Instructor’s Resource Manual
to accompany
Introductory
Circuit Analysis
Eleventh Edition
Robert L. Boylestad
Upper Saddle River, New Jersey
Columbus, Ohio
Contents
CHAPTER 1
1
CHAPTER 2
9
CHAPTER 3
13
CHAPTER 4
22
CHAPTER 5
29
CHAPTER 6
39
CHAPTER 7
52
CHAPTER 8
65
CHAPTER 9
86
CHAPTER 10
106
CHAPTER 11
124
CHAPTER 12
143
CHAPTER 13
150
CHAPTER 14
157
CHAPTER 15
170
CHAPTER 16
195
CHAPTER 17
202
CHAPTER 18
222
CHAPTER 19
253
CHAPTER 20
266
CHAPTER 21
280
CHAPTER 22
311
CHAPTER 23
318
CHAPTER 24
333
CHAPTER 25
342
TEST ITEM FILE
353
iii
Chapter 1
1.
2.
−
−
3.
−
4.
υ=
5.
6.
⎡ 1h ⎤
4 min ⎢
⎥ = 0.067 h
⎣ 60 min ⎦
31 mi
d
υ= =
= 29.05 mph
t 1.067 h
a.
b.
c.
7.
8.
d 20,000 ft ⎡ 1 mi ⎤ ⎡ 60 s ⎤ ⎡ 60 min ⎤
=
= 1363.64 mph
t
10 s ⎢⎣ 5,280 ft ⎥⎦ ⎢⎣1 min ⎥⎦ ⎢⎣ 1 h ⎥⎦
−
95 mi ⎡ 5,280 ft ⎤ ⎡ 1 h ⎤ ⎡1 min ⎤
= 139.33 ft/s
h ⎢⎣ mi ⎥⎦ ⎢⎣ 60 min ⎥⎦ ⎢⎣ 60 s ⎥⎦
60 ft
d
t= =
= 0.431 s
υ 139.33 ft/s
d 60 ft ⎡ 60 s ⎤ ⎡ 60 min ⎤ ⎡ 1 mi ⎤
= 40.91 mph
υ= =
t
1 s ⎢⎣1 min ⎥⎦ ⎢⎣ 1 h ⎥⎦ ⎢⎣ 5,280 ft ⎥⎦
−
9.
−
10.
MKS, CGS, °C =
5
5
5
(°F − 32) = (68 − 32) = (36) = 20°
9
9
9
SI: K = 273.15 + °C = 273.15 + 20 = 293.15
11.
⎡ 0.7378 ft - lb ⎤
1000 J ⎢
⎥ = 737.8 ft-lbs
1J
⎦
⎣
12.
⎡ 3 ft ⎤ ⎡12 in. ⎤ ⎡ 2.54 cm ⎤
0.5 yd ⎢
⎥⎢
⎥⎢
⎥ = 45.72 cm
⎣1 yd ⎦ ⎣ 1 ft ⎦ ⎣ 1 in. ⎦
13.
a.
104
14.
a.
15 × 103
e.
Chapter 1
b. 106
4.02 × 10−4
c. 103
b.
30 × 10−3
f. 2 × 10−10
d. 10−3
c. 2.4 × 106
e. 100
f. 10−1
d. 150 × 103
1
4.2 × 103 + 48.0 × 103 = 52.2 × 103 = 5.22 × 104
90 × 103 + 360 × 103 = 450 × 103 = 4.50 × 105
50 × 10−5 − 6 × 10−5 = 44 × 10−5 = 4.4 × 10−4
1.2 × 103 + 0.05 × 103 − 0.6 × 103 = 0.65 × 103 = 6.5 × 102
15.
a.
b.
c.
d.
16.
a.
b.
c.
d.
e.
f.
17.
a.
b.
c.
d.
18.
a.
b.
c.
d.
e.
f.
102/104 = 10−2 = 10 × 10−3
10−2/103 = 10−5 = 10 × 10−6
104/10−3 = 107 = 10 × 106
10−7/102 = 1.0 × 10−9
1038/10−4 = 1.0 × 1042
100 / 10 −2 = 101/10−2 = 1 × 103
19.
a.
b.
c.
d.
(2 × 103)/(8 × 10−5) = 0.25 × 108 = 2.50 × 107
(4 × 10−3)/(60 × 104) = 4/60 × 10−7 = 0.667 × 10−7 = 6.67 × 10−8
(22 × 10−5)/(5 × 10−5) = 22/5 × 100 = 4.4
(78 × 1018)/(4 × 10−6) = 1.95 × 1025
20.
a.
c.
21.
a.
b.
c.
d.
22.
a.
b.
c.
d.
2
(102)(103) = 105 = 100 × 103
(10−2)(103) = 101 = 10
(103)(106) = 1 × 109
(102)(10−5) = 1 × 10−3
(10−6)(10 × 106) = 10
(104)(10−8)(1028) = 1 × 1024
(50 × 103)(3 × 10−4) = 150 × 10−1 = 1.5 × 101
(2.2 × 103)(2 × 10−3) = 4.4 × 100 = 4.4
(82 × 106)(2.8 × 10−6) = 229.6 = 2.296 × 102
(30 × 10−4)(4 × 10−3)(7 × 108) = 840 × 101 = 8.40 × 103
(102)3 = 1.0 × 106
(104)8 = 100.0 × 1030
b.
d.
(10−4)1/2 = 10.0 × 10−3
(10−7)9 = 1.0 × 10−63
(4 × 102)2 = 16 × 104 = 1.6 × 105
(6 × 10−3)3 = 216 × 10−9 = 2.16 × 10−7
(4 × 10−3)(6 × 102)2 = (4 × 10−3)(36 × 104) = 144 × 101 = 1.44 × 103
((2 × 10−3)(0.8 × 104)(0.003 × 105))3 = (4.8 × 103)3 = (4.8)3 × (103)3
= 110.6 × 109 = 1.11 × 1011
−3 2
−6
(−10 ) = 1.0 × 10
(10 2 )(10 −4 )
= 10−2/103 = 1.0 × 10−5
3
10
(10 −3 ) 2 (10 2 ) (10 −6 )(10 2 ) 10 −4
=
= 4 = 1.0 × 10−8
10 4
10 4
10
3
4
(10 )(10 )
= 107/10−4 = 1.0 × 1011
10 − 4
Chapter 1
e.
[(10 )(10 )]
[(10 ) ][10 ]
(1 × 10−4)3(102)/106 = (10−12)(102)/106 = 10−10/106 = 1.0 × 10−16
−2
2
f.
2 2
23.
a.
b.
c.
d.
e.
f.
−3
−3
=
1
1
= 1.0 × 10−1
=
−3
(10 )(10 ) 10
4
(3 × 10 2 ) 2 (10 2 )
= (9 × 104)(102)/(3 × 104) = (9 × 106)/(3 × 104) = 3 × 102 = 300
3 × 10 4
(4 × 10 4 ) 2 16 × 10 8
=
= 2 × 105 = 200.0 × 103
(20) 3
8 × 10 3
(6 × 104 ) 2
36 × 108
= 9.0 × 1012
=
−2 2
−4
( 2 × 10 )
4 × 10
(27 × 10 −6 )1 / 3 3 × 10 −2
= 1.5 × 10−7 = 150.0 × 10−9
=
2 × 10 5
2 × 10 5
(4 × 10 3 ) 2 (3 × 10 2 ) (16 × 10 6 )(3 × 10 2 ) 48 × 10 8
= 24.0 × 1012
=
=
−4
−4
−4
2 × 10
2 × 10
2 × 10
(16 × 10−6)1/2(105)5(2 × 10−2) = (4 × 10−3)(1025)(2 × 10−2) = 8 × 1020
= 800.0 × 1018
−3 3
−4
2
2
⎡
⎤⎡
⎤ ⎡
⎤
⎣(3 × 10 ) ⎦ ⎣1.60 × 10 ⎦ ⎣(2 × 10 )(8 × 10 ) ⎦
(7 × 10−5 ) 2
2
g.
1/ 2
(27 × 10−9 )(2.56 × 104 )(16 × 10−2 )1/ 2
49 × 10−10
(69.12 × 10−5 )(4 × 10−1 ) 276.48 × 10−6
=
=
49 × 10−10
49 × 10−10
= 5.64 × 104 = 56.4 × 103
=
+3
24.
a.
6 × 103 = 0.006 × 10+6
−3
b.
4 × 10−3 = 4000 × 10−6
+3
c.
−2
+3
+3
50 × 105 = 5000 × 103 = 5 × 106 = 0.005 × 109
+2
Chapter 1
−3
−3
−3
3
−3
+5
d.
30 × 10−8 = 0.0003 × 10−3 = 0.3 × 10−6 = 300 × 10−9
−5
25.
a.
−3
+3
−3
+3
0.05 × 100 s = 50 × 10−3 s = 50 ms
+3
+3
b.
2000 × 10−6 s = 2 × 10−3 s = 2 ms
−3
c.
−3
0.04 × 10−3 s = 40 × 10−6 s = 40 μs
+3
+6
d.
8400 × 10−12 s ⇒ 0.0084 × 10−6 s = 0.0084 μs
−6
e.
−3
4 × 10−3 × 103 m = 4 × 100 m = 4000 × 10−3 m = 4000 mm
+3
increase by 3
100
f.
260 × 103 × 10−3 m = 0.26 × 103 m = 0.26 km
−3
26.
a.
b.
c.
d.
4
⎡ 60 s ⎤
1.5 min ⎢
⎥ = 90 s
⎣1 min ⎦
⎡ 60 min ⎤ ⎡ 60 s ⎤
0.04 h ⎢
⎥⎢
⎥ = 144 s
⎣ 1 h ⎦ ⎣1 min ⎦
⎡ 1 μs ⎤
0.05 s ⎢ −6 ⎥ = 0.05 × 106 μs = 50 × 103 μs
⎣10 s ⎦
⎡ 1 mm ⎤
0.16 m ⎢ −3 ⎥ = 0.16 × 103 mm = 160 mm
⎣10 m ⎦
Chapter 1
27.
28.
e.
⎡ 1 ns ⎤
1.2 × 10−7 s ⎢ −9 ⎥ = 1.2 × 102 ns = 120 ns
⎣10 s ⎦
f.
⎡1 min ⎤ ⎡ 1 h ⎤ ⎡1 day ⎤
3.62 × 106 s ⎢
⎥ = 41.90 days
⎥⎢
⎥⎢
⎣ 60 s ⎦ ⎣ 60 min ⎦ ⎣ 24 h ⎦
a.
⎡10 −6 F ⎤ ⎡ 1 pF ⎤
−6
12
5
0.1 μF ⎢
⎥ ⎢ −12 ⎥ = 0.1 × 10 × 10 pF = 10 pF
1
F
μ
10
F
⎦
⎣
⎦⎣
b.
⎡100 cm ⎤
= 8000 × 10−3 cm = 8 cm
80 × 10−3 m ⎢
⎥
⎣ 1m ⎦
c.
⎡ 1 m ⎤ ⎡ 1 km ⎤
−5
60 cm ⎢
⎥⎢
⎥ = 60 × 10 km
⎣100 cm ⎦ ⎣1000 m ⎦
d.
⎡ 60 min ⎤
3.2 h ⎢
⎥
⎣ 1h ⎦
e.
⎡10−3 m ⎤ ⎡ 1 μ m ⎤
3
0.016 mm ⎢
⎥ ⎢ − 6 ⎥ = 0.016 × 10 μm = 16 μm
⎣ 1 mm ⎦ ⎣10 m ⎦
f.
⎡ 1m ⎤⎡ 1m ⎤
−4
2
60 cm2 ⎢
⎥ ⎢100 cm ⎥ = 60 × 10 m
100
cm
⎣
⎦⎣
⎦
a.
⎡ 1m ⎤
100 in. ⎢
⎥ = 2.54 m
⎣ 39.37 in. ⎦
b.
⎡12 in. ⎤ ⎡ 1 m ⎤
4 ft ⎢
⎥ = 1.22 m
⎥⎢
⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦
c.
⎡ 4.45 N ⎤
6 lb ⎢
⎥ = 26.7 N
⎣ 1 lb ⎦
d.
⎡ 1 N ⎤ ⎡ 1 lb ⎤
60 × 103 dynes ⎢ 5
⎥⎢
⎥ = 0.13 lb
⎣10 dynes ⎦ ⎣ 4.45 N ⎦
e.
⎡ 1 in. ⎤
150,000 cm ⎢
⎥
⎣ 2.54 cm ⎦
f.
⎡ 5280 ft ⎤ ⎡12 in. ⎤ ⎡ 1 m ⎤
0.002 mi ⎢
⎥⎢
⎥⎢
⎥ = 3.22 m
⎣ 1 mi ⎦ ⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦
Chapter 1
⎡ 1 ms ⎤
6
⎢10− 3 s ⎥ = 11.52 × 10 ms
⎦
⎣
⎡ 60 s ⎤
⎢1 min ⎥
⎦
⎣
⎡ 1 ft ⎤
⎢12 in. ⎥ = 4921.26 ft
⎣
⎦
5
29.
⎡1 yd ⎤
5280 ft ⎢
⎥ = 1760 yds
⎣ 3 ft ⎦
⎡ 12 in. ⎤ ⎡ 1 m ⎤
5280 ft ⎢
⎥⎢
⎥ = 1609.35 m, 1.61 km
⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦
5280 ft,
m ⎡ 39.37 in. ⎤ ⎡ 1 ft ⎤ ⎡ 1 mi ⎤ ⎡ 60 s ⎤ ⎡ 60 min ⎤
s ⎢⎣ 1 m ⎥⎦ ⎢⎣12 in. ⎥⎦ ⎢⎣ 5280 ft ⎥⎦ ⎢⎣1 min ⎥⎦ ⎢⎣ 1 h ⎥⎦
= 670,615,288.1 mph ≅ 670.62 × 106 mph
30.
299,792,458
31.
⎡ 3 ft ⎤
100 yds ⎢
⎥
⎣1 yd ⎦
⎡ 1 mi ⎤
⎢ 5,280 ft ⎥ = 0.0568 mi
⎣
⎦
60 mi ⎡ 1 h ⎤ ⎡1 min ⎤
= 0.0167 mi/s
h ⎢⎣ 60 min ⎥⎦ ⎢⎣ 60 s ⎥⎦
t=
32.
33.
34.
υ
d
=
0.0568 mi
= 3.40 s
0.0167 mi/s
30 mi ⎡ 5280 ft ⎤ ⎡12 in. ⎤ ⎡ 1 m ⎤ ⎡ 1 h ⎤ ⎡ 1 min ⎤
= 13.41 m/s
h ⎣⎢ 1 mi ⎦⎥ ⎣⎢ 1 ft ⎦⎥ ⎣⎢ 39.37 in. ⎦⎥ ⎣⎢ 60 min ⎦⎥ ⎣⎢ 60 s ⎦⎥
50 yd ⎡ 60 min ⎤ ⎡ 3 ft ⎤ ⎡ 1 mi ⎤
= 1.705 mi/h
⎢
⎥
min ⎢⎣ 1 h ⎥⎦ ⎣1 yd ⎦ ⎢⎣ 5,280 ft ⎥⎦
3000 mi
d
⎡1 day ⎤
t= =
= 1760 h ⎢
⎥ = 73.33 days
υ 1.705 mi/h
⎣ 24 h ⎦
⎡1000 m ⎤ ⎡ 39.37 in. ⎤ ⎡ 1 ft ⎤ ⎡ 1 mi ⎤
10 km ⎢
⎥⎢
⎥⎢
⎥⎢
⎥
⎣ 1 km ⎦ ⎣ 1 m ⎦ ⎣12 in. ⎦ ⎣ 5280 ft ⎦
1 mi
d 6.214 mi
= 40.39 min
υ=
,t= =
1 mi
6.5 min
υ
6.5 min
= 6.214 mi
35.
⎡ 3 ft ⎤ ⎡12 in. ⎤
100 yds ⎢
⎥⎢
⎥ = 3600 in ⇒ 3600 quarters
⎣1 yd ⎦ ⎣ 1 ft ⎦
36.
60 mph:
6
=
100 mi
= 1.67 h = 1 h:40.2 min
υ 60 mi/h
d 100 mi
75 mph:
t= =
= 1.33 h = 1h:19.98 min
υ 75 mi/h
difference = 20.22 minutes
t=
d
Chapter 1
37.
cm ⎤
⎡
[0.016 h ] ⎡⎢ 60 min ⎤⎥
d = υt = ⎢600
⎥
s ⎦
⎣ 1h ⎦
⎣
38.
⎡ 14 ft ⎤
d = 86 stories ⎢
⎥
⎣ story ⎦
υ=
39.
40.
⎡ 1m ⎤
⎢100 cm ⎥ = 345.6 m
⎦
⎣
⎡
⎤
⎢1 step ⎥
⎢ 9 ⎥ = 1605 steps
⎢ ft ⎥
⎣⎢ 12 ⎦⎥
d 1605 steps
d
⎡ 1 minute ⎤
⇒t = =
= 802.5 seconds ⎢
⎥ = 13.38 minutes
2 steps
t
υ
⎣ 60 seconds ⎦
second
⎡ 14 ft ⎤
⎡ 1 mile ⎤
d = (86 stories) ⎢
⎥ = 1204 ft ⎢
⎥ = 0.228 miles
story
⎣ 5,280 ft ⎦
⎦
⎣
min 10.7833 min
=
= 47.30 min/mile
mile 0.228 miles
5 min
1 mile ⎡ 5,280 ft ⎤ 1056 ft
,
=
⇒
mile
5 min ⎢⎣ 1mile ⎥⎦ minute
υ=
41.
⎡ 60 s ⎤
⎢1 min ⎥
⎦
⎣
1204 ft
d
d
= 1.14 minutes
⇒t = =
υ 1056 ft
t
min
a.
⎡ 1 Btu ⎤
−3
5 J⎢
⎥ = 4.74 × 10 Btu
⎣1054.35 J ⎦
b.
⎡ 1 gallon ⎤
24 ounces ⎢
⎥
⎣128 ounces ⎦
c.
d.
⎡ 14 ft ⎤
distance = 86 stories ⎢
⎥ = 1204 ft
⎣ story ⎦
⎤
⎡
1 m3
−4 3
⎥ = 7.1 × 10 m
⎢
⎣ 264.172 gallons ⎦
⎡ 86,400 s ⎤
5
1.4 days ⎢
⎥ = 1.21 × 10 s
1
day
⎣
⎦
⎡ 264.172 gallons ⎤ ⎡ 8 pints ⎤
1 m3 ⎢
⎥ ⎢1 gallon ⎥ = 2113.38 pints
1 m3
⎣
⎦⎣
⎦
42.
6(4 + 8) = 72
43.
(20 + 32)/4 = 13
(82 + 122 ) = 14.42
44.
45.
MODE = DEGREES: cos 50° = 0.64
46.
MODE = DEGREES: tan−1(3/4) = 36.87°
Chapter 1
7
47.
48.
49.
50.
8
( 400 /(6
205 × 10−6
2
)
+ 10) = 2.95
1.20 × 1012
6.667 × 106 + 0.5 × 106 = 7.17 × 106
Chapter 1
Chapter 2
1.
−
2.
a.
F= k
b.
F=k
c.
F= k
d.
Exponentially,
a.
r = 1 mi:
b.
r =10 ft:
c.
1 in. ⎡ 1 m ⎤
= 1.59 mm
16 ⎢⎣ 39.37 in. ⎥⎦
3.
Q1Q2 (9 × 109 )(1 C)(2 C)
= 18 × 109 N
=
r2
(1 m)2
(9 × 109 )(1 C)(2 C )
Q1Q2
= 2 × 109 N
=
r2
(3 m)2
Q1Q2 (9 × 109 )(1 C)(2 C)
=
= 0.18 × 109 N
2
2
r
(10 m)
r3 10 m
F
18 × 109 N
= 100
=
= 10 while 1 =
F2 0.18 × 109 N
r1 1 m
⎡ 5280 ft ⎤ ⎡12 in. ⎤ ⎡ 1 m ⎤
1 mi ⎢
⎥ = 1609.35 m
⎥⎢
⎥⎢
⎣ 1 mi ⎦ ⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦
kQ1Q2 (9 × 109 )(8 × 10−6 C)(40 × 10-6 C) 2880 × 10−3
=
=
F=
(1609.35 m) 2
2.59 × 106
r2
= 1.11 μN
⎡12 in. ⎤ ⎡ 1 m ⎤
10 ft ⎢
⎥⎢
⎥ = 3.05 m
⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦
kQ1Q2 2880 × 10−3 2880 × 10−3
=
=
= 0.31 N
F=
9.30
r2
(3.05 m) 2
kQ1Q2
2880 × 10−3
2880 × 10−3
=
=
= 1138.34 × 103 N
(1.59 × 10−3 m) 2 2.53 × 10−6
r2
= 1138.34 kN
F=
4.
−
5.
F=
Chapter 2
kQ1Q2
⇒r=
r2
kQ1Q2
=
F
(9 × 109 )(20 × 10 −6 ) 2
= 10 mm
3.6 × 104
9
6.
F=
a.
b.
kQ1Q2
kQ Q
⇒ 1.8 = 1 22 ⇒ kQ1Q2 = 4(1.8) = 7.2
2
r
(2 m)
kQ1Q2
7 .2
= 72 mN
=
2
r
(10) 2
Q1/Q2 = 1/2 ⇒ Q2 = 2Q1
7.2 = kQ1Q2 = (9 × 109)(Q1)(2Q1) = 9 × 109 2Q12
F=
7 .2
7 .2
= Q12 ⇒ Q1 =
= 20 μC
9
18 × 10
18 × 109
Q2 = 2Q1 = 2(2 × 10−5 C) = 40 μC
W
1.2 J
=
= 3 kV
Q 0.4 mC
7.
V=
8.
W = VQ = (60 V)(8 mC) = 0.48 J
9.
Q=
W 96 J
=
=6C
V 16 V
10.
Q=
W 72 J
=
=8C
V
9V
11.
I=
Q 12 mC
=
= 4.29 mA
t
2.8 s
12.
I =
Q
312 C
=
= 2.60 A
t (2)(60 s)
13.
Q = It = (40 mA)(0.8)(60 s) = 1.92 C
14.
Q = It = (250 mA)(1.2)(60 s) = 18.0 C
15.
t=
16.
17.
10
( )
Q 6 mC
=
=3s
I
2 mA
1C
⎤
⎡
21.847 × 1018 electrons ⎢
18
⎥ = 3.5 C
⎣ 6.242 × 10 electrons ⎦
Q 3.5 C
I=
=
= 0.29 A
12 s
t
Q = It = (4 mA)(90 s) = 360 mC
⎡ 6.242 × 1018 electrons ⎤
18
360 mC ⎢
⎥ = 2.25 × 10 electrons
1
C
⎦
⎣
Chapter 2
18.
19.
20.
I=
86 C
Q
=
= 1.194 A > 1 A (yes)
(1.2)(60 s)
t
1C
⎤
⎡
0.84 × 1016 electrons ⎢
18
⎥ = 1.346 mC
⎣ 6.242 × 10 electrons ⎦
Q 1.346 mC
=
= 22.43 mA
I=
60 ms
t
a.
Q = It = (2 mA)(0.01 μs) = 2 × 10−11 C
⎡ 6.242 × 1018 electrons ⎤ ⎡ 1 ¢ ⎤
2 × 10−11 C ⎢
⎥⎢
⎥
1C
⎣
⎦ ⎣ electron ⎦
= 1.25 × 108 ¢ = $1.25 × 106 = 1.25 million
Q = It = (100 μA)(1.5 ns) = 1.5 × 10−13 C
⎡ 6.242 × 1018 electrons ⎤ ⎡ $1 ⎤
1.5 × 10−13 C ⎢
⎥⎢
⎥ = 0.94 million
1C
⎦ ⎣ electron ⎦
⎣
(a) > (b)
b.
21.
22.
23.
Q = It = (200 × 10−3 A)(30 s) = 6 C
W 40 J
V=
=
= 6.67 V
Q 6C
⎡ 420 C ⎤
Q = It = ⎢
⎥ (0.5 min) = 210 C
⎣ min ⎦
742 J
W
=
= 3.53 V
V=
Q 210 C
W 0.4 J
=
= 0.0167 C
V
24 V
Q 0.0167 C
I=
=
= 3.34 A
t 5 × 10−3 s
Q=
Ah rating 200 Ah
=
=5A
40 h
t (hours)
24.
I=
25.
Ah = (0.8 A)(75 h) = 60.0 Ah
26.
t(hours) =
Chapter 2
Ah rating 32 Ah
= 25 h
=
I
1.28 A
11
27.
⎡ 60 min ⎤ ⎡ 60 s ⎤
6
40 Ah(for 1 h): W1 = VQ = V⋅I⋅t = (12 V)(40 A)(1 h) ⎢
⎥ ⎢1 min ⎥ = 1.728 × 10 J
1
h
⎦
⎦⎣
⎣
60
min
60
s
⎤
⎤⎡
⎡
6
60 Ah(for 1 h): W2 = (12 V)(60 A)(1 h) ⎢
⎥ ⎢1 min ⎥ = 2.592 × 10 J
1
h
⎦
⎦⎣
⎣
Ratio W2/W1 = 1.5 or 50% more energy available with 60 Ah rating.
⎡1 min ⎤ ⎡ 1 h ⎤
= I(16.67 × 10−3 h)
For 60 s discharge: 40 Ah = It = I [60 s] ⎢
⎥
⎢
⎥
⎣ 60 s ⎦ ⎣ 60 min ⎦
40 Ah
= 2400 A
and I =
16.67 × 10-3 h
⎡1 min ⎤ ⎡ 1 h ⎤
= I(16.67 × 10−3 h)
60 Ah = It = I [60 s] ⎢
⎥
⎢
⎥
⎣ 60 s ⎦ ⎣ 60 min ⎦
60 Ah
= 3600 A
and I =
16.67 × 10-3 h
I2/I1 = 1.5 or 50 % more starting current available at 60 Ah
28.
I=
29.
−
30.
31.
32.
33.
34.
3 Ah
= 500 mA
6.0 h
⎡ 60 min ⎤ ⎡ 60 s ⎤
Q = It = (500 mA)(6 h) ⎢
⎥ = 10.80 kC
⎥⎢
⎣ 1 h ⎦ ⎣1 min ⎦
W = QV = (10.8 kC)(12 V) ≅ 129.6 kJ
−
−
−
−
−
35.
⎡ 60 s ⎤
4 min ⎢
⎥ = 240 s
⎣1 min ⎦
Q = It = (2.5 A)(240 s) = 600 C
36.
Q = It = (10 × 10−3 A)(20 s) = 200 mC
W = VQ = (12.5 V)(200 × 10−3 C) = 2.5 J
12
Chapter 2
Chapter 3
1.
a.
0.5 in. = 500 mils
b.
⎡1000 mils ⎤
0.02 in. ⎢
⎥ = 20 mils
⎣ 1 in. ⎦
c.
1
⎡1000 mils ⎤
in. = 0.25 in. ⎢
⎥ = 250 mils
4
⎣ 1 in. ⎦
d.
1 in. = 1000 mils
e.
2.
3.
3
⎡12 in. ⎤ ⎡10 mils ⎤
0.02 ft ⎢
⎢
⎥ = 240 mils
⎥
⎣ 1 ft ⎦ ⎣ 1 in. ⎦
⎡1000 mils ⎤
⎢ 1 in. ⎥ = 39.37 mils
⎦
⎣
f.
⎡ 1 in. ⎤
0.1 cm ⎢
⎥
⎣ 2.54 cm ⎦
a.
ACM = (30 mils)2 = 900 CM
b.
0.016 in. = 16 mils, ACM = (16 mils)2 = 256 CM
c.
1"
= 0.125" = 125 mils, ACM = (125 mils)2 = 15.63 × 103 CM
8
d.
⎡ 1 in. ⎤ ⎡1000 mils ⎤
= 393.7 mils, ACM = (393.7 mils)2 = 155 × 103 CM
1 cm ⎢
⎥
⎢
⎥
⎣ 2.54 cm ⎦ ⎣ 1 in. ⎦
e.
⎡12 in. ⎤ ⎡1000 mils ⎤
= 240 mils, ACM = (240 mils)2 = 57.60 × 103 CM
0.02 ft ⎢
⎥
⎢
⎥
⎣ 1 ft ⎦ ⎣ 1 in. ⎦
f.
⎡ 39.37 in. ⎤
2
3
0.0042 m ⎢
⎥ = 0.1654 in. = 165.4 mils, ACM = (165.4 mils) = 27.36 × 10 CM
⎣ 1m ⎦
ACM = (dmils )2 → dmils =
ACM
a.
d = 1600 CM = 40 mils = 0.04 in.
b.
d=
820 CM = 28.64 mils = 0.029 in.
c.
d=
40,000 CM = 200 mils = 0.2 in.
Chapter 3
13
4.
d.
d=
625 CM = 25 mils = 0.025 in.
e.
d=
6.25 CM = 2.5 mils = 0.0025 in.
f.
d = 100 CM = 10 mils = 0.01 in.
0.01 in. = 10 mils, ACM = (10 mils)2 = 100 CM
l
(200′)
= 20.74 Ω
R = ρ = (10.37)
A
100 CM
5.
ACM = (4 mils)2 = 16 CM, R = ρ
6.
a.
A= ρ
b.
d=
7.
8.
10.
14
⎛ 80′ ⎞
l
⎟⎟ = 544 CM
= 17⎜⎜
R
⎝ 2 .5 Ω ⎠
ACM = 544 CM = 23.32 mils = 23.3 × 10−3 in.
1 "
= 0.03125" = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM
32
l
RA (2.2 Ω)(976.56 CM)
=
R=ρ ⇒l=
= 3.58 ft
R
ρ
600
a.
ACM = ρ
d=
9.
(150 ft)
l
= 92.81 Ω
= (9.9)
A
16 CM
l
(10.37)(300′)
=
= 942.73 CM
3.3 Ω
A
942.73 CM = 30.70 mils = 30.7 × 10−3 in.
b.
larger
c.
smaller
a.
Rsilver > Rcopper > Raluminum
b.
Silver: R = ρ
ρ =
l
(9.9)(10 ft)
= 99 Ω { ACM = (1 mil)2 = 1 CM
=
1 CM
A
l
(10.37)(50 ft)
Copper: R = ρ =
= 5.19 Ω { ACM = (10 mils)2 = 100 CM
100 CM
A
l
(17)(200 ft)
Aluminum: R = ρ =
= 1.36 Ω { ACM = (50 mils)2 = 2500 CM
2500 CM
A
RA (500 Ω)(94 CM)
= 47 ⇒ nickel
=
l
1000′
Chapter 3
11.
a.
3" = 3000 mils, 1/2" = 0.5 in. = 500 mils
Area = (3 × 103 mils)(5 × 102 mils) = 15 × 105 sq. mils
⎡ 4 / π CM ⎤
5
15 × 105 sq mils ⎢
⎥ = 19.108 × 10 CM
1
sq
mil
⎣
⎦
R= ρ
12.
13.
l
(10.37)(4′)
=
= 21.71 μΩ
A 19.108 × 105 CM
b.
l
(17)(4′)
=
= 35.59 μΩ
A 19.108 × 105 CM
Aluminum bus-bar has almost 64% higher resistance.
c.
increases
d.
decreases
R= ρ
l2 = 2l1, A2 = A1/4, ρ2 = ρ1
ρ 2 l2
2l A
ρl A
R2
A
= 2 = 22 1 = 1 1 =8
l
ρ
ρ1l1 A2 l1 A1 / 4
R1
11
A1
and R2 = 8R1 = 8(0.2 Ω) = 1.6 Ω
ΔR = 1.6 Ω − 0.2 Ω = 1.4 Ω
A=
πd 2
4
⇒d=
π
4A
=
4(0.04 in.2 )
π
= 0.2257 in.
dmils = 225.7 mils
ACM = (225.7 mils)2 = 50,940.49 CM
l
ρ1 1
R1
lA
ρlA
A1
(ρ1 = ρ2)
=
= 11 2 = 1 2
R2 ρ l2
ρ 2l2 A1 l2 A1
2
A2
R l A (800 mΩ)(300 ft)(40,000 CM)
and R2 = 1 2 1 =
= 942.28 mΩ
l1 A2
(200 ft)(50,940.49 CM)
14.
a.
⎡1.260 Ω ⎤
#11: 450 ft ⎢
⎥ = 0.567 Ω
⎣ 1000 ft ⎦
⎡ 2.525 Ω ⎤
#14: 450 ft ⎢
⎥ = 1.136 Ω
⎣ 1000 ft ⎦
b.
Resistance: #14:#11 = 1.136 Ω:0.567 Ω ≅ 2:1
c.
Area:
Chapter 3
#14:#11 = 4106.8 CM:8234.0 CM ≅ 1:2
15
15.
a.
⎡ 0.6282 Ω ⎤
#8: R = 1800 ft ⎢
⎥ = 1.13 Ω
⎣ 1000 ft ⎦
⎡ 6.385 Ω ⎤
#18: R = 1800 ft ⎢
⎥ = 11.49 Ω
⎣ 1000 ft ⎦
b.
16.
17.
18.
#18:#8 = 11.49 Ω:1.13 Ω = 10.17:1 ≅ 10:1
c.
#18:#8 = 1624.3 CM:16,509 CM = 1:10.16 ≅ 1:10
a.
A= ρ
b.
A= ρ
a.
A/CM = 230 A/211,600 CM = 1.09 mA/CM
b.
⎡
⎤
1.09 mA ⎢ 1 CM ⎥ ⎡1000 mils ⎤ ⎡1000 mils ⎤
= 1.39 kA/in.2
⎢π
⎥⎢
⎥
⎢
⎥
CM ⎢ sq mils ⎥ ⎣ 1 in. ⎦ ⎣ 1 in. ⎦
⎣4
⎦
c.
⎡ 1 in.2 ⎤
2
5 kA ⎢
⎥ = 3.6 in.
⎣1.39 kA ⎦
l (10.37)(30′) 311.1 CM
=
= 51,850 CM ⇒ #3
=
R
6 mΩ
6 × 10−3
but 110 A ⇒ #2
l (10.37)(30′) 311.1 CM
=
= 103,700 CM ⇒ #0
=
R
3 mΩ
3 × 10−3
1
⎛ 2.54 cm ⎞
in. = 0.083 in. ⎜
⎟ = 0.21 cm
12
⎝ 1 in. ⎠
πd2
=
(3.14)(0.21 cm) 2
= 0.035 cm2
4
4
RA (2 Ω)(0.035 cm 2 )
=
l=
= 40,603 cm = 406.03 m
ρ
1.724 × 10−6
A=
19.
a.
b.
16
1 " ⎡ 2.54 cm ⎤
⎡ 2.54 cm ⎤
= 1.27 cm, 3 in. ⎢
⎢
⎥
⎥ = 7.62 cm
2 ⎣ 1" ⎦
⎣ 1 in. ⎦
⎡12 in. ⎤ ⎡ 2.54 cm ⎤
4 ft ⎢
⎥⎢
⎥ = 121.92 cm
⎣ 1 ft ⎦ ⎣ 1 in. ⎦
l (1.724 x 10 -6)(121.92 cm)
R= ρ
= 21.71 μΩ
A
(1.27 cm)(7.62 cm)
R= ρ
A
=
(2.825 × 10−6 )(121.92 cm)
= 35.59 μΩ
(1.27 cm)(7.62 cm)
Chapter 3
c.
increases
d.
decreases
20.
Rs =
ρ
21.
R l (150 Ω)(1/ 2 in.)
= 0.15 in.
R = Rs l ⇒ w = s =
R
500 Ω
w
22.
a.
d = 1 in. = 1000 mils
ACM = (103 mils)2 = 106 CM
RA (1 mΩ)(106 CM)
ρ1 =
=
= 1 CM-Ω/ft
l
103 ft
b.
1 in. = 2.54 cm
π d 2 π (2.54 cm)2
=
= 5.067 cm2
A=
4
4
⎡12 in. ⎤ ⎡ 2.54 cm ⎤
l = 1000 ft ⎢
⎥⎢
⎥ = 30,480 cm
⎣ 1 ft ⎦ ⎣ 1 in. ⎦
d
= 100 ⇒ d =
ρ2 =
c.
23.
24.
25.
k=
ρ
100
=
250 × 10−6
= 2.5 μcm
100
RA (1 mΩ)(5.067 cm 2 )
=
= 1.66 × 10−7 Ω-cm
l
30,480 cm
ρ 2 1.66 × 10−7 Ω-cm
=
= 1.66 × 10−7
1 CM-Ω / ft
ρ1
234.5 + 10 234.5 + 80
=
,
2Ω
R2
R2 =
(314.5)(2 Ω)
= 2.57 Ω
244.5
236 + 0 236 + 100
=
0.02 Ω
R2
(0.02 Ω)(336)
= 0.028 Ω
R2 =
236
5
5
(°F − 32) = (32 − 32) = 0° (=32°F)
9
9
5
C = (70 − 32) = 21.11° (=70°F)
9
234.5° + 21.11° 234.5° + 0°
=
4Ω
R2
(234.5)(4 Ω)
= 3.67 Ω
R2 =
255.61
C=
Chapter 3
17
26.
27.
28.
234.5 + 30 234.5 − 40
=
0.76 Ω
R2
(194.5)(0.76 Ω)
= 0.56 Ω
R2 =
264.5
243 + (−30) 243 + 0
=
0.04 Ω
R2
(243)(40 mΩ)
= 46 mΩ
R2 =
213
a.
b.
29.
a.
b.
30.
a.
c.
18
68°F = 20°C, 32°F = 0°C
234.5 + 20 234.5 + 0
=
R2
0.002
(234.5)(2 m Ω)
R2 =
= 1.84 mΩ
254.5
212°F = 100 °C
234.5 + 20 234.5 + 100
=
2 mΩ
R2
(334.5)(2 m Ω)
= 2.63 mΩ
R2 =
254.5
ΔR 2.63 m Ω − 2 m Ω 0.63 m Ω
=
=
= 7.88 μΩ/°C or 7.88 × 10-5 Ω/10°C
100° C − 20° C
80° C
ΔT
234.5 + 4 234.5 + t2
=
,
1Ω
1.1 Ω
234.5 + 4 234.5 + t2
=
,
1Ω
0.1 Ω
t2 = 27.85°C
t2 = −210.65°C
K = 273.15 + °C
50 = 273.15 + °C
°C = −223.15°
234.5 + 20 234.5 − 223.15
=
10 Ω
R2
11.35
(10 Ω) = 0.446 Ω
R2 =
254.5
F=
b.
9
9
°C + 32 = (−273.15°) + 32 = −459.67°
5
5
K = 273.15 + °C
38.65 = 273.15 + °C
°C = −234.5°
234.5 + 20 234.5 − 234.5
=
10 Ω
R2
(0)10 Ω
=0Ω
R2 =
254.5
Recall: −234.5° =
Inferred absolute zero
R=0Ω
Chapter 3
31.
a.
b.
32.
33.
α20 =
1
1
1
=
=
= 0.003929 ≅ 0.00393
Ti + 20°C 234.5 + 20 254.5
R = R20[1 + α20(t − 20°C)]
1 Ω = 0.8 Ω[1 + 0.00393(t − 20°)]
1.25 = 1 + 0.00393t − 0.0786
1.25 − 0.9214 = 0.00393t
0.3286 = 0.00393t
0.3286
= 83.61°C
t=
0.00393
R = R20[1 + α20(t − 20°C)]
= 0.4 Ω[1 + 0.00393(16 − 20)] = 0.4 Ω[1 − 0.01572] = 0.39 Ω
Table: 1000′ of #12 copper wire = 1.588 Ω @ 20°C
5
5
C° = (F° − 32) = (115 − 32) = 46.11°C
9
9
R = R20[1 + α20(t − 20°C)]
= 1.588 Ω[1 + 0.00393(46.11 − 20)]
= 1.75 Ω
22 Ω
Rnominal
(200)(65° − 20°) = 0.198 Ω
(PPM)( ΔT) =
106
106
R = Rnominal + ΔR = 22.198 Ω
34.
ΔR =
35.
ΔR =
36.
−
Rnominal
100 Ω
(100)(50° − 20°) = 0.30 Ω
(PPM)( ΔT) =
6
10
106
R = Rnominal + ΔR = 100 Ω + 0.30 Ω = 100.30 Ω
37.
−
38.
#12: Area = 6529 CM
⎡ 2.54 cm ⎤
6529 CM = 80.8 mils = 0.0808 in. ⎢
⎥ = 0.205 cm
⎣ 1 in. ⎦
πd 2 π (0.205 cm)2
A=
=
= 0.033 cm2
4
4
1 MA
I=
[0.033 cm 2 ] = 33 kA >>> 20 A
cm2
d=
39.
−
40.
a.
41.
10 kΩ − 3.5 kΩ = 6.5 kΩ
Chapter 3
2 times larger
b.
4 times larger
19
42.
43.
6.25 kΩ and 18.75 kΩ
−
44.
a.
b.
c.
45.
a.
b.
c.
d.
46.
47.
560 kΩ ± 5%, 560 kΩ ± 28 kΩ, 532 kΩ ↔ 588 kΩ
220 Ω ± 10%, 220 Ω ± 22 Ω, 198 Ω ↔ 242 Ω
100 Ω ± 20%, 100 Ω ± 20 Ω, 80 Ω ↔ 120 Ω
120 Ω = Brown, Red, Brown, Silver
8.2 Ω = Gray, Red, Gold, Silver
6.8 kΩ = Blue, Gray, Red, Silver
3.3 MΩ = Orange, Orange, Green, Silver
10 Ω ± 20% ⇒ 8 Ω − 12 Ω ⎫
⎬ no overlap, continuance
15 Ω ± 20% ⇒ 12 Ω − 18 Ω ⎭
10 Ω ± 10% ⇒ 10 Ω ± 1 Ω = 9 Ω − 11 Ω
⎫
⎬ No overlap
15 Ω ± 10% ⇒ 15 Ω ± 1.5 Ω = 13.5 Ω − 16.5 Ω ⎭
48.
a.
b.
c.
d.
621 = 62 × 101 Ω = 620 Ω = 0.62 kΩ
333 = 33 × 103 Ω = 33 kΩ
Q2 = 3.9 × 102 Ω = 390 Ω
C6 = 1.2 × 106 Ω = 1.2 MΩ
49.
a.
G=
1
1
=
= 8.33 mS
R 120 Ω
b.
G=
1
= 0.25 mS
4 kΩ
c.
G=
50.
20
a.
1
= 0.46 μS
2 .2 M Ω
Ga > Gb > Gc vs. Rc > Rb > Ra
Table 3.2, Ω/1000′ = 1.588 Ω
1
1
= 629.72 mS
G= =
R 1.588 Ω
A 6529.9 CM (Table 3.2)
or G =
=
= 629.69 mS (Cu)
ρl
(10.37)(1000′)
b.
G=
6529.9 CM
= 384.11 mS (Al)
(17)(1000′)
c.
G=
6529.9 CM
= 88.24 mS (Fe)
(74)(1000′)
Chapter 3
51.
l
2
5
⎛ 2⎞
A1 = A1, l2 = ⎜1 − ⎟ l1 = 1 , ρ2 = ρ1
3
3
3
⎝ 3⎠
A
⎛ l1 ⎞
ρ1 1
A
G1
l1
ρ 2l2 A1 ⎜⎝ 3 ⎟⎠ 1 1
=
=
=
=
G2 ρ A2 ρ1l1 A2
⎛5 ⎞ 5
l1 ⎜ A1 ⎟
2
l2
⎝3 ⎠
A2 = 1
G2 = 5G1 = 5(100 S) = 500 S
52.
53.
−
−
54.
−
55.
a.
56.
57.
−50°C specific resistance ≅ 105 Ω-cm
50°C specific resistance ≅ 500 Ω-cm
200°C specific resistance ≅ 7 Ω-cm
b.
negative
c.
No
d.
ρ=
a.
Log scale:
10 fc ⇒ 3 kΩ
100 fc ⇒ 0.4 kΩ
b.
c.
no—log scales imply linearity
d.
a.
b.
c.
Chapter 3
ΔΩ − cm 300 − 30 270 Ω − cm
=
=
≅ 3.6 Ω-cm/°C
125 − 50
75° C
ΔT
@ 0.5 mA, V ≅ 195 V
@ 1 mA, V ≅ 200 V
@ 5 mA, V ≅ 215 V
negative
1 kΩ ⇒ ≅ 30 fc
10 kΩ ⇒ ≅ 2 fc
10 k Ω − 1 k Ω
ΔR
= 321.43 Ω/fc
=
30 fc − 2 fc
Δfc
and
ΔR
= −321.43 Ω/fc
Δ fc
ΔVtotal = 215 V − 195 V = 20 V
5 mA:0.5 mA = 10:1
compared to
215 V: 200 V = 1.08:1
21
Chapter 4
1.
V = IR = (2.5 A)(47 Ω) = 117.5 V
2.
I=
V 12 V
=
= 1.76 A
R 6.8 Ω
3.
R=
V
6V
=
= 4 kΩ
I 1.5 mA
4.
I=
V
12 V
=
= 300 A
R 40 × 10−3 Ω
5.
V = IR = (3.6 μA)(0.02 MΩ) = 0.072 V = 72 mV
6.
I=
V
62 V
=
= 4.13 mA
R 15 kΩ
7.
R=
V 120 V
=
= 54.55 Ω
I 2.2 A
8.
I=
V
120 V
=
= 16 mA
R 7.5 kΩ
9.
R=
V 120 V
=
= 28.57 Ω
I 4.2 A
10.
R=
V
4.5 V
=
= 36 Ω
I 125 m A
11.
R=
V 24 mV
=
= 1.2 kΩ
I 20 μ A
12.
V = IR = (15 A)(0.5 Ω) = 7.5 V
13.
a.
14.
15.
22
R=
V 120 V
=
= 12.63 Ω
I 9.5 A
V = IR = (2.4 μA)(3.3 MΩ) = 7.92 V
b.
⎡ 60 min ⎤ ⎡ 60 s ⎤
t = 1 h⎢
⎥⎢
⎥ = 3600 s
⎣ 1 h ⎦ ⎣1 min ⎦
W = Pt = VIt
= (120 V)(9.5 A)(3600 s)
= 4.1 × 106 J
−
Chapter 4
16.
b.
17.
−
18.
(0.13 mA)(500 h) = 65 mAh
−
19.
−
20.
P=
21.
t=
22.
a.
⎡ 60 min ⎤ ⎡ 60 s ⎤
8 h⎢
⎥⎢
⎥ = 28,800 s
⎣ 1 h ⎦ ⎣1 min ⎦
W = Pt = (2 W)(28,000 s) = 57.6 kJ
b.
kWh =
23.
24.
W
=
t
420 J
420 J
=
= 1.75 W
⎡ 60 s ⎤ 240 s
4 min ⎢
⎣1 min ⎥⎦
W 640 J
=
= 16 s
P 40 J/s
(2 W)(8 h)
= 16 × 10−3 kWh
1000
Q 300 C ⎡1 min ⎤
=
= 5 C/s = 5 A
t 1 min ⎢⎣ 60 s ⎥⎦
P = I2R = (5 A)2 10 Ω = 250 W
I=
P = VI = (3 V)(1.4 A) = 4.20 W
W
12 J
=
= 2.86 s
t=
P 4.2 W
25.
48 C ⎡1 min ⎤
= 0.8 A
min ⎢⎣ 60 s ⎥⎦
P = EI = (6 V)(0.8 A) = 4.8 W
26.
P = I2R = (7.2 mA)2 4 kΩ = 207.36 mW
27.
P = I2R ⇒ I =
28.
I=
P
240 mW
= 10.44 mA
=
R
2.2 kΩ
P
2W
= 129.10 mA
=
R
120 Ω
V = IR = (129.10 mA)(120 Ω) = 15.49 V
I=
Chapter 4
23
29.
E
12 V
=
= 2.14 mA
R 5.6 kΩ
P = I2R = (2.14 mA)2 5.6 kΩ = 25.65 mW
⎛ ⎡ 60 min ⎤ ⎡ 60 s ⎤ ⎞
W = P ⋅ t = (25.65 mW) ⎜ 1 h ⎢
⎥⎢
⎥ ⎟ = 92.34 J
⎝ ⎣ 1 h ⎦ ⎣1 min ⎦ ⎠
I=
30.
E=
31.
I=
P 324 W
=
= 120 V
2.7 A
I
P
1W
=
= 461.27 μA
R
4.7 MΩ
no
PR = (42 mW)(2.2 kΩ) = 92.40 = 9.61 V
32.
V=
33.
P = EI = (9 V)(45 mA) = 405 mW
34.
P 100 W
=
= 0.833 A
V 120 V
V
120 V
R= =
= 144.06 Ω
I 0.833 A
P = VI, I =
P 450 W
=
= 120 V
I 3.75 A
V 120 V
= 32 Ω
R= =
I 3.75 A
35.
V=
36.
a.
P = EI and I =
b.
Ah rating = (0.13 mA)(500 h) = 66.5 mAh
P 0.4 × 10−3 W
=
= 0.13 mA
E
3V
100 W
P
=
= 5 × 10−3 = 70.71 mA
20 kΩ
R
V = PR = (100 W)(20 kΩ) = 1.42 kV
37.
I=
38.
a.
⎛V 2
W = Pt = ⎜
⎝ R
b.
Energy doubles, power the same
24
⎞ ⎛ 12 V ⎞ 2
⎟t = ⎜
⎟ 60 s = 864 J
⎠ ⎝ 10 Ω ⎠
Chapter 4
39.
⎡ 1
⎤
4 weeks ⎥
12 h ⎢ 3
⎢
⎥ [5 months] = 260 h
week ⎢ 1 month ⎥
⎣
⎦
(230 W)(260 h)
= 59.80 kWh
kWh =
1000
Pt
(1000)(kWh) (1000)(12 kWh)
⇒t =
=
=8h
P
1000
1500 W
40.
kWh =
41.
kWh =
42.
a.
kWh =
b.
I=
(24 W)(3 h)
= 72 × 10−3 kWh
1000
(72 × 10−3 kWh)(9¢/kWh) = 0.65¢
c.
43.
P 120 × 103 W
=
= 576.92 A
E
208 V
Plost = Pi − Po = Pi − ηPi = Pi(1 − η) = 120 kW(1 − 0.82) = 21.6 kW
Pt
(21.6 kW)(10 h)
=
= 216 kWh
kWhlost =
1000
1000
$1.00
= 11.11
9¢
Pt
(kWh)(1000) (11.11)(1000)
⇒t =
=
= 44.44 h
kWh =
P
1000
.250
#kWh =
t=
44.
Pt
(1000)(kWh) (1000)(1200 kWh)
⇒P=
=
= 120 kW
1000
10 h
P
(kWh)(1000) (11.11)(1000)
=
= 2.32 h
P
4800
a.
W = Pt = (60 W)(1 h) = 60 Wh
b.
⎛ ⎡ 60 min ⎤ ⎡ 60 s ⎤ ⎞
W = Pt = (60 W) ⎜1 h ⎢
⎥⎢
⎥ ⎟ = 216 kWs
⎝ ⎣ 1 h ⎦ ⎣1 min ⎦ ⎠
c.
1 kJ = 1 Ws, ∴216 kJ
d.
W=
Chapter 4
(60 W)(1 h)
Pt
=
= 60 × 10−3 kWh
1000
1000
25
45.
46.
a.
P = EI = (9 V)(0.455 A) = 4.1 W
b.
R=
c.
W = Pt = (4.1 W)(21,600 s) = 88.56 kJ
⎡ 60 min ⎤ ⎡ 60 s ⎤
6 h⎢
⎥⎢
⎥ = 21,600 s
⎣ 1 h ⎦ ⎣1 min ⎦
E
9V
=
= 19.78 Ω
I 0.455 A
a.
P = EI = (120 V)(100 A) = 12 kW
b.
⎡ 746 W ⎤
PT = 5 hp ⎢
⎥ + 3000 W + 2400 W + 1000 W
⎣ hp ⎦
= 10,130 < 12,000 W (Yes)
c.
47.
W = Pt = (10.13 kW)(2 h) = 20.26 kWh
⎛
⎛ 1 h ⎞⎞
(860 W)(6 h) + (4800 W)(1/2 h) + (900 W) ⎜ 20 min ⎜
⎟ ⎟ + (110 W)(3.5 h)
⎝ 60 min) ⎠ ⎠
⎝
kWh =
1000
5160 Wh + 2400 Wh + 300 Wh + 385 Wh
=
= 8.245 kWh
1000
(8.245 kWh)(9¢/kWh) = 74.21¢
⎛
48.
kWh =
⎛
⎛ 1 h ⎞⎞
⎛ 1 h ⎞⎞
⎟ ⎟ + (70 W)(1.5 h) + (150 W) ⎜130 min ⎜
⎟⎟
⎝ 60 min ⎠ ⎠
⎝ 60 min ⎠ ⎠
⎝
(200 W)(4 h) + (1200 W) ⎜ 20 min ⎜
⎝
1000
800 Wh + 400 Wh + 105 Wh + 325 Wh
= 1.63 kWh
=
1000
(1.63 kWh)(9¢/kWh) = 14.67¢
49.
⎡ 746 W ⎤
(0.5 hp) ⎢
⎥
P
⎣ hp ⎦ × 100% = 373 × 100% = 94.43%
η = o × 100% =
395 W
395
Pi
50.
η=
51.
η=
26
P (1.8 hp)(746 W/hp)
Po
= 1960.29 W
, Pi = o =
Pi
0.685
η
P 1960.29 W
Pi = EI, I = i =
= 16.34 A
E
120 V
Po
746 W
746
× 100% =
× 100% =
× 100% = 84.77%
Pi
(4 A)(220 V)
880
Chapter 4
52.
a.
b.
Pi = EI = (120 V)(2.4 A) = 288 W
Pi = Po + Plost, Plost = Pi − Po = 288 W − 50 W = 238 W
η% =
Po
50 W
× 100% =
× 100% = 17.36%
288 W
Pi
Po
⇒I=
η
=
η
Po (3.6 hp)(746 W/hp)
= 16.06 A
=
ηE
(0.76)(220 V)
53.
Pi = EI =
54.
a.
Pi =
b.
Pi = EI = 1657.78 W
(110 V)I = 1657.78 W
1657.78 W
= 15.07 A
I=
110 V
c.
Pi =
Po
(2 hp)(746 W/hp)
= 1657.78 W
0.9
(2 hp)(746 W/hp)
= 2131.43 W
0.7
η
Pi = EI = 2131.43 W
(110 V)I = 2131.43 W
2131.43 W
= 19.38 A
I=
110 V
Po
=
(15 hp)(746 W/hp)
= 12,433.33 W
η
0.9
P 12, 433.33 W
= 56.52 A
I= i=
E
220 V
Po
55.
Pi =
56.
ηT = η1 ⋅ η2
=
0.75 = 0.85 × η2
η2 = 0.88
57.
58.
ηT = η1 ⋅ η2 = (0.87)(0.75) = 0.6525 ⇒ 65.25%
η1 = η2 = .08
ηT = (η1)(η2) = (0.8)(0.8) = 0.64
ηT =
59.
Wo
⇒ Wo = ηT Wi = (0.64)(60 J) = 38.4 J
Wi
ηT = η1 ⋅ η2 = 0.72 = 0.9η2
Chapter 4
η2 =
0.72
= 0.8 ⇒ 80%
0.9
27
60.
a.
b.
61.
ηT = η1 ⋅ η2 ⋅ η3 = (0.98)(0.87)(0.21) = 0.1790 ⇒ 17.9%
ηT = η1 ⋅ η2 ⋅ η3 = (0.98)(0.87)(0.90) = 0.7673 ⇒ 76.73%
76.73% − 17.9%
× 100% = 328.66%
17.9%
ηT =
Po
= η1 ⋅ η2 = η1 ⋅ 2η1 = 2η12
Pi
η12 =
Po
⇒ η1 =
2 Pi
Po
128 W
=
= 0.4
2 Pi
2(400 W)
η2 = 2η1 = 2(0.4) = 0.8
∴ η2 = 40%, η2 = 80%
28
Chapter 4
Chapter 5
1.
a.
b.
c.
d.
E and R1
R1 and R2
E and R1
E and R1, R3 and R4
2.
a.
b.
c.
d.
RT = 0.1 kΩ + 0.39 kΩ + 1.2 kΩ = 1.69 kΩ
RT = 1.2 Ω + 2.7 Ω + 8.2 Ω = 12.1 Ω
RT = 8.2 kΩ + 10 kΩ + 9.1 kΩ + 1.8 kΩ + 2.7 kΩ = 31.8 kΩ
RT = 47 Ω + 820 Ω + 91 Ω + 1.2 kΩ = 2158.0 Ω
3.
a.
b.
RT = 1.2 kΩ + 1 kΩ + 2.2 kΩ + 3.3 kΩ = 7.7 kΩ
RT = 1 kΩ + 2 kΩ + 3 kΩ + 4.7 kΩ + 6.8 kΩ = 17.5 kΩ
4.
a.
b.
c.
1 MΩ
100 Ω, 1 kΩ
RT = 100 Ω + 1 kΩ + 1 MΩ + 200 kΩ = 1.2011 MΩ vs. 1.2 MΩ for part b.
5.
a.
b.
c.
d.
RT = 105 Ω = 10 Ω + 33 Ω + R, R = 62 Ω
RT = 10 kΩ = 2.2 kΩ + R + 2.7 kΩ + 3.3 kΩ, R = 1.8 kΩ
RT = 138 kΩ = R + 56 kΩ + 22 kΩ + 33 kΩ, R = 27 kΩ
RT = 91 kΩ = 24 kΩ + R1 + 43 kΩ + 2R1 = 67 kΩ + 3R1, R1 = 8 kΩ
R2 = 16 kΩ
6.
a.
b.
c.
d.
1.2 kΩ
3.3 kΩ + 4.3 kΩ = 7.6 kΩ
0Ω
∞Ω
7.
a.
b.
c.
8.
a.
b.
c.
9.
a.
b.
Chapter 5
RT = 10 Ω + 12 Ω + 18 Ω = 40 Ω
E 120 V
=
Is =
=3A
40 Ω
RT
V1 = I1R1 = (3 A)(10 Ω) = 30 V, V2 = I2R2 = (3 A)(12 Ω) = 36 V,
V3 = I3R3 = (3 A)(18 Ω) = 54 V
the most: R3, the least: R1
R3, RT = 1.2 kΩ + 6.8 kΩ + 82 kΩ = 90 kΩ
45 V
E
=
= 0.5 mA
Is =
RT 90 kΩ
V1 = I1R1 = (0.5 mA)(1.2 kΩ) = 0.6 V, V2 = I2R2 = (0.5 mA)(6.8 kΩ) = 3.4 V,
V3 = I3R3 = (0.5 mA)(82 kΩ) = 41 V, results agree with part (a)
RT = 12 kΩ + 4 kΩ + 6 kΩ = 22 kΩ
E = IRT = (4 mA)(22 kΩ) = 88 V
RT = 18 Ω + 14 Ω + 8 Ω + 40 Ω = 80 Ω
E = IRT = (250 mA)(80 Ω) = 20 V
29
10.
a.
b.
b.
c.
d.
V 5.2 V
=
=4A
R 1.3 Ω
E = IRT = (4 A)(9 Ω) = 36 V
RT = 9 Ω = 4.7 Ω + 1.3 Ω + R,
V4.7 Ω = (4 A)(4.7 Ω) = 18.8 V
V1.3 Ω = (4 A)(1.3 Ω) = 5.2 V
V3 Ω = (4 A)(3 Ω) = 12 V
a.
I=
a.
b.
c.
d.
11.
12.
I=
R=3Ω
6.6 V
V
=
= 3 mA
R 2.2 kΩ
V3.3 kΩ = (3 mA)(3.3 kΩ) = 9.9 V
E = 6.6 V + 9 V + 9.9 V = 25.5 V
9V
V
=
= 3 kΩ
R=
I 3 mA
V2.2 kΩ = 6.6 V, V3 kΩ = 9 V, V3.3 kΩ = 9.9 V
E
36 V
1
1
= 8.18 mA, V = E = (36 V) = 18 V
=
RT 4.4 kΩ
2
2
a.
I=
b.
RT = 1 kΩ + 2.4 kΩ + 5.6 kΩ = 9 kΩ
22.5 V
E
=
I=
= 2.5 mA, V = 2.5 mA(2.4 kΩ + 5.6 kΩ) = 20 V
9 kΩ
RT
c.
RT = 10 kΩ + 22 kΩ + 33 kΩ + 10 MΩ = 10.065 MΩ
E
100 V
= 9.94 μA
=
I=
RT 10.065 MΩ
V = (9.935 μA)(10 MΩ) = 99.35 V
a.
RT = 3 kΩ + 1 kΩ + 2 kΩ = 6 kΩ
E 120 V
=
= 20 mA
Is =
6 kΩ
RT
VR1 = (20 mA)(3 kΩ) = 60 V
VR2 = (20 mA)(1 kΩ) = 20 V
VR3 = (20 mA)(2 kΩ) = 40 V
b.
PR1 = I12 R1 = (20 mA)2 ⋅ 3 kΩ = 1.2 W
PR2 = I 22 R2 = (20 mA)2 ⋅ 1 kΩ = 0.4 W
PR3 = I 32 R3 = (20 mA)2 ⋅ 2 kΩ = 0.8 W
30
c.
PT = PR1 + PR2 + PR3 = 1.2 W + 0.4 W + 0.8 W = 2.4 W
d.
PT = EIs = (120 V)(20 mA) = 2.4 W
Chapter 5
13.
e.
the same
f.
R1 − the largest
g.
dissipated
h.
R1: 2 W, R2 : 1/2 W, R3: 1 W
a.
RT = 22 Ω + 10 Ω + 47 Ω + 3 Ω = 82.0 Ω
E 20.5 V
=
= 250 mA
Is =
RT 82.0 Ω
VR1 = I1R1 = (250 mA)(22 Ω) = 5.50 V
VR2 = I2R2 = (250 mA)(10 Ω) = 2.50 V
VR3 = I3R3 = (250 mA)(47 Ω) = 11.75 V
VR4 = I4R4 = (250 mA)(3 Ω) = 0.75 V
b.
PR1 = I12 R1 = (250 mA)2 ⋅ 22 Ω = 1.38 W
PR2 = I 22 R2 = (250 mA)2 ⋅ 10 Ω = 625.00 mW
PR3 = I 32 R3 = (250 mA)2 ⋅ 47 Ω = 2.94 W
PR4 = I 42 R4 = (250 mA)2 ⋅ 3 Ω = 187.50 mW
14.
c.
PT = PR1 + PR2 + PR3 + PR4 = 1.38 W + 625.00 mW + 2.94 W + 187.50 mW = 5.13 W
d.
P = EIs = (20.5 V)(250 mA) = 5.13 W
e.
the same
f.
47 Ω − the largest
g.
dissipated
h.
R1: 2 W; R2: 1/2 W, R3: 5 W, R4: 1/2 W
a.
b.
Chapter 5
P = 21 W = (1 A)2⋅R, R = 21 Ω
V1 = I1R1 = (1 A)(2 Ω) = 2 V, V2 = I2R2 = (1 A)(1 Ω) = 1 V
V3 = I3R3 = (1 A)(21 Ω) = 21 V
E = V1 + V2 + V3 = 2 V + 1 V + 21 V = 24 V
P = 4 W = I2⋅1 Ω, I = 4 = 2 A
P = 8 W = I2R1 = (2 A)2R1, R1 = 2 Ω
RT = 16 Ω = 2 Ω + R2 + 1 Ω = 3 Ω + R2,
E = IRT = (2 A)(16 Ω) = 32 V
R2 = 13 Ω
31
15.
a.
⎛ 1 ⎞
RT = NR1 = 8⎜ 28 Ω ⎟ = 225 Ω
⎝ 8 ⎠
120 V
E
= 0.53 A
=
I=
RT 225 Ω
b.
⎛ 8 ⎞ ⎛ 1 ⎞ ⎛ 64 ⎞⎛ 225 ⎞
P = I R = ⎜ A ⎟ ⎜ 28 Ω ⎟ = ⎜
⎟ =8W
⎟⎜
⎝ 15 ⎠ ⎝ 8 ⎠ ⎝ 225 ⎠⎝ 8 ⎠
c.
⎛ 8 ⎞⎛ 225 ⎞
V = IR = ⎜ A ⎟⎜
Ω ⎟ = 15 V
⎠
⎝ 15 ⎠⎝ 8
d.
All go out!
2
16.
2
Ps = PR1 + PR2 + PR3
E⋅I = I2R1 + I2R2 + 24
(R1 + R2)I2 − E⋅I + 24 = 0
6I2 − 24 I + 24 = 0
I2 − 4 I + 4 = 0
− (−4) ± (−4) 2 − 4(1)(4) 4 ± 16 − 16 4
=
= =2A
2(1)
2
2
24 Ω
P = 24 W = (2 A)2R, R =
=6Ω
4
I=
17.
a.
b.
c.
Vab = −4 V − 8 V + 12 V = 0 V
Vab = −4 V − 8 V + 6 V = −6 V
Vab = −10 V + 18 V − 6 V + 12 V = 14 V
18.
a.
ET = 16 V − 4 V − 8V = 4 V, I =
b.
19.
a.
4V
= 388.35 mA (CCW)
10.3 Ω
2V
ET = 18 V − 12 V − 4 V = 2 V, I =
= 173.91 mA (CW)
11.5 Ω
P = 8 mW = I2R, R =
I=
b.
32
8 mW
8 mW
=
= 2 kΩ
2
(2 mA)2
I
E
20 V − E
=
= 2 mA (CW),
RT 3 kΩ + 2 kΩ
E = 10 V
16 V
12 V
V
=
= 8 mA, R =
= 1.5 kΩ
2 kΩ
I 8 mA
E
E − 4 V − 10 V E − 14 V
= 8 mA (CCW)
=
=
I=
RT
2 kΩ + 1.5 kΩ
3.5 kΩ
E = 42 V
I=
Chapter 5
20.
a.
b.
c.
21.
a.
b.
22.
a.
b.
23.
a.
b.
24.
25.
+10 V + 4 V − 3 V − V = 0
V = 14 V − 3 V = 11 V
+30 V + 20 V − 8 V − V = 0
V = 50 V − 8 V = 42 V
+16 V − 10 V − 4 V − V + 60 V = 0
V = 76 V − 14 V = 62 V
+60 V − 12 V − V − 20 V = 0
V = 60 V − 32 V = 28 V
+E − 14 V − 6 V − 2 V + 18 V = 0
E = 22 V − 18 V = 4 V
+10 V − V2 = 0
V2 = 10 V
+10 V − 6 V − V1 = 0
V1 = 4 V
+24 V − 10 V − V1 = 0
V1 = 14 V
+10 V − V2 + 8 V = 0
V2 = 18 V
+20 V − V1 − 10 V − 1 V = 0, V1 = 9 V
+10 V − 2 V − V2 = 0, V2 = 8 V
+ 10 V − V1 + 6 V − 2 V − 3 V = 0, V1 = 11 V
+10 V − V2 − 3 V = 0, V2 = 7 V
1 V 50 V
(50 V)(2 Ω)
=
, R2 =
= 100 Ω
2Ω
R2
1V
(100 V)(2 Ω)
1 V 100 V
= 200 Ω
=
, R3 =
2Ω
R3
1V
a.
b.
c.
d.
Chapter 5
8.2 kΩ
V3: V2 = 8.2 kΩ:1 kΩ = 8.2:1
V3: V1 = 8.2 kΩ:100 Ω = 82:1
(8.2 kΩ)(60 V)
RE
V3 = 3 =
= 52.90 V
0.1 kΩ + 1 kΩ + 8.2 kΩ
RT
( R + R3 ) E (1 kΩ + 8.2 kΩ)(60 V)
V′ = 2
=
= 59.35 V
RT
9 .3 k Ω
33
26.
a.
V=
b.
V=
c.
27.
a.
40 Ω(30 V)
= 20 V
40 Ω + 20 Ω
(2 kΩ + 3 kΩ)(40 V)
(5 kΩ)(40 V)
=
= 20 V
4 kΩ + 1 kΩ + 2 kΩ + 3 kΩ
10 kΩ
(1.5 Ω + 0.6 Ω + 0.9 Ω)(0.72 V)
(3 Ω)(0.72 V)
=
= 0.36 V
(2.5 Ω + 1.5 Ω + 0.6 Ω + 0.9 Ω + 0.5 Ω)
6 kΩ
20 V
(6 Ω)(20 V)
V1
= 60 V
=
, V1 =
6Ω
2Ω
2Ω
V2
20 V
(4 Ω)(20 V)
=
, V2 =
= 40 V
4Ω
2Ω
2Ω
E = V1 + 20 V + V2 = 60 V + 20 V + 40 V = 120 V
b.
c.
d.
28.
120 V − V1 − 80 V = 0, V1 = 40 V
80 V − 10 V − V3 = 0, V3 = 70 V
1000 V
V
2 Ω(1000 V)
= 2 , V2 =
= 20 V
100 Ω
2Ω
100 Ω
1000 V
V
1 Ω(1000 V)
= 10 V
= 1 , V1 =
100 Ω
100 Ω 1 Ω
E = V1 + V2 + 1000 V = 10 V + 20 V + 1000 V = 1030 V
16 V − V1 − 6 V = 0, V1 = 10 V
6V
=3V
V2 =
2
2V
2 kΩ( 2 V)
V
=4V
= 2 , V2 =
1 kΩ 2 kΩ
1 kΩ
V
2V
3 kΩ(2 V)
= 4 , V4 =
=6V
1 kΩ 3 kΩ
1 kΩ
2V
= 2 mA
1 kΩ
E = 2 V + 4 V + 12 V + 6 V = 24 V
I=
34
Chapter 5
29.
30.
4V=
b.
100 V =
a.
b.
c.
31.
32.
R(20 V)
,
2 kΩ + 6 kΩ
a.
R = 1.6 kΩ
(6 Ω + R )140 V
3 Ω+6 Ω+R
300 Ω + 600 Ω + 100R = 840 Ω + 140 R
140R − 100R = − 840 Ω + 900 Ω
40R = 60 Ω
60 Ω
R=
= 1.5 Ω
40
V
4V
20 Ω(4 V)
= 2 ⇒ V2 =
=8V
10 Ω 20 Ω
10 Ω
V3 = E − V1 − V2 = 40 V − 4 V − 8 V = 28 V
(28 V)(10 Ω)
4 V 28 V
=
⇒ R3 =
= 70 Ω
10 Ω
R3
4V
8V
= 160 Ω
50 mA
R (12 V) 160 Ω(12 V)
, Rx = 80 Ω in series with the bulb
=
Vbulb = 8 V = bulb
160 Ω + Rx
Rbulb + Rx
a.
Rbulb =
b.
VR = 12 V − 8 V = 4 V, P =
V 2 (4 V) 2
=
= 0.2 W, ∴1/4 W okay
80 Ω
R
VR1 + VR2 = 72 V
1
VR + VR2 = 72 V
5 2
72 V
⎡ 1⎤
= 60 V
VR2 ⎢1 + ⎥ = 72 V, VR2 =
1.2
⎣ 5⎦
VR
VR
60 V
72 V − 60 V 12 V
R2 = 2 =
=
= 15 kΩ, R1 = 1 =
= 3 kΩ
4 mA
4 mA
4 mA
I R2
I R1
33.
RT = R1 + R2 + R3 = 2R3 + 7R3 + R3 = 10R3
R (60 V)
VR3 = 3
= 6 V, VR1 = 2VR3 = 2 (6 V) = 12 V, VR2 = 7VR3 = 7(6 V) = 42 V
10 R3
34.
a.
VR3 = 4VR2 = 4(3VR1 ) = 12VR1
E = VR1 + 3VR1 + 12VR1 ∴RT = R1 + 3R1 + 12R1 = 16R1 =
R1 =
Chapter 5
64 V
= 6.4 kΩ
10 mA
6.4 kΩ
= 400 Ω, R2 = 3R1 = 1.2 kΩ, R3 = 12R1 = 4.8 kΩ
16
35
b.
35.
36.
37.
a.
b.
Va = +10 V + 3 V = 13 V
Vb = +6 V
Vab = Va − Vb = 13 V − 6 V = 7 V
a.
I(CW) =
b.
70 V − 10 V
60 V
=1A
=
10 Ω + 20 Ω + 30 Ω
60 Ω
V = IR = (1 A)(10 Ω) = 10 V
a.
I=
Va = 20 V − 6 V = 14 V
Vb = +4 V
Vab = Va − Vb = 14 V − 4 V = 10 V
80 V − 26 V
54 V
=
=6A
6Ω +3Ω
9Ω
V = IR = (6 A)(3 Ω) = 18 V
I(CW) =
16 V − 4 V 12 V
=
= 0.4 A (CW)
10 Ω + 20 Ω 30 Ω
Va = 16 V − I(10 Ω) = 16 V − (0.4 A)(10 Ω) = 16 V − 4 V = 12 V
V1 = IR = (0.4 A)(20 Ω) = 8 V
12 V + 10 V + 8 V
30 V
=
= 5.455 mA
2.2 k Ω + 3.3 k Ω 5.5 k Ω
Va = 12 V − I(2.2 kΩ) + 10 V
= 12 V − (5.455 mA)(2.2 kΩ) + 10 V
= 12 V − 12 V + 10 V = 10 V
V1 = I(2.2 kΩ) = (5.455 mA)(2.2 kΩ) = 12 V
I=
47 V − 20 V
27 V
=
= 3 mA (CCW)
2kΩ+3kΩ+ 4kΩ 9kΩ
V2kΩ = 6 V, V3kΩ = 9 V, V4kΩ = 12 V
I=
a.
b.
c.
36
Va = +12 V − 8 V = 4 V
Vb = −8 V
Vab = Va − Vb = 4 V − (−8 V) = 12 V
c.
b.
38.
