fc =
16.
(a)
Vout = 7.07 V
189
(c)
50 Hz is borderline.
Chapter 18
(b)
(c)
(d)
17.
1
= 720 Hz;
2(47 )(4.7 F)
1
= 10.5 kHz;
fc =
2(5 mH/330 )
1
fc =
= 19.9 kHz;
2(80 H/10 )
fc =
Vout = 7.07 V
Vout = 7.07 V
Vout = 7.07 V
See Figure 18-2.
720 Hz
Figure 18-2
18.
Position 1:
RT = 1 k + 3.3 k + 1 k = 5.3 k
1
= 2.00 kHz
fc =
2(5.3 k)(0.015 F)
Position 2:
RT = 3.3 k + 1 k = 4.3 k
1
= 0.006 F
CT =
1
1

0.015 F 0.01 F
1
= 6.17 kHz
fc =
2(4.3 k)(0.006 F)
Position 3:
RT = 860  + 1 k = 1.86 k
1
= 5.70 kHz
fc =
2(1.86 k)(0.015 F)
Position 4:
RT = 2.2 k + 3.3 k + 1 k = 6.5 k
1
fc =
= 1.63 kHz
2(6.5 k)(0.015 F)
190
Chapter 18
Section 18-3 Band-Pass Filters
19.
20.
21.
1

1
(a)
f0 =
(b)
f0 =
(a)
RT = 10  + 75  = 85 
1
1
f0 =
= 14.5 kHz

2 LC 2 (12 mH)(0.01 F)
XL = 2(14.5 kHz)(12 mH) = 1.10 k
X
1.1 k
Q= L 
= 13
RT
85 
f
14.5 kHz
BW = 0 
= 1.12 kHz
Q
13
(b)
RT = 10  + 22  = 32 
1
1
f0 =
= 24.0 kHz

2 LC 2 (2 mH)(0.022 F)
XL = 2(24.0 kHz)(2 mH) = 302 
X
302 
Q= L 
= 9.44
RT
32 
f
24.0 kHz
BW = 0 
= 2.54 kHz
Q
9.44
2 LC
1
2 LC

2 (12 mH)(0.01 F)
1
2 (2 mH)(0.022 F)
= 14.5 kHz
= 24.0 kHz
Using the results of Problems 19 and 20:
BW
1.12 kHz
 14.5 kHz 
(a)
f2 = f0 +
= 14.5 kHz + 560 kHz = 15.06 kHz
2
2
BW
1.12 kHz
 14.5 kHz 
f1 = f0 
= 14.5 kHz  560 Hz = 13.94 kHz
2
2
(b)
BW
2.54 kHz
 24.0 kHz 
= 24.0 kHz + 1.27 kHz = 25.3 kHz
2
2
BW
2.24 kHz
f1 = f0 
= 24.0 kHz  1.27 kHz = 22.7 kHz
 24.0 kHz 
2
2
f2 = f0 +
RW2 C
L
Center frequency = f0 =
2 LC
1
22.
Since RW is assumed to be zero, f0 =
1
2 LC
.
191
Chapter 18
(a)
f0 =
(b)
f0 =
1
2 (1 H)(10 F)
1
= 50.3 Hz
2 (2.5 H)(25 pF)
= 20.1 MHz
RW2 C
(4 ) 2 (10 F)
1
L 
1H
f0 =
= 50.3 Hz
2 (1 H)(10 F)
2 LC
XL = 2(50.3 Hz)(1 H) = 316 
X
316 
Q= L 
= 79
RW
4
Ztank = RW(Q2 + 1) = 4 (792 + 1) = 24,968 


24,968 
120 V = 117 V
Vout = 
 24,968   680  
1
23.
(a)
(b)
24.
1

1
= 10.1 MHz
2 LC 2 (2 H)(25 pF)
XL = 2(20.1 MHz)(2.5 H) = 316 
X
316 
Q= L 
= 79
RW
4
Ztank = RW(Q2 + 1) = 4 (792 + 1) = 24,968 


24,968 
120 V = 115 V
Vout = 
 24,968   1000  
f0 =
Position 1:
1
1
f0 =

= 712 kHz
2 LC 2 (50 H)(1000 pF)
Position 2:
1
1

= 159 kHz
f0 =
2 LC 2 (100 H)(0.01 F)
Position 3:
1
1

= 306 kHz
f0 =
2 LC 2 (270 H)(0.001 F)
f12 = 712 kHz  159 kHz = 553 kHz
f23 = 306 kHz  159 kHz = 147 kHz
f13 = 712 kHz  159 kHz = 405 kHz
Responses do not overlap.
192
Chapter 18
25.
f0 = (BW) Q = (500 Hz)40 = 20 kHz
2.5 V
XC =
= 125 
20 mA
1
= 0.064 F
C=
2f 0 X C
X
Q = L = 40
RW
X
RW = L = 0.025XL = 0.025(2f0L)
40
RW2
L
f0 =
2 LC
1
RW2
(0.025)(2f 0 L)) 2 C
1
2
1  (0.025(2 ) 2 f 0 ) LC
2
L
L
f0  2


4 LC
42 LC
4 2 LC
2
Note: in the above derivation, (0.025(2)) = 0.025
f 02 4 2 LC = 1  0.25 f 02 LC
1
f 02 LC ( 4 2  0.025) = 1
1
L= 2
= 989 H
2
f 0 C (4  0.025)
Section 18-4 Band-Stop Filters
26.
(a)
(b)
27.
1
1