6.4 MΩ
64 V
= 6.4 MΩ, R1 =
= 400 kΩ, R2 = 1.2 MΩ, R3 = 4.8 MΩ
10 μA
16
R ′ 400 kΩ
I1 10 mA
=
= 103 and 1 =
= 103 also
R1
400 Ω
I ′ 10 μ A
RT =
Va = 20 V, Vb = 20 V + 6 V = 26 V, Vc = 20 V + 6 V + 9 V = 35 V
Vd = −12 V, Ve = 0 V
Vab = −6 V, Vdc = −47 V, Vcb = 9 V
Vac = −15 V, Vdb = −47 V + 9 V = −38 V
Chapter 5
39.
40.
VR
4 V + 4 V 8V
12 V − 4 V 8 V
=
=
= 8 Ω,
= 1 A, R1 = 1 =
I
8Ω
8Ω
1A
1A
VR
8V −4V 4V
=
=4Ω
R3 = 3 =
1A
1A
I
I R2 =
VR2 = 48 V − 12 V = 36 V
36 V
= 2.25 kΩ
16 mA
I
= 12 V − 0 V = 12 V
R2 =
VR3
R3 =
VR4
VR2
VR3
=
=
I
= 20 V
12 V
= 0.75 kΩ
16 mA
20 V
= 1.25 kΩ
16 mA
I
VR1 = E − VR2 − VR3 − VR4
R4 =
VR4
=
= 100 V − 36 V − 12 V − 20 V = 32 V
VR
32 V
= 2 kΩ
R1 = 1 =
I
16 mA
41.
44 V − 20 V
24 V
=
= 2 mA (CW)
2 k Ω + 4 k Ω + 6 k Ω 12 k Ω
V2kΩ = IR = (2 mA)(2 kΩ) = 4 V
V4kΩ = IR = (2 mA)(4 kΩ) = 8 V
V6kΩ = IR = (2 mA)(6 kΩ) = 12 V
I=
a.
b.
c.
42.
Va = 44 V, Vb = 44 V − 4 V = 40 V, Vc = 44 V − 4 V − 8 V = 32 V
Vd = 20 V
Vab = V2kΩ = 4 V, Vcb = −V4kΩ = −8 V
Vcd = V6kΩ = 12 V
Vad = Va − Vd = 44 V − 20 V = 24 V
Vca = Vc − Va = 32 V − 44 V = −12 V
V0 = 0 V
V4 = −12 V + 2 V = 0, V4 = +10 V
V7 = 4 V
V10 = 20 V
V23 = +6 V
V30 = −8 V
V67 = 0 V
V56 = −6 V
V
V
6V
= 1.5 A↑
I = 4 = 23 =
4Ω 4Ω 4Ω
Chapter 5
37
43.
44.
45.
46.
38
V0 = 0 V, V03 = V0 − V3 = 0 V, V2 = (2 mA)(3 kΩ + 1 kΩ) = (2 mA)(4 kΩ) = 8 V
V23 = V2 − V3 = 8 V − 0 V = 8 V, V12 = 20 V - 8 V = 12 V,
ΣIi = ΣIo ⇒ Ii = 2 mA + 5 mA + 10 mA = 17 mA
a.
VL = ILRL = (2 A)(28 Ω) = 56 V
Vint = 60 V − 56 V = 4 V
4V
V
=2Ω
Rint = int =
I
2A
60 V − 56 V
VNL − VFL
× 100% = 7.14%
× 100% =
56 V
VFL
b.
VR =
a.
VL =
b.
VR =
c.
Is = IL =
a.
I=
12 V
12 V
E
=
=
= 1.2 mA
RT 2 kΩ + 8 kΩ 10 kΩ
b.
I=
12 V
12 V
E
= 1.17 mA
=
=
RT 10 kΩ + 0.25 kΩ 10.25 kΩ
c.
not for most applications.
3.3 Ω(12 V)
= 11.82 V
3.3 Ω + 0.05 Ω
VNL − VFL
12 V − 11.82 V
× 100% = 1.52%
× 100% =
11.82 V
VFL
11.82 V
= 3.58 A
3.3 Ω
Ps = EIs = (12 V)(3.58 A) = 42.96 W
Pint = I2Rint = (3.58 A)2 0.05 Ω = 0.64 W
Chapter 5
Chapter 6
1.
a.
b.
c.
d.
e.
f.
g.
R2 and R3
E and R3
E and R1
R2, R3 and R4
E, R1, R2, R3, and R4
E, R1, R2, and R3
R2 and R3
2.
a.
b.
R3 and R4, R5 and R6
E and R1
3.
a.
RT =
b.
RT =
c.
RT =
d.
e.
4.
1
1
=
−3
1
1
1
1 × 10 S + 0.5 × 10−3 S + 0.333 × 10−3 S
+
+
1 kΩ 2 kΩ 3 kΩ
1
= 545.55 Ω
=
1.833 × 10−3 S
1
1
1
=
=
−3
−3
−3
1
1
1
10 × 10 S + 1 × 10 S + 0.1 × 10 S 11.1 × 10− 3 S
+
+
100 Ω 1 kΩ 10 kΩ
= 90.09 Ω
18 kΩ
= 6 kΩ
3
(6 kΩ)(6 MΩ)
= 5.99 kΩ
RT =
6 kΩ + 6 MΩ
RT′ =
22 Ω
10 Ω
=5Ω
= 5.5 Ω, RT ′′ =
4
2
(5.5 Ω)(5 Ω)
= 2.62 Ω
RT =
5 .5 Ω + 5 Ω
RT′ =
1
1
=
−3
1
1
1
1000 × 10 S + 1 × 10− 3 S + 0.001 × 10−3 S
+
+
1 Ω 1 kΩ 1 MΩ
1
= 0.99 Ω
=
1001.001 × 10− 3 S
f.
RT =
a.
RT =
Chapter 6
(9.1 Ω)(18 Ω)
= 6.04 Ω
9.1 Ω + 18 Ω
1
1
=
−
3
1
1
1
1 × 10 S + 0.833 × 10−3 S + 3.333 × 10− 3 S
+
+
1 kΩ 1.2 kΩ 0.3 kΩ
1
= 193.57 Ω
=
5.166 × 10−3 S
39
b.
5.
a.
b.
c.
d.
e.
1
1
1
1
1
+
+
+
1 kΩ 1.2 kΩ 2.2 kΩ 1 kΩ
1
= 304.14 Ω
=
3.288 × 10− 3 S
RT =
=
1
1 × 10 S + 0.833 × 10 S + 0.455 × 10− 3 S + 1 × 10− 3 S
−3
−3
RT′ = 3 Ω || 6 Ω = 2 Ω
(2 Ω)( R )
, R=8Ω
RT = 1.6 Ω =
2Ω+R
6 kΩ
= 2 kΩ
RT′ =
3
(2 kΩ)( R )
RT = 1.8 kΩ =
, R = 18 kΩ
2 kΩ + R
(20 kΩ)( R )
, R = 20 kΩ
RT = 10 kΩ =
20 kΩ + R
1
1
RT = 628.93 Ω =
=
1
1
1
1
833.33 × 10−3 S + + 454.55 × 10−3 S
+ +
1.2 kΩ R 2.2 kΩ
R
628.93
Ω
=1
811.32 × 10−3 +
R
628.93 Ω
R=
= 3.3 kΩ
1 − 811.32 × 10−3
R′ = R1 || R2 =
R1
R
, R3 = 1
2
2
⎛ R1 ⎞⎛ R1 ⎞
⎜ 2 ⎟⎜ 2 ⎟ R
R′R3
RT = 1.6 kΩ =
= ⎝ ⎠⎝ ⎠ = 1
R1 R1
4
R′ + R3
+
2
2
R1 = 4(1.6 kΩ) = 6.4 kΩ = R2
R3 =
6.
a.
b.
1.2 kΩ
about 1 kΩ
c.
1
1
1
1
+
+
+
1.2 kΩ 22 kΩ 220 kΩ 2.2 MΩ
1
=
−6
−6
833.333 × 10 S + 45.455 × 10 S + 4.545 × 10− 6 S + 0.455 × 10− 6 S
1
=
= 1.131 kΩ
883.788 × 10− 6 S
(1.2 kΩ)(22 kΩ)
= 1.138 kΩ
220 kΩ, 2.2 MΩ: RT =
1.2 kΩ + 22 kΩ
RT reduced.
d.
e.
40
6 .4 k Ω
= 3.2 kΩ
2
RT =
1
Chapter 6
7.
8.
a.
RT =
b.
∞Ω
c.
∞Ω
d.
RT =
(2 Ω)(8 Ω)
= 1.6 Ω
2 Ω+8 Ω
1
1
1
= 1.18 Ω
=
=
1
1
1
0.25
S
+
0.50
S
0.10
S
0.85
S
+
+
+
4 Ω 2 Ω 10 Ω
1
1
1
1
=
+
+
RT R1 R2 R3
⎡ 1 ⎤ 1⎡ 1 ⎤
⎡1⎤
⎡1⎤
1
1
1
1
=
+
+
= 1 ⎢ ⎥ + ⎢ ⎥ + 2 ⎢ ⎥ = 3.2 ⎢ ⎥
R
20 Ω R1 5 R1
1
⎣ R1 ⎦ 5 ⎣ R1 ⎦
⎣ R1 ⎦
⎣ R1 ⎦
2
and R1 = 3.2(20 Ω) = 64 Ω
R2 = 5R1 = 5(64 Ω) = 320 Ω
1
64 Ω
= 32 Ω
R3 = R1 =
2
2
9.
10.
24 Ω || 24 Ω = 12 Ω
1
1
1
1
=
+
+
RT R1 12 Ω 120 Ω
1
+ 0.08333 S + 0.00833 S
0.1 S =
R1
1
0.1 S =
+ 0.09167 S
R1
1
= 0.1 S − 0.09167 S = 0.00833 S
R1
1
= 120 Ω
R1 =
0.00833 S
a.
b.
c.
d.
Chapter 6
(8 Ω)(24 Ω)
=6Ω
8 Ω + 24 Ω
VR1 = VR2 = 36 V
RT =
E 36 V
=
=6A
RT
6Ω
VR 36 V
I1 = 1 =
= 4.5 A
8Ω
R1
VR
36 V
I2 = 2 =
= 1.5 A
R2 24 Ω
Is = I1 + I2
6 A = 4.5 A + 1.5 A = 6 A (checks)
Is =
41
11.
a.
b.
d.
18 V
E
=
= 8.5 A
RT 2.12 Ω
VR 18 V
VR 18 V
VR 18 V
I1 = 1 =
= 6 A, I 2 = 2 =
= 2 A, I 3 = 3 =
= 0.5 A
R3 36 Ω
R1
R2
3Ω
9Ω
Is = 8.5 A = 6 A + 2 A + 0.5 A = 8.5 A (checks)
a.
RT =
c.
12.
13.
Is =
b.
1
1
=
−
6
1
1
1
100 × 10 S + 833.333 × 10− 6 S + 147.059 × 10− 6 S
+
+
10 kΩ 1.2 kΩ 6.8 kΩ
1
=
= 925.93 Ω
1.080 × 10− 3 S
VR1 = VR2 = VR3 = 24 V
c.
Is =
24 V
E
=
= 25.92 mA
RT 925.93 Ω
I R1 =
VR2
I R3 =
VR3
R1
=
VR
24 V
24 V
= 20 mA,
= 2.4 mA, I R2 = 2 =
R 2 1.2 kΩ
10 kΩ
d.
24 V
= 3.53 mA
=
R3 6.8 kΩ
IT = 25.92 mA = 2.4 mA + 20 mA + 3.53 mA = 25.93 mA (checks)
a.
RT ≅ 1 kΩ
b.
c.
d.
42
1
1
=
1
1
1
0.333 S + 0.111 S + 0.028 S
+
+
3 Ω 9 Ω 36 Ω
1
=
= 2.12 Ω
472 × 10− 3 S
VR1 = VR2 = VR3 = 18 V
RT =
1
1
1
1
1
+
+
+
10 kΩ 22 kΩ 1.2 kΩ 56 kΩ
1
=
−6
−6
100 × 10 S + 45.46 × 10 S + 833.333 × 10− 6 S + 17.86 × 10− 6 S
1
=
= 1.003 kΩ, very close
996.65 × 10− 6 S
I3 the most, I4 the least
VR
VR
44 V
44 V
I R1 = 1 =
= 4.4 mA, I R2 = 2 =
= 2 mA
R1 10 kΩ
R2 22 kΩ
VR
VR
44 V
44 V
= 0.79 mA
= 36.67 mA, I R4 = 4 =
I R3 = 3 =
56 kΩ
R3 1.2 kΩ
R4
RT =
Chapter 6
e.
f.
14.
15.
E
44 V
=
= 43.87 mA
RT 1.003 kΩ
Is = 43.87 mA = 4.4 mA + 2 mA + 36.67 mA + 0.79 mA = 43.86 mA (checks)
always greater
Is =
−
RT′ = 3 Ω || 6 Ω = 2 Ω, RT = RT′ || R3 = 2 Ω || 2 Ω = 1 Ω
E 12 V
=
= 12 A
Is = I ′ =
RT
1Ω
E 12 V
=
I R1 =
=4A
R1 3 Ω
I ′′ = I ′ − I R1 = 12 A − 4 A = 8 A
V = E = 12 V
16.
(20 Ω)(10.8 A)
=9A
20 Ω + 4 Ω
E = VR3 = I 3 R3 = (9 A)(4 Ω) = 36 V
I3 =
I R1 = 12.3 A − 10.8 A = 1.5 A
R1 =
17.
I R1
=
36 V
= 24 Ω
1.5 A
RT = 20 Ω || 5 Ω = 4 Ω
E 30 V
=
= 7.5 A
Is =
4Ω
RT
5 Ω Is
1
= (7.5 A) = 1.5 A
CDR: I1 =
5 Ω + 20 Ω 5
a.
b.
18.
VR1
a.
b.
c.
Chapter 6
10 kΩ || 10 kΩ = 5 kΩ
RT = 1 kΩ || 5 kΩ = 0.833 kΩ
8V
E
=
Is =
= 9.6 mA
RT 0.833 kΩ
RT′ = 10 kΩ || 1 kΩ = 0.9091 kΩ
RT′ I s
(0.9091 kΩ)(9.6 mA) 8.727 mA
I1 =
=
=
= 0.8 mA
RT′ + 10 kΩ
0.9091 kΩ + 10 kΩ
10.9091
24 V − 8 V 16 V
=
= 4 mA
4 kΩ
4 kΩ
V = 24 V
24 V
24 V
Is =
+ 4 mA +
= 2.4 mA + 4 mA + 12 mA = 18.4 mA
10 kΩ
2 kΩ
I=
43
19.
a.
1
1
=
−
6
1
1
1
1000 × 10 S + 30.303 × 10− 6 S + 121.951 × 10− 6 S
+
+
1 kΩ 33 kΩ 8.2 kΩ
1
= 867.86 Ω
=
1.152 × 10− 3 S
VR 100 V
VR
100 V
I R1 = 1 =
= 100 mA, I R2 = 2 =
= 3.03 mA
R1
R2 33 kΩ
1 kΩ
RT =
I R3 =
b.
VR3
=
100 V
= 12.2 mA
R3 8.2 kΩ
PR1 = VR1 ⋅ I R1 = (100 V)(100 mA) = 10 W
PR2 = VR2 ⋅ I R2 = (100 V)(3.03 mA) = 0.30 W
PR3 = VR3 ⋅ I R3 = (100 V)(12.2 mA) = 1.22 W
c.
d.
e.
20.
a.
b.
c.
d.
e.
f.
21.
44
E
100 V
=
= 115.23 mA
RT 867.86 Ω
Ps = EsIs = (100 V)(115.23 mA) = 11.52 W
Ps = 11.52 W = 10 W + 0.30 W + 1.22 W = 11.52 W (checks)
R1 = the smallest parallel resistor
Is =
E
120 V
= 66.667 mA
=
Rbulb 1.8 kΩ
R 1.8 kΩ
=
RT =
= 225 Ω
N
8
E 120 V
=
= 0.533 A
Is =
RT 225 Ω
Ibulb =
V 2 (120 V)2
=8W
=
R
1 .8 k Ω
Ps = 8(8 W) = 64 W
none, Is drops by 66.667 mA
P=
1
1
=
−3
1
1
1
200 × 10 S + 100 × 10−3 S + 50 × 10−3 S
+
+
5 Ω 10 Ω 20 Ω
1
=
= 2.86 Ω
350 × 10−3 S
60 V
E
=
= 20.98 A
Is =
RT 2.86 Ω
P = E ⋅ Is = (60 V)(20.98 A) = 1.26 kW
RT =
Chapter 6
22.
a.
P1 = 10(60 W) = 600 W = E ⋅ I1 = 120 V ⋅ I1, I1 =
P2 = 400 W = 120 V ⋅ I2, I2 =
b.
c.
d.
23.
a.
b.
c.
400 W
= 3.33 A
120 V
200 W
P3 = 200 W = 120 V ⋅ I3, I3 =
= 1.67 A
120 V
110 W
= 0.92 A
P4 = 110 W = 120 V ⋅ I4 , I4 =
120 V
Is = 5 A + 3.33 A + 1.67 A + 0.92 A = 10.92 A (no)
120 V
E
=
= 10.99 Ω
RT =
I s 10.92 A
Ps = E ⋅ Is = (120 V)(10.92 A) = 1.31 kW
Ps = 1.31 kW = 600 W + 400 W + 200 W + 110 W = 1.31 kW (the same)
8 Ω || 12 Ω = 4.8 Ω, 4.8 Ω || 4 Ω = 2.182 Ω
24 V + 8 V
= 14.67 A
I1 =
2.182 Ω
V 2 (24 V + 8 V) 2
P4 =
=
= 256 W
R
4Ω
I2 = I1 = 14.67 A
24.
I1 = 12.6 mA − 8.5 mA = 4.1 mA
I2 = 8.5 mA − 4 mA = 4.5 mA
25.
a.
b.
26.
a.
b.
Chapter 6
600 W
=5A
120 V
9 A + 2 A + I1 = 12 A,
I1 = 12 A − 11 A = 1 A
I2 + 1 A = 1 A + 3 A,
I2 = 4 A − 1 A = 3 A
6 A = 2 A + I1, I1 = 6 A − 2 A = 4 A
4 A + 5 A = I2, I2 = 9 A
9 A = I3 + 3 A, I3 = 9 A − 3 A = 6 A
3 A + 10 A = I4, I4 = 13 A
I1 = 3 mA
I1 + 5 mA = 8 mA,
5 mA = I2 + 3.5 mA,
I2 = 1.5 mA
I1 = 3 mA = I3 + 1 mA, I3 = 2 mA
I4 = 5 mA
I3 = 1.5 μA + 0.5 μA = 2.0 μA
6 μA = I2 + I3 = I2 + 2 μA, I2 = 4 μA
I2 + 1.5 μA = I4, I4 = 4 μA + 1.5 μA = 5.5 μA
I1 = 6 μA
45
27.
I R2 = 5 mA − 2 mA = 3 mA
E = VR2 = (3 mA)(4 kΩ) = 12 V
R1 =
R3 =
RT =
28.
a.
b.
VR1
I R1
VR3
I R3
=
=
12 V
12 V
= 3 kΩ
=
(9 mA − 5 mA) 4 mA
12 V
= 6 kΩ
2 mA
E
12 V
=
= 1.33 kΩ
I T 9 mA
E 10 V
=
=5Ω
2A
I1
I2 = I − I1 = 3 A − 2 A = 1 A
E 10 V
=
R=
= 10 Ω
1A
I2
R1 =
E = I1R1 = (2 A)(6 Ω) = 12 V
E 12 V
=
= 1.33 A
I2 =
R2
9Ω
P 12 W
=1A
=
V 12 V
E 12 V
=
= 12 Ω
R3 =
I3
1A
I = I1 + I2 + I3 = 2 A + 1.33A + 1 A = 4.33 A
I3 =
c.
46
64 V
= 64 mA
1 kΩ
64 V
I3 =
= 16 mA
4 kΩ
Is = I1 + I2 + I3
I2 = Is − I1 − I3 = 100 mA − 64 mA − 16 mA = 20 mA
64 V
E
=
R=
= 3.2 kΩ
I 2 20 mA
I = I2 + I3 = 20 mA + 16 mA = 36 mA
I1 =
Chapter 6
d.
V12
⇒ V1 = PR1 = (30 W)(30 Ω) = 30 V
R1
E = V1 = 30 V
E 30 V
=
I1 =
=1A
R1 30 Ω
Because R3 = R2, I3 = I2 , and Is = I1 + I2 + I3 = I1 + 2I2
2 A = 1 A + 2I2
1
I2 = (1 A) = 0.5 A
2
I3 = 0.5 A
E 30 V
=
R2 = R3 =
= 60 Ω
I 2 0.5 A
P=
PR2 = I 22 R2 = (0.5 A)2 ⋅ 60 Ω = 15 W
29.
30.
4Ω
1
I1 = I1 = 2 A
12 Ω
3
4Ω
I1 = 2 I1 = 12 A
I3 =
2Ω
4Ω
1
I4 =
I1 = I1 = 0.6 A
40 Ω
10
IT = I1 + I2 + I3 + I4 = 6 A + 2 A + 12 A + 0.6 A = 20.6 A
I2 =
a.
8 kΩ(20 mA)
= 16 mA
2 kΩ + 8 kΩ
I2 = 20 mA − 16 mA = 4 mA
b.
RT =
Chapter 6
I1 =
1
1
=
−6
1
1
1
454.55 × 10 S + 833.33 × 10− 6 S + 5000 × 10− 6 S
+
+
2.2 kΩ 1.2 kΩ 0.2 kΩ
1
=
= 159.03 Ω
6,288 × 10− 6 S
159.03 Ω
R
(18 mA) = 1.30 mA
Ix = T I , I1 =
2.2 kΩ
Rx
159.03 Ω
I2 =
(18 mA) = 2.39 mA
1.2 kΩ
159.03 Ω
(18 mA) = 14.31 mA
I3 =
0.2 kΩ
I4 = 18 mA
47
c.
d.
31.
20 Ω(9 A)
=6A
20 Ω + 10 Ω
I3 = 9 A − I1 = 9 A − 6 A = 3 A
I4 = 9 A
I1 = I2 =
a.
I1 ≅
b.
I1/I2 = 10 Ω/1 Ω = 10,
c.
I1/I3 = 1 kΩ/1 Ω = 1000, I3 = I1/1000 = 9 A/1000 ≅ 9 mA
d.
e.
f.
48
1
1
=
−
3
1
1
1
250 × 10 S + 125 × 10−3 S + 83.333 × 10−3 S
+
+
4 Ω 8 Ω 12 Ω
1
= 2.18 Ω
=
458.333 × 10− 3
R
2.18 Ω
Ix = T I,
I1 =
(6 A) = 3.27 A
4Ω
Rx
2.18 Ω
I2 =
(6 A) = 1.64 A
8Ω
2.18 Ω
(6 A) = 1.09 A
I3 =
12 Ω
I4 = 6 A
RT =
9
(10 A) = 9 A
10
I2 =
I1 9 A
=
≅ 0.9 A
10 10
I1/I4 = 100 kΩ/1 Ω = 100,000, I4 = I1/100,000 = 9 A/100,000 ≅ 90 μA
very little effect, 1/100,000
1
RT =
1
1
1
1
+
+
+
1 Ω 10 Ω 1 kΩ 100 kΩ
1
=
1 S + 0.1 S + 1 × 10− 3 S + 10 × 10− 6 S
1
=
= 0.91 Ω
1.10 S
R
0.91 Ω
(10 A) = 9.1 A excellent (9 A)
Ix = T I , I1 =
1Ω
Rx
g.
I2 =
h.
I3 =
i.
I4 =
0.91 Ω
(10 A) = 0.91 A excellent (0.9 A)
10 Ω
0.91 Ω
(10 A) = 9.1 mA excellent (9 mA)
1 kΩ
0.91 Ω
(10 A) = 91 μA excellent (90 μA)
100 kΩ
Chapter 6
32.
a.
b.
33.
a.
b.
2ΩI
=1A
2Ω+6Ω
1 A(8 Ω)
I=
= 4 A = I2
2Ω
I1 = I − 1 A = 3 A
CDR: I6Ω =
I3 = I = 7 μA
By inspection: I2 = 2 μA
I1 = I − 2(2 μA) = 7 μA − 4 μA = 3 μA
VR = (2 μA)(9 Ω) = 18 μV
V
18 μ V
R= R =
=6Ω
3 μA
IR
R = 3(2 kΩ) = 6 kΩ
6 kΩ(32 mA)
I1 =
= 24 mA
6 k Ω + 2 kΩ
I
24 mA
I2 = 1 =
= 8 mA
3
3
34.
84 mA = I1 + I2 + I3 = I1 + 2I1 + 2I2 = I1 + 2I1 + 2(2I1)
84 mA = I1 + 2I1 + 4I1 = 7I1
84 mA
and I1 =
= 12 mA
7
I2 = 2I1 = 2(12 mA) = 24 mA
I3 = 2I2 = 2(24 mA) = 48 mA
VR
24 V
R1 = 1 =
= 2 kΩ
I1 12 mA
VR
24 V
R2 = 2 =
= 1 kΩ
24 mA
I2
VR
24 V
R3 = 3 =
= 0.5 kΩ
I 3 48 mA
35.
a.
b.
c.
PL = VLIL
72 W = 12 V ⋅ IL
72 W
=6A
IL =
12 V
I
6A
I1 = I2 = L =
=3A
2
2
Psource = EI = (12 V)(3 A) = 36 W
Ps1 + Ps 2 = 36 W + 36 W = 72 W (the same)
d.
Idrain = 6 A (twice as much)
Chapter 6
49
36.
RT = 8 Ω || 56 Ω = 7 Ω
E 12 V
=
= 1.71 A
I2 = I3 =
RT
7Ω
1
1
I1 = I 2 = (1.71 A) = 0.86 A
2
2
16 V
= 2 A, I = 5 A − 2 A = 3 A
8Ω
16 V
V
IR = 5 A + 3 A = 8 A, R = R =
=2Ω
8A
IR
37.
I8 Ω =
38.
a.
c.
E
12 V
12 V
=
=
= 1.188 mA
RT 0.1 kΩ + 10 kΩ 10.1 kΩ
VL = IsRL = (1.19 mA)(10 kΩ) = 11.90 V
12 V
= 120 mA
Is =
100 Ω
VL = E = 12 V
a.
VL =
b.
39.
b.
c.
40.
41.
Is =
4.7 kΩ(9 V)
42.3V
= 6.13 V
=
4.7 kΩ + 2.2 kΩ
6.9
VL = E = 9 V
VL = E = 9 V
b.
c.
20 V
= 5 A, I2 = 0 A
4Ω
V1 = 0 V, V2 = 20 V
Is = I1 = 5 A
a.
V2 =
a.
b.
c.
d:
I1 =
RT′ = 11 MΩ || 22 kΩ = 21.956 kΩ
21.956 kΩ(20 V)
V2 =
= 16.47 V (very close to ideal)
21.956 kΩ + 4.7 kΩ
Rm = 20 V[20,000 Ω/V] = 400 kΩ
RT′ = 400 kΩ || 22 kΩ = 20.853 kΩ
20.853 kΩ(20 V)
V2 =
= 16.32 V (still very close to ideal)
20.853 kΩ + 4.7 kΩ
a.
b.
50
22 kΩ(20 V)
= 16.48 V
22 kΩ + 4.7 kΩ
V2 =
200 kΩ(20 V)
= 13.33 V
200 kΩ + 100 kΩ
RT′ = 200 kΩ || 11 MΩ = 196.429 kΩ
(196.429 kΩ)(20 V)
V2 =
= 13.25 V (very close to ideal)
196.429 kΩ + 100 kΩ
Chapter 6
c.
42.
e.
DMM level of 11 MΩ not a problem for most situations
VOM level of 400 kΩ can be a problem for some situations.
a.
Vab = 20 V
b.
Vab =
c.
43.
44.
Rm = 400 kΩ
RT′ = 400 kΩ || 200 kΩ = 133.333 kΩ
(133.333 kΩ)(20 V)
V2 =
= 11.43 V (a 1.824 V drop from Rint = 11 MΩ level)
133.333 kΩ + 100 kΩ
11 MΩ(20 V)
= 18.33 V
11 MΩ + 1 MΩ
Rm = 200 V[20,000 Ω/V] = 4 MΩ
4 MΩ(20 V)
Vab =
= 16.0 V (significant drop from ideal)
4 MΩ + 1 MΩ
Rm = 20 V[20,000 Ω/V] = 400 kΩ
400 kΩ(20 V)
= 5.71 V (significant error)
Vab =
400 kΩ + 1 MΩ
not operating properly, 6 kΩ not connected at both ends
6V
= 1.71 kΩ
RT =
3.5 mA
RT = 3 kΩ || 4 kΩ = 1.71 kΩ
Vab = E + I4 kΩ ⋅ R4 kΩ
12 V − 4 V
8V
=
I4 kΩ =
= 1.6 mA
1 kΩ + 4 kΩ 5 kΩ
Vab = 4 V + (1.6 mA)(4 kΩ) = 4 V + 6.4 V = 10.4 V
4 V supply connected in reverse so that
12 V + 4 V 16 V
I=
=
= 3.2 mA
1 kΩ + 4 kΩ 5 kΩ
and Vab = 12 V − (3.2 mA)(1 kΩ) = 12 V − 3.2 V = 8.8 V obtained
Chapter 6
51
Chapter 7
1.
a.
b.
c.
d.
e.
f.
2.
a.
b.
c.
d.
3.
a.
b.
c.
d.
e.
f.
g.
4.
a.
b.
c.
d.
5.
a.
b.
c.
52
E and R1 in series; R2, R3 and R4 in parallel
E and R1 in series; R2, R3 and R4 in parallel
R1 and R2 in series; E, R3 and R4 in parallel
E and R1 in series, R4 and R5 in series; R2 and R3 in parallel
E and R1 in series, R2 and R3 in parallel
E, R1 and R4 in parallel; R6 and R7 in series; R2 and R5 in parallel
RT = 4 Ω + 10 Ω + 4 Ω = 18 Ω
10 Ω
= 10 Ω + 5 Ω = 15 Ω
RT = 10 Ω +
2
RT = 4 Ω || (4 Ω + 4 Ω) + 10 Ω = 4 Ω || 8 Ω + 10 Ω = 2.67 Ω + 10 Ω = 12.67 Ω
RT = 10 Ω
yes
I2 = Is − I1 = 10 A − 4 A = 6 A
yes
V3 = E − V2 = 14 V − 8 V = 6 V
RT′ = 4 Ω || 2 Ω = 1.33 Ω , RT′′ = 4 Ω || 6 Ω = 2.4 Ω
RT = RT′ + RT′′ = 1.33 Ω + 2.4 Ω = 3.73 Ω
20 Ω
= 10 Ω, RT = RT′ + RT′′ = 10 Ω + 10 Ω = 20 Ω
RT′ = RT′′ =
2
E
20 V
=
=1A
Is =
RT 20 Ω
Ps = EIs = Pabsorbed = (20 V)(1 A) = 20 W
12 Ω
6Ω
= 6 Ω, RT′′ = R2 || RT′ =
=3Ω
2
2
RT = R1 + RT′′ = 4 Ω + 3 Ω = 7 Ω
E
14 V
1
2A
=
= 2 A, I2 = I s =
=1A
Is =
RT
7Ω
2
2
1A
= 0.5 A
I3 =
2
I5 = 1 A
V2 = I2R2 = (1 A)(6 Ω) = 6 V
V4 = V2 = 6 V
RT′ = R3 || R4 =
RT′ = R1 || R2 = 10 Ω || 15 Ω = 6 Ω
RT = RT′ || (R3 + R4) = 6 Ω || (10 Ω + 2 Ω) = 6 Ω || 12 Ω = 4 Ω
E 36 V
E
36 V
Is =
=
= 9 A, I1 =
=6A
=
RT
4Ω
RT′
6Ω
E
36 V
36 V
I2 =
=
=
=3A
R3 + R4 10 Ω + 2 Ω 12 Ω
V4 = I4 R4 = I2R4 = (3 A)(2 Ω) = 6 V
Chapter 7
6.
a.
b.
c.
7.
a.
b.
c.
8.
a.
b.
c.
RT′ = 1.2 kΩ + 6.8 kΩ = 8 kΩ, RT′′ = 2 kΩ || RT′ = 2 kΩ || 8 kΩ = 1.6 kΩ
RT′ ′′ = RT′′ + 2.4 kΩ = 1.6 kΩ + 2.4 kΩ = 4 kΩ
RT = 1 kΩ || RT′′′ = 1 kΩ || 4 kΩ = 0.8 kΩ
E
48 V
=
= 60 mA
Is =
RT 0.8 kΩ
RT′′E
(1.6 kΩ)(48 V)
V=
= 19.2 V
=
RT′′ + 2.4 k Ω 1.6 kΩ + 2.4 kΩ
RT = (R1 || R2 || R3) || (R6 + R4 || R5)
= (12 kΩ || 12 kΩ || 3 kΩ) || (10.4 kΩ + 9 kΩ || 6 kΩ)
= (6 kΩ || 3 kΩ) || (10.4 kΩ + 3.6 kΩ)
= 2 kΩ || 14 kΩ = 1.75 kΩ
28 V
28 V
E
E
Is =
=
=
= 16 mA, I2 =
= 2.33 mA
RT 1.75 kΩ
R2 12 kΩ
R′ = R1 || R2 || R3 = 2 kΩ
R′′ = R6 + R4 || R5 = 14 kΩ
R′( I s )
2 kΩ(16 mA)
=
I6 =
= 2 mA
′
′
′
R +R
2 kΩ + 14 kΩ
V1 = E = 28 V
R′ = R4 || R5 = 6 kΩ || 9 kΩ = 3.6 kΩ
V5 = I6 R′ = (2 mA)(3.6 kΩ) = 7.2 V
VR2
(28 V) 2
= 261.33 mW
P= 3 =
R3
3 kΩ
R′ = R4 || R5 || (R7 + R8) = 4 Ω || 8 Ω || (6 Ω + 2 Ω) = 4 Ω || 8 Ω || 8 Ω
= 4 Ω || 4 Ω = 2 Ω
R′′ = (R3 + R′ ) || (R6 + R9) = (8 Ω + 2 Ω) || (6 Ω + 4 Ω)
= 10 Ω || 10 Ω = 5 Ω
RT = R1 || (R2 + R′′ ) = 10 Ω || (5 Ω + 5 Ω) = 10 Ω || 10 Ω = 5 Ω
E 80 V
I=
=
= 16 A
RT
5Ω
I R2 =
I 16 A
=
=8A
2
2
8A
I3 = I9 =
=4A
2
I8 =
( R4 R5 )( I 3 )
( R4 R5 ) + ( R7 + R8 )
(4 Ω 8 Ω)(4 A)
(4 Ω 8 Ω) + (6 Ω + 2 Ω)
(2.67)(4 A)
=
=1A
2.67 Ω + 8 Ω
=
Chapter 7
53
d.
9.
20 V
=4A
5Ω
RT = 16 Ω || 25 Ω = 9.756 Ω
7V
I2 =
= 0.72 A
9.756 Ω
I1 =
10.
a, b.
11.
a.
b.
12.
54
−I8R8 − Vx + I9R9 = 0
Vx = I9R9 − I8R8 = (4 A)(4 Ω) − (1 A)(2 Ω) = 16 V − 2 V = 14 V
a.
24 V
8V
= 0.8 A ↑
= 6 A ↓, I3 =
4Ω
10 Ω
24 V + 8 V 32 V
=
I2 =
= 16 A
2Ω
2Ω
I = I1 + I2 = 6 A + 16 A = 22 A ↓
I1 =
R′ = R4 + R5 = 14 Ω + 6 Ω = 20 Ω
R″ = R2 || R′ = 20 Ω || 20 Ω = 10 Ω
R′″ = R″ + R1 = 10 Ω + 10 Ω = 20 Ω
RT = R3 || R′″ = 5 Ω || 20 Ω = 4 Ω
E 20 V
=5A
Is =
=
RT 4 Ω
20 V
20 V
20 V
=
=
=1A
I1 =
′′
R1 + R 10 Ω + 10 Ω 20 Ω
20 V
I3 =
=4A
5Ω
I
1A
= 0.5 A
I4 = 1 = (since R′ = R2) =
2
2
Va = I3R3 − I4R5 = (4 A)(5 Ω) − (0.5 A)(6 Ω) = 20 V − 3 V = 17 V
⎛I ⎞
Vbc = ⎜ 1 ⎟ R2 = (0.5 A)(20 Ω) = 10 V
⎝2⎠
E
20 V
=
R1 + R4 ( R2 + R3 R5 ) 3 Ω + 3 Ω (3 Ω + 6 Ω 6 Ω)
20 V
20 V
20 V
=
=
=
3 Ω + 3 Ω (3 Ω + 3 Ω) 3 Ω + 3 Ω 6 Ω 3 Ω + 2 Ω
=4A
I1 =
Chapter 7
b.
c.
13.
14.
CDR: I2 =
R4 ( I1 )
3 Ω(4 A)
=
R4 + R2 + R3 R5 3 Ω + 3 Ω + 6 Ω 6 Ω
12 A
= 1.33 A
=
6+3
I
I3 = 2 = 0.67 A
2
I4 = I1 − I2 = 4 A − 1.33 A = 2.67 A
Va = I4R4 = (2.67 A)(3 Ω) = 8 V
Vb = I3R3 = (0.67 A)(6 Ω) = 4 V
VE
2V
=
= 2 mA
RE 1 kΩ
IC = IE = 2 mA
a.
IE =
b.
IB =
c.
VB = VBE + VE = 2.7 V
VC = VCC − ICRC = 8 V − (2 mA)(2.2 kΩ) = 8 V − 4.4 V = 3.6 V
VCC − (VBE + VE ) 8V − (0.7 V + 2 V)
=
RB
RB
220 k Ω
8 V − 2.7 V
5.3 V
=
= 24 μA
=
220 kΩ
220 kΩ
VRB
=
d.
VCE = VC − VE = 3.6 V − 2 V = 1.6 V
a.
IG = 0 ∴ VG =
b.
c.
d.
Chapter 7
VBC = VB − VC = 2.7 V − 3.6 V = −0.9 V
270 kΩ(16 V)
= 1.9 V
270 kΩ + 2000 kΩ
VG − VGS − VS = 0
VS = VG − VGS = 1.9 V − (−1.75 V) = 3.65 V
16 V
= 7.05 μA
270 kΩ + 2000 kΩ
V
3.65 V
ID = IS = S =
= 2.43 mA
RS 1.5 kΩ
I1 = I2 =
VDS = VDD − IDRD − ISRS = VDD − ID(RD + RS) since ID = IS
= 16 V − (2.43 mA)(4 kΩ) = 16 V − 9.72 V = 6.28 V
VDD − IDRD − VDG − VG = 0
VDG = VDD − IDRD − VG
= 16 V − (2.43 mA)(2.5 kΩ) − 1.9 V = 16 V − 6.08 V − 1.9 V = 8.02 V
55
15.
a.
Network redrawn:
100 Ω + 220 Ω = 320 Ω
400 Ω || 600 Ω = 240 Ω
400 Ω || 220 Ω = 141.94 Ω
240 Ω + 141.94 Ω = 381.94 Ω
RT = 320 Ω || 381.94 Ω = 174.12 Ω
Va =
c.
V1 = 32 V − Va = 32 V − 11.89 V = 20.11 V
d.
V2 = Va = 11.89 V
e.
16.
17.
141.94 Ω(32 V)
= 11.89 V
141.94 Ω + 240 Ω
b.
20.11 V
= 33.52 mA
600 Ω
11.89 V
I220Ω =
= 54.05 mA
220 Ω
I + I600Ω = I220Ω
I = I200Ω − I600Ω
= 54.05 mA − 33.52 mA
= 20.53 mA →
I600Ω =
E1
9V
= 0.6 A
=
7Ω +8Ω
R2 + R3
a.
I=
b.
E1 − V1 + E2 = 0
V1 = E1 + E2 = 9 V + 19 V = 28 V
a.
R8 "shorted out"
R′ = R3 + R4 || R5 + R6 || R7
= 10 Ω + 6 Ω || 6 Ω + 6 Ω || 3 Ω
= 10 Ω + 3 Ω + 2 Ω
= 15 Ω
RT = R1 + R2 || R′
= 10 Ω + 30 Ω || 15 Ω = 10 Ω + 10 Ω
= 20 Ω
E 100 V
=
=5A
20 Ω
RT
R′( I )
(15 Ω)(5 A)
=
= 1.67 A
I2 =
R′ + R2 15 Ω + 30 Ω
I=
56
Chapter 7
I3 = I − I2 = 5 A − 1
2
1
A= 3 A
3
3
⎛ 10 ⎞
3 Ω⎜ A ⎟
R7 I 3
⎝ 3 ⎠ = 1.11 A
=
I6 =
R7 + R6
3Ω + 6Ω
I8 = 0 A
b.
18.
19.
⎛ 10 ⎞
A ⎟ (3 Ω) = 10 V
V4 = I3(R4 || R5) = ⎜
⎝ 3
⎠
V8 = 0 V
8 Ω || 8 Ω = 4 Ω
30 V
30 V
=
=3A
I=
4 Ω + 6 Ω 10 Ω
V = I(8 Ω || 8 Ω) = (3 A)(4 Ω) = 12 V
a.
All resistors in parallel (between terminals a & b)
RT = 16 Ω || 16 Ω || 8 Ω || 4 Ω || 32 Ω
8 Ω || 8 Ω || 4 Ω || 32 Ω
4 Ω || 4 Ω || 32 Ω
2 Ω || 32 Ω = 1.88 Ω
b.
All in parallel. Therefore, V1 = V4 = E = 32 V
c.
I3 = V3/R3 = 32 V/4 Ω = 8 A ←
d.
Chapter 7
Is = I1 + I2 + I3 + I4 + I5
32 V 32 V 32 V 32 V 32 V
+
+
+
+
=
16 Ω 8 Ω
4 Ω 32 Ω 16 Ω
=2A+4A+8A+1A+2A
= 17 A
E 32 V
RT =
=
= 1.88 Ω as above
I s 17 A
57
20.
a.
KVL: +6 V − 20 V + Vab = 0
Vab = +20 V − 6 V = 14 V
b.
21.
a.
20 V
=4A
5Ω
V
14 V
I2Ω = ab =
=7A
2Ω 2Ω
6V
=2A
I3Ω =
3Ω
I′ + I3Ω = I2Ω
and I′ = I2Ω − I3Ω = 7 A − 2 A = 5 A
I = I′ + I5Ω = 5 A + 4 A = 9 A
I5Ω =
Applying Kirchoff's voltage law in the CCW direction in the upper "window":
+18 V + 20 V − V8Ω = 0
V8Ω = 38 V
38 V
= 4.75 A
8Ω
18 V
18 V
=
=2A
I3Ω =
3Ω + 6Ω 9Ω
I8Ω =
KCL: I18V = 4.75 A + 2 A = 6.75 A
b.
22.
V = (I3Ω)(6 Ω) + 20 V = (2 A)(6 Ω) + 20 V = 12 V + 20 V = 32 V
I2R2 = I3R3 and I2 =
I1 = I2 + I3 =
I 3 R3 2 R3 R3
=
=
(since the voltage across parallel elements is the same)
R2
20 10
R3
+2
10
⎛R
⎞
KVL: 120 = I112 + I3R3 = ⎜ 3 + 2 ⎟ 12 + 2R3
⎝ 10
⎠
and 120 = 1.2R3 + 24 + 2R3
3.2R3 = 96 Ω
96 Ω
= 30 Ω
R3 =
3.2
58
Chapter 7
23.
Assuming Is = 1 A, the current Is will divide as determined by the load appearing in each
branch. Since balanced Is will split equally between all three branches.
10
⎛1 ⎞
V1 = ⎜ A ⎟ (10 Ω) = V
3
⎝3 ⎠
10
⎛1 ⎞
V2 = ⎜ A ⎟ (10 Ω) = V
6
⎝6 ⎠
10
⎛1 ⎞
V3 = ⎜ A ⎟ (10 Ω) = V
3
⎝3 ⎠
10
10
10
E = V1 + V2 + V3 =
V + V + V = 8.33 V
3
6
3
E 8.33 V
= 8.33 Ω
RT = =
1A
I
24.
25.
36 kΩ || 6 kΩ || 12 kΩ = 3.6 kΩ
3.6 k Ω(45 V)
= 16.88 V ≠ 27 V. Therefore, not operating properly!
V=
3.6 k Ω + 6 k Ω
6 kΩ resistor "open"
R′(45V) 9 kΩ(45 V)
=
R′ = 12 kΩ || 36 kΩ = 9 kΩ, V =
= 27 V
R′ + 6 kΩ 9 kΩ + 6 kΩ
a.
b.
Chapter 7
R′T = R5 || (R6 + R7) = 6 Ω || 3 Ω = 2 Ω
R″T = R3 || (R4 + R′T) = 4 Ω || (2 Ω + 2 Ω) = 2 Ω
RT = R1 + R2 + R″T = 3 Ω + 5 Ω + 2 Ω = 10 Ω
240 V
= 24 A
I=
10 Ω
4 Ω( I )
4 Ω(24 A)
=
= 12 A
4Ω+4Ω
8Ω
6 Ω(12 A) 72 A
=
=8A
I7 =
6Ω + 3Ω
9
I4 =
59
c.
d.
26.
27.
a.
V3 = I3R3 = (I − I4)R3 = (24 A − 12 A)4 Ω = 48 V
V5 = I5R5 = (I4 − I7)R5 = (4 A)6 Ω = 24 V
V7 = I7R7 = (8 A)2 Ω = 16 V
P = I 72 R7 = (8 A)22 Ω = 128 W
P = EI = (240 V)(24 A) = 5760 W
R′T = R4 || (R6 + R7 + R8) = 2 Ω || 7 Ω = 1.56 Ω
R″T = R2 || (R3 + R5 + R′T) = 2 Ω || (4 Ω + 1 Ω + 1.56 Ω) = 1.53 Ω
RT = R1 + R″T = 4 Ω + 1.53 Ω = 5.53 Ω
b.
I = 2 V/5.53 Ω = 361.66 mA
c.
I3 =
2 Ω( I )
2 Ω(361.66 mA)
=
= 84.50 mA
2 Ω + 6.56 Ω
2 Ω + 6.56 Ω
2 Ω(84.5 mA)
= 18.78 mA
I8 =
2Ω+7Ω
The 12 Ω resistors are in parallel.
Network redrawn:
RT = 12 Ω
E 24 V
=
Is =
=2A
RT 12 Ω
I
2A
=1A
I2Ω = s =
2
2
24 Ω( I 2 Ω ) 2
I12Ω =
= A
24 Ω + 12 Ω 3
2
⎛2 ⎞
P10Ω = (I10Ω)2 10 Ω = ⎜ A ⎟ ⋅ 10 Ω = 4.44 W
⎝3 ⎠
28.
60
a.
R10 + R11 || R12 = 1 Ω + 2 Ω || 2 Ω = 2 Ω
R4 || (R5 + R6) = 10 Ω || 10 Ω = 5 Ω
R1 + R2 || (R3 + 5 Ω) = 3 Ω + 6 Ω || 6 Ω = 6 Ω
RT = 2 Ω || 3 Ω || 6 Ω = 2 Ω || 2 Ω = 1 Ω
I = 12 V/1 Ω = 12 A
b.
I1 = 12 V/6 Ω = 2 A
6 Ω(2 A)
=1A
I3 =
6Ω+6Ω
1A
= 0.5 A
I4 =
2
d.
I10 =
c.
I6 = I4 = 0.5 A
12 A
=6A
2
Chapter 7
29.
a.
E = (40 mA)(1.6 kΩ) = 64 V
b.
RL2 =
c.
I R1 = 72 mA − 40 mA = 32 mA
48 V
= 4 kΩ
12 mA
24 V
RL3 =
= 3 kΩ
8 mA
I R2 = 32 mA − 12 mA = 20 mA
I R3 = 20 mA − 8 mA = 12 mA
R1 =
R2 =
R3 =
30.
VR1
I R1
VR2
I R2
VR3
I R3
=
=
=
64 V − 48 V 16 V
=
= 0.5 kΩ
32 mA
32 mA
48 V − 24 V
24 V
=
= 1.2 kΩ
20 mA
20 mA
24 V
= 2 kΩ
12 mA
I R1 = 40 mA
I R2 = 40 mA − 10 mA = 30 mA
I R3 = 30 mA − 20 mA = 10 mA
I R5 = 40 mA
I R4 = 40 mA − 4 mA = 36 mA
R1 =
R2 =
R3 =
R4 =
R5 =
Chapter 7
VR1
I R1
VR2
I R2
VR3
I R3
VR4
I R4
VR5
I R5
=
=
120 V − 100 V
20 V
=
= 0.5 kΩ
40 mA
40 mA
100 V − 40 V
60 V
=
= 2 kΩ
30 mA
30 mA
=
40 V
= 4 kΩ
10 mA
=
36 V
= 1 kΩ
36 mA
=
60 V − 36 V
24 V
=
= 0.6 kΩ
40 mA
40 mA
61
P1 = I12 R1 = (40 mA)20.5 kΩ = 0.8 W (1 watt resistor)
P2 = I 22 R2 = (30 mA)22 kΩ = 1.8 W (2 watt resistor)
P3 = I 32 R3 = (10 mA)24 kΩ = 0.4 W (1/2 watt or 1 watt resistor)
P4 = I 42 R4 = (36 mA)21 kΩ = 1.3 W (2 watt resistor)
P5 = I 52 R5 = (40 mA)20.6 kΩ = 0.96 W (1 watt resistor)
All power levels less than 2 W. Four less than 1 W.
31.
a.
yes, RL
Rmax (potentiometer)
b.
VDR: VR2 = 3 V =
c.
VR1 = E − VL = 12 V − 3 V = 9 V (Chose VR1 rather than VR2
R2 (12 V) R2 (12 V)
=
1kΩ
R1 + R2
3 V(1 k Ω)
= 0.25 kΩ = 250 Ω
R2 =
12 V
R1 = 1 kΩ − 0.25 kΩ = 0.75 kΩ = 750 Ω
R1 (12 V)
R1 + ( R2 RL )
9R1 + 9(R2 || RL) = 12R1
R1 = 3( R2 RL )
R1 + R2 = 1 k Ω
VR1 = 9 V =
RL
since numerator of VDR
equation "cleaner")
⎫
⎬ 2 eq. 2 unk( RL = 10 kΩ)
⎭
3R2 RL
3R2 10 kΩ
⇒
R2 + RL
R2 + 10 kΩ
and R1(R2 + 10 kΩ) = 30 kΩ R2
R1R2 + 10 kΩ R1 = 30 kΩ R2
R1 + R2 = 1 kΩ: (1 kΩ − R2)R2 + 10 kΩ (1 kΩ − R2) = 30 kΩ R2
R22 + 39 kΩ R2 − 10 kΩ2 = 0
R2 = 0.255 kΩ, −39.255 kΩ
R2 = 255 Ω
R1 = 1 kΩ − R2 = 745 Ω
R1 =
32.
a.
b.
80 Ω(40 V)
= 32 V
100 Ω
Vbc = 40 V − 32 V = 8 V
Vab =
80 Ω || 1 kΩ = 74.07 Ω
20 Ω || 10 kΩ = 19.96 Ω
74.07 Ω(40 V)
= 31.51 V
Vab =
74.07 Ω + 19.96 Ω
Vbc = 40 V − 31.51 V = 8.49 V
62
Chapter 7
33.
c.
(31.51 V ) 2 (8.49 V ) 2
+
= 12.411 W + 3.604 W = 16.02 W
P=
80 Ω
20 Ω
d.
P=
a.
ICS = 1 mA
b.
Rshunt =
(32 V ) 2 (8 V ) 2
+
= 12.8 W + 3.2 W = 16 W
80 Ω
20 Ω
The applied loads dissipate less than 20 mW of power.
Rm I CS
(100 Ω)(1 mA) 0.1
≅
Ω = 5 mΩ
=
20
I max − I CS
20 A − 1 mA
(1 k Ω)(50 μ A)
≅2Ω
25 mA − 0.05 mA
(1 k Ω)(50 μ A)
50 mA: Rshunt =
=1Ω
50 mA − 0.05 mA
100 mA: Rshunt ≅ 0.5 Ω
34.
25 mA: Rshunt =
35.
a.
Rs =
b.
Ω/V = 1/ICS = 1/50 μA = 20,000
36.
37.
38.
Vmax − VVS 15 V − (50 μ A)(1 kΩ)
=
= 300 kΩ
I CS
50 μ A
5 V − (1 mA)(100 Ω)
= 4.9 kΩ
1 mA
50 V − 0.1 V
= 49.9 kΩ
50 V: Rs =
1 mA
500 V − 0.1 V
500 V: Rs =
= 499.9 kΩ
1 mA
5 V: Rs =
10 MΩ = (0.5 V)(Ω/V) ⇒ Ω/V = 20 × 106
1
= 0.05 μA
ICS = 1/(Ω/V) =
20 × 106
a.
Chapter 7
Rs =
zero adjust
3V
2 kΩ
E
=
− Rm −
− 1 kΩ −
= 28 kΩ
2
2
100 μ A
Im
63
b.
xIm =
Runk =
E
Rseries
+ Rm +
zero adjust
+ Runk
2
E
zero adjust ⎞
⎛
− ⎜ Rseries + Rm +
⎟
xI m
2
⎝
⎠
30 × 103
3V
− 30 × 103
− 30 kΩ ⇒
x
x100 μ A
3
1
1
x = , Runk = 10 kΩ; x = , Runk = 30 kΩ; x = , Runk = 90 kΩ
4
2
4
=
39.
−
40.
a.
Carefully redrawing the network will reveal that all three resistors are in parallel
R 12 Ω
=
=4Ω
and RT =
3
N
b.
Again, all three resistors are in parallel and RT =
64
R 18 Ω
=
=6Ω
N
3
Chapter 7
Chapter 8
1.
a.
b.
c.
2.
a.
b.
c.
I2 = I3 = 10 mA
V1 = I1R1 = (10 mA)(1 kΩ) = 10 V
RT = 1 kΩ + 2.2 kΩ + 0.56 kΩ = 3.76 kΩ
Vs = IRT = (10 mA)(3.76 kΩ) = 37.6 V
10 kΩ( 4 A)
Rs ( I )
=
= 3.996 A, I2 ≅ I
Rs + R1 + R2 10 kΩ + 10 Ω
V2 = I2R2 = (3.996 A)(6 Ω) = 23.98 V
Vs = I2(R1 + R2) = (3.996 A)(10 Ω) = 39.96 V
I2 =
VR1 = IR1 = (6 A)(3 Ω) = 18 V
3.
E + VR1 − Vs = 0, Vs = E + VR1 = 10 V + 18 V = 28 V
4.
a.
b.
c.
5.
6.
Vs = E = 24 V
24 V
24 V
E
=
=
=6A
I2 =
4Ω
R1 + R2 1 Ω + 3 Ω
I + Is = I2, Is = I2 − I = 6 A − 2 A = 4 A
V1 = V2 = Vs = IRT = 0.6 A[6 Ω || 24 Ω || 24 Ω] = 0.6 A[6 Ω || 12 Ω] = 2.4 V
2.4 V
V
I2 = 2 =
= 0.1 A
R2 24 Ω
16 Ω(2.4 V)
R3Vs
=
= 1.6 V
V3 =
24 Ω
R3 + R4
a.
E 24 V
E
24 V
24 Ω
=
= 12 A, I R2 =
=
=
=3A
R1 2 Ω
R2 + R3 6 Ω + 2 Ω 8 Ω
KCL: I + Is − I1 − I R2 = 0
I1 =
I s = I1 + I R2 − I = 12 A + 3 A − 4 A = 11 A
b.
VDR: V3 =
7.
R3 E
2 Ω(24 V) 48 V
=
=
=6V
R2 + R3 6 Ω + 2 Ω 8 Ω
Vs = E = 24 V
a.
I=
E 18 V
=
= 3 A, Rp = Rs = 6 Ω
Rs 6 Ω
b.
I=
9V
E
=
= 4.09 mA, Rp = Rs = 2.2 kΩ
Rs 2.2 kΩ
CHAPTER 8
65
8.
9.
10.
11.
a.
E = IRs = (1.5 A)(3 Ω) = 4.5 V, Rs = 3 Ω
b.
E = IRs = (6 mA)(4.7 kΩ) = 28.2 V, Rs = 4.7 kΩ
a.
CDR: IL =
b.
Es = IR = (12 A)(100 Ω) = 1.2 kV
Rs = 100 Ω
Es
1.2 kV
=
I=
= 11.76 A
Rs + RL 100 Ω + 2 Ω
a.
E = IR2 = (2 A)(6.8 Ω) = 13.6 V, R = 6.8 Ω
b.
I1 (CW) = (12 V + 13.6 V)/(10 Ω + 6.8 Ω + 39 Ω) =
c.
V ab = I1R3 = (458.78 mA)(39 Ω) = 17.89 V
a.
IT = 6.8 A − 1.2 A − 3.6 A = 2 A
b.
12.
13.
Rs ( I ) 100 Ω(12 A)
=
= 11.76 A, IL ≅ I
Rs + RL 100 Ω + 2 Ω
25.6 V
= 458.78 mA
55.8 Ω
+−
Vs = IT ⋅ R = (2 A)(4 Ω) = 8 V
IT↑ = 7 A − 3 A = 4 A
R ( I ) 6 Ω(4 A)
= 2.4 A
CDR: I1 = 2 T =
R1 + R2 4 Ω + 6 Ω
V2 = I1R1 = (2.4 A)(4 Ω) = 9.6 V
a.
b.
Conversions: I1 = E1/R1 = 9 V/3 Ω = 3 A, R1 = 3 Ω
I2 = E2/R2 = 20 V/2 Ω = 10 A, R2 = 2 Ω
IT↓ = 10 A − 3A = 7 A, RT = 3 Ω || 6 Ω || 2 Ω || 12 Ω
= 2 Ω || 2 Ω || 12 Ω
= 1 Ω || 12 Ω
= 0.92 Ω
V ab Vab = −IT RT = −(7 A)(0.92 Ω) = −6.44 V
+−
14.
66
c.
I3 ↑ =
a.
I=
6.44 V
= 1.07 A
6Ω
E
12 V
=
= 5.45 mA, Rp = 2.2 kΩ
R2 2.2 k Ω
CHAPTER 8
b.
15.
IT↑ = 8 mA + 5.45 mA − 3 mA = 10.45 mA
R′ = 6.8 kΩ || 2.2 kΩ = 1.66 kΩ
V1 = ITR′ = (10.45 mA)(1.66 kΩ) = 17.35 V
c.
V1 = V2 + 12 V ⇒ V2 = V1 − 12 V = 17.35 V − 12 V
= 5.35 V
d.
I2 =
V2 5.35 V
=
= 2.43 mA
R2 2.2 k Ω
a.
I1 = − A, I 2 =
5
4
A, I 3 = A
7
7
1
5
4
= I1 = − A, I R2 = I 2 = A, I R3 = I 3 = A
7
7
7
1
7
I R1
b.
4 − 4I1 − 8I3 = 0
6 − 2I2 − 8I3 = 0
I1 + I2 = I3
────────────
10 + 12 − 3I3 − 4I1 = 0
12 − 3I3 − 12I2 = 0
I1 + I2 = I3
───────────────
I1 = 3.06 A
I2 = 0.19 A
I3 = 3.25 A
I R1 = I1 = 3.06 A, I R2 = I 2 = 0.19 A
I R3 = I 3 = 3.25 A
16.
(I):
10 − I1 5.6 kΩ − I3 2.2 kΩ + 20 = 0
−20 + I3 2.2 kΩ + I2 3.3 kΩ − 30 = 0
I1 + I2 = I3
─────────────────────────
I1 = I R1 = 1.45 mA, I2 = I R2 = 8.51 mA, I3 = I R3 = 9.96 mA
(II):
−1.2 kΩ I1 + 9 − 8.2 kΩ I3 = 0
−10.2 kΩ I2 + 8.2 kΩ I3 + 6 = 0
I2 + I3 = I1
──────────────────────
I1 = 2.03 mA, I2 = 1.23 mA, I3 = 0.8 mA
I R1 = I1 = 2.03 mA
I R2 = I3 = 0.8 mA
I R3 = I R4 = I2 = 1.23 mA
CHAPTER 8
67
17.
−25 − 2I1 − 3I3 + 60 = 0
−60 + 3I3 + 6 − 5I2 − 20 = 0
I1 = I2 + I3
──────────────
I2 = −8.55 A
(I):
V ab = 20 V − I2 5 Ω = 20 V − (8.55 A)(5 Ω) = −22.75 V
+−
(II): Source conversion: E = IR1 = (3 A)(3 Ω) = 9 V, R1 = 3 Ω
9 + 6 − 3I2 − 4I2 − 6I4 = 0
+ 6I4 − 8I3 − 4 = 0
I2 = I3 + I4
────────
I2 = 1.27 A
V ab = I2 4 Ω − 6 V = (1.27 A)4 Ω − 6 V = −0.92 V
+−
18.
I1 = I R1 (CW), I2 = I R2 (down), I3 = I R3 (right), I4 = I R4 (down)
I5 = I R5 (CW)
a.
c.
d.
68
E1 − I1R1 − I2R2 = 0
I2R2 − I3R3 − I4R4 = 0
I4R4 − I5R5 − E2 = 0
I1 = I2 + I3
I3 = I4 + I5
────────────────
b.
E1 − I2(R1 + R2) − I3R1 = 0
I2R2 − I3(R3 + R4) + I5R4 = 0
I3R4 − I5(R4 + R5) − E2 = 0
───────────────────
I2(R1 + R2) + I3R1
+ 0
= E1
I2(R2)
− I3(R3 + R4) + I5R4
=0
0
+ I3R4
− I5(R4 + R5) = E2
───────────────────────────
3I2 + 2I3 + 0 = 10
1I2 − 9I3 + 5I5 = 0
0 + 5I3 − 8I5 = 6
─────────────
I3 = I R 3 = −63.69 mA
CHAPTER 8
19.
a.
b.
c.
20.
20 V − IB(270 kΩ) − 0.7 V − IE(0.51 kΩ) = 0
IE(0.51 kΩ) + 8 V + IC(2.2 kΩ) − 20 V = 0
IE = IB + IC
──────────────────────────────
IB = 63.02 μA, IC = 4.42 mA, IE = 4.48 mA
VB = 20 V − IB(270 kΩ) = 20 V − (63.02 μA)(270 kΩ) = 20 V − 17.02 V = 2.98 V
VE = IERE = (4.48 mA)(510 Ω) = 2.28 V
VC = 20 V − IC(2.2 kΩ) = 20 V − (4.42 mA)(2.2 kΩ) = 20 V −9.72 V = 10.28 V
β ≅ IC/IB = 4.42 mA/63.02 μA = 70.14
a.
b.
4 − 4I1 − 8(I1 − I2) = 0
−8(I2 − I1) − 2I2 − 6 = 0
───────────────
1
5
I1 = − A , I2 = − A
7
7
1
I R1 = I1 = − A
7
5
I R2 = I2 = − A
7
4
⎛ 1 ⎞ ⎛ 5 ⎞
A (dir. of I1 )
I R3 = I1 − I2 = ⎜ − A ⎟ − ⎜ − A ⎟ =
7
⎝ 7 ⎠ ⎝ 7 ⎠
−10 − 4I1 − 3(I1 − I2) − 12 = 0
12 − 3(I2 − I1) − 12I2 = 0
─────────────────
I1 = −3.06 A, I2 = 0.19 A
I R1 = I1 = −3.06 A
I R3 = I2 = 0.19 A
I R2 = I1 − I2 = (−3.06 A) − (0.19 A) = −3.25 A
21.
(I):
10 − I1(5.6 kΩ) − 2.2 kΩ(I1 − I2) + 20 = 0
−20 − 2.2 kΩ(I2 − I1) − I2 3.3 kΩ − 30 = 0
────────────────────────────
I1 = 1.45 mA, I2 = 8.51 mA
I R1 = I1 = 1.45 mA, I R2 = I2 = 8.51 mA
I R3 = I2 − I1 = 7.06 mA (direction of I2)
CHAPTER 8
69
(II):
−I1(1.2 kΩ) + 9 − 8.2 kΩ(I1 − I2) = 0
−I2(1.1 kΩ) + 6 − I2 (9.1 kΩ) − 8.2 kΩ(I2 − I1) = 0
──────────────────────────────────
I1 = 2.03 mA, I2 = 1.23 mA
I R1 = I1 = 2.03 mA, I R3 = I R4 = I 2 = 1.23 mA
I R2 = I1 − I 2 = 2.03 mA − 1.23 mA = 0.80 mA (direction of I1)
22.
(I):
−25 − 2I1 − 3(I1 − I2) + 60 = 0
−60 − 3(I2 − I1) + 6 − 5I2 − 20 = 0
───────────────────────
I1 = 1.87 A, I2 = −8.55 A
V ab = 20 − I2 5 = 20 − (8.55 A)(5) = 20 V − 42.75 V
+−
= −22.75 V
(II): Source conversion: E = 9 V, R = 3 Ω
9 − 3I2 − 4I2 + 6 − 6(I2 − I3) = 0
−6(I3 − I2) − 8I3 − 4 = 0
──────────────────────
I2 = 1.27 A, I3 = 0.26 A
V ab = I2 4 − 6 = (1.27 A)(4 Ω) − 6 V
+−
23.
(a):
10 − I12 − 1(I1 − I2) = 0
−1(I2 − I1) − I2 4 − 5(I2 − I3) = 0
−5(I3 − I2) − I3 3 − 6 = 0
──────────────────────
3I1 − 1I 2 + 0 = 10
= 5.08 V − 6 V
= −0.92 V
−1I1 + 10 I 2 − 5 I 3 = 0
0 − 5I 2 + 8I 3 = − 6
I2 = I R3 = −63.69 mA
24.
a.
−1I1 − 4 − 5I1 + 6 − 1(I1 − I2) = 0
−1(I2 − I1) − 6 − 3I2 − 15 − 10I2 = 0
────────────────────────
I1 = I5Ω = 72.16 mA
Va = −4 − (72.16 mA)(6 Ω)
= −4 − 0.433 V
= −4.43 V
70
CHAPTER 8
b.
Network redrawn:
−6I1 − 4(I1 − I2) − 12 = 0
12 − 4(I2 − I1) − 5I2 − 2(I2 − I3) + 16 = 0
−16 − 2(I3 − I2) − 3I3 = 0
───────────────────────────
I2 = I5Ω = 1.95 A
Va = (I3)(3 Ω)
= (−2.42 mA)(3 Ω)
= −7.26 V
25.
I1(2.2 kΩ + 9.1 kΩ) − 9.1 kΩ I2 = 18
I2(9.1 kΩ + 7.5 kΩ + 6.8 kΩ) − 9.1 kΩ I1 − 6.8 kΩ I3 = −18
I3(6.8 kΩ + 3.3 kΩ) − I2 6.8 kΩ = −3
───────────────────────────────────────
I1 = 1.21 mA, I2 = −0.48 mA, I3 = −0.62 mA
(I):
16 − 4I1 − 3(I1 − I2) − 12 − 4(I1 − I3) = 0
12 − 3(I2 − I1) − 10 I2 − 15 − 4(I2 − I3) = 0
−16 − 4(I3 − I1) − 4(I3 − I2) − 7I3 = 0
─────────────────────────────
I1 = −0.24 A, I2 = −0.52 A, I3 = −1.28 A
(II):
26.
−6.8 kΩ I1 − 4.7 kΩ(I1 − I2) + 6 − 2.2 kΩ(I1 − I4) = 0
−6 − 4.7 kΩ(I2 − I1) − 2.7 kΩ I2 − 8.2 kΩ (I2 − I3) = 0
−1.1 kΩ I3 − 22 kΩ(I3 − I4) − 8.2 kΩ(I3 − I2) − 9 = 0
5 − 1.2 kΩ I4 − 2.2 kΩ(I4 − I1) − 22 kΩ(I4 − I3) = 0
────────────────────────────────────
I1 = 0.03 mA, I2 = −0.88 mA, I3 = −0.97 mA, I4 = −0.64 mA
a.
b.
Network redrawn:
−2I1 − 6 − 4I1 + 4I2 = 0
−4I2 + 4I1 − 1I2 + 1I3 − 6 = 0
−1I3 + 1I2 + 6 − 8I3 = 0
───────────────────
I1 = −3.8 A, I2 = −4.20 A, I3 = 0.20 A
CHAPTER 8
71
27.
a.
24 V − 6I1 − 4I2 − 10I1 + 12 V = 0
and 16I1 + 4I2 = 36
I1 − I2 = 6 A
───────────────────
I1 = I2 + 6 A
16[I2 + 6 A] + 4I2 = 36
16I2 + 96 + 4I2 = 36
20I2 = −60
I2 = −3 A
I1 = I2 + 6 A = −3 A + 6 A = 3 A
I24V = I6Ω = I10Ω = I12V = 3 A (CW)
I4Ω = 3 A (CCW)
b.
20 V − 4I1 − 6(I1 − I2) − 8(I3 − I2) − 1I3 = 0
10I1 − 14I2 + 9I3 = 20
I3 − I1 = 3 A
I2 = 8 A
────────────────────────
10I1 − 14(8 A) + 9[I1 + 3 A] = 20
19I1 = 105
I1 = 5.526 A
I3 = I1 + 3 A = 5.526 A + 3 A = 8.526 A
I2 = 8 A
I20V = I4Ω = 5.53 A (dir. of I1)
I6Ω = I2 − I1 = 2.47 A (dir. of I2)
I8Ω = I3 − I2 = 0.53 A (dir. of I3)
I1Ω = 8.53 A (dir. of I3)
72
CHAPTER 8
28.
(4 + 8)I1 − 8I2 = 4
(8 + 2)I2 − 8I1 = −6
─────────────
1
5
I1 = − A , I2 = − A
7
7
a.
b.
(4 + 3)I1 − 3I2 = −10 − 12
(3 + 12)I2 − 3I1 = 12
─────────────────
I1 = −3.06 A, I2 = 0.19 A
29.
(I):
a.
b.
c.
I1(5.6 kΩ + 2.2 kΩ) − 2.2 kΩ (I2) = 10 + 20
I2(2.2 kΩ + 3.3 kΩ) − 2.2 kΩ (I1) = −20 − 30
──────────────────────────────
I1 = 1.45 mA, I2 = −8.51 mA
I R1 = I1 = 1.45 mA, I R2 = I2 = −8.51 mA
I R3 = I1 + I2 = 8.51 mA + 1.44 mA = 9.96 mA (direction of I1)
(II):
a.
30.
I1(1.2 kΩ + 8.2 kΩ) − 8.2 kΩ I2 = 9
I2(8.2 kΩ + 1.1 kΩ + 9.1 kΩ) − 8.2 kΩ I1 = 6
──────────────────────────────
b.
I1 = 2.03 mA, I2 = 1.23 mA
c.
I R1 = I1 = 2.03 mA, I R3 = I R4 = I2 = 1.23 mA
(2 + 3)I1 − 3I2 = −25 + 60
(3 + 5)I2 − 3I1 = −60 + 6 − 20
────────────────────
(I):
b.
c.
(II): a.
CHAPTER 8
I R2 = I1 − I2 = 2.03 mA − 1.23 mA = 0.80 mA (direction of I1)
I1 = 1.87 A, I2 = −8.55 A
I R1 = I1 = 1.87 A, I R2 = I2 = −8.55 A
I R3 = I1 − I2 = 1.87 A − (−8.55 A) = 10.42 A (direction of I1)
(3 + 4 + 6)I2 − 6I3 = 9 + 6
(6 + 8)I3 − 6I2 = −4
──────────────────
73
b.
I2 = 1.27 A, I3 = 0.26 A
c.
I R2 = I2 = 1.27 A, I R3 = I3 = 0.26 A
I R4 = I2 − I3 = 1.27 A − 0.26 A = 1.01 A
I R1 = 3 A − I2 = 3 A − 1.27 A = 1.73 A
31.
I1(2 + 1) − 1I2 = 10
I2(1 + 4 + 5) − 1I1 − 5I3 = 0
I3(5 + 3) − 5I2 = −6
───────────────────
I2 = I R3 = −63.69 mA (exact match with problem 18)
32.
From Sol. 24(b)
33.
(I):
I1(6 + 4) − 4I2 = −12
I2(4 + 5 + 2) − 4I1 − 2I3 = 12 + 16
I3(2 + 3) − 2I2 = −16
─────────────────
I5Ω = I2 = 1.95 A
I3 = −2.42 A, ∴ Va = (I3)(3 Ω) = (−2.42 A)(3 Ω) = −7.26 V
(2.2 kΩ + 9.1 kΩ)I1 − 9.1 kΩI2 = 18
(9.1 kΩ + 7.5 kΩ + 6.8 kΩ)I2 − 9.1 kΩ I1 − 6.8 kΩI3 = −18
(6.8 kΩ + 3.3 kΩ)I3 − 6.8 kΩI2 = −3
───────────────────────────
I1 = 1.21 mA, I2 = −0.48 mA, I3 = −0.62 mA
(II):
34.
a.
b.
74
(4 Ω + 4 Ω + 3 Ω)I1 − 3 Ω I2 − 4 Ω I3 = 16 − 12
(4 Ω + 3 Ω + 10 Ω)I2 − 3I1 − 4 Ω I3 = 12 − 15
(4 Ω + 4 Ω + 7 Ω)I3 − 4I1 − 4I2 = −16
────────────────────────
I1 = −0.24 A, I2 = −0.52 A, I3 = −1.28 A
I1(6.8 kΩ + 4.7 kΩ + 2.2 kΩ) − 4.7 kΩ I2 − 2.2 kΩ I4 = 6
I2(2.7 kΩ + 8.2 kΩ + 4.7 kΩ) − 4.7 kΩ I1 − 8.2 kΩ I3 = −6
I3(8.2 kΩ + 1.1 kΩ + 22 kΩ) − 22 kΩ I4 − 8.2 kΩ I2 = −9
I4(2.2 kΩ + 22 kΩ + 1.2 kΩ) − 2.2 kΩ I1 − 22 kΩ I3 = 5
───────────────────────────
I1 = 0.03 mA, I2 = −0.88 mA, I3 = −0.97 mA, I4 = −0.64 mA
From Sol. 26(b):
I1(2 + 4) − 4I2 = −6
I2(4 + 1) − 4I1 − 1I3 = −6
I3(1 + 8) − 1I2 = 6
──────────────────
I1 = 3.8 A, I2 = −4.20 A, I3 = 0.20 A
CHAPTER 8
35.
a.
⎡1 1 1⎤ 1
V1 ⎢ + + ⎥ − V2 = 5
⎣2 5 2⎦ 2
V1 = 8.08 V
V2 = 9.39 V
⎡1 1⎤ 1
V2 ⎢ + ⎥ − V1 = 3
⎣2 4⎦ 2
──────────────────
Symmetry is present
b.
⎡1 1⎤ 1
V1 ⎢ + ⎥ − V2 = 4 − 2
⎣2 4⎦ 4
V1 = 4.80 V
V2 = 6.40 V
⎡1 1 1⎤ 1
+ ⎥ − V1 = 2
V2 ⎢ +
⎣ 4 20 5 ⎦ 4
───────────────────
Symmetry is present
36.
(I):
⎡1 1 1 ⎤ 1
V1 ⎢ + + ⎥ − V2 = − 5 − 3
⎣3 6 4⎦ 4
⎡1 1 ⎤ 1
V2 ⎢ + ⎥ − V1 = 3 − 4
⎣8 4 ⎦ 4
──────────────────────
V1 = −14.86 V, V2 = −12.57 V
VR1 = VR4 = −14.86 V
VR2 = −12.57 V
+
(II):
VR3 − = 12 V + 12.57 V − 14.86 V = 9.71 V
1
⎡1 1 1 ⎤ 1
V1 ⎢ + + ⎥ − V2 − V2 = − 6
2
⎣5 3 2⎦ 3
1
⎡1 1 1 1⎤ 1
V2 ⎢ + + + ⎥ − V1 − V1 = 7
2
⎣2 3 4 8⎦ 3
─────────────────
V1 = −2.56 V, V2 = 4.03 V
VR1 = −2.56 V
VR2 = VR5 = 4.03 V
VR4 = VR3 = 4.03 V + 2.56 V = 6.59 V
CHAPTER 8
75
37.
(I): a.
1
1 ⎤
1
⎡ 1
V1 ⎢
V2 = − 1.98 mA
+
+
−
⎥
⎣ 2.2 kΩ 9.1 kΩ 7.5 kΩ ⎦ 7.5 kΩ
1
1 ⎤
1
⎡ 1
V2 ⎢
V1 = 0.91 mA
+
+
−
⎥
⎣ 7.5 kΩ 6.8 kΩ 3.3 kΩ ⎦ 7.5 kΩ
──────────────────────────────────────
b.
V1 = −2.65 V, V2 = 0.95 V
c.
VR3 = V1 = −2.65 V, VR5 = V2 = 0.95 V, VR4 = V2 − V1 = 3.60 V
(+)
R1
+
VR1 = 18 V − 2.65 V = 15.35 V
R2
−
VR2 = 3 V − 0.95 V = 2.05 V
(−)
−
+
1
⎡1 1 1⎤ 1
V1 ⎢ + + ⎥ − V2 − V3 = 4
4
⎣4 4 7⎦ 4
(II): a.
1
⎡1 1 1 ⎤ 1
V2 ⎢ + + ⎥ − V1 − V3 = 4 + 1.5
3
⎣ 4 3 10 ⎦ 4
1
⎡1 1 1⎤ 1
V3 ⎢ + + ⎥ − V1 − V3 = − 4 − 4
3
⎣4 3 4⎦ 4
──────────────────────────
b.
V1 = 8.88 V, V2 = 9.83 V, V3 = −3.01 V
c.
VR6 = V1 = 8.88 V, VR4 = V3 = −3.01 V, VR5 =
− VR1 +
R1
− VR2 +
R2
R3
76
+
VR1 = 16 V − V1 + V3 = 4.12 V
(+)
V2
( −)
= 0.95 V
V1
VR2 = V2 − V3 − 12 V = 0.84 V
V R 3 = 15 V − V2 = 5.17 V
CHAPTER 8
38.
(I):
1
⎡1 1 1 ⎤ 1
V1 ⎢ + + ⎥ − V2 − V3 = 5
3
6
6
6
6
⎣
⎦
1
⎡1 1 1⎤ 1
V2 ⎢ + + ⎥ − V1 − V 3 = − 3
5
⎣6 4 5⎦ 6
1
⎡1 1 1 ⎤ 1
V3 ⎢ + + ⎥ − V2 − V1 = 0
6
⎣6 5 7⎦ 5
───────────────────────
V1 = 7.24 V, V2 = −2.45 V, V3 = 1.41 V
(II): Source conversion: I = 4 A, R = 4 Ω
1⎤ 1
1
⎡1 1
+ ⎥ − V2 − V3 = − 2
V1 ⎢ +
20
⎣ 9 20 20 ⎦ 20
1
1⎤ 1
1
⎡1
+ ⎥ − V1 − V3 = 0
V2 ⎢ +
20
⎣ 20 20 18 ⎦ 20
1 1⎤ 1
1
⎡1
+ ⎥ − V2 − V1 = 4
V3 ⎢ +
20
⎣ 20 20 4 ⎦ 20
──────────────────────────
V1 = −6.64 V, V2 = 1.29 V, V3 = 10.66 V
39.
(I)
1
⎡1 1⎤
⎢ 2 + 2 ⎥ V1 − 2 V2 + 0 = − 5
⎣
⎦
1
1
⎡1 1 1 1⎤
⎢ 2 + 9 + 7 + 2 ⎥ V2 − 2 V1 − 2 V3 = 0
⎣
⎦
1
⎡1 1 1⎤
⎢ 2 + 2 + 4 ⎥ V3 − 2 V2 = 5
⎣
⎦
────────────────────────
V1 = −5.31 V, V2 = −0.62 V, V3 = 3.75 V
CHAPTER 8
77
(II)
⎡1 1⎤ 1
V1 ⎢ + ⎥ − V3 = − 5
⎣2 6⎦ 6
⎡1⎤
V2 ⎢ ⎥ = 5 − 2
⎣4⎦
⎡1 1⎤ 1
V3 ⎢ + ⎥ − V 1 = 2
⎣6 5⎦ 6
───────────────────
V1 = −6.92 V, V2 = 12 V, V3 = 2.3 V
40.
a.
Node V1:
ΣIi = ΣIo
2A=
V1 V1 − V2
+
6 Ω 10 Ω
Supernode V2, V3:
V
V −V
V
0= 2 1 + 2 + 3
10 Ω
4 Ω 12 Ω
Independent source:
V2 − V3 = 24 V or V3 = V2 − 24 V
2 eq. 2 unknowns:
V1 V1 − V2
+
=2A
6 Ω 10 Ω
V2 − V1 V2 V2 − 24 V
+
+
=0
10 Ω
4Ω
12 Ω
─────────────────────
0.267V1 − 0.1V2 = 2
+0.1V1 − 0.433V2 = −2
────────────────
V1 = 10.08 V, V2 = 6.94 V
V3 = V2 − 24 V = −17.06 V
78
CHAPTER 8
b.
Supernode:
ΣIi = ΣIo
3A+4A=3A+
V1
V
⎧
+ 2
⎪4 A =
2 eq. 2 unk. ⎨
20 Ω 40 Ω
⎪V2 − V1 = 16 V
⎩
V1
V
+ 2
20 Ω 40 Ω
Subt. V2 = 16 V + V1
V
(16 V + V1 )
4A= 1 +
20 Ω
40 Ω
and V1 = 48 V
V2 = 16 V + V1 = 64 V
──────────────────
41.
a.
⎡1 1 1⎤ 1
V1 ⎢ + + ⎥ − V2 = 5
⎣2 5 2⎦ 2
⎡1 1⎤ 1
V2 ⎢ + ⎥ − V1 = 3
⎣2 4⎦ 2
──────────────
V1 = 8.08 V, V2 = 9.39 V
Symmetry present
b.
⎡1 1⎤ 1
V1 ⎢ + ⎥ − V2 = 4 − 2
⎣2 4⎦ 4
⎡1 1 1⎤ 1
V2 ⎢ +
+ ⎥ − V1 = 2
⎣ 4 20 5 ⎦ 4
───────────────
V1 = 4.8 V, V2 = 6.4 V
Symmetry present
CHAPTER 8
79
42.
(I): a.
b.
c.
Note the solution to problem 36(I).
V1 = −14.86 V, V2 = −12.57 V
VR1 = VR4 = V1 = −14.86 V, VR2 = V2 = −12.57 V
+ VR3 −
R3
(II): a.
b.
c.
Note the solution to problem 36(II).
V1 = −2.56 V, V2 = 4.03 V
VR1 = V1 = −2.56 V, VR2 = VR5 = V2 = 4.03 V
VR3 = VR4
43.
(I): a.
VR3 = V1 − V2 + 12 V = (−14.86 V) − (−12.57 V) + 12 V = 9.71 V
(+) (−)
= V2 − V1 = 6.59 V
Source conversion: I = 5 A, R = 3 Ω
1
⎡1 1 1 ⎤ 1
V1 ⎢ + + ⎥ − V2 − V3 = 5
6
⎣3 6 6⎦ 6
1
⎡1 1 1⎤ 1
V2 ⎢ + + ⎥ − V1 − V3 = − 3
5
⎣6 4 5⎦ 6
b.
1
⎡1 1 1 ⎤ 1
V3 ⎢ + + ⎥ − V2 − V1 = 0
6
⎣6 5 7⎦ 5
──────────────────
V1 = 7.24 V, V2 = −2.45 V, V3 = 1.41 V
c.
R1
−
VR1 = 15 V − 7.24 V = 7.76 V
+
VR2 = V2 = −2.45 V, VR3 = V3 = 1.41 V
(+) (−)
VR4 = V3 − V2 = 1.41 V − (−2.45 V) = 3.86 V
VR5 = V1 − V2 = 7.24 V − (−2.45 V) = 9.69 V
VR6 = V1 − V3 = 7.24 V − 1.41 V = 5.83 V
80
CHAPTER 8
(II): a.
Source conversion: I = 4 A, R = 4 Ω
1⎤ 1
1
⎡1 1
+ ⎥ − V2 − V3 = − 2
V1 ⎢ +
20
⎣ 9 20 20 ⎦ 20
1
1⎤ 1
1
⎡1
+ ⎥ − V1 − V3 = 0
V2 ⎢ +
20
⎣ 20 20 18 ⎦ 20
1 1⎤ 1
1
⎡1
+ ⎥ − V2 − V1 = 4
V3 ⎢ +
20
⎣ 20 20 4 ⎦ 20
──────────────────────────
b.
V1 = −6.64 V, V2 = 1.29 V, V3 = 10.66 V
c.
VR1 = V1 = −6.64 V, R2
−
VR2 = 16 V − 10.66 V = 5.34 V
+
VR3 = V2 = 1.29 V, VR4 = V2 − V1 = 1.29 V − (−6.64 V) = 7.93 V
(+)
(−)
VR5 = V3 − V2 = 10.66 V − 1.29 V = 9.37 V
(+)
(−)
VR6 = V3 − V1 = 10.66 V − (−6.64 V) = 17.30 V
(+)
44.
(−)
a.
Note the solution to problem 39(I).
V1 = −5.31 V, V2 = −0.62 V, V3 = 3.75 V
V5A = V1 = −5.31 V
b.
Note the solution to problem 39(II).
V1 = −6.92 V, V2 = 12 V, V3 = 2.3 V
V2A = V2 − V3 = 9.7 V, V5A = V2 − V1 = 18.92 V
(+)
45.
(+)
(−)
I1(6 + 5 + 10) − 5I2 − 10I3 = 6
I2(5 + 5 + 5) − 5I1 − 5I3 = 0
I3(5 + 10 + 20) − 10I1 − 5I2 = 0
─────────────────────
I1 = 0.39 A, I2 = 0.18 A, I3 = 0.14 A
a.
b.
(−)
I5 = I2 − I3 = 40 mA (direction of I2)
c, d.
CHAPTER 8
no
81
46.
Source conversion: I = 1 A, R = 6 Ω
1
1
⎡1 1 1⎤
⎢ 6 + 5 + 5 ⎥ V1 − 5 V2 − 5 V3 = 1
⎣
⎦
1
1
⎡1 1 1 ⎤
⎢ 5 + 5 + 20 ⎥ V2 − 5 V1 − 5 V3 = 0
⎣
⎦
1
1
⎡1 1 1 ⎤
⎢ 5 + 5 + 10 ⎥ V3 − 5 V1 − 5 V2 = 0
⎣
⎦
──────────────────
VR5 = 196.70 mV, no
47.
I1(2 kΩ + 33 kΩ + 3.3 kΩ) − 33 kΩ I2 − 3.3 kΩ I3 = 24
I2(33 kΩ + 56 kΩ + 36 kΩ) − 33 kΩ I1 − 36 kΩ I3 = 0
I3(3.3 kΩ + 36 kΩ + 5.6 kΩ) − 36 kΩ I2 − 3.3 kΩ I1 = 0
─────────────────────────────────────
I1 = 0.97 mA, I2 = I3 = 0.36 mA
a.
b.
I5 = I2 − I3 = 0.36 mA − 0.36 mA = 0
c, d. yes
48.
Source conversion: I = 12 A, R = 2 kΩ
1
1 ⎤
1
1
⎡ 1
⎢ 2 kΩ + 33 kΩ + 56 kΩ ⎥ V1 − 56 kΩ V2 − 33 kΩ V3 = 12
⎣
⎦
1
1 ⎤
1
1
⎡ 1
⎢ 56 kΩ + 36 kΩ + 5.6 kΩ ⎥ V2 − 56 kΩ V1 − 36 kΩ V3 = 0
⎣
⎦
1
1 ⎤
1
1
⎡ 1
⎢ 33 kΩ + 3.3 kΩ + 36 kΩ ⎥ V3 − 33 kΩ V1 − 36 kΩ V2 = 0
⎣
⎦
────────────────────────────
I R5 = 0 A, yes
49.
Source conversion: I = 9 mA, R = 1 kΩ
1
1 ⎤
1
1
⎡ 1
V1 ⎢
V2 −
V3 = 4 mA
+
+
−
⎥
200 kΩ
⎣1 kΩ 100 kΩ 200 kΩ ⎦ 100 kΩ
1
1 ⎤
1
1
⎡ 1
V2 ⎢
+
+
⎥ − 100 kΩ V1 − 1 kΩ V3 = − 9 mA
100
k
200
k
1
k
Ω
Ω
Ω
⎣
⎦
1
1 ⎤
1
1
⎡ 1
V3 ⎢
V1 −
V2 = 9 mA
+
+
−
⎥
1 kΩ
⎣ 200 kΩ 100 kΩ 1 kΩ ⎦ 200 kΩ
────────────────────────────────────────────
82
CHAPTER 8
50.
a.
(1 kΩ + 2 kΩ + 2 kΩ)I1 − 2 kΩ I2 − 2 kΩ I3 = 10
(2 kΩ + 2 kΩ + 2 kΩ)I2 − 2 kΩ I1 − 2 kΩ I3 = 0
(2 kΩ + 2 kΩ + 2 kΩ)I3 − 2 kΩ I1 − 2 kΩ I2 = 0
─────────────────────────────────
I1 = I10V = 3.33 mA
Source conversion: I = 10 V/1 kΩ = 10 mA, R = 1 kΩ
1
1 ⎤
1
1
⎡ 1
+
+
−
V1 ⎢
V2 −
V3 = 10 mA
⎥
2 kΩ
⎣1 kΩ 2 kΩ 2 kΩ ⎦ 2 kΩ
1
1 ⎤
1
1
⎡ 1
+
+
−
V2 ⎢
V1 −
V3 = 0
⎥
2 kΩ
⎣ 2 kΩ 2 kΩ 2 kΩ ⎦ 2 kΩ
1
1 ⎤
1
1
⎡ 1
+
+
−
V3 ⎢
V2 −
V1 = 0
⎥
2 kΩ
⎣ 2 kΩ 2 kΩ 2 kΩ ⎦ 2 kΩ
──────────────────────────────────────
V1 = 6.67 V = E − IRs = 10 V − I(1 kΩ)
10 − 6.67 V
I=
= 3.33 mA
1kΩ
b.
Source conversion: E = 20 V, R = 10 Ω
(10 + 10 + 20)I1 − 10I2 − 20I3 = 20
(10 + 20 + 20)I2 − 10I1 − 20I3 = 0
(20 + 20 + 10)I3 − 20I1 − 20I2 = 0
──────────────────────
I1 = I20V = 0.83 A
V = 20 V − 8.3 V
= 11.7 V
Is =
V 11.70 V
=
= 1.17A
10 Ω
Rs
1
1⎤ ⎡1⎤
⎡1
⎡1⎤
V1 ⎢ + + ⎥ − ⎢ ⎥ V2 − ⎢ ⎥ V3 = 2
⎣10 10 20 ⎦ ⎣ 20 ⎦
⎣10 ⎦
1
1⎤ ⎡1⎤
⎡1
⎡1⎤
+ ⎥ − ⎢ ⎥ V1 − ⎢ ⎥ V3 = 0
V2 ⎢ +
⎣ 20 20 10 ⎦ ⎣ 20 ⎦
⎣ 20 ⎦
1
1⎤ ⎡1⎤
⎡1
⎡1⎤
+ ⎥ − ⎢ ⎥ V1 − ⎢ ⎥ V2 = 0
V3 ⎢ +
⎣10 20 20 ⎦ ⎣10 ⎦
⎣ 20 ⎦
───────────────────────
V
I Rs = 1 = 1.17 A
Rs
CHAPTER 8
83
51.
a.
I=
20 V
4
⎡2
⎤ ⎡2
⎤
Ω + ⎢ Ω + 3 Ω⎥ ⎢ Ω + 4 Ω⎥
5
⎣5
⎦ ⎣5
⎦
20 V
4
Ω + (3.14 Ω) (4.4 Ω)
5
= 7.36 A
=
b.
RT = 2.27 kΩ + [4.7 kΩ + 2.27 kΩ] [1.1 kΩ + 2.27 kΩ]
= 2.27 kΩ + [6.97 kΩ] [3.37 kΩ]
= 2.27 kΩ + 2.27 kΩ
= 4.54 kΩ
8V
= 1.76 mA
I=
4.54 kΩ
52.
(Y-Δ conversion)
400 V
400 V
=
I=
12 kΩ 12 kΩ 6 kΩ
3 kΩ
= 133.33 mA
a.
b.
I=
42 V
42 V
=
(18 Ω 18 Ω) [ (18 Ω 18 Ω) + (18 Ω 18 Ω) ] 9 Ω [9 Ω + 9 Ω ]
= 7 A (Y−Δ conversion)
84
CHAPTER 8
53.
3 kΩ
3 kΩ
7.5 kΩ = 2.14 kΩ
15 kΩ = 2.5 kΩ
R′T = 2.14 kΩ (2.5 kΩ + 2.5 kΩ) = 1.5 kΩ
(1.5 kΩ)(5 A)
= 2.14 A
CDR:
I=
1.5 kΩ + 2 kΩ
54.
I s1 =
10 V
5V
15 V
+
=
= 0.83 mA
18 k Ω 18 k Ω 18 k Ω
55.
R′ = R1 + 1 kΩ = 3 kΩ
R″ = R2 + 1 kΩ = 3 kΩ
3kΩ
= 1.5 kΩ
R′T =
2
RT = 1 kΩ + 1.5 kΩ + 1 kΩ = 3.5 kΩ
E
20 V
=
Is =
= 5.71 mA
RT 3.5 k Ω
56.
c − g: 27 Ω 9 Ω
27 Ω = 5.4 Ω
a − h: 27 Ω
27 Ω = 5.4 Ω
RT = 5.4 Ω
= 5.4 Ω
= 4.2 Ω
CHAPTER 8
9Ω
(13.5 Ω + 5.4 Ω)
18.9Ω
85
Chapter 9
1.
a.
E1 :
30 V
12 Ω + 6 Ω 6 Ω
30 V
=
=2A
12 Ω + 3 Ω
I′
I′2 = I′3 = 1 = 1 A
2
60 V
60 V
=
I″3 =
6 Ω + 6 Ω 12 Ω 6 Ω + 4 Ω
=6A
6 Ω( I 3′′)
= 2A
I″1 =
6 Ω + 12 Ω
12 Ω( I 3′′)
=4A
I″2 =
12 Ω + 6 Ω
I′1 =
E2 :
I1 = I′1 + I″1 = 2 A + 2 A = 4 A (dir. of I1′ )
I2 = I 2′′ − I 2′ = 4 A − 1 A = 3 A (dir. of I 2′′ )
I3 = I′3 + I″3 = 1 A + 6 A = 7 A (dir. of I 3′ )
2.
b.
E1: P′1 = I1′2 R1 = (2 A)2 12 Ω = 48 W
c.
P1 = I12 R1 = (4 A)2 12 Ω = 192 W
d.
P′1 + P″1 = 48 W + 48 W = 96 W ≠ 192 W = P1
E2: P″1 = I1′′2 R1 I″R1 = (2 A)2 12 Ω = 48 W
I:
I′ =
8 Ω(9 A)
=4A
8 Ω + 10 Ω
I″ =
18 V
=1A
10 Ω + 8 Ω
E:
I = I′ − I″ = 4 A − 1 A = 3 A (dir of I′)
86
CHAPTER 9
3.
E1:
42 V
= 1.944 A
18 Ω + 3.6 Ω
9 Ω( IT ) 9 Ω(1.944 A)
=
I1 =
9Ω+6Ω
15 Ω
= 1.17 A
IT =
E2:
E2 24 V
=
=2A
RT 12 Ω
IT =
I24V = IT + I1 = 2 A + 1.17 A = 3.17 A (dir. of I1)
4.
I1′ =
3.3 kΩ(5 mA) 16.5 mA
=
2.2 kΩ + 3.3 kΩ
5.5
= 3 mA
I″1 =
8V
8V
=
3.3 k Ω + 2.2 k Ω 5.5 k Ω
= 1.45 mA
I1 = I′1 + I″1 = 3 mA + 1.45 mA = 4.45 mA
5.
E1 :
I1 =
CHAPTER 9
E1
12 V
=
= 1.03 A
RT 6 Ω + 5.88 Ω
87
30 Ω( I1 )
30 Ω(1.03 A)
=
30 Ω + 7 Ω
37 Ω
= 835.14 mA
Vs′ = I′(4 Ω) = (835.14 mA)(4 Ω)
= 3.34 V
I′ =
I:
8 Ω(6 A)
=4A
8Ω + 4Ω
Vs′′ = I′(4 Ω) = 4 A(4 Ω) = 16 V
I ′=
E2 :
R′T = 12 Ω || (4 Ω + 5 Ω) = 12 Ω || 9 Ω = 5.14 Ω
E
8V
= 0.875 A
I′ = 2 =
RT 4 Ω + 5.14 Ω
12 Ω( I ′)
12 Ω(0.875 A)
=
= 0.5 A
I″ =
12 Ω + 9 Ω
21 Ω
Vs′′′= I ′′(4 Ω) = 0.5 A(4 Ω) = 2 V
Vs = Vs′′− Vs′ − Vs′′′ = 16 V − 3.34 V − 2 V = 10.66 V
88
CHAPTER 9
6.
E:
V′2 =
6.8 kΩ(36 V)
= 13.02 V
6.8 kΩ + 12 kΩ
I:
I2 =
12 kΩ(9 mA)
= 5.75 mA
12 kΩ + 6.8 kΩ
V2′′ = I 2 R2 = (5.75 mA)(6.8 kΩ) = 39.10 V
V2 = V2′ + V2′′ = 13.02 V + 39.10 V = 52.12 V
7.
a.
b.
8.
RTh = R3 + R1 || R2 = 4 Ω + 6 Ω || 3 Ω = 4 Ω + 2 Ω = 6 Ω
R2 E
3 Ω(18 V)
=
=6 V
ETh =
R2 + R1 3 Ω + 6 Ω
I1 =
ETh
6V
=
= 0.75 A
RTh + R 6 Ω + 2 Ω
6V
I2 =
= 166.67 mA
6 Ω + 30 Ω
6V
I3 =
= 56.60 mA
6 Ω + 100 Ω
a.
RTh = 2 Ω + 12 Ω = 14 Ω
ETh = IR = (3 A)(12 Ω) = 36 V
CHAPTER 9
89
⎛ ETh ⎞
⎛ 36 V ⎞
R = 2 Ω: P = ⎜
⎟ R=⎜
⎟ 2 Ω = 10.13 W
+
R
R
⎝ 14 Ω + 2 Ω ⎠
⎝ Th
⎠
2
b.
2
36 V
⎛
⎞
R = 100 Ω: P = ⎜
⎟ 100 Ω = 9.97 W
Ω
+
Ω
14
100
⎝
⎠
2
9.
a.
RTh:
← RTh = 5 Ω + 5 Ω || 5 Ω = 7.5 Ω
ETh:
ETh =
⎛ ETh ⎞
⎛ 10 V
⎞
R = 2 Ω: P = ⎜
⎟R =⎜
⎟ 2 Ω = 2.22 W
⎝ 7.5 Ω + 2 Ω ⎠
⎝ RTh + R ⎠
2
b.
20 V
= 10 V
2
2
10 V
⎛
⎞
R = 100 Ω: P = ⎜
⎟ 100 Ω = 0.87 W
⎝ 7.5 Ω + 100 Ω ⎠
2
10.
RTh:
RTh = 6 Ω || 3 Ω = 2 Ω
ETh:
6 Ω(18 V)
= 12 V
6Ω +3Ω
ETh = 72 V + 12 V = 84 V
V6 Ω =
90
CHAPTER 9
11.
RTh:
RTh = 5.6 kΩ || 2.2 kΩ = 1.58 kΩ
ETh: Superposition:
I:
E′Th = IRT
= 8 mA(5.6 kΩ || 2.2 kΩ)
= 8 mA(1.579 kΩ)
= 12.64 V
E:
5.6 kΩ(16 V)
5.6 kΩ + 2.2 kΩ
= 11.49 V
E″Th =
+
ETh = 11.49 V − 12.64 V = −1.15 V
−
12.
a.
RTh:
RTh = 25 Ω || 16 Ω = 9.76 Ω
ETh: Superposition:
E:
E′Th = −20 V
I:
E″Th = (3 A)(9.76 Ω) = 29.28 V
ETh = E″Th − E′Th = 29.28 V − 20 V = 9.28 V
CHAPTER 9
91
b.
RTh:
RTh = 4 Ω || (2 Ω + 6 Ω || 3 Ω) = 2 Ω
ETh:
72 V
=9A
6 Ω + 3 Ω (2 Ω + 4 Ω)
3 Ω( IT ) 3 Ω(9 A)
I2 =
=
=3A
3Ω + 6Ω
9Ω
ETh = V6 + V2 = (IT)(6 Ω) + I2(2 Ω)
= (9 A)(6 Ω) + (3 A)(2 Ω) = 60 V
IT =
13.
(I): RTh:
←RTh = 25 Ω + 60 Ω || 30 Ω = 45 Ω
ETh:
ETh = V30Ω − 10 V − 0
30 Ω(15 V)
− 10 V
=
30 Ω + 60 Ω
= 5 V − 10 V = −5 V
(II): RTh:
RTh = 2.7 kΩ || (4.7 kΩ + 3.9 kΩ) = 2.7 kΩ || 8.6 kΩ = 2.06 kΩ
92
CHAPTER 9
ETh:
3.9 kΩ(18 mA)
= 6.21 mA
3.9 kΩ + 7.4 kΩ
ETh = I′(2.7 kΩ) = (6.21 mA)(2.7 kΩ) = 16.77 V
I′ =
14.
(I): RTh:
←RTh = 2 Ω + 8 Ω = 10 Ω
ETh:
ETh = V16Ω
20 V
= 825.08 mA
20 Ω + 4.24 Ω
5 Ω( I T )
5 Ω(825.08 mA)
I′ =
=
= 125.01 mA
5 Ω + 28 Ω
33 Ω
ETh = V16Ω = (I′)(16 Ω) = (125.01 mA)(16 Ω) = 2 V
IT =
CHAPTER 9
93
(II): RTh:
←RTh = 3. 3 kΩ + 2.2 kΩ || 1.1 kΩ
= 3.3 kΩ + 0.73 kΩ
= 4.03 kΩ
ETh: Superposition:
E1 :
E′Th = V2.2kΩ =
=8V
2.2 kΩ(12 V)
2.2 kΩ + 1.1 kΩ
ETh″ = E2 = 4 V
ETh = E′Th + E″Th = 8 V + 4 V = 12 V
15.
RTh:
←RTh = 2.2 kΩ || 5.6 kΩ = 1.58 kΩ
R′ = 1.58 kΩ + 3.3 kΩ
= 4.88 kΩ
R″ = 4.88 kΩ || 6.8 kΩ = 2.84 kΩ
RTh = 1.2 kΩ + R″ = 1.2 kΩ + 2.84 kΩ = 4.04 kΩ
ETh: Source conversions:
22 V
= 10 mA, Rs = 2.2 kΩ
I1 =
2.2 kΩ
12 V
= 2.14 mA, Rs = 5.6 kΩ
I2 =
5.6 kΩ
Combining parallel current sources: I′T = I1 − I2 = 10 mA − 2.14 mA = 7.86 mA
2.2 kΩ || 5.6 kΩ = 1.58 kΩ
94
CHAPTER 9
Source conversion:
E = (7.86 mA)(1.58 kΩ) = 12.42 V
R′ = Rs + 3.3 kΩ = 1.58 kΩ + 3.3 kΩ = 4.88 kΩ
I=
12.42 V − 6 V
6.42 V
=
= 549.66 μA
4.88 kΩ + 6.8 kΩ 11.68 kΩ
V6.8kΩ = I(6.8 kΩ) = (549.66 μA)(6.8 kΩ) = 3.74 V
ETh = 6 V + V6.8kΩ = 6 V + 3.74 V = 9.74 V
16.
a.
RTh:
←RTh = 51 kΩ || 10 kΩ = 8.36 kΩ
ETh:
ETh =
b.
10 kΩ(20 V)
= 3.28 V
10 kΩ + 51 kΩ
IERE + VCE + ICRC = 20 V
but IC = IE
and IE(RC + RE) + VCE = 20 V
20 V − VCE
20 V − 8 V
12 V
= 4.44 mA
or IE =
=
=
RC + RE
2.2 kΩ + 0.5 kΩ 2.7 kΩ
c.
ETh − IBRTh − VBE − VE = 0
CHAPTER 9
95
and IB =
d.
17.
a.
VC = 20 V − ICRC = 20 V − (4.44 mA)(2.2 kΩ)
= 20 V − 9.77 V
= 10.23 V
ETh = 20 V
I = 1.6 mA =
b.
c.
18.
ETh − VBE − VE 3.28 V − 0.7 V − (4.44 mA)(0.5 kΩ)
=
RTh
8.36 kΩ
2.58 V − 2.22 V 0.36 V
= 43.06 μA
=
=
8.36 kΩ
8.36 kΩ
ETh 20 V
20 V
=
= 12.5 kΩ
, RTh =
RTh
RTh
1.6 mA
ETh= 60 mV, RTh = 2.72 kΩ
ETh = 16 V, RTh = 2.2 kΩ
RN:
RN = 2 Ω + 12 Ω = 14 Ω
IN:
IN =
19.
a.
12 Ω(3 A)
= 2.57 A
12 Ω + 2 Ω
RN:
← RN = 5 Ω +
5Ω
= 7.5 Ω
2
IN:
20 V
= 2.67 A
5Ω
5Ω +
2
I
IN = T = 1.34 A
2
IT =
b.
96
ETh = INRN = (1.34 A)(7.5 Ω) = 10.05 V ≅ 10 V, RTh = RN = 7.5 Ω
CHAPTER 9
20.
RN:
RN = 5.6 kΩ || 2.2 kΩ = 1.58 kΩ
IN:
I:
I′N = 8 mA
E:
I″N =
16 V
= 7.27 mA
2.2 kΩ
IN↑ = 8 mA − 7.27 mA = 0.73 mA
21.
(I): (a)
RN = 25 Ω || 16 Ω = 9.76 Ω
CHAPTER 9
97
IN:
Superposition:
I:
I′N = 3 A
E:
I″N =
20 V
= 2.05 A
9.76 Ω
IN = I′N − I″N = 3 A − 2.05 A = 0.95 A (direction of I′N)
b.
(II): a.
ETh = INRN = (0.95 A)(9.76 Ω) = 9.27 V ≅ 9.28 V, RTh = RN = 9.76 Ω
RN:
← RN = 4 Ω || (2 Ω + 2 Ω) = 2 Ω
IN:
72 V
4 Ω (3 Ω + 6 Ω 2 Ω)
72 V
=
≅ 34 A
2.118 Ω
4 Ω( I )
= 16 A
I1 =
4 Ω + 4.5 Ω
2 Ω( I1 )
=4A
I2 =
2Ω+6Ω
I=
IN = I − I2 = 34 A − 4 A = 30 A
b.
22.
(I)
ETh = INRN = (30 A)(2 Ω) = 60 V, RTh = RN = 2 Ω
RN:
← RN = 2 Ω + 16 Ω || (12 Ω + 4 Ω)
= 2 Ω + 16 Ω || 16 Ω
= 2 Ω + 8 Ω = 10 Ω
98
CHAPTER 9
IN:
20 V
E
= 0.845 A
=
RT 20 Ω + 3.67 Ω
5 Ω(0.845 A)
I1 =
= 0.225 A
5 Ω + 13.78 Ω
16 Ω(0.225 A)
IN =
= 0.2 A
16 Ω + 2 Ω
I=
(II): RN:
← RN = 3.3 kΩ + 1.1 kΩ || 2.2 kΩ
= 3.3 kΩ + 0.733 kΩ
= 4.03 kΩ
IN: Superposition:
E1 :
12 V
1.1 kΩ + 1.32 kΩ
= 4.96 mA
IT =
2.2 kΩ(4.96 mA)
2.2 kΩ + 3.3 kΩ
= 1.98 mA
I′N =
E2 :
I″N =
4V
3.3 kΩ + 0.73 kΩ
= 0.99 mA
IN = I′N + I″N = 1.98 mA + 0.99 mA = 2.97 mA
CHAPTER 9
99
23.
a.
RN:
RN = 4 Ω || 12 Ω = 3 Ω
E = 12 V:
I′N =
12 V
=3A
4Ω
I = 2 A:
I″N = 2 A
IN = I′N + I″N = 3 A + 2 A = 5 A
b.
RN:
← RN = (2 Ω + 4 Ω || 4 Ω) || 4 Ω
= (2 Ω + 2 Ω) || 4 Ω
4Ω
=2Ω
=
2
IN:
100
V4Ω 6 V − 2 V 4 V
=
=
4Ω
4Ω
4Ω
=1A
(4 Ω 2 Ω)(2 V)
V=
(4 Ω 2 Ω) + 4 Ω
= 0.5 V
V 0.5 V
= 0.25 A
I2 = =
2Ω
R
I=
CHAPTER 9
24.
IN = I − I2 = 1 A − 0.25 A = 0.75 A
(I): (a) R = RTh = 9.76 Ω (from problem 12)
(II): (a) R = RTh = 2 Ω (from problem 12)
2
(I): (b) Pmax = ETh
/ 4 RTh = (9.28 V)2/4(9.76 Ω) = 2.21 W
2
(II): (b) Pmax = ETh
/ 4 RTh = (60 V)2/4(2 Ω) = 450 W
25.
(I): (a) R = RTh = 10 Ω (from problem 14)
(II): (a) R = RTh = 4.03 kΩ (from problem 14)
2
(I): (b) Pmax = ETh
/ 4 RTh = (2 V)2/4(10 Ω) = 100 mW
2
(II): (b) Pmax = ETh
/ 4 RTh = (12 V)2/4(4.03 kΩ) = 8.93 mW
26.
27.
RL = RTh = 4.04 kΩ (from problem 15)
ETh = 9.74 V (from problem 15)
2
Pmax = ETh
/ 4 RTh = (9.74 V)2/4(4.04 kΩ) = 5.87 mW
a.
← R = RTh = 4 Ω || 4 Ω = 2 Ω
b.
ETh:
E:
E′Th =
24 V
= 12 V
2
I:
E″Th = IRT = (5 A)(4 Ω || 4 Ω)
= (5 A)(2 Ω) = 10 V
ETh = E′Th + E″Th = 12 V + 10 V = 22 V
E2
(22 V) 2
Pmax = Th =
= 60.5 W
4 RTh 4(2 Ω)
CHAPTER 9
101
⎛ ETh ⎞
P=IR= ⎜
⎟R
⎝ RTh + R ⎠
1
R = (2 Ω) = 0.5 Ω, P = 38.72 W
4
1
R = (2 Ω) = 1 Ω, P = 53.78 W
2
3
R = (2 Ω) = 1.5 Ω, P = 59.27 W
4
R = 2 Ω, P = 60.5 W
5
R = (2 Ω) = 2.5 Ω, P = 59.75 W
4
3
R = (2 Ω) = 3 Ω, P = 58 W
2
7
R = (2 Ω) = 3.5 Ω, P = 56 W
4
R = 2(2 Ω) = 4 Ω, P = 53.78 W
c.
2
2
⎛ ETh ⎞
Pmax = ⎜
⎟ R4
⎝ RTh + R4 ⎠
with R1 = 0 Ω
ETh is a maximum and RTh a minimum
∴ Pmax a maximum
2
28.
29.
a.
b.
V, and therefore V4, will be its largest value when R2 is as
large as possible. Therefore choose R2 = open-circuit (∞ Ω).
V2
Then P4 = 4 will be a maximum.
R4
No, examine each individually.
30.
102
CHAPTER 9
Since RL fixed, maximum power to RL when VRL a maximum as defined by PL =
∴ R = 500 Ω and Pmax =
RL
2
(12 V)
= 1.44 W
100 Ω
IT↑ = 4 A + 7 A = 11 A
RT = 10 Ω || 6 Ω || 3 Ω = 1.67 Ω
VL = ITRT = (11 A)(1.67 Ω) = 18.37 V
V
18.37 V
IL = L =
= 6.12 A
RL
3Ω
31.
32.
VR2L
−5 V / 2.2 kΩ + 20 V / 8.2 kΩ
= 0.2879 V
1/ 2.2 kΩ + 1/ 8.2 kΩ
1
= 1.7346 kΩ
Req =
1/ 2.2 kΩ + 1/ 8.2 kΩ
Eeq
0.2879 V
=
IL =
= 39.3 μA
Req + RL 1.7346 kΩ + 5.6 kΩ
Eeq =
VL = ILRL = (39.3 μA)(5.6 kΩ) = 220 mV
33.
34.
IT↓ = 5 A − 0.4 A − 0.2 A = 4.40 A
RT = 200 Ω || 80 Ω || 50 Ω || 50 Ω = 17.39 Ω
VL = ITRT = (4.40 A)(17.39 Ω) = 75.52 V
V
76.52 V
= 0.38 A
IL = L =
RL
200 Ω
(4 A)(4.7 Ω) + (1.6 A)(3.3 Ω) 18.8 V + 5.28 V
= 3.01 A
=
4.7 Ω + 3.3 Ω
8Ω
Req = 4.7 Ω + 3.3 Ω = 8 Ω
Req ( I eq ) 8 Ω(3.01 A)
= 2.25 A
IL =
=
Req + RL 8 Ω + 2.7 Ω
Ieq =
VL = ILRL = (2.25 A)(2.7 Ω) = 6.08 V
CHAPTER 9
103
35.
←
(4 mA)(8.2 kΩ) + (8 mA)(4.7 kΩ) − (10 mA)(2 kΩ)
I eq =
8.2 kΩ + 4.7 kΩ + 2 kΩ
32.8 V + 37.6 V − 20 V
= 3.38 mA
=
14.9 kΩ
Req = 8.2 kΩ + 4.7 kΩ + 2 kΩ = 14.9 kΩ
Req I eq
(14.9 kΩ)(3.38 mA)
=
IL =
= 2.32 mA
14.9 kΩ + 6.8 kΩ
Req + RL
VL = ILRL = (2.32 mA)(6.8 kΩ) = 15.78 V
36.
15 kΩ || (8 kΩ + 7 kΩ) = 15 kΩ || 15 kΩ = 7.5 kΩ
7.5 kΩ(60 V)
= 45 V
Vab =
7.5 kΩ + 2.5 kΩ
45 V
Iab =
= 3 mA
15 kΩ
37.
10 V − 8 V
2 kΩ + 0.51 kΩ + 1.5 kΩ
= 498.75 μA
V0.51kΩ = (498.75 μA)(0.51 kΩ)
= 0.25 V
Vab = 10 V − 0.25 V = 9.75 V
Iba =
+−
38.
Vab = 0 V (short)
Iab = 0 A (open)
R2 any resistive value
∴ R2 = short-circuit, open-circuit, any value
104
CHAPTER 9
39.
a.
Is =
I
24 V
= 1.5 mA, I = s = 0.5 mA
24 kΩ
3
8 kΩ +
3
b.
c.
40.
24 V
= 0.83 mA
24 kΩ + 8 kΩ 12 kΩ
12 kΩ( I s )
I=
= 0.5 mA
12 kΩ + 8 kΩ
Is =
yes
(a)
10 V
4 kΩ 8 kΩ + 4 kΩ 4 kΩ
10 V
=
2.67 kΩ + 2 kΩ
10 V
= 2.14 mA
=
4.67 kΩ
IT =
8 Ω( IT )
= 1.43 mA, I2 = IT/2 = 1.07 mA
8Ω + 4Ω
I = I1 − I2 = 1.43 mA − 1.07 mA = 0.36 mA
I1 =
(b)
41.
a.
R1 ( I )
3 Ω(6 A)
=
=2A
R1 + R2 + R3 3 Ω + 2 Ω + 4 Ω
V = I R2 R2 = (2 A)(2 Ω) = 4 V
b.
R2 ( I )
2 Ω(6 A)
=
= 1.33 A
R1 + R2 + R3 3 Ω + 2 Ω + 4 Ω
V = I R1 R1 = (1.33 A)(3 Ω) = 4 V
CHAPTER 9
(8 kΩ 4 kΩ)(10 V)
8 kΩ 4 kΩ + 4 kΩ 4 kΩ
= 5.72 V
V
I1 = 1 = 0.71 mA
8 kΩ
V2 = E − V1 = 10 V − 5.72 V
= 4.28 V
V2
= 1.07 mA
I2 =
4 kΩ
I = I2 − I1 = 1.07 mA − 0.71 mA
= 0.36 mA
V1 =
I R2 =
I R1 =
105
Chapter 10
1.
Q1 (9 × 109 )(4 μ C)
= 9 × 103 N/C
=
2
2
r
(2 m)
(a)
E= k
(b)
E = k
Q1
r
2
=
(9 × 109 )(4 μ C)
2
= 36 × 109 N/C
(1 mm)
E (1 mm): E (2 m) = 4 × 106: 1
kQ
(9 × 109 )(2 μ C)
kQ
= 15.81 m
r
=
=
⇒
E
72 N/C
r2
2.
E =
3.
C=
4.
Q = CV = (0.15 μF)(45 V) = 6.75 μC
5.
E=
6.
⎡10−3 in. ⎤ ⎡ 1 m ⎤
d = 10 mils ⎢
⎥⎢
⎥ = 0.254 mm
⎣ 1 mil ⎦ ⎣ 39.37 in. ⎦
E =
7.
8.
9.
Q 1200 μ C
=
= 120 μF
10 V
V
V 100 mV
=
= 50 V/m
2 mm
d
V
100 mV
= 393.70 V/m
=
d 0.254 mm
Q 160 μ C
= 40 V
=
C
4 μF
V 40 V
E = =
= 8 × 103 V/m
d 5 mm
V=
C = 8.85 × 10−12εr
A
(0.1 m 2 )
= 8.85 × 10−12(1)
= 442.50 pF
d
2 mm
(0.1 m 2 )
A
= 1.11 nF
C = 8.85 × 10−12 εr = 8.85 × 10−12(2.5)
2 mm
d
10.
C = 8.85 × 10−12εr
11.
C = εrCo ⇒ εr =
106
A
8.85 × 10−12 (4)(0.15 m 2 )
= 2.66 μm
⇒d=
d
2 μF
6 nF
C
=
= 5 (mica)
Co 1200 pF
CHAPTER 10
12.
13.
14.
a.
C = 8.85 × 10-12(1)
b.
E =
c.
Q = CV = (3.54 nF)(200 V) = 0.71 μC
a.
E=
b.
Q = εE A = εrεoE A = (7)(8.85 ×10−12)(106 V/m)(0.08 m2) = 4.96 μC
c.
C=
a.
b.
c.
d.
15.
16.
17.
18.
d=
(0.08 m 2 )
= 3.54 nF
0.2 mm
V
200 V
=
= 106 V/m
d 0.2 mm
V
200 V
=
= 106 V/m
d 0.2 mm
Q 4.96 μ C
= 24.80 nF
=
200 V
V
1
(5 μF) = 2.5 μF
2
C = 2(5 μF) = 10 μF
C = 20(5 μF) = 100 μF
⎛1⎞
(4)⎜ ⎟
⎝ 3 ⎠ (5 μF) = 26.67 μF
C=
⎛1⎞
⎜ ⎟
⎝4⎠
C=
8.85 × 10−12 ε r A (8.85 × 10−12 )(5)(0.02 m 2 )
=
= 0.1475 mm = 147.5 μm
0.006 μF
C
⎡10−3 m ⎤ ⎡ 39.37 in. ⎤ ⎡1000 mils ⎤
d = 0.1475 mm ⎢
⎥⎢
⎥⎢
⎥ = 5.807 mils
⎣ 1 mm ⎦ ⎣ 1 m ⎦ ⎣ 1 in. ⎦
⎡ 5000 V ⎤
5.807 mils ⎢
⎥ = 29.04 kV
⎣ mil ⎦
1200 V
⎡ mil ⎤
mica:
= 1200 V ⎢
⎥ = 0.24 mils
5000 V
⎣ 5000 V ⎦
mil
1′′
⎤ ⎡ 1m ⎤
⎡
0.24 mils ⎢
⎥ = 6.10 μm
⎥⎢
⎣1000 mils ⎦ ⎣ 39.37 in. ⎦
200
(22 μF)/°C = 4400 pF/°C
1 × 106
4400 pF
4400 pF
[ΔT] =
[80°C] = 0.35 μF
°C
°C
J = ±5%, Size ⇒ 40 pF ± 2 pF, 38 pF → 42 pF
CHAPTER 10
107
19.
M = ±20%, Size ⇒ 220 μF ± 44 μF, 176 μF → 264 μF
20.
K = ±10%, Size ⇒ 33,000 pF ± 3300 pF, 29,700 pF → 36,300 pF
21.
a.
b.
τ = RC = (105 Ω)(5.1 μF) = 0.51 s
υC = E(1 − e−t/τ) = 20 V(1 − e−t/0.51 s)
c.
1τ = 0.632(20 V) = 12.64 V, 3τ = 0.95(20 V) = 19 V
5τ = 0.993(20 V) = 19.87 V
d.
iC =
20 V −t/τ
e = 0.2 mAe−t/0.51 s
100 kΩ
υR = Ee−t/τ = 20 Ve−t/0.51 s
e.
22.
23.
108
a.
τ = RC = (106 Ω)(5.1 μF) = 5.1 s
b.
c.
1τ = 12.64 V, 3τ = 19 V, 5τ = 19.87 V
d.
e.
Same as problem 21 with 5τ = 25 s and Im = 20 μA
a.
τ = RC = (2.2 kΩ + 3.3 kΩ)1 μF = (5.5 kΩ)(1 μF) = 5.5 ms
b.
υC = E(1 − e−t/τ) = 100 V(1 − e−t/5.5 ms)
c.
1τ = 63.21 V, 3τ = 95.02 V, 5τ = 99.33 V
d.
iC =
υC = E(1 − e−t/τ) = 20 V(1 − e−t/5.1s)
20 V −t/τ
e = 20 μA e−t/5.1s
1MΩ
υR = Ee−t/τ = 20V e−t/5.1s
iC =
E −t/τ 100 V −t/τ
e = 18.18 mAe−t/5.5 ms
e =
5.5 kΩ
RT
3.3 kΩ(100 V)
= 60 V
VR2 =
3.3 kΩ + 2.2 kΩ
υR = υ R2 = 60 Ve−t/5.5 ms
CHAPTER 10
e.
24.
a.
τ = RC = (56 kΩ)(0.1 μF) = 5.6 ms
c.
iC =
b. υC = E(1 − e−t/τ) = 25 V(1 − e−t/5.6ms)
E −t/τ
25 V −t/τ
e = 0.45 mAe−t/5.6ms
e =
56 k Ω
R
d.
25.
a.
b.
5 ms
υC = 60 mV(1 − e−2ms/5ms) = 60 mV(1 − e−0.4) = 60 mV(1 − 0.670)
υC = 60 mV(1 − e−100ms/5ms) = 60 mV(1 − e−20) = 60 mV(1 − 2.06 × 10−9)
= 60 mV(0.330) = 19.8 mV
c.
26.
a.
b.
c.
27.
≅ 60 mV(1) = 60 mV
τ = 40 ms, 5τ = 5(40 ms) = 200 ms
τ 40 ms
= 4 kΩ
τ = RC, R = =
C 10 μF
υC (20 ms) = 12 V(1 − e−20 ms/40ms) = 12 V(1 − e−0.5)
= 12 V(1 − 0.607) = 12 V(0.393) = 4.72 V
d.
e.
f.
υC = 12 V(1 − e−10) = 12 V(1 − 45 × 10−6) ≅ 12.0 V
Q = CV = (10 μF)(12 V) = 120 μC
τ = RC = (1000 × 106 Ω)(10 μF) = 10 × 103 s
a.
τ = RC = (2 kΩ)(100 μF) = 200 ms
b.
⎡1 min ⎤ ⎡ 1 h ⎤
5τ = 50 × 103 s ⎢
⎥ = 13.89 h
⎥⎢
⎣ 60 s ⎦ ⎣ 60 min ⎦
υC = E(1− e−t/τ) = 8 V(1 − e−t/200ms)
iC =
c.
8 V −t / 200 ms
E −t / τ
e
e
=
= 4 mAe−t/200ms
2 kΩ
R
υC(1 s) = 8 V(1 − e−1s/200ms) = 8 V(1 − e−5)
= 8 V(1 − 6.738 × 10−3) = 8 V(0.9933) = 7.95 V
iC (1 s) = 4 mAe−5 = 4 mA(6.738 × 10−3) = 26.95 μA
CHAPTER 10
109
d.
υC = 7.95 Ve−t/200ms
iC =
7.95 V − t / 200ms
e
= 3.98 mAe−t/200ms
2 kΩ
e.
28.
a.
τ = RC = (3 kΩ + 2 kΩ)(2 μF) = 10 ms
υC = 50 V(1 − e−t/10ms)
iC =
υ R1
b.
c.
50 V −t / 10ms
e
= 10 mA−t/10ms
5 kΩ
= iC ⋅ R1 = (10 mA)(3 kΩ)e−t/10ms = 30 Ve−t/10ms
100ms: e−10 = 45.4 × 10−6
υC = 50 V(1 − 45.4 × 10−6) = 50 V
iC = 10 mA(45.4 × 10−6) = 0.45 μA
υ R1 = 30 V(45.4 × 10−6) = 1.36 mV
200 ms: τ′ = R2C = (2 kΩ)(2 μF) = 4 ms
υC = 50 Ve−t/4ms
50 V −t / 4 ms
iC = −
= −25 mAe−t/4ms
e
2 kΩ
υ R2 = υC = −50 Ve−t/4ms
d.
110
CHAPTER 10
29.
a.
τ = RC = (5 kΩ)(20 μF) = 100 ms
υC = 50 V(1 − e−t/100ms)
iC =
υ R1
b.
c.
50 V −t / 100ms
e
= 10 mAe−t/100ms
5 kΩ
= iC ⋅ R1 = (10 mA)(3 kΩ) e−t/100ms = 30 Ve−t/100ms
100 ms: e−1 = 0.368
υC = 50 V(1 − 0.368) = 50 V(0.632) = 31.6 V
iC = 10 mA(0.368) = 3.68 mA
υ R1 = 30 V(0.368) = 11.04 V
200 ms: τ′ = R2C = (2 kΩ)(20 μF) = 40 ms
υC = 31.6 Ve−t/40ms
31.6 V −t/ 40ms
e
= −15.8 mAe−t/40ms
2 kΩ
= −υC = −31.6 Ve−t/40ms
iC = −
υ R2
d.
CHAPTER 10
111
30.
a.
τ = R1C = (105 Ω)(10 pF) = 1 μs
υC = 80 V 1 − e − t/1μs
(
iC =
b.
υC = 80 Ve−t/τ′ = 80 Ve − t/4.9×10
c.
112
80 V −t / τ
= 0.8 mAe − t/1 μs
e
100 kΩ
τ′ = R′C = (490 kΩ)(10 pF) = 4.9 μs
iC =
31.
)
−6
−6
80 V −t/τ ′
e = 0.16 mAe − t/4.9×10
490 kΩ
a.
τ = RC = (2 mΩ)(1000 μF) = 2 μs
5τ = 10 μs
b.
Im =
c.
yes
6V
V
= 3 kA
=
R 2 mΩ
CHAPTER 10
32.
a.
b.
υC = Vf + (Vi − Vf)e−t/τ
τ = RC = (10 kΩ)(2 μF) = 20 ms, Vf = 50 V, Vi = −10 V
υC = 50 V + (−10 V − (+50 V))e−t/20ms
υC = 50 V − 60 Ve−t/20ms
Initially VR = E + υC = 50 V + 10 V = 60 V
60 V − t / 20ms
V
e
iC = R e −t / τ =
= 6 mA e−t/20ms
10 kΩ
R
c.
33.
τ = RC = (2.2 kΩ)(2000 μF) = 4.4 s
υC = VCe−t/τ = 40 Ve−t/4.4 s
VC −t /τ
40 V −t / 4.4 s
e
e
=
= 18.18 mAe−t/4.4 s
2.2 kΩ
R
υR = υC = 40 Ve−t/4.4s
IC =
34.
υC = Vf + (Vi −Vf)e−t/τ
τ = RC = (860 Ω)(4000 pF) = 3.44 μs, Vf = −30 V, Vi = 20 V
υC = −30 V + (20 V − (−30 V))e−t/3.44μs
υC = −30 V + 50 Ve−t/3.44μs
20 V + 30 V
= 58.14 mA
860 Ω
iC = −58.14 mAe−t/3.44μs
Im =
CHAPTER 10
113
35.
τ = RC = (18.2 kΩ)(6.8 μF) = 123.8 ms
υC = Vf + (Vi − Vf) e−t/τ
a.
= 52 V + (12 V − 52 V)e−t/123.8 ms
υC = 52 V − 40 Ve−t/123.8 ms
υR(0+) = 52 V − 12 V = 40 V
40 V −t/123.8 ms
iC =
e
18.2 kΩ
= 2.20 mAe−t/123.8 ms
b.
36.
a.
υC = 12 V(1 − e−10μs/20 μs) = 12 V(1 − e−0.5) = 12 V(1 − 0.607)
= 12 V(0.393) = 4.72 V
b.
c.
114
υC = 12 V(1 − e−10τ/τ) = 12 V(1 − e−10) = 12 V(1 − 45.4 × 10−6)
≅ 12 V
6 V = 12 V(1 − e−t/20 μs)
0.5 = 1 − e−t/20 μs
−0.5 = −e−t/20 μs
0.5 = e−t/20 μs
loge 0.5 = loge e−t/20 μs
−0.693 = −t/20 μs
t = 0.693 (20 μs) = 13.86 μs
CHAPTER 10
d.
37.
υC = 11.98 V = 12 V(1 − et/20 μs)
0.998 = 1 − e−t/20 μs
−0.002 = −e−t/20 μs
0.002 = −e−t/20 μs
loge 0.002 = −t/20 μs
−6.215 = −t/20 μs
t = (6.215)(20 μs) = 124.3 μs
τ = RC = (33 kΩ)(20 μF) = 0.66 s
υC = 12 V(1 − e−t/0.66 s)
8 V = 12 V(1 − e−t/0.66 s)
8 V = 12 V − 12 Ve−t/0.66 s
−4 V = −12 Ve−t/0.66 s
0.333 = e−t/0.66 s
loge 0.333 = −t/0.66 s
−1.0996 = −t/0.66 s
t = 1.0996(0.66 s) = 0.73 s
⎛ υ ⎞
t = −τ loge ⎜1 − C ⎟
E⎠
⎝
⎛ 12 V ⎞
10 s = −τ loge ⎜1 −
⎟
⎝ 20 V ⎠
38.
.4
τ=
−916.29 × 10−3
10 s
= 10.92 s
0.916
τ 10.92 s
τ = RC ⇒ R = =
= 54.60 kΩ
C 200 μ F
39.
a.
τ = (R1 + R2)C = (20 kΩ)(6 μF) = 0.12 s
υC = E(1 − e−t/τ)
60 V = 80 V(1 − e−t/0.12s)
0.75 = 1 − e−t/0.12s
0.25 = e−t/0.12s
t = −(0.12 s)(−1.39)
= 166.80 ms
b.
E −t / τ
e
R
166.80 ms
80 V − 0.12s
e
= 4 mAe−1.39
iC =
20 kΩ
= (4 mA)(249.08 × 10−3)
≅ 1 mA
iC =
CHAPTER 10
115
c.
40.
a.
is = iC = 4 mAe−t/τ = 4 mAe−2τ/τ = 4 mAe−2
= 4 mA(135.34 × 10−3)
= 0.54 mA
Ps = EIs = (80 V)(0.54 mA)
= 43.20 mW
τ = RC = (1 MΩ)(0.2 μF) = 0.2 s
υC = 60 V(1 − e−t/0.2s)
E − t / τ 60 V −t / 0.2s
=
e
e
= 60 μAe−t/0.2s
1 MΩ
R
υ R1 = Ee−t/τ = 60 Ve−t/0.2s
iC =
υC: 0.5 s = 55.07 V
1 s = 59.58 V
iC: 0.5 s = 4.93 V
1 s = 0.40 V
b.
8 μA
0.667
loge 0.667
−0.41
t
= 12 μAe−t
= e−t
= −t
= −t
= 0.41 s
τ′ = RC = (1 MΩ + 4 MΩ)(0.2 μF)
= (5 MΩ)(0.2 μF)
=1s
60 V −t
e = 12 μAe−t
iC =
5 MΩ
υC = 60 Ve−tτ′
10 V = 60 Ve−t
0.167 = e−t
loge 0.167 = −t
−1.79 = −t
t = 1.79 s
Longer = 1.79 s − 0.41 s = 1.38 s
41.
a.
υm = υR = Ee−t/τ = 60 Ve−1τ/τ = 60 Ve−1
= 60 V(0.3679)
= 22.07 V
b.
116
E −t / τ
60 V − 2τ / τ
= 6 μAe−2
e
=
e
10 MΩ
R
= 6 μA(0.1353)
= 0.81 μA
iC =
CHAPTER 10
42.
υC = E(1 − e−t/τ)
c.
50 V = 60 V(1 − e )
0.8333 = 1 − e−t/2 s
loge 0.1667 = −t/2 s
t = −(2 s)(−1.792)
= 3.58 s
a.
Thevenin’s theorem:
RTh:
−t/2 s
τ = RC = (10 MΩ)(0.2 μF) = 2 s
ETh:
←RTh = 8 kΩ || 24 kΩ
= 6 kΩ
ETh =
24 kΩ(20 V)
= 15 V
24 kΩ + 8 kΩ
τ = RC = (10 kΩ)(15 μF) = 0.15 s
υC = E(1 − e−t/τ)
= 15 V(1 − e−t/0.15 s)
iC =
E −t / τ
15 V −t / 0.15
e
e
= 1.5 mAe−t/0.15 s
=
10 kΩ
R
b.
43.
a.
Source conversion and combining series resistors:
τ = RC = (8.3 kΩ)(2.2 μF) = 18.26 ms
υC = Vf + (Vi − Vf)e−t/τ
= 27.2 V + (2 V − 27.2 V)e−t/18.26 ms
υC = 27.2 V − 25.2 Ve−t/18.26 ms
υR(0+) = 27.2 V − 2 V = 25.2 V
25.2 V −t / 18.26 ms
iC =
e
8.3 kΩ
iC = 3.04 mAe−t/18.26 ms
b.
CHAPTER 10
117
44.
a.
RTh = 3.9 kΩ + 0 Ω || 1.8 kΩ = 3.9 kΩ
ETh = 36 V
τ = RC = (3.9 kΩ)(20 μF) = 78 ms
υC = Vf + (Vi − Vf)e−t/τ
= 36 V + (−4 V − 36 V)e−t/78 ms
υC = 36 V − 40 Ve−t/78 ms
υR(0+) = 36 V + 4 V = 40 V
40 V − t/78 ms
e
iC =
3.9 kΩ
iC = 10.26 mAe−t/78 ms
b.
45.
Source conversion:
E = IR1 = (5 mA)(0.56 kΩ) = 2.8 V
R′ = R1 + R2 = 0.56 kΩ + 3.9 kΩ = 4.46 kΩ
RTh = 4.46 kΩ || 6.8 kΩ = 2.69 kΩ
4 V − 2.8 V
1.2 V
I=
=
= 0.107 mA
6.8 kΩ + 4.46 kΩ 11.26 kΩ
ETh = 4 V − (0.107 mA)(6.8 kΩ)
= 4 V − 0.727 V
= 3.27 V
υC = 3.27 V(1 − e−t/τ)
τ = RC = (2.69 kΩ)(20 μF)
= 53.80 ms
υC = 3.27 V(1 − e−t/53.80 ms)
3.27 V −t /τ
e
iC =
2.69 kΩ
= 1.22 mA e−t/53.80 ms
118
CHAPTER 10
46.
τ = RC = (6.8 kΩ)(39 μF) = 265.2 ms
υC = Vf + (Vi − Vf)e−t/τ
a.
= 20 V + (3 V − 20 V)e−t/265.2 ms
υC = 20 V − 17 Ve−t/265.2 ms
υR(0 +) = 20 V − 3 V = 17 V
17 V −t / 265.2 ms
e
iC =
6.8 kΩ
iC = 2.5 mAe−t/265.2 ms
b.
47.
a.
τ = RThC = (1.67 MΩ)(1 μF) = 1.67 s
RTh = 2 MΩ || 10 MΩ = 1.67 MΩ
10 MΩ(24 V)
= 20 V
ETh =
10 MΩ + 2 MΩ
υC = ETh(1 − e−t/τ)
= 20 V(1 − e−4τ/τ)
= 20 V(1 − e−4)
= 20 V(1 − 0.0183)
= 19.63 V
E −t / τ
e
R
20 V −t /1.67s
e
3 μA =
1.67 MΩ
0.25 = e−t/1.67s
loge 0.25 = −t/1.67 s
t = −(1.67 s)(−1.39)
= 2.32 s
iC =
c.
υmeter = υC
υC = ETh(1 − e−t/τ)
10 V = 20 V(1 − e−t/1.67s)
0.5 = 1 − e−t/1.67s
−0.5 = −e−t/1.67s
loge 0.5 = −t/1.67 s
t = −(1.67 s)(−0.69)
= 1.15 s
CHAPTER 10
119
48.
iCao = C
ΔυC
Δt
0 → 4 ms: iC = 2 × 10−6
(20 V)
= 10 mA
4 ms
(0 V)
= 0 mA
4 → 6 ms: iC = 2 × 10−6
2 ms
(20 V)
6 → 7 ms: iC = 2 × 10−6
= 40 mA
1 ms
(0 V)
7 → 9 ms: iC = 2 × 10−6
= 0 mA
2 ms
(40 V)
= −40 mA
9→ 11 ms: iC = −2 × 10−6
2 ms
49.
iCao = C
ΔυC
Δt
( −5 V )
= −1.18 A
20 μs
(0 V )
=0A
20 → 30 μs: iC = 4.7 μF
10 μs
(+10 V )
30 → 50 μs: iC = 4.7 μF
= +2.35 A
20 μs
(−15 V )
50 → 80 μs: iC = 4.7 μF
= −2.35 A
30 μs
(+10 V )
80 → 90 μs: iC = 4.7 μF
= +4.7 A
10 μs
(0 V )
=0A
90 μs → 100 μs: iC = 4.7 μF
10 μs
0 → 20 μs: iC = 4.7 μF
120
CHAPTER 10
Δt
ΔυC
⇒ ΔυC = (iC )
C
Δt
0 → 2 ms: iC = 0 mA ΔυC = 0 V
(2 ms)
2 → 6 ms: iC = −80 mA ΔυC =
(−80 mA) = −8 V
20 μF
(10 ms)
6 → 16 ms: iC = +40 mA ΔυC =
(40 mA) = +20 V
20 μF
16 → 18 ms: iC = 0 mA ΔυC = 0 V
(2 ms)
(−120 mA) = −12 V
18 → 20 ms: iC = −120 mA ΔυC =
20 μF
20 → 25 ms: iC = 0 mA ΔυC = 0 V
50.
iC = C
51.
CT = 6 μF + 4 μF + 3 μF || 6 μF = 10 μF + 2 μF = 12 μF
52.
CT′ = 6 μF || 12 μF = 4 μF
CT′′ = CT′ + 12 μF = 4 μF + 12 μF = 16 μF
6 μF ⋅ CT′′ (6 μF)(16 μF)
CT = 6 μF || CT′′ =
=
= 4.36 μF
6 μF + CT′′
6 μF + 16 μF
CHAPTER 10
121
53.
54.
V1 = 10 V, Q1 = V1C1 = (10 V)(6 μF) = 60 μC
CT = 6 μF || 12 μF = 4 μF, QT = CTE = (4 μF)(10 V) = 40 μC
Q2 = Q3 = 40 μC
Q
40 μ C
= 6.67 V
V2 = 2 =
C2
6μ F
Q
40μ C
= 3.33 V
V3 = 3 =
C3 12 μ F
CT = 1200 pF || (200 pF + 400 pF) || 600 pF
= 1200 pF || 600 pF || 600 pF = 1200 pF || 300 pF
= 240 pF
QT = CTE = (240 pF)(40 V) = 9.60 nC
Q1 = Q4 = QT = 9.60 nC
Q 9.60 nC
Q
9.60 nC
= 8.00 V, V4 = 4 =
= 16.00 V
V1 = 1 =
C1 1200 pF
C4 600 pF
V2 = V3 = E − V1 − V4 = 40 V − 8 V − 16 V = 16 V
Q2 = C2V2 = (200 pF)(16 V) = 3.20 nC, Q3 = C3V3 = (400 pF)(16 V) = 6.40 nC
(9 μF)(72 μF)
= 8 μF
9 μF + 72 μF
8 μF + 10 μF = 18 μF
(9 μF)(18 μF)
= 6 μF
9 μF + 18 μF
55.
CT =
56.
122
Q Q
=
⇒ Q = CTE = (6 μF)(24 V) = 144 μC
V E
Q1 = 144 μC
Q 144 μC
V1 = 1 =
= 16 V
C1
9 μF
V2 = E − V1 = 24 V − 16 V = 8 V
Q2 = C2V2 = 10 μF(8 V) = 80 μC
Q3−4 = C′V = (8 μF)(8 V) = 64 μC
Q3 = Q4 = 64 μC
Q
64 μC
V3 = 3 =
= 7.11 V
C3
9 μF
Q
64 μC
= 0.89 V
V4 = 4 =
C4 72 μF
4 kΩ(48 V)
= 32 V = V0.08μF
4 kΩ + 2 kΩ
Q0.08μF = (0.08 μF)(32 V) = 2.56 μC
V0.04μF = 48 V
Q0.04μF = (0.04 μF)(48 V) = 1.92 μC
V4kΩ =
CHAPTER 10
1
1
CV 2 = (120 pF)(12 V)2 = 8,640 pJ
2
2
57.
WC =
58.
W=
Q2
⇒Q=
2C
59.
a.
V6μF = V12μF =
b.
60.
2CW = 2(6 μF)(1200 J) = 0.12 C
3 kΩ(24 V)
=8V
3 kΩ + 6 kΩ
1
1
W6μF = CV 2 = (6 μF)(8 V)2 = 0.19 mJ
2
2
1
1
W12μF = CV 2 = (12 μF)(8 V)2 = 0.38 mJ
2
2
(6 μF)(12 μF)
= 4 μF
6 μF + 1 2 μF
QT = CTV = (4 μF)(8 V) = 32 μC
Q6μF = Q12μF = 32 μC
Q 32 μC
=
V6μF =
= 5.33 V
C
6 μF
Q 32 μC
=
= 2.67 V
V12μF =
C 12 μF
1
1
W6μF = CV 2 = (6 μF)(5.33 V)2 = 85.23 μJ
2
2
1
1
W12μF = CV 2 = (12 μF)(2.67 V)2 = 42.77 μJ
2
2
CT =
1
1
CV 2 = (1000 μF)(100 V)2 = 5 pJ
2
2
a.
WC =
b.
Q = CV = (1000 μF)(100 V) = 0.1 C
c.
I = Q/t = 0.1 C/(1/2000) = 200 A
d.
P = VavIav = W/t = 5 J(1/2000 s) = 10,000 W
e.
t = Q/I = 0.1 C/10 mA = 10 s
CHAPTER 10
123
Chapter 11
1.
a.
b.
c.
d.
2.
A=
L=
3.
Φ 4 × 10−4 Wb
=
= 4 × 10−2 Wb/m2 = 0.04 Wb/m2
2
A
0.01 m
0.04 T
F = NI = (40 t)(2.2 A) = 88 At
⎡104 gauss ⎤
3
0.04 T ⎢
⎥ = 0.4 × 10 gauss
1
T
⎣
⎦
B=
πd 2
4
N 2μ A
L = N2
5.
L=
a.
b.
c.
d.
a.
b.
124
π (5 mm)2
= 19.63 × 10−6 m2
4
(200 t) 2 (4π × 10−7 )(19.63 × 10−6 m 2 )
=
= 9.87 μH
100 mm
⎡ 1m ⎤
d = 0.2 in. ⎢
⎥ = 5.08 mm
⎣ 39.37 in. ⎦
π d 2 (π )(5.08 mm) 2
A=
=
= 20.27 × 10−6 m2
4
4
⎛ 1m ⎞
= 1.6 in. ⎜
⎟ = 40.64 mm
⎝ 39.37 in. ⎠
N 2 μ r μo A (200 t) 2 (500)(4π × 10−7 )(20.27 × 10−6 m 2 )
L=
= 12.54 mH
=
40.64 mm
4.
6.
=
μ r μo
=
(200 t) 2 (1000)(4π × 10−7 )(1.5 × 10−4 m 2 )
= 50.27 mH
0.15 m
N 2 μ r μo A
L′ = (3)2Lo = 9Lo = 9(5 mH) = 45 mH
1
1
L′ = Lo = (5 mH) = 1.67 mH
3
3
2
(2)(2)
L′ =
Lo = 16 (5 mH) = 80 mH
1
2
2
⎛1⎞ 1
⎜ ⎟ (1500) Lo
2 2
= 375(5 mH) = 1875 mH
L′ = ⎝ ⎠
1
2
12 × 103 μH ± 5% ⇒ 12,000 μH ± 600 μH ⇒ 11,400 μH →12,600 μH
47 μH ± 10% ⇒ 47 μH ± 4.7 μH ⇒ 42.3 μF →51.7 μF
CHAPTER 11
7.
e= N
8.
e= N
9.
10.
dφ
= (50 t)(120 mWb/s) = 6.0 V
dt
dφ
dφ e 20 V
⇒
= =
= 100 mWb/s
dt
dt N 200 t
dφ
⇒N=
e= N
dt
a.
b.
⎛
⎞
⎜ 1 ⎟
1
⎛
⎞
e⎜
⎟ = 42 mV ⎜
⎟ = 14 turns
φ
d
3
m
Wb/s
⎝
⎠
⎜⎜
⎟⎟
⎝ dt ⎠
diL
= (5 H)(1 A/s) = 5 V
dt
di
e = L L = (5 H)(60 mA/s) = 0.3 V
dt
diL
⎡ 0.5 A ⎤ ⎡1000 ms ⎤
e= L
= (5 H) ⎢
⎥⎢
⎥ = 2.5 kV
dt
⎣ ms ⎦ ⎣ 1 s ⎦
e= L
⎛ 0.1 mA ⎞
diL
⎟⎟ = 5 V
= (50 mH) ⎜⎜
dt
⎝ μs ⎠
11.
e= L
12.
a.
τ=
b.
iL =
c.
υL = Ee−t/τ = 40 mVe−t/12.5 μs
υR = iRR = iLR = E(1 − e−t/τ) = 40 mV(1 − e−t/12.5 μs)
d.
L 250 mH
=
= 12.5 μs
R
20 kΩ
E
40 mV
(1 − e −t/τ ) =
(1 − e−t/τ)
20 kΩ
R
= 2 μA(1 − e−t/12.5μs)
iL: 1τ = 1.26 μA, , 3τ = 1.9 μA, 5τ = 1.99 μA
υL: 1τ = 14.72 V, 3τ = 1.99 V, 5τ = 0.27 V
e.
CHAPTER 11
125
13.
a.