= 339 kHz
2 (100 H)(0.0022 F)
1

= 10.4 kHz
f0 =
2 LC 2 (5 mH)(0.047 F)
f0 =
(a)
f0 
(b)
f0 
2 LC
1
1
2 LC
1
2 LC


1
2 (0.5 H)(6.8 F)
1
2 (10 H)(47 pF)
= 86.3 Hz
= 7.34 MHz
RW2 C
L = 86.3 Hz
f0 =
2 LC
XL = 2(86.3 Hz)(0.5 H) = 271 
X
271 
Q= L 
= 33.9
RW
8
Ztank = RW(Q2 + 1) = 8 ((33.9)2 +1)= 9.20 k
 1 k 
50 V = 4.90 V
Vout = 
 10.2 k 
1
28.
(a)
193
Chapter 18
1
(b)
f0 =
(8 ) 2 (47 pF)
10 H
2 (10 H)(47 pF)
= 7.34 MHz
XL = 2(7.34 MHz)(10 H) = 461 
461 
X
Q= L 
= 57.6
RW
8
Ztank = RW(Q2 + 1) = 8 (57.62 + 1) = 26.6 k
 2 .2 k  
50 V = 3.82 V
Vout = 
 28.8 k 
29.
For the pass band, f0 = 1200 kHz:
1
f0 =
2 L1C
1
4 L1C
1
1
 2
= 0.08 H
L1 =
2 2
4 f 0 C 4 (1200 kHz)2 (0.22 F)
For the stop band, f0 = 456 kHz:
1
f0 =
2 L2C
f 02 
L2 =
2
1
1
 2
= 0.554 H
2
4 f 0 C 4 (456 kHz) 2 (0.22 F)
2
Multisim Troubleshooting and Analysis
30.
C1 is open.
31.
C2 is leaky.
32.
R3 is open.
33.
C1 is shorted.
34.
L2 is open.
35.
No fault.
36.
fr = 107.637 kHz
37.
BW  88.93 MHz
194
Chapter 19
Circuit Theorems in AC Analysis
Section 19-1 The Superposition Theorem
1.
Z1 = R2  R3 = 6880 
Z2 = R1 + Z1 = 16880 
(16880 )(290 k)
Z3 = XL  Z2 =
= 1.2940.2 k = 985  + j833 
261749.8 
ZT1 = XC + Z3 = j1 k + 985  + 833  = 985   j167  = 9999.6 
V
20 V
IT1 = 1 
= 29.6 mA
Z T1 999  9.6 
 XL 
 20 k 
I T1  
29.6 mA = 1.5349.8 mA
IR1 = 
 2.6249.8 k 
 XL  Z 2 
 R2 
 10 k 
I R1  
1.5349.8 mA = 46949.8 A = 303 A + j358 A
IR3(V1) = 

 3.20 k 
 R2  R3 
With V1 reduced to zero (shorted):
(1  90 k)(290 k)
Z1 = XC  XL =
= 290 k
190 k
Z2 = R1 + Z1 = 1 k  j2 k = 2.2463.4 k
(2.20 k)(2.2463.4 k)
Z3 = R3  Z2 =
3.2 k  j2 k
4.93  63.4 k
=
= 1.3131.4 k = 1.12 k  j0.68 k
3.77  32 k
ZT2 = R2 + Z3 = 2.12 k  j0.68 k = 2.2317.8 k
V
330 V
= 1.3547.8 mA
IT2 = 2 
Z T2 2.23  17.8 
 2.24  63.4 k 
1.3547.8 mA = 80216.4 A = 769 A + j226 A
IR3(V2) = 
 3.77  32 k 
IR3(tot) = IR3(V1) + IR3(V2) = 1.07 mA + j584 A = 1.22 28.6 mA
195
Chapter 19
2.
Use the results of Problem 1:
With V2 reduced to zero (shorted):
IR1 = 1.53 49.8 mA
 R3 
2 .2 k  
I R1  
IR2(V1) = 
1.5349.8 mA = 1.0549.8 mA = 678 A + j802 A

 3 .2 k  
 R2  R3 
With V1 reduced to zero (shorted):
IR2(V2) = IT2 = 1.3547.8 mA = 907 A = j1 mA
The total current through R2 is:
IR2 = IR2(V2) + IR2(V2) = 1.59 mA + j1.80 mA = 2.448.5 mA
VR2 = IR2R2 = (2.448.5 mA)(10k) = 2.448.5 V
The total voltage across the R2 branch is:
VT = V2 + VR2 = 330 V + 2.448.5 V
= (2.6 V + j1.5 V) + (1.59 V + 1.8 V) = 4.19 V + j3.3 V = 5.3338.2 V
3.
With Vs reduced to zero (shorted):
XL = 1.9 k, XC = 2.41 k
 (1.80 k)(4.7 k  j1.9 k) 

Z1 = R1  (R2 + XL) = 
6.5 k  j1.9 k


=
(1.80 k)(5.122 k)
= 1.365.7 k = 1.35 k  j0.135 k
6.7716.3 k
 XC 
 2.41  90 k 
I S  
1000 mA
IZ1 = 

 1.35 k  j2.28 k 
 X C  Z1 
 2.41  90 k 
1000 mA = 90.930.6 mA
= 
 2.65  59.4 k 
 R 2  XC 
I Z1
IR1(I) = 