τ=
L
5 mH
=
= 2.27 μs
R 2.2 kΩ
b.
iL =
E
12 V
(1 − e −t/τ ) =
(1 − e−t/τ) = 5.45 mA(1 − e−t/2.27 μs)
2.2 kΩ
R
c.
υL = Ee−t/τ = 12 Ve−t/2.27 μs
υR = iRR = iLR = E(1 − e−t/τ) = 12 V(1 − e−t/2.27 μs)
d.
iL: 1τ = 3.45 mA, , 3τ = 5.18 mA, 5τ = 5.41 mA
υL: 1τ = 4.42 V, 3τ = 0.60 V, 5τ = 0.08 V
e.
14.
a.
τ=
b.
υL = −E(1 − e−t/τ) = −12 V(1 − e−t/66.67ms)
L
2H
2H
=
=
= 66.67 ms
R (20 Ω + 10 Ω) 30 Ω
iL = −
E − t/τ
12 V −t/ 66.67ms
e =−
e
= −400 mAe−t/66.67 ms
30 Ω
R
c.
15.
126
a.
iL = If + (Ii − If)e−t/τ
36 V
E
L 120 mH
Ii = 8 mA, If =
= 9.23 mA, τ =
= 30.77 μs
=
=
R 3.9 kΩ
R 3.9 kΩ
iL = 9.23 mA + (8 mA − 9.23 mA)e−t/30.77 μs
iL = 9.23 mA − 1.23 mAe−t/30.77 μs
+E − υ L − υ R = 0 and υ L = E − υ R
υ R = iRR = iLR = (8 mA)(3.9 kΩ) = 31.2 V
υ L = E − υ R = 36 V − 31.2 V = 4.8 V
υL = 4.8 Ve−t/30.77 μs
CHAPTER 11
b.
16.
a.
Ii = −8 mA, If = 9.23 mA, τ =
L 120 mH
= 30.77 μs
=
R 3.9 kΩ
iL = If + (Ii − If)e−t/τ
= 9.23 mA + (−8 mA − 9.23 mA)e−t/30.77 μs
iL = 9.23 mA − 17.23 mA e−t/30.77 μs
+E − υ L − υ R = 0 (at t = 0−)
but, υ R = iRR = −iLR = (−8 mA)(3.9 kΩ) = −31.2 V
υ L = E − υ R = 36 V − (−31.2 V) = 67.2 V
υL = 67.2 V e−t/30.77 μs
b.
17.
c.
Final levels are the same. Transition period defined by 5τ is also the same.
a.
Source conversion:
τ=
2H
L
= 588.2 μs
=
R 3.4 kΩ
iL = If + (Ii − If)e−t/τ
6V
If =
= 1.76 mA
3.4 kΩ
iL = 1.76 mA + (−3 mA − 1.76 mA)e−t/588.2μs
iL = 1.76 mA − 4.76 mA e−t/588.2μs
υR(0 +) = 3 mA(3.4 kΩ) = 10.2 V
KVL: +6 V + 10.2 V − υL(0+) = 0
υL(0+) = 16.2 V
υL = 16.2 Ve−t/588.2μs
CHAPTER 11
127
b.
18.
If = −
7.2 V
= −0.69 mA
10.4 kΩ
L
200 mH
=
= 19.23 μs
τ=
10.4 kΩ
R
iL = If + (Ii − If )e−t/τ
= −0.69 mA + (−3 mA − (−0.69 mA))e−t/19.23 μs
iL = −0.69 mA − 2.31 mAe−t/19.23 μs
a.
KVL: −7.2 V + 31.2 V − υL(0+) = 0
υL(0+) = 24 V
υL = 24 Ve-t/19.23 μs
b.
19.
a.
τ=
υL = 20 Ve−t/1μs, iL =
128
L
10 mH
=
= 1 μs
R
10 kΩ
E
(1 − e −t/τ ) = 2 mA(1 − e−t/1μs)
R
CHAPTER 11
b.
20.
a.
b.
5τ ⇒ steady state
L 10 mH
τ′ = =
= 1 μs
R 10 kΩ
iL = Ime−t/τ′ = 2 mAe−t/1μs
υL = −(2 mA)(20 kΩ)e−t/τ = −40 Ve−t/1μs
τ=
L 1 mH
= 0.5 μs
=
R 2 kΩ
E
12 V
(1 − e −t / τ ) = 6 mA(1 − e−t/0.5μs)
iL = (1 − e− t / τ ) =
R
2 kΩ
υL = Ee−t/τ = 12 V e−t/0.5μs
iL = 6 mA(1 − e−t/0.5μs) = 6 mA(1 − e−1μs/0.5μs)
= 6 mA(1 − e−2) = 5.19 mA
L 1 mH
= 83.3 ns
τ′ = =
iL = I′me−t/τ′
R 12 kΩ
iL = 5.19 mAe−t/83.3ns
t = 1 μs: υL = 12 Ve−t/0.5μs = 12 Ve−2 = 12 V(0.1353) = 1.62 V
VL′ = (5.19 mA)(12 kΩ) = 62.28 V
υL = −62.28 Ve−t/83.3ns
c.
CHAPTER 11
129
21.
a.
RTh = 6.8 kΩ
ETh = 6 V
τ=
L
5 mH
= 0.74 μs
=
R 6.8 kΩ
E
6V
(1 − e −t/τ ) =
(1 − e − t/τ ) = 0.88 mA(1 − e−t/0.74μs)
6.8 kΩ
R
υL = Ee−t/τ = 6 Ve−t/0.74μs
iL =
b.
Assume steady state and IL = 0.88 mA
τ′ =
L 5 mH
=
= 0.33 μs
R 15 kΩ
iL = Ime−t/τ′ = 0.88 mA e−t/0.33μs
υL = −Vme−t/τ′
Vm = ImR = (0.88 mA)(15 kΩ) = 13.23 V
υL = −13.23 Ve−t/0.33μs
130
CHAPTER 11
c.
d.
22.
a.
b.
VR2 max = ImR2 = (0.88 mA)(8.2 kΩ) = 7.22 V
RTh = 2 kΩ + 3 kΩ || 6 kΩ = 2 kΩ + 2 kΩ = 4 kΩ
6 kΩ(12 V)
L 100 mH
= 8 V, τ = =
= 25 μs
ETh =
4 kΩ
R
6 kΩ + 3 kΩ
E
8V
= 2 mA
If = Th =
RTh 4 kΩ
iL = 2 mA(1 − e−t/25μs)
υL = 8 Ve−t/25μs
iL = 2 mA(1 − e−1) = 1.26 mA
υL = 8 Ve−1 = 2.94 V
CHAPTER 11
131
23.
a.
Source conversion: E = IR = (4 mA)(12 kΩ) = 48 V
τ=
iL =
L 2 mH
= 55.56 ns
=
R 36 kΩ
E
48 V
(1 − e − t/τ ) =
(1 − e − t/τ ) = 1.33 mA(1 − e−t/55.56ns)
R
36 kΩ
υL = Ee−t/τ = 48 Ve−t/55.56ns
b.
t = 100 ns:
iL = 1.33 mA(1 − e−100ns/55.56ns) = 1.33 mA(1 − e−1.8) = 1.11 mA
υL = 48 Ve−1.8 = 7.93 V
24.
0.165
RTh = 2.2 kΩ || 4.7 kΩ = 1.50 kΩ
4.7 kΩ(8 V)
ETh =
= 5.45 V
4.7 kΩ + 2.2 kΩ
L 10 mH
τ= =
= 6.67 μs
R 1.50 kΩ
a.
E
5.45 V
(1 − e −t/τ ) =
(1 − e −t/τ ) = 3.63 mA(1 − e−t/6.67μs)
1.5 kΩ
R
υL = Ee−t/τ = 5.45 Ve−t /6.67μs
iL =
b.
c.
132
t = 10 μs:
iL = 3.63 mA(1 − e−10μs/6.67μs) = 3.63 mA(1 − e−1.4)
= 2.74 mA
υL = 5.45 V(0.246) = 1.34 V
L 10 mH
τ′ = =
= 2.13 μs
R 4.7 kΩ
iL = 2.74 mAe−t/2.13μs
At t = 10 μs
VL = (2.74 mA)(4.7 kΩ) = 12.88 V
υL = −12.88 Ve−t/2.13μs
0.246
CHAPTER 11
25.
a.
υL = Ee−t/τ
τ=
b.
υL = 36 Ve−1 ms/5 ms = 36 Ve−0.2 = 36 V(0.819) = 29.47 V
c.
υ R1 = iR1 R1 = iL R1 = ⎜
L
0.6 H
0 .6 H
= 5 ms
=
=
R1 + R3 100 Ω + 20 Ω 120 Ω
υL = 36 Ve−t/5 ms
υL = 36 Ve−25 ms/5 ms = 36 Ve−5 = 36 V(0.00674) = 0.24 V
⎛
⎞
E
(1 − e −t/τ ) ⎟ R1
⎝ R1 + R3
⎠
⎛ 36 V
⎞
=⎜
(1 − e −t/5ms ) ⎟ 100 Ω
⎝ 120 Ω
⎠
= (300 mA(1 − e−t/5 ms))100 Ω
= 30 V(1 − e−5 ms/5 ms) = 30 V(1 − e−1)
= 30 V(1 − 0.368) = 18.96 V
d.
26.
a.
iL = 300 mA(1 − e−t/5 ms)
100 mA = 300 mA(1 − e−t/5 ms)
0.333 = 1 − e−t/5 ms
0.667 = e−t/5 ms
loge 0.667 = −t/5 ms
0.405 = t/5 ms
t = 0.405(5 ms) = 2.03 ms
16 V
= 2 mA
4.7 kΩ + 3.3 kΩ
t = 0 s: Thevenin:
RTh = 3.3 kΩ + 1 kΩ || 4.7 kΩ = 3.3 kΩ + 0.82 kΩ = 4.12 kΩ
1 kΩ(16 V)
ETh =
= 2.81 V
1 k Ω + 4 .7 k Ω
iL = If + (Ii − If)e−t/τ
Ii =
CHAPTER 11
133
2.81 V
L
2H
= 0.49 ms
= 0.68 mA, τ = =
4.12 kΩ
R 4.12 kΩ
iL = 0.68 mA + (2 mA − 0.68 mA)e−t/0.49 ms
iL = 0.68 mA + 1.32 mAe−t/0.49 ms
υR(0+) = 2 mA(4.12 kΩ) = 8.24 V
KVL(0+): 2.81 V − 8.24 V − υL = 0
υL = −5.43 V
υL = −5.43 Ve−t/0.49 ms
If =
b.
27.
a.
Redrawn:
Source conversions:
IT = 5 mA − 2 mA = 3 mA↑
1
1
1
1
1
=
+
+
+
RT 12 kΩ 3 kΩ 4 kΩ 1.5 kΩ
and RT = 0.75 kΩ
Source conversion:
ET = ITRT = (3 mA)(0.75 kΩ) = 2.25 V
τ=
L
5 mH
=
= 6.67 μs
R 0.75 kΩ
2.25 V
(1 − e−t/τ) = 3 mA(1 − e−t/6.67 μs)
0.75 kΩ
υL = 2.25 Ve−t/6.67 μs
iL =
134
CHAPTER 11
b.
2τ: 0.865 Im, 0.135 Vm
iL: 0.865(3 mA) = 2.60 mA
υL: 0.135(2.25 V) = 0.30 V
τ′ =
L 5 mH
=
= 3.33 μs
R 1.5 k Ω
iL = 2.60 mA e−t/3.33 μs
iL(0+) = 2.60 mA
υR(0+) = (2.60 mA)(1.5 kΩ) = 3.90 V
υL = −3.90 Ve−t/3.33 μs
c.
d.
28.
a.
RTh = 2 MΩ || 10 MΩ = 1.67 MΩ
10 MΩ(24 V)
= 20 V
ETh =
10 MΩ + 2 MΩ
I L (0− ) =
ETh
20 V
= 12 μA
=
RTh 1.67 M Ω
5H
L
τ′ =
=
= 5μ s
Rmeter 10 MΩ
iL = 12 μAe−t/5 μs
10 μA = 12 μAe−t/5 μs
0.833 = e−t/5 μs
loge 0.833 = −t/5 μs
CHAPTER 11
135
0.183 = t/5 μs
t = 0.183(5 μs) = 0.92 μs
b.
29.
υL (0+) = iL(0+)Rm = (12 μA)(10 MΩ) = 120 V
υL = 120 Ve−t/5μs = 120 Ve−10μs/5μs = 120 Ve−2 = 120 V(0.135) = 16.2 V
c.
υL = 120 Ve−5τ/τ = 120 Ve−5 = 120 V(6.74 × 10−3) = 0.81 V
a.
Ii = −
24 V
= −10.91 mA
2.2 k Ω
24 V
24 V
Switch open: If = −
= −
= −3.48 mA
2.2 k Ω + 4.7 k Ω
6.9 k Ω
iL = If + (Ii − If)e−t/τ
L
1.2 H
τ=
=
= 173.9 μs
6.9 k Ω
R
iL = −3.48 mA + (−10.91 mA − (−3.48 mA))e−t/173.9 μs
iL = −3.48 mA − 7.43 mAe−t/173.9 μs
υR(0+) = (10.91 mA)(6.9 kΩ) = 75.28 V
t = 0+:
KVL: −24 V + 75.28 V − υL = 0
υL = 51.28 V
υL = 51.28 Ve-t/173.9 μs
30.
a.
b.
c.
136
iL = 100 mA(1 − e−1ms/20ms) = 100 mA(1 − e−1/20)
= 100 mA(1 − e−0.05) = 100 mA(1 − 951.23 ×10−3) = 100 mA(48.77 × 10−3)
= 4.88 mA
iL = 100 mA(1 − e−100ms/20ms) = 100 mA(1 − e−5)
= 99.33 mA
50 mA = 100 mA(1 − e−t/τ)
0.5 = 1 − e−t/τ
−0.5 = −e−t/τ
0.5 = e−t/τ
loge 0.5 = −t/τ
t = −(τ)(loge 0.5) = −(20 ms)(loge 0.5) = −(20 ms)(−693.15 × 10−3)
= 13.86 ms
CHAPTER 11
d.
31.
a.
99 mA = 100 mA(1 − e−t/20 ms)
0.99 = 1 − e−t/20ms
−0.01 = −e−t/20ms
0.01 = e−t/20ms
loge 0.01 = −t/20 ms
t = −(20 ms)(loge 0.01) = −(20 ms)(−4.605) = 92.1ms
L ⇒ open circuit equivalent
10 MΩ(24 V)
VL =
= 20 V
10 MΩ + 2 MΩ
b.
I Lfinal =
c.
d.
RTh = 2 MΩ || 10 MΩ = 1.67 MΩ
10 MΩ(24 V)
ETh =
= 20 V
10 MΩ + 2 MΩ
ETh
20 V
=
= 12 μA
RTh 1.67 MΩ
iL = 12 μA(1 − e−t/3 μs)
10 μA = 12 μA(1 − e−t/3 μs)
0.8333 = 1 − e−t/3 μs
0.1667 = e−t/3 μs
loge(0.1667) = −t/3 μs
1.792 = t/3 μs
t = 1.792(3 μs) = 5.38 μs
τ=
L
5H
=
= 3 μs
R 1.67 MΩ
υL = 20 V e−t/3 μs = 20 V e−12 μs/3 μs = 20 V e−4
= 20 V(0.0183) = 0.37 V
32.
eL = L
Δi
:
Δt
0 − 3 ms, eL = 0 V
⎛ 40 × 10 −3 A ⎞
⎟ = 1.6 V
3 − 8 ms, eL = (200 mH) ⎜⎜
−3
⎟
⎝ 5 × 10 s ⎠
⎛ 40 × 10 −3 A ⎞
⎟ = −1.6 V
8 − 13 ms, eL = −(200 mH) ⎜⎜
−3
⎟
⎝ 5 × 10 s ⎠
13 − 14 ms, eL = 0 V
⎛ 40 × 10 −3 A ⎞
⎟ =8V
14 − 15 ms, eL = (200 mH) ⎜⎜
−3
⎟
⎝ 5 × 10 s ⎠
CHAPTER 11
137
15 − 16 ms, eL = −8 V
16 − 17 ms, eL = 0 V
33.
34.
138
υL = L
ΔiL
Δt
⎛ 20 mA ⎞
0 → 2 ms: υL = (5 mH) ⎜ −
⎟ = −50 mV
⎝ 2 ms ⎠
2 → 5 ms: ΔiL = 0 mA, υL = 0 V
⎛ + 30 mA ⎞
5 → 11 ms: υL = (5 mH) ⎜
⎟ = +25 mV
⎝ 6 ms ⎠
⎛ − 10 mA ⎞
11 → 18 ms: υL = (5 mH) ⎜
⎟ = −7.14 V
⎝ 7 ms ⎠
18 → : Δ iL = 0 mA, υL = 0 V
L = 10 mH, 4 mA at t = 0 s
Δi
Δt
υL = L ⇒ Δi = υ L
Δt
L
0 − 5 μs: υL = 0 V, ΔiL = 0 mA and iL = 4 mA
5 μs
5 − 10 μs: ΔiL =
(−24 V) = −12 mA
10 mH
CHAPTER 11
10 − 12 μs: ΔiL =
2 μs
(+60 V) = +12 mA
10 mH
12 − 16 μs: υL = 0 V, ΔiL = 0 mA and iL = 4 mA
8 μs
16 − 24 μs: ΔiL =
(−5 V) = −4 mA
10 mH
35.
a.
b.
36.
37.
38.
LT = L1 + L2 || (L3 + L4) = 6 H + 6 H || (6 H + 6 H)
= 6 H + 6 H || 12 H = 6 H + 4 H = 10 H
LT = (L1 + L2 || L3) || L4 = (4 H + 4 H || 4 H) || 4 H
= (4 H + 2 H) || 4 H = 6 H || 4 H = 2.4 H
L′T = 6 H || (1 H + 2 H) = 6 H || 3 H = 2 H
LT′ = 6 mH + 14 mH || 35 mH = 6 mH + 10 mH = 16 mH
CT′ = 9 μ F + 10 μF || 90 μF = 9 μF + 9 μF = 18 μF
16 mH in series with 18 μF
a.
RT′ = 2 kΩ || 8 kΩ = 1.6 kΩ, LT′ = 4 H || 6 H = 2.4 H
L′
2.4 H
= 1.5 ms
τ= T =
RT′ 1.6 kΩ
E
(1 − e − t / τ )
iL =
RT′
36 V
(1 − e − t / 1.5ms ) = 22.5 mA(1 − e−t/1.5ms)
1.6 kΩ
υL = Ee−t/τ = 36 Ve −t/1.5ms
=
CHAPTER 11
139
b.
39.
a.
Source conversion: E = 16 V, Rs = 2 kΩ
RTh = 2 kΩ + 2 kΩ || 8 kΩ = 2 kΩ + 16 kΩ = 3.6 kΩ
8 kΩ(16 V)
= 12.8 V
ETh =
8 kΩ + 2 kΩ
L 30 mH
E
12.8 V
= 8.33 μs
= 3.56 mA, τ =
=
Im = Th =
R 3.6 kΩ
RTh 3.6 kΩ
iL = 3.56 mA(1 − e−t/8.33μs)
υL = EThe−t/τ = 12.8 Ve−t/8.33μs
b.
140
CHAPTER 11
40.
a.
←RTh = 5 kΩ || 20 kΩ = 4 kΩ
20 kΩ(20 V)
= 16 V
ETh =
20 kΩ + 5 kΩ
LT = 5 H + 6 H || 30 H = 5 H + 5 H = 10 H
LT 10 H
=
= 2.5 ms
τ=
R 4 kΩ
υL = 16Ve−t/2.5 ms
16 V
iL =
(1 − e−t/τ) = 4 mA(1 − e−t/2.5 ms)
4 kΩ
b.
c.
41.
υ L3 =
υL
2
= 8 V e−t/2.5 ms
I R1 =
E 20 V
=5A
=
4Ω
R1
E
20 V
20 V
=2A
I2 = I R2 =
=
=
R2 + R3 6 Ω + 4 Ω 10 Ω
I1 = I R1 + I2 = 5 A + 2 A = 7 A
42.
I1 = I2 = 0 A
V1 = V2 = E = 60 V
43.
I1 =
12 V
= 3 A, I2 = 0 A
4Ω
V1 = 12 V, V2 = 0 V
CHAPTER 11
141
44.
V1 =
(3 Ω + 3 Ω 6 Ω)(50 V)
(3 Ω + 3 Ω 6 Ω) + 20 Ω
=
(3 Ω + 2 Ω)(50 V)
= 10 V
(3 Ω + 2 Ω) + 20 Ω
RT = 20 Ω + 3 Ω + 3 Ω || 6 Ω = 23 Ω + 2 Ω = 25 Ω
50 V
Is = I1 =
=2A
25 Ω
6 Ω( I s )
6 Ω( 2 A )
=
= 1.33 A
I5Ω = 0 A, ∴ I2 =
6 Ω+3Ω 6 Ω+3Ω
V2 =
(3 Ω 6 Ω)(50 V)
(3 Ω 6 Ω) + 20 Ω + 3 Ω
=
2 Ω(50 V)
2 Ω + 23 Ω
=4V
142
CHAPTER 11
Chapter 12
1.
Φ: CGS: 5 × 104 Maxwells, English: 5 × 104 lines
B: CGS: 8 Gauss, English: 51.62 lines/in.2
2.
Φ: SI 6 × 10−4 Wb, English 60,000 lines
B: SI 0.465 T, CGS 4.65 × 103 Gauss, English 30,000 lines/in.2
3.
a.
B=
4.
a.
R=
l
0.06 m
300
=
=
−4
2
μ A μ 2 × 10 m
μm
b.
R=
l
0.0762 m
152.4
=
=
−4
2
μ A μ 5 × 10 m
μm
c.
R=
l
0.1 m
1000
=
=
−4
2
μ A μ1 × 10 m
μm
Φ 4 × 10−4 Wb
= 0.04 T
=
A
0.01 m 2
from the above R (c) > R (a) > R (b)
5.
R=
400 At
F
= 952.4 × 103 At/Wb
=
Φ 4.2 × 10−4 Wb
6.
R=
F
120 gilberts
=
= 1.67 × 10−3 rels (CGS)
Φ 72,000 maxwells
7.
8.
9.
⎡ 1m ⎤
= 0.1524 m
6 in. ⎢
⎣ 39.37 in. ⎥⎦
F
400 At
H= =
= 2624.67 At/m
l 0.1524 m
μ=
2 B 2(1200 × 10−4 T)
=
= 4 × 10−4 Wb/Am
H
600 At/m
Φ 10 × 10−4 Wb
= 0.33 T
=
A
3 × 10−3 m 2
Fig. 12.7: H ≅ 800 At/m
NI = Hl ⇒ I = Hl/N = (800 At/m)(0.2 m)/75 t = 2.13 A
B=
CHAPTER 12
143
10.
11.
12.
Φ 3 × 10−4 Wb
= 0.6 T
=
A 5 × 10−4 m 2
Fig. 12.7, Hiron = 2500 At/m
Fig. 12.8, Hsteel = 70 At/m
NI = Hl(iron) + Hl(steel)
(100 t)I = (Hiron + Hsteel)l
(100 t)I = (2500 At/m + 70 At/m)0.3 m
771 A
= 7.71 A
I=
100
B=
a.
N1I1 + N2I2 = Hl
Φ 12 × 10−4 Wb
B= =
=1T
A 12 × 10−4 m 2
Fig. 12.7: H ≅ 750 At/m
N1(2 A) + 30 At = (750 At/m)(0.2 m)
N1 = 60 t
b.
μ=
a.
B
1T
=
= 13.34 × 10−4 Wb/Am
H 750 At/m
⎡ 1 Wb ⎤
80,000 lines ⎢ 8
= 8 × 104 × 10−8 Wb = 8 × 10−4 Wb
⎥
⎣10 lines ⎦
⎡ 1m ⎤
l(cast steel) = 5.5 in. ⎢
= 0.14 m
⎣ 39.37 in. ⎥⎦
⎡ 1m ⎤
= 0.013 m
l(sheet steel) = 0.5 in. ⎢
⎣ 39.37 in. ⎥⎦
⎡ 1m ⎤ ⎡ 1m ⎤
Area = 1 in.2 ⎢
= 6.45 × 10−4 m2
⎥
⎢
⎥
⎣ 39.37 in. ⎦ ⎣ 39.37 in. ⎦
8 × 10−4 Wb
Φ
=
= 1.24 T
A 6.45 × 10−4 m 2
Fig 12.8: Hsheet steel ≅ 460 At/m, Fig. 12.7: Hcast steel ≅ 1275 At/m
NI = Hl(sheet steel) + Hl(cast iron)
= (460 At/m)(0.013 m) + (1275 At/m)(0.14 m)
= 5.98 At + 178.50 At
NI = 184.48 At
B=
b.
144
B
1.24 T
=
= 9.73 × 10−4 Wb/Am
H 1275 At/m
B
1.24 T
Sheet steel: μ =
=
= 26.96 × 10−4 Wb/Am
H 460 At/m
Cast steel: μ =
CHAPTER 12
13.
Hl
+ Hl
cast steel cast iron
(20 t)I + (30 t)I = "
(50 t)I = "
N1I + N2 =
B=
Φ
⎡ 1m ⎤ ⎡ 1m ⎤
with 0.25 in.2 ⎢
= 1.6 × 10−4 m2
⎥
⎢
⎥
A
⎣ 39.37 in. ⎦ ⎣ 39.37 in. ⎦
0.8 × 10−4 Wb
= 0.5 T
1.6 × 10−4 m 2
Fig. 12.8: Hcast steel ≅ 280 At/m
Fig. 12.7: Hcast iron ≅ 1500 At/m
⎡ 1m ⎤
lcast steel = 5.5 in. ⎢
= 0.14 m
⎣ 39.37 in. ⎥⎦
B=
⎡ 1m ⎤
= 0.064 m
lcast iron = 2.5 in. ⎢
⎣ 39.37 in. ⎥⎦
(50 t)I = (280 At/m)(0.14 m) + (1500 At/m)(0.064 m)
50I = 39.20 + 96.00 = 135.20
I = 2.70 A
14.
15.
a.
lab = lef = 0.05 m, laf = 0.02 m, lbc = lde = 0.0085 m
NI = 2Hablab + 2Hbclbc + Hfalfa + Hglg
Φ 2.4 × 10−4 Wb
= 1.2 T ⇒ H ≅ 360 At/m (Fig. 12.8)
B= =
A
2 × 10−4 m 2
100I = 2(360 At/m)(0.05 m) + 2(360 At/m)(0.0085 m)
+ (360 At/m)(0.02 m) + 7.97 × 105(1.2 T)(0.003 m)
= 36 At + 6.12 At + 7.2 At + 2869 At
100I = 2918.32 At
I ≅ 29.18 A
b.
air gap: metal = 2869 At:49.72 At = 58.17:1
B
1.2 T
=
= 3.33 × 10−3 Wb/Am
μsheet steel =
H 360 At/m
μair = 4π × 10−7 Wb/Am
μsheet steel: μair = 3.33 × 10−3 Wb/Am:4π × 10−7 ≅ 2627:1
⎡ 1m ⎤
4 cm ⎢
= 0.04 m
⎣100 cm ⎥⎦
(8 × 10−4 Wb − 0.5 × 10−4 Wb) 36(7.5 × 10−4 )
1
dφ 1
=
f = NI
= (80 t)(0.9 A)
1
2
dx 2
0.02
(0.04 m)
2
= 1.35 N
CHAPTER 12
145
16.
C = 2πr = (6.28)(0.3 m) = 1.88 m
Φ 2 × 10−4 Wb
= 1.54 T
B= =
A 1.3 × 10−4 m 2
Fig. 12.7: Hsheet steel ≅ 2100 At/m
Hg = 7.97 × 105Bg = (7.97 × 105)(1.54 T) = 1.23 × 106 At/m
N1I1 + N2I2 = Hglg + Hl(sheet steel)
(200 t)I1 + (40 t)(0.3 A) = (1.23 × 106 At/m)(2 mm) + (2100 At/m)(1.88 m)
I1 = 31.98 A
17.
a.
⎡ 1m ⎤
= 2 × 10−3 m
0.2 cm ⎢
⎣100 cm ⎥⎦
π d 2 (3.14)(0.01 m) 2
=
= 0.79 × 10−4 m2
A=
4
4
NI = Hglg, Hg = 7.96 × 105 Bg
⎡
⎛ 0.2 × 10−4 Wb ⎞ ⎤
2 × 10−3 m
(200 t)I = ⎢(7.96 × 105 ) ⎜
2 ⎟⎥
−4
⎢⎣
⎝ 0.79 × 10 m ⎠ ⎥⎦
I = 2.02 A
b.
18.
2 × 10−4 Wb
Φ
= 0.25 T
=
A 0.79 × 10−4 m 2
2
1 Bg A 1 (0.25 T ) 2(0.79 × 10−4 m 2 )
=
F≅
2 μo
2
4π × 10−7
≅ 2N
Bg =
Table:
Section
a−b, g−h
b−c, f−g
c−d, e−f
Φ(Wb)
2 × 10
−4
2 × 10−4
a−h
b−g
d−e
146
2 × 10−4
A(m2)
5 × 10
5 × 10
−4
−4
5 × 10−4
5 × 10
2 × 10
−4
−4
5 × 10−4
B(T)
H
l(m)
Hl
0.2
0.1
0.099
0.2
0.2
0.002
CHAPTER 12
Φ 2 × 10−4 Wb
=
= 0.4 T
A 5 × 10−4 m 2
Air gap: Hg = 7.97 × 105(0.4 T) = 3.19 × 105 At/m
Hglg = (3.19 × 105 At/m)(2 mm) = 638 At
Fig 12.8: Hbc = Hcd = Hef = Hfg = 55 At/m
Hbclbc = Hfglfg = (55 At/m)(0.1 m) = 5.5 At
Hcdlcd = Heflef = (55 At/m)(0.099 m) = 5.45 At
Bbc = Bcd = Bg = Bef = Bfg =
For loop 2:
∑F
=0
Hbclbc + Hcdlcd + Hglg + Heflef + Hfglfg − Hgblgb = 0
5.5 At + 5.45 At + 638 At + 5.45 At + 5.50 At − Hgblgb = 0
Hgblgb = 659.90 At
659.90 At
and Hgb =
= 3300 At/m
0.2 m
Fig 12.7: Bgb ≅ 1.55 T
with Φ2 = BgbA = (1.55 T)(2 × 10−4 m2) = 3.1 × 10−4 Wb
ΦT = Φ1 + Φ2
= 2 × 10−4 Wb + 3.1 × 10−4 Wb
= 5.1 × 10−4 Wb = Φab = Φha = Φgh
Φ
5.1 × 10−4 Wb
= 1.02 T
Bab = Bha = Bgh = T =
A
5 × 10−4 m 2
B−H curve: (Fig 12.8):
Hab = Hha = Hgh ≅ 180 At/m
Hablab = (180 At/m)(0.2 m) = 36 At
Hhalha = (180 At/m)(0.2 m) = 36 At
Hghlgh = (180 At/m)(0.2 m) = 36 At
which completes the table!
Loop #1:
∑F
=0
NI = Hablab + Hbglbg + Hghlgh + Hahlah
(200 t)I = 36 At + 659.49 At + 36 At + 36 At
(200 t)I = 767.49 At
I ≅ 3.84 A
19.
NI = Hl
l = 2πr = (6.28)(0.08 m) = 0.50 m
(100 t)(2 A) = H(0.50 m)
H = 400 At/m
Fig. 12.8: B ≅ 0.68 T
Φ = BA = (0.68 T)(0.009 m2)
Φ = 6.12 mWb
CHAPTER 12
147
20.
NI = Hab(lab + lbc + lde + lef + lfa) + Hglg
300 At = Hab(0.8 m) + 7.97 × 105 Bg(0.8 mm)
300 At = Hab(0.8 m) + 637.6 Bg
Assuming 637.6 Bg
Hab(0.8 m)
then 300 At = 637.6 Bg
and Bg = 0.47 T
Φ = BA = (0.47 T)(2 × 10−4 m2) = 0.94 × 10−4 Wb
Bab = Bg = 0.47 T ⇒ H ≅ 270 At/m (Fig. 12.8)
300 At = (270 At/m)(0.8 m) + 637.6(0.47 T)
300 At ≠ 515.67 At
∴ Poor approximation!
300 At
× 100% ≅ 58%
515.67 At
Reduce Φ to 58%
0.58(0.94 × 10−4 Wb) = 0.55 × 10−4 Wb
Φ 0.55 × 10−4 Wb
= 0.28 T ⇒ H ≅ 190 At/m (Fig. 12.8)
B= =
A
2 × 10−4 m 2
300 At = (190 At/m)(0.8 m) + 637.6(0.28 T)
300 At ≠ 330.53 At
Reduce Φ another 10% = 0.55 × 10−4 Wb − 0.1(0.55 × 10−4 Wb)
= 0.495 × 10−4 Wb
Φ 0.495 × 10−4 Wb
B= =
= 0.25 T ⇒ H ≅ 175 At/m (Fig. 12.7)
A
2 × 10−4 m 2
300 At = (175 At/m)(0.8) + 637.6(0.28 T)
300 At ≠ 318.53 At but within 5% ∴ OK
Φ ≅ 0.55 × 10−4 Wb
21.
a.
1τ = 0.632 Tmax
Tmax ≅ 1.5 T for cast steel
0.632(1.5 T) = 0.945 T
At 0.945 T, H ≅ 700 At/m (Fig. 12.7)
∴ B = 1.5 T(1 − e−H/700 At/m)
b.
H = 900 At/m:
B = 1.5 T (1 − e
−
900 At/m
700 At/m
Graph: ≅ 1.1 T
H = 1800 At/m:
(
) = 1.09 T
1800 At/m
−
)
B = 1.5 T 1 − e 700 At/m = 1.39 T
Graph: ≅ 1.38 T
H = 2700 At/m:
(
−
)
B = 1.5 1 − e 700 At/m = 1.47 T
Graph: ≅ 1.47 T
Excellent comparison!
148
2700 At/m
CHAPTER 12
c.
B = 1.5 T(1 − e−H/700 At/m) = 1.5 T − 1.5 Te−H/700 At/m
B − 1.5 T = −1.5 Te−H/700 At/m
1.5 − B = 1.5 Te−H/700 At/m
1.5 T − B
= e−H/700 At/m
1.5 T
−H
B ⎞
⎛
loge ⎜ 1 −
⎟ = 700 At/m
⎝ 1.5 T ⎠
B ⎞
⎛
and H = −700 loge ⎜ 1 −
⎟
⎝ 1.5 T ⎠
d.
B = 1 T:
1T ⎞
⎛
H = −700 loge ⎜ 1 −
⎟ = 769.03 At/m
⎝ 1.5 T ⎠
Graph: ≅ 750 At/m
B = 1.4 T:
⎛ 1.4 T ⎞
H = −700 loge ⎜ 1 −
⎟ = 1895.64 At/m
⎝ 1.5 T ⎠
Graph: ≅ 1920 At/m
e.
B ⎞
⎛
H = −700 loge ⎜ 1 −
⎟
⎝ 1.5 T ⎠
⎛ 0.2 T ⎞
= −700 loge ⎜ 1 −
⎟
⎝ 1.5 T ⎠
= 100.2 At/m
Hl (100.2 At/m)(0.16 m)
I=
=
= 40.1 mA
400 t
N
vs 44 mA for Ex. 12.1
CHAPTER 12
149
Chapter 13
1.
a.
b.
c.
d.
e.
20 mA
15 ms: −20 mA, 20 ms: 0 mA
40 mA
20 ms
2.5 cycles
2.
a.
b.
c.
d.
e.
40 V
5 μs: 40 V, 11 μs: −40 V
80 V
4 μs
3 cycles
3.
a.
b.
c.
d.
8 mV
3 μs: −8 mV, 9 μs: 0 mV
16 mV
4.5 μs
10 μS
= 2.22 cycles
4.5 μS/cycle
e.
4.
a.
b.
c.
d.
5.
a.
b.
c.
d.
6.
150
T=
1
f
1
T=
f
1
T=
f
1
T=
f
T=
=
1
= 40 ms
25 Hz
1
=
= 25 ns
40 mHz
1
=
= 40 μs
25 kHz
1
=
=1s
1 Hz
1
0
=
= 60 Hz
1
T
s
60
1
1
= 100 Hz
f= =
T 0.01 s
1
1
f= =
= 25 Hz
T 40 ms
1
1
= 40 kHz
f= =
T 25 μs
f=
1
= 0.05 s, 5(0.05 s) = 0.25 s
20 Hz
CHAPTER 13
7.
8.
9.
10.
24 ms
= 0.3 ms
80 cycles
42 cycles
= 7 Hz
f=
6s
T=
a.
b.
T = ( 4 div.)(10 μs/div.) = 40 μs
c.
f=
a.
b.
c.
d.
11.
a.
b.
c.
d.
12.
a.
b.
c.
d.
13.
Vpeak = (3 div.)(50 mV/div) = 150 mV
a.
b.
c.
d.
1
1
=
= 25 kHz
T 40 μs
π
⎛ π ⎞
Radians = ⎜
⎟45° = rad
4
⎝ 180° ⎠
π
⎛ π ⎞
Radians = ⎜
⎟60° = rad
3
⎝ 180° ⎠
⎛ π ⎞
Radians = ⎜
⎟270° = 1.5π rad
⎝ 180° ⎠
⎛ π ⎞
Radians = ⎜
⎟170° = 0.94π rad
⎝ 180° ⎠
⎛ 180° ⎞ π
Degrees = ⎜
⎟ = 45°
⎝ π ⎠4
⎛ 180° ⎞ π
Degrees = ⎜
⎟ = 30°
⎝ π ⎠6
⎛ 180° ⎞ 1
Degrees = ⎜
⎟ π = 18°
⎝ π ⎠ 10
⎛ 180° ⎞
Degrees = ⎜
⎟ 0.6 π = 108°
⎝ π ⎠
2π 2π
=
= 3.14 rad/s
T
2s
2π
ω=
= 20.94 × 103 rad/s
−3
0.3 × 10 s
2π
ω=
= 1.57 × 106 rad/s
4 × 10− 6 s
2π
ω=
= 157.1 rad/s
1 / 25 s
ω=
ω = 2π f = 2π (50 Hz) = 314.16 rad/s
ω = 2π f = 2π (600 Hz) = 3769.91 rad/s
ω = 2π f = 2π (2 kHz) = 12.56 × 103 rad/s
ω = 2π f = 2π (0.004 MHz) = 25.13 × 103 rad/s
CHAPTER 13
151
14.
15.
2π
ω
⇒f=
T
2π
2π 1
T=
=
ω
f
ω 754 rad/s
f=
=
= 120 Hz, T = 8.33 ms
2π
2π
a.
ω = 2π f =
b.
f=
c.
f=
d.
f=
ω 8.4 rad/s
=
= 1.34 Hz, T = 746.27 ms
2π
2π
ω 6000 rad/s
=
= 954.93 Hz, T = 1.05 ms
2π
2π
ω 1 / 16 rad/s
= 9.95 × 10−3 Hz, T = 100.5 ms
=
2π
2π
⎛ π ⎞ π
(45°) ⎜
⎟ = radians
⎝ 180° ⎠ 4
θ π / 4 rad π / 4 rad
1
1
=
=
=
=
= 2.08 ms
t=
ω
2πf
2π (60 Hz) (8)(60) 480
16.
α
π /6
⎛ π ⎞ π
(30°) ⎜
= 104.7 rad/s
⎟ = , α = ωt ⇒ ω = =
°
180
6
t
5 × 10 −3 s
⎝
⎠
17.
a.
b.
c.
d.
18.
−
ω 377 rad/s
=
= 60 Hz
2π
2π
ω 754 rad/s
=
= 120 Hz
Amplitude = 5, f =
2π
2π
ω 10,000 rad/s
Amplitude = 106, f =
=
= 1591.55 Hz
2π
2π
ω 942 rad/s
=
= 149.92 Hz
Amplitude = −6.4, f =
2π
2π
Amplitude = 20, f =
19.
−
20.
T=
21.
i = 0.5 sin 72° = 0.5(0.9511) = 0.48 A
152
2π
ω
=
2π
1
= 40 ms,
cycle = 20 ms
157
2
CHAPTER 13
22.
23.
24.
⎛ 180° ⎞
1.2π ⎜
⎟ = 216°
⎝ π ⎠
υ = 20 sin 216° = 20(−0.588) = −11.76 V
6 × 10−3 = 30 × 10−3 sin α
0.2 = sin α
α = sin−1 0.2 = 11.54° and 180° − 11.54° = 168.46°
υ = Vm sin α
30° 1 ms
=
360°
T
⎛ 360 ⎞
T = 1 ms ⎜
⎟ = 12 ms
⎝ 30 ⎠
1
1
= 83.33 Hz
f= =
T 12 × 10−3 s
ω = 2π f = (2π)(83.33 Hz) = 523.58 rad/s
40 = Vm sin 30° = Vm (0.5)
40
= 80 V
∴Vm =
0 .5
and υ = 80 sin 523.58t
25.
−
26.
−
27.
a.
b.
28.
a.
b.
29.
30.
31.
32.
ω = 2π f = 377 rad/s
υ = 25 sin (ωt + 30°)
2
π
π=
= 60°, ω = 2πf = 6.28 × 103 rad/s
3
3
i = 3 × 10−3 sin(6.28 × 103t − 60°)
π−
ω = 2π f = 2π (40 Hz) = 251.33 rad/s
υ = 0.01 sin (251.33t − 110°)
ω = 2π f = 2π (10 kHz) = 62.83 × 103 rad/s,
i = 2 × 10−3 sin (62.83 × 103t + 135°)
υ leads i by 10°
i leads υ by 70°
i leads υ by 80°
υ = 2 sin (ωt − 30° + 90°)
i = 5 sin(ωt + 60°)
33.
3 ⎛ 180° ⎞
π⎜
⎟ = 135°
4 ⎝ π ⎠
+60°
in phase
υ = 4 sin(ωt + 90° + 90° + 180° = 4 sinωt
i = sin(ωt + 10° + 180°) = sin(ωt + 190°)
CHAPTER 13
i leads υ by 190°
153
34.
T=
t1 =
35.
1
1
=
= 1 ms
f 1000 Hz
120° ⎛ T ⎞ 2 ⎛ 1 ms ⎞ 1
⎟ = ms
⎜ ⎟= ⎜
180° ⎝ 2 ⎠ 3 ⎝ 2 ⎠ 3
ω = 2π f = 50,000 rad/s
50,000
= 7957.75 Hz
2π
1
= 125.66 μs
T=
f
f=
t1 =
36.
37.
38.
a.
T = ( 8 div.)(1 ms/div.) = 8 ms (both waveforms)
b.
f=
c.
Peak = (2.5 div)(0.5 V/div.) = 1.25 V
Vrms = 0.707(1.25 V) = 0.884 V
d.
Phase shift = 4.6 div., T = 8 div.
4.6 div.
× 360° = 207° i leads e
θ=
8 div.
or e leads i by 153°
G=
G=
40.
a.
b.
c.
154
1
1
= 125 Hz (both)
=
T 8 ms
(6 V)(1 s) + (3 V)(1 s) − (3 V)(1 s) 6 V
=
=2V
3s
3
⎡1
⎤
⎢ 2 (4 ms)(20 mA)⎥ − ( 2 ms)(5 mA) 40 mA − 10 mA 30 mA
⎦
= 3.87 mA
G= ⎣
=
=
8 ms
8
8
39.
41.
40° ⎛ T ⎞
⎜ ⎟ = 0.222(62.83 μs) = 13.95 μs
180° ⎝ 2 ⎠
2 Am − (5 mA)(π ) 2(20 mA) − (5 mA)(π ) 40 mA − 15.708 mA
=
=
= 3.87 mA
2π
2π
2π
T = ( 2 div.)(50 μs) = 100 μs
1
1
= 10 kHz
f= =
T 100 μs
Average = (−1.5 div.)(0.2 V/div.) = −0.3 V
a.
T = (4 div.)(10 μs/div.) = 40 μs
b.
f=
1
1
=
= 25 kHz
T 40 μs
CHAPTER 13
c.
G=
( 2.5 div.)(1.5 div.) + (1 div.)(0.5 div.) + (1 div.)(0.6 div.) + (2.5 div.)(0.4 div.)+(1 div.)(1 div.)
3.75 div. + 0.5 div. + 0.6 div. + 1 div. + 1 div.
4
6.85 div.
= 1.713 div.
=
4
1.713 div.(10 mV/div.) = 17.13 mV
4 div.
=
42.
a.
b.
c.
43.
a.
b.
c.
Vrms = 0.7071(140 V) = 98.99 V
Irms = 0.7071(6 mA) = 4.24 mA
Vrms = 0.7071(40 μV) = 28.28 μV
υ = 14.14 sin 377t
i = 70.7 × 10−3 sin 377t
υ = 2.83 × 103 sin 377t
44.
Vrms =
45.
Vrms =
46.
G=
(3 V) 2 ( 2 s) + (2 V ) 2 (2 s) + (1 V)2 (2 s) + (−1 V) 2 (2 s) + (−3 V)2 ( 2 s) + (−2 V)2 (2 s)
12 s
= +2.16 V
(10 V)(4 ms) − (10 V)(4 ms)
0
=
=0V
8 ms
8 ms
Vrms =
47.
⎛1 ⎞
(2 V) 2 ( 4 s) + (−2 V) 2 (1 s) + (3 V) 2 ⎜ s ⎟
⎝2 ⎠
= 1.43 V
12 s
a.
(10 V)2 (4 ms) + (−10 V)2 ( 4 ms)
= 10 V
8 ms
T = (4 div.)(10 μs/div.) = 40 μs
1
1
f= =
= 25 kHz
T 40 μs
Av. = (1 div.)(20 mV/div.) = 20 mV
Peak = (2 div.)(20 mV/div.) = 40 mV
rms =
CHAPTER 13
V02 +
2
(40 mV)2
Vmax
= (20 mV)2 +
= 34.64 mV
2
2
155
b.
T = (2 div.)(50 μs) = 100 μs
1
1
= 10 kHz
f= =
T 100 μs
Av. = (−1.5 div.)(0.2 V/div.) = −0.3 V
Peak = (1.5 div.)(0.2 V/div.) = 0.3 mV
rms =
48.
156
V02 +
2
(.3 V)2
Vmax
= (.3 V)2 +
= 367.42 mV
2
2
a.
Vdc = IR = (4 mA)(2 kΩ) = 8 V
Meter indication = 2.22(8 V) = 17.76 V
b.
Vrms = 0.707(16 V) = 11.31 V
CHAPTER 13
Chapter 14
1.
−
2.
−
3.
a.
(377)(10) cos 377t = 3770 cos 377t
b.
(754)(0.6) cos(754t + 20°) = 452.4 cos(754t + 20°)
c.
( 2 20)(157) cos(157t − 20°) = 4440.63 cos(157t − 20°)
4.
5.
d.
(−200)(1) cos(t + 180°) = −200 cos(t + 180°) = 200 cos t
a.
Im = Vm/R = 150 V/5 Ω = 30 A, i = 30 sin 200t
b.
Im = Vm/R = 30 V/5 Ω = 6 A, i = 6 sin(377t + 20°)
c.
Im = Vm/R = 40 V/5 Ω = 8 A, i = 8 sin(ωt + 100°)
d.
Im = Vm/R = 80 V/5 Ω = 16 A, i = 16 sin(ωt + 220°)
a.
Vm = ImR = (0.1 A)(7 × 103 Ω) = 700 V
υ = 700 sin 1000t
b.
Vm = ImR = (2 × 10−3 A)((7 × 103 Ω) = 14.8 V
υ = 14.8 sin(400t − 120°)
c.
d.
6.
a.
b.
i = 6 × 10−6 sin(ωt − 2° + 90°) = 6 × 10−6 sin(ωt + 88°)
Vm = ImR = (6 × 10−6 A)((7 × 103 Ω) = 42 × 10−3 V
υ = 42 × 10−3 sin(ωt + 88°)
i = 0.004 sin(ωt + 90° + 90° + 180°) = 0.004 sin(ωt + 360°) = 0.0004 sin ωt
Vm = ImR = (4 × 10−3 A)((7 × 103 Ω) = 28 V
υ = 28 sin ωt
0Ω
XL = 2πfL = 2πLf = (6.28)(2 H)f = 12.56f = 12.56(10 Hz) = 125.6 Ω
c.
XL = 12.56f = 12.56(60 Hz) = 753.6 Ω
d.
XL = 12.56f = 12.56(2000 Hz) = 25.13 kΩ
e.
XL = 12.56f = 12.56(105 Hz) = 1.256 MΩ
CHAPTER 14
157
7.
8.
a.
L=
XL
20 Ω
=
= 1.59 H
2π f 2π (2 Hz)
b.
L=
XL
1000 Ω
=
= 2.65 H
2π f 2π (60 Hz)
c.
L=
XL
5280 Ω
=
= 1.68 H
2π f 2π (500 Hz)
a.
XL = 2πfL ⇒ f =
f=
b.
c.
d.
9.
a.
b.
c.
d.
10.
a.
b.
c.
158
XL
XL
X
= L
=
2π L (6.28)(10 H) 62.8
100 Ω
= 1.59 Hz
62.8
3770 Ω
f= XL =
= 60.03 Hz
62.8
62.8
15,700 Ω
f= XL =
= 250 Hz
62.8
62.8
243 Ω
f= XL =
= 3.87 Hz
62.8 62.8
Vm = ImXL = (5 A)(20 Ω) = 100 V
υ = 100 sin(ωt + 90°)
Vm = ImXL = (40 × 10−3 A)(20 Ω) = 0.8 V
υ = 0.8 sin(ωt + 150°)
i = 6 sin(ωt + 150°), Vm = ImXL = (6 A)(20 Ω) = 120 V
υ = 120 sin(ωt + 240°) = 120 sin(ωt − 120°)
i = 3 sin(ωt + 100°), Vm = ImXL = (3 A)(20 Ω) = 60 V
υ = 60 sin(ωt + 190°)
XL = ωL = (100 rad/s)(0.1 H) = 10 Ω
Vm = ImXL = (10 A)(10 Ω) = 100 V
υ = 100 sin(100t + 90°)
XL = ωL = (377 rad/s)(0.1 H) = 37.7 Ω
Vm = ImXL = (6 × 10−3 A)(37.7 Ω) = 226.2 mV
υ = 226.2 × 10−3 sin(377t + 90°)
XL = ωL = (400 rad/s)(0.1 H) = 40 Ω
Vm = ImXL = (5 × 10−6 A)(40 Ω) = 200 μV
υ = 200 × 10−6 sin(400t + 110°)
CHAPTER 14
11.
12.
d.
i = 4 sin(20t + 200°)
XL = ωL = (20 rad/s)(0.1 H) = 2 Ω
Vm = ImXL = (4 A)(2 Ω) = 8 V
υ = 8 sin(20t + 290°) = 8 sin(20t − 70°)
a.
Im =
Vm 120 V
=
= 2.4 A, i = 2.4 sin(ωt − 90°)
X L 50 Ω
b.
Im =
Vm 30 V
=
= 0.6 A, i = 0.6 sin(ωt − 70°)
X L 50 Ω
c.
υ = 40 sin(ωt + 100°)
V
40 V
Im = m =
= 0.8 A, i = 0.8 sin(ωt + 10°)
X L 50 Ω
d.
υ = 80 sin(377t + 220°)
V
80 V
Im = m =
= 1.6 A, i = 1.6 sin(377t + 130°)
X L 50 Ω
a.
b.
c.
d.
13.
XL = ωL = (60 rad/s)(0.2 H) = 12 Ω
Im = Vm/XL = 1.5 V/12 Ω = 0.125 A
i = 0.125 sin(60t − 90°)
XL = ωL = (10 rad/s)(0.2 H) = 2 Ω
Im = Vm/XL = 16 mV/2 Ω = 8 mA
i = 8 × 10−3 sin(t + 2° − 90°) = 8 × 10−3 sin(t − 88°)
υ = 4.8 sin(0.05t + 230°)
XL = ωL = (0.05 rad/s)(0.2 H) = 0.01 Ω
Im = Vm/XL = 4.8 V/0.01 Ω = 480 A
i = 480 sin(0.05t + 230° − 90°) = 480 sin(0.05t + 140°)
υ = 9 × 10−3 sin(377t + 90°)
XL = ωL = (377 rad/s)(0.2 H) = 75.4 Ω
Im = Vm/XL = 9 mV/75.4 Ω = 0.119 mA
i = 0.119 × 10−3 sin 377t
a.
XC =
1
1
=
=∞Ω
2π fC 2π (0 Hz)(5 × 10−6 F)
b.
XC =
1
1
=
= 530.79 Ω
2π fC 2π (60 Hz)(5 × 10−6 F)
c.
XC =
1
1
=
= 265.39 Ω
2π fC 2π (120 Hz)(5 × 10−6 F)
CHAPTER 14
159
14.
15.
16.
d.
XC =
1
1
=
= 15.92 Ω
2π fC 2π (2 kHz)(5 × 10−6 F)
e.
XC =
1
1
=
= 62.83 Ω
6
2π fC 2π (2 × 10 Hz)(5 × 10−6 F)
a.
C=
1
1
=
= 10.62 μF
2π fX C 6.28(60 Hz)(250 Ω)
b.
C=
1
1
=
= 9.28 μF
2π fX C 6.28(312 Hz)(55 Ω)
c.
C=
1
1
=
= 636.94 μF
2π fX C 6.28(25 Hz)(10 Ω)
a.
f=
1
1
= 31.83 Hz
=
−6
2π CX C 2π (50 × 10 F)(100 Ω)
b.
f=
1
1
=
= 4.66 Hz
−6
2π CX C 2π (50 × 10 F)(684 Ω)
c.
f=
1
1
=
= 9.31 Hz
−6
2π CX C 2π(50 × 10 F)(342 Ω)
d.
f=
1
1
= 1.59 Hz
=
−6
2π CX C 2π (50 ×10 F)(2000 Ω)
Im = Vm/XC = 120 V/2.5 Ω = 48 A
i = 48 sin(ωt + 90°)
a.
b.
c.
d.
160
Im = Vm/XC = 0.4 V/2.5 Ω = 0.16 A
i = 0.16 sin(ωt + 110°)
υ = 8 sin(ωt + 100°)
Im = Vm/XC = 8 V/2.5 Ω = 3.2 A
i = 3.2 sin(ωt + 190°)
υ = −70 sin(ωt + 40°) = 70 sin(ωt + 220°)
Im = Vm/XC = 70 V/2.5 Ω = 28 A
i = 28 sin(ωt + 310°) = 28 sin(ωt − 50°)
CHAPTER 14
17.
a.
υ = 30 sin 200t, XC =
Im =
b.
1
1
=
= 2.67 kΩ
ωC (374)(1 × 10−6 )
Vm
120 V
=
= 44.94 mA, i = 44.94 × 10−3 sin(374t + 300°)
X C 2,670 Ω
1
1
=
= 1.25 kΩ
ωC (800)(1 × 10−6 )
Vm
70 V
=
= 56 mA, i = 56 × 10−3 sin(ωt + 160°)
X C 1250 Ω
Vm = ImXC = (50 × 10−3 A)(10 Ω) = 0.5 V
υ = 0.5 sin(ωt − 90°)
a.
b.
c.
19.
−3
V m 60 × 10 V
= 22.64 μA, i = 22.64 × 10−6 sin(377t + 90°)
=
2,650 Ω
XC
υ = 70 sin(800t + 70°), XC =
Im =
18.
1
1
=
= 2.65 kΩ
ωC (377)(1 × 10−6 )
υ = 120 sin(374t + 210°), XC =
Im =
d.
Vm 30 V
=
= 6 mA, i = 6 × 10−3 sin(200t + 90°)
X C 5 kΩ
υ = 60 × 10−3 sin 377t, XC =
Im =
c.
1
1
=
= 5 kΩ
ωC (200)(1 × 10−6 )
Vm = ImXC = (2 × 10−6)(10 Ω) = 20 μV
υ = 20 × 10−6 sin(ωt − 30°)
i = −6 sin(ωt − 30°) = 6 sin(ωt + 150°)
Vm = ImXC = (6 A)(10 Ω) = 60 V
υ = 60 sin(ωt + 60°)
d.
i = 3 sin(ωt + 100°)
Vm = ImXC = (3 A)(10 Ω) = 30 V
υ = 30 sin(ωt + 10°)
a.
i = 0.2 sin 300t, XC =
1
1
=
= 6.67 kΩ
ωC (300)(0.5 × 10−6 )
Vm = ImXC = (0.2 A)(6,670 Ω) = 1334 V, υ = 1334 sin(300t − 90°)
b.
i = 8 × 10−3 sin 377t, XC =
1
1
=
= 5.31 kΩ
ωC (377)(0.5 × 10−6 )
Vm = ImXC = (8 × 10−3 A)(5.31 × 103 Ω) = 42.48 V
υ = 42.48 sin(377t − 90°)
CHAPTER 14
161
c.
i = 60 × 10−3 sin(754t + 90°), XC =
1
1
=
= 2.65 kΩ
ωC (754)(0.5 × 10-6 )
Vm = ImXC = (60 × 10−3 A)(2.65 × 103 Ω) = 159 V
υ = 159 sin 754t
d.
i = 80 × 10−3 sin(1600t − 80°), XC =
1
1
=
= 1.25 kΩ
ωC (1600)(0.5 × 10−6 )
Vm = ImXC = (80 × 10−3 A)(1.25 × 103 Ω) = 100 V
υ = 100 sin(1600t − 170°)
20.
a.
b.
c.
21.
a.
b.
c.
22.
23.
162
υ leads i by 90° ⇒ L, XL = Vm/Im = 550 V/11 A = 50 Ω
X
50 Ω
= 132.63 mH
L= L=
ω 377 rad/s
υ leads i by 90° ⇒ L, XL = Vm/Im = 36 V/4 A = 9 Ω
1
1
L=
= 147.36 μH
=
ωX L (754 rad/s)(9 Ω)
υ and i are in phase ⇒ R
V
10.5 V
R= m=
=7Ω
I m 1.5 A
i = 5 sin(ωt + 90°) ⎫
⎬ i leads υ by 90° ⇒ C
υ = 2000 sin ωt
⎭
Vm 2000 V
XC =
=
= 400 Ω
Im
5A
i = 2 sin(157t + 60°)
⎫
⎬ υ leads i by 90° ⇒ L
υ = 80 sin(157t + 150°) ⎭
V
80 V
40 Ω
XL = m =
= 254.78 mH
= 40 Ω, L = X L =
ω 157 rad/s
Im 2 A
υ = 35 sin(ωt − 20°) ⎫
⎬ in phase ⇒ R
i = 7 sin(ωt − 20°)
⎭
Vm 35 V
=
=5Ω
R=
Im 7 A
−
−
CHAPTER 14
1
1
1
1
=
=R ⇒ f =
=
3
−6
2π fC
2π RC 2π (2 × 10 Ω)(1 × 10 F) 12.56 × 10−3
≅ 79.62 Hz
24.
XC =
25.
XL = 2πfL = R
10,000 Ω
R
=
L=
= 318.47 mH
2π f 2π (5 × 103 Hz)
26.
XC = XL
27.
XC = XL
28.
29.
30.
1
= 2π fL
2π fC
1
f2 =
4π 2 LC
1
1
=
and f =
= 1.59 kHz
2π LC 2π (10 × 10−3 H)(1 × 10−6 F)
1
1
1
= 2π fL ⇒ C = 2 2 =
= 5.07 nF
2π fC
4π f L 4(9.86)(2500 × 106 )(2 × 10−3 )
a.
P=
Vm I m
(550 V)(11 A)
cos θ =
cos 90° = ( )(0) = 0 W
2
2
b.
P=
Vm I m
(36 V)(4 A)
cos θ =
cos 90° = ( )(0) = 0 W
2
2
c.
P=
Vm I m
(10.5 V)(1.5 A)
cos θ =
cos 0° = 7.88 W
2
2
a.
P=
Vm I m
(5 A)(2000 V)
cos θ =
cos 90° = 0 W
2
2
b.
cos θ = 0 ⇒ 0 W
c.
P=
(35 V)(7 A)
cos 0° = 122.5 W
2
a.
P=
(60 V)(15 A)
cos 30° = 389.7 W, Fp = 0.866
2
b.
P=
(50 V)(2 A)
cos 0° = 50 W, Fp = 1.0
2
CHAPTER 14
163
c.
d.
(50 V)(3 A)
cos 10° = 73.86 W, Fp = 0.985
2
(75 V)(0.08 A)
P=
cos 40° = 2.30 W, Fp = 0.766
2
P=
⎛8A⎞
V m 48 V
=
= 6 Ω, P = I2R = ⎜
⎟ 6 Ω = 192 W
Im 8A
⎝ 2⎠
V I
(48 V)(8 A)
cos 0° = 192 W
P = m m cos θ =
2
2
⎛ 48 V ⎞⎛ 8 A ⎞
P = VI cos θ = ⎜
⎟⎜
⎟ cos 0° = 192 W
⎝ 2 ⎠⎝ 2 ⎠
All the same!
2
31.
32.
33.
34.
R=
P = 100 W: Fp = cos θ = P/VI = 100 W/(150 V)(2 A) = 0.333
P = 0 W: Fp = cos θ = 0
300
=1
P = 300 W: Fp =
300
V mI m
cos θ
2
(50 V) I m
500 W =
(0.5) ⇒ Im = 40 A
2
i = 40 sin(ωt − 50°)
P=
a.
Im = Em/R = 30 V/6.8 Ω = 4.41 A, i = 4.41 sin(377t + 20°)
b.
⎛ 4.41 A ⎞
P = I2R = ⎜
⎟ 3 Ω = 29.18 W
2 ⎠
⎝
c.
T =
a.
Im =
2
35.
b.
c.
164
2π
6.28
=
= 16.67 ms
ω 377 rad/s
6(16.67 ms) = 100.02 ms ≅ 0.1 s
Vm 100 V
=
= 4 A, i = 4 sin(314t − 30°)
X L 25 Ω
25 Ω
L= XL=
= 79.62 mH
ω 314 rad/s
L⇒0W
CHAPTER 14
36.
37.
a.
Em = ImXC = (30 × 10−3 A)(2.4 kΩ) = 72 V
e = 72 sin(377t − 20° − 90°) = 72 sin(377t − 110°)
b.
C=
c.
P=0W
a.
1
1
=
= 1.11 μF
ωX C (377 rad/s)(2.4 kΩ)
1
1
1
=
=
= 50 Ω
2π f C1 ωC1 (104 rad/s)(2 μ F)
1
1
= 10 Ω
X C2 =
=
4
ωC 2 (10 )(10 μ F )
120 V ∠60°
E
E = 100 V ∠60°
I1 =
= 2.4 A ∠150°
=
Z C1 50 Ω ∠ − 90°
X C1 =
I2 =
i1 =
i2 =
b.
38.
a.
E
ZC 2
=
120 V ∠60°
= 12 A ∠150°
10 Ω ∠ − 90°
2 2.4 sin(10 t + 150°) = 3.39 sin(104t + 150°)
2 12 sin(104t + 150°) = 16.97 sin(104t + 150°)
4
Is = I1 + I2 = 2.4 A ∠150° + 12 A ∠150° = 14.4 A ∠150°
is = 2 14.4 sin(104t + 150°) = 20.36 sin(104t + 150°)
L1 || L2 = 60 mH || 120 mH = 40 mH
3
X LT = 2πfLT = 2π(10 Hz)(40 mH) = 251.33 Ω
Vm = I m X LT =
and υs =
(
)
2 24 A (251.33 Ω) = 2 6.03 kV
2 6.03 kV sin(103t + 30° + 90°)
or υs = 8..53 × 103 sin(103t + 120°)
b.
I m1 =
I m1 =
Vm
, X L1 = 2πfL1 = 2π(103 Hz)(60 mH) = 376.99 Ω
X L1
8.53 × 103
= 22.63 A
376.99 Ω
and i1 = 22.63 sin(103t + 30°)
3
X L 2 = 2πfL2 = 2π(10 Hz)(120 mH) = 753.98 Ω
I m2 =
8.53 × 103
= 11.31 A
753.98 Ω
and i2 = 11.31 sin(103t + 30°)
CHAPTER 14
165
39.
a.
c.
e.
4,326.66 ∠123.69°
l.
25.5 × 10−3 ∠−78.69°
a.
5.196 + j3.0
b.
6.946 + j39.39
c.
2530.95 + j6953.73
d.
e.
j0.04
f.
6.91 × 10−3 + j6.22 × 10−3
−469.85 − j171.01
h.
k.
a.
c.
−56.29 + j32.50
−4.31 − j6.16
15.03 ∠86.19
0.30 ∠88.09°
j.
l.
b.
d.
8.94 ∠−153.43°
101.73 ∠−39.94°
3.96 × 10−4 + j5.57 × 10−5
−0.85 + j0.85
5177.04 − j3625.0
−6.93 × 10−3 − j4.00 × 10−3
60.21 ∠4.76°
223.61 ∠−63.43°
86.18 ∠93.73°
f.
38.69 ∠−94.0°
a.
12.95 + j1.13
b.
8.37 + j159.78
e.
75.82 − j5.30
d.
f.
−34.51 − j394.49
a.
11.8 + j7.0
151.90 + j49.90
d.
5.20 + j1.60
7.00 × 10−6 + j2.44 × 10−7
−8.69 + j0.46
c.
4.72 ×10−6 + j71
b.
e.
209.30 + j311.0
f.
g.
6 ∠20° + 8 ∠80° = (5.64 + j2.05) + (1.39 + j7.88) = 7.03 + j9.93
h.
a.
c.
e.
g.
166
61.85 ∠−104.04°
h.
e.
c.
44.
6.58 × 10−3 ∠81.25°
k.
i.
43.
1.0 × 103 ∠84.29°
j.
g.
42.
1077.03 ∠21.80°
d.
2.83 ∠45°
11.78 ∠−49.82°
i.
41.
17.09 ∠69.44°
b.
f.
g.
40.
5.0 ∠36.87°
−21.20 + j12.0
(29.698 + j29.698) + (31.0 + j53.69) − (−35 + j60.62) = 95.7 + j22.77
−12.0 + j34.0
56. × 10−3 − j 8 × 10−3
8.00 ∠20°
40 × 10−3 ∠40°
b.
d.
f.
h.
86.80 + j312.40
698.00 ∠−114°
49.68 ∠−64.0°
−16,740 ∠160°
CHAPTER 14
45.
a.
c.
e.
f.
g.
h.
46.
6.0 ∠−50°
b.
109 ∠−170.0°
d.
4∠0°
200 × 10−6 ∠60°
76.47 ∠−80
5.93 ∠−134.47°
(0.05 + j0.25)/(8 −j60) = 0.255∠78.69°/60.53 ∠−82.41° = 4.21 × 10−3 ∠161.10°
9.30 ∠−43.99°
a.
10 − j 5
= 10.0 − j5.0
1 + j0
b.
8 ∠60°
8 ∠60°
=
= 19.38 × 10−3 ∠−15.69°
102 + j 400 412.80 ∠75.69°
c.
(6 ∠20°)(120 ∠ − 40°)(8.54 ∠69.44°) 6.15 × 103 ∠49.44°
=
= 3.07 × 103 ∠79.44°
2 ∠ − 30°
2 ∠ − 30°
d.
(0.16 ∠120°)(300 ∠40°)
48 ∠160°
= 5.06 ∠88.44°
=
9.487 ∠71.565°
9.487 ∠71.565°
e.
⎞
1
1
⎛
⎞⎛ 8 ⎞⎛
⎟
⎜
⎟⎜
2 ⎟⎜
−4
36
j
30
−
⎝ 4 × 10 ∠20° ⎠ ⎝ j ( j ) ⎠ ⎝
⎠
⎛ 8 ⎞⎛
1
⎞
⎟⎜
⎟
j
46.861
39.81
−
∠
−
°
⎠
⎝ ⎠⎝
( 2500 ∠ − 20° ) ⎜
(2500 ∠−20°)(8j)(0.0213 ∠39.81°) = 426 ∠109.81°
47.
a.
b.
x + j4 + 3x + jy − j7 = 16
(x + 3x) + j(4 + y − 7) = 16 + j0
x + 3x = 16
4+y−7=0
4x = 16
y = +7 − 4
x=4
y=3
(10 ∠20°)(x ∠−60°) = 30.64 − j25.72
10x ∠−40° = 40 ∠−40°
10 x = 40
x=4
CHAPTER 14
167
c.
5x + j10
2 − jy
──────
10x + j20 − j5xy − j210y = 90 − j70
(10x + 10y) + j(20 − 5xy) = 90 − j70
10x + 10y = 90
x+y=9
x=9−y⇒
20 − 5xy = −70
20 − 5(9 − y)y = −70
5y(9 − y) = 90
y2 − 9y + 18 = 0
−( −9) ± (−9) 2 − 4(1)(18)
2
9±3
= 6, 3
y=
2
y=
For y = 6, x = 3
y = 3, x = 6
(x = 3, y = 6) or (x = 6, y = 3)
d.
48.
a.
c.
e.
49.
a.
c.
e.
50.
51.
52.
168
80 ∠0°
= 4 ∠−θ = 3.464 − j2 = 4 ∠−30°
40 ∠θ
θ = 30°
160.0 ∠30°
70.71 ∠−90°
4.24 × 10−6 ∠90°
56.57 sin(377t + 20°)
11.31 × 10−3 sin(377t + 120°)
1696.8 sin(377t − 50°)
b.
d.
f.
b.
d.
f.
25 × 10−3 ∠−40°
14.14 ∠0°
2.55 × 10−6 ∠70°
169.68 sin (377t + 10°)
7.07 sin(377t + 90°)
6000 sin(377t − 180°)
(Using peak values)
ein = υa + υb ⇒ υa = ein − υb
= 60 V ∠20° − 20 V ∠−20°
= 46.49 V ∠ 36.05°
and ein = 46.49 sin (377t + 36.05°)
is = i1 + i2 ⇒ i1 = is − i2
(Using peak values) = (20 × 10−6 A ∠60°) − (6 × 10−6 A ∠−30°) = 20.88 × 10−6 A ∠ 76.70°
i1 = 20.88 × 10−6 sin (ωt + 76.70°)
e = υa + υb + υc
= 60 V ∠30° + 30 V ∠60° + 40 V ∠120°
= 102.07 V ∠62.61°
and e = 102.07 sin(ωt + 62.61°)
CHAPTER 14
53.
(Using effective values)
Is = I1 + I2 + I3 = 4.24 mA ∠180° + 5.66 mA ∠−180° + 11.31 mA ∠−180°
= −4.24 mA − 5.66 mA − 11.31 mA
= 21.21 × 10−3 sin (377t + 180°)
is = −21.21 × 10−3 sin 377t
CHAPTER 14
169
Chapter 15
1.
a.
b.
c.
d.
e.
2.
170
R ∠ 0° = 6.8 Ω ∠ 0° = 6.8 Ω
XL = ωL = (377 rads/s)(1.2 H) = 452.4 Ω
XL ∠ 90° = 452.4 Ω ∠ 90° = +j452.4 Ω
XL = 2πfL = (6.28)(50 Hz)(0.05 H) = 15.7 Ω
XL ∠ 90° = 15.7 Ω ∠ 90° = +j15.7 Ω
1
1
=
= 1 kΩ
ωC (100 rad/s)(10 × 10−6 F)
XC ∠ −90° = 1 kΩ ∠ −90° = −j1 kΩ
XC =
1
1
=
= 318.47 Ω
3
2π fC 2π (10 × 10 Hz)(0.05 × 10−6 F)
XC ∠ −90° = 318.47 Ω ∠ −90° = −j318.47 Ω
XC =
f.
R ∠ 0° = 220 Ω ∠ 0° = 220 Ω
a.
V = 10.61 V ∠ 10°, I =
V ∠θ 10.61 V ∠10°
= 3.54 A ∠ 10°
=
R∠0°
3 Ω ∠0°
b.
V = 39.60 V ∠ 10°, I =
39.60 V ∠10°
V ∠θ
=
= 5.66 A ∠ −80°
7 Ω ∠90°
X L ∠90°
c.
V = 17.68 V ∠ −20°, I =
d.
V = 2.828 mV ∠ −120°, I =
e.
V = 11.312 V ∠ 60°, I =
f.
V = 84.84 V ∠ 0°, XC =
i = 5 sin (ωt + 10°)
i = 8 sin (ωt − 80°)
i = 0.25 sin (ωt + 70°)
17.68 V ∠ − 20°
V ∠θ
= 0.1768 A ∠ 70°
=
X C ∠ − 90° 100 Ω ∠ − 90°
V ∠θ 2.828 mV ∠ − 120°
= 0.555 μA ∠−120°
=
5.1 kΩ ∠0°
R ∠0°
i = 0.785 × 10−6 sin (ωt − 120°)
11.312 V ∠60°
V ∠θ
=
= 150.03 mA ∠ −30°
X L ∠90° (377 rad/s)(0.2 H ∠90°)
i = 106.09 × 10−3 sin (377t − 30°)
1
1
=
= 15.924 Ω
2π fC 2π (5 kHz)(2 μ F)
84.84 V ∠0°
V ∠θ
I=
=
= 5.328 A ∠90°
X C ∠ − 90° 15.924 Ω ∠ − 90°
i = 7.534 sin (ωt + 90°)
CHAPTER 15
3.
a.
b.
I = (0.707)(4 mA ∠ 0°) = 2.828 mA ∠ 0°
V = (I ∠ 0°)(R ∠ 0°) = 2.828 mA ∠ 0°)(22 Ω ∠ 0°) = 62.216 mV ∠ 0°
υ = 88 × 10−3 sin ωt
I = (0.707)(1.5 A ∠ 60°) = 1.061 A ∠ 60°
XL = ωL = (1000 rad/s)(0.016 H) = 16 Ω
V = (I ∠ θ)(XL ∠ 90°) = (1.061 A ∠ 60°)(16 Ω ∠ 90°) = 16.98 V ∠ 150°
υ = 16.98 sin(1000t + 150°)
c.
I = (0.707)(2 mA ∠ 40°) = 1.414 mA ∠ 40°
1
1
=
= 127.39 kΩ
XC =
ωC (157rad/s)(0.05 × 10−6 F)
V = (I ∠ θ)(XC ∠ −90°) = 1.414 mA ∠ 40°)(127.39 kΩ ∠ −90°) = 180.13 V ∠ −50°
Vp = 2(180.13 V) = 254.7 V
and υ = 254.7 sin (157t − 50°)
4.
a.
b.
c.
5.
a.
b.
c.
6.
ZT = 6.8 Ω + j8.2 Ω = 10.65 Ω ∠ 50.33°
ZT = 2 Ω − j6 Ω + 10 Ω = 12 Ω − j6 Ω = 13.42 Ω ∠ −26.57°
ZT = 1 kΩ + j3 kΩ + 4 kΩ + j7 kΩ = 5 kΩ + j10 kΩ = 11.18 kΩ ∠ 63.44°
ZT = 3 Ω + j4 Ω − j5 Ω = 3 Ω − j1 Ω = 3.16 Ω ∠ −18.43°
ZT = 1 kΩ + j8 kΩ − j4 kΩ = 1 kΩ + j4 kΩ = 4.12 kΩ ∠ 75.96°
LT = 240 mH
XL = ωL = 2πfL = 2π(103 Hz)(240 × 10−3 H) = 1.51 kΩ
1
1
XC =
=
= 1.59 kΩ
3
2π fC 2π (10 Hz)(0.1 × 10−6 F)
= 470 Ω + j1.51 kΩ − j1.59 kΩ
= 470 Ω − j80 Ω = 476.76 Ω ∠ −9.66°
a.
ZT =
b.
ZT =
c.
ZT =
CHAPTER 15
E 120 V ∠0°
=
= 2 Ω ∠ −70° = 0.684 Ω − j1.879 Ω = R − jXC
I 60 A ∠70°
E 80 V ∠320°
=
= 4 kΩ ∠ 280° = 4 kΩ ∠ −80° = 0.695 kΩ − j3.939 Ω
I 20 mA ∠40°
= R − jXC
8 kV ∠0°
E
=
= 40 kΩ ∠ 60° = 20 kΩ + j34.64 kΩ = R + jXL
I 0.2 A ∠ − 60°
171
7.
ZT = 8 Ω + j6 Ω = 10 Ω ∠ 36.87°
a.
I = E/ZT = 100 V ∠ 0°/10 Ω ∠ 36.87° = 10 A ∠ −36.87°
VR = (I ∠ θ)(R ∠ 0°) = (10 A ∠ −36.87°)(8 Ω ∠ 0°) = 80 V ∠ −36.87°
VL = (I ∠ θ)(XL ∠ 90°) = (10 A ∠ −36.87°)(6 Ω ∠ 90°) = 60 V ∠ 53.13°
c.
P = I2R = (10 A)2 8 Ω = 800 W
f.
Fp = cos θT = R/ZT = 8 Ω/10 Ω = 0.8 lagging
g.
υR = 113.12 sin(ωt − 36.87°)
υL = 84.84 sin(ωt + 53.13°)
i = 14.14 sin (ωt − 36.87°)
h.
8.
a.
ZT = 6 Ω − j30 Ω = 30.59 Ω ∠−78.69°
c.
I=
120 V ∠20°
E
=
= 3.92 A ∠98.69°
ZT 30.59 Ω ∠ − 78.69°
VR = (I ∠θ)(R ∠0°) = (3.92 A ∠98.69°)(6 Ω ∠0°) = 23.52 V ∠98.69°
VC = (I ∠θ)(XC ∠−90°) = (3.92 A ∠98.69°)(30 Ω ∠−90°) = 117.60 V ∠8.69°
f.
g.
h.
9.
a.
P = I2R = (3.92 A)2 6 Ω = 92.2 W
Fp = R/ZT = 6 Ω/30.59 Ω = 0.196 leading
i = 5.54 sin(377t + 98.69°)
υR = 33.26 sin(377t + 98.69°)
υC = 166.29 sin(377t + 8.69°)
XC =
1
1
=
= 795.77 Ω
3
2π fC 2π (10 Hz)(0.2 × 10−6 F)
ZT = 2.2 kΩ − j795.77 Ω = 2.34 kΩ ∠−19.89°
b.
c.
d.
172
I = E/ZT = 14.14 V ∠0°/2.34 kΩ ∠−19.89° = 6.04 mA ∠19.89°
VR = (I ∠θ)(R ∠0°) = (6.04 mA ∠ 19.89°)(2.2 × 103 Ω ∠0°) = 13.29 V ∠19.89°
VC = (I ∠θ)(XC ∠−90°) = (6.04 mA ∠19.89°)(795.77 Ω ∠−90°)
= 4.81 V ∠−70.11°
P = I2R = (6.04 mA)2 2.2 kΩ = 80.26 mW
Fp = cos θT = cos 19.89° = 0.94 leading
CHAPTER 15
10.
a.
c.
d.
f.
6Ω
= 16 mH
ω 377 rad/s
1
1
1
=
XC =
⇒ C=
= 265 μF
ωC
ω X C (377 rad/s)(10 Ω)
XL = ωL ⇒ L =
XL
=
50 V ∠0°
E
=
= 8.83 A ∠45°
5.66 Ω ∠ − 45°
ZT
VR = (I ∠θ)(R ∠0°) = (8.83 A ∠45°)(4 Ω ∠0°) = 35.32 V ∠45°
VL = (I ∠θ)(XL ∠90°) = (8.83 A ∠45°)(6 Ω ∠90°) = 52.98 V ∠135°
VC = (I ∠θ)(XC ∠−90°) = (8.83 A ∠45°)(10 Ω ∠−90°) = 88.30 V ∠−45°
I=
E = VR + VL + VC
50 V ∠0° = 35.32 V ∠45° + 52.98 V ∠135° + 88.30 V ∠−45°
50 V ∠0° = 49.95 V ∠0° ≅ 50 V ∠ 0°
g.
P = I2R = (8.83 A)2 4 Ω = 311.88 W
h.
Fp = cos θT =
i.
11.
ZT = 4 Ω + j6 Ω − j10 Ω = 4 Ω − j4 Ω = 5.66 Ω ∠−45°
R
ZT
= 4 Ω/5.66 Ω = 0.707 leading
i = 12.49 sin(377t + 45°)
e = 70.7 sin 377t
υR = 49.94 sin(377t + 45°)
υL = 74.91 sin(377t + 135°)
υC = 124.86 sin(377t − 45°)
a.
ZT = 1.8 kΩ + j2 kΩ − j0.6 kΩ = 1.8 kΩ + j1.2 kΩ = 2.16 kΩ ∠33.69°
c.
XC =
1
1
1
=
⇒ C=
= 5.31 μF
ωC
ω X C (314 rad/s)(0.6 kΩ)
2 × 103 Ω
= 6.37 H
XL = ωL ⇒ L = X L =
ω 314 rad/s
d.
I = E/ZT = 4.242 V ∠60°/2.16 kΩ ∠33.69° = 1.96 mA ∠26.31°
VR = (I ∠θ)(R ∠0°) = (1.96 mA ∠26.31°)(1.8 kΩ ∠0°) = 3.53 V ∠26.31°
VL = (I ∠θ)(XL ∠90°) = (1.96 mA ∠26.31°)(2 kΩ ∠90°) = 2.68 V ∠116.31°
VC = (I ∠θ)(XC ∠−90°) = (1.96 mA ∠26.31°)(0.6 kΩ ∠−90°) = 1.18 V ∠−63.69°
g.
P = I2R = (1.96 mA)2 1.8 kΩ = 6.91 mW
h.
Fp = cos θT = cos 33.69° = 0.832 lagging
CHAPTER 15
173
i.
12.
13.
i = 2.77 × 10−3 sin(ωt + 26.31°)
υR = 4.99 sin(ωt + 26.31°)
υL = 3.79 sin(ωt + 116.31°)
υC = 1.67 sin(ωt − 63.69°)
⎛ 45.27 V ⎞
V80Ω(rms) = 0.7071 ⎜
⎟ = 16 V
2
⎝
⎠
80 Ω (20 V)
= 16 V
Vscope =
80 Ω + R
1600 = 1280 + 16 R
320
= 20 Ω
R=
16
a.
⎛ 21.28 V ⎞
VL(rms) = 0.7071 ⎜
⎟ = 7.524 V
2
⎝
⎠
V
7.524 V
XL = L =
= 251.303 Ω
I L 29.94 mA
XL
251.303 Ω
XL = 2πfL ⇒ L =
=
= 39.996 mH ≅ 40 mH
2π (1 kHz)
2π f
b.
E2 = VR2 + VL2
VR =
=
R=
14.
E 2 − VL2
(100 V) − (56.611) =
43.389 = 6.587 V
VR
6.587 V
=
= 220 Ω
29.94 mA
IR
⎛ 8.27 V ⎞
VR(rms) = 0.7071 ⎜
⎟ = 2.924 V
⎝ 2 ⎠
VC =
E 2 − VR2
= 144 − 8.55 = 135.45 = 11.638 V
2.924 V
= 292.4 μA
10 k Ω
V
11.638 V
= 39.802 kΩ
XC = C =
292.4 μ A
IC
1
1
1
XC =
⇒C=
=
= 99.967 pF ≅ 100 pF
2π fC
2π f X C 2π (40 kHz)(39.802 kΩ)
IC = IR =
174
CHAPTER 15
15.
a.
b.
16.
a.
b.
17.
a.
(2 k Ω ∠0°)(120 V ∠60°) 240 V ∠60°
=
= 29.09 V ∠−15.96°
2 k Ω + j8 k Ω
8.25 ∠75.96°
(8 k Ω ∠90°)(120 V ∠60°)
= 116.36 V ∠74.04°
V2 =
8.25 k Ω ∠75.96°
V1 =
( 40 Ω ∠90°)(60 V ∠5°) 2400 V ∠95°
=
= 48.69 V ∠40.75°
6.8 Ω + j 40 Ω + 22 Ω
28.8 + j 40
(22 Ω ∠0°)(60 V ∠5°)
1.32 kV ∠5°
=
V2 =
= 26.78 V ∠−49.25°
49.29 Ω ∠54.25°
49.29 Ω ∠54.25°
V1 =
(20 Ω ∠90°)(20 V ∠70°)
= 14.14 V ∠−155°
20 Ω + j 20 Ω − j 40 Ω
(40 Ω ∠ − 90°)(20 V ∠70°)
V2 =
= 28.29 V ∠25°
28.28 Ω ∠ − 45°
V1 =
ZT = 4.7 kΩ + j30 kΩ + 3.3 kΩ − j10 kΩ = 8 kΩ + j20 kΩ = 21.541 kΩ ∠68.199°
ZT′ = 3.3 kΩ + j30 kΩ − j10 kΩ = 3.3 kΩ + j20 kΩ = 20.27 kΩ ∠80.631°
Z′ E (20.27 kΩ ∠80.631°)(120 V ∠0°)
= 112.92 V ∠12.432°
V1 = T =
ZT
21.541 kΩ ∠68.199°
Z′′ E
ZT′′ = 3.3 kΩ − j10 kΩ = 10.53 kΩ ∠−71.737°
V2 = T
ZT
(10.53 kΩ ∠ − 71.737°)(120 V ∠0°)
=
= 58.66 V ∠−139.94°
21.541 kΩ ∠68.199°
XL = ωL = (377 rad/s)(0.4 H) = 150.8 Ω
1
1
= 663 Ω
XC =
=
ωC (377 rad/s)(4 μ F)
ZT = 30 Ω + j150.8 Ω − j663 Ω = 30 Ω − j512.2 Ω = 513.08 Ω ∠−86.65°
20 V ∠40°
E
=
= 39 mA ∠126.65°
I=
ZT 513.08 Ω ∠ − 86.65°
VR = (I ∠ θ)(R ∠ 0°) = (39 mA ∠126.65°)(30 Ω ∠0°) = 1.17 V ∠126.65°
VC = (39 mA ∠126.65°)(0.663 kΩ ∠−90°) = 25.86 V ∠36.65°
30 Ω
R
=
= 0.058 leading
ZT 513.08 Ω
b.
cos θT =
c.
P = I2R = (39 mA)2 30 Ω = 45.63 mW
f.
VR =
g.
ZT = 30 Ω − j512.2 Ω = R − jXC
(30 Ω ∠0°)(20 V ∠40°)
600 V ∠40°
=
= 1.17 V ∠126.65°
513.08 Ω ∠ − 86.65°
ZT
(0.663 k Ω ∠ − 90°)(20 V ∠40°)
VC =
= 25.84 V ∠36.65°
513.08 Ω ∠ − 86.65°
CHAPTER 15
175
18.
a.
b.
Fp = cos θT = R/ZT = 30 Ω/141.95 Ω = 0.211 lagging
c.
P = I2R = (140.89 mA)2 30 Ω = 595.50 mW
f.
g.
19.
XL = ωL = (377 rad/s)(0.4 H) = 150.8 Ω
1
1
=
XC =
= 12.06 Ω
ωC (377 rad/s)(220 × 10−6 F)
ZT = 30 Ω + j150.8 Ω − j12.06 Ω
= 30 Ω + j138.74 Ω = 141.95 Ω ∠77.80°
I = E/ZT = 20 V ∠40°/141.95 Ω ∠77.80° = 140.89 mA ∠−37.80°
VR = (I ∠ θ)(R ∠0°) = (140.89 mA ∠−37.80°)(30 Ω ∠0°) = 4.23 V ∠−37.80°
VC = (I ∠ θ)(XC ∠−90°) = (140.89 mA ∠−37.80°)(12.06 Ω ∠−90°)
= 1.70 V ∠−127.80°
(30 Ω ∠0°)(20 V ∠40°)
= 4.23 V ∠−37.80°
141.95 Ω ∠77.80°
(12.06 Ω ∠ − 90°)(20 V ∠40°)
= 1.70 V ∠−127.80°
VC =
141.95 Ω ∠77.80°
VR =
ZT = 30 Ω + j138.74 Ω = R + jXL
P = VI cos θ ⇒ 8000 W = (200 V)(I)(0.8)
8000 A
= 50 A
I=
160
0.8 = cos θ
θ = 36.87°
V = 200 V ∠0°, I = 50 A ∠−36.87°
200 V ∠0°
V
=
ZT =
= 4 Ω ∠36.87° = 3.2 Ω + j2.4 Ω
I 50 A ∠ − 36.87°
20.
176
P = VI cos θ ⇒ 300 W = (120 V)(3 A) cos θ
cos θ = 0.833 ⇒ θ = 33.59°
V = 120 V ∠0°, I = 3 A ∠−33.59°
V
120 V ∠0°
=
ZT =
= 40 Ω ∠33.59° = 33.34 Ω + j22.10 Ω
I 3 A ∠ − 33.59°
RT = 33.34 Ω = 2 Ω + R ⇒ R = 31.34 Ω
CHAPTER 15
21.
a.
b.
ZT =
R 2 + X L2 ∠tan−1XL/R
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
ZT
1.0 kΩ
1.008 kΩ
1.181 kΩ
1.606 kΩ
2.134 kΩ
2.705 kΩ
XLE
ZT
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
VL
0.0 V
0.623 V
2.66 V
3.888 V
4.416 V
4.646 V
θT
0.0°
7.16°
32.14°
51.49°
62.05°
68.3°
VL =
c.
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
CHAPTER 15
θL = 90° − tan−1 XL/R
90.0°
82.84°
57.85°
38.5°
27.96°
21.7°
177
d.
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
22.
a.
ZT =
│ZT│ =
−1
2
2
R + X C ∠−tan XC/R
R 2 + X C2 , θT = −tan−1XC/R
f
0 kHz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
178
VR = RE/ZT
5.0 V
4.96 V
4.23 V
3.11 V
2.34 V
1.848 V
│ZT│
∞Ω
333.64 Ω
145.8 Ω
118.54 Ω
109.85 Ω
104.94 Ω
θT
−90.0°
−72.56°
−46.7°
−32.48°
−24.45°
−17.66°
CHAPTER 15
b.
VC =
XC E
( X C ∠ − 90°)(E ∠0°)
=
∠−90° + tan−1XC/R
2
2
R − jX C
R + XC
│VC│ =
XC E
R 2 + X C2
│VC│
10.0 V
9.54 V
7.28 V
5.37 V
4.14 V
3.03 V
f
0 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
c.
θC = −90° + tan−1 XC/R
θC
0.0°
−17.44°
−43.3°
−57.52°
−65.55°
−72.34°
f
0 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
d.
│VR│ =
R + X C2
f
0 Hz
1 kHz
3 kHz
5 kHz
7 kHz
10 kHz
CHAPTER 15
RE
2
│VR│
0.0 V
3.0 V
6.86 V
8.44 V
9.10 V
9.53 V
179
23.
a.
R 2 + ( X L − X C ) 2 ∠ tan −1 ( X L − X C ) / R
ZT =
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
b.
ZT
∞Ω
19,793.97 Ω
3,496.6 Ω
1,239.76 Ω
1,145.47 Ω
1,818.24 Ω
│VC│ =
XC E
ZT
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
c.
180
θT
−90.0°
−87.1°
−73.38°
−36.23°
+29.19°
+56.63°
E
ZT
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
│VC│
120.0 V
120.61 V
136.55 V
192.57 V
138.94 V
65.65 V
I =
I
0.0 mA
6.062 mA
34.32 mA
96.79 mA
104.76 mA
66.0 mA
CHAPTER 15
24.
1
1
1
=
= 1.54 kHz
=R⇒f=
2π fC
2π RC 2π (220 Ω)(0.47 μ F)
a.
XC =
b.
Low frequency: XC very large resulting in large ZT
High frequency: XC approaches zero ohms and ZT approaches R
c.
f = 100 Hz: XC =
1
1
=
= 3.39 kΩ
2π fC 2π (100 Hz)(0.47 μ F)
f = 10 kHz: XC =
1
1
=
= 33.86 Ω
2π fC 2π (10 kHz)(0.47 μ F)
ZT ≅ XC
ZT ≅ R
d.
−
e.
f = 40 kHz: XC =
θ = −tan−1
25.
a.
b.
c.
d.
e.
f.
1
1
=
= 8.47 kΩ
2π fC 2π (40 kHz)(0.47 μ F)
XC
8.47 Ω
= − tan −1
= −2.2°
R
220 Ω
ZT = 91 Ω ∠0° = R ∠0°, YT = 10.99 mS ∠0° = G ∠0°
ZT = 200 Ω ∠90° = XL ∠90°, YT = 5 mS ∠−90° = BL ∠−90°
ZT = 0.2 kΩ ∠−90° = XC ∠−90°, YT = 5.00 mS ∠90° = BC ∠90°
(10 Ω∠0°)(60 Ω ∠90°)
= 9.86 Ω ∠9.46° = 9.73 Ω + j1.62 Ω = R + jXL
10 Ω + j 60 Ω
YT = 0.10 S ∠−9.46° = 0.1 S − j0.02 S = G − jBL
ZT =
22 Ω || 2.2 Ω = 2 Ω
(2 Ω∠0°)(6 Ω ∠ − 90°)
12 Ω ∠ − 90°
=
= 1.90 Ω ∠−18.43°
ZT =
2 Ω − j6 Ω
6.32 Ω ∠ − 71.57°
= 1.80 Ω − j0.6 Ω = R − jXC
YT = 0.53 S ∠18.43° = 0.5 S + j0.17 S = G + jBC
1
1
1
+
+
3 k Ω ∠0° 6 k Ω ∠90° 9 k Ω ∠ − 90°
= 0.333 × 10−3 ∠0° + 0.167 × 10−3 ∠−90° + 0.111 × 10−3 ∠90°
= 0.333 × 10−3 S − j0.056 × 10−3 S = 0.34 mS ∠−9.55°
= G − jBL
1
= 2.94 kΩ ∠9.55° = 2.90 kΩ + j0.49 kΩ
ZT =
YT
YT =
CHAPTER 15
181
26.
27.
a.
ZT = 4.7 Ω + j8 Ω = 9.28 Ω ∠ 59.57°, YT = 0.108 S ∠−59.57°
b.
ZT = 33 Ω + 20 Ω − j70 Ω = 53 Ω − j70 Ω = 87.80 Ω ∠−52.87°
YT = 54.7 mS − j93.12 mS = G − jBL
YT = 11.39 mS ∠52.87° = 6.88 mS + j9.08 mS = G + jBC
c.
ZT = 200 Ω + j500 Ω − j600 Ω = 200 Ω − j100 Ω = 223.61 Ω ∠−26.57°
a.
YT =
b.
YT =
YT = 4.47 mS ∠26.57° = 4 mS + j2 mS = G + jBC
I 60 A ∠70°
= 0.5 S ∠70° = 0.171 S + j0.470 S = G + jBC
=
E 120 V ∠0°
1
1
R=
= 5.85 Ω, XC =
= 2.13 Ω
G
BC
I 20 mA ∠40°
=
= 0.25 mS ∠−280° = 0.25 mS ∠80°
E 80 V ∠320°
= 0.043 mS + j0.246 mS = G + jBC
1
1
= 23.26 kΩ, XC =
= 4.07 kΩ
R=
G
BC
28.
c.
YT =
a.
YT =
c.
1
1
+
= 0.1 S − j0.05 S = 111.8 mS ∠−26.57°
10 Ω ∠0° 20 Ω ∠90°
E = Is/YT = 2 A ∠0°/111.8 mS ∠−26.57°= 17.89 V ∠26.57°
E ∠θ
= 17.89 V ∠26.57°/10 Ω ∠0° = 1.79 A ∠26.57°
IR =
R ∠0°
E ∠θ
IL =
= 17.89 V ∠26.57°/20 Ω ∠90° = 0.89 A ∠−63.43°
X L ∠90°
f.
P = I2R = (1.79 A)2 10 Ω = 32.04 W
g.
Fp =
h.
182
I 0.2 A ∠ − 60°
=
= 0.25 mS ∠−60° = 0.0125 mS − j0.02165 mS = G − jBL
8 kV ∠0°
E
1
1
= 80 kΩ, XL =
= 46.19 kΩ
R=
G
BL
G
YT
=
0.1 S
= 0.894 lagging
111.8 mS
e = 25.30 sin(377t + 26.57°)
iR = 2.53 sin(377t + 26.57°)
iL = 1.26 sin(377t − 63.43°)
is = 2.83 sin 377t
CHAPTER 15
29.
a.
c.
e.
1
1
= 0.1 mS ∠0° + 0.05 mS ∠−90°
+
10 k Ω ∠0° 20 k Ω ∠ − 90°
= 0.112 mS ∠26.57°
2 mA ∠20°
= 17.89 V ∠−6.57°
E = Is =
0.1118 mS ∠26.565°
YT
E
17.89 V ∠ − 6.57°
IR =
= 1.79 mA ∠−6.57°
=
10 k Ω ∠0°
ZR
E
17.89 V ∠ − 6.57°
= 0.90 mA ∠83.44°
IC =
=
20 k Ω ∠ − 90°
ZC
Is = IR + IC
2 mA ∠20° = 1.79 mA ∠−6.57° + 0.90 mA ∠83.44°
= 1.88 mA + j0.69 mA
2 mA ∠20° ✓ 2 mA ∠20.15°
=
f.
P = I2R = (1.79 mA)2 10 kΩ = 32.04 mW
g.
Fp =
h.
30.
YT =
a.
c.
G
YT
=
0.1 mS
= 0.894 leading
0.1118 mS
ω = 2πf = 377 rad/s
is = 2.83 × 10−3 sin(ωt + 20°)
iR = 2.53 × 10−3 sin(ωt − 6.57°)
iC = 1.27 × 10−3 sin(ωt + 83.44°)
e = 25.3 sin(ωt − 6.57°)
YT =
1
1
= 0.083 S − j0.1 S = 129.96 mS ∠−50.31°
+
12 Ω ∠0° 10 Ω ∠90°
Is = EYT = (60 V ∠0°)(0.13 S∠−50.31°) = 7.8 A ∠−50.31°
E ∠θ
IR =
= 60 V ∠0°/12 Ω ∠0° = 5 A ∠0°
R ∠ 0°
E ∠θ
IL =
= 60 V ∠0°/10 Ω ∠90° = 6 A ∠−90°
X L ∠ 90°
f.
P = I2R = (5 A)2 12 Ω = 300 W
g.
Fp = G/YT = 0.083 S/0.13 S = 0.638 lagging
h.
e = 84.84 sin 377t
iR = 7.07 sin 377t
iL = 8.484 sin(377t − 90°)
is = 11.03 sin(377t − 50.31°)
CHAPTER 15
183
31.
YT =
c.
XC =
d.
(0.707)(3 A) ∠60°
2.121 A ∠60°
E = Is =
= 2.40 V ∠79.81°
=
0.885 S ∠ − 19.81° 0.885 S ∠ − 19.81°
YT
E ∠θ 2.397 V ∠79.81°
IR =
= 2.00 A ∠79.81°
=
1.2 Ω ∠0°
R ∠0°
E ∠θ
2.397 V ∠79.81°
IL =
= 1.20 A ∠−10.19°
=
2 Ω ∠90°
X L ∠90°
E ∠θ
2.397 V ∠79.81°
IC =
=
= 0.48 A ∠169.81°
5 Ω ∠ − 90°
X C ∠ − 90°
g.
h.
Fp =
i.
a.
c.
184
1
1
1
⇒ C=
= 531 μF
=
ωC
ω X C (377 rad/s)(5 Ω)
2Ω
XL = ωL ⇒ L = X L =
= 5.31 mH
ω 377 rad/s
Is = IR + IL + IC
2.121 A ∠60° = 2.00 A ∠79.81° + 1.20 A ∠−10.19° + 0.48 A ∠169.81°
✓
2.121 A ∠60° = 2.13 A ∠60.01°
P = I2R = (2.00 A)2 1.2 Ω = 4.8 W
f.
32.
1
1
1
+
+
1.2 Ω ∠0° 2 Ω ∠90° 5 Ω ∠ − 90°
= 0.833 S ∠0° + 0.5 S ∠−90° + 0.2 S ∠90°
= 0.833 S − j0.3 S = 0.89 S ∠−19.81°
ZT = 1.12 Ω ∠ 19.81°
a.
G
YT
=
0.833 S
= 0.941 lagging
0.885 S
e = 3.39 sin(377t + 79.81°)
iR = 2.83 sin(377t + 79.81°)
iL = 1.70 sin(377t − 10.19°)
iC = 0.68 sin(377t + 169.81°)
1
1
1
+
+
3 k Ω ∠0° 4 k Ω ∠90° 8 k Ω ∠ − 90°
= 0.333 mS ∠0° + 0.25 mS ∠−90° + 0.125 mS ∠90°
= 0.333 mS + j0.125 mS = 0.356 mS ∠20.57°
YT =
XL = ωL ⇒ L = XL/ω = 4000 Ω/377 rad/s = 10.61 H
1
1
1
⇒ C=
= 0.332 μF
=
XC =
ωC
ω X C (377 rad/s)(8 kΩ)
CHAPTER 15
d.
g.
P = I2R = (3.31 mA)23 kΩ = 32.87 mW
h.
Fp = G/YT = 0.333 mS/0.356 mS = 0.935 leading
i.
33.
E = I/YT = 3.535 mA ∠−20°/0.356 mS ∠20.57° = 9.93 V ∠−40.57°
E ∠θ
IR =
= 9.93 V∠−40.57°/3 kΩ ∠0° = 3.31 mA ∠−40.57°
R ∠0°
E ∠θ
IL =
= 9.93 V∠−40.57°/4 kΩ ∠90° = 2.48 mA ∠−130.57°
X L ∠90°
E ∠θ
IC =
= 9.93 V∠−40.57°/8 kΩ ∠−90° = 1.24 mA ∠49.43°
X C ∠ − 90°
e = 14.04 sin(377t − 40.57°)
iR ≅ 4.68 × 10−3 sin(377t − 40.57°)
iL ≅ 3.51 × 10−3 sin(377t − 130.57°)
iC = 1.75 × 10−3 sin(377t + 49.43°)
1
1
1
+
+
5 Ω ∠ − 90° 22 Ω ∠0° 10 Ω ∠90°
= 0.2 S ∠90° + 0.045 S ∠0° + 0.1 S ∠−90°
= 0.045 S + j0.1 S = 0.110 S ∠65.77°
ZT = 9.09 Ω ∠ −65.77°
a.
YT =
c.
C=
d.
f.
g.
1
1
= 636.9 μF
=
ω X C (377 rad/s)(5 Ω)
10 Ω
L= XL=
= 31.8 mH
ω 314 rad/s
E = (0.707)(35.4 V) ∠60° = 25.03 V ∠60°
Is = EYT = (25.03 V ∠60°)(0.11 S ∠65.77°) = 2.75 A ∠125.77°
E ∠θ
25.03 V ∠60°
= 5 A ∠150°
IC =
=
5 ∠ − 90°
X C ∠ − 90°
E ∠θ 25.03 V ∠60°
= 1.14 A ∠60°
IR =
=
22 Ω ∠0°
R ∠0°
E ∠θ
25.03 V ∠60°
= 2.50 A ∠−30°
IL =
=
10 Ω ∠90°
X L ∠90°
Is = IC + IR + IL
2.75 A ∠125.77° = 5 A ∠150° + 1.14 A ∠60° + 2.50 A ∠−30°
= (−4.33 + j2.5) + (0.57 + j0.9) + (2.17 − j1.25)
= −1.59 + j2.24
✓
= 2.75 ∠125.4°
P = I2R = (1.14 A)2 22 Ω = 28.59 W
CHAPTER 15
185
h.
i.
34.
35.
Fp =
G
YT
0.045 S
= 0.409 leading
0.110 S
e = 35.4 sin(314t + 60°)
is = 3.89 sin(314t + 125.77°)
iC = 7.07 sin(314t + 150°)
iR = 1.61 sin(314t + 60°)
iL = 3.54 sin(314t − 30°)
( 80 Ω ∠90°)(20 A ∠40°) 1600 A ∠130°
= 19.28 A ∠55.38°
=
22 Ω + j80 Ω
82.97 ∠74.62°
(22 Ω ∠0°)(20 A ∠40°)
440 A ∠40°
=
= 5.30 A ∠−34.62°
I2 =
82.97 Ω ∠74.62°
82.97 ∠74.62°
a.
I1 =
b.
I1 =
a.
ZT =
(12 Ω − j 6 Ω)(6 A ∠30°) (13.42 ∠ − 26.57°)(6 A ∠30°)
=
12 Ω − j 6 Ω + j 4 Ω
12 − j 2
80.52 A∠3.43°
=
= 6.62 A ∠12.89°
12.17∠ − 9.46°
(4 Ω ∠90°)(6 A ∠30°)
24 A ∠120°
=
I2 =
= 1.97 A ∠129.46°
12.17 Ω∠ − 9.46°
12.17 ∠ − 9.46°
(R ∠0° )( X C ∠ − 90° )
= RX C ∠−90° + tan−1XC/R
2
2
R − jX C
R + XC
│ZT│ =
f
0 Hz
1 kHz
2 kHz
3 kHz
4 kHz
5 kHz
10 kHz
20 kHz
186
=
RX C
R +
2
X C2
│ZT│
40.0 Ω
35.74 Ω
28.22 Ω
22.11 Ω
17.82 Ω
14.79 Ω
7.81 Ω
3.959 Ω
θT = −90° + tan−1XC/R
θT
0.0°
−26.67°
−45.14°
−56.44°
−63.55°
−68.30°
−78.75°
−89.86°
CHAPTER 15
b.
IRX C
│VC│ =
R 2 + X C2
│VC│
2.0 V
1.787 V
1.411 V
1.105 V
0.891 V
0.740 V
0.391 V
0.198 V
f
0 kHz
1 kHz
2 kHz
3 kHz
4 kHz
5 kHz
10 kHz
20 kHz
c.
│IR│ =
VC
R
│IR│
50.0 mA
44.7 mA
35.3 mA
27.64 mA
22.28 mA
18.50 mA
9.78 mA
4.95 mA
f
0 kHz
1 kHz
2 kHz
3 kHz
4 kHz
5 kHz
10 kHz
20 kHz
36.
a.
ZRZL
( R ∠0°)( X L ∠90°)
=
=
ZR + ZL
R + jX L
ZT =
│ZT│ =
CHAPTER 15
= I[ZT(f)]
RX L
R +
2
X L2
RX L
R +
2
X L2
∠90° − tan−1XL/R
θT = 90° − tan−1XL/R
f
| ZT |
0 Hz
1 kHz
5 kHz
7 kHz
10 kHz
0.0 kΩ
1.22 kΩ
3.91 kΩ
4.35 kΩ
4.65 kΩ
θT
90.0°
75.86°
38.53°
29.6°
21.69°
187
│IL│ =
b.
E
XL
f
| IL |
∞
31.75 mA
6.37 mA
4.55 mA
3.18 mA
0 Hz
1 kHz
5 kHz
7 kHz
10 kHz
c.
37.
IR =
YT =
E 40 V
=
= 8 mA (constant)
R 5kΩ
R 2 + X C2
RX C
∠90° − tan−1XC/R
f
│YT│
0 Hz 25.0 mS
1 kHz 27.98 mS
2 kHz 35.44 mS
3 kHz 45.23 mS
4 kHz 56.12 mS
5 kHz 67.61 mS
10 kHz 128.04 mS
20 kHz 252.59 mS
38.
YT =
0.0°
26.67°
45.14°
56.44°
63.55°
68.30°
78.75°
89.86°
1
(use data of Prob. 36), θ TY = − θTZ
ZT
f
0 Hz
1 kHz
5 kHz
7 kHz
10 kHz
188
θT
YT
∞
0.82 mS
0.256 mS
0.23 mS
0.215 mS
θT
−90.0°
−75.86°
−38.53°
−29.6°
−21.69°
CHAPTER 15
39.
a.
YT = G ∠0° + BL ∠−90° + BC ∠90°
B − BL
= G 2 + ( BC − BL ) 2 ∠tan−1 C
G
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
f
0 Hz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
b.
ZT =
1
YT
f
0 kHz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
CHAPTER 15
│YT│
XL ⇒ 0 Ω, ZT = 0 Ω,
YT = ∞ Ω
1.857 mS
1.018 mS
1.004 mS
1.036 mS
1.086 mS
│θT│
−90.0°
−57.42°
−10.87°
+5.26°
+15.16°
+22.95°
,θ T Z = − θ T Y
ZT
0.0 Ω
538.5 Ω
982.32 Ω
996.02 Ω
965.25 Ω
921.66 Ω
θT
90.0°
57.42°
10.87°
−5.26°
−15.16°
−22.95°
189
c.
VC(f) = I[ZT(f)]
│VC│
0.0 V
5.39 V
9.82 V
9.96 V
9.65 V
9.22 V
f
0 kHz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
d.
IL =
VC (f)
XL
f
0 kHz
1 kHz
5 kHz
10 kHz
15 kHz
20 kHz
40.
a.
Rp =
Xp =
b.
Rp =
Xp =
41.
a.
IL
10.0 mA
8.57 mA
3.13 mA
1.59 mA
1.02 mA
0.733 mA
Rs2 + X s2 (20 Ω) 2 + (40 Ω) 2
=
= 100 Ω (R)
Rs
20 Ω
Rs2 + X s2 2000 Ω
=
= 50 Ω (C)
40
Xs
Rs2 + X s2 (2 k Ω) 2 + (3 k Ω) 2
= 6.5 kΩ (R)
=
Rs
2 kΩ
Rs2 + X s2 (2 k Ω) 2 + (3 k Ω) 2
=
= 4.33 kΩ (C)
Xs
3kΩ
(8.2 k Ω)(20 k Ω) 2
=
Rs = 2
= 7.02 kΩ
X p + R p2 (20 k Ω) 2 + (8.2 k Ω) 2
R p X p2
Xs =
R p2 X p
X p2 + R p2
=
(8.2 k Ω) 2(20 k Ω)
= 2.88 kΩ
467.24 k Ω
ZT = 7.02 kΩ − j2.88 kΩ
b.
Rs =
Xs =
R p X p2
X p2 + R p2
R p2 X p
X p2 + R p2
=
(68 Ω)(40 Ω) 2
= 17.48 Ω
(40 Ω) 2 + (68 Ω) 2
=
(68 Ω) 2(40 Ω)
= 29.72 Ω
6224 Ω 2
ZT = 17.48 Ω + j29.72 Ω
190
CHAPTER 15
42.
a.
CT = 2 μF
1
1
XC =
=
= 79.62 Ω
3
ωC 2π (10 Hz)(2 μ F)
XL = ωL = 2π(103 Hz)(10 mH) = 62.80 Ω
YT =
1
1
1
+
+
220 Ω ∠0° 79.62 Ω ∠ − 90° 62.8 Ω ∠90°
= 4.55 mS ∠0° + 12.56 mS ∠90° + 15.92 mS ∠−90°
= 4.55 mS − j3.36 mS = 5.66 mS ∠−36.44°
E = I/YT = 1 A ∠0°/5.66 mS ∠−36.44° = 176.68 V ∠36.44°
E ∠θ
= 176.68 V ∠36.44°/220 Ω ∠0° = 0.803 A ∠36.44°
IR =
R ∠0°
E ∠θ
= 176.68 V ∠36.44°/62.80 ∠90° = 2.813 A ∠−53.56°
IL =
X L ∠90°
43.
b.
Fp = G/YT = 4.55 mS/5.66 mS = 0.804 lagging
e.
P = I2R = (0.803 A)2 220 Ω = 141.86 W
f.
Is = IR + 2IC + IL
I −I −I
and IC = s R L
2
1 A ∠0° − 0.803 A ∠36.44° − 2.813 A ∠ − 53.56°
=
2
1 − (0.646 + j 0.477) − (1.671 − j 2.263) −1.317 + j1.786
=
=
2
2
IC = −0.657 + j0.893 = 1.11 A ∠126.43°
g.
ZT =
a.
1
1
= 176.7 Ω ∠36.44°
=
YT
5.66 mS ∠ − 36.44°
= 142.15 Ω + j104.96 Ω = R + jXL
(R = 220 Ω) || (L = 1 H) || (C = 2 μF)
1
1
=
XC =
= 79.62 Ω
3
ωC 2π (10 Hz)(2 μ F)
XL = ωL = 2π(103 Hz)(1 H) = 6.28 kΩ
1
1
1
YT =
+
+
3
220 Ω ∠0° 6.28 × 10 Ω ∠90° 79.62 Ω ∠ − 90°
= 0.0045 − j0.1592 × 10−3 + j0.0126
= 4.5 × 10−3 − j0.1592 × 10−3 + j12.6 × 10−3
= 4.5 mS + j12.44 mS = 13.23 mS ∠70.11°
CHAPTER 15
191
E = I/YT = 1 A ∠0°/13.23 mS ∠70.11° = 75.6 V ∠−70.11°
E ∠θ
= 75.6 V ∠−70.11°/220 Ω ∠0° = 0.34 A ∠−70.11°
IR =
R ∠0°
E ∠θ
IL =
= 75.6 V ∠−70.11°/6.28 kΩ ∠90° = 12.04 mA ∠−160.11°
X L ∠90°
b.
c.
f.
g.
44.
192
Fp =
G
YT
=
4.5 mS
= 0.340 leading
13.23 mS
P = I2R = (0.34A)2 220 Ω = 25.43 W
2IC = Is − IR − IL
− −
1 A ∠0° − 0.34 A ∠ − 70.11° − 12.04 mA ∠ − 160.11°
IC = I s I R I L =
2
2
1 − (0.12 − j 0.32) − (−11.32 × 10−3 − j 4.1 × 10−3 )
=
2
0.89 + j 0.32
=
2
IC = 0.45 + j0.16 = 0.47 A ∠19.63°
ZT =
1
= 75.59 Ω ∠−70.11° = 25.72 Ω − j71.08 Ω
Y T 13.23 mS ∠70.11°
R = 25.72 Ω, XC = 71.08 Ω
1
=
P = VI cos θ = 3000 W
3000 W
3000 W
3000
cos θ =
=
=
= 0.75 (lagging)
VI
(100 V)(40 A) 4000
θ = cos−1 0.75 = 41.41°
40 A ∠ − 41.41°
I
YT =
=
= 0.4 S ∠−41.41° = 0.3 S − j0.265 S = GT − jBL
E
100 V ∠0°
1
1
1
+ = 0.05 S +
GT = 0.3 S =
R′
20 Ω R′
1
and R′ =
=4Ω
0.25 S
1
1
XL =
= 3.74 Ω
=
B L 0.265 S
CHAPTER 15
45.
a.
b.
c.
46.
a.
e and υ R2
b.
e and is
c.
iL and iC
CHAPTER 15
193
47.
(I):
(a)
(b)
(II):
υ1: peak-to-peak = (5 div.)(0.5 V/div.) = 2.5 V
⎛ 2.5 V ⎞
V1(rms) = 0.7071 ⎜
⎟ = 0.88 V
⎝ 2 ⎠
υ2: peak-to-peak = (2.4 div.)(0.5 V/div.) = 1.2 V
⎛ 1.2 V ⎞
V2(rms) = 0.7071 ⎜
⎟ = 0.42 V
⎝ 2 ⎠
(c)
T = (4 div.)(0.2 ms/div.) = 0.8 ms
1
1
f=
=
= 1.25 kHz (both)
T
0.8 ms
(a)
θdiv. = 2.2 div., θT = 6 div.
2.2 div.
θ=
× 360° = 132°
6 div.
υ1 leads υ2 by 132°
(b)
(c)
194
θdiv. = 0.8 div., θT = 4 div.
0.8 div.
× 360° = 72°
θ=
4 div.
υ1 leads υ2 by 72°
υ1: peak-to-peak = (2.8 div.)(2 V/div.) = 5.6 V
⎛ 5.6 V ⎞
V1(rms) = 0.7071 ⎜
⎟ = 1.98 V
⎝ 2 ⎠
υ2: peak-to-peak = (4 div.)(2 V/div.) = 8 V
⎛8V⎞
V2(rms) = 0.7071 ⎜
⎟ = 2.83 V
⎝ 2 ⎠
T = (6 div.)(10 μs/div.) = 60 μs
1
1
f=
= 16.67 kHz
=
T
60 μ s
CHAPTER 15
Chapter 16
1.
2.
a.
ZT = 2 Ω + j6 Ω + 8 Ω ∠−90° || 12 Ω ∠−90°
(8 Ω ∠ − 90°)(12 Ω ∠ − 90°)
96 Ω ∠ − 180°
= 2 Ω + j6 Ω +
= 2 Ω + j6 Ω +
20 ∠ − 90°
− j8Ω − j12Ω
= 2 Ω + j6 Ω + 4.8 Ω ∠−90° = 2 Ω + j6 Ω − j4.8 Ω = 2 Ω + j 1.2 Ω
ZT = 2.33 Ω ∠30.96°
12 V ∠0°
E
=
= 5.15 A ∠−30.96°
2.33 Ω ∠30.96°
ZT
b.
I=
c.
I1 = I = 5.15 A ∠−30.96°
d.
(CDR)I2 =
e.
VL = (I ∠θ)(XL ∠90°) = (5.15 A ∠−30.96°)(6 Ω ∠90°) = 30.9 V ∠59.04°
a.
(12 Ω ∠ − 90°)(5.15 A ∠ − 30.96°) 61.80 A ∠ − 120.96°
=
= 3.09 A ∠−30.96°
− j12 Ω − j8 Ω
20 ∠ − 90°
(8 Ω ∠ − 90°)(5.15 A ∠ − 30.96°) 41.2 A ∠ − 120.96°
I3 =
=
= 2.06 A ∠−30.96°
20 Ω ∠ − 90°
20 ∠ − 90°
ZT = 3 Ω + j6 Ω + 2 Ω ∠0° || 8 Ω ∠−90°
= 3 Ω + j6 Ω + 1.94 Ω ∠−14.04°
= 3 Ω + j6 Ω + 1.88 Ω − j0.47 Ω
= 4.88 Ω + j5.53 Ω = 7.38 Ω ∠48.57°
b.
Is =
c.
IC =
=
d.
3.
a.
30 V ∠0°
E
= 4.07 A ∠−48.57°
=
ZT 7.38 Ω ∠48.57°
Z R2 I s
Z R2 + ZC
=
(2 Ω ∠0°)(4.07 A ∠ − 48.57°)
2 Ω − j8 Ω
8.14 A ∠ − 48.57°
= 0.987 A ∠27.39°
8.25 ∠ − 75.96°
Z L E (6 Ω ∠90°)(30 V ∠0°)
180 V ∠90°
=
=
ZT
7.38 Ω ∠48.57°
7.38 Ω ∠48.57°
= 24.39 V ∠41.43°
VL =
ZT = 12 Ω ∠90° || (9.1 Ω − j12 Ω) = 12 Ω ∠90° || 15.06 Ω ∠−52.826°
180.72 Ω∠37.17°
=
9.10∠0°
= 19.86 Ω ∠37.17°
1
1
YT =
=
= 50.35 mS ∠−37.17°
19.86 Ω ∠37.17°
ZT
CHAPTER 16
195
b.
4.
Is =
E
ZT
=
60 V ∠30°
= 3.02 A ∠−7.17°
19.86 Ω ∠37.17°
(12 Ω ∠90°)(3.02 A ∠ − 7.17°) 36.24 A ∠82.83°
=
j12 Ω + 9.1 Ω − j12 Ω
9.1 ∠0°
= 3.98 A ∠82.83°
c.
(CDR) I2 =
d.
(VDR) VC =
e.
P = EI cos θ = (60 V)(3.02 A)cos(30° − 7.17°)
= 181.20(0.922) = 167.07 W
a.
(12 Ω ∠ − 90°)(60 V ∠30°)
720 V ∠ − 60°
=
9.1 Ω − j12 Ω
15.06 ∠ − 52.826°
= 47.81 V ∠−7.17°
ZT = 2 Ω +
(4 Ω ∠ − 90°)(6Ω ∠90°) (4 Ω ∠0°)(3 Ω ∠90°)
+
4 Ω + j3 Ω
− j4 Ω + j6 Ω
24 Ω ∠0° 12 Ω ∠90°
+
2 ∠90°
5 ∠36.87°
= 2 Ω + 12 Ω ∠−90° + 2.4 ∠53.13°
= 2 Ω − j12 Ω + 1.44 Ω + j1.92 Ω
= 3.44 Ω − j10.08 Ω = 10.65 Ω ∠−71.16°
=2Ω+
b.
c.
5.
196
a.
V2 = I(2.4 Ω ∠53.13°) = (5 A ∠0°)(2.4 Ω ∠53.13°) = 12 V ∠53.13°
(4 Ω ∠0°)(I ) (4 Ω ∠0°)(5 A ∠0°) 20 A ∠0°
= 4 A ∠−36.87°
IL =
=
=
4 Ω + j3 Ω
5 Ω ∠36.87°
5 ∠36.87°
Fp =
R
3.44 Ω
=
= 0.323 (leading)
ZT 10.65 Ω
400 Ω ∠ − 90°
= 200 Ω ∠−90°
2
Z′ = 200 Ω − j200 Ω = 282.843 Ω ∠−45°
Z′′ = −j200 Ω + j560 Ω = +j360 Ω = 360 Ω ∠90°
(282.843 Ω ∠ − 45°)(360 Ω ∠90°) 101.83 kΩ∠45°
ZT = Z′ || Z′′ =
=
(200 Ω − j 200 Ω) + j 360 Ω
256.12 ∠38.66°
= 397.59 Ω ∠6.34°
E
100 V ∠0°
=
I=
= 0.25 A ∠−6.34°
397.59 Ω ∠6.34°
ZT
400 Ω ∠−90° || 400 Ω ∠−90° =
(200 Ω ∠ − 90°)(100 V ∠0°) 20,000 V ∠ − 90°
=
= 70.71 V ∠−45°
200 Ω − j 200 Ω
282.843 ∠ − 45°
b.
VC =
c.
P = EI cos θ = (100 V)(0.25 A) cos 6.34°
= (25)(0.994) = 24.85 W
CHAPTER 16
6.
a.
Z1 = 3 Ω + j4 Ω = 5 Ω ∠53.13°
E 120 V ∠60°
=
I1 =
= 24 A ∠6.87°
Z1 5 Ω ∠53.13°
(13 Ω ∠ − 90°)(120 V ∠60°) 1560 V ∠ − 30°
=
= 260 V ∠60°
− j13 Ω + j 7 Ω
6 ∠ − 90°
b.
VC =
c.
VR1 = (I1 ∠θ)R ∠0° = (24 A ∠6.87°)(3 Ω ∠0°) = 72 V ∠6.87°
Vab + VR1 − VC = 0
Vab = VC − VR1 = 260 V ∠60° − 72 V ∠6.87°
+-
+-
= (130 V + j225.167 V) − (71.483 V + j8.612 V)
= 58.52 V + j216.56 V = 224.33 V ∠74.88°
7.
Z1 = 10 Ω ∠0°
Z2 = 80 Ω ∠90° || 20 Ω ∠0°
1600 Ω ∠90°
1600 Ω ∠90°
=
=
20 + j80
82.462 ∠75.964°
= 19.403 Ω ∠14.036°
Z3 = 60 Ω ∠−90°
a.
ZT = (Z1 + Z2) || Z3
= (10 Ω + 18.824 Ω + j4.706 Ω) || 60 Ω ∠−90°
1752.36 Ω ∠ − 80.727°
= 29.206 Ω ∠9.273° || 6 Ω ∠−90° =
28.824 + j 4.706 − j 60
1752.36 Ω ∠ − 80.727°
= 28.103 Ω ∠−18.259°
=
62.356 ∠ − 62.468°
40 V ∠0°
E
=
I1 =
= 1.42 A ∠18.26°
ZT 28.103 Ω ∠ − 18.259°
Z2E
(19.403 Ω ∠14.036°)(40 V ∠0°) 776.12 V ∠14.036°
=
=
29.206 Ω ∠9.273°
29.206 ∠9.273°
Z 2 + Z1
= 26.57 V ∠4.76°
b.
V1 =
c.
P = EI cos θ = (40 V)(1.423 A)cos 18.259°
= 54.07 W
CHAPTER 16
197
8.
a.
Z1 = 2 Ω + j1 Ω = 2.236 Ω ∠26.565°, Z2 = 3 Ω ∠0°
Z3 = 16 Ω + j15 Ω − j7 Ω = 16 Ω + j8 Ω = 17.889 Ω ∠26.565°
1
1
1
1
1
1
+
+
=
+
+
YT =
Z1 Z 2 Z3 2.236 Ω ∠26.565° 3 Ω ∠0° 17.889 Ω ∠26.565°
= 0.447 S ∠−26.565° + 0.333 S ∠0° + 0.056 S ∠−26.565°
= (0.4 S − j0.2 S) + (0.333 S) + (0.05 S − j0.025 S)
= 0.783 S − j0.225 S = 0.82 S ∠−16.03°
1
1
=
= 1.23 Ω ∠16.03°
ZT =
0.82 S ∠ − 16.03°
YT
60 V ∠0°
E
=
= 26.83 A ∠−26.57°
Z1 2.236 Ω ∠26.565°
E 60 V ∠0°
=
= 20 A ∠0°
I2 =
Z 2 3 Ω ∠0°
60 V ∠0°
E
=
= 3.35 A ∠−26.57°
I3 =
Z3 17.889 Ω ∠26.565°
b
I1 =
c.
Is =
60 V ∠0°
E
=
= 48.9 A ∠−16.03°
ZT 1.227 Ω ∠16.032°
?
Is = I1 + I2 + I3
?
48.9 A ∠−16.03° = 26.83 A ∠−26.57° + 20 A ∠0° + 3.35 A ∠−26.57°
= (24 A − j12 A) + (20 A) + (3 A − j1.5 A)
✓
= 47 A + j13.5 A = 48.9 A ∠−16.03° (checks)
d.
9.
a.
Fp =
G 0.783 S
=
= 0.955 (lagging)
YT 0.820 S
Z′ = 3 Ω ∠0° || 4 Ω ∠−90° =
12 Ω ∠ − 90°
3 − j4
12 Ω ∠ − 90°
= 2.4 Ω ∠−36.87°
5 ∠ − 53.13°
Z3 = 2 Z′ + j7 Ω
= 4.8 Ω ∠−36.87° + j7 Ω
= 3.84 Ω − j2.88 Ω + j7 Ω
= 3.84 Ω + j4.12 Ω
= 5.63 Ω ∠47.02°
=
198
CHAPTER 16
ZT = Z1 + Z2 || Z3 = 8.8 Ω + 8.2 Ω ∠0° || 5.63 Ω ∠47.02°
46.18 Ω ∠47.02°
46.18 Ω ∠47.02°
= 8.8 Ω +
= 8.8 Ω +
8.2 + 3.84 + j 4.12
12.73 ∠18.89°
= 8.8 Ω + 3.63 Ω ∠28.13° = 8.8 Ω + 3.20 Ω + j1.71 Ω
= 12 Ω + j1.71 Ω = 12.12 Ω ∠8.11°
1
= 82.51 mS ∠−8.11°
YT =
ZT
b.
10.
V1 = IZ1 = (3 A ∠30°)(6.8 Ω ∠0°) = 20.4 V ∠30°
V2 = I(Z2 || Z3) = (3 A ∠30°)(3.63 Ω ∠28.13°)
= 10.89 V ∠58.13°
V2
10.89 V ∠58.13°
=
= 1.93 A ∠11.11°
Z3
5.63 Ω ∠47.02°
c.
I3 =
a.
X L1 = ωL1 = 2π(103 Hz)(0.1 H) = 628 Ω
X L2 = ωL2 = 2π(103 Hz)(0.2 H) = 1.256 kΩ
1
1
=
= 0.159 kΩ
3
ωC 2π (10 Hz)(1 μ F)
ZT = R ∠0° + X L1 ∠90° + XC ∠−90° || X L2 ∠90°
XC =
= 300 Ω + j628 Ω + 0.159 kΩ ∠−90° || 1.256 kΩ ∠90°
= 300 Ω + j628 Ω − j182 Ω
= 300 Ω + j446 Ω = 537.51 Ω ∠56.07°
1
1
=
= 1.86 mS ∠−56.07°
YT =
537.51 Ω ∠56.07°
ZT
50 V ∠0°
E
=
= 93 mA ∠−56.07°
ZT 537.51 Ω ∠56.07°
b.
Is =
c.
(CDR):
I1 =
Z L2 I s
Z L2 + ZC
=
(1.256 k Ω ∠90°)(93 mA ∠ − 56.07°)
+ j1.256 k Ω − j 0.159 k Ω
116.81 mA ∠33.93°
= 106.48 mA ∠−56.07°
1.097 ∠90°
ZC I s
(0.159 k Ω ∠ − 90°)(93 mA ∠ − 56.07°)
I2 =
=
Z L2 + ZC
1.097 k Ω ∠90°
=
14.79 mA ∠ − 146.07°
= 13.48 mA ∠−236.07°
1.097 ∠90°
= 13.48 mA ∠123.93°
=
CHAPTER 16
199
d.
V1 = (I2 ∠θ)( X L2 ∠90°) = (13.48 mA ∠123.92°)(1.256 kΩ ∠90°)
= 16.93 V ∠213.93°
Vab = E − (Is ∠θ)(R ∠0°) = 50 V ∠0° − (93 mA ∠−56.07°)(300 Ω ∠0°)
+-
= 50 V − 27.9 V ∠−56.07°
= 50 V − (15.573 V − j23.149 V)
= 34.43 V + j23.149 V = 41.49 V ∠33.92°
e.
P = I s2 R = (93 mA)2300 Ω = 2.595 W
f.
Fp =
300 Ω
R
=
= 0.558 (lagging)
ZT 537.51 Ω
Z1 = 2 Ω − j2 Ω = 2.828 Ω ∠−45°
Z2 = 3 Ω − j9 Ω + j6 Ω
= 3 Ω − j3 Ω = 4.243 Ω ∠−45°
Z3 = 10 Ω ∠0°
11.
1
1
1
1
1
1
+
+
=
+
+
Z1 Z 2 Z3
2.828 Ω ∠ − 45° 4.243 Ω ∠ − 45° 10 Ω ∠0°
= 0.354 S ∠45° + 0.236 S ∠45° + 0.1 S ∠0° = 0.59 S ∠45° + 0.1 S ∠0°
= 0.417 S + j0.417 S + 0.1 S
YT = 0.517 S + j 0.417 S = 0.66 S ∠38.89°
1
1
ZT =
=
= 1.52 Ω ∠−38.89°
YT
0.66 S ∠38.89°
50 V ∠0°
E
I=
=
= 32.89 A ∠38.89°
ZT 1.52 Ω ∠ − 38.89°
YT =
12.
200
Z′ = 12 Ω − j20 Ω = 23.32 Ω ∠−59.04°
XL ∠90° || Z′ = 20 Ω ∠90° || 23.32 Ω ∠−59.04° = 33.34 Ω ∠19.99°
Z′′ = R3 ∠0° + XL ∠90° || Z′ = 12 Ω + 33.34 Ω ∠19.99°
= 12 Ω + (31.33 Ω − j11.40 Ω)
= 43.33 Ω − j11.40 Ω = 44.80 Ω ∠14.74°
′′
R2 ∠0° || Z = 20 Ω ∠0° || 44.80 Ω ∠ 14.74° = 13.93 Ω ∠4.54°
ZT = R1 ∠0° + R2 ∠0° || Z′′ = 12 Ω + 13.93 ∠4.54°
= 12 Ω + (13.89 Ω + j1.10 Ω)
= 25.89 Ω + j1.10 Ω = 25.91 Ω ∠2.43°
100 V ∠0°
E
Is =
=
= 3.86 A ∠−2.43°
25.91 Ω ∠2.43°
ZT
I R1 = I
CHAPTER 16
R2 ∠0° I s
(20 Ω ∠0°)(3.86 A ∠ − 2.43°) 77.20 A ∠ − 2.43°
=
=
20 Ω + 43.33 Ω + j11.40 Ω
64.35 ∠10.20°
R2 ∠0° + Z "
63.33 Ω
= 1.20 A ∠−12.63°
I R3 =
X L ∠90°I R3
X L ∠90° + Z′
= 2.00 A ∠77.37°
I4 =
13.
(20 Ω ∠90°)(1.20 A ∠ − 12.63°) 24.00 A ∠77.37°
=
12 ∠0°
j 20 Ω + 12 Ω − j 20 Ω
R3 + R4 = 2.7 kΩ + 4.3 kΩ = 7 kΩ
R′ = 3 kΩ || 7 kΩ = 2.1 kΩ
Z′ = 2.1 kΩ − j10 Ω
(CDR)
(CDR)
14.
=
(40 k Ω ∠0°)(20 mA ∠0°)
40 k Ω + 2.1 k Ω − j10 Ω
= 19 mA ∠+0.014° as expected since R1
I′ (of 10 Ω cap.) =
Z′
(3 k Ω ∠0°)(19 mA ∠0.014°) 57 mA ∠0.014°
=
3kΩ + 7 kΩ
10
= 5.7 mA ∠0.014°
P = I2R = (5.7 mA)2 4.3 kΩ = 139.71 mW
I4 =
Z′ = X C2 ∠ − 90° R1 ∠0° = 2 Ω ∠−90° || 1 Ω ∠0°
2 Ω ∠ − 90°
2 Ω ∠ − 90°
=
1 − j2
2.236 ∠ − 63.435°
= 0.894 Ω ∠−26.565°
′′
Z = X L2 ∠90° + Z′ = +j8 Ω + 0.894 Ω ∠−26.565°
=
IXL =
2
X C1 ∠ − 90°I
= +j8 Ω + (0.8 Ω − j4 Ω)
= 0.8 Ω + j4 = 4.079 Ω ∠78.69°
X C1 ∠ − 90° + Z′′
=
(2 Ω∠ − 90°)(0.5 A ∠0°) 1 A ∠ − 90°
=
0.8 + j 2
− j 2 Ω + (0.8 Ω + j 4 Ω)
1 A ∠ − 90°
= 0.464 A ∠ −158.99°
2.154 ∠68.199°
X C2 ∠ − 90° I X C
(2 Ω ∠ − 90°)(0.464 A ∠ − 158.99°) 0.928 A ∠ − 248.99°
2
I1 =
=
=
− j2 Ω +1 Ω
X C2 ∠ − 90° + R1
2.236 ∠ − 63.435°
=
= 0.42 A ∠174.45°
CHAPTER 16
201
Chapter 17
1.
−
2.
a.
b.
3.
a.
b.
4.
5.
Z = 2.2 Ω + 5.6 Ω + j8.2 Ω = 7.8 Ω + j8.2 Ω = 11.32 Ω ∠46.43°
20 V ∠20°
E
I= =
= 1.77 A ∠−26.43°
Z 11.32 Ω ∠46.43°
Z = −j5 Ω + 2 Ω ∠0° || 5 Ω ∠90° = −j5 Ω + 1.72 Ω + j0.69 Ω = 4.64 Ω ∠−68.24°
60 V ∠30°
E
I=
= 12.93 A ∠98.24°
=
Z 4.64 Ω ∠ − 68.24°
Z = 15 Ω − j16 Ω = 21.93 Ω ∠−46.85°
E = IZ = (0.5 A ∠60°)(21.93 Ω ∠−46.85°)
= 10.97 V ∠13.15°
Z = 10 Ω ∠0° || 6 Ω ∠90° = 5.15 Ω ∠59.04°
E = IZ = (2 A ∠120°)(5.15 Ω ∠59.04°)
= 10.30 V ∠179.04°
μV
16 V
= 4 × 10−3 V
=
3
R
4 × 10
Z = 4 kΩ ∠0°
a.
I=
b.
V = (hI)(R) = (50 I)(50 kΩ) = 2.5 × 106 I
Z = 50 kΩ ∠0°
a.
Clockwise mesh currents:
E − I1Z1 − I1Z2 + I2Z2 = 0
−I2Z2 + I1Z2 − I2Z3 − E2 = 0
────────────────────
[Z1 + Z2]I1 − Z2I2 = E1
−Z2I1 + [Z2 + Z3]I2 = −E2
──────────────────
−Z 2
E1
− E2
[ Z 2 + Z3 ]
I R1 = I1 =
[ Z1 + Z 2 ] −Z 2
−Z 2
202
[ Z 2 + Z3 ]
Z1 = R1 ∠0° = 4 Ω ∠0°
Z2 = XL ∠90° = 6 Ω ∠90°
Z3 = XC ∠−90° = 8 Ω ∠−90°
E1 = 10 V ∠0°, E2 = 40 V ∠60°
=
[ Z2 + Z3 ] E1
− Z 2 E2
Z1Z 2 + Z1Z3 + Z 2 Z3
= 5.15 A ∠−24.5°
CHAPTER 17
b.
By interchanging the right two branches, the general configuration of part (a) will result
and
I50Ω = I1 =
[ Z2
+ Z3 ] E1 − Z 2 E2
Z1Z 2 + Z1Z3 + Z 2 Z3
= 0.44 A ∠143.48°
6.
Z1 = R1 = 50 Ω ∠0°
Z2 = XC ∠−90° = 60 Ω ∠−90°
Z3 = XL ∠90° = 20 Ω ∠90°
E1 = 5 V ∠30°, E2 = 20 V ∠0°
Z1 = 12 Ω + j12 Ω = 16.971 Ω ∠45°
Z2 = 3 Ω ∠0°
Z3 = −j1 Ω
E1 = 20 V ∠50°
E2 = 60 V ∠70°
E3 = 40 V ∠0°
a.
I1[Z1 + Z2] − Z2I2 = E1 − E2
I2[Z2 + Z3] − Z2I1 = E2 − E3
───────────────────
(Z1 + Z2)I1 − Z2I2 = E1 − E2
−Z2I1 + (Z2 + Z3)I2 = E2 − E3
─────────────────────
Using determinants:
(E − E2 )(Z 2 + Z3 ) + Z 2 (E2 − E3 )
I R1 = I1 = 1
= 2.55 A ∠132.72°
Z1Z 2 + Z1Z3 + Z 2 Z3
b.
Source conversion:
E1 = IZ = (6 A ∠0°)(2 Ω ∠0°)
= 12 V ∠0°
Z1 = 2 Ω + 20 Ω + j20 Ω = 22 Ω + j20 Ω
= 29.732 Ω ∠42.274°
Z2 = −j10 Ω = 10 Ω ∠−90°
Z3 = 10 Ω ∠0°
I1[Z1 + Z2] − Z2I2 = E1
I2[Z2 + Z3] − Z2I1 = −E2
─────────────────
(Z1 + Z2)I1 − Z2I2 = E1
−Z2I1 + (Z2 + Z3)I2 = −E2
──────────────────
E ( Z + Z3 ) − Z 2 E2
I R1 = I1 = 1 2
= 0.495 A ∠72.26°
Z1Z 2 + Z1Z3 + Z 2 Z3
CHAPTER 17
203
7.
a.
Z1 = 4 Ω + j3 Ω, Z2 = −j1 Ω
Z3 = +j6 Ω, Z4 = !j2 Ω
Z5 = 8 Ω
E1 = 60 V ∠0°, E2 = 120 V ∠120°
Clockwise mesh currents:
E1 − I1Z1 − I1Z2 + I2Z2 = 0
−I2Z2 + I1Z2 − I2Z3 − I2Z4 + I3Z4 = 0
−I3Z4 + I2Z4 − I3Z5 − E2 = 0
─────────────────────────
− Z2 I 2
[Z1 + Z 2 ]I1
+0
−Z 2
= E1
I1 + [Z 2 + Z 3 + Z 4] I 2
− Z4 I3 = 0
− Z 4 I 2 + [Z 4 + Z 5]I 3 = − E 2
0
───────────────────────────────────
I R1 = I 3 =
[ Z 2Z 4] E1 + ⎡⎣ Z 22 − [ Z1 + Z 2] [ Z 2 + Z3 + Z 4]⎤⎦ E2
[ Z1 + Z 2][ Z 2 + Z3 + Z 4][ Z 4 + Z5] − [ Z1 + Z 2] Z 24 − [ Z 4 + Z5] Z 22
= 13.07 A ∠ − 33.71°
b.
Z1 = 15 Ω ∠0°, Z2 = 15 Ω ∠0°
Z3 = −j10 Ω = 10 Ω ∠−90°
Z4 = 3 Ω + j4 Ω = 5 Ω ∠53.13°
E1 = 220 V ∠0°
E2 = 100 V ∠90°
I1(Z1 + Z3) − I2Z3 − I3Z1 = E1
I2(Z2 + Z3) − I1Z3 − I3Z2 = −E2
I3(Z1 + Z2 + Z4) − I1Z1 − I2Z2 = 0
───────────────────────
I1(Z1 + Z3) − I2Z3
− I3Z1
= E1
+ I2(Z2 + Z3) − I3Z2
= −E2
−I1Z3
−I1Z1
− I2Z2
+ I3(Z1 + Z2 + Z4) = 0
───────────────────────────────────
Applying determinants:
I3 =
[
−( Z1 + Z 3)( Z 2 )E 2 − Z1Z 3E 2 + E1 [ Z 2 Z 3 + Z1( Z 2 + Z 3) ]
]
( Z1 + Z 3) ( Z 2 + Z 3)( Z1 + Z 2 + Z 4 ) − Z 22 + Z 3 [ Z 3( Z1 + Z 2 + Z 4 ) − Z1Z 2 ] − Z1 [ − Z 2 Z 3 − Z1( Z 2 + Z 3) ]
= 48.33 A ∠−77.57°
−
or I3 = E1 E2 if one carefully examines the network!
Z4
204
CHAPTER 17
8.
a.
Z1 = 5 Ω ∠0°, Z2 = 5 Ω ∠90°
Z3 = 4 Ω ∠0°, Z4 = 6 Ω ∠−90°
Z5 = 4 Ω ∠0°, Z6 = 6 Ω + j8 Ω
E1 = 20 V ∠0°, E2 = 40 V ∠60°
I1(Z1 + Z2 + Z4) − I2Z2 − I3Z4 = E1
I2(Z2 + Z3 + Z5) − I1Z2 − I3Z5 = −E2
I3(Z4 + Z5 + Z6) − I1Z4 − I2Z5 = 0
─────────────────────────
(Z1 + Z2 + Z4) I1
− Z2I2
− Z4I3 = E1
−Z2I1 + (Z2 + Z3 + Z5)I2
− Z5I3 = −E2
−Z4I1
− Z5I2 + (Z4 + Z5 + Z6)I3 = 0
─────────────────────────────────────────
Using Z′ = Z1 + Z2 + Z4, Z″ = Z2 + Z3 + Z5, Z′′′ = Z4 + Z5 + Z6 and determinants:
E1 (Z′′Z′′′ − Z52 ) − E2 (Z 2 Z′′′ + Z 4 Z5 )
I R1 = I1 =
Z′(Z′′Z′′′ − Z52 ) − Z 2( Z 2 Z′′′ + Z 4 Z5 ) − Z 4 (Z 2 Z5 + Z 4 Z′′)
= 3.04 A ∠169.12°
b.
Z1 = 10 Ω + j20 Ω
Z3 = 80 Ω ∠0°
Z5 = 15 Ω ∠90°
Z7 = 5 Ω ∠0°
E1 = 25 V ∠0°
Z2 = −j20 Ω
Z4 = 6 Ω ∠0°
Z6 = 10 Ω ∠0°
Z8 = 5 Ω − j20 Ω
E2 = 75 V ∠20°
I1(Z4 + Z6 + Z7) − I2Z4 − I4Z6 = E1
I2(Z1 + Z2 + Z4) − I1Z4 − I3Z2 = 0
I3(Z2 + Z3 + Z5) − I2Z2 − I4Z5 = −E2
I4(Z5 + Z6 + Z8) − I1Z6 − I3Z5 = 0
─────────────────────────
(Z4 + Z6 + Z7) I1
− Z4 I2
+0
− Z6I4 = E1
−Z4I1 + (Z1 + Z2 + Z4)I2
− Z2I3
+0 =0
0
− Z2 I2 + (Z2 + Z3 + Z5)I3
− Z5I4 = −E2
−Z6I1
+0
− Z5I3 + (Z5 + Z6 + Z7)I4 = 0
────────────────────────────────────────────────────
Applying determinants:
I R1 = I 80 Ω = 0.68 A ∠−162.9°
CHAPTER 17
205
Z1 = 5 kΩ ∠0°
Z2 = 10 kΩ ∠0°
Z3 = 1 kΩ + j4 kΩ = 4.123 kΩ ∠75.96°
9.
I1(Z1 + Z2) − Z2I2 = −28 V
I2(Z2 + Z3) − Z2I1 = 0
──────────────────
(Z1 + Z2)I1 − Z2I2 = −28 V
−Z2I1 + (Z2 + Z3)I2 = 0
───────────────────
−Z 2 28 V
IL = I2 =
= −3.17 × 10−3 V ∠137.29°
Z1Z 2 + Z1Z3 + Z 2 Z3
10.
a.
Source Conversion:
E = (I ∠θ)(Rp ∠0°)
= (50 I)(40 kΩ ∠0°)
= 2 × 106 I ∠0°
Z1 = Rs = Rp = 40 kΩ ∠0°
Z2 = −j0.2 kΩ
Z3 = 8 kΩ ∠0°
Z4 = 4 kΩ ∠90°
I1(Z1 + Z2 + Z3) − Z3I2 = −E
I2(Z3 + Z4) − Z3I1 = 0
────────────────────
(Z1 + Z2 + Z3)I1 − Z3I2 = −E
−Z3I1 + (Z3 + Z4)I2 = 0
────────────────────
− Z3 E
IL = I2 =
= 42.91 I ∠149.31°
(Z1 + Z 2 + Z3 )(Z3 + Z 4 ) − Z 23
11.
6Vx − I1 1 kΩ − 10 V ∠0° = 0
10 V∠0° − I2 4 kΩ − I2 2 kΩ = 0
───────────────────────
Vx = I2 2 kΩ
−I1 1 kΩ + I2 12 kΩ = 10 V ∠0°
−I2 6 kΩ = −10 V ∠0°
──────────────────────
10 V ∠0°
I2 = I 2kΩ =
= 1.67 mA ∠0° = I2kΩ
6 kΩ
−I1 1 kΩ + (1.667 mA ∠0°)(12 kΩ) = 10 V ∠0°
−I1 1 kΩ + 20 V ∠0° = 10 V ∠0°
−I1 1 kΩ = −10 V ∠0°
206
CHAPTER 17
I1 = I1kΩ =
12.
10 V ∠0°
= 10 mA ∠0°
1kΩ
E1 − I1Z1 − Z2(I1 − I2) = 0
−Z2(I2 − I1) + E2 − I3Z3 = 0
───────────────────
I3 − I2 = I
E1 = 5 V ∠0°
E2 = 20 V ∠0°
Z1 = 2.2 kΩ ∠0°
Z2 = 5 kΩ ∠90°
Z3 = 10 kΩ ∠0°
I = 4 mA ∠0°
Substituting, we obtain:
I1(Z1 + Z2) − I2Z2 = E1
I1Z2 − I2(Z2 + Z3) = IZ3 − E2
────────────────────
Determinants:
I1 = 1.39 mA ∠−126.48°, I2 = 1.341 mA ∠−10.56°, I3 = 2.693 mA ∠−174.8°
I10kΩ = I3 = 2.693 mA ∠−174.8°
Z1 = 1 kΩ ∠0°
Z2 = 4 kΩ + j6 kΩ
E = 10 V ∠0°
13.