 R 1  R 2  XC 
 5.2827.1 k 
90.9  30.6 mA = 69.323.8 mA = 63.4 mA  j28.0 mA
= 
 6.9320.3 k 
With Is reduced to zero (opened):
( 2.41  90 k)(5.122 k)
ZT = R1 + (XC  (R2 + XL)) = 1.80 k +
4.7 k  j0.51 k
= 1.8 k + 2.6059.2 k = 3.8535.5 k
V
750 V
IR1(V) = s 
= 19.535.5 k = 15.9 mA + j11.3 mA
Z T 3.85  35.5 k
The total current through R1 is:
IR1(tot) = IR1(I) + IR1(V) = 79.3 mA  j16.7 mA = 81.011.9 mA
196
Chapter 19
4.
(a)
With Is2 zeroed (open), there is no current through RL due to Is1, so IL(1) = 0 A.
With Is1 zeroed (open), the current through RL due to Is2 is:
 XC 
 2  90 k 
 2  90 k 
I s2  
10 mA  
10 A
IL(2) = 

 4.7 k  j2 k 
 5.1  23.1 k 
 R L  XC 
= 39266.7 mA
IL = IL(1) + IL(2) = 0 A + 39266.9 A = 39266.9 mA
(b)
1
= 637 k
2(2.5 kHz)(100 pF)
With V2 zeroed (shorted), the impedance “seen” by V1 is developed as follows:
ZA = RL  jXC3 = 5 M  j637 k = 5.047.26 M
(10 M)(5.04  7.26 M)
ZB = R2  ZA =
= 8351.2 k = 835 k  j17.5 k
6.03  6.06 M
ZC = XC2 + ZB = j637 k + 835 k  j17.5 k = 835 k  j654 k = 1.0638 M
(10 M)(1.06  38 M)
= 54518.5 k = 517 k  j172 k
ZD = R1  ZC =
1.95  19.6 M
ZT(1) = XC1 + ZD = j637 k + 517 k  j172 k = 517 k  j809 k = 96057.4 k
V1
4060 V
IT(1) =

= 41.7117.4 A
Z T(1) 960  57.4 k
XC1 = XC2 = XC3 =
 R1 


10 M
I T(1)  
41.7117.4 A = 21.4137 A
IC2(1) = 

 1.94  19.8 M 
 R1  ZC 
 R2 


10 M
I C2(1)  
21.4137 A = 3.55143 A
IL(1) = 
 6.03  6.06 M 
 R 2  ZA 
With V1 zeroed (shorted), the impedance “seen” by V2 is developed as follows:
(10 M)(637  90 k)
ZA = R1  XC1 =
= 53557.5 k = 287 k  j450 k
1.19  32.5 M
ZB = XC2 + ZA = 287 k + j1.09 M = 1.1375.2 M
(5 M  j637 k)(1.13  75.2 M)
ZD = (RL + XC3)  ZB =
5.29 M  j1.73 M
(4.74  7.7 M)(1.13  75.2 M)
=
= 1.0264.4 M = 442 k  j921 k
5.56  18.1 M
ZT(2) = R2 + ZC = 1.44 M  j921 k = 1.7132.6M
IT(2) =
V2
2030 V

= 11.762.6 A
Z T2 1.71  32.6 M


 1.13  75.2 M 
ZB
I T(2)  
11.762.6 A = 2.375.46 A
IL(2) = 

 5.56  18.1 M 
 Z B  R L  XC3 
IL(tot) = IL(1) + IL(2) = 3.55143 A + 2.375.46 A
= (2.88 A + j2.13 A) + (2.36 A  j0.225 A) = 0.478 A + j2.35 A
= 2.40101.5 A
197
Chapter 19
5.
See Figure 19-1(a).
RT = R2 + R3 + R4 + R5 = 1 k + 3.9 k + 10 k + 5.1 k = 20 k
20 V
IT =
= 1 mA
20 k
IR3 = ITR3 = (1 mA)(3.9 k) = 3.9 V
VB(dc) = 20 V  3.9 V = 16.1 V VD(dc) = 0 V
VC(dc) = VB(dc)  ITR2 = 16.1 V  (1 mA)(1 k) = 15.1 V
VA(dc) = 0 V
See Figure 19-1(b).
VA(peak) = 9 V
RT = R1 + R3  (R2 + R4  R6) = 1.2 k + 2.36 k = 3.56 k
9V
IT(peak) =
= 2.53 mA
3.56 k
VB(peak) = VA(peak)  IT(peak)R1 = 9 V  (2.53 mA)(1.2 k) = 9 V  3.04 V = 5.96 V


R3
 3.9 k 
I
IR2(peak) = 

2.53 mA = 1 mA
 R  R  R R  T ( peak )  9.9 k 
3
2
4
6


VC(peak) = VD(peak) = VB(peak)  IR2(peak)R2 = 5.96 V  (1 mA)(1 k) = 4.96 V
198
Chapter 19
Figure 19-1
6.
With the current source zeroed (see Figure 19-2(a)):
(2090 )(31.671.6 )
ZT = 1890  +
5178.7 
= 1890  + 12.482.9  = j18  + 1.53  + j12.3  = 1.53   j5.69 
= 5.9074.9 
1230 V
= 2.04105 A = 522 mA + j1.97 A
IT = IC(Vs) =
5.90  74.9 
With the voltage source zeroed (see Figure 19-2(b)):
Impedance of the L1, L2, C branch:
3600 
(2090 )(18  90 )
= j30   j180  = j150 
= j30  +
Z = j30  +
290 
j2 
 100  