−Z1(I2 − I1) + E − I3Z3 = 0
I1 = 6 mA ∠0°, 0.1 Vs = I3 − I2, Vs = (I1 − I2)Z1
─────────────────────────────────
Substituting:
(1 kΩ)I2 + (4 kΩ + j6 kΩ)I3 = 16 V ∠0°
(99 Ω)I2 +
I3 = 0.6 V ∠0°
────────────────────────────
Determinants:
I3 = I6 kΩ = 1.38 mA ∠−56.31°
CHAPTER 17
207
14.
a.
Z1 = 4 Ω ∠0°
Z2 = 5 Ω ∠90°
Z3 = 2 Ω ∠−90°
I1 = 3 A ∠0°
I2 = 5 A ∠30°
I1 = I3 + I4
⎡1
1 ⎤
⎡1 ⎤
V1 V1 − V 2
+
⇒ V 1 ⎢ + ⎥ − V 2 ⎢ ⎥ = I1
Z1
Z2
⎣ Z1 Z 2 ⎦
⎣ Z2 ⎦
or V1[Y1 + Y2] − V2[Y2] = I1
I1 =
I4 = I5 + I2
⎡1
1⎤
⎡1 ⎤
V1 − V 2 V 2
=
+ I 2 ⇒ V 2 ⎢ + ⎥ − V1 ⎢ ⎥ = − I 2
Z2
Z3
⎣ Z2 ⎦
⎣ Z 2 Z3 ⎦
or V2[Y2 + Y3] − V1[Y2] = −I2
[Y1 + Y2]V1
− Y2V2 = I1
−Y2V1 + [Y2 + Y3]V2 = −I2
──────────────────────
[Y2 + Y3 ] I1 − Y2 I 2
V1 =
= 14.68 V ∠68.89°
Y1 Y2 + Y1 Y3 + Y2 Y3
−[Y1 + Y2 ] I 2 + Y2 I1
V2 =
= 12.97 V ∠155.88°
Y1 Y2 + Y1 Y3 + Y2 Y3
b.
Z1 = 3 Ω + j4 Ω = 5 ∠53.13°
Z2 = 2 Ω ∠0°
Z3 = 6 Ω ∠0° || 8 Ω ∠−90°
= 4.8 Ω ∠−36.87°
I1 = 0.6 A ∠20°
I2 = 4 A ∠80°
0 = I1 + I3 + I4 + I2
V V − V2
0 = I1 + 1 + 1
+ I2
Z1
Z2
⎡1
1 ⎤
⎡1 ⎤
V1 ⎢ + ⎥ − V 2 ⎢ ⎥ = − I1 − I 2
⎣ Z1 Z 2 ⎦
⎣ Z2 ⎦
or
V1[Y1 + Y2] − V2[Y2] = −I1 − I2
──────────────────────
I2 + I4 = I5
V − V2 V2
=
I2 + 1
Z2
Z3
208
CHAPTER 17
⎡ 1
⎡ 1 ⎤
1 ⎤
+
V2 ⎢
⎥ − V1 ⎢ ⎥ = + I 2
⎣ Z2 ⎦
⎣ Z 2 Z3 ⎦
or
V2[Y2 + Y3] − V1[Y2] = I2
──────────────────
and
[Y1 + Y2]V1 − Y2V2 = −I1 − I2
−Y2V1 + [Y2 + Y3]V2 = I2
Applying determinants:
−[Y 2 + Y 3][I1 + I 2] + Y 2I 2
= 5.12 V ∠−79.36°
Y1Y 2 + Y1Y 3 + Y 2 Y 3
Y1I 2 − I1Y 2
V2 =
= 2.71 V ∠39.96°
Y1Y 2 + Y1Y 3 + Y 2 Y 3
V1 =
15.
Z1 = 5 Ω ∠0°
Z2 = 6 Ω ∠90°
Z3 = 4 Ω ∠−90°
Z4 = 2 Ω ∠0°
E = 30 V ∠50°
I = 0.04 A ∠90°
a.
I1 = I2 + I3
⎡1
1
1⎤ V
E1 − V1 = V1 + ( V1 − V 2) ⇒
+ ⎥ − 2 = E1
V1 ⎢ +
Z1
Z1
Z2
Z3
⎣ Z1 Z 2 Z 3 ⎦ Z 3
or V1[Y1 + Y2 + Y3] − Y3V2 = E1Y1
I3 + I = I4
V1 − V 2
+I=
Z3
⎡1
1 ⎤ V
V2
⇒ V2 ⎢ + ⎥ − 1 = I
Z4
⎣ Z3 Z 4 ⎦ Z3
or V2[Y3 + Y4] − V1Y3 = I
resulting in
V1[Y1 + Y2 + Y3] − V2Y3 = E1Y1
−V1[Y3] + V2[Y3 + Y4] = +I
──────────────────────
Using determinants:
V1 = 19.86 V ∠43.8° and V2 = 8.94 V ∠106.9°
CHAPTER 17
209
b.
Z1 = 10 Ω ∠0°
Z2 = 10 Ω ∠0°
Z3 = 4 Ω ∠90°
Z4 = 2 Ω ∠0°
Z5 = 8 Ω ∠−90°
E = 50 V ∠120°
I = 0.8 A ∠70°
I1 = I2 + I5
⎡1
⎡1
1
1
1⎤
1⎤ E
E − V1
( V1 − V 2 )
V
V − V2
= 1 +
+ 1
⇒ V1 ⎢ +
+
+ ⎥ − V2 ⎢
+
⎥=
Z1
Z2
Z5
Z3
⎣ Z1 Z 2 Z3 Z 5 ⎦
⎣ Z 3 Z5 ⎦ Z1
or V1[Y1 + Y2 + Y3 + Y5] − V2[Y3 + Y5] = E1Y1
I3 + I5 = I4 + I
⎡1
⎡1
1
1⎤
1⎤
V1 − V 2 V1 − V 2 V 2
+ ⎥ − V1 ⎢
+
+
=
+ I ⇒ V2 ⎢ +
⎥ = −I
Z3
Z5
Z4
⎣ Z3 Z 4 Z5 ⎦
⎣ Z3 Z5 ⎦
or V2[Y3 + Y4 + Y5] − V1[Y3 + Y5] = −I
resulting in
V1[Y1 + Y2 + Y3 + Y5] − V2[Y3 + Y5] = E1Y1
−V1[Y3 + Y5] + V2[Y3 + Y4 + Y5] = −I
───────────────────────────────
Applying determinants:
V1 = 19.78 V ∠132.48° and V2 = 13.37 V ∠98.78°
16.
V1 V1 − V 2
+
Z1
Z2
V2 − V1 V2 V2 − E
+
+
0=
Z2
Z3
Z4
──────────────────
Rearranging:
⎡1
1 ⎤ 1
V1 ⎢ + ⎥ −
V2 = I
⎣ Z1 Z 2 ⎦ Z 2
⎡1
1
1 ⎤
− V1
E
+ V2 ⎢ +
+ ⎥ =
Z4
Z2
⎣ Z 2 Z3 Z 4 ⎦
───────────────────────────
Determinants and substituting:
I=
Z1 = 2 Ω ∠0°
Z2 = 20 Ω + j 20 Ω
Z3 = 10 Ω ∠−90°
Z4 = 10 Ω ∠0°
I = 6 A ∠ 0°
E = 30 V ∠0°
V1 = 11.74 V ∠−4.61°, V2 = 22.53 V ∠−36.48°
210
CHAPTER 17
∑ Ii = ∑ I o
0 = I1 + I2 + I3
− V1
−
V2
+
+ V 2 V3
= V2
15 Ω
10 Ω ∠ − 90°
15 Ω
Through manipulation:
17.
(Note that 3 + j4 branch has no effect
on nodal voltages)
V2[2 + j1.5] − V1 − V3 = 0
but V1 = 220 V ∠0° and V3 = 100 V ∠90°
220 + j100
= 96.66 V∠−12.43°
and V2 =
2 + j1.5
with V3 = 0V ∠0°
Z1 = 5 Ω ∠0°
Z2 = 6 Ω ∠−90°
Z3 = 5 Ω ∠90°
Z4 = 4 Ω ∠0°
Z5 = 4 Ω ∠0°
Z6 = 6 Ω + j8 Ω
E1 = 20 V ∠0°
E2 = 40 V ∠60°
18.
V1 − E1 V1 − V 3 V1 − E 2 − V 2
=0
+
+
Z1
Z3
Z5
V 2 + E 2 − V1 V 2 − V 3 V 2
+
+
=0
node V2:
Z5
Z4
Z6
−
−
node V3: V 3 + V 3 V1 + V 3 V 2 = 0
Z2
Z3
Z4
─────────────────────────
Rearranging:
⎛ 1
1
1 ⎞ V V
V1 ⎜ +
+ ⎟ − 2 − 3 = E1 + E 2
Z1 Z 5
⎝ Z1 Z 3 Z 5 ⎠ Z 5 Z 3
node V1:
⎛ 1
1
1 ⎞ V V
V2 ⎜
+
+ ⎟ − 1 − 3 = − E2
Z5
⎝ Z5 Z 4 Z6 ⎠ Z5 Z 4
⎛ 1
1
1 ⎞ V1 V 2
=0
−
V3 ⎜
+
+
⎟ −
⎝ Z 2 Z3 Z 4 ⎠ Z3 Z 4
─────────────────────────────────
Determinants: V1 = 5.84 V ∠29.4°, V2 = 28.06 V ∠−89.15°, V3 = 31.96 V ∠−77.6°
CHAPTER 17
211
19.
E1 = 25 V ∠0°
E2 = 75 V ∠20°
Z1 = 10 Ω + j20 Ω
Z2 = 6 Ω ∠0°
Z3 = 5 Ω ∠0°
Z4 = 20 Ω ∠−90°
Z5 = 10 Ω ∠0°
Z6 = 80 Ω ∠0°
Z7 = 15 Ω ∠90°
Z8 = 5 Ω − j20 Ω
V1 − V 2 V1 − V 4 V1 − E1
=0
+
+
Z1
Z2
Z3
V 2 − V1 V 2 − V 4 V 2 − E 2 − V 3
+
+
V2:
=0
Z1
Z4
Z6
V 3 + E2 − V 2 V 3 − V 4 V 3
=0
V3:
+
+
Z6
Z7
Z8
V 4 − V1 V 4 − V 2 V 4 − V 3 V 4
=0
+
+
+
V4:
Z2
Z4
Z7
Z5
───────────────────────
Rearranging:
⎛ 1
1
1 ⎞ V
V
V1 ⎜ +
+ ⎟ − 2 − 4 = E1
Z1 Z 2
Z3
⎝ Z1 Z 2 Z 3 ⎠
V1:
⎛ 1
1
1 ⎞ V V
V
V2 ⎜ +
+ ⎟ − 1 − 4 − 3 = E2
Z6
Z1 Z 4 Z 6
⎝ Z1 Z 4 Z 6 ⎠
⎛ 1
1
1 ⎞
+
+ ⎟ − V 2 − V 4 = − E2
V3 ⎜
Z6 Z7
Z6
⎝ Z 6 Z 7 Z8 ⎠
⎛ 1
1
1
1 ⎞ V V
V
V4 ⎜
+
+
+ ⎟ − 1− 2− 3 =0
⎝ Z 2 Z 4 Z7 Z5 ⎠ Z 2 Z 4 Z 7
Setting up and then using determinants:
V1 = 14.62 V ∠−5.86°, V2 = 35.03 V ∠−37.69°
V3 = 32.4 V ∠−73.34°, V4 = 5.67 V ∠23.53°
212
CHAPTER 17
20.
a.
1
4 Ω ∠0°
= 0.25 S ∠0°
1
Y2 =
1 Ω ∠90°
= 1 S ∠−90°
1
Y3 =
5 Ω ∠0°
= 0.2 S ∠0°
1
Y4 =
4 Ω ∠ − 90°
= 0.25 S ∠90°
1
Y5 =
8 Ω ∠90°
= 0.125 S ∠−90°
I1 = 2 A ∠30°
I2 = 3 A ∠150°
Y1 =
V1[Y1 + Y2] − Y2V2 = I1
V2[Y2 + Y3 + Y4] − Y2V1 − Y4V3 = −I2
V3[Y4 + Y5] − Y4V2 = I2
───────────────────────────
[Y1 + Y2]V1
− Y2 V2
+ 0 = I1
−Y2V1 + [Y2 + Y3 + Y4]V2
− Y4 V3 = −I2
0
− Y4 V2 + [Y4 + Y5]V3 = I2
────────────────────────────────────
2
I1 ⎡⎣( Y 2 + Y 3 + Y 4)(Y 4 + Y 5) − Y 4 ⎤⎦ − I 2[Y 2 Y 5]
V1 =
[ Y1 + Y 2] ⎡⎣(Y 2 + Y3 + Y 4)(Y 4 + Y5) − Y 24 ⎤⎦ − Y 22(Y 4 + Y5) = Y Δ
= 5.74 V ∠122.76°
( + )−
( + )
V2 = I1Y 2 Y 4 Y 5 I 2 Y 5 Y1 Y 2 = 4.04 V ∠145.03°
YΔ
2
I 2 ⎡( Y1 + Y 2)(Y 3 + Y 4) − Y 2 ⎤⎦ − Y 2Y 4I1
V3 = ⎣
= 25.94 V ∠78.07°
YΔ
CHAPTER 17
213
b.
1
4 Ω ∠0°
= 0.25 S ∠0°
1
Y2 =
6 Ω ∠0°
= 0.167 S ∠0°
1
Y3 =
8 Ω ∠0°
= 0.125 S ∠0°
1
Y4 =
2 Ω ∠ − 90°
V1[Y1 + Y2 + Y3] − Y2V2 − Y3V3 = I1
= 0.5 S ∠90°
V2[Y2 + Y4 + Y5] − Y2V1 − Y4V3 = 0
1
Y5 =
V3[Y3 + Y4 + Y6] − Y3V1 − Y4V2 = −I2
5 Ω ∠90°
───────────────────────────
= 0.2 S ∠−90°
− Y2V2
− Y3V3 = I1
[Y1 + Y2 + Y3]V1
1
−Y2V1 + [Y2 + Y4 + Y5]V2
− Y4V3 = 0
Y6 =
4 Ω ∠90°
−Y3V1
− Y4V2 + [Y3 + Y4 + Y6]V3 = −I2
= 0.25 S ∠−90°
──────────────────────────────────────────
I1 = 4 A ∠0°
I2 = 6 A ∠90°
I [ ( Y + Y + Y )( Y + Y + Y ) − Y ] − I [ Y Y + Y ( Y + Y + Y ) ]
V1 =
Y = ( Y + Y + Y ) [ ( Y + Y + Y )( Y + Y + Y ) − Y ] − Y [ Y ( Y + Y + Y )+ Y Y ] − Y [ Y Y + Y ( Y + Y + Y ) ]
Y1 =
2
1
Δ
2
4
5
3
4
6
4
2
2
4
3
3
4
5
2
= 15.13 V ∠1.29°
I1[ (Y 2)(Y 3 + Y 4 + Y 6)+ Y 3Y 4 ] +I 2 [ Y 4( Y1 + Y 2 + Y 3) − Y 2Y 3]
= 17.24 V ∠3.73°
V2 =
YΔ
2
I1[ (Y 3)(Y 2 + Y 4 + Y 5) + Y 2Y 4 ] + I 2 ⎡⎣ Y 2 - (Y1 + Y 2 + Y 3)(Y 2 + Y 4 + Y 5) ⎤⎦
V3 =
YΔ
= 10.59 V ∠−0.11°
21.
Left node:
1
2
3
2
4
5
3
4
6
4
2
2
3
4
6
3
4
3
2
4
3
2
4
5
V1
∑ Ii = ∑ I o
−
4Ix = Ix + 5 mA ∠0° + V1 V 2
2 kΩ
Right node: V2
∑ Ii = ∑ I o
V 2 + V 2 − V 1 + 4I
8 mA ∠0° =
x
1kΩ
2 kΩ
V1
Insert Ix =
4 k Ω ∠ − 90°
Rearrange, reduce and 2 equations with 2 unknowns result:
V1[1.803 ∠123.69°] + V2 = 10
V1[2.236 ∠116.57°] + 3 V2 = 16
──────────────────────
214
CHAPTER 17
Determinants:
V1 = 4.37 V ∠−128.66°
V2 = V1kΩ = 2.25 V ∠17.63°
22.
Z1 = 1 kΩ ∠0°
Z2 = 2 kΩ ∠90°
Z3 = 3 kΩ ∠−90°
I1 = 12 mA ∠0°
I2 = 4 mA ∠0°
E = 10 V ∠0°
∑ Ii = ∑ I o
V1 V 2
+
+ I2
Z1 Z 3
V V
and 1 + 2 = −I1 − I2
Z1 Z3
with V2 − V1 = E
0 = I1 +
Substituting and rearranging:
⎡1
1⎤
E
V1 ⎢ + ⎥ = −I1 − I2 −
Z3
⎣ Z1 Z 3 ⎦
and solving for V1:
V1 = 15.4 V ∠ 178.2°
with V2 = VC = 5.41 V ∠ 174.87°
23.
Left node: V1
∑ Ii = ∑ I o
2 mA ∠0° = 12 mA ∠0° +
and 1.5 V1 − V2 = −10
Right node: V2
∑ Ii = ∑ I o
V1
V − V2
+ 1
2 kΩ
1kΩ
V 2 − V1 V 2 − 6 V x
−
1kΩ
3.3 k Ω
and 2.7 V1 − 3.7 V2 = −6.6
0 = 2 mA ∠0° +
Using determinants:
CHAPTER 17
V1 = V2kΩ = −10.67 V ∠0° = 10.67 V ∠180°
V2 = −6 V ∠0° = 6 V ∠180°
215
Z1 = 2 kΩ ∠0°
Z2 = 1 kΩ∠0°
Z3 = 1 kΩ ∠0°
I = 5 mA ∠0°
24.
V1
V
+ 3I1 + 2
Z1
Z3
V -V
with I1 = 1 2
Z2
and V2 − V1 = 2Vx = 2V1 or V2 = 3V1
V1: I =
Substituting will result in:
or
and
with
25.
⎡1
⎡1
3⎤
3⎤
V1 ⎢ + ⎥ + 3 V1 ⎢ − ⎥ = I
⎣ Z1 Z 2 ⎦
⎣ Z3 Z 2 ⎦
⎡1
6
3⎤
+ ⎥ =I
V1 ⎢ −
⎣ Z1 Z 2 Z3 ⎦
V1 = Vx = −2 V ∠0°
V2 = −6 V ∠0°
Ei ∠θ
= 1 × 10−3 Ei
R1 ∠0°
1
= 0.02 mS ∠0°
Y1 =
50 kΩ
1
= 1 mS ∠0°
Y2 =
1 kΩ
Y3 = 0.02 mS ∠0°
I2 = (V1 − V2)Y2
I1 =
V1(Y1 + Y2) − Y2V2 = −50I1
V2(Y2 + Y3) − Y2V1 = 50I2 = 50(V1 − V2)Y2 = 50Y2V1 − 50Y2V2
────────────────────────────────────────────
(Y1 + Y2)V1 − Y2V2 = −50I1
−51Y2V1 + (51Y2 + Y3)V2 = 0
───────────────────────
−(50)(51)Y2 I1
VL = V2 =
= −2451.92 Ei
(Y1 + Y2 )(51Y2 + Y3 ) − 51Y22
216
CHAPTER 17
26.
a.
b.
yes
Z1 = Z 2
Z3
Z4
3
8 × 103 ∠0°
5 × 10 ∠0°
=
2.5 × 103 ∠90°
4 × 103 ∠90°
2 ∠−90° = 2 ∠−90° (balanced)✓
Z1 = 5 kΩ ∠0°, Z2 = 8 kΩ ∠0°
Z3 = 2.5 kΩ ∠90°, Z4 = 4 kΩ ∠90°
Z5 = 5 kΩ ∠−90°, Z6 = 1 kΩ ∠0°
I1[Z1 + Z3 + Z6] − Z1I2 − Z3I3 = E
I2[Z1 + Z2 + Z5] − Z1I1 − Z5I3 = 0
I3[Z3 + Z4 + Z5] − Z3I1 − Z5I2 = 0
───────────────────────
[Z1 + Z3 + Z6]I1
− Z1I2
− Z3I3 = E
−Z1I1 + [Z1 + Z2 + Z5]I2
− Z5I3 = 0
−Z3I1
− Z5I2 + [Z3 + Z4 + Z5]I3 = 0
───────────────────────────────────────
I2 =
I3 =
Z Δ = ( Z1 + Z 3 + Z 6 )[( Z1 + Z 2 + Z 5 )( Z 3 + Z 4 + Z 5 ) − Z 5 ] − Z1[ Z1( Z 3 + Z 4 + Z 5 ) − Z 3 Z 5 ] − Z 3[ Z1Z 5 + Z 3 ( Z1 + Z 2 + Z 5 )]
E[ Z1Z5 + Z3 ( Z1 + Z 2 + Z5 )]
I Z5 = I2 − I3 =
CHAPTER 17
E[ Z1( Z 3 + Z 4 + Z 5 ) + Z 3 Z 5 ]
ZΔ
E[ Z1Z 4 − Z3 Z 2 ]
ZΔ
=
2
E ⎡⎣ 20 × 106 ∠90° − 20 × 106 ∠90°⎤⎦
ZΔ
=0A
217
c.
V1[Y1 + Y2 + Y6] − Y1V2 − Y2V3 = I
V2[Y1 + Y3 + Y5] − Y1V1 − Y5V3 = 0
V3[Y2 + Y4 + Y5] − Y2V1 − Y5V2 = 0
─────────────────────────
− Y1V2
− Y2V3 = I
[Y1 + Y2 + Y6]V1
− Y5V3 = 0
−Y1V1 + [Y1 + Y3 + Y5]V2
−Y2V1
− Y5V2 + [Y2 + Y4 + Y5]V3 = 0
─────────────────────────────────────────
V2 =
V3 =
I[ Y 1( Y 2 + Y 4 + Y 5 ) + Y 2 Y 5 ]
10 V ∠0°
I = Es =
1 k Ω ∠0°
Rs
= 10 mA ∠0°
1
Y1 =
5 k Ω ∠0°
= 0.2 mS ∠0°
1
Y2 =
8 k Ω ∠0°
= 0.125 mS ∠0°
1
Y3 =
2.5 k Ω ∠90°
= 0.4 mS ∠−90°
1
Y4 =
4 k Ω ∠90°
= 0.25 mS ∠−90°
1
Y5 =
5 k Ω ∠ − 90°
= 0.2 mS ∠90°
1
Y6 =
1 k Ω ∠0°
V2 = 1 mS ∠0°
2
Y Δ = ( Y 1 + Y 2 + Y 6 )[( Y 1 + Y 3 + Y 5 )( Y 2 + Y 4 + Y 5 ) − Y 5 ] − Y 1[ Y 1( Y 2 + Y 4 + Y 5 ) + Y 2 Y 5 ] − Y 2[ Y 1Y 5 + Y 2 ( Y 1+ Y 3 + Y 5)]
I[ Y1Y5 + Y2 (Y1 +Y3 +Y5 ) ]
YΔ
V Z 5 = V2 − V3 =
I[ Y1Y4 − Y4 Y3 ]
YΔ
=
I ⎡⎣0.05 × 10−3 ∠ − 90° − 0.05 × 10−3 ∠ − 90° ⎤⎦
YΔ
=0V
27.
a.
Z1 = Z 2
Z3
Z4
3
4 × 10 ∠0°
4 × 103 ∠0°
?
3
4 × 10 ∠90°
4 × 103 ∠ − 90°
1 ∠−90° ≠ 1 ∠90° (not balanced)
b.
The solution to 26(b) resulted in
E(Z1Z5 + Z3 (Z1 + Z 2 + Z5 )
I3 = I X C =
ZΔ
where
and
and
218
ZΔ = (Z1 + Z3 + Z6)[(Z1 + Z2 + Z5)(Z3 + Z4 + Z5) − Z52 ]
− Z1[Z1(Z3 + Z4 + Z5) − Z3Z5] − Z3[Z1Z5 + Z3(Z1 + Z2 + Z5)]
Z1 = 5 kΩ ∠0°, Z2 = 8 kΩ ∠0°, Z3 = 2.5 kΩ ∠90°
Z4 = 4 kΩ ∠90°, Z5 = 5 kΩ ∠−90°, Z6 = 1 kΩ ∠0°
I X C = 1.76 mA ∠−71.54°
CHAPTER 17
c.
The solution to 26(c) resulted in
I[ Y1Y5 + Y2 (Y1 +Y3 +Y5 ) ]
V3 = VX C =
YΔ
YΔ = (Y1 + Y2 + Y6)[(Y1 + Y3 + Y5)(Y2 + Y4 + Y5) − Y52 ]
− Y1 [Y1(Y2 + Y4 + Y5) + Y2Y5]
− Y2[Y1Y5 + Y2(Y1 + Y3 + Y5)]
Y1 = 0.2 mS ∠0°, Y2 = 0.125 mS ∠0°, Y3 = 0.4 mS ∠−90°
Y4 = 0.25 mS ∠−90°, Y5 = 0.2 mS ∠90°
where
with
Y6 = 1 mS ∠0°, I = 10 mA ∠0°
V3 = 7.03 V ∠−18.46°
Source conversion:
and
28.
(
Z1Z4 = Z3Z2
)
(R1 − jXC) Rx + jX Lx = R3R2
(
)
XC =
1
1
= 1 kΩ
=
3
ωC (10 rad/s)(1 μ F)
(1 kΩ − j1 kΩ) Rx + jX Lx = (0.1 kΩ)(0.1 kΩ) = 10 kΩ
and Rx + jX Lx =
∴ Rx = 5 Ω, Lx =
29.
10 × 103 Ω
10 × 103
=
= 5 Ω + j5 Ω
1 × 103 − j1 × 103 1.414 × 103 ∠ − 45°
X Lx
ω
=
5Ω
= 5 mH
10 rad/s
3
1
1
1
=
= kΩ
ωC (1000 rad/s)(3 μ F) 3
1
Z1 = R1 || X C1 ∠−90° = (2 kΩ ∠0°) || 2 kΩ ∠−90° = 328.8 Ω ∠−80.54°
3
Z2 = R2 ∠0° = 0.5 kΩ ∠0°, Z3 = R3 ∠0° = 4 kΩ ∠0°
Z4 = Rx + j X Lx = 1 kΩ + j6 kΩ
X C1 =
Z1 = Z 2
Z3
Z4
328.8 Ω ∠ − 80.54°
0.5 kΩ ∠0°
?
4 kΩ ∠0°
6.083 Ω ∠80.54°
82.2 ∠−80.54° 9 82.2 ∠−80.54° (balanced)
30.
Apply Eq. 17.6.
CHAPTER 17
219
31.
For balance:
R1(Rx + j X Lx ) = R2(R3 + j X L3 )
R1Rx + jR1 X Lx = R2R3 + jR2 X L3
∴ R1Rx = R2R3 and Rx =
R1 X Lx = R2 X L3
R2 R3
R1
and R1ωLx = R2ωL3
so that Lx =
32.
R2 L3
R1
Z1 = 8 Ω ∠−90° = −j8 Ω
Z2 = 4 Ω ∠90° = +j4 Ω
Z3 = 8 Ω ∠90° = +j8 Ω
Z4 = 6 Ω ∠−90° = −j6 Ω
Z5 = 5 Ω ∠0°
a.
Z1Z 2
= 5 Ω ∠38.66°
+
Z1 Z 2 + Z5
Z1Z5
Z7 =
= 6.25 Ω ∠−51.34°
Z1 + Z 2 + Z5
Z 2Z5
= 3.125 Ω ∠128.66°
Z8 =
Z1 + Z 2 + Z5
Z′ = Z7 + Z3 = 3.9 Ω + j3.12 Ω = 4.99 Ω ∠38.66°
Z″ = Z8 + Z4 = −1.95 Ω − j3.56 Ω = 4.06 Ω ∠−118.71°
Z′ || Z″ = 10.13 Ω ∠−67.33°= 3.90 Ω − j9.35 Ω
ZT = Z6 + Z′ || Z″ = 7.80 Ω − j6.23 Ω = 9.98 Ω ∠−38.61°
120 V ∠0°
E
= 12.02 A ∠38.61°
I=
=
ZT 9.98 Ω ∠ − 38.61°
Z6 =
b.
220
12 Ω − j 9 Ω
ZY = Z Δ =
= 4 Ω − j3 Ω
3
3
CHAPTER 17
ZT = 2 Ω + 4 Ω + j3 Ω + [4 Ω − j3 Ω + j3 Ω] || [4 Ω − j3 Ω + j3 Ω]
= 6 Ω − j3 Ω + 2 Ω
= 8 Ω − j3 Ω = 8.544 Ω ∠−20.56°
60 V ∠0°
E
I=
=
= 7.02 A ∠20.56°
8.544 Ω ∠ − 20.56°
ZT
33.
ZΔ = 3ZY = 3(3 Ω ∠90°) = 9 Ω ∠90°
Z = 9 Ω ∠90° || (12 Ω − j16 Ω)
= 9 Ω ∠90° || 20 Ω ∠53.13°
= 12.96 Ω ∠67.13°
a.
ZT = Z || 2Z =
2Z 2
2
2
= Z = [12.96 Ω ∠67.13°] = 8.64 Ω ∠67.13°
3
Z + 2Z 3
E
100 V ∠0°
I=
= 11.57 A ∠−67.13°
=
ZT 8.64 Ω ∠67.13°
b.
ZΔ = 3ZY = 3(5 Ω) = 15 Ω
Z1 = 15 Ω ∠0° || 5 Ω ∠−90°
= 4.74 Ω ∠−71.57°
Z2 = 15 Ω ∠0° || 6 Ω ∠90°
= 5.57 Ω ∠68.2° = 2.07 Ω + j5.17 Ω
Z3 = Z1 = 4.74 Ω ∠−71.57°
= 1.5 Ω − j4.5 Ω
ZT = Z1|| (Z2 + Z3) = (4.74 Ω ∠−71.57°) || (2.07 Ω + j5.17 Ω + 1.5 Ω − j4.5 Ω)
= (4.74 Ω ∠−7.57°) || (3.63 Ω ∠10.63°)
= 2.71 Ω ∠−23.87°
100 V ∠0°
E
=
I=
= 36.9 A ∠23.87°
ZT 2.71 Ω ∠ − 23.87°
CHAPTER 17
221
Chapter 18
1.
a.
Z1 = 3 Ω ∠0°, Z2 = 8 Ω ∠90°, Z3 = 6 Ω ∠−90°
Z2 || Z3 = 8 Ω ∠90° || 6 Ω ∠−90° = 24 Ω ∠−90°
30 V ∠30°
E1
= 1.24 A ∠112.875°
=
Z1 + Z 2 Z 3 3 Ω − j 24 Ω
Z3 I
(6 Ω ∠ − 90°)(1.24 A ∠112.875°)
=
I′ =
= 3.72 A ∠−67.125°
Z 2 + Z3
2 Ω ∠90°
I=
Z1|| Z2 = 3 Ω ∠0° || 8 Ω ∠90° = 2.809 Ω ∠20.556°
60 V ∠10°
E2
=
I=
Z 3 + Z1 Z 2 − j 6 Ω + 2.630 Ω + j 0.986 Ω
= 10.597 A ∠72.322°
I″ =
Z1 I
(3 Ω ∠0°)(10.597 A ∠72.322°)
=
= 3.721 A ∠2.878°
Z1 + Z 2
3 Ω + j8 Ω
I L1 = I′ + I″ = 3.72 A ∠−67.125° + 3.721 A ∠2.878°
= 1.446 A − j3.427 A + 3.716 A + j0.187 A
= 5.162 A − j3.24 A
= 6.09 A ∠−32.12°
b.
Z1 = 8 Ω ∠90°, Z2 = 5 Ω ∠−90°
I = 0.3 A ∠60°, E = 10 V ∠0°
I
(5 Ω ∠ − 90°)(0.3 A ∠60°)
I′ = Z 2 =
+ j8 Ω − j 5 Ω
Z 2 + Z1
= 0.5A ∠−120°
10 V ∠0°
E
= 3.33 A ∠−90°
I″ =
=
Z1 + Z 2 3 Ω ∠90°
I Z1 = I L1 = I′ + I″
= 0.5 A ∠−120° + 3.33 A ∠−90°
= −0.25 A − j0.433 A − j3.33 A
= −0.25 A − j3.763 A
= 3.77 A ∠−93.8°
222
CHAPTER 18
2.
a.
E1:
E1 = 20 V ∠0°,
Z1 = 4 Ω + j3 Ω = 5 Ω∠36.87°
Z2 = 1 Ω ∠0°
E1
20 V ∠0°
I′ =
=
4 Ω + j3 Ω + 1 Ω
Z1 + Z 2
= 3.43 A ∠−30.96°
E2:
I″ =
I:
b.
E2
120 V ∠0°
=
5.83 Ω ∠30.96°
Z1 + Z 2
= 20.58 A ∠−30.96°
I′′′ =
Z2I
(1 Ω ∠0°)(0.5 A ∠60°)
=
5.83 Ω ∠30.96°
Z 2 + Z1
= 0.0858 A ∠29.04°
↑IL = I′ − I″ − I′′′
= (3.43 A ∠−30.96°) − (20.58 A ∠−30.96°) − (0.0858 A ∠29.04°)
= 17.20 A ∠149.30° or 17.20 A ∠−30.70°↓
E:
Z1 = 3 Ω ∠90°, Z2 = 7 Ω ∠−90°
E = 10 V ∠90°
Z3 = 6 Ω ∠−90°, Z4 = 4 Ω ∠0°
Z′ = Z1|| (Z3 + Z4)
= 3 Ω ∠90° || (4 Ω − j6 Ω)
= 3 Ω ∠90° || 7.21 Ω ∠−56.31°
= 4.33 Ω ∠70.56°
Z′E
V1 =
′
Z + Z2
(4.33 Ω ∠70.56°)(10 V ∠90°)
=
(1.44 Ω + j 4.08 Ω) − j 7Ω
43.3 V ∠160.56°
=
= 13.28 V ∠224.31°
3.26 ∠ − 63.75°
13.28 V ∠224.31°
V
I′ = 1 =
3 Ω ∠90°
Z1
= 4.43 A ∠134.31°
CHAPTER 18
223
I:
CDR:
Z″ = Z3 + Z1|| Z2
= −j6 Ω + 3 Ω ∠90° || 7 Ω ∠−90°
= −j6 Ω + 5.25 Ω ∠90°
= −j6 Ω + j5.25 Ω
= −j0.75 Ω = 0.75 Ω∠−90°
Z4 I
(4 Ω ∠0°)(0.6 A ∠120°)
2.4 A ∠120°
I3 =
=
=
4 Ω − j 0.75 Ω
4.07 ∠ − 10.62°
Z 4 + Z′′
= 0.59 A ∠130.62°
Z2I3
(7 Ω ∠ − 90°)(0.59 A ∠130.62°) 4.13 A ∠40.62°
=
=
I″ =
− j 7 Ω + j3 Ω
4 ∠ − 90°
Z 2 + Z1
= 1.03 A ∠130.62°
IL = I′ − I″ (direction of I′)
= 4.43 A ∠134.31° − 1.03 A ∠130.62°
= (−3.09 A + j3.17 A) − (−0.67 A + j0.78 A) = −2.42 A + j2.39 A
= 3.40 A ∠135.36°
3.
DC:
IDC =
4V
= 0.5 A
8Ω
AC:
Z2 = R2 + jXL = 8 Ω + j4 Ω
= 8.944 Ω ∠26.565°
E
10 V ∠0°
I= 1 =
Z 2 8.944 Ω ∠26.565°
= 1.118 A ∠−26.565°
I = 0.5 A + 1.118 A ∠−26.57°
i = 0.5 A + 1.58 sin(ωt − 26.57°)
224
CHAPTER 18
4.
DC:
AC:
(6 Ω ∠0°)(I )
6 Ω + 3 Ω − j1 Ω
(6 Ω ∠0°)(4 A ∠0°)
=
9 Ω − j1 Ω
24 A ∠0°
=
9.055∠ − 6.34°
= 2.65 A ∠6.34°
VC = ICXC = (2.65 A ∠6.34°)(1 Ω ∠−90°) = 2.65 V ∠−83.66°
= 12 V + 2.65 V ∠−83.66°
υC = 12 V + 3.75 sin(ωt − 83.66°)
IC =
E = 20 V ∠0°
Z1 = 10 kΩ ∠0°
Z2 = 5 kΩ − j5 kΩ
= 7.071 kΩ ∠−45°
Z3 = 5 kΩ ∠90°
I = 5 mA ∠0°
5.
Z′ = Z1 || Z2 = 10 kΩ ∠0° || 7.071 kΩ ∠−45° = 4.472 kΩ ∠−26.57°
(CDR)
Z′I
(4.472 kΩ ∠ − 26.57°)(5 mA ∠0°) 22.36 mA ∠ − 26.57°
=
=
′
Z + Z3
4 kΩ − j 2 kΩ + j 5 kΩ
5 ∠36.87°
= 4.472 mA ∠−63.44°
I′ =
Z″ = Z2 || Z3
= 7.071 kΩ ∠−45° || 5 kΩ ∠90°
= 7.071 kΩ ∠45°
CHAPTER 18
225
Z′′E
(7.071 kΩ ∠45°)(20 V ∠0°) 141.42 V ∠45°
=
=
Z′′+ Z1
(5 kΩ + j 5 kΩ) + (10 kΩ)
15.81 ∠18.435°
= 8.945 V ∠26.565°
V ′ 8.945 V ∠26.565°
=
I″ =
= 1.789 mA ∠−63.435° = 0.8 mA − j1.6 mA
5 kΩ ∠90°
Z3
I = I′ + I″ = (2 mA − j4 mA) + (0.8 mA − j1.6 mA) = 2.8 mA − j5.6 mA
= 6.26 mA ∠−63.43°
(VDR)
V′ =
Z1 = 20 kΩ ∠0°
Z2 = 10 kΩ ∠90°
I = 2 mA ∠0°
E = 10 V ∠0°
6.
I′ =
Z1 (hI )
(20 k Ω ∠0°)(100)(2 mA ∠0°)
=
= 0.179 A ∠−26.57°
20 kΩ + j10 kΩ
Z1 + Z 2
E
10 V ∠0°
=
22.36 k Ω ∠26.57°
Z1 + Z 2
= 0.447 mA ∠−26.57°
IL = I′ − I″ (direction of I′)
= 179 mA ∠−26.57° − 0.447 mA ∠−26.57°
= 178.55 mA ∠−26.57°
I″ =
μV:
7.
Z1 = 5 kΩ ∠0°, Z2 = 1 kΩ ∠−90°
Z3 = 4 kΩ ∠0°
V = 2 V ∠0°, μ = 20
V′L =
−Z 3( μ V )
−(4 kΩ ∠0°)(20)(2 V ∠0°)
= −17.67 V ∠6.34°
=
5 kΩ − j1 kΩ + 4 kΩ
Z1 + Z 2 + Z 3
I:
Z1I
Z1 + Z 2 + Z3
(5 kΩ ∠0°)(2 mA ∠0°)
=
9.056 kΩ ∠ − 6.34°
= 1.104 mA ∠6.34°
CDR: I′ =
V″L = −I′Z3 = −(1.104 mA ∠6.34°)(4 kΩ ∠0°) = −4.416 V ∠6.34°
VL = V′L + V″L = −17.67 V ∠6.34° − 4.416 V ∠6.34° = −22.09 V ∠6.34°
226
CHAPTER 18
8.
Z1 = 20 kΩ ∠0°
Z2 = 5 kΩ + j5 kΩ
I′ =
Z1 (hI )
(20 kΩ ∠0°)(100)(1 mA ∠0°)
=
= 78.45 mA ∠−11.31°
20 kΩ + 5 kΩ + j 5 kΩ
Z1 + Z 2
μV
(20)(10 V ∠0°)
=
25.495 kΩ ∠11.31°
Z1 + Z 2
= 7.845 mA ∠−11.31°
I″ =
IL = I′ − I″ (direction of I′)
= 78.45 mA ∠−11.31° − 7.845 mA ∠−11.31°
= 70.61 mA ∠−11.31°
9.
⎡V − E⎤
VL = −(h + 1)IZ2 = −(h + 1) ⎢ L
⎥ Z2
⎣ Z1 ⎦
Subt. for Z1, Z2
Z1 = 2 kΩ ∠0°, Z2 = 2 kΩ ∠0°
VL = −ILZ2
IL = hI + I = (h + 1)I
VL = −(h + 1)IZ2
and by KVL: VL = IZ1 + E
V −E
so that I = L
Z1
VL = −(h + 1)(VL − E)
VL(2 + h) = E(h + 1)
(h + 1)
51
E=
(20 V ∠53°) = 19.62 V ∠53°
VL =
(h + 2)
52
10.
I1:
I1 = 1 mA ∠0°
Z1 = 2 kΩ ∠0°
Z2 = 5 kΩ ∠0°
KVL: V1 − 20 V − V = 0
I′ =
Z
V1
21 V
∴ I′ =
or V = 1 I ′
Z1
21
Z1
V1 = 21 V
CHAPTER 18
227
V = I5Z2 = [I1 − I′]Z2
Z1
I′ = I1Z2 − I′Z2
21
⎡Z
⎤
I′⎢ 1 + Z 2 ⎥ = I1Z 2
⎣ 21
⎦
Z2
[I1] =
and I′ =
Z1
+ Z2
21
( 5 k Ω ∠0°)(1 mA ∠0°)
= 0.981 mA ∠0°
⎛ 2 k Ω ∠0° ⎞
+
5
k
0
Ω
∠
°
⎜
⎟
21
⎝
⎠
I2:
V1 = 20 V + V = 21 V
Z
V 21 V
⇒ V = 1 I′′
I″ = 1 =
21
Z1
Z1
Z1
V
I5 =
I ′′
=
Z 2 21 Z 2
I″ = I2 − I5 = I2 −
Z1
I′′
21 Z 2
⎡
Z ⎤
I″ ⎢1+ 1 ⎥ = I2
⎣ 21 Z 2 ⎦
2 mA ∠0°
I2
= 1.963 mA ∠0°
I″ =
=
Z1
2 kΩ
1+
1+
21(5 kΩ)
21 Z 2
I = I′ + I″ = 0.981 mA ∠0° + 1.963 mA ∠0°
= 2.94 mA ∠0°
11.
E1:
10 V ∠0° − I 10 Ω − I 2 Ω − 4 Vx = 0
with Vx = I 10 Ω
Solving for I:
10 V ∠0°
= 192.31 mA ∠0°
I=
52 Ω
Vs′ = 10 V ∠0° − I(10 Ω) = 10 V − (192.31 mA ∠0°)(10 Ω ∠0°) = 8.08 V ∠0°
228
CHAPTER 18
I:
∑ Ii = ∑ Io
5 A ∠0 ° +
Vx 5 Vx
=0
+
10 Ω 2 Ω
5 A + 0.1 Vx + 2.5 Vx = 0
2.6 Vx = −5 A
5
V = −1.923 V
Vx = −
2.6
Vs′′ = −Vx = −(−1.923 V) = 1.923 V ∠0°
Vs = Vs′ + Vs′′ = 8.08 V ∠0° + 1.923 V ∠0° = 10 V ∠0°
12.
a.
ZTh:
Z1 = 3 Ω ∠0°, Z2 = 4 Ω ∠90°
E = 100 V ∠0°
ZTh = Z1 || Z2 = (3 Ω ∠0° || 4 Ω ∠90°)
= 2.4 Ω ∠36.87° = 1.92 Ω + j1.44 Ω
ETh:
Z2E
(4 Ω ∠90°)(100 V ∠0°)
=
5 Ω ∠53.13°
Z 2 + Z1
= 80 V ∠36.87°
ETh =
b.
ZTh:
ETh:
ZTh = Z3 + Z1 || Z2
= +j6 kΩ + (2 kΩ ∠0° || 3 kΩ ∠−90°)
= +j6 kΩ + 1.664 kΩ ∠−33.69°
= +j6 kΩ + 1.385 kΩ −j0.923 kΩ
= 1.385 kΩ + j5.077 kΩ
= 5.26 kΩ ∠74.74°
Z2E
(3 kΩ ∠ − 90°)(20 V ∠0°)
=
Z 2 + Z1
2 kΩ − j 3 kΩ
60 V ∠ − 90°
=
= 16.64 V ∠−33.69°
3.606 ∠ − 56.31°
ETh =
CHAPTER 18
229
13.
a.
From #27. ZTh = Z1 || Z2
ZTh = ZN = 21.31 Ω ∠32.2°
ETh = IZ′ = IZTh
= (0.1 A ∠0°)(21.31 Ω ∠32.12°)
= 2.13 V ∠32.2°
b.
From #27. ZTh = ZN = 6.81 Ω ∠−54.23° = 3.98 Ω − j5.53 Ω
Z1 = 2 Ω ∠0°, Z3 = 8 Ω ∠−90°
Z2 = 4 Ω ∠90°, Z4 = 10 Ω ∠0°
E = 50 V ∠0°
ETh = V2 + V4
Z2 E
V2 =
Z 2 + Z1 (Z3 + Z 4 )
( 4 Ω ∠90°)(50 V ∠0°)
=
+ j 4 Ω + 2 Ω ∠0° (10 Ω − j8 Ω)
= 47.248 V ∠24.7°
V1 = E − V2 = 50 V ∠0° − 47.248 V ∠24.7° = 20.972 V ∠−70.285°
Z 4 V1
(10 Ω ∠0°)(20.972 V ∠ − 70.285°)
=
V4 =
= 16.377 V ∠−31.625°
10 Ω − j8 Ω
Z 4 + Z3
ETh = V2 + V4 = 47.248 V ∠24.7° + 16.377 V ∠−31.625°
= (42.925 V + j19.743 V) + (13.945 V − j8.587 V)
= 56.870 V + j11.156 V = 57.95 V ∠11.10°
14.
a.
ZTh:
Z1 = 10 Ω ∠0°, Z2 = 8 Ω ∠90°
Z3 = 8 Ω ∠−90°
ZTh = Z3 + Z1 || Z2
= −j8 Ω + 10 Ω ∠0°∠ || 8 Ω ∠90°
= −j8 Ω + 6.247 Ω ∠51.34°
= −j8 Ω + 3.902 Ω + j4.878 Ω
= 3.902 Ω − j3.122 Ω
= 5.00 Ω ∠−38.66°
230
CHAPTER 18
ETh: Superposition:
(8 Ω ∠90°)(120 V ∠0°)
10 Ω + j8 Ω
960 V ∠90°
=
12.806 ∠38.66°
= 74.965 V ∠51.34°
(E1)
E′Th =
(I)
E″Th = VZ 2 + VZ3
= IZ3 + I(Z1 || Z2)
= I(Z3 + Z1 || Z2)
= (0.5 A ∠60°)(−j8 Ω + 10 Ω ∠0° || 8 Ω ∠90°)
= (0.5 A ∠60°)(−j8 Ω + 3.902 Ω + j4.878 Ω)
= (0.5 A ∠60°)(3.902 Ω − j3.122 Ω)
= (0.5 A ∠60°)(4.997 Ω ∠−38.663°)
= 2.499 V ∠21.337°
ETh = E′Th + E″Th
= 74.965 V ∠51.34° + 2.449 V ∠21.337°
= (46.83 V + j58.538 V) + (2.328 V + j0.909 V)
= 49.158 V + j59.447 V = 77.14 V ∠50.41°
b.
ZTh:
ZTh = Z = 10 Ω − j10 Ω = 14.14 Ω ∠−45°
ETh:
CHAPTER 18
ETh = E − VZ
= 20 V ∠40° − IZ
= 20 V ∠40° − (0.6 A ∠90°)(14.14 Ω ∠−45°)
= 20 V ∠40° − 8.484 V ∠45°
= (15.321 V + j12.856 V) − (6 V + j6 V)
= 9.321 V + j6.856 V
= 11.57 V ∠36.34°
231
15.
a.
Z1 = 6 Ω − j2 Ω = 6.325 Ω ∠−18.435°
Z2 = 4 Ω ∠90°
ZTh = Z2 = 4 Ω ∠90°
By inspection:
ETh = E2 + E1
= 4 V + 10 V ∠0°
DC
AC
b.
I=
E1
E2
+
R2 R2 + jX L
4 V 10 V ∠0°
+
8 Ω 8 Ω + j4 Ω
10 V ∠0°
= 0.5 A +
8.944 Ω ∠26.565°
= 0.5 A + 1.118 A ∠−26.565°
(dc)
(ac)
i = 0.5 + 1.58 sin(ωt − 26.57°)
=
16.
a.
ZTh:
←ZTh = Z R1 + Z R2 = 6 Ω + 3 Ω = 9 Ω
DC:
E′Th = 12 V
AC:
←E″Th = IZ R1 = (4 A ∠ 0°)(6 Ω ∠0°) = 24 V ∠ 0°
ETh = 12 V + 24 V ∠ 0°
(DC) (AC)
232
CHAPTER 18
b.
DC: VC = 12 V
ZC E
AC: VC =
ZC + Z RTh
(1 Ω ∠ − 90°)(24 V ∠0°)
− j1 Ω + 9 Ω
24 V ∠ − 90°
=
9.055 ∠ − 6.34°
VC = 2.65 V ∠−83.66°
=
υC = 12 V+ 2.65 V ∠−83.66°
= 12 V + 3.75 sin(ωt − 83.66°)
17.
a.
ZTh:
Z1 = 10 kΩ ∠0°
Z2 = 5 kΩ − j5 kΩ
= 7.071 kΩ ∠−45°
ZTh = Z1 || Z2 = (10 kΩ ∠0°) || (7.071 kΩ ∠−45°) = 4.47 kΩ ∠−26.57°
Source conversion:
E1 = (I∠θ)(R1∠0°) = (5 mA ∠0°)(10 kΩ ∠0°) = 50 V ∠0°
Z 2 (E + E1 )
Z 2 + Z1
(7.071 k Ω ∠ − 45°)(20 V ∠0° + 50 V ∠0°)
=
(5 k Ω − j 5 k Ω) + (10 k Ω)
(7.071 k Ω ∠ − 45°)(70 V ∠0°)
=
(15 k Ω − j 5 k Ω)
494.97 V ∠ − 45°
=
15.811 ∠ − 18.435°
= 31.31 V ∠−26.57°
ETh =
b.
ETh
31.31 V ∠ − 26.565°
=
ZTh + Z L 4.472 k Ω ∠ − 26.565° + 5 k Ω ∠90°
31.31 V ∠ − 26.565°
31.31 V ∠ − 26.565°
=
=
4 k Ω − j 2 k Ω + j5 k Ω
4 k Ω + j3 k Ω
31.31 V ∠ − 26.565°
=
= 6.26 mA ∠63.44°
5 k Ω ∠36.87°
I=
CHAPTER 18
233
Z1 = 10 kΩ ∠ 0°
Z2 = 10 kΩ ∠ 0°
Z3 = 1 kΩ ∠ −90°
18.
ZTh = Z3 + Z1 || Z2 = 5 kΩ − j1 kΩ ≅ 5.1 kΩ ∠−11.31°
ETh: (VDR)
19.
ETh =
Z 2 (20 V ) (10 k Ω ∠0°)(20 V )
=
= 10 V
Z 2 + Z1
10 k Ω + 10 k Ω
ZTh:
Z1 = 40 kΩ ∠0°
2 Z 2 = 0.2 kΩ ∠−90°
Z3 = 5 kΩ ∠0°
ZTh = Z3 || (Z1 + Z2) = 5 kΩ ∠0° || (40 kΩ − j0.2 kΩ) = 4.44 kΩ ∠−0.03°
I′ =
Z1 (100 I )
Z1 + Z 2 + Z3
(40 k Ω ∠0°)(100 I)
=
45 k Ω ∠ − 0.255°
= 88.89 I ∠0.255°
ETh = −I′Z3 = −(88.89 I ∠0.255°)(5 kΩ ∠0°) = −444.45 × 103 I ∠0.26°
20.
ZTh:
←
ZTh = Z1 = 20 kΩ ∠0°
ETh:
E′Th = −(hI)(Z1)
= −(100)(2 mA ∠ 0°)(20 kΩ ∠ 0°)
= −4 kV ∠0°
234
CHAPTER 18
E:
ETh = E′Th + E″Th
= −4 kV ∠0° + 10 V ∠0°
= −3990 V ∠0°
21.
ZTh:
Z1 = 5 kΩ ∠0°
Z2 = −j1
←ZTh = Z1 + Z2 = 5 kΩ − j1 kΩ
= 5.10 kΩ ∠−11.31°
ETh:
22.
ZTh:
ETh = − ⎡⎣ μ V + VZ1 ⎤⎦
= − μ V − IZ1
= −(20)(2 V ∠0°) − (2 mA ∠0°)(5 kΩ ∠0°)
= −50 V ∠0°
Z1 = 20 kΩ ∠ 0°
Z2 = 5 kΩ ∠ 0°
←ZTh = Z1 + Z2 = 25 kΩ ∠ 0°
ETh:
ETh = μV − (hI)(Z1)
= (20)(10 V ∠ 0°) − (100)(1 mA ∠0°)(20 kΩ ∠ 0°)
= −1800 V ∠0°
CHAPTER 18
235
23.
ETh: (Eoc)
hI = −I
Z1 = 2 kΩ ∠0°
∴I = 0
and hI = 0
with Eoc = ETh = E = 20 V ∠53°
Isc:
Isc = −(h + 1)I
= −(h + 1)(10 mA ∠53°)
= −510 mA ∠53°
ZTh =
24.
Eoc
20 V ∠53°
= −39.22 Ω ∠0°
=
I sc −510 mA ∠53°
ETh:
Z1 = 5 kΩ ∠0°
E′oc = 21 V
V = I1Z1 = (1 mA ∠0°)(5 kΩ ∠0°)
= 5 V ∠0 °
E′oc = E′Th = 21(5 V ∠0°)
= 105 V ∠0°
V = I2Z1
= (2 mA ∠0°)(5 kΩ ∠0°)
= 10 V ∠0°
E″oc = E″Th = V + 20 V = 21 V = 210 V ∠0°
Isc:
I′sc = I1
20 V = V ∴ V = 0 V
and I′ = 0 A
∴ I″sc = I2
Isc = I′sc + I″sc = 3 mA ∠0°
Eoc = E′oc + E″oc = 315 V ∠0° = ETh
315 V ∠0°
ZTh = Eoc =
= 105 kΩ ∠0°
3 mA ∠0°
I sc
236
CHAPTER 18
25.
Eoc:
(ETh)
KVL: −6 Ix(2 kΩ) − Ix(1 kΩ) + 8 V ∠0° − Ix(3.3 kΩ) = 0
8 V ∠0°
= 0.491 mA ∠0°
Ix =
16.3 k Ω
Eoc = ETh = Ix(3.3 kΩ) = 1.62 V ∠0°
Isc:
8V
= 2.667 mA ∠0°
3kΩ
E
1.62 V ∠0°
ZTh = oc =
= 607.42 Ω ∠0°
2.667 mA ∠0°
I sc
Isc =
26.
a.
From Problem 12(a): ZN = ZTh = 1.92 Ω + j1.44 Ω = 2.4 Ω ∠36.87°
IN:
Z1 = 3 Ω ∠0°, Z2 = 4 Ω ∠90°
E 100 V ∠0°
=
Isc = IN =
Z1
3 Ω ∠0°
= 33.33 A ∠0°
CHAPTER 18
237
b.
From Problem 12(b): ZN = ZTh = 5.263 kΩ ∠74.74° = 1.39 kΩ + j5.08 kΩ
IN:
Z1 = 2 kΩ ∠0°, Z2 = 3 kΩ ∠−90°
Z3 = 6 kΩ∠90°
ZT = Z1 + Z2 || Z3
= 2 kΩ + 3 kΩ ∠−90° || 6 kΩ ∠90°
= 2 kΩ + 6 kΩ ∠−90°
= 2 kΩ − j6 kΩ
= 6.325 kΩ ∠−71.565°
20 V ∠0°
E
=
6.325 kΩ ∠ − 71.565°
ZT
= 3.162 mA ∠71.565°
Z2I s
(3 kΩ ∠ − 90°)(3.162 mA ∠71.565°)
Isc = IN =
=
Z 2 + Z3
− j 3 kΩ + j 6 kΩ
9.486 mA ∠ − 18.435°
= 3.16 mA ∠−108.44°
=
3 ∠90°
Is =
27.
a.
Z1 = 20 Ω + j20 Ω = 28.284 Ω ∠45°
Z2 = 68 Ω ∠0°
←ZN = Z1 || Z2
= (28.284 Ω ∠45°) || (68 Ω ∠0°)
= 21.31 Ω ∠32.2°
←Isc = I = IN = 0.1 A ∠0°
b.
Z1 = 2 Ω ∠0°, Z2 = 4 Ω ∠90°
Z3 = 8 Ω ∠−90°, Z4 = 10 Ω ∠0°
E = 50 V ∠0°
238
CHAPTER 18
ZN:
Z′ = Z1 || Z2 = 2 Ω ∠0° || 4 Ω ∠90°
= 1.789 Ω ∠26.565° = 1.6 Ω + j0.8 Ω
Z′ + Z4 = 1.6 Ω + j0.8 Ω + 10 Ω = 11.6 Ω + j0.8 Ω = 11.628 Ω ∠3.945°
ZN = Z3 || (Z′ + Z4) = (8 Ω ∠−90°) || (11.628 Ω ∠3.945°) = 6.81 Ω ∠−54.23°
= 3.98 Ω − j5.53 Ω
50 V ∠0°
= 6.250 A ∠90°
Z 3 8 Ω ∠ − 90°
Z′ = Z2 || Z4 = 4 Ω ∠90° || 10 Ω ∠0°
= 3.714 Ω ∠68.2°
(3.714 Ω ∠68.2°)(50 V ∠0°)
Z′E
=
V2 =
′
1.378 Ω + j 3.448 Ω + 2 Ω
Z + Z1
I3 =
=
E
=
185.7 V ∠68.2°
= 38.471 V ∠22.612°
4.827 ∠45.588°
38.471 V ∠22.612°
I4 = V 2 =
= 3.847 A ∠22.612°
10 Ω ∠0°
Z4
IN = I3 + I4 = 6.250 A ∠90° + 3.847 A ∠22.612°
= +j6.25 A + 3.551 A + j1.479 A = 3.551 A + j7.729 A
= 8.51 A ∠65.32°
28.
a.
From Problem 14(a): ZN = ZTh = 5.00 Ω ∠−38.66°
IN:
Superposition:
ZT = Z1 + Z2 || Z3
= 10 Ω + 8 Ω ∠90° || 8 Ω ∠−90°
64 Ω ∠0°
= 10 Ω +
0
= very large impedance
E
=0A
Is =
ZT
and V Z1 = 0 V
with V Z2 = V Z3 = E1 = 120 V ∠0°
120 V ∠0°
so that I′sc = E1 =
8 Ω ∠ − 90°
Z3
= 15 A ∠90°
CHAPTER 18
239
(I)
I″sc = I = 0.5 A ∠60°
IN = I′sc + I″sc = + j15 A + 0.5 A ∠60° = + j15 A + 0.25 A + j0.433 A
= 0.25 A + j15.433 A = 15.44 A ∠89.07°
b.
From Problem 14(b):
ZN = ZTh = 10 Ω − j10 Ω = 14.14 Ω ∠−45°
IN = I′ − I
E
= −I
Z
20 V ∠40°
− 0.6 A ∠90°
=
14.142 Ω ∠ − 45°
= 1.414 A ∠85° − j0.6 A
= 0.123 A + j1.409 A − j0.6 A
= 0.123 A + j0.809 A
= 0.82 A ∠81.35°
IN:
29.
a.
ZN:
E = 20 V ∠0°, I2 = 0.4 A ∠20°
Z1 = 6 Ω + j8 Ω = 10 Ω ∠53.13°
Z2 = Ω − j12 Ω = 15 Ω ∠−53.13°
ZN = Z1 || Z2 = (10 Ω ∠53.13°) || (15 Ω ∠−53.13°)
= 9.66 Ω ∠14.93°
IN:
(E)
I′sc = E/Z1 = 20 V ∠0°/10 Ω ∠53.13°
(I2)
I″sc = I2 = 0.4 A ∠20°
= 2 A ∠−53.13°
IN = I′sc + I″sc = 2 A ∠−53.13° + 0.4 A ∠20°
= 2.15 A ∠−42.87°
240
CHAPTER 18
b.
ZN:
E1 = 120 V ∠30°, Z1 = 3 Ω ∠0°
Z2 = 8 Ω − j8 Ω, Z3 = 4 Ω ∠90°
IN:
ZN = Z3 + Z1 || Z2
= 4 Ω ∠90° + (3 Ω ∠0°) || (8 Ω − j8 Ω)
= 4.37 Ω ∠55.67° = 2.47 Ω + j3.61 Ω
E1
120 V ∠30°
=
ZT
Z1 +Z 2 Z3
120 V ∠30°
=
3 Ω + (8 Ω − j8 Ω) 4 Ω ∠90°
120 V ∠30°
=
6.65 Ω ∠46.22°
= 18.05 A ∠−16.22°
I=
Isc = IN =
30.
a.
(8 Ω − j8 Ω)(18.05 A ∠ − 16.22°)
Z 2 (I )
=
= 22.83 A ∠−34.65°
8 Ω − j8 Ω + j 4 Ω
Z 2 + Z3
ZN = 8 Ω ∠0°
IN:
DC
AC
Z1 = 6 Ω − j2 Ω
Z2 = 8 Ω ∠ 0°
E1 10 V ∠0°
= 1.25 A ∠0°
=
Z 2 8 Ω ∠0°
IN = 0.5 A + 1.25 A ∠0°
Isc =
CHAPTER 18
241
b.
ZN = 8 Ω ∠0°
ZL = 4 Ω ∠90°
DC:
AC:
31.
a.
( ) (8 Ω ∠0°)(1.25 A ∠0°)
= 1.118 A ∠−26.57°
I = ZN I2 =
8 Ω + j4 Ω
ZN + ZL
I8Ω = 0.5 A + 1.118 A ∠−26.57°
(dc)
(ac)
i = 0.5 + 1.58 sin(ωt − 26.57°)
From #16
ZN = ZTh = 9 Ω ∠0°
DC:
I′N =
E
RT
=
12 V
= 1.33 A
9Ω
AC:
R1 I
(6 Ω ∠0°)(4 A ∠0°)
=
R1 + R2
9 Ω ∠0°
24 V ∠0°
= 2.67 A ∠0°
=
9 Ω ∠0°
I″N =
IN = 1.33 A + 2.67 A ∠0°
242
CHAPTER 18
b.
DC: VC = IR
= (1.33 A)(9 Ω)
= 12 V
AC: Z′ = 9 Ω ∠0° || 1 Ω ∠−90°
= 0.994 Ω ∠−83.66°
VC = IZ′ = (2.667 A ∠0°)(0.994 Ω ∠−83.66°)
= 2.65 V ∠−83.66°
VC = 12 V + 2.65 V ∠−83.66°
32.
a.
Note Problem 17(a):
ZN = ZTh = 4.47 kΩ ∠−26.57°
Using the same source conversion: E1 = 50 V ∠0°
Defining ET = E1 + E = 50 V ∠0° + 20 V ∠0° = 70 V ∠0°
Z1 = 10 kΩ ∠0°
Z2 = 5 kΩ − j5 kΩ = 7.071 kΩ ∠−45°
Isc =
ET
70 V ∠0°
=
= 7 mA ∠0°
Z1 10 k Ω ∠0°
IN = Isc = 7 mA ∠0°
b.
( ) (4.472 k Ω ∠ − 26.565°)(7 mA ∠0°)
I = ZN IN =
Z N + Z L 4.472 k Ω ∠ − 26.565° + 5 k Ω ∠90°
31.30 mA ∠ − 26.565° 31.30 mA ∠ − 26.565°
=
=
4 − j 2 + j5
4 + j3
31.30 mA ∠ − 26.565°
= 6.26 mA ∠63.44° as obtained in Problem 17.
=
5 ∠36.87°
CHAPTER 18
243
33.
Z1 = 10 kΩ ∠0°, Z2 = 10 kΩ ∠0
Z3 = −j1 kΩ
ZN = Z3 + Z1 || Z2 = 5 kΩ − j1 kΩ
= 5.1 kΩ ∠−11.31°
ZN:
IN:
V2 =
−(Z 2 Z3 )20 V
(Z 2 Z3 ) + Z1
−(0.995 k Ω ∠ − 84.29°)(20 V)
0.1 k Ω − j 0.99 k Ω + 10 k Ω
V2 = −1.961 V ∠−78.69°
=
IN = Isc =
34.
ZN:
IN:
V2 −1.961 V ∠ − 78.69°
=
= −1.96 × 10−3 V ∠11.31°
1 k Ω ∠ − 90°
Z3
Z1 = 40 kΩ ∠0°, Z2 = 0.2 kΩ ∠−90°
Z3 = 5 kΩ ∠0°
ZN = Z3 || (Z1 + Z2)
= 5 kΩ ∠0° || (40 kΩ − j0.2 kΩ)
= 4.44 kΩ ∠−0.03°
Z1 (100 I )
Z1 + Z 2
( 40 kΩ ∠0°)(100 I )
=
40 kΩ ∠ − 0.286°
= 100 I ∠0.29°
IN = Isc =
244
CHAPTER 18
35.
ZN:
Z1 = 5 kΩ ∠0°, Z2 = 1 kΩ ∠−90°
← ZN = Z1 + Z2 = 5 kΩ − j1 kΩ
= 5.1 kΩ ∠−11.31°
.
IN:
μV
(20)(2 V ∠0°)
Z1 + Z 2 5.1 kΩ ∠ − 11.31°
= 7.843 mA ∠11.31°
I′sc =
(I):
=
Z1 (I )
Z1 + Z 2
(5 kΩ ∠0°)(2 mA ∠0°)
=
5.1 kΩ ∠ − 11.31°
= 1.96 mA ∠11.31°
I″sc =
IN = I′sc + I″sc = 7.843 mA ∠11.31° + 1.96 mA ∠11.31°
= 9.81 mA ∠11.31°
36.
ZN:
Z1 = 20 kΩ ∠0°, Z2 = 5 kΩ ∠0°
V = 10 V ∠0°, μ = 20, h = 100
I = 1 mA ∠0°
ZN = Z1 + Z2 = 25 kΩ ∠0°
IN: (hI)
Z1 (hI )
Z1 + Z 2
(20 kΩ ∠0°)(hI )
=
20 kΩ ∠0° + 5 kΩ∠0°
= 80 mA ∠0°
I′sc =
(μV)
I″sc =
μV
=
Z1 + Z 2
= 8 mA ∠0°
(20)(10 V ∠0°)
25 kΩ
IN (direction of I′sc) = I′sc − I″sc = 80 mA ∠0° − 8 mA ∠0° = 72 mA ∠0°
CHAPTER 18
245
37.
Z1 = 1 kΩ ∠0°
Z2 = 3 kΩ ∠0°
Z3 = 4 kΩ ∠0°
V2 = 21 V = Eoc ⇒ V =
I2 =
and
⎡ 1
Eoc
Eoc
E
1 ⎤
+ oc = Eoc ⎢
, I = I1 + I2 =
+
⎥
Z2
Z2
21 Z1
⎣ 21 Z1 Z 2 ⎦
⎡ + 21 Z1 ⎤
I = Eoc ⎢ Z 2
⎥
⎣ 21 Z1Z 2 ⎦
Eoc
21
Eoc
V
I = I1 + I2, I1 =
=
Z1
21 Z1
21 Z1Z 2I
(21)(1 k Ω ∠0°)(3 k Ω ∠0°)(2 mA ∠0°)
=
Z 2 + 21Z1
3 k Ω + 21(1 k Ω ∠0°)
ETh = Eoc = 5.25 V ∠0°
Eoc =
V3 21 V
Z
⇒ V = 3 Isc
=
21
Z3
Z3
V = I1Z1
I = I1 + I′
Isc =
Isc =
⎛ Z + Z3 ⎞
Z 2 I′
⇒ I′ = ⎜ 2
⎟ I sc
Z 2 + Z3
⎝ Z2 ⎠
⎡ Z3
Z + Z3 ⎤
V ⎛ Z 2 + Z3 ⎞
+ 2
+⎜
⎟ I sc = ⎢
⎥ I sc
Z2 ⎦
Z1 ⎝ Z 2 ⎠
⎣ 21 Z1
I
2 mA ∠0°
= 0.79 mA ∠0°
Isc =
=
4 kΩ 7 kΩ
Z3
Z3 + Z 2
+
+
21 k Ω 3 k Ω
21 Z1
Z2
I = I1 + I′ =
∴
246
IN = 0.79 mA ∠0°
E
5.25 V ∠0°
ZN = oc =
= 6.65 kΩ ∠0°
I sc 0.79 mA ∠0°
CHAPTER 18
38.
Z1 = 2 kΩ ∠0°
Z2 = 5 kΩ ∠0°
I2 = I3 + I5
V = I5Z2 = (I2 − I3)Z2
Eoc = ETh = 21 V = 21(I2 − I3)Z2
⎛
E ⎞
= 21⎜ I 2 − oc ⎟ Z 2
Z1 ⎠
⎝
⎡
Z2 ⎤
Eoc ⎢1 + 21 ⎥ = 21 Z2I2
Z1 ⎦
⎣
21 Z 2 I 2 21(5 k Ω ∠0°)(2 mA ∠0°)
Eoc =
=
Z
⎛ 5 k Ω ∠0 ° ⎞
2
1 + 21
1 + 21⎜
⎟
Z1
⎝ 2 k Ω ∠0° ⎠
ETh = Eoc = 3.925 V ∠0°
20 V ≠ −V ∴ V = 0
and IN = Isc = I2 = 2 mA ∠0°
3.925 V ∠0°
ZN = Eoc =
= 1.96 kΩ
2 mA ∠0°
I sc
39.
a.
Z1 = 3 Ω + j4 Ω, Z2 = −j6 Ω
← ZTh = Z1 || Z2
= 5 Ω ∠53.13° || 6 Ω ∠−90°
= 8.32 Ω ∠−3.18°
ZL = 8.32 Ω ∠3.18° = 8.31 Ω − j0.46 Ω
CHAPTER 18
247
ETh =
Z2 E
Z 2 + Z1
(6 Ω ∠ − 90°)(120 V ∠0°)
3.61 Ω ∠ − 33.69°
= 199.45 V ∠−56.31°
2
(3.124 V ) 2
E
Pmax = Th =
= 1198.2 W
4RTh
4(8.31 Ω)
=
Z1 = 3 Ω + j4 Ω = 5 Ω ∠53.13°
Z2 = 2 Ω ∠0°
← ZN = ZTh = Z1 || Z2
= 5 Ω ∠53.13° || 2 Ω ∠0°
10 Ω ∠53.13°
=
2 + 3 + j4
10 Ω ∠53.13°
=
5 + j4
10 Ω ∠53.13°
=
6.403 ∠38.66°
= 1.56 Ω ∠14.47°
ZTh = 1.56 Ω ∠14.47°
= 1.51 Ω + j0.39 Ω
ZL = 1.51 Ω − j0.39 Ω
b.
ETh = I(Z1|| Z2)
= (2 A ∠30°)(1.562 Ω ∠14.47°)
= 3.12 V ∠44.47°
2
(3.12 V ) 2
E
= 1.61 W
Pmax = Th =
4RTh 4(1.51 Ω)
40.
a.
ZTh:
Z1 = 4 Ω ∠90°, Z2 = 10 Ω ∠0°
Z3 = 5 Ω ∠−90°, Z4 = 6 Ω ∠−90°
E = 60 V ∠60°
ZTh = Z4 + Z3 || (Z1 + Z2) = −j6 Ω + (5 Ω ∠−90°) || (10 Ω + j4 Ω)
= 2.475 Ω − j4.754 Ω
= 11.04 Ω ∠−77.03°
ZL = 11.04 Ω ∠77.03°
248
CHAPTER 18
ETh:
Z 3 (E)
Z3 + Z1 + Z 2
( 5 Ω ∠ − 90°)(60 V ∠60°)
=
− j 5 Ω + j 4 Ω + 10 Ω
= 29.85 V ∠−24.29°
ETh =
2
/ 4 RTh = (29.85 V)2/4(2.475 Ω) = 90 W
Pmax = E Th
b.
Z1 = 3 Ω + j4 Ω = 5 Ω ∠53.13°
Z2 = −j8 Ω
Z3 = 12 Ω + j9 Ω
ZTh = Z2 + Z1 || Z3 = −j8 Ω + (5 Ω ∠53.13°) || (15 Ω ∠36.87°)
= 5.71 Ω ∠−64.30° = 2.475 Ω − j5.143 Ω
ZL = 5.71 Ω ∠64.30° = 2.48 Ω + j5.15 Ω
ETh + V Z3 − E2 = 0
ETh = E2 − VZ3
Z3 (E 2 − E1 )
Z3 + Z1
= 168.97 V ∠112.53°
VZ3 =
ETh = E2 − VZ3 = 200 V ∠90° − 168.97 V ∠112.53° = 78.24 V ∠34.16°
2
/ 4 RTh = (78.24 V)2/4(2.475 Ω) = 618.33 W
Pmax = E Th
41.
E ∠0° 1 V ∠0°
= 1 mA ∠0°
=
R1∠0° 1 k Ω ∠0°
ZTh = 40 kΩ ∠0°
ETh = (50 I)(40 kΩ ∠0°) = (50)(1 mA ∠0°)(40 kΩ ∠0°) = 2000 V ∠0°
2
( 2 kV ) 2
E
Pmax = Th =
= 25 W
4 RTh 4(40 k Ω)
I=
CHAPTER 18
249
42.
a.
b.
43.