100 
500120 mA  
500120 mA = 33.3206 mA
IL2 = 
 10   j150  
 150.3  86.2  
 2090  
33.3206 mA = 333206 mA = 298 mA  j147 mA
IC(Is) = 
 290  
The total capacitor current is:
IC(tot) = IC(Vs) + IC(Is) = 821 mA  j1.82 A = 2.00114 A
199
Chapter 19
Figure 19-2
7.
From Problem 6
With current source zeroed (open): IT(VS) = 2.04105 A
 jX L1  ( R  jX L 2 ) 
 12.497.7 
 IT(VS)  
 2.04105  1.26112.7A
jX L1
 2090 


IR(VS) = 
With the voltage source zeroed (shorted): Impedance of LC branch is ZLC = 150-90 
Ζ 
R  Ζ LC 1500  90

 100
150  90
Z LC

 100 
 IS  
 0.5120A = 0.5120A

 100 
IRT = IR(VS) – IR(IS) = 1.26112.7A  0.5120A  -0.24 A + j0.727 A
= 766  71.7A
 ZT
 R
IR(IS) = 
Section 19-2 Thevenin’s Theorem
8.
From Problem 5, VD(peak) = VC(peak) = 4.96 V
Vth = VD(rms) = 0.707(4.96 V) = 3.51 V
Rth = R4  (R2 + R1  R3) = 10 k  (1.0 k + 1.2 k  3.9 k) = 10 k  2.14 k = 1.76 k
9.
(a)
(b)
(c)

 XC
 75  90  
Vs  
250 V = 1553.1 V
Vth = 

 100   j75  
 R1  jXC 
(1000 )(75  90 )
R 1 XC
 270 
Zth = R2 +
= 63   j48 
125  36.9 
R 1  jXC
= 79.237.3 
 XL1

 40090  
Vs  
30 V = 1.220 V
Vth = 
 98090  
 XL1  XL2 
(40090 )(58090 )
XL1 XL2

Zth =
= 23790  = j237  = 23790 
98090 
XL1  XL2
VT = V1 + V2 = 15 V + 8.66 V + j5 V = 24.211.9 V
200
Chapter 19
 R2 
 1000 k 
VT  
24.211.9 V = 12.111.9 V
Vth = 
 2000 k 
 R1  R 2 
Zth = XC + R1  R2 = 50 k  j20 k = 53.921.8 k
10.
1
= 33.86 k
2(100 Hz)(0.047 F)
Find Zth looking from the open terminals after removing RL:
(220 k)(33.86  90 k)
= 18.4433.0 k = 15.5 k  j10 k
ZA = R1  XC1 =
40.4  57.0 k
ZB = R2 + ZA = 22 k + 15.5 k  j10 k = 37.5 k  j10 k = 38.814.9 k
(33.86  90 k)(38.8  14.9 k)
ZC = XC2  ZB =
= 22.7755.4 k
57.7  49.5 k
= 12.9 k  j18.7 k
ZD = R3 + ZC = 34.9 k  j18.7 k = 39.628.2 k
Zth = ZD = 39.628.2 k = 34.9 k  j18.7 k
XC1 = XC2 =
Find Vth looking from the source after removing RL:
ZA = R2  jXC2 = 22 k  j33.86 k = 40.457 k
(33.86  90 k)(40.4  57 k)
= 19.275 k = 4.97 k  j18.6 k
ZB = XC1  ZA =
71.2  72 k
ZT = R1 + ZB = 26.97 k  j18.6 k = 32.734.5 k
320 V
V
IT = s 
= 0.9834.5 mA
Z T 32.7  34.5 k
 XC1 
 33.86  90 k 
I T  
0.9834.5 mA = 0.4716.5 mA
IR2 = 

 22 k  j67.7 k 
 XC1  Z A 
IC2 = IR2 = 0.4716.5 mA
Vth = VC2 = IC2XC2 = (0.4716.5)(33.8690 k) = 15.973.5 V
The Thevenin equivalent circuit with RL connected is shown in Figure 19-3.
The current through RL is:
IL =
15.9  73.5 V
Vth
= 11765.6 A

R L  Z th 136.2  7.9 k
Figure 19-3
201
Chapter 19
11.
The circuit is redrawn in Figure 19-4(a) for easier analysis.
Combining R1, R2, and XL:
(3.30 k)(390 k)
= 2.49 k + j1.64 k
ZA = R1 + R2  XL = 10 k +
3.3 k  j3 k
= 2.9833.4 k
Combining R3 and ZA:
(100 k)(2.9833.4 k)
ZB = R3  ZA =
= 2.3725.9 k
12.67.5 k
Figure 19-4
Combining XC and ZB:
(5  90 k)(2.3725.9 k)
= 2.621.8 k  2.62 k  j0.082 k
Zth = XC  ZB =
4.52  62.3 k
 ZB 
 2.3725.9 k 
Vs  
500 V = 26.387.6 V
Vth = 

 4.5  61.7 k 
 XC  Z B 
The Thevenin circuit with R4 connected is shown in Figure 19-4(b).

 R4

4.70 k 
Vth  
26.387.6 V = 16.988.2 V
VR4 = 

 7.32  0.64 k 
 R 4  Z th 
12.
Refer to Figure 19-5 (note that R3 has been removed).
(1000 )(9090 )
= 66.948  = 44.8  + j49.7 
ZA = R1  XL =
134.542 
ZB = R2 + ZA = 194.8  + j49.7  = 20114.3 
(120  90 )(20114.3 )
= 11755.9 
Zth = XC  ZB =
207  19.8 
Looking from Vs:
(1000 )(192  38.7 )
ZT = XL + R2  (R1 + XC) = j90  +
277  25.6 
= j90  + 69.313.1  = j90  + 67.5   j15.7  = 67.5  + j74.3  = 10047.7 
X
VL =  L
 ZT

 9090  
Vs  
750 V = 67.542.3 V
 10047.7  

202
Chapter 19
 R2 
 1500  
(Vs  VL )  
(750 V  67.542.3 V)
VR2 = 

 150   j120  
 R 2  XC 
 1500  
= 
 (75 V  49.9 V  j45.4 V) = (0.78138.7)(51.961.1 V)
 192  38.7 
= 40.522.4 V
Vth = Vab = VR2 + VL
= (37.4 V  15.4 V) + (49.9 V + j45.4 V) = 87.2 V + j30 V = 92.219 V
Figure 19-5
Section 19-3 Norton’s Theorem
13.
14.
Using Zth and Vth from Problem 9 in each part:
15  53.1 V
V
(a)
In = th 
= 18915.8 mA
Z th 79.2  37.3 
Zn = Zth = 79.237.3 
Vth 1.220 V

= 5.1590 mA
Z th 23790 
Zn = Zth = 23790 
(b)
In =
(c)
In =
12.111.9 V
Vth

= 22433.7 A
Z th 53.9  21.8 k
Zn = Zth = 53.921.8 k
From Problem 10, Zn = Zth = 39.628.2 k
The total impedance seen by the source with the terminals shorted is the same (in this case) as Zn.
320 V
V
IT = s 
= 80828.2 A
Z T 39.6  28.2 k
203
Chapter 19




XC1
I   33.9  90 80828.2 A = 47512.3 A
IR2 = 
 X  R  R X  T  57.7  49.5 k 


2
3
C2 
 C1
 XC2 
 33.9  90 
I R2  
475  12.3 A = 39945.3 A
In = IR3 = 

 40.4  57 k 
 R 3  XC2 
The Norton equivalent circuit with RL connected is shown in Figure 19-6.
 Zn

 39.6  28.2 k 
I n  
399  45.3 A = 11665.7 A
IRL = 

 136.2  7.9 k 
 R L  Zn 
39.628.2 k
39945.3 A
Figure 19-6
15.
First remove R4 and determine Zn looking in at the resulting open terminals.
(100 k)(5  90 k)
ZA = R3  XC =
= 4.4663.4 = 2 k  j4 k
11.2  26.6 k
(3.30 k)(390 k)
ZB = R1 + R2  XL = 10 k +
4.4642.3 k
= 10 k + 2.2247.7 k = 2.49 k + j1.64 k = 2.9833.4 k
(4.46  63.4 k)(2.9833.4 k)
Zn = ZA  ZB =
= 2.612.30 k =2.61 k  j0.105 k
5.1  27.7 k
Looking from the source with R4 shorted:
ZT = XC = 590 k
500 V
V
In = s 
= 1090 mA
Z T 5  90 k
The Norton equivalent circuit with R4 connected is shown in Figure 19-7.
 Zn 
 2.61  2.30 k 
I n  
1090 mA = 3.5788.5 mA
IR4 = 
 7.31  0.823 k 
 R 4  Zn 
VR4 = IR4R4 = (3.5788.5 mA)(4.70 k) = 16.888.5 V
Figure 19-7
204
Chapter 19
Section 19-4 Maximum Power Transfer Theorem
16.
(a)
1
1
= 11.3 k

2fC 2(3 kHz)(0.0047 F)
ZL = RL + jXL = 6.8 k + j11.3 k
X
11.3 k
L= L 
= 599 mH
2f 2(3 kHz)
XC =
(b)
ZL = 8.2 k + j5 k
(c)
XL = 75.4 , XC = 60.3 
Zth = R + XC  XL = 500  +
(75.490 )(60.3  90 )
= 50   j301 
15.190 
ZL = 50  + j301 
301 
L=
= 0.4 H
2(120 Hz)
17.
For maximum load power, ZL equals the complex conjugate of Zth.
ZA = R1  jXC1 = 8.2   j10  = 12.950.6 
(180 )(4  90 )
= 3.9177.5 = 0.846   j3.82 
ZB = R2  XC2 =
18.4  12.5 
(12.9  50.6 )(3.91  77.5 )
= 3.0671.4 = 0.976   j2.90 
ZC = ZB  ZA =
16.5  56.7 
Zth = R3 + ZC = 9.18   j2.90 
ZL = 9.18  + j2.90 
18.
First convert the delta to a wye:
(12 )(12 )
X1 = X2 = X3 =
=4
12   12   12 
The circuit is redrawn in Figure 19-8(a). Remove ZL and Thevenize:
(6.8   j4 )( j4 )
(7.930.5 )(490 )
 j4  
6.8   j8 
10.549.6 
= j4  + 370.9  = j4  + 0.98  + j2.8  = 0.98  + j6.8 
 7.930.5  
 6.8   j4  
100 V = 7.519.1 V
100 V  
Vth = 
 10.549.6  
 6.8   j8  
For maximum power to ZL:
ZL = 0.98   j6.8 
7.5  19.1 V
Vth