ZTh = ZN = 8 Ω ∠0° (Problem 30(b))
ZL = 8 Ω ∠0°
ETh = I N ⋅ Z N :
DC: ETh = I′N ⋅ Z N = (0.5 A)(8 Ω) = 4 V
AC: ETh = I′N ⋅ Z N = (1.25 A ∠0°)(8 Ω ∠0°) = 10 V
2
(4 V) 2 (10 V) 2
+
= 0.5 W + 3.13 W = 3.63 W
Pmax = E Th =
4 RTh 4(8 Ω) 4(8 Ω)
From #16, ZTh = 9 Ω, ETh = 12 V + 24 V ∠0°
a.
b.
∴ ZL = 9 Ω
2
(12 V ) 2 (24 V ) 2
Pmax = E Th =
= 4 W + 16 W = 20 W
+
4 RTh 4(9 Ω)
4(9 Ω)
or ETh = V 02 + V1eff2
= 26.833 V
2
2
(26.833 V )
and Pmax = E Th =
= 20 W
4 RTh
4(9 Ω)
44.
45.
a.
Problem 17(a):
ZTh = 4.47 kΩ ∠−26.57° = 4 kΩ − j2 kΩ
ZL = 4 kΩ + j2 kΩ
ETh = 31.31 V ∠−26.57°
b.
2
/ 4 RTh = (31.31 V)2/4(4 kΩ) = 61.27 mW
Pmax = E Th
a.
ZTh = 2 kΩ ∠0° || 2 kΩ ∠−90° = 1 kΩ − j1 kΩ
R L = R Th + ( X Th + X Load )
2
2
= (1 kΩ) 2 + (−1 kΩ + 2 kΩ) 2
= (1 kΩ) 2 + (1 kΩ) 2
= 1.41 kΩ
b.
250
Rav = (RTh + RLoad)/2 = (1 kΩ + 1.41 kΩ)/2 = 1.21 kΩ
2
(50 V ) 2
E
= 516.53 mW
Pmax = Th =
4 Rav 4(1.21 kΩ)
CHAPTER 18
46.
a.
ZTh:
1
1
=
2π fC 2π (10 kHz)(4 nF)
≅ 3978.87 Ω
XL = 2πfL = 2π(10 kHz)(30 mH)
≅ 1884.96 Ω
Z1 = 1 kΩ ∠0°, Z2 = 1884.96 Ω ∠90°
Z3 = 3978.87 Ω ∠−90°
ZTh = (Z1 + Z2) || Z3 = (1 kΩ + j1884.96 Ω) || 3978.87 Ω ∠−90°)
= 2133.79 Ω ∠62.05° || 3978.87 Ω ∠−90°)
= 3658.65 Ω ∠36.52°
XC =
∴ ZL = 3658.65 Ω ∠−36.52° = 2940.27 Ω − j2177.27 Ω
1
1
= 7.31 nF
C=
=
2π fX C 2π (10 kHz)(2177.27 Ω)
b.
RL = RTh = 2940.27 Ω
c.
ETh =
(3978.87 Ω ∠ − 90°)(2 V∠0°)
Z 3 ( E)
=
= 3.43 V∠−25.53°)
Z3 + Z1 + Z 2 1 kΩ + j1884.96 Ω − j 3978.87Ω
2
Pmax = ETh
/ 4 RTh = (3.43 V) 2 /4(2940.27 Ω) = 1 mW
(4 kΩ ∠0°)(4 mA ∠0°)
= 1.33 mA ∠0°
4 kΩ + 8 kΩ
Vab = (Iab)(8 kΩ ∠0°) = 10.67 V ∠0°
47.
Iab =
48.
a.
4 kΩ ( E )
1
= (20 V ∠0°)
4 kΩ + 12 kΩ 4
= 5 V ∠0°
5 V ∠0°
I=
= 0.83 mA ∠0°
6 kΩ
V=
b.
6 kΩ(E)
1
= (20 V ∠0°)
2
6 kΩ + 6 kΩ
= 10 V ∠0°
10 V ∠0°
= 0.83 mA ∠0°
I=
12 kΩ
V=
CHAPTER 18
251
49.
100 V ∠0°
= 50 mA ∠0°
2 k Ω ∠0°
50 V ∠0°
I2 =
4 k Ω ∠90°
= 12.5 mA ∠−90°
Z1 = 2 kΩ ∠0°
Z2 = 4 kΩ ∠90°
Z3 = 4 kΩ ∠−90°
IT = I1 − I2 = (50 mA ∠0° − 12.5 mA ∠−90°) = 50 mA + j12.5 mA
= 51.54 mA ∠14.04°
Z′ = Z1 || Z2 = (2 kΩ ∠0°) || (4 kΩ ∠90°) = 1.79 kΩ ∠26.57°
Z′IT
(1.79 k Ω ∠26.57°)(51.54 mA ∠14.04°)
=
IC =
1.6 kΩ + j 0.8 kΩ − j 4 kΩ
Z′ + Z 3
= 25.77 mA ∠104.04°
I1 =
252
CHAPTER 18
Chapter 19
1.
a.
PT = 60 W + 20 W + 40 W = 120 W
b.
QT = 0 VARS, ST = PT = 120 VA
c.
ST = EIs, Is =
S T 120 VA
=
= 0.5 A
240 V
E
60 W
= 240 Ω
(0.5 A ) 2
V = IsR = (0.5 A)(240 Ω) = 120 V
V1 = V2 = E − V = 240 V − 120 V = 120 V
2
2
(120 V ) 2
= 720 Ω
P1 = V 1 , R 1 = V 1 =
20 W
R1
P1
d.
P = I s2 R, R =
P
2
Is
(120 V )
= 360 Ω
P2 = V 2 , R2 = V 2 =
40 W
R2
P2
V 120 V
V
120 V
= 0.33 A
= 0.17 A, I2 = 2 =
I1 = 1 =
R2 360 Ω
R1 720 Ω
2
e.
2.
a.
=
2
2
ZT = 3 Ω − j5 Ω + j9 Ω = 3 Ω + j4 Ω = 5 Ω ∠53.13°
50 V ∠0°
E
I=
=
= 10 A ∠−53.13°
ZT 5 Ω ∠53.13°
R:
L:
C:
P = I2R = (10 A)2 3 Ω = 300 W
P=0W
P=0W
b.
R:
C:
L:
Q = 0 VAR
QC = I2XC = (10 A)2 5 Ω = 500 VAR
QL = I2XL = (10 A)2 9 Ω = 900 VAR
c.
R:
C:
L:
S = 300 VA
S = 500 VA
S = 900 VA
d.
PT = 300 W
QT = QL − QC = 400 VAR(L)
ST =
2
2
PT + QT = EI = (50 V)(10 A) = 500 VA
300 W
= 0.6 lagging
Fp = P T =
S T 500 VA
e.
−
CHAPTER 19
253
f.
WR =
⎡ VI ⎤
⎡ VI ⎤ VI
VI
: WR = 2 ⎢ ⎥ = 2 ⎢
⎥=
f1
⎣ f2 ⎦
⎣ 2 f1 ⎦ f1
V = IR = (10 A)(3 Ω) = 30 V
(30 V)(10 A)
WR =
=5J
60 Hz
g.
3.
a.
VC = IXC = (10 A)(5 Ω) = 50 V
VI (50 V)(10 A)
=
= 1.33 J
WC =
ω1 (2π )(60 Hz)
VL = IXL = (10 A)(9 Ω) = 90 V
VI (90 V)(10 A)
WL =
= 2.39 J
=
ω1
376.8
PT = 0 + 100 W + 300 W = 400 W
QT = 200 VAR(L) − 600 VAR(C) + 0 = −400 VAR(C)
ST =
Fp =
b.
a.
PT = EIs cos θT
400 W = (100 V)Is(0.7071)
400 W
Is =
= 5.66 A
70.71 V
Is = 5.66 A ∠135°
PT = 600 W + 500 W + 100 W = 1200 W
QT = 1200 VAR(L) + 600 VAR(L) − 600 VAR(C) = 1200 VAR(L)
ST =
b.
5.
2
2
2
2
PT + QT = (1200 W) + (1200 VAR) = 1697 VA
1200 W
Fp = P T =
= 0.7071 (lagging)
S T 1697 VA
c.
−
d.
Is =
a.
PT = 200 W + 200 W + 0 + 100 W = 500 W
QT = 100 VAR(L) + 100 VAR(L) − 200 VAR(C) − 200 VAR(C) = −200 VAR(C)
ST =
254
PT
400 W
=
= 0.707 (leading)
ST 565.69 VA
−
c.
4.
PT2 + QT2 = 565.69 VA
S T = 1697 VA
= 8.485 A, 0.7071 ⇒ 45° (lagging)
E
200 V
Is = 8.49 A ∠−45°
2
2
PT + QT = 538.52 VA
CHAPTER 19
b.
c.
6.
500 W
Fp = PT =
= 0.928 (leading)
S T 538.52 VA
−
d.
PT = EIs cos θT
500 W = (50 V)Is(0.928)
500 W
= 10.776 A
Is =
46.4 V
Is = 10.78 A ∠21.88°
a.
IR =
b.
IL =
c.
PT = 180 W + 400 W = 580 W
QT = 600 VAR(L) + 360 VAR(L) = 960 VAR(L)
60 V ∠30°
= 3 A ∠30°
20 Ω ∠0°
P = I2R = (3 A)2 20 Ω = 180 W
QR = 0 VAR
S = P = 180 VA
60 V ∠30°
= 6 A ∠−60°
10 Ω ∠90°
PL = 0 W
QL = I2XL = (6 A)2 10 Ω = 360 VAR(L)
S = Q = 360 VA
ST =
(580 W)2 + (960 VAR) 2 = 1121.61 VA
580 W
Fp = PT =
= 0.517 (lagging) θ = 58.87°
1121.61
VA
ST
d.
ST = EIs
1121.61 VA
S
= 18.69 A
Is = T =
E
60 V
θ I s = 30° − 58.87° = −28.87°
Is = 18.69 A ∠−28.87°
2
2
7.
a.
b.
(20 V)
= 200 W
R: P = E =
2Ω
R
PL,C = 0 W
R:
Q = 0 VAR
C:
2
(20 V) 2
QC = E =
= 80 VAR(C)
5Ω
XC
L:
(20 V )
= 100 VAR(L)
QL = E =
4Ω
XL
2
CHAPTER 19
2
255
c.
R:
C:
L:
d.
PT = 200 W + 0 + 0 = 200 W
QT = 0 + 80 VAR(C) + 100 VAR(L) = 20 VAR(L)
ST =
S = 200 VA
S = 80 VA
S = 100 VA
(200 W) 2 + (20 VAR) 2 = 200 VA
200 W
Fp = PT =
= 0.995 (lagging) ⇒ 5.73°
200.998
VA
ST
e.
f.
8.
a.
−
200.998 VA
Is = S T =
= 10.05 A
20 V
E
Is = 10.05 A∠−5.73°
50 V ∠60°
= 10 A ∠6.87°
5 Ω ∠53.13°
PR = I2R = (10 A)2 3 Ω = 300 W
PL = 0 W
PC = 0 W
R − L:
I=
b.
QR = 0 VAR
QL = I2XL = (10 A)2 4 Ω = 400 VAR
50 V ∠60°
= 5 A ∠150°
IC =
10 Ω ∠ − 90°
QC = I2XC = (5 A)2 10 Ω = 250 VAR
c.
SR = P = 300 VA
SL = QL = 400 VA
SC = QC = 250 VA
d.
PT = PR = 300 W
QT = 400 VAR(L) − 250 VAR(C) = 150 VAR(L)
ST =
(300 W)2 + (150 VAR) 2 = 335.41 VA
300 W
= 0.894 (lagging)
Fp = PT =
S T 335.41 VA
e.
f.
256
−
S T 335.41 VA
=
= 6.71 A
E
50 V
0.894 ⇒ 26.62° lagging
θ = 60° − 26.62° = 33.38°
Is = 6.71 A ∠33.38°
Is =
CHAPTER 19
9. a−c.
XL = ωL = (400 rad/s)(0.1 H) = 40 Ω
1
1
=
XC =
ωC (400 rad/s)(100 μ F)
= 25 Ω
Z1 = 40 Ω ∠90°, Z2 = 25 Ω ∠−90°
Z3 = 30 Ω ∠0°
ZT = Z1 + Z2 || Z3 = +j40 Ω + (25 Ω ∠−90°) || (30 Ω ∠0°)
= +j40 Ω + 19.21 Ω ∠−50.19°
= +j40 Ω + 12.3 Ω − j14.76 Ω
= 12.3 Ω + j25.24 Ω
= 28.08 Ω ∠64.02°
Is =
E
=
50 V ∠0°
= 1.78 A ∠−64.02°
28.08 Ω ∠64.02°
ZT
V2 = Is(Z2 || Z3) = (1.78 A ∠−64.02°)(19.21 Ω ∠−50.19°)
= 34.19 V ∠−114.21°
34.19 V ∠ − 114.21°
V
= 1.37 A ∠−24.21°
I2 = 2 =
25 Ω ∠ − 90°
Z2
34.19 V ∠ − 114.21°
V
= 1.14 A ∠−114.21°
I3 = 2 =
30 Ω ∠0°
Z3
P = 0 W, QL = I s2 X L = (1.78 A)2 40 Ω = 126.74 VAR(L), S = 126.74 VA
Z1:
P = 0 W, QC = I 22 X C = (1.37 A)2 25 Ω = 46.92 VAR(C), S = 46.92 VA
Z2:
P = I 32 R = (1.14 A)2 30 Ω = 38.99 W, QR = 0 VAR, S = 38.99 VA
Z3:
d.
PT = 0 + 0 + 38.99 W = 38.99 W
QT = +126.74 VAR(L) − 46.92 VAR(C) + 0 = 79.82 VAR(L)
2
2
PT + QT = 88.83 VA
ST =
38.99 W
Fp = PT =
= 0.439 (lagging)
S T 88.83 VA
e.
−
f.
WR =
f1 =
g.
VR I R V2 I 3 (34.19 V)(1.14 A)
=
=
= 0.31 J
2f1
2 f1
2(63.69 Hz)
ω1 400 rad/s
=
= 63.69 Hz
2π
6.28
(1.78 A) 2 40 Ω
= 0.32 J
ω1
ω1
ω1
400 rad/s
V I
(34.19 V)(1.37 A)
VI
WC = C C = 2 2 =
= 0.12 J
ω1
ω1
400 rad/s
WL =
CHAPTER 19
VL I L
=
(Is X L )Is
=
I s2 X L
=
257
10.
11.
12.
S T 10,000 VA
=
= 50 A
200 V
E
0.5 ⇒ 60° leading
∴ Is leads E by 60°
200 V ∠0°
E
ZT =
=
= 4 Ω ∠−60° = 2 Ω − j3.464 Ω = R − jXC
50 A ∠60°
Is
a.
Is =
b.
Fp = PT ⇒ PT = FpST = (0.5)(10,000 VA) = 5000 W
ST
a.
5000 VA
I = ST =
= 41.67 A
E
120 V
Fp = 0.8 ⇒ 36.87° (lagging)
E = 120 V ∠0°, I = 41.67 A ∠−36.87°
120 V ∠0°
E
Z= =
= 2.88 Ω ∠36.87° = 2.30 Ω + j1.73 Ω = R + jXL
I 41.67 A ∠ − 36.87°
b.
P = S cos θ = (5000 VA)(0.8) = 4000 W
a.
PT = 0 + 300 W = 300 W
QT = 600 VAR(C) + 200(L) = 400 VAR(C)
2
2
PT + QT = 500 VA
P
300 W
Fp = T =
= 0.6 (leading)
ST 500 VA
ST =
ST 500 VA
=
= 16.67 A
30 V
E
Fp = 0.6 ⇒ 53.13°
Is = 16.67 A ∠53.13°
b.
Is =
c.
−
d.
Load: 600 VAR(C), 0 W
R = 0, L = 0, QC = I2XC ⇒ XC =
QC
2
=
600 VAR
= 2.159 Ω
(16.67 A) 2
I
Load: 200 VAR(L), 300 W
C = 0, R = P/I2 = 300 W/(16.67 A)2 = 1.079 Ω
Q
200 VAR
= 0.7197 Ω
XL = 2L =
(16.67 A) 2
I
ZT = !j2.159 Ω + 1.0796 Ω + j0.7197 Ω
= 1.08 Ω − j1.44 Ω
258
CHAPTER 19
13.
a.
PT = 0 + 300 W + 600 W = 900 W
QT = 500 VAR(C) + 0 + 500 VAR(L) = 0 VAR
ST = PT = 900 VA
Fp = PT = 1
ST
b.
Is =
c.
−
S T 900 VA
=
= 9 A, Is = 9 A ∠0°
E
100 V
d.
V
V
10
=
⇒ XC=
= 20 Ω
XC
QC 500
2
Z1:
QC =
Z3:
14.
a.
4
100 V ∠0°
= 5A ∠90°
Z1 20 Ω ∠ − 90°
I2 = Is − I1 = 9 A − j5 A = 10.296 A ∠−29.05°
P
300 W
300
= 2.83 Ω
=
R = 2=
2
106
(10.296
A)
I
XL,C = 0 Ω
P
600 W
R= 2=
= 5.66 Ω
I 2 (10.296 A) 2
Q
500
XL = 2 =
= 4.72 Ω, XC = 0 Ω
I 2 (10.296 A) 2
I1 =
Z2:
2
E
=
PT = 200 W + 30 W + 0 = 230 W
QT = 0 + 40 VAR(L) + 100 VAR(L) = 140 VAR(L)
ST =
2
2
PT + QT = 269.26 VA
230 W
Fp = PT =
= 0.854 (lagging) ⇒ 31.35°
S T 269.26 VA
b.
S T 269.26 VA
=
= 2.6926 A
100 V
E
Is = 2.69 A ∠−31.35°
Is =
CHAPTER 19
259
V
10
=
= 50 Ω
P 200
XL,XC = 0 Ω
100 V ∠0°
I1 =
= 2 A ∠0°
50 Ω ∠0°
I2 = Is − I1
= 2.6926 A ∠−31.35° − 2 A ∠0°
= 2.299 A − j1.40 A − 2.0 A
= 0.299 A − j1.40 A
= 1.432 A ∠−77.94°
c.
15.
a.
2
Z1:
4
R=
P
30 W
Q
40 VAR
=
= 14.63 Ω, XL = 2 =
= 19.50 Ω
2
2
I 2 (1.432 A)
I 2 (1.432 A) 2
XC = 0 Ω
Z2:
R=
Z3:
XL =
Q 100 VAR
=
= 48.76 Ω, R = 0 Ω, XC = 0 Ω
I 22 (1.432 A) 2
PT = 100 W + 1000 W = 1100 W
QT = 75 VAR(C) + 2291.26 VAR(C) = 2366.26 VAR(C)
ST =
2
2
PT + QT = 2609.44 VA
1100 W
Fp = PT =
= 0.422 (leading) ⇒ 65.04°
S T 2609.44 VA
b.
2609.44 VA
ST = EI ⇒ E = S T =
= 521.89 V
I
5A
E = 521.89 V ∠−65.07°
c.
S S 125 VA
= =
= 0.2395 A
V1 E 521.89 V
S S 2500 VA
I Z2 = = =
= 4.79 A
V2 E 521.89 V
I Z1 =
260
CHAPTER 19
Z1:
R=
100 W
P
=
= 1743.38 Ω
2
I Z1 (0.2395) 2
Q = I Z21 X C ⇒ X C =
Z2:
R=
XC =
16.
a.
b.
P
I Z21 X C
=
Q
I Z21 X C
75 VAR
Q
= 1307.53 Ω
=
2
I Z1 (0.2395 A) 2
1000 W
= 43.59 Ω
(4.790 A) 2
=
2291.26 VAR
= 99.88 Ω
(4.790 A) 2
0.7 ⇒ 45.573°
P = S cos θ = (10 kVA)(0.7) = 7 kW
Q = S sin θ = (10 kVA)(0.714) = 7.14 kVAR(L)
QC = 7.14 kVAR =
V2
XC
(208 V) 2
V2
=
= 6.059 Ω
Q C 7.14 kVAR
1
1
1
= 438 μF
XC =
=
⇒ C=
2π fX C (2π)(60 Hz)(6.059 Ω)
2π fC
XC =
c.
Uncompensated:
10,000 VA
Is = S T =
= 48.08 A
E
208 V
Compensated:
7,000 W
Is = S T = P T =
= 33.65 A
208 V
E
E
d.
CHAPTER 19
cos θ = 0.9
θ = cos−10.9 = 25.842°
x
tan θ =
7 kW
x = (7 kW)(tan 25.842°)
= (7 kW)(0.484)
= 3.39 kVAR
y = (7.14 − 3.39) kVAR
= 3.75 kVAR
261
QC = 3.75 kVAR =
XC =
C=
V2
XC
(208 V) 2
V2
=
= 11.537 Ω
QC 3.75 kVAR
1
1
= 230 μF
=
2π fX C (2π)(60 Hz)(11.537 Ω)
Uncompensated:
Is = 48.08 A
Compensated:
(7 kW) 2 + (3.39 kVAR) 2 = 7.778 kVA
ST =
Is =
17.
a.
S T 7.778 kVA
=
= 37.39 A
E
208 V
PT = 5 kW, QT = 6 kVAR(L)
ST =
2
2
PT + QT = 7.81 kVA
b.
5 kW
Fp = P T =
= 0.640 (lagging)
S T 7.81 kVA
c.
Is =
d.
XC =
S T 7,810 VA
=
= 65.08 A
E
120 V
2
(120 V) 2
1
, QC = I2XC = E =
XC
XC
2π fC
and
e.
(120 V) 2 14, 400
=
= 2.4 Ω
6000
QC
1
1
C=
= 1105 μF
=
2π fX C (2π )(60 Hz)(2.4 Ω)
XC =
ST = EIs = PT
5000 W
∴ Is = P T =
= 41.67 A
120 V
E
262
CHAPTER 19
18.
a.
Load 1:
Load 2:
P = 20,000 W, Q = 0 VAR
θ = cos−10.7 = 45.573°
x
10 kW
x = (10 kW)tan 45.573°
= (10 kW)(1.02)
= 10,202 VAR(L)
tan θ =
Load 3:
θ = cos−10.85 = 31.788°
x
5 kW
x = (5 kW)tan 31.788°
= (5 kW)(0.62)
= 3098.7 VAR(L)
tan θ =
PT = 20,000 W + 10,000 W + 5,000 W = 35 kW
QT = 0 + 10,202 VAR + 3098.7 VAR = 13,300.7 VAR(L)
ST =
b.
QC = QL = 13,300.7 VAR
2
(103 V) 2
= 75.184 Ω
XC = E =
Q C 13,300.7 VAR
C=
c.
2
2
PT + QT = 37,442 VA = 37.442 kVA
1
1
=
= 35.28 μF
2π fX C (2π)(60 Hz)(75.184 Ω)
Uncompensated:
37.442 kVA
S
Is = T =
= 37.44 A
1 kV
E
Compensated:
ST = PT = 35 kW
35 kW
S
= 35 A
Is = T =
1 kV
E
19.
a.
ZT = R1 + R2 + R3 + jXL − jXC
= 2 Ω + 3 Ω + 1 Ω + j3 Ω − j12 Ω = 6 Ω − j9 Ω = 10.82 Ω ∠−56.31°
50 V ∠0°
E
I=
= 4.62 A ∠56.31°
=
ZT 10.82 Ω ∠ − 56.31°
P = VI cos θ = (50 V)(4.62 A) cos 56.31° = 128.14 W
CHAPTER 19
263
20.
21.
a-b:
b-c:
a-c:
a-d:
c-d:
d-e:
f-e:
a.
ST = 660 VA = EIs
660 VA
= 5.5 A
Is =
120 V
θ = cos−10.6 = 53.13°
∴E = 120 V ∠0°, Is = 5.5 A ∠−53.13°
P = EI cos θ = (120 V)(5.5 A)(0.6) = 396 W
Wattmeter = 396 W, Ammeter = 5.5 A, Voltmeter = 120 V
b.
ZT =
a.
R=
XL =
L=
b.
R=
c.
R=
XL =
L=
22.
P = I2R = (4.62 A)2 2 Ω = 42.69 W
P = I2R = (4.62 A)2 3 Ω = 64.03 W
42.69 W + 64.03 W = 106.72 W
106.72 W
0W
0W
P = I2R = (4.62 A)2 1 Ω = 21.34 W
b.
a.
120 V ∠0°
E
=
= 21.82 Ω ∠53.13° = 13.09 Ω + j17.46 Ω = R + jXL
I
5.5 A ∠ − 53.13°
P
I
2
=
80 W
E 200 V
= 5 Ω, ZT = =
= 50 Ω
2
4A
I
(4 A )
ZT2 − R 2 = (50 Ω) 2 − (5 Ω) 2 = 49.75 Ω
XL
49.75 Ω
= 132.03 mH
=
2π f (2π)(60 Hz)
P
I
2
P
I
2
=
90 W
= 10 Ω
(3 A) 2
=
60 W
E 200 V
= 15 Ω, ZT = =
= 100 Ω
2
2A
I
(2 A)
ZT2 − R 2 = (100 Ω) 2 − (15 Ω) 2 = 98.87 Ω
XL
98.87 Ω
=
= 262.39 mH
2π f
376.8
XL = 2πfL = (6.28)(50 Hz)(0.08 H) = 25.12 Ω
R 2 + X L2 = (4 Ω) 2 + (25.12 Ω) 2 = 25.44 Ω
ZT =
60 V
= 2.358 A
Z T 25.44 Ω
P = I2R = (2.358 A)2 4 Ω = 22.24 W
I=
264
E
=
CHAPTER 19
b.
P
30 W
=
= 2.07 A
7Ω
R
E
60 V
= 28.99 Ω
ZT = =
I 2.07 A
I=
XL =
L=
c.
(28.99 Ω) 2 − (7 Ω) 2 = 28.13 Ω
XL
28.13 Ω
=
= 89.54 mH
2π f (2π)(50 Hz)
P = I2R = (1.7 A)2 10 Ω = 28.9 W
E 60 V
ZT = =
= 35.29 Ω
I 1.7 A
XL =
L=
CHAPTER 19
(35.29 Ω) 2 − (10 Ω) 2 = 33.84 Ω
XL
38.84 Ω
= 107.77 mH
=
2π f
314
265
Chapter 20
1.
a.
ωs =
fs =
b.
ωs =
fs =
c.
ωs =
fs =
2.
a.
d.
e.
3.
ωs 250 rad/s
=
= 39.79 Hz
2π
2π
1
= 3535.53 rad/s
(0.5 H)(0.16 μ F)
ωs 3535.53 rad/s
=
= 562.7 Hz
2π
2π
1
= 21,880 rad/s
(0.28 mH)(7.46 μ F)
ωs 21,880 rad/s
=
= 3482.31 Hz
2π
2π
XC = 30 Ω
b.
ZTs = 10 Ω
c.
VR = IR = (5 mA)(10 Ω) = 50 mV = E
VL = IXL = (5 mA)(30 Ω) = 150 mV
VC = IXC = (5 mA)(30 Ω) = 150 mV
VL = VC
30 Ω
Qs = X L =
= 3 (low Q)
R 10 Ω
a.
XL = 40 Ω
b.
I=
c.
266
1
1
= 250 rad/s
=
LC
1 H)(16 μ F)
f.
I=
E 50 mV
= 5 mA
=
10 Ω
ZTs
P = I2R = (5 mA)2 10 Ω = 0.25 mW
E 20 mV
=
= 10 mA
2Ω
ZTs
VR = IR = (10 mA)(2 Ω) = 20 mV = E
VL = IXL = (10 mA)(40 Ω) = 400 mV
VC = IXC = (10 mA)(40 Ω) = 400 mV
VL = VC = 20 VR
d.
40 Ω
Qs = X L =
= 20 (high Q)
R
2Ω
e.
XL = 2πfL, L =
XL
40 Ω
=
= 1.27 mH
2π f 2π (5 kHz)
1
1
1
=
,C=
= 795.77 nF
XC =
2π fX C 2π (5 kHz)(40 Ω)
2π fC
CHAPTER 20
f.
g.
4.
a.
b.
c.
BW =
f s 5 kHz
=
= 250 Hz
20
Qs
BW
0.25 kHz
= 5 kHz +
= 5.13 kHz
2
2
BW
0.25 kHz
= 5 kHz −
= 4.88 kHz
f1 = fs −
2
2
f2 = fs +
fs =
1
2π LC
⇒ L=
1
1
= 3.91 mH
=
2
(2π f s ) C (2π 1.8 kHz ) 2 2 μ F
XL = 2πfL = 2π(1.8 kHz)(3.91 mH) = 44.2 Ω
1
1
XC =
=
= 44.2 Ω
2π fC 2π (1.8 kHz)(2 μ F)
XL = XC
Erms = (0.707)(20 mV) = 14.14 mV
E
14.14 mV
= 3.01 mA
Irms = rms =
R
4.7 Ω
d.
P = I2R = (3.01 mA)2 4.7 Ω = 42.58 μW
e.
ST = PT = 42.58 μVA
g.
h.
44.2 Ω
Qs = X L =
= 9.4
4.7 Ω
R
f
1.8 kHz
BW = s =
= 191.49 Hz
9.4
Qs
f.
Fp = 1
2
1 ⎡ R 1 ⎛R⎞
4 ⎤
⎢ +
⎥
+
⎜ ⎟
2π ⎢ 2 L 2 ⎝ L ⎠ LC ⎥
⎣
⎦
2
⎤
1 ⎡ 4.7 Ω
1 ⎛ 4.7 Ω ⎞
4
⎢
⎥
=
+
+
⎜
⎟
2π ⎢ 2(3.91 mH) 2 ⎝ 3.91 mH ⎠ (3.91 mH)(2 μ F) ⎥
⎣
⎦
1
⎡601.02 + 11.324 × 103 ⎤
=
⎦
2π ⎣
= 1897.93 Hz
2
1 ⎡ R 1 ⎛R⎞
4 ⎤
⎢−
⎥
f1 =
+
+
⎜ ⎟
2π ⎢ 2 L 2 ⎝ L ⎠ LC ⎥
⎣
⎦
1
⎡ −601.02 + 11.324 × 103 ⎤
=
⎦
2π ⎣
= 1.71 kHz
1
1
PHPF = Pmax = (42.58 μW) = 21.29 μW
2
2
f2 =
CHAPTER 20
267
5.
6.
a.
BW = fs/Qs = 6000 Hz/15 = 400 Hz
b.
f2 = fs +
c.
Qs =
d.
PHPF =
a.
b.
7.
268
BW
= 6000 Hz + 200 Hz = 6200 Hz
2
BW
f1 = f s −
= 6000 Hz − 200 Hz = 5800 Hz
2
XL
⇒ XL = QsR = (15)(3 Ω) = 45 Ω = XC
R
1
1
1
Pmax = (I2R) = (0.5 A)2 3Ω = 375 mW
2
2
2
XL
200 Ω
=
= 3.185 mH
2π f 2π (104 Hz)
R
5Ω
=
≅ 250 Hz
BW =
2π L 2π (3.185 mH)
200 Ω
f
10,000 Hz
= 40, BW = s =
= 250 Hz
or Qs = X L = X C =
5Ω
40
R
R
Qs
L=
f2 = fs + BW/2 = 10,000 Hz + 250 Hz/2 = 10,125 Hz
f1 = fs − BW/2 = 10,000 Hz − 125 Hz = 9,875 Hz
c.
200 Ω
= 40
Qs = X L =
5Ω
R
d.
I=
e.
P = I2R = (6 A)2 5 Ω = 180 W
a.
BW =
b.
Qs = X L ⇒ XL = QsR = (10)(2 Ω) = 20 Ω
R
c.
L=
E ∠0° 30 V ∠0°
=
= 6 A ∠0°, VL = (I ∠0°)(XL ∠90°)
R∠0° 5 Ω ∠0°
= (6 A ∠0°)(200 Ω ∠90°)
= 1200 V ∠90°
VC = (I ∠0°)(XC ∠−90°) = 1200 V ∠−90°
fs
⇒ Qs = fs/BW = 2000 Hz/200 Hz = 10
Qs
XL
20 Ω
= 1.59 mH
=
2π f (6.28)(2 kHz)
1
1
C=
= 3.98 μF
=
2π fX C (6.28)(2 kHz)(20 Ω)
CHAPTER 20
d.
8.
a.
BW = 6000 Hz − 5400 Hz = 600 Hz
b.
BW = fs/Qs ⇒ fs = QsBW = (9.5)(600 Hz) = 5700 Hz
c.
Qs =
d.
9.
f2 = fs + BW/2 = 2000 Hz + 100 Hz = 2100 Hz
f1 = fs − BW/2 = 2000 Hz − 100 Hz = 1900 Hz
XL
⇒ XL = XC = QsR = (9.5)(2 Ω) = 19 Ω
R
XL
19 Ω
=
= 0.53 mH
2π f 2π (5700 Hz)
1
1
C=
= 1.47 μF
=
2π fX C 2π (5.7 kHz)(19 Ω)
L=
E
5V
E
= 10 Ω
⇒R=
=
500 mA
R
IM
BW = fs/Qs ⇒ Qs = fs/BW = 8400 Hz/120 Hz = 70
X
Qs = L ⇒ XL = QsR = (70)(10 Ω) = 700 Ω
R
XC = XL = 700 Ω
XL
700 Ω
=
= 13.26 mH
L=
2π f (2π)(8.4 kHz)
1
1
C=
= 27.07 nF
=
2π fX C (2π)(8.4 kHz)(0.7 k Ω)
IM =
f2 = fs + BW/2 = 8400 Hz + 120 Hz/2 = 8460 Hz
f1 = fs − BW/2 = 8400 Hz − 60 Hz = 8340 Hz
10.
Qs = X L ⇒ XL = QsR = 20(2 Ω) = 40 Ω = XC
R
f
BW = s ⇒ fs = QsBW = (20)(400 Hz) = 8 kHz
Qs
40 Ω
= 795.77 μH
L= XL =
2π f 2π (8 kHz)
1
1
=
= 497.36 nF
C=
2π fX C 2π (8 kHz)(40 Ω)
f2 = fs + BW/2 = 8000 Hz + 400 Hz/2 = 8200 Hz
f1 = fs − BW/2 = 8000 Hz − 200 Hz = 7800 Hz
11.
a.
b.
fs =
ωs 2π × 106 rad/s
= 1 MHz
=
2π
2π
f 2 − f1
= 0.16 ⇒ BW = f2 − f1 = 0.16 fs = 0.16(1 MHz) = 160 kHz
fs
CHAPTER 20
269
c.
d.
12.
a.
2
2
(120 V) 2
V
VR
= 720 Ω
⇒R= R=
P
R
20 W
R
R
720 Ω
BW =
⇒ L=
= 0.716 mH
=
2π L
2π BW (6.28)(160 kHz)
1
1
1
= 2 6
⇒C=
= 35.38 pF
fs =
2 2
4π f s L 4π (10 Hz ) 2(0.716 mH)
2π LC
P=
2π f s L 2π (106 Hz)(0.716 mH)
X
Q = X L = 80 ⇒ RΡ = L =
=
= 56.23 Ω
80
80
80
R
XL
R
X
2π fL 2π (1MHz)(100 μ H)
=
= 50.27 Ω
R = L=
Q
Q
12.5
f 2 − f1 1
=
= 0.2
fs
Qs
Q =
X
1
2π fL 2π (1 MHz)(100 μ H) 628.32 Ω
=
= 5= L =
=
R
R
R
R
0.2
628.32 Ω
R=
= 125.66
5
R = Rd + R
Qs =
125.66 Ω = Rd + 50.27 Ω
and Rd = 125.66 Ω − 50.27 Ω = 75.39 Ω
13.
1
= XL
2π fC
1
1
C=
= 253.3 pF
=
2π fX C 2π (1 MHz)(628.32 Ω)
c.
XC =
a.
fp =
2π LC
IL =
4V
4V
VL
= 40 mA
=
=
X L 2π f p L 100 Ω
IC =
4V
4V
VL
= 40 mA
=
=
X C 1/ 2π f pC 100 Ω
1
=
2
= 159.16 kHz
2π (0.1 mH)(10 nF)
b.
c.
270
CHAPTER 20
14.
d.
2 kΩ
2 kΩ
=
Qp = R s =
= 20
X Lp 2π f p L 100 Ω
a.
fs =
b.
Q =
c.
Since Q ≥ 10, fp ≅ fs = 41.09 kHz
d.
1
2π LC
=
1
= 41.09 kHz
2π (0.5 mH)(30 nF)
X L 2π fL 2π (41.09 kHz)(0.5 mH)
=
=
= 16.14 ≥ 10 (yes)
R
R
8Ω
XL = 2πfpL = 2π(41.09 kHz)(0.5 mH) = 129.1 Ω
2
2
=
= 129.1 Ω
XC =
2π f p C 2π (41.09 kHz)(30 nF)
XL = XC
15.
e.
ZTp = Q 2 R = (16.14)2 8 Ω = 2.084 kΩ
f.
VC = IZ T p = (10 mA)(2.084 kΩ) = 20.84 V
g.
Q ≥ 10, Qp = Q = 16.14
f p 41.09 kHz
=
= 2545.85 Hz
BW =
16.14
Qp
h.
IL = IC = Q IT = (16.14)(10 mA) = 161.4 mA
a.
fs =
b.
Q =
c.
2
C
(4 Ω) 2 2 μ F
= 11,253.95 Hz(0.825)
fp = fs 1 − R = 11,253.95 Hz 1 −
0.1 mH
L
= 9,280.24 Hz
1
2π LC
=
1
= 11,253.95 Hz
2π (0.1 mH)(2 μ F)
X L 2π f s L 2π (11, 253.95 Hz)(0.1 mH)
=
=
= 1.77 (low Q )
4Ω
R
R
1 ⎡ 2C ⎤
1 ⎡ (4 Ω) 2 2 μ F ⎤
fm = fs 1 − ⎢ R ⎥ = 11,253.95 Hz 1 − ⎢
⎥
4⎣ L ⎦
4 ⎣ 0.1 mH ⎦
= 11,253.95 Hz(0.996) = 10,794.41 Hz
CHAPTER 20
271
d.
XL = 2πfpL = 2π(9,280.24 Hz)(0.1 mH) = 5.83 Ω
1
1
XC =
=
= 8.57 Ω
2π f pC 2π (9,280.24 Hz)(2 μ F)
XL ≠ XC, XC > XL
e.
16.
2
⎛ 2+ 2 ⎞
+ 2 (4 Ω) 2 + (5.83 Ω) 2
= 12.5 Ω
ZTp = Rs || Rp = Rs || ⎜ R X L ⎟ = R X L =
4Ω
R
⎝ R
⎠
f.
VC = IZ T p = (2 mA)(12.5 Ω) = 25 mV
g.
Since Rs = ∞ Ω
h.
IC =
V C 25 mV
= 2.92 mA
=
X C 8.57 Ω
IL =
VC
VL
25 mV
25 mV
= 3.54 mA
=
=
=
Z R − L R + jX L 4 Ω + j 5.83 Ω 7.07 Ω
a.
Q =
∴
b.
c.
2π f p L 2π (9,280.24 Hz)(0.1 mH)
=
Qp = Q = X L =
= 1.46
4Ω
R
R
f
9,280.24 Hz
BW = p =
= 6.36 kHz
1.46
Qp
X L 100 Ω
= 5 ≤ 10
=
20 Ω
RL
1
R 2 + X 2 (20 Ω) 2 + (100 Ω) 2
XL
=
=
⇒
X
=
= 104 Ω
C
XL
100 Ω
R 2 + X L2 X C
2
+ 2
10, 400 Ω
ZT = Rs || Rp = Rs || R X L = 1000 Ω
= 342.11 Ω
R
20
E = IZTp = (5 mA ∠0°)(342.11 Ω ∠0°) = 1.711 V ∠0°
IC =
E
1.711 V ∠0°
=
= 16.45 mA ∠90°
X C ∠ − 90° 104 Ω ∠ − 90°
ZL = 20 Ω + j100 Ω = 101.98 Ω ∠78.69°
E
1.711 V ∠0°
=
= 16.78 mA ∠−78.69°
IL =
Ω ∠78.69°
101.98
ZL
d.
272
100 Ω
L= XL =
= 795.77 μH
2π f 2π (20 kHz)
1
1
= 76.52 nF
C=
=
2π fX C 2π (20 kHz)(104 Ω)
CHAPTER 20
17.
R 342.11 Ω
=
= 3.29
XC
104 Ω
BW = fp/Qp = 20,000 Hz/3.29 = 6079.03 Hz
e.
Qp =
a.
Q =
b.
ZTp = Rs || Q 2 R = 450 Ω || (15)2 2 Ω = 450 Ω || 450 Ω = 225 Ω
c.
E = I ZTp = (80 mA ∠0°)(225 Ω ∠0°) = 18 V ∠0°
E
18 V ∠0°
=
= 0.6 A ∠90°
X C ∠ − 90° 30 Ω ∠ − 90°
IC =
E
IL =
d.
18 V ∠0°
18 V ∠0°
≅ 0.6 A ∠−86.19°
=
2 Ω + j 30 Ω 30.07 Ω ∠86.19°
=
Z R-L
XL = 2πfpL, L =
XC =
18.
X L 30 Ω
=
= 15 (use approximate approach): XC = XL = 30 Ω
2Ω
R
e.
Qp =
a.
fs =
30 Ω
XL =
= 0.239 mH
2π f p 2π (20 ×103 Hz)
1
1
1
= 265.26 nF
,C=
=
2π f p C
2π f p X C 2π (20 ×103 Hz)(30 Ω)
Z Tp
XL
1
=
f
225 Ω
20,000 Hz
= 7.5, BW = p =
= 2.67 kHz
30 Ω
7.5
Qp
2π LC
=
1
= 102.73 kHz
2π (80 μ H)(0.03 μ F)
2
(1.5 Ω) 2 0.03 μ F
C
= 102.73 kHz(.99958)
fp = fs 1 − R = 102.73 kHz 1 −
L
80 μ H
= 102.69 kHz
1 ⎡ 2C ⎤
fm = fs 1 − ⎢ R ⎥ = 102.73 kHz(0.99989) = 102.72 kHz
4⎣ L ⎦
Since fs ≅ fp ≅ fm ⇒ high Qp
b.
XL = 2πfpL = 2π(102.69 kHz)(80 μH) = 51.62 Ω
1
1
=
= 51.66 Ω
XC =
2π f p C 2π (102.69 kHz)(0.03 μ F)
XL ≅ XC
CHAPTER 20
273
c.
ZTp = Rs || Q 2 R
51.62 Ω
= 34.41
Q = XL=
R
1.5 Ω
ZTp = 10 k Ω (34.41) 21.5 Ω = 10 k Ω 1.776 k Ω = 1.51 kΩ
R s Q R Z T p 1.51 k Ω
=
=
= 29.25
XL
X L 51.62 Ω
f p 102.69 kHz
=
= 3.51 kHz
BW =
Qp
29.25
2
d.
Qp =
e.
IT =
R s I s = 10 k Ω(10 mA) = 8.49 mA
2
R s + Q R 10 k Ω + 1.78 k Ω
IC = IL ≅ Q IT = (34.41)(8.49 mA) = 292.14 mA
19.
f.
VC = IZTp = (10 mA)(1.51 kΩ) = 15.1 V
a.
fs =
1
2π LC
f p = fs 1 −
=
1
= 7.12 kHz
2π (0.5 mH)(1 μ F)
R 2C
(8 Ω) 2 (1 μ F)
= 7.12 kHz 1 −
= 7.12 kHz(0.9338) = 6.65 kHz
0.5 mH
L
1 ⎡ R 2C ⎤
1 ⎡ (8 Ω) 2 (1 μ F) ⎤
f m = fs 1 − ⎢
⎥ = 7.12 kHz 1 − ⎢
⎥ = 7.12 kHz (0.9839)
4 ⎣ 0.5 mH ⎦
4⎣ L ⎦
= 7.01 kHz
Low Qp
b.
XL = 2πfpL = 2π(6.647 kHz)(0.5 mH) = 20.88 Ω
1
1
= 23.94 Ω
XC =
=
2π fC 2π (6.647 kHz)(1 μ F)
XC > XL (low Q)
c.
2
(8 Ω) 2 + (20.88 Ω) 2
+ 2
ZTp = Rs || Rp = Rs || R X L = 500 Ω ||
= 500 Ω || 62.5 Ω
8Ω
R
= 55.56 Ω
d.
Qp =
BW =
274
ZT p
=
55.56 Ω
= 2.32
23.94 Ω
=
6.647 kHz
= 2.87 kHz
2.32
X Lp
fp
Qp
CHAPTER 20
e.
One method: VC = IZTp = (40 mA)(55.56 Ω) = 2.22 V
IC =
IL =
20.
VC
2.22 V
=
= 92.73 mA
X C 23.94 Ω
VC
2.22 V
2.22 V
=
= 99.28 mA
=
8
+
20.88
22.36
Ω
+
jX
j
R
L
f.
VC = 2.22 V
a.
R 2 + X L2
ZTp =
= 50 kΩ
R
(50 Ω)2 + X L2 = (50 kΩ)(50 Ω)
XL =
21.
250 × 104 − 2.5 × 103 = 1580.3 Ω
X L 1580.3
=
= 31.61 ≥ 10
R
50
b.
Q=
c.
XL = 2πfpL ⇒ fp =
d.
XC =
a.
Q = 20 > 10 ∴ fp = fs =
b.
Q =
∴ XC = XL = 1580.3 Ω
XL
1580.3 Ω
=
= 15.72 kHz
2π L 2π(16 mH)
1
1
1
⇒ C=
= 6.4 nF
=
2π f p C
2π f s X C 2π (15.72 kHz)(1580.3 Ω)
1
2π LC
=
1
= 3558.81 Hz
2π (200 mH)(10 nF)
X L 2π fL
2π fL 2π (3558.81 Hz)(0.2 H)
=
⇒ R =
=
= 223.61 Ω
20
R
R
Q
ZTp = Rs || R p = Rs || Q 2 R = 40 kΩ || (20)2 223.61 Ω
ZTp = 27.64 kΩ
VC = IZTp = (5 mA)(27.64 kΩ) = 138.2 V
c.
P = I2R = (5 mA)227.64 kΩ = 691 mW
d.
Qp =
BW =
CHAPTER 20
27.64 k Ω
R Rs R p
=
=
= 6.18
2π (3558.81 Hz)(0.2 H)
XL
XL
f
p
Qp
=
3558.81 Hz
= 575.86 Hz
6.18
275
22.
a.
Ratio of XC to R suggests high Q system.
∴ XL = 400 Ω = XC
b.
Q =
c.
Qp =
BW =
d.
e.
23.
a.
X L 400 Ω
= 50
=
8Ω
R
R
XL
fp
Qp
=
Rs R p
=
Rs Q 2R
XL
XL
=
20 k Ω (50) 2 8 Ω 10 k Ω
= 25
=
400 Ω
400 Ω
⇒ fp = QpBW = (25)(1000 Hz) = 25 kHz
V C max = IZTp = (0.1 mA)(10 kΩ) = 1 V
1 kHz
= 25.5 kHz
2
1 kHz
= 24.5 kHz
f1 = fp − BW/2 = 25 kHz −
2
f2 = fp + BW/2 = 25 kHz +
XC =
R 2 + X L2
⇒ X L2 − XLXC + R 2 = 0
XL
X L2 − 100 XL + 144 = 0
XL =
−( − 100) ± (100) 2 − 4(1)(144)
2
104 − 576
= 50 Ω ± 48.54 Ω
2
XL = 98.54 Ω or 1.46 Ω
= 50 Ω ±
b.
c.
Q =
Qp =
X L 98.54 Ω
= 8.21
=
12 Ω
R
Rs R p
X Lp
40 kΩ
=
R 2 + X L2
(12 Ω) 2 + (98.54 Ω) 2
40 kΩ
R
12 Ω
=
XC
100 Ω
40 kΩ 821.18 Ω 804.66 Ω
=
= 8.05
100 Ω
100 Ω
BW = fp/Qp ⇒ fp = QpBW = (8.05)(1 kHz) = 8.05 kHz
=
d.
e.
276
VCmax = IZTp = (6 mA)(804.66 Ω) = 4.83 V
1 kHz
= 8.55 kHz
2
1 kHz
= 7.55 kHz
f1 = fp − BW/2 = 8.05 kHz −
2
f2 = fp + BW/2 = 8.05 kHz +
CHAPTER 20
24.
a.
fs =
1
2π LC
=
1
= 41.09 kHz
2π (0.5 mH)(30 nF)
R 2C
(6 Ω) 2 30 nF
= 41.09 kHz(0.9978) = 41 kHz
= 41.09 kHz 1 −
0.5 mH
L
fp = fs 1 −
1 ⎡ R 2C ⎤
1 ⎡ (6 Ω) 2(30 nF) ⎤
−
=
41.09
kHz
fm = fs 1 − ⎢
1
⎢
⎥ = 41.09 kHz(0.0995)
⎥
4 ⎣ 0.5 mH ⎦
4⎣ L ⎦
= 41.07 kHz
High Qp
b.
80 V ∠0°
= 4 mA ∠0°, Rs = 20 kΩ
20 k Ω ∠0°
X
2π fL 2π (41 kHz)(0.5 mH)
Q = L=
= 21.47 (high Q coil)
=
6Ω
R
R
I=
2
2
R +XL
(6 Ω) 2 + (128.81 Ω) 2
20 k Ω
R Rp
R
6Ω
Qp = s
=
=
2
2
(6 Ω) 2 + (128.81 Ω) 2
R +XL
X Lp
128.81 Ω
XL
20 k Ω 2.771 k Ω 2.434 k Ω
=
=
= 18.86 (high Qp)
129.09 Ω
129.09 Ω
Rs
c.
ZTp = Rs || Rp = 20 kΩ || 2.771 kΩ = 2.43 kΩ
d.
VC = IZTp = (4 mA)(2.43 kΩ) = 9.74 V
e.
BW =
f.
XC =
fp
Qp
=
1
1
=
= 129.39 Ω
2π fC 2π (41 kHz)(30 nF)
V
9.736 V
= 75.25 mA
IC = C =
X C 129.39 Ω
IL =
25.
41 kHz
= 2.17 kHz
18.86
9.736 V
9.736 V
VC
= 75.50 mA
=
=
R + jX L 6 Ω + j128.81 Ω 128.95 Ω
2π f p L
2π f p L 2π (20 kHz)(2 mH)
=
= 3.14 Ω
Q = XL=
⇒ R =
Q
80
R
R
BW = fp/Qp ⇒ Qp = fp/BW = 20 kHz/1.8 kHz = 11.11
1
1
1
= 2
⇒C=
= 31.66 nF
High Q∴ fp ≅ fs =
2 2
4π f p L 4π (20 kHz) 2 2 mH
2π LC
Qp =
R
XC
CHAPTER 20
⇒ R = QpXC =
Qp
2π f pC
=
11.11
= 2.79 kΩ
2π (20 kHz)(31.66 nF)
277
Rp = Q 2 R = (80)2 3.14 Ω = 20.1 kΩ
Rp R
(20.1 kΩ)(2.793 kΩ)
Rs R p
⇒ Rs =
= 3.24 kΩ
=
R = Rs || Rp =
R p − R 20.1 kΩ − 2.793 kΩ
Rs + R p
26.
1.8 V
= 9 kΩ
I
0.2 mA
R p 9 kΩ
R Rs R p R p
=
Qp =
⇒ XL =
= 300 Ω = XC
=
=
XL
XL
XL
30
Qp
VCmax = IZTp ⇒ ZTp =
BW =
fp
Qp
VCmax
⇒ fp = QpBW = (30)(500 Hz) = 15 kHz
L=
XL
300 Ω
= 3.18 mH
=
2π f 2π (15 kHz)
C=
1
1
= 35.37 nF
=
2π fX C 2π (15 kHz)(300 Ω)
Qp = Q (Rs= ∞ Ω) =
27.
a.
fs =
1
2π LC
Q =
b.
=
=
XL
X
300 Ω
= 10 Ω
⇒ R = L=
R
30
Qp
1
2π (200 μ H)(2 nF)
= 251.65 kHz
X L 2π (251.65 kHz)(200 μ H)
=
= 15.81 ≥ 10
20 Ω
R
∴ fp = fs = 251.65 kHz
ZTp = Rs || Q 2 R = 40 kΩ || (15.81)2 20 Ω = 4.44 kΩ
c.
Rs Q 2 R 4.444 kΩ
=
= 14.05
Qp =
XL
316.23 Ω
d.
BW =
e.
20 μH, 20 nF
fs the same since product LC the same
fs = 251.65 kHz
X
2π (251.65 kHz)(20 μ H)
= 1.581
Q = L=
R
20 Ω
Low Q :
fp
Qp
=
251.65 kHz
= 17.91 kHz
14.05
fp = fs 1 −
R 2C
(20 Ω) 2 (20 nF)
= (251.65 kHz) 1 −
20 μ H
L
= (251.65 kHz)(0.775) = 194.93 kHz
278
CHAPTER 20
XL = 2πfpL = 2π(194.93 kHz)(20 μH) = 24.496 Ω
R 2 + X L2 (20 Ω) 2 + (24.496 Ω) 2
Rp =
=
= 50 Ω
R
20 Ω
ZTp = Rs || Rp = 40 kΩ || 50 Ω = 49.94 Ω
R
49.94 Ω
=
= 2.04
X L 24.496 Ω
f
194.93 kHz
= 95.55 kHz
BW = p =
2.04
Qp
Qp =
f.
0.4 mH, 1 nF
fs = 251.65 kHz since LC product the same
2π (251.65 kHz)(0.4 mH)
= 31.62 ≥ 10
Q = XL=
20 Ω
R
∴ fp = fs = 251.65 kHz
ZTp = Rs || Q 2 R = 40 kΩ || (31.62)2 20 Ω = 40 kΩ || (≅ 20 kΩ) ≅ 13.33 kΩ
Rs Q 2 R 13.33 k Ω
=
= 21.08
XL
632.47 Ω
f
251.65 kHz
BW = p =
= 11.94 kHz
Qp
21.08
Qp =
g.
h.
L 200 μ H
=
= 100 × 103
C
2 nF
L 20 μ H
= 1 × 103
part (e) =
C 20 nF
L 0.4 mH
= 400 × 103
part (f) =
C
1 nF
Network
L
ratio increased BW decreased.
C
Also, Vp = IZTp and for a fixed I, ZTp and therefore Vp will increase with increase in the
Yes, as
L/C ratio.
CHAPTER 20
279
Chapter 21
1.
3
″ = 0.1875″, d2 = 1″
16
Value = 103 × 100.1875"/1"
= 103 × 1.54
= 1.54 kHz
3
right: d1 = ″ = 0.75″, d2 = 1″
4
Value = 103 × 100.75"/1"
= 103 × 5.623
= 5.62 kHz
a.
left:
b.
bottom:
top:
d1 =
5
15
″ = 0.3125″, d2 =
″ = 0.9375″
16
16
Value = 10−1 × 100.3125"/ 0.9375" = 10−1 × 100.333
= 10−1 × 2.153
= 0.22 V
11
″ = 0.6875″, d2 = 0.9375″
d1 =
16
Value = 10−1 × 100.6875"/ 0.9375" = 10−1 × 100.720
= 10−1 × 5.248
= 0.52 V
d1 =
a.
5
b.
−4
c.
8
d.
e.
1.30
f.
3.94
g.
4.75
h.
a.
1000
b.
1012
c.
1.59
d.
1.1
e.
1010
f.
1513.56
g.
10.02
h.
1,258,925.41
4.
a.
11.51
b.
−9.21
c.
5.
log10 48 = 1.68
log10 8 + log10 6 = 0.903 + 0.778 = 1.68
2.
3.
6.
7.
8.
280
2.996
d.
−6
−0.498
9.07
log10 0.2 = −0.699
log10 18 − log10 90 = 1.255 − 1.954 = −0.699
log10 0.5 = −0.30
−log10 2 = −(0.301) = −0.30
log10 27 = 1.43
3 log10 3 = 3(0.4771) = 1.43
CHAPTER 21
9.
P2
280 mW
= log10
= log10 70 = 1.85
P1
4 mW
a.
bels = log10
b.
dB = 10 log10
P2
= 10(log10 70) = 10(1.845) = 18.45
P1
10.
dB = 10 log10 P 2
P1
100 W
6 dB = 10 log10
P1
0.6 = log10 x
100 W
x = 3.981 =
P1
100 W
= 25.12 W
P1 =
3.981
11.
dB = 10 log10
12.
dBm = 10 log10
dBm = 10 log10
P2
40 W
= 10 log10 20 = 13.01
= 10 log10
2W
P1
P
1 mW
120 mW
= 10 log10 120 = 20.79
1 mW
13.
dBv = 20 log10
V2
8.4 V
= 20 log10
= 20 log 10 84 = 38.49
V1
0.1 V
14.
dBυ = 20 log10
V2
V1
22 = 20 log10
Vo
20 mV
1.1 = log10 x
Vo
20 mV
Vo = 251.79 mV
x = 12.589 =
15.
P
0.0002 μ bar
0.001 μ bar
= 13.98
dBs = 20 log10
0.0002 μ bar
0.016 μ bar
dBs = 20 log10
= 38.06
0.0002 μ bar
Increase = 24.08 dBs
dBs = 20 log10
CHAPTER 21
281
16.
60 dBs ⇒ 90 dBs
quiet
loud
60 dBs = 20 log10
P1
= 20 log10x
0.002 μ bar
3 = log10x
x = 1000
90 dBs = 20 log10
P2
= 20 log10y
0.002 μ bar
4.5 = log10y
y = 31.623 × 103
P1
103
x 0.002 μ bar
=
= P1 =
P2
y
P 2 31.623 × 103
0.002 μ bar
and P2 = 31.62 P1
18.
a.
b.
19.
a.
V2
0.775 V
V2
0.4 = log10
0.775 V
8 dB = 20 log10
V2
= 2.512
0.775 V
V2 = (2.512)(0.775 V) = 1.947 V
2
(1.947 V) 2
V
= 6.32 mW
P=
=
600 Ω
R
V2
−5 dB = 20 log10
0.775 V
V2
−0.25 = log10
0.775 V
V2
= 0.562
0.775 V
V2 = (0.562)(0.775 V) = 0.436 V
2
(0.436 V) 2
V
= 0.32 mW
P=
=
600 Ω
R
Aυ =
Vo
=
Vi
XC
R +
2
X C2
∠−90° + tan−1 XC/R =
1
⎛ R ⎞
⎜
⎟ +1
⎝ XC ⎠
2
∠−tan−1 R/XC
1
1
=
= 3617.16 Hz
2π RC 2π (2.2 kΩ)(0.02 μ F)
V
f = fc:
Aυ = o = 0.707
Vi
f = 0.1fc:
At fc, XC = R = 2.2 kΩ
fc =
282
CHAPTER 21
XC =
Aυ =
f = 0.5fc =
1
fc :
2
f = 2fc:
XC =
Aυ =
b.
θ = −tan−1 R/XC
f = fc:
a.
⎛ R ⎞
⎜
⎟ +1
⎝ XC ⎠
2
=
1
⎛ 2.2 k Ω ⎞
⎜
⎟ +1
⎝ 22 k Ω ⎠
2
=
1
(.1) 2 + 1
= 0.995
1
=
2π fC
1
⎛ 2.2 k Ω ⎞
⎜
⎟ +1
⎝ 1.1 k Ω ⎠
2
=
1
(2) 2 + 1
= 0.447
1
1 ⎡ 1 ⎤ 1
= ⎢
⎥ = [ 2.2 k Ω ] = 0.22 kΩ
2π (10 f c )C 10 ⎣ 2π f cC ⎦ 10
1
⎛ 2.2 k Ω ⎞
⎜
⎟ +1
⎝ 0.22 k Ω ⎠
2
=
1
(10) 2 + 1
θ = −tan−1 = −45°
f = 0.1fc:
θ = −tan−1 2.2 kΩ/22 kΩ = −tan−1
f = 0.5fc:
θ = −tan−1 2.2 kΩ/4.4 kΩ = −tan−1
f = 2fc:
f = 10fc:
20.
1
⎡ 1 ⎤
1
= 2⎢
⎥ = 2 [ 2.2 k Ω] = 4.4 kΩ
2π f cC ⎦
⎛ fc ⎞
⎣
2π ⎜ ⎟ C
⎝ 2⎠
1
1
= 0.894
Aυ =
=
2
2
(0.5
+
1
)
⎛ 2.2 k Ω ⎞
⎜
⎟ +1
⎝ 4.4 k Ω ⎠
1
1⎡ 1 ⎤ 1
XC =
= ⎢
⎥ = [ 2.2 k Ω ] = 1.1 kΩ
2π (2f c )C 2 ⎣ 2π f cC ⎦ 2
XC =
Aυ =
f = 10fc:
1
1
1 ⎡ 1 ⎤
=
=
⎢
⎥ = 10 [ 2.2 k Ω ] = 22 kΩ
2π fC 2π 0.1 f c C 0.1 ⎣ 2π f c C ⎦
= 0.0995
1
= −5.71°
10
1
= −26.57°
2
θ = −tan−1 2.2 kΩ/1.1 kΩ = −tan−1 2 = −63.43°
θ = −tan−1 2.2 kΩ/0.22 kΩ = −tan−1 10 = −84.29°
1
1
= 15.915 kHz
=
2π RC 2π ( 1 k Ω)(0.01 μ F)
f = 2fc = 31.83 kHz
1
1
XC =
= 500 Ω
=
2π fC 2π (31.83 kHz)(0.01 μ F)
500 Ω
V
XC
=
Aυ = o =
= 0.4472
2
2
Vi
(1 k Ω) 2 + (0.5 k Ω) 2
R + XC
fc =
Vo = 0.4472Vi = 0.4472(10 mV) = 4.47 mV
CHAPTER 21
283
b.
1
1
f c = (15,915 kHz) = 1.5915 kHz
10
10
1
1
XC =
= 10 kΩ
=
2π fC 2π (1.5915 kHz)(0.01 μ F)
10 k Ω
V
XC
=
Aυ = o =
= 0.995
2
2
Vi
(1 k Ω) 2 + (10 k Ω) 2
R + XC
f=
Vo = 0.995Vi = 0.995(10 mV) = 9.95 mV
c.
21.
Yes, at f = fc, Vo = 7.07 mV
1
at f =
f c , Vo = 9.95 mV (much higher)
10
at f = 2fc, Vo = 4.47 mV (much lower)
1
1
=
2π RC 2π (1.2 k Ω)C
1
1
=
= 0.265 μF
C=
2π Rf c 2π (1.2 k Ω)(500 Hz)
fc = 500 Hz =
Aυ =
Vo
1
=
2
Vi
⎛ R ⎞
⎜
⎟ +1
⎝ XC ⎠
At f = 250 Hz, XC = 2402.33 Ω and Aυ = 0.895
At f = 1000 Hz, XC = 600.58 Ω and Aυ = 0.4475
θ = −tan−1R/XC
1
At f = 250 Hz = fc, θ = −26.54°
2
At f = 1 kHz = 2fc, θ = −63.41°
284
CHAPTER 21
22.
a.
b.
fc =
1
1
= 67.73 kHz
=
2π RC 2π (4.7 k Ω)(500 pF)
f = 0.1 fc = 0.1(67.726 kHz) ≅ 6.773 kHz
1
1
= 46.997 kΩ
=
XC =
2π fC 2π (6.773 kHz)(500 pF)
V
46.997 k Ω
XC
=
Aυ = o =
= 0.995 ≅ 1
2
2
Vi
R +X
(4.7 k Ω) 2 + (46.997 k Ω) 2
C
c.
f = 10fc = 677.26 kHz
1
1
XC =
=
≅ 470 Ω
2π fC 2π (677.26 kHz)(500 pF)
V
470 Ω
XC
=
Aυ = o =
= 0.0995 ≅ 0.1
2
2
Vi
R + XC
(4.7 k Ω) 2 + (470 Ω) 2
d.
Aυ =
Vo
= 0.01 =
Vi
XC
2
R + X C2
X C = 100 X
2
2
R + XC =
C
0.01
R2 + X C2 = 104 X C2
R2 = 104 X C2 − X C2 = 9,999 X C2
4.7 k Ω
R
XC =
≅ 47 Ω
=
9,999 99.995
1
1
1
= 6.77 MHz
⇒f=
XC =
=
2π fC
2π X CC 2π (47 Ω)(500 pF)
23.
a.
Aυ =
Vo
=
Vi
fc =
f = fc :
f = 2 fc :
R
R +
2
1
⎛X ⎞
1+ ⎜ C ⎟
⎝ R ⎠
2
∠tan−1 XC/R
1
1
= 3.62 kHz
=
2π RC 2π (2.2 kΩ)(20 nF)
V
Aυ = o = 0.707
Vi
At fc, XC = R = 2.2 kΩ
1
1
1⎡ 1 ⎤ 1
=
= ⎢
XC =
⎥ = [ 2.2 k Ω ] = 1.1 kΩ
2π fC 2π (2 f c )C 2 ⎣ 2π f cC ⎦ 2
Aυ =
CHAPTER 21
X C2
∠tan−1 XC/R =
1
⎛ 1.1 k Ω ⎞
1+ ⎜
⎟
⎝ 2.2 k Ω ⎠
2
= 0.894
285
f=
1
fc :
2
f = 10fc:
⎡ 1 ⎤
1
= 2⎢
⎥ = 2 [ 2.2 k Ω] = 4.4 kΩ
f
2π f cC ⎦
⎛ c⎞
⎣
2π ⎜ ⎟ C
⎝ 2⎠
1
Aυ =
= 0.447
2
⎛ 4.4 k Ω ⎞
1+ ⎜
⎟
⎝ 2.2 k Ω ⎠
1
1 ⎡ 1 ⎤ 2.2 k Ω
= 0.22 kΩ
= ⎢
XC =
⎥=
2π (10 f c )C 10 ⎣ 2π f cC ⎦
10
XC =
1
⎛ 0.22 k Ω ⎞
1+ ⎜
⎟
⎝ 2.2 k Ω ⎠
⎡ 1 ⎤
1
= 10 ⎢
XC =
⎥ = 10 [ 2.2 k Ω ] = 22 kΩ
⎛f ⎞
⎣ 2π f cC ⎦
2π ⎜ c ⎟ C
⎝ 10 ⎠
1
= 0.0995
Aυ =
2
22
k
Ω
⎛
⎞
1+ ⎜
⎟
2.2
k
Ω
⎝
⎠
Aυ =
f=
b.
1
fc :
10
f = fc ,
f = 2 fc ,
f=
1
fc ,
2
f = 10fc,
f=
24.
1
fc ,
10
a.
f = fc: Aυ =
b.
fc =
2
= 0.995
θ = 45°
θ = tan−1 (XC/R) = tan−1 1.1 kΩ/2.2 kΩ = tan−1
θ = tan−1
θ = tan−1
θ = tan−1
4.4 k Ω
= tan−1 2 = 63.43°
2.2 k Ω
0.22 k Ω
= 5.71°
2.2 k Ω
22 k Ω
= 84.29°
2.2 k Ω
1
= 26.57°
2
Vo
= 0.707
Vi
1
1
= 15.915 kHz
=
2π RC 2π (10 k Ω)(1000 pF)
f = 4fc = 4(15.915 kHz) = 63.66 kHz
1
1
=
= 2.5 kΩ
XC =
2π fC 2π (63.66 kHz)(1000 pF)
V
R
10 k Ω
= 0.970 (significant rise)
Aυ = o =
=
2
2
Vi
R + XC
(10 k Ω) 2 + (2.5 k Ω) 2
286
CHAPTER 21
c.
d.
25.
Aυ =
fc =
f = 100fc = 100(15.915 kHz) = 1591.5 kHz ≅ 1.592 MHz
1
1
=
= 99.972 Ω
XC =
2π fC 2π (1.592 MHz)(1000 pF)
R
10 k Ω
Aυ =
= 0.99995 ≅ 1
=
2
2
(10 k Ω) 2 + (99.972 Ω) 2
R +XC
At f = fc, Vo= 0.707Vi = 0.707(10 mV) = 7.07 mV
2
(7.07 mV ) 2
V
≅ 5 nW
Po = o =
R
10 k Ω
Vo
=
Vi
1
⎛X ⎞
1+ ⎜ C ⎟
⎝ R ⎠
2
∠tan−1 XC/R
1
1
1
⇒ R=
= 795.77 Ω
=
2π RC
2π f cC 2π ( 2 kHz)(0.1 μ F)
R = 795.77 Ω ⇒ 750 Ω + 47 Ω = 797 Ω
nominal values
1
∴ fc =
= 1996.93 Hz using nominal values
2π (797 Ω)(0.1 μ F)
At
f = 1 kHz, Aυ = 0.458
f = 4 kHz, Aυ ≅ 0.9
θ = tan−1 X C
R
f = 1 kHz, θ = 63.4°
f = 4 kHz, θ = 26.53°
CHAPTER 21
287
26.
a.
b.
c.
d.
fc =
1
1
= 79.58 kHz
=
2π RC 2π (100 k Ω)(20 pF)
f = 0.01fc = 0.01(79.577 kHz) = 0.7958 kHz ≅ 796 Hz
1
1
= 9.997 MΩ
=
XC =
2π fC 2π (796 Hz)(20 pF)
100 k Ω
R
V
=
Aυ = o =
= 0.01 ≅ 0
2
2
Vi
(100 k Ω) 2 + (9.997 M Ω) 2
R +XC
f = 100fc = 100(79.577 kHz) ≅ 7.96 MHz
1
1
= 999.72 Ω
XC =
=
2π fC 2π (7.96 MHz)(20 pF)
V
R
100 k Ω
Aυ = o =
= 0.99995 ≅ 1
=
2
2
Vi
(100 k Ω) 2 + (999.72 Ω) 2
R + XC
Aυ =
Vo
R
= 0.5 =
Vi
R 2 + X C2
R 2 + X C2 = 2R
R2 + X C2 = 4R2
X C2 = 4R2 − R2 = 3R2
3R 2 = 3R = 3 (100 kΩ) = 173.2 kΩ
1
1
1
⇒f=
=
XC =
2π fC
2π X C C 2π (173.2 k Ω)(20 pF)
f = 45.95 kHz
XC =
27.
a.
1
1
= 795.77 Hz
=
2π RC 2π (0.1 k Ω)(2 μ F)
1
1
f c2 =
=
= 1989.44 Hz
2π RC 2π (10 k Ω)(8 nF)
low-pass section:
f c1 =
high-pass section:
For the analysis to follow, it is assumed (R2 + jX C 2 ) || R1 ≅ R1 for all frequencies of
interest.
At f c1 = 795.77 Hz:
VR1 = 0.707 Vi
X C2 =
|Vo | =
1
= 25 kΩ
2π fC2
25 k Ω(VR1 )
(10 k Ω) 2 + (25 k Ω) 2
= 0.9285 VRi
Vo = (0.9285)(0.707 Vi) = 0.66 Vi
288
CHAPTER 21
At f c2 = 1989.44 Hz:
Vo = 0.707 VR1
1
= 40 Ω
2π fC1
R1Vi
100 Ω(Vi )
VR1 =
=
= 0.928 Vi
R12 + X C21
(100 Ω) 2 + (40 Ω) 2
X C1 =
|Vo | = (0.707)(0.928 Vi) = 0.66 Vi
(1989.44 Hz − 795.77 Hz)
= 1392.60 Hz
2
X C 1 = 57.14 Ω, X C 2 = 14.29 kΩ
At f = 795.77 Hz +
VR1 =
Vo =
100 Ω(Vi )
(100 Ω) 2 + (57.14 Ω) 2
( )
14.29 k Ω VR1
= 0.868 Vi
(10 k Ω) + (14.29 k Ω) 2
2
= 0.8193 VR1
Vo = 0.8193(0.868 Vi) = 0.71Vi
V
and Aυ = o = 0.71 (≅ maximum value)
Vi
After plotting the points it was determined that the gain should also be determined at
f = 500 Hz and 4 kHz:
f = 500 Hz:
X C1 = 159.15 Ω, X C2 = 39.8 kΩ,
VR1 = 0.532 Vi, Vo = 0.97 VR1
f = 4 kHz:
Vo = 0.52 Vi
X C1 = 19.89 Ω, X C2 = 4.97 kΩ,
VR1 = 0.981 Vi, Vo = 0.445 VR1
Vo = 0.44 Vi
b.
Using 0.707(.711) = 0.5026 ≅ 0.5 to define the bandwidth
BW = 3.4 kHz − 0.49 kHz = 2.91 kHz
and BW ≅ 2.9 kHz
⎛ 2.9 kHz ⎞
with fcenter = 490 Hz + ⎜
⎟ = 1940 Hz
2 ⎠
⎝
CHAPTER 21
289
28.
1
= 4 kHz
2π R1C1
Choose R1 = 1 kΩ
1
1
C1 =
= 39.8 nF ∴ Use 39 nF
=
2π f1R1 2π (4 kHz)(1 k Ω)
f1 =
1
= 80 kHz
2π R2C2
Choose R2 = 20 kΩ
1
1
=
C2 =
= 99.47 pF ∴ Use 100 pF
2π f 2 R2 2π (80 kHz)(20 k Ω)
f2 =
80 kHz − 4 kHz
= 42 kHz
2
At f = 42 kHz, X C1 = 97.16 Ω, X C2 = 37.89 kΩ
Center frequency = 4 kHz +
Assuming Z2
Z1
R1 (Vi )
= 0.995Vi
|VR1 |=
R12 + X C21
|Vo |=
X C2 (VR1 )
R22 + X C21
= 0.884Vi
Vo = 0.884 VR1 = 0.884(0.995Vi) = 0.88 Vi
as f = f1: VR1 = 0.707Vi, X C2 = 221.05 kΩ
and Vo = 0.996 VR1
so that Vo = 0.996 VR1 = 0.996(0.707Vi) = 0.704Vi
Although Aυ = 0.88 is less than the desired level of 1, f1 and f2 do define a band of frequencies
for which Aυ ≥ 0.7 and the power to the load is significant.