= 3.8319.1 A
IL =
Z th  Z L
1.960 
Zth = j4  + (6.8  + j4 )  j4  = j4  +
PL(true) = I L2 RL = (3.83 A)2(0.98 ) = 14.4 W
205
Chapter 19
Figure 19-8
19.
The circuit is redrawn in Figure 19-9 to determine Zth. The load impedance (real part) must
equal the Thevenin impedance (real part) and the reactive parts must be equal in magnitude but
opposite in sign. That is, the impedances must be complex conjugates.
(1000 )(9090 ) (2200 )(120  90 )
= 6748  + 105.361.4 

100   j90 
220   j120 
= 44.8  + j49.8  + 50.4   j92.5  = 95.2   j42.7 
ZL = 95.2  + j42.7 
Zth =
Figure 19-9
Multisim Troubleshooting and Analysis
20.
R2 is open.
21.
C2 is leaky.
22.
C1 is open.
23.
No fault.
24.
VTH = 750.281.40 mV
ZTH = 11.970 k
25.
IN = 30.142113.1 A
ZN = 30.364.28 k
206
Chapter 20
Time Response of Reactive Circuits
Section 20-1 The RC Integrator
1.
 = RC = (2.2 k)(0.047 F) = 103 s
2.
(a)
(b)
(c)
(d)
5RC = 5(56 )(47 F) = 13.2 ms
5RC = 5(3300 k)(0.015 F) = 248 s
5RC = 5(22 k)(100 pF) = 11 s
5RC = 5(5.6 M)(10 pF) = 280 s
Section 20-2 Response of an Integrator to a Single Pulse
3.
VC  0.632(20 V) = 12.6 V
4.
(a)
(b)
(c)
(d)
5.
See Figure 20-1.
v
v
v
v




0.865(20 V) = 17.3 V
0.950(20 V) = 19.0 V
0.982(20 V) = 19.6 V
0.993(20 V) = 19.9 V (considered full charge of 20 V)
Figure 20-1
6.
 = RC = (1 k)(1 F) = 1 ms
vout = 0.632(8 V) = 5.06 V
See Figure 20-2 for output waveform.
The time to reach steady-state with repetitive pulses is 5 ms.
Figure 20-2
207
Chapter 20
7.
(a)
(b)
Looking from the capacitor, the Thevenin resistance is R1  R2 = 5 k.
 = (5 k)(4.7 F) = 23.5 ms
 10 k 
Vout(max) = 
20 V = 10 V
 20 k 
See Figure 20-3.
Figure 20-3
8.
See Figure 20-4.
Figure 20-4
9.
From Problem 7 23.5 ms
The input pulse width equals one time constant, therefore
Vout = 0.632(10V) = 6.32 V
See Figure 20-5.
6.32 V
23.5 ms
Figure 20-5
Section 20-3 Response of RC Integrators to Repetitive Pulses
10.
Transient time = 5RC = 5(4.7 k)(10 F) = 235 ms
208
Chapter 20
11.
 = (4.7 k)(10 F) = 47 ms
5 = 5(47 ms)= 235 ms
See Figure 20-6.
Figure 20-6
12.
See Figure 20-7.
Figure 20-7
13.
1
1

= 100 s
f 10 kHz
tW = 0.25(100 s) = 25 s
1st pulse: 0.632(1 V) 632 mV
Between 1st and 2nd pulses: 0.05(0.632 V) = 31.6 mV
2nd pulse: 0.632(1 V  0.0316 V) + 0.0316 V = 644 mV
Between 2nd and 3rd pulses: 0.05(0.644 V) = 32.2 mV
3rd pulse: 0.632(1 V  0.0322 V) + 0.0322 V = 644 mV
See Figure 20-8.
T=
Figure 20-8
209
Chapter 20
14.
The steady-state output equals the average value of the square wave input which is
Vin 30 V
= 15 V (with a small ripple voltage)

2
2
Section 20-4 Response of RC Differentiators to a Single Pulse
15.
See Figure 20-9.
Figure 20-9
16.
 = (1 k)(1 F) = 1 ms
Steady-state is reached in 5 = 5 ms.
At 1 ms, V  (0.368)(8 V) = 2.94 V
See Figure 20-10.
Figure 20-10
17.
(a)
(b)
Looking from the source and capacitor:
( 2.2 k)(1 k  1 k)
= 1.05 k
RT =
4 .2 k 
 = RTC = (1.05 k)(470 pF) = 493.5 ns
5 = 5(493.5 ns) = 2.467 s
 1 k 
Vout(max) = 
10 V = 5 V
 2 k 
See Figure 20-11.
Figure 20-11
Section 20-5 Response of RC Differentiators to Repetitive Pulses
18.
 = (1 k)(1 F) = 1 ms
See Figure 20-12.
Figure 20-12
210
Chapter 20
19.
Since 5>> tW, the output shape is an approximate reproduction of the input but with a zero
average value.
Section 20-6 Response of RL Integrators to Pulse Inputs
20.
10 mH
= 1 ms
10 
5 = 5 ms
Vout(max) = 0.637(8 V) = 5.06 V
See Figure 20-13.
=
Figure 20-13
50 mH
= 50 ms
1
5 = 250 ms
See Figure 20-14.
21.
=
22.
LT = 8 H + 4 H = 12 H
(100 )(156 )
RT =
= 60.9 
256 
L
12 H
= 197 ns
= T 
RT 60.9 
This circuit is an integrator.
Figure 20-14
Section 20-7 Response of RL Differentiators to Pulse Inputs
23.
100 H
= 4.55 s
22 
(a)
=
(b)
See Figure 20-15.
Figure 20-15
211
Chapter 20
24.
100 H
= 4.55 s
22 
(a)
=
(b)
See Figure 20-16.
Figure 20-16
Section 20-8 Relationship of Time Response to Frequency Response
0.35
, 5 = 50 s
tr
0.9 = 1(1  et/RC)
t2 = RCln(0.1) = (10 s)ln(0.1) = 23 s
t1 = RCln(0.9) = (10 s)ln(0.9) = 1.05 s
tr = 23 s  1.05 s = 22.0 s
0.35
fh =
= 15.9 kHz
22.0 s
25.
fh =
26.
fh =
0.35 0.35

= 8.33 MHz
tf
42 ns
Section 20-9 Troubleshooting
27.
(b)
(c)
(d)
Vout = Vin: C is open or R could be shorted.
C is leaky or C is greater than 0.22 F or R is greater than 3.3 k.
Resistor open or capacitor shorted.
28.
(a)
(b)
(c)
No problem since 5<tW.
C is leaky.
C is open or R is shorted.
Multisim Troubleshooting and Analysis
29.
C1 open or R1 shorted.
30.
No fault.
31.
R1 or R2 open.
32.
L1 or L2 open.
212
Chapter 21
Three-Phase Systems in Power Applications
Section 21-1 Generators in Power Applications
V
100 V

= 376 mA
Z 265.8 
1.
IL =
2.
 = tan1 
3.
220 A = 1.88 A + j0.684 A
3140 A = 2.3 A + j1.93 A
1.5100 A = 0.26 A  j1.48 A
In = (1.88 A + j0.684 A) + (2.3 A + j1.93 A) + (0.26 A  j1.48 A)
= (1.88 A  2.3 A  0.26 A) + j(0.684 A + 1.93 A  1.48 A) = 0.68 A + j1.134 A
= 1.32121 A
 175  
 = 41.2
 200  
Section 21-2 Types of Three-Phase Generators
4.
VL(ba) = 600120 V  6000 V = 300 V + j520 V  600 V = 900 V + j520 V = 1.04150 kV
VL(ca) = 600120 V  6000 V = 300 V  j520 V  600 V = 900 V  j520 V
= 1.04150 kV
VL(cb) = 600120 V  600120 V = 300 V  j520 V + 300 V  j520 V = j1.04 kV
= 104 90 kV
5
ILa = Ia  Ib = 50 A  5120 A = 5 A  (2.5 A + j4.33 A) = 7.5 A  j4.33 A = 8.6630 A
ILb = 3 (590 A) = 8.6690 A
ILc =
6.
3 (5  150 A) = 8.66150 A
See Figure 21-1.
IL1 = Ia  Ib = 50 A  5120 A = 5 A + 2.5 A  j4.33 A = 7.5 A  j4.33 A = 8.6630 A
IL2 = Ib  Ic = 5120 A  5120 A = 2.5 A + j4.33 A + 2.5 A  j4.33 A = j8.66 A
= 8.6690 A
IL3 = Ic  Ia = 5120 A  50 A = 2.5 A  j4.33 A  5 A = 7.5 A  j4.33 A
= 8.66150 A
213
Chapter 21
Figure 21-1
Section 21-3 Three-Phase Source/Load Analysis
7.
(a)
Line voltages:
VL(ab) = 3 Va(30) = 3 (500(030)) V = 86630 V
VL(ca) = 3 Vc(30) = 3 (500(12030)) V = 866150 V
VL(bc) = 3 Vb(30) = 3 (500(12030)) V = 86690 V
(b)
Phase currents:
5000 V
= 50032 mA
132 k
500120 V
= 50088 mA
Ib = IZb =
132 k
500  120 V
Ic = IZc =
= 500152 mA
132 k
Ia = IZa =
(c)
Line currents:
ILa = 50032 mA
ILb = 50088 mA
ILc = 500152 mA
(d)
Load currents:
IZa = 50032 mA
IZb = 50088 mA
IZc = 500152 mA
214
(e)
Load voltages:
VZa = Va = 5000 V
VZb = Vb = 500120 V
VZc = Vc = 500120 V
Chapter 21
8.
9.
(a)
Line voltages:
VL(ab) = 3 Va(30) = 3 (100(030)) V = 17330 V
VL(ca) =
3 Vc(30) = 3 (100(12030)) V = 173150 V
VL(bc) =
3 Vb(30) = 3 (100(12030)) V = 17390 V
(b)
Phase currents:
1000 V
= 74145 mA
Ia =
13545 
100120 V
= 160 A
Ib =
10060 
100  120 V
Ic =
= 500140 mA
20020 
(c)
Line currents:
(d)
ILa = Ia = 74145 mA
ILb = Ib = 160 A
ILc = Ic = 500140 mA
(f)
Neutral current:
In = IZa + IZb + IZc = 74145mA + 160 A + 500140 mA
= (524 mA  j524 mA) + (383 mA  j321 mA) + (500 mA + j866 mA)
= 641 mA  j20.9 mA = 6411.86 mA
(a)
Line voltages:
VL(ab) = 3 Va(30) = 3 (50(030)) V = 86.630 V
(b)
Load currents:
(e)
IZa = Ia = 74145 mA
IZb = Ib = 160 A
IZc = Ic = 500140 mA
Load voltages:
VZa = Va = 1000 V
VZb = Vb = 100120 V
VZc = Vc = 100120 V
VL(ca) =
3 Vc(30) = 3 (50(12030)) V = 86.6150 V
VL(bc) =
3 Vb(30) = 3 (50(12030)) V = 86.690 V
Phase currents:
First find the load currents:
VL ( ca ) 86.6  150 V

IZa =
= 144220 mA = 110 mA + j92.6 mA
Za
60070 
VL (bc ) 86.690 V

IZb =
= 14420 mA = 135 mA + j49.3 mA
Zb
60070 
VL ( ab ) 86.6  30 V

IZc =
= 144100 mA = 25.0 mA + j142 mA
Zc
60070 
Ia = IZa  IZc = (110 mA + j92.6 mA)  (25.0 mA  j142 mA)
= 85 mA + j235 mA = 250110 mA
Ib = IZc  IZb = (25.0 mA  j142 mA)  (135 mA + j49.3 mA)
= 160 mA + j191.3 mA = 250130 mA
Ic = IZb  IZa = (135 mA + j49.3 mA)  (110 mA + j92.6 mA)
= 245 mA  j43.3 mA = 25010 mA
215
Chapter 21
10.
(c)
Line currents:
ILa = Ia = 250110 mA
ILb = Ib = 250130 mA
ILc = Ic = 25010 mA
(d)
Load currents were found in part (b).
(e)
Load voltages:
VZa = VL(ca) = 86.6150 V
VZb = VL(bc) = 86.690 V
VZc = VL(ab) = 86.630 V
(a)
Line voltages:
(b)
VL(ab) = 120120 V
VL(ca) = 1200 V
VL(bc) = 120120 V
11.
Phase currents:
1200 V
Ia = IZa =
= 1250 A
1050 
120  120 V
= 12170 A
Ib = Ib =
1050 
120120 V
Ic = Ic =
= 1270 A
1050 
(c)
Line currents:
IL1 = 120 A  12120 A = 12 A  (6 A + j10.4 A) = 18 A  j10.4 A = 20.830 A
IL2 = 12120 A  120 A = (6 A  j10.4 A)  12 A = 18 A  j10.4 A = 20.8150 A
IL3 = 12120 A  12120 A = (6 A + j10.4 A)  (6 A  j10.4 A) = 20.890 A
(d)
Line currents:
IZa = 1250 A
IZb = 12170 A
IZc = 1270 A
(a)
Line voltages:
VL(ab) = Va = 330120 V
VL(ca) = Vc = 330120 V
VL(bc) = Vb = 3300 V
(b)
Load currents:
First find the load voltages:
V 
VZa =  a (120  30) V = 19190 V
 3
V 
VZb =  b (0  30) V = 19130 V
 3
V 
VZc =  c (120  30) V = 191150 V
 3
V
191  90 V
IZa = Za 
= 38.2150 A
Za
560 
(e)
Load voltages:
VZa = 1200 V
VZb = 120120 V
VZc = 120120 V
216
Chapter 21
VZb 19130 V

= 38.230 A
Zb
560 
191150 V
V
IZc = Zc 
= 38.290 A
Zc
560 
IZb =
Section 21-4 Three-Phase Power
12.
PT = 3(1200 W) = 3.6 kW
13.
Figure 21-34 in text:
500 V
IZ =
= 500 mA
1 k
PL = 3VZIZcos = 2(500 V)(500 mA)cos 32 = 636 W
Figure 21-35 in text:
100 V
IZa =
= 741 mA
135 
PZa = VZaIZacos = (100 V)(741 mA)cos 45 = 52.4 W
100 V
IZb =
= 500 mA
200 
PZb = VZbIZbcos = (100 V)(500 mA)cos 20 = 47.0 W
100 V
=1A
IZc =
100 
PZc = VZcIZccos = (100 V)(1 A)cos 60 = 50.0 W
PL = PZa + PZb + PZc = 52.4 W + 47.0 W + 50.0 W = 149 W
Figure 21-36 in text:
VZ = 3 (50 V) = 86.6 V
86.6 V
IZ =
= 144 mA
600 
PL = 3VZIZcos = 3(86.65 V)(144 mA)cos 70 = 12.8 W
Figure 21-37 in text:
120 V
IZ =
= 12 A
10 
PL = 3VZIZcos = 3(120 V)(12 A)cos 50 = 2.78 kW
Figure 21-38 in text:
330 V
VZ =
= 191 V
3
191 V
IZ =
= 38.2 A
5
PL = 3VZIZcos = 3(191 V)(38.2 A)cos 60 = 10.9 kW
217
Chapter 21
14.
VZ =
IZ =
120 V
3
VZ

Z
= 69.3 V
69.3 V
(100 )  (100 )
2
2

69.3 V
= 490 mA
141.4 
 = 45
PL = 3VZIZcos = 3(69.3 V)(490 mA)cos 45 = 72 W
15.
ZL = 141.445 
IL = IZ
120 V
V
VZ = L 
= 69.8 V
3
3
69.8 V
V
IZ = Z 
= 494 mA
Z L 141.4 
PT =
3 VLILcos
(120 V)(494 mA)
V I 
Peach =  L L  cos 
cos 45 = 24.2 W
3
 3 
16.
P1 = VLILcos( + 30)
P2 = VLILcos(  30)
 100  
 = 45
 = tan 1 
 100  
P1 = (120 V)(494 mA)cos(45 + 30) = 15.3 W
P2 = (120 V)(494 mA)cos(45  30) = 56.8 W
218