290
CHAPTER 21
29.
a.
b.
c.
fs =
1
2π LC
=
1
= 100.66 kHz
2π (5 mH)(500 pF)
XL
2π (100.658 kHz)(5 mH)
= 18.39
=
160 Ω + 12 Ω
R+ R
f
100.658 kHz
= 5,473.52 Hz
BW = s =
Qs
18.39
Qs =
R
160 Ω(1 V)
V
= 0.93 V and Aυ = o = 0.93
Vi=
Vi
172 Ω
R+R
BW
5, 473.52 Hz
Since Qs ≥ 10,
f1 = fs −
= 100.658 kHz −
= 97,921.24 Hz
2
2
BW
= 103,394.76 Hz
f 2 = fs +
2
At f = 95 kHz:
XL = 2πfL = 2π(95 × 103 Hz)(5 mH) = 2.98 kΩ
1
1
=
= 3.35 kΩ
XC =
2π fC 2π (95 × 103 Hz)(500 pF)
160 Ω(1 V ∠0°)
160 V ∠0°
=
Vo =
172 + j 2.98 k Ω − j 3.35 k Ω 172 − j 370
160 V ∠0°
= 0.39 V∠65.07°
=
480 ∠ − 65.07°
At f = fs: Vomax =
At f = 105 kHz:
d.
XL = 2πfL = 2π(105 kHz)(5 mH) = 3.3 kΩ
1
1
XC =
= 3.03 kΩ
=
2π fC 2π (105 kHz)(500 pF)
160 (1 V ∠0°)
160 V ∠0°
=
Vo =
172 + j 3.3 k Ω − j 3.03 k Ω 172 + j 270
160 V ∠0°
=
= 0.5∠−57.50°
320 ∠57.50°
f = fs: Vomax = 0.93 V
f = f1 = 97,921.24 Hz, Vo = 0.707(0.93 V) = 0.66 V
f = f2 = 103,394.76 Hz, Vo = 0.707(0.93 V) = 0.66 V
30.
a.
R 2C
≅ 159.15 kHz
L
2π LC
2π f p L 2π (159.15 kHz)(1 mH)
X
=
Q = L=
= 62.5
R
R
16 Ω
fp =
1
1−
ZTp = Q 2 R = (62.5)2 16 Ω = 62.5 kΩ
and Vo ≅ Vi at resonance.
CHAPTER 21
10
4 kΩ
291
However, R = 4 kΩ affects the shape of the resonance curve and BW = fp/ Q cannot be
applied.
V
For Aυ = o = 0.707, | X | = R for the following configuration
Vi
For frequencies near fp, XL
and X = XL || XC.
R and ZL = R + jXL ≅ XL
For frequencies near fp but less than fp
XC X L
X=
XC − X L
and for Aυ = 0.707
XC X L
=R
XC − X L
1
and XL = 2πf1L
2π f1C
the following equation can be derived:
Substituting XC =
f12 +
1
1
=0
f1 − 2
2π RC
4π LC
For this situation:
1
1
= 39.79 × 103
=
2π RC 2π (4 k Ω)(0.001 μ F)
1
1
= 2.53 × 1010
= 2
2
4π LC 4π (1 mH)(0.001 μ F)
and solving the quadratic equation, f1 = 140.4 kHz
BW
= 159.15 kHz − 140.4 kHz = 18.75 kHz
and
2
with BW = 2(18.75 kHz) = 37.5 kHz
b.
292
Qp =
fp
BW
=
159.15 kHz
= 4.24
37.5 kHz
CHAPTER 21
31.
a.
b.
Qs =
X L = 5000 Ω = 5000 Ω
= 12.2
R + R 400 Ω + 10 Ω 410 Ω
f s 5000 Hz
=
= 409.84 Hz
Qs
12.2
409.84 Hz
f1 = 5000 Hz −
= 4.80 kHz
2
410 Hz
= 5.20 kHz
f2 = 5000 Hz +
2
BW =
c.
At resonance
10 Ω(Vi )
Vo =
10 Ω + 400 Ω
= 0.024 Vi
32.
d.
At resonance,
a.
Q =
10 Ω || 2 kΩ = 9.95 Ω
9.95 Ω(Vi )
≅ 0.024 Vi as above!
Vo =
9.95 Ω + 400 Ω
X L 400 Ω
=
= 40
R
10 Ω
2
2
Z T p = Q R = (40) 20 Ω = 32 kΩ
At resonance, Vo =
1 kΩ
32 k Ω Vi
= 0.97Vi
32 k Ω + 1 k Ω
Vo
= 0.97
Vi
For the low cutoff frequency note solution to Problem 30:
1
1
f12 +
f1 − 2
=0
2π RC
4π LC
1
1
=
= 19.9 nF
C=
2π fX C 2π (20 kHz)(400 Ω)
and Aυ =
L=
XL
400 Ω
=
= 3.18 mH
2π f 2π (20 kHz)
Substituting into the above equation and solving
f1 = 16.4 kHz
BW
= 20 kHz − 16.4 kHz = 3.6 kHz
with
2
and BW = 2(3.6 kHz) = 7.2 kHz
fp
20 kHz
Qp =
=
= 2.78
BW 7.2 kHz
CHAPTER 21
293
b.
c.
−
At resonance
Z T p = 32 kΩ || 100 kΩ = 24.24 kΩ
24.24 k Ω V i
= 0.96Vi
24.24 k Ω + 1 k Ω
V
and Aυ = o = 0.96 vs 0.97 above
Vi
with Vo =
At frequencies to the right and left of fp, the impedance Z T p will decrease and be
affected less and less by the parallel 100 kΩ load. The characteristics, therefore, are
only slightly affected by the 100 kΩ load.
d.
At resonance
Z T p = 32 kΩ || 20 kΩ = 12.31 kΩ
with Vo =
12.31 k Ω Vi
= 0.925Vi vs 0.97 Vi above
12.31 k Ω + 1 k Ω
At frequencies to the right and left of fp, the impedance of each frequency will actually
be less due to the parallel 20 kΩ load. The effect will be to narrow the resonance curve
and decrease the bandwidth with an increase in Qp.
33.
a.
fp =
1
2π LC
(
1
= 726.44 kHz (band-stop)
2π (400 μ H)(120 pF)
=
)
X Ls ∠90° + X L p ∠90° X C ∠ − 90° = 0
jX Ls +
jX Ls +
(X
)
∠90° ( X C ∠ − 90° )
jX L p − jX C
X Lp X C
j X Lp − X C
jX Ls − j
X Ls −
(
Lp
(X
X Lp X C
Lp
− XC
X Lp X C
X Lp − X C
)
)
=0
=0
=0
=0
X Ls X C − X Ls X L p + X L p X C = 0
ω Lp
ω Ls
− ω Ls +
=0
ωC
ωC
294
CHAPTER 21
1
⎡ Ls + L p ⎤⎦ = 0
C⎣
L + Lp
ω= s
CLs L p
LsLpω2 −
f=
34.
a.
Ls + L p
1
2π
CLs L p
c.
fc =
⇒ Ls =
f=
f=
1
fc:
2
1
fc:
10
f = 10fc:
f=
1
fc:
2
f = 2fc:
CHAPTER 21
2
XL
24.15 k Ω
=
= 128.19 mH
2π f 2π (30 kHz)
1
1
= 772.55 Hz
=
2π RC 2π (0.47 k Ω)(0.05 μ F)
f = 2fc:
d.
460 × 10−6
= 2.01 MHz (pass-band)
28.8 × 10−19
1
2π
1
1
= 2
= 12.68 mH
2
4π f s C 4π (100 kHz ) 2(200 pF)
2π LC
XL = 2πfL = 2π(30 kHz)(12.68 mH) = 2388.91 Ω
1
1
XC =
= 26.54 kΩ
=
2π fC 2π (30 kHz)(200 pF)
XC − XL = 26.54 kΩ − 2388.91 Ω = 24.15 kΩ(C)
X Lp = X C(net) = 24.15 kΩ
1
fs =
Lp =
35. a, b.
=
Aυ dB = 20 log10
Aυ dB = 20 log10
1 + ( f c /f )
1
1
1 + (0.5) 2
1
Aυ dB = 20 log10
1 + (10) 2
Aυ dB = 20 log10
1
Aυ =
Aυ =
1 + ( f c /f )
1
1 + (0.5) 2
2
=
1
1 + (2) 2
= −7 dB
= −0.969 dB
= −20.04 dB
1 + (0.1) 2
1
2
= 20 log10
= −0.043 dB
1
1+ (2) 2
= 0.447
= 0.894
295
e.
36.
a.
fc =
1
1
1
=
=
= 1989.44 Hz
2π RC 2π (6 k Ω 12 k Ω)0.01 μ F 2π (4 k Ω)(0.01 μ F)
1
Vo
=
Vi
1+ ( f /f ) 2
⎛
⎞
1
⎟Vi
and Vo = ⎜
⎜ 1+ ( f /f ) 2 ⎟
c
⎝
⎠
c
b.
c. & d.
296
CHAPTER 21
e.
Remember the log scale! 1.5fc is not midway between fc and 2fc
Aυ dB = 20 log10 Aυ
−1.5 = 20 log10 Aυ
−0.075 = log10 Aυ
V
Aυ = o = 0.84
Vi
f.
37. a, b.
θ = tan−1 fc/f
Aυ =
fc =
c.
Vo
= Aυ ∠θ =
Vi
1
1 + (f/f c )
f = fc/2 = 6.63 kHz
AυdB = 20 log10
AυdB = 20 log10
1
1 + (0.5) 2
1
1 + (2) 2
AυdB = 20 log10
1
1 + (0.1) 2
f = 10fc = 132.6 kHz
Aυ dB = 20 log10
f = fc/2:
Aυ =
f = 2fc:
Aυ =
1
1 + (10) 2
1
1 + (0.5) 2
1
1 + (2) 2
= −0.97 dB
= −6.99 dB
f = fc/10 = 1.326 kHz
CHAPTER 21
∠−tan−1f/fc
1
1
=
= 13.26 kHz
2π RC 2π (12 k Ω)(1 nF)
f = 2fc = 26.52 kHz
d.
2
= −0.04 dB
= −20.04 dB
= 0.894
= 0.447
297
e.
38.
a.
θ = tan−1 f/fc
f = fc/2:
θ = −tan−1 0.5 = −26.57°
θ = −tan−1 1 = −45°
f = f c:
f = 2fc:
θ = −tan−1 2 = −63.43°
R2 || XC =
( R2 )(− jX C )
R2 X C
=− j
R2 − jX C
R2 − jX C
⎛ − jR2 X C ⎞
⎜
⎟ Vi
R2 − jX C ⎠
R2 X C Vi
⎝
Vo =
=− j
R2 X C
R1 ( R2 − jX C ) − jR2 X C
R1 − j
R2 − jX C
=
=
− jR2 X C Vi
-jR2 X C Vi
=
R1R2 − jR1 X C − jR2 X C R1R2 − j ( R1 + R2 ) X C
R2 X C Vi
R2 Vi
=
jR1R2 + ( R1 + R2 ) X C j R1R2 + ( R + R )
1
2
XC
⎛ R2 ⎞
⎜
⎟ Vi
R1 + R2 ⎠
R2 Vi
⎝
=
=
RR
R1 + R2 + j 1 2 1 + j ⎛⎜ R1R2 ⎞⎟ 1
XC
⎝ R1 + R2 ⎠ X C
R2
V
R1 + R2
and Aυ = o =
Vi
⎛ RR ⎞
1 + jω ⎜ 1 2 ⎟ C
⎝ R1 + R2 ⎠
or Aυ =
298
R2
R1 + R2
⎡
⎤
1
⎢
⎥
⎣1+ j 2π f ( R1 R2 )C ⎦
CHAPTER 21
defining fc =
2π ( R1 R2 )C
1
⎡
⎤
1
⎢
⎥
⎣1 + j f/f c ⎦
⎤
1
R2 ⎡
⎢
∠ − tan −1 f/f c ⎥
and Aυ =
R1 + R2 ⎢ 1+ ( f/f c ) 2
⎥⎦
⎣
⎤
1
R2 ⎡
⎢
⎥|Vi |
with |Vo |=
R1 + R2 ⎢ 1+ (f/f c ) 2 ⎥
⎣
⎦
R2
27 k Ω
for f fc, Vo =
Vi =
Vi = 0.852Vi
R1 + R2
4.7 k Ω + 27 k Ω
Aυ =
R2
R1 + R2
at f = fc: Vo = 0.852[0.707]Vi = 0.602Vi
1
= 994.72 Hz
fc =
2π ( R1 R2 )C
b.
c. & d.
−20 log10
CHAPTER 21
R1 + R2
4.7 k Ω + 27 k Ω
= −20 log10
R2
27 k Ω
= −20 log10 1.174 = −1.39 dB
299
e.
AυdB ≅ −1.39 dB − 0.5 dB = −1.89 dB
AυdB = 20 log10 Aυ
−1.89 = 20 log10 Aυ
0.0945 = log10 Aυ
V
Aυ = o = 0.80
Vi
f.
θ = −tan−1 f/fc
a.
From Section 21.11,
39.
Aυ =
j f/f1
Vo
=
Vi 1 + jf/f c
1
1
=
= 663.15 Hz
2π R2′ C
2π (24 k Ω)(0.01 μ F)
1
1
fc =
= 468.1 Hz
=
2π ( R1 + R2′ ) C
2π (10 k Ω + 24 k Ω)(0.01 μ F)
f1 =
20log10
f
f
= 20log10 c
f1
f1
468.1 Hz
663.15 Hz
= −3.03 dB
= 20 log10
300
CHAPTER 21
b.
θ = 90° − tan−1
f
f
= + tan−1 1
f1
f
θ = 45°
θ = 54.78°
f = f1:
f = fc:
1
f = f1 = 331.58 Hz, θ = 63.43°
2
1
f=
f1 = 66.31 Hz, θ = 84.29°
10
f = 2f1 = 1,326.3 Hz, θ = 26.57°
f = 10f1 = 6,631.5 Hz, θ = 5.71°
40.
a.
VTh =
12 k Ω Vi
= 0.667 Vi
12 k Ω + 6 k Ω
RTh = 6 kΩ || 12 kΩ = 4 kΩ
f = ∞ Hz: (C ⇒ short circuit)
8 k Ω (0.667 Vi )
Vo =
= 0.445 Vi
8kΩ + 4 kΩ
CHAPTER 21
301
0.667 R2 Vi
R2 (0.667 Vi )
=
R1 + R2 − jX C R1 + R2 − jX C
V
0.667 R2
j 2π f (0.667 R2 )C
and Aυ = o =
=
Vi R1 + R2 − jX C 1 + j 2π f ( R1 + R2 )C
j f/f1
1
1
so that Aυ =
with f1 =
=
1 + j f/f c
2π 0.667R2C 2π 0.667(8 k Ω)(100 nF)
= 298.27 Hz
1
1
=
and fc =
2π ( R1 + R2 )C 2π (4 k Ω + 8 k Ω)(100 nF)
= 132.63 Hz
voltage-divider rule: Vo =
132.63 Hz
298.27 Hz
= 20 log10 0.445 = −7.04 dB
20 log10 f/f1 = 20 log10
b.
θ = 90° − tan−1 f/fc = +tan−1 fc/f = tan−1 132.6 Hz/f
or
302
CHAPTER 21
1+ j
41.
a.
f
f1
Aυ =
f
1+ j
fc
1
1
=
= 19,894.37 Hz
2π R2C 2π (10 kΩ)(800 pF)
1
1
fc =
=
2π ( R1 + R2 )C 2π (10 kΩ + 90 kΩ)(800 pF)
= 1,989.44 Hz
f1 =
−20log10
R1 + R2
= −20log10 10 = 20 dB
R2
b.
θ = tan−1 f/f1 − tan−1 f/fc
f = 10 kHz
10 kHz
10 kHz
θ = tan−1
− tan−1
= 26.69° − 78.75° = −52.06°
19.89 kHz
1.989 kHz
f = fc: (f1 = 10 fc)
fc
f
− tan −1 c = tan −1 0.1tan −1 1 = 5.71° − 45° = −39.29°
θ = tan−1
10 f c
fc
42.
a.
R1 no effect!
Note Section 21.12.
Aυ =
Vo 1 + j ( f/f1 )
=
Vi 1 + j ( f/f c )
1
= 2652.58 Hz
2π (6 k Ω)(0.01 μ F)
1
fc =
= 884.19 Hz
2π (12 k Ω + 6 k Ω)(0.01 μ F)
f1 =
Note Fig. 21.65.
Asymptote at 0 dB from 0 → fc
−6 dB/octave from fc to f1
12 k Ω + 6 k Ω
⎛
⎞
−9.54 dB from f1 on ⎜ −20 log
= − 9.54 dB ⎟
6 kΩ
⎝
⎠
CHAPTER 21
303
(b)
Note Fig. 21.67.
From 0° to −26.50° at fc and f1
θ = tan−1 f/f1 − tan−1 f/fc
At f = 1500 Hz (between fc and f1)
θ = tan−1 1500 Hz/2652.58 Hz − tan−1 1500 Hz/884.19 Hz
= 29.49° − 59.48° = −30°
43.
a.
V o 1 − jf1/f
=
V i 1 − jf c /f
1
1
f1 =
=
= 964.58 Hz
2π R1C 2π (3.3 kΩ)(0.05 μ F)
1
1
=
fc =
= 7,334.33 Hz
2π (R1 R2 )C 2π (3.3 k Ω 0.5 k Ω)(0.05 μ F)
Aυ =
0.434 k Ω
−20 log10
R1 + R2
3.3 k Ω + 0.5 k Ω
= − 20 log10 7.6 = −17.62 dB
= − 20 log10
0.5 k Ω
R2
b.
θ = −tan−1
f = 1.3 kHz:
44.
304
a.
f
f1
+ tan−1 c
f
f
964.58 kHz
7334.33 Hz
+ tan −1
1.3 kHz
1.3 kHz
= −36.57° + 79.95° = 43.38°
θ = −tan−1
Note Section 21.13.
1 − j ( f1 /f )
Aυ =
1 − j (f c /f )
f1 =
1
1
=
= 964.58 Hz
2π R1C 2π (3.3 k Ω)(0.05 μ F)
fc =
2π ( R1 R2 )C
1
=
1
= 7334.33 Hz
2π (3.3 k Ω 0.5 k Ω) 0.05 μ F
0.434 kΩ
CHAPTER 21
Note Fig. 21.72.
−20 log10
R1 + R2
3.3 k Ω + 0.5 k Ω
= −17.62 dB
= −20 log10
R2
0.5 k Ω
Asymptote at −17.62 dB from 0 → f1
+6 dB/octave from f1 to fc
0 dB from fc on
b.
θ = −tan−1 f1/f + tan−1 fc/f
Test at 3 kHz
θ = −tan−1 964.58 Hz/3.0 kHz + tan−1 7334.33 Hz/3.0 kHz
= −17.82° + 67.75° = 49.93° ≅ 50°
Therefore rising above 45° at and near the peak
50 kHz vs 23 kHz → drop about 1 dB at 23 kHz due to 50 kHz break.
Ignore effect of break frequency at 10 Hz.
Assume −2 dB drop at 68 Hz due to break frequency at 45 Hz.
Rough sketch suggests low cut-off frequency of 90 Hz.
Checking: Ignoring upper terms
⎛ 10 Hz ⎞
⎛ 45 Hz ⎞
⎛ 68 Hz ⎞
1+ ⎜
⎟ − 20 log10 1 + ⎜
⎟ − 20 log10 1 + ⎜
⎟
⎝ f ⎠
⎝ f ⎠
⎝ f ⎠
= −0.0532 dB − 0.969 dB − 1.96 dB
= −2.98 dB (excellent)
Aυ′ dB = −20 log10
2
2
2
High frequency cutoff: Try 20 kHz
⎛ f ⎞
⎛ f ⎞
Aυ′ dB = − 20log10 1 + ⎜
⎟ − 20 log10 1 + ⎜
⎟
⎝ 23 kHz ⎠
⎝ 50 kHz ⎠
= −2.445 dB − 0.6445 dB
= −3.09 dB (excellent
2
2
∴ BW = 20 kHz − 90 Hz = 19,910 Hz ≅ 20 kHz
f1 = 90 Hz, f2 = 20 kHz
CHAPTER 21
305
Testing: f = 100 Hz
10 Hz
45 Hz
68 Hz
f
f
+ tan −1
+ tan −1
− tan −1
− tan −1
23 kHz
50 kHz
f
f
f
−1
−1
−1
−1
−1
0.1
+
tan
0.45
+
tan
0.68
−
tan
0.00435
−
tan
.002
= tan
= 5.71° + 24.23° + 34.22° − 0.249° − 0.115°
= 63.8° vs about 65° on the plot
θ = tan −1
45.
a.
1
Aυ
=
100 Hz ⎞⎛
130 Hz ⎞ ⎛
f ⎞⎛
f ⎞
Aυ max ⎛
⎟⎜ 1 + j
⎟
⎜1 − j
⎟⎜1 − j
⎟ ⎜1 + j
20 kHz ⎠⎝
50 kHz ⎠
f ⎠⎝
f ⎠⎝
⎝
Proximity of 100 Hz to 130 Hz will raise lower cutoff frequency above 130 Hz:
Testing: f = 180 Hz: (with lower terms only)
⎛ 100 ⎞
⎛ 130 ⎞
= −20 log10 1 + ⎜
⎟ − 20 log10 1 + ⎜
⎟
⎝ f ⎠
⎝ f ⎠
2
Aυ dB
⎛ 100 ⎞
⎛ 130 ⎞
= −20 log10 1 + ⎜
⎟ − 20 log10 1 + ⎜
⎟
⎝ 180 ⎠
⎝ 180 ⎠
= 1.17 dB − 1.82 dB = −2.99 dB ≅ −3 dB
2
2
2
Proximity of 50 kHz to 20 kHz will lower high cutoff frequency below 20 kHz:
Testing: f = 18 kHz: (with upper terms only)
Aυ dB = −20 log10
⎛ f ⎞
⎛ f ⎞
1+ ⎜
⎟ − 20 log10 1 + ⎜
⎟
⎝ 20 kHz ⎠
⎝ 50 kHz ⎠
2
⎛ 18 kHz ⎞
⎛ 13 kHz ⎞
= −20 log10 1 + ⎜
⎟ − 20 log10 1 + ⎜
⎟
⎝ 20 kHz ⎠
⎝ 20 kHz ⎠
= −2.576 dB − 0.529 dB = −3.105 dB
2
306
2
2
CHAPTER 21
b.
Testing:
47.
100
130
1.8 kHz
1.8 kHz
+ tan −1
− tan −1
− tan −1
1.8 kHz
1.8 kHz
20 kHz
50 kHz
= 3.18° + 4.14° − 5.14° − 2.06°
= 0.12° ≅ 0°
θ = tan−1
flow = fhigh − BW = 36 kHz − 35.8 kHz = 0.2 kHz = 200 Hz
Aυ =
48.
f = 1.8 kHz:
−120
⎛
50 ⎞⎛
200 ⎞ ⎛
f ⎞
⎟
⎜1 − j ⎟⎜ 1 − j
⎟ ⎜1 + j
36 kHz ⎠
f ⎠⎝
f ⎠⎝
⎝
+jf
0.05
1
1
=
=
=
100
100
2000 +jf + 2000
0.05 − j
1− j
1− j
f
f
0.05 f
f
+j
2000 and f = 2000 Hz
=
1
f
1+ j
2000
Aυ =
CHAPTER 21
307
200
1
1
=
=
200 + j 0.1 f 1 + j 0.1 f 1 + j f
200
2000
1
f
,
= 1 and f = 2 kHz
Aυ dB = 20 log 20
2
⎛ f ⎞ 2000
1+ ⎜
⎟
⎝ 2000 ⎠
49.
Aυ =
50.
Aυ =
51.
f ⎞⎛
f ⎞
⎛
⎜1 + j
⎟⎜1 + j
⎟
1000 ⎠⎝
2000 ⎠
Aυ = ⎝
2
f ⎞
⎛
⎜1 + j
⎟
3000 ⎠
⎝
jf/ 1000
(1+ jf/ 1000)(1 + jf/10,000)
⎛ f ⎞
⎛ f ⎞
1 + ⎜ 1 ⎟ + 20 log10 1 + ⎜ 2 ⎟ + 40 log10
⎝ 1000 ⎠
⎝ 2000 ⎠
2
AυdB = 20 log10
308
2
1
⎛ f ⎞
1+ ⎜ 3 ⎟
⎝ 3000 ⎠
2
CHAPTER 21
52.
2π f
jω
f
jω
f
f
,
= j
= j
= j
= j
1000
1000
1000
159.16 Hz 5000
795.78 Hz
2π
53.
a.
Woofer − 400 Hz:
XL = 2πfL = 2π(400 Hz)(4.7 mH) = 11.81 Ω
1
1
=
= 10.20 Ω
XC =
2π (400 Hz)(39 μ F)
2π fC
R || XC = 8 Ω ∠0° || 10.20∠−90° = 6.3 Ω ∠−38.11°
( R X C )(Vi )
(6.3 Ω ∠ − 38.11°)(Vi )
=
Vo =
( R X C ) + jX L
(6.3 Ω ∠ − 38.11°) + j 11.81 Ω
Vo = 0.673 ∠−96.11° Vi
V
and Aυ = o = 0.673 vs desired 0.707 (off by less than 5%)
Vi
Tweeter − 5 kHz:
XL = 2πfL = 2π(5 kHz)(0.39 mH) = 12.25 Ω
1
1
XC =
=
= 11.79 Ω
2π fC 2π ( 5 kHz)(2.7 μ F)
R || XL = 8 Ω ∠0° || 12.25 Ω ∠90° = 6.7 Ω ∠33.15°
(6.7 Ω ∠33.15°)(Vi )
Vo =
(6.7 Ω ∠33.15°) − j 11.79 Ω
Vo = 0.678 ∠88.54° Vi
CHAPTER 21
309
Vo
= 0.678 vs 0.707 (off by less than 5%)
Vi
Woofer − 3 kHz:
XL = 2πfL = 2π(3 kHz)(4.7 mH) = 88.59 Ω
1
1
XC =
=
= 1.36 Ω
2π fC 2π ( 3 kHz)(39 μ F)
R || XC = 8 Ω ∠0° || 1.36 Ω ∠−90° = 1.341 Ω ∠−80.35°
( R X C )(Vi )
(1.341 Ω ∠ − 80.35°)(Vi )
=
Vo =
( R X C ) + jX L (1.341 Ω ∠ − 80.35°) + j 88.59 Ω
Vo = 0.015 ∠−170.2° Vi
V
and Aυ = o = 0.015 vs desired 0 (excellent)
Vi
and Aυ =
b.
Tweeter − 3 kHz:
XL = 2πfL = 2π(3 kHz)(0.39 mH) = 7.35 Ω
1
1
=
= 19.65 Ω
XC =
2π fC 2π (3 kHz)(2.7 μ F)
R || XL = 8 Ω ∠0° || 7.35 Ω ∠90° = 5.42 Ω ∠47.42°
( R X L )(Vi )
(5.42 Ω ∠47.42°)(Vi )
=
Vo =
( R X L ) + jX C (5.42 Ω ∠47.42°) − j 19.65 Ω
Vo = 0.337 ∠124.24° Vi
V
and Aυ = o = 0.337 (acceptable since relatively close to cut frequency for tweeter)
Vi
c.
Mid-range speaker − 3 kHz:
Z′ = 7.41 Ω ∠ − 22.15°
Z′′ = 8.24 Ω ∠ 33.58°
Z′′′ = 7.816 Ω ∠ 37.79°
(7.816 Ω ∠37.79°) Vi
Z′′′Vi
= 1.11 ∠8.83° Vi
=
Z′′′ − jX C 7.816 Ω ∠37.79° − j1.36 Ω
(7.41 Ω ∠ − 22.15°)Vi
Z′V1
=
Vo =
= 0.998 ∠−46.9° Vi
Z′ + jX L 7.41 Ω ∠ − 22.15° + j 7.35 Ω
V1 =
and
310
Aυ =
Vo
= 0.998 (excellent)
Vi
CHAPTER 21
Chapter 22
1.
a.
M = k L p Ls ⇒ Ls =
b.
ep = N p
es = kNs
c.
ep = L p
es = M
2.
a.
dφ p
dt
dφ p
dt
di p
dt
di p
dt
2
(80 mH) 2
M
=
= 0.2 H
2
(50 mH)(0.8) 2
L pk
= (20)(0.08 Wb/s) = 1.6 V
= (0.8)(80 t)(0.08 Wb/s) = 5.12 V
= (50 mH)(0.03 × 103 A/s) = 15 V
= (80 mH)(0.03 × 103 A/s) = 24 V
k=1
2
b.
(a)
2
( 80 mH )
Ls = M 2 =
= 128 mH
(50 mH)(1) 2
L pk
(b)
ep = 1.6 V, es = kNs
(c)
ep = 15 V, es = 24 V
dφ p
= (1)(80 t)(0.08 Wb/s) = 6.4 V
dt
k = 0.2
2
(a)
2
(80 mH )
= 3.2 H
Ls = M 2 =
(50 mH)(0.2) 2
L pk
(b)
ep = 1.6 V, es = kNs
(c)
ep = 15 V, es = 24 V
dφ p
dt
= (0.2)(80 t)(0.08 Wb/s) = 1.28 V
2
3.
a.
2
( 80 mH )
Ls = M 2 =
= 158.02 mH
(50
mH)(0.9) 2
L pk
b.
ep = N p
es = kNs
c.
dφ p
dt
dφ p
dt
= (300 t)(0.08 Wb/s) = 24 V
= (0.9)(25 t)(0.08 Wb/s) = 1.8 V
ep and es the same as problem 1: ep = 15 V, es = 24 V
CHAPTER 22
311
4.
5.
Ns
64 t
Ep =
(25 V) = 200 V
8t
Np
a.
Es =
b.
Φmax =
a.
Es =
b.
Φm(max) =
6.
Ep =
7.
f=
8.
a.
25 V
Ep
=
= 11.73 mWb
4.4 fN p 4.44(60 Hz)(8 t)
30 t
Ns
(25 V) = 3.13 V
E p=
240 t
Np
25 V
Ep
= 391.02 μWb
=
4.44 fN p (4.44)(60 Hz)(240 t)
60 t
Np
Es =
(240 V) = 20 V
720 t
Ns
25 V
Ep
=
= 56.31 Hz
(4.44) N p Φ m (max) (4.44)(8 t)(12.5 mWb)
⎛1⎞
IL = aIp = ⎜ ⎟ (2 A) = 0.4 A
⎝5⎠
⎛2 ⎞
VL = ILZL = ⎜ A ⎟ (2 Ω) = 0.8 V
⎝5 ⎠
⎛1⎞
Zin = a2ZL = ⎜ ⎟ 2 Ω = 0.08 Ω
⎝5⎠
2
b.
9.
10.
11.
Zp =
Vg
Ip
=
1600 V
= 400 Ω
4A
⎛1⎞
Vg = aVL = ⎜ ⎟ (1200 V) = 300 V
⎝4⎠
300 V
V
= 75 A
Ip = g =
4Ω
Zi
IL = Is =
VL 240 V
= 12 A
=
Z L 20 Ω
I s = a = N p ⇒ 12 A = N p
0.05 A 50
Ns
Ip
50(12)
Np =
= 12,000 turns
0.05
312
CHAPTER 22
12.
a.
a=
1
N p 400 t
=
=
3
1200
t
Ns
⎛1⎞
Zi = a ZL = ⎜ ⎟ [9 Ω + j12 Ω] = 1 Ω + j1.333 Ω = 1.667 Ω ∠53.13°
⎝ 3⎠
Ip = Vg/Zi = 100 V/1.667 Ω = 60 A
2
2
13.
IL = aIp =
a.
Zp = a2ZL ⇒ a =
Zp =
b.
36 Ω
=3
4Ω
1
1
1
Vs Ns 1
=
= ⇒ V s = V p = (10 V) = 3 V
3
3
3
Vp Np 3
P=
a.
b.
Zp
ZL
10 V
Vp
= 36 Ω
=
20
V/ 72 Ω
Ip
a=
14.
1
(60 A) = 20 A, VL = ILZL = (20 A)(15 Ω) = 300 V
3
b.
Vs2 (3.33 V ) 2
= 2.78 W
=
4Ω
Zs
Re = Rp + a2Rs = 4 Ω + (4)2 1 Ω = 20 Ω
Xe = Xp + a2Xs = 8 Ω + (4)2 2 Ω = 40 Ω
c.
120 V ∠0°
120 V ∠0°
Vg
= 0.351 A ∠−6.71°
=
=
Z p 20 Ω + 320 Ω + j 40 Ω 340 Ω + j 40 Ω
d.
Ip =
e.
aVL =
2
or
f.
−
g.
VL =
CHAPTER 22
a R LV g
= Ipa2RL
2
( R e + a R L) + j X e
VL = aIpRL∠0° = (4)(0.351 A ∠−6.71°)(20 Ω ∠0°) = 28.1 V ∠−6.71°
1
Ns
Vg = (120 V) = 30 V
4
Np
313
15.
4t
=4
Ns 1t
Re = Rp + a2Rs = 4 Ω + (4)2 1 Ω = 20 Ω
Xe = Xp + a2Xs = 8 Ω + (4)2 2 Ω = 40 Ω
Zp = Z Re + Z X e + a 2 Z X L = 20 Ω + j40 Ω + j(4)2 20 Ω
= 20 Ω + j40 Ω + j320 Ω = 20 Ω + j360 Ω = 360.56 Ω ∠86.82°
Np
a.
a=
b.
Ip =
c.
VRe = (I∠θ)(Re∠0°) = (332.82 mA ∠−86.82°)(20 Ω ∠0°)
=
120 V ∠0°
Vg
=
= 332.82 mA ∠−86.82°
Z p 360.56 Ω ∠86.82°
= 6.66 V ∠−86.82°
VX e = (I∠θ)(Xe∠90°) = (332.82 mA ∠−86.32°)(40 Ω ∠90°)
= 13.31 V ∠3.18°
VX L = I(a2 Z X L ) = (332.82 mA ∠−86.82°)(320 Ω ∠90°)
= 106.50 V ∠3.18°
16.
a.
a = Np/Ns = 4 t/1 t = 4, Re = Rp + a2Rs = 4 Ω + (4)2 1 Ω = 20 Ω
Xe = Xp + a2Xs = 8 Ω + (4)2 2 Ω = 40 Ω
Zp = Re + jXe − ja2XC = 20 Ω + j40 Ω − j(4)2 20 Ω
= 20 Ω − j280 Ω = 280.71 Ω ∠−85.91°
120 V ∠0°
Vg
= 0.43 A ∠85.91°
=
280.71 Ω ∠ − 85.91°
Zp
b.
Ip =
c.
V Re = (Ip∠θ)(Re∠0°) = (0.427 A ∠85.91°)(20 Ω ∠0°) = 8.54 V ∠85.91°
V X e = (Ip∠θ)(Xe∠90°) = (0.427 A ∠85.91°)(40 Ω ∠90°) = 17.08 V ∠175.91°
2
V X C = (Ip∠θ)(a XC∠−90°) = (0.427 A ∠85.91°)(320 Ω ∠−90°) = 136.64 V ∠−4.09°
17.
18.
19.
−
Coil 1:
L1 − M12
Coil 2:
L2 − M12
LT = L1 + L2 − 2M12 = 4 H + 7 H − 2(1 H) = 9 H
LT ( + ) = L1 + L2 + 2M 12
M12 = k L1 L2 = (0.8) (200 mH)(600 mH) = 277 mH
LT (+) = 200 mH + 600 mH + 2(277 mH) = 1.35 H
314
CHAPTER 22
20.
M23 = k L2 L3 = 1 (1 H)(4 H) = 2 H
Coil 1:
Coil 2:
Coil 3:
21.
L1 + M12 − M13 = 2 H + 0.2 H − 0.1 H = 2.1 H
L2 + M12 − M23 = 1 H + 0.2 H − 2 H = −0.8 H
L3 − M23 − M13 = 4 H − 2 H − 0.1 H = 1.9 H
LT = 2.1 H − 0.8 H + 1.9 H = 3.2 H
E1 − I1[ Z R1 + Z L1 ] − I2[Zm] = 0
I2[ Z L2 + Z RL ] + I1[Zm] = 0
──────────────────────
I1( Z R1 + Z L1 ) + I2(Zm) = E1
I1(Zm) + I2( Z L2 + Z RL ) = 0
───────────────────────
22.
Zi = Zp +
Xm = −ωM ∠90°
(ω M ) 2
(ω M ) 2
= Rp + j X L p +
Zs + ZL
Rs + jX Ls + RL
Rp = 2 Ω, X L p = ωLp = (103 rad/s)(8 H) = 8 kΩ
Rs = 1 Ω, X Ls = ωLs = (103 rad/s)(2 H) = 2 kΩ
M = k L p L s = 0.05 (8 H)(2 H) = 0.2 H
Zi = 2 Ω + j8 kΩ +
(103 rad/s ⋅ 0.2 H ) 2
1 Ω + j 2 k Ω + 20 Ω
4 ×104 Ω
21 + j 2 ×103
= 2 Ω + j8 kΩ + 0.21 Ω − j19.99 Ω = 2.21 Ω + j7980 Ω
Zi = 7980 Ω ∠89.98°
= 2 Ω + j8 kΩ +
23.
Np
Vp
2400 V
= 20
120 V
a.
a=
b.
10,000 VA = VsIs ⇒ Is =
c.
Ip =
d.
a=
Is =
CHAPTER 22
Ns
=
Vs
=
10,000 VA 10,000 VA
=
= 83.33 A
120 V
Vs
10,000 VA 10,000 VA
= 4.17 A
=
2400 V
Vp
Vp
Vs
=
120 V
1
= 0.05 =
2400 V
20
10,000 VA
= 4.17 A, Ip = 83.33 A
2400 V
315
24.
Is = I1 = 2 A, Ep = VL = 40 V
Es = Vs − VL = 200 V − 40 V = 160 V
200 V
VgI1 = VLIL ⇒ IL = Vg/VL ⋅ I1 =
(2 A) = 10 A
40 V
Ip + I1 = IL ⇒ Ip = IL − I1 = 10 A − 2A = 8 A
25.
a.
Es =
Ns
Ep
Np
25 t
(100 V ∠0°) = 25 V ∠0° = VL
100 t
E
25 V ∠0°
Is = s =
= 5 A ∠0° = IL
Z L 5 Ω ∠0 °
=
b.
⎛ Np ⎞
⎛ 100 t ⎞
2
Zi = a ZL = ⎜
⎟ ZL = ⎜
⎟ 5 Ω ∠0° = (4) 5 Ω ∠0° = 80 Ω ∠0°
⎝ 25 t ⎠
⎝ Ns ⎠
c.
1
1
Z1/ 2 = Z i = (80 Ω ∠0°) = 20 Ω ∠0°
4
4
2
26.
a.
b.
2
2
15 t
N2
E1 =
(60 V ∠0°) = 10 V ∠0°
90 t
N1
45 t
N
E3 = 3 E1 =
(60 V ∠0°) = 30 V ∠0°
90 t
N1
10 V ∠0°
I2 = E 2 =
= 1.25 A ∠0°
Z 2 8 Ω ∠0°
30 V ∠0°
= 6 A ∠0°
I3 = E3 =
Z 3 5 Ω ∠0°
E2 =
1
=
R1
1
1
+
2
( N1 / N 2 ) R2 ( N1 / N 3 ) 2R3
1
1
+
2
(90 t /15 t ) 8 Ω (90 t / 45 t ) 2 5 Ω
1
1
1
=
+
= 0.05347 S
R1 288 Ω 20 Ω
R1 = 18.70 Ω
=
27.
a.
N2
⎛ 40 t ⎞
E1 = ⎜
⎟ (120 V ∠60°) = 40 V ∠60°
N1
⎝ 120 t ⎠
E
40 V ∠60°
I2 = 2 =
= 3.33 A ∠60°
Z 2 12 Ω ∠0°
E2 =
N3
⎛ 30 t ⎞
E1 = ⎜
⎟ (120 V ∠60°) = 30 V ∠60°
N1
⎝ 120 t ⎠
E
30 V ∠60°
= 3 A ∠60°
I3 = 3 =
Z3 10 Ω ∠0°
E3 =
316
CHAPTER 22
b.
1
=
R1
1
1
+
2
( N1 / N 2 ) R2 ( N1 / N 3 ) 2R3
1
1
+
2
(120 t / 40 t ) 12 Ω (120 t / 30 t ) 210 Ω
1
1
1
+
= 0.0155 S
=
108
160
Ω
Ω
R1
1
R1 =
= 64.52 Ω
0.0155 S
=
28.
ZM = Z M12 = ωM12 ∠90°
E − I1Z1 − I1 Z L1 − I1(−Zm) − I2(+Zm) − I1 Z L2 + I2 Z L2 − I1(−Zm) = 0
E − I1(Z1 + Z L1 − Zm + Z L2 − Zm) − I2(Zm − Z L2 ) = 0
or
or
I1(Z1 + Z L1 + Z L2 − 2 Zm) + I2(Zm − Z L2 ) = E
──────────────────────────────────────────────
−I2Z2 − Z L2 (I2 − I1) − I1(+Zm) = 0
I1(Zm − Z L2 ) + I2(Z2 + Z L2 ) = 0
──────────────────────────
E1 − I1Z1 − I1 Z L1 − I2(− Z M12 ) − I3(+ Z M13 ) = 0
29.
or
or
or
∴
E1 − I1[Z1 + Z L1 ] + I2 Z M12 − I3 Z M13 = 0
────────────────────────────────
−I2(Z2 + Z3 + Z L2 ) + I3Z2 − I1(− Z M12 ) = 0
−I2(Z2 + Z3 + Z L2 ) + I3Z2 + I1 Z M12 = 0
────────────────────────────────
−I3(Z2 + Z4 + Z L3 ) + I2Z2 − I1(+ Z M13 ) = 0
−I3(Z2 + Z4 + Z L3 ) + I2Z2 − I1 Z M13 = 0
─────────────────────────────
[Z1 + Z L1 ]I1 −
Z M12 I2 +
Z M12 I1 − [Z2 + Z3 + Z L2 ]I2 +
Z M13 I3 = E1
Z2I3 = 0
Z M13 I1
Z2I2 + [Z2 + Z4 + Z L3 ]I3 = 0
────────────────────────────────────────
CHAPTER 22
317
Chapter 23
1.
2.
a.
Eφ = EL/ 3 = 208 V/1.732 = 120.1 V
b.
Vφ = Eφ = 120.1 V
c.
Iφ =
d.
IL = Iφ = 12.01 A
a.
Eφ = EL/ 3 = 208 V/1.732 = 120.1 V
b.
Vφ = Eφ = 120.1 V
c.
Zφ = 12 Ω − j16 Ω = 20 Ω ∠−53.13°
d.
IL = I φ = 6 A
b.
Vφ = 120.1 V
Iφ =
3.
a.
c.
4.
5.
V φ 120.1 V
= 12.01 A
=
10 Ω
Rφ
V φ 120.1 V
=
≅ 6A
20 Ω
Zφ
Eφ = 120.1 V
Zφ = (10 Ω ∠0° || (10 Ω ∠−90°) = 7.071 Ω ∠−45°
120.1 V
V
= 16.98 A
Iφ = φ =
Z φ 7.071 Ω
d.
IL = 16.98 A
a.
θ2 = −120°, θ3 = 120°
b.
Van = 120 V ∠0°, Vbn = 120 V ∠−120°, Vcn = 120 V ∠120°
c.
Ian =
d.
IL = Iφ = 6A
a.
θ2 = −120°, θ3 = +120°
b.
c.
V an 120 V ∠0°
= 6 A ∠0°
=
Z an 20 Ω ∠0°
120 V ∠ − 120°
Ibn = V bn =
= 6 A ∠−120°
20 Ω ∠0°
Zbn
120 V ∠120°
V
= 6 A ∠120°
Icn = cn =
20 Ω ∠0°
Z cn
e.
VL =
3 Vφ =
3 (120 V) = 207.8 V
Van = 120 V ∠0°, Vbn = 120 V ∠−120°, Vcn = 120 V ∠120°
Zφ = 9 Ω + j12 Ω = 15 Ω ∠53.13°
120 V ∠0°
120 V ∠ − 120°
= 8 A ∠−173.13°
= 8 A ∠−53.13°, Ibn =
15 Ω ∠53.13°
15 Ω ∠53.13°
120 V ∠120°
= 8 A ∠66.87°
Icn =
15 Ω ∠53.13°
Ian =
318
CHAPTER 23
e.
6.
7.
a, b.
IL = Iφ = 8 A
f.
EL =
3 Eφ = (1.732)(120 V) = 207.85 V
The same as problem 4.
Zφ = 6 Ω ∠0° || 8 Ω ∠−90° = 4.8 Ω ∠−36.87°
c.
120 V ∠0°
= 25 A ∠36.87°
Ian = V an =
Z an 4.8 Ω ∠ − 36.87°
120 V ∠ − 120°
V
Ibn = bn =
= 25 A ∠−83.13°
Zbn 4.8 Ω ∠ − 36.87°
120 V ∠120°
V
Icn = cn =
= 25 A ∠156.87°
Z cn 4.8 Ω ∠ − 36.87°
d.
IL = Iφ = 25 A
e.
VL =
3 Vφ =
3 (120 V) = 207.84 V
220 V
= 127.0 V
3 1.732
Zφ = 10 Ω − j10 Ω = 14.42 Ω ∠−45°
Vφ = Van = Vbn = Vcn =
VL
=
127 V
Vφ
= 8.98 A
=
Z φ 14.142 Ω
IL = IAa = IBb = ICc = Iφ = 8.98 A
Iφ = Ian = Ibn = Icn =
8.
Zφ = 12 Ω + j16 Ω = 20 Ω∠53.13°
V φ 50 V
= 2.5 A
=
Z φ 20 Ω
ZT φ = 13 Ω + j16 Ω = 20.62 Ω ∠50.91°
Iφ =
Vφ = Iφ Z T φ = (2.5 A)(20.62 Ω) = 51.55 V
9.
VL =
3 Vφ =
a.
EAN =
( 3 ) (51.55 V) = 89.29 V
22 kV
∠ − 30° = 12.7 kV ∠−30°
3
22 kV
EBN =
∠ − 150° = 12.7 kV ∠−150°
3
22 kV
ECN =
∠90° = 12.7 kV ∠90°
3
CHAPTER 23
319
b, c. IAa = Ian =
E AN
12.7 kV ∠ − 30°
=
Z AN (30 Ω + j 40 Ω) + (0.4 k Ω + j1 k Ω)
12.7 kV ∠ − 30°
12.7 kV ∠ − 30°
=
430 Ω + j1040 Ω 1125.39 Ω ∠67.54°
= 11.29 A ∠−97.54°
12.7 kV ∠ − 150°
=
= 11.29 A ∠−217.54°
1125.39 Ω ∠67.54°
12.7 kV ∠90°
= 11.29 A ∠22.46°
=
1125.39 Ω ∠67.54°
=
IBb = Ibn = E BN
Z BN
ICc = Icn = E CN
Z CN
d.
10.
11.
a.
Eφ = EL/ 3 = 208 V/1.732 = 120.1 V
c.
Iφ =
a.
Eφ = EL/ 3 = 208 V/1.732 = 120.1 V
c.
d.
12.
13.
V φ 208 V
=
= 10.4 A
Z φ 20 Ω
b.
Vφ = EL = 208 V
d.
IL = 3 Iφ = (1.732)(10.4 A) = 18 A
b.
Vφ = EL = 208 V
b.
Vφ = 208 V
Zφ = 6.8 Ω + j14 Ω = 15.564 Ω ∠64.09°
208 V
V
Iφ = φ =
= 13.36 A
Z φ 15.564 Ω
IL = 3 Iφ = (1.732)(13.36 A) = 23.14 A
Zφ = 18 Ω ∠0° || 18 Ω ∠−90° = 12.728 Ω ∠−45°
a.
Eφ = VL/ 3 = 208 V/ 3 = 120.09 V
c.
Iφ =
d.
IL = 3 Iφ = (1.732)(16.34 A) = 28.30 A
a.
θ2 = −120°, θ3 = +120°
b.
320
Van = IanZan = (11.29 A ∠−97.54°)(400 + j1000)
= (11.29 A ∠−97.54°)(1077.03 Ω ∠68.2°)
= 12.16 kV ∠−29.34°
Vbn = IbnZbn = (11.29 A ∠−217.54°)(1077.03 ∠68.2°)
= 12.16 kV ∠−149.34°
Vcn = IcnZcn = (11.29 A ∠22.46°)(1077.03 ∠68.2°)
= 12.16 kV ∠90.66°
208 V
Vφ
= 16.34 A
=
Z φ 12.728 Ω
Vab = 208 V ∠0°, Vbc = 208 V ∠−120°, Vca = 208 V ∠120°
CHAPTER 23
c.
d.
14.
−
V ab 208 V ∠0°
=
= 9.46 A ∠0°
Z ab 22 Ω ∠0°
208 V ∠120°
V
Ibc = bc =
= 9.46 A ∠−120°
22 Ω ∠0°
Zbc
208 V ∠120°
V
Ica = ca =
= 9.46 A ∠120°
22 Ω ∠0°
Z ca
Iab =
e.
IL = 3 Iφ = (1.732)(9.46 A) = 16.38 A
f.
Eφ = EL/ 3 = 208 V/1.732 = 120.1 V
a.
θ2 = −120°, θ3 = +120°
b.
c.
d.
Vab = 208 V ∠0°, Vbc = 208 V ∠−120°, Vca = 208 V ∠120°
−
Zφ = 100 Ω − j100 Ω = 141.42 Ω∠−45°
208 V ∠0°
V
= 1.47 A ∠45°
Iab = ab =
141.42
Ω ∠ − 45°
Z ab
208 V ∠ − 120°
V
Ibc = bc =
= 1.47 A ∠−75°
Zbc 141.42 Ω ∠ − 45°
208 V ∠120°
V
Ica = ca =
= 1.47 A ∠165°
Z ca 141.42 Ω ∠ − 45°
e.
IL = 3 Iφ = (1.732)(1.471 A) = 2.55 A
f.
Eφ = EL/ 3 = 208 V/1.732 = 120.1 V
15. a, b.
c.
d.
The same as problem 13.
−
Zφ = 3 Ω ∠0° || 4 Ω ∠90° = 2.4 Ω ∠36.87°
208 V ∠0°
= 86.67 A ∠−36.87°
Iab = V ab =
Z ab 2.4 Ω ∠36.87°
208 V ∠ − 120°
V
= 86.67 A ∠−156.87°
Ibc = bc =
Zbc 2.4 Ω ∠36.87°
208 V ∠120°
V
= 86.67 A ∠83.13°
Ica = ca =
Z ca 2.4 Ω ∠36.87°
CHAPTER 23
321
e.
IL =
3 Iφ = (1.732)(86.67 A) = 150.11 A
16.
Vab = Vbc = Vca = 220 V
Zφ = 10 Ω + j10 Ω = 14.142 Ω∠45°
220 V
V
Iab = Ibc = Ica = φ =
= 15.56 A
Ω
14.142
Zφ
17.
a.
b.
c.
f.
Eφ = 120.1 V
16 kV ∠0°
16 kV ∠0°
V ab
=
=
Z ab 300 Ω + j1000 Ω 1044.03 Ω ∠73.30°
Iab = 15.33 A ∠−73.30°
16 kV ∠ − 120°
= 15.33 A ∠−193.30°
Ibc = V bc =
Zbc 1044.03 Ω ∠73.30°
16 kV ∠120°
V
Ica = ca =
= 15.33 A ∠46.7°
Z ca 1044.03 Ω ∠73.30°
Iab =
IAa − Iab + Ica = 0
IAa = Iab − Ica = 15.33 A ∠−73.30° − 15.33 A ∠46.7°
= (4.41 A − j14.68 A) − (10.51 A + j11.16 A)
= 4.41 A − 10.51 A − j(14.68 A + 11.16 A)
= −6.11 A − j25.84 A = 26.55 A ∠−103.30°
IBb + Iab = Ibc
IBb = Ibc − Iab = 15.33 A ∠−193.30° − 15.33 A ∠−73.30°
= 26.55 A ∠136.70°
ICc + Ibc = Ica
ICc = Ica − Ibc = 15.33 A ∠46.7° − 15.33 A ∠−193.30°
= 26.55 A ∠16.70°
EAB = IAa(10 Ω + j20 Ω) + Vab − IBb(22.361 Ω ∠63.43°)
= (26.55 A ∠−103.30°)(22.361 Ω ∠63.43°) + 16 kV ∠0°
− (26.55 A ∠136.70°)(22.361 Ω ∠63.43°)
= (455.65 V − j380.58 V) + 16,000 V − (−557.42 V − j204.32 V)
= 17.01 kV − j176.26 V
= 17.01 kV ∠−0.59°
EBC = IBb(22.361 Ω ∠63.43°) + Vbc − ICc(22.361 Ω ∠63.53°)
= (26.55 A ∠136.70°)(22.361 Ω ∠63.53°) + 16 kV ∠−120°
− (26.55 A ∠16.70°)(22.361 Ω ∠63.53°)
= 17.01 kV ∠−120.59°
ECA = ICc(22.361 Ω ∠63.43°) + Vca − IAa(22.361 Ω ∠63.43°)
= 17.01 kV ∠119.41°
322
CHAPTER 23
18.
19.
a.
Eφ = EL = 208 V
c.
Iφ =
a.
Eφ = EL = 208 V
c.
Iφ =
20. a, b.
c.
21.
22.
V φ 120.09 V
= 7.08 A
=
Z φ 16.971 Ω
208 V
= 120.1 V
Vφ = E L =
3 1.732
d.
IL = Iφ ≅ 4 A
b.
Vφ = EL 3 = 120.09 V
d.
IL = Iφ = 7.08 A
The same as problem 18.
Zφ = 15 Ω ∠0° || 20 Ω ∠−90° = 12 Ω ∠−36.87°
Iφ =
d.
V φ 120.1 V
= 4.00 A
=
30 Ω
Zφ
b.
V φ 120.1 V
≅ 10 A
=
12 Ω
Zφ
IL = Iφ ≅ 10 A
120 V 120 V
= 69.28 V
=
1.732
3
69.28 V
Ian = Ibn = Icn =
= 2.89 A
24 Ω
IAa = IBb = ICc = 2.89 A
Van = Vbn = Vcn =
120 V
= 69.28 V
3
Zφ = 10 Ω + j20 Ω = 22.36 Ω∠63.43°
Van = Vbn = Vcn =
Ian = Ibn = Icn =
Vφ
Zφ
=
69.28 V
= 3.10 A
22.36 Ω
IAa = IBb = ICc = Iφ = 3.10 A
23.
Van = Vbn = Vcn = 69.28 V
Zφ = 20 Ω ∠0° || 15 Ω ∠−90° = 12 Ω ∠−53.13°
69.28 V
= 5.77 A
12 Ω
IAa = IBb = ICc = 5.77 A
Ian = Ibn = Icn =
24.
a.
Eφ = EL = 440 V
c.
Iφ =
CHAPTER 23
V φ 440 V
=
=2A
Z φ 220 Ω
b.
Vφ = EL = Eφ = 440 V
d.
IL = 3 Iφ = (1.732)(2 A) = 3.46 A
323
25.
a.
Eφ = EL = 440 V
c.
Zφ = 12 Ω − j9 Ω = 15 Ω ∠−36.87°
26. a, b.
c.
The same as problem 24.
Zφ = 22 Ω ∠0° || 22 Ω ∠90° = 15.56 Ω ∠45°
440 V
Vφ
=
= 28.28 A
Z φ 15.56 Ω
3 Iφ = (1.732)(28.28 A) = 48.98 A
d.
IL =
a.
θ2 = −120°, θ3 = +120°
b.
c.
d.
28.
3 Iφ = (1.732)(29.33 A) = 50.8 A
IL =
Iφ =
27.
Vab = 100 V ∠0°, Vbc = 100 V ∠−120°, Vca = 100 V ∠120°
−
V ab 100 V ∠0°
= 5 A ∠0°
=
Z ab 20 Ω ∠0°
100 V ∠ − 120°
V
Ibc = bc =
= 5 A ∠−120°
20 Ω ∠0°
Zbc
100 V ∠120°
V
Ica = ca =
= 5 A ∠120°
20 Ω ∠0°
Z ca
Iab =
e.
IAa = IBb = ICc =
a.
θ2 = −120°, θ3 = +120°
b.
c.
d.
Vφ = EL = 440 V
V φ 440 V
=
= 29.33 A
Z φ 15 Ω
Iφ =
d.
b.
3 (5 A) = 8.66 A
Vab = 100 V ∠0°, Vbc = 100 V ∠−120°, Vca = 100 V ∠120°
−
Zφ = 12 Ω + j16 Ω = 20 Ω ∠53.13°
100 V ∠0°
V ab
=
= 5 A ∠−53.13°
Z ab 20 Ω ∠53.13°
100 V ∠ − 120°
V
= 5 A ∠−173.13°
Ibc = bc =
Zbc 20 Ω ∠53.13°
Iab =
324
CHAPTER 23
Ica =
29.
V ca 100 V ∠120°
= 5 A ∠66.87°
=
Z ca 20 Ω ∠53.13°
3 Iφ = (1.732)(5 A) = 8.66 A
e.
IAa = IBb = ICc =
a.
θ2 = −120°, θ3 = 120°
b.
c.
d.
Vab = 100 V ∠0°, Vbc = 100 V ∠−120°, Vca = 100 V ∠120°
−
Zφ = 20 Ω ∠0° || 20 Ω ∠−90° = 14.14 Ω ∠−45°
100 V ∠0°
= 7.07 A ∠45°
14.14 Ω ∠ − 45°
100 V ∠ − 120°
= 7.07 A ∠−75°
Ibc =
14.14 Ω ∠ − 45°
100 V ∠120°
Ica =
= 7.07 A ∠165°
14.14 Ω ∠ − 45°
Iab =
e.
30.
IAa = IBb = ICc =
( 3 ) (7.07 A) = 12.25 A
PT = 3Iφ2 Rφ = 3(6 A)2 12 Ω = 1296 W
QT = 3Iφ2 X φ = 3(6 A)2 16 Ω = 1728 VAR(C)
ST =
2
2
PT + QT = 2160 VA
1296 W
Fp = P T =
= 0.6 (leading)
2160
VA
ST
31.
Vφ = 120 V, Iφ = 120 V/20 Ω = 6 A
PT = 3Iφ2 Rφ = 3(6 A)2 20 Ω = 2160 W
QT = 0 VAR
ST = PT = 2160 VA
2160 W
=1
Fp = P T =
S T 2160 VA
32.
PT = 3Iφ2 Rφ = 3(8.98 A)2 10 Ω = 2419.21 W
QT = 3Iφ2 X φ = 3(8.98 A)2 10 Ω = 2419.21 VAR(C)
ST =
Fp =
PT2 + Q T2 = 3421.28 VA
PT
2419.21 W
=
= 0.7071 (leading)
ST 3421.28 VA
CHAPTER 23
325
33.
Vφ = 208 V
⎛V 2 ⎞
(208 V ) 2
= 7210.67 W
PT = 3 ⎜ φ ⎟ = 3 ⋅
18 Ω
⎝ Rφ ⎠
⎛V2 ⎞
(208 V ) 2
QT = 3 ⎜ φ ⎟ = 3 ⋅
= 7210.67 VAR(C)
18 Ω
⎝ Xφ ⎠
2
2
PT + QT = 10,197.42 VA
ST =
7210.67 W
= 0.707 (leading)
Fp = P T =
10,197.42
VA
ST
34.
PT = 3Iφ2 Rφ = 3(1.471 A)2 100 Ω = 649.15 W
QT = 3Iφ2 X φ = 3(1.471 A)2 100 Ω = 649.15 VAR(C)
PT2 + Q T2 = 918.04 VA
ST =
Fp =
35.
PT
649.15 W
=
= 0.7071 (leading)
ST 918.04 VA
PT = 3Iφ2 Rφ = 3(15.56 A)2 10 Ω = 7.26 kW
QT = 3Iφ2 X φ = 3(15.56 A)2 10 Ω = 7.26 kVAR
PT2 + Q T2 = 10.27 kVA
ST =
Fp =
PT
7.263 kW
=
= 0.7071 (lagging)
ST 10.272 kVA
V φ 3(120.1 V )
=
= 2884.80 W
15 Ω
Rφ
2
36.
PT = 3
V φ 3(120.1 V )
=
= 2163.60 VAR(C)
20 Ω
Xφ
2
QT = 3
ST =
Fp =
37.
2
2
2
2
PT + QT = 3605.97 VA
PT
2884.80 W
=
= 0.8 (leading)
ST 3605.97 VA
Zφ = 10 Ω + j20 Ω = 22.36 Ω ∠63.43°
120 V
= 69.28 V
Vφ = V L =
3 1.732
69.28 V
V
Iφ = φ =
= 3.098 A
Z φ 22.36 Ω
PT = 3Iφ2 Rφ = 3(3.098 A)2 10 Ω = 287.93 W
326
CHAPTER 23
QT = 3Iφ2 X φ = 3(3.098 A)2 20 Ω = 575.86 VAR
PT2 + Q T2 = 643.83 VA
ST =
Fp =
PT
287.93 W
=
= 0.447 (lagging)
ST 643.83 VA
V φ 3(440 V )
=
= 26.4 kW
22 Ω
Rφ
QT = PT = 26.4 kVAR(L)
2
38.
ST =
Fp =
39.
2
PT = 3
PT2 + Q T2 = 37.34 kVA
PT
26.4 kW
=
= 0.707 (lagging)
ST 37.34 kVA
Zφ = 12 Ω + j16 Ω = 20 Ω ∠53.13°
Iφ =
V φ 100 V
=
=5A
Z φ 20 Ω
PT = 3Iφ2 Rφ = 3(5 A)2 12 Ω = 900 W
QT = 3Iφ2 X φ = 3(5 A)2 16 Ω = 1200 VAR(L)
2
2
PT + QT = 1500 VA
P
900 W
= 0.6 (lagging)
Fp = T =
ST 1500 VA
ST =
40.
PT =
3 ELIL cos θ
4800 W = (1.732)(200 V)IL (0.8)
IL = 17.32 A
I
17.32 A
Iφ = L =
= 10 A
1.732
3
θ = cos−1 0.8 = 36.87°
200 V ∠0°
V
Zφ = φ =
= 20 Ω ∠36.87° = 16 Ω + j12 Ω
Iφ 10 A ∠ − 36.87°
41.
PT =
3 ELIL cos θ
1200 W = 3 (208 V)IL(0.6) ⇒ IL = 5.55 A
V
208 V
= 120.1 V
Vφ = L =
3 1.732
θ = cos−1 0.6 = 53.13° (leading)
120.1 V ∠0°
V
Zφ = φ =
= 21.64 Ω ∠−53.13° = 12.98 Ω − j 17.31 Ω
Iφ 5.55 A ∠53.13°
R
XC
CHAPTER 23
327
42.
Δ:
Zφ = 15 Ω + j20 Ω = 25 Ω∠53.13°
Vφ
Iφ =
Zφ
=
125 V
=5A
25 Ω
PT = 3Iφ2 Rφ = 3(5 A)2 15 Ω = 1125 W
QT = 3Iφ2 X φ = 3(5 A)2 20 Ω = 1500 VAR(L)
Y:
Vφ = VL/ 3 = 125 V/1.732 = 72.17 V
Zφ = 3 Ω − j4 Ω = 5 Ω∠−53.13°
Iφ =
V φ 72.17 V
=
= 14.43 A
5Ω
Zφ
PT = 3Iφ2 Rφ = 3(14.43 A)2 3 Ω = 1874.02 W
QT = 3Iφ2 X φ = 3(14.43 A)2 4 Ω = 2498.7 VAR
PT = 1125 W + 1874.02 W = 2999.02 W
QT = 1500 VAR(L) − 2498.7 VAR(C) = 998.7 VAR(C)
ST =
Fp =
43.
a.
c.
d.
e.
PT2 + QT2 = 3161 VA
PT 2999.02 W
=
= 0.949 (leading)
ST
3161 VA
Eφ =
16 kV
= 9,237.6 V
3
b.
IL = Iφ = 80 A
1200 kW
= 400 kW
3
P4Ω = (80 A)24 Ω = 25.6 kW
PT = 3Pφ = 3(25.6 kW + 400 kW) = 1276.8 kW
Pφ L =
Fp =
PT
, ST =
ST
Fp =
1, 276.8 kW
= 0.576 lagging
2, 217.025 kVA
3 VLIL =
3 (16 kV)(80 A) = 2,217.025 kVA
θL = cos−1 0.576 = 54.83° (lagging)
E AN∠0° ⇒ 80A ∠ − 54.83°
IAa =
Z T ∠54.83°
given
↑
for entire load
328
CHAPTER 23
f.
Van = EAN − IAa(4 Ω + j20 Ω)
= 9237.6 V ∠0° − (80 A ∠−54.83°)(20.396 Ω ∠78.69°)
= 9237.6 V ∠0° − 1631.68 V ∠23.86°
= 9237.6 V − (1492.22 V + j660 V)
= 7745.38 V − j660 V
= 7773.45 V ∠−4.87°
Zφ =
h.
Fp(entire system) = 0.576 (lagging)
Fp(load) = 0.643 (lagging)
−
1276.8 kW − 3(25.6 kW)
= P o = P i P lost =
= 0.9398 ⇒ 93.98%
1276.8 kW
Pi
Pi
i.
44.
V an 7773.45 V ∠ − 4.87°
=
= 97.168 Ω ∠49.95°
80 A ∠ − 54.83°
I Aa
= 62.52 Ω + j 74.38 Ω
R
XC
g.
a.
−
b.
Vφ =
220 V
= 127.02 V, Zφ = 10 Ω − j10 Ω = 14.14 Ω∠−45°
3
127.02 V
V
Iφ = φ =
= 8.98 A
Z φ 14.14 Ω
PT = 3Iφ2 Rφ = 3(8.98 A)2 10 Ω = 2419.2 W
Each wattmeter:
2419.2 W
= 806.4 W
3
45.
b.
PT = 5899.64 W, Pmeter = 1966.55 W
46.
a.
−
b.
PT = P + Ph = 85 W + 200 W = 285 W
c.
0.2 ⇒ P = 0.5
Ph
100 W
Ph = P =
= 200 W
0.5
0.5
PT = Ph − P = 200 W − 100 W = 100 W
CHAPTER 23
329
48.
a.
b.
c.
208 V ∠0°
Iab = E AB =
= 20.8 A ∠0°
R∠0° 10 Ω ∠0°
208 V ∠ − 120°
Ibc = E BC =
= 20.8 A ∠−120°
R∠0°
10 Ω ∠0°
208 V ∠120°
Ica = ECA =
= 20.8 A ∠120°
R∠0°
10 Ω ∠0°
IAa + Ica − Iab = 0
IAa = Iab − Ica
= 20.8 A ∠0° − 20.8 A ∠120°
= 20.8 A − (−10.4 A + j18.01 A)
= 31.2 A − j18.01 A
= 36.02 A ∠−30°
IBb + Iab − Ibc = 0
IBb = Ibc − Iab
= 20.8 A ∠−120° − 20.8 A ∠0°
= (−10.4 A − j18.01 A) − 20.8 A
= −31.2 A − j18.01 A
= 36.02 A ∠−150°
ICc + Ibc − Ica = 0
ICc = Ica − Ibc
= 20.8 A ∠120° − 20.8 A ∠−120°
= (−10.4 A + j18.01 A) − (−10.4 A − j18.01 A)
= −10.4 A + 10.4 A + j18.01 A + j18.01 A
= 32.02 A ∠90°
P1 = VacIAa cos IVAaca , Vac = Vca ∠ θ − 180° = 208 V ∠120° − 180°
= 208 V ∠−60°
IAa = 36.02 A ∠−30°
= (208 V)(36.02 A) cos 30°
= 6488.4 W
P2 = VbcIBb cos IVBbbc , Vbc = 208 V ∠−120°, IBb = 36.02 A ∠−150°
= (208 V)(36.02 A) cos 30°
= 6488.4 W
d.
330
PT = P1 + P2 = 6488.4 W + 6488.4 W
= 12,976.8 W
CHAPTER 23
49.
a.
b.
c.
Vφ = Eφ = E L = 120.09 V
3
V an 120.09 V
=
= 8.49 A
Z an 14.142 Ω
120.09 V
V
Ibn = bn =
= 7.08 A
Z bn 16.971 Ω
120.09 V
V
= 42.47 A
Icn = cn =
Z cn 2.828 Ω
Ian =
2
2
2
PT = I an
10 Ω + I bn
12 Ω + I cn
2Ω
= (8.49 A)2 10 Ω + (7.08 A)2 12 Ω + (42.47 A)2 2 Ω
= 720.80 W + 601.52 W + 3.61 kW
= 4.93 kW
QT = PT = 4.93 kVAR(L)
ST =
Fp =
d.
e.
50.
PT2 + Q T2 = 6.97 kVA
PT
= 0.707 (lagging)
ST
Ean = 120.09 V∠−30°, Ebn = 120.09 V∠−150°, Ecn = 120.09 V∠90°
120.09 V ∠ − 30° 120.09 V ∠ − 30°
= 8.49 A ∠−75°
Ian = E an =
=
10 Ω + j10 Ω
14.142 Ω ∠45°
Z an
120.09 V ∠ − 150° 120.09 V ∠ − 150°
Ibn = Ebn =
=
= 7.08 A ∠−195°
12 Ω + j12 Ω
16.971 Ω ∠45°
Z bn
120.09 V ∠90° 120.09 V ∠90°
= 42.47 A ∠45°
Icn = Ecn =
=
2 Ω + j2 Ω
2.828 Ω ∠45°
Z cn
IN = Ian + Ibn + Icn
= 8.49 A ∠−75° + 7.08 A ∠−195° + 42.47 A∠45°
= (2.02 A − j8.20 A) + (−6.84 A + j1.83 A) + (30.30 A + j30.30 A)
= 25.66 A − j23.93 A
= 35.09 A ∠−43.00°
Z1 = 12 Ω − j16 Ω = 20 Ω ∠−53.13°, Z2 = 3 Ω + j4 Ω = 5 Ω ∠53.13°
Z3 = 20 Ω ∠0°
EAB = 200 V∠0°, EBC = 200 V ∠−120°, ECA = 200 V ∠120°
ZΔ = Z1Z2 + Z1Z3 + Z2Z3
= (20 Ω ∠−53.13°)(5 Ω ∠53.13°) + (20 Ω ∠−53.13°)(20 Ω ∠0°)
+ (5 Ω ∠53.13°)(20 Ω ∠0°)
= 100 Ω ∠0° + 400 Ω ∠−53.13° + 100 Ω ∠53.13°
= 100 Ω + (240 Ω − j320 Ω) + (60 Ω + j80 Ω)
= 400 Ω − j240 Ω
= 466.48 Ω ∠−30.96°
CHAPTER 23
331
Ian =
E AB Z3 − ECA Z 2 (200 V ∠0°)(20 Ω ∠0°) − (200 V ∠120°)(5 Ω ∠53.13°)
=
ZΔ
ZΔ
4000 A ∠0° − 1000 A ∠173.13°
= 10.71 A ∠29.59°
466.48 ∠ − 30.96°
E Z − E AB Z3 (200 V ∠ − 120°)(20 Ω ∠ − 53.13°) − (200 V ∠0°)(20 Ω ∠0°)
Ibn = BC 1
=
ZΔ
ZΔ
4000 A ∠ − 173.13° − 4000 A ∠0°
= 17.12 A ∠−145.61°
=
466.48 ∠ − 30.96°
E Z − E BC Z1 (200 V ∠120°)(5 Ω∠53.13°) − (200 V ∠ − 120°)(20 Ω∠ − 53.13°)
Icn = CA 2
=
ZΔ
ZΔ
1000 A ∠173.13° − 4000 A ∠ − 173.13°
= 6.51 A ∠42.32°
=
466.48 ∠ − 30.96°
2
2
2
PT = I an
12 Ω + I bn
4 Ω + I cn
20 Ω
= 1376.45 W + 1172.38 W + 847.60 W = 3396.43 W
2
2
QT = I an
16 Ω + I bn
3 Ω = 1835.27 VAR(C) + 879.28 VAR(L) = 955.99 VAR(C)
=
ST =
Fp =
332
PT2 + Q T2 = 3508.40 VA
PT
3396.43 W
=
= 0.968 (leading)
ST 3508.40 VA
CHAPTER 23
Chapter 24
1.
a.
positive-going
d.
Amplitude = 8 V − 2 V = 6 V
e.
2.
c.
tp = 0.2 ms
V1 −V 2
× 100%
V
8V + 7.5 V
= 7.75 V
V=
2
8 V − 7.5 V
% tilt =
× 100% = 6.5%
7.75 V
negative-going
d.
−8 mV ( from base line level)
f.
Vb = 2 V
% tilt =
a.
e.
3.
b.
b.
c.
+7 mV
3 μs
−8 mV − 7 mV −15 mV
=
= −7.5 mV
2
2
−
−8 mV − (−7 mV)
× 100%
% Tilt = V 1 V 2 × 100% =
−7.5 mV
V
−1 mV
× 100% = 13.3%
=
−7.5 mV
V=
T = 15 μs − 7 μs = 8 μs
1
1
prf = =
= 125 kHz
T 8 μs
3 μs
tp
× 100% =
× 100% = 37.5%
T
8 μs
g.
Duty cycle =
a.
positive-going
d.
Amplitude = (30 − 10)mV = 20 mV
e.
% tilt =
b.
Vb = 10 mV
c.
⎛8⎞
tp = ⎜ ⎟ 4 ms = 3.2 ms
⎝ 10 ⎠
V1 −V 2
× 100%
V
30 mV + 28 mV
V=
= 29 mV
2
30 mV − 28 mV
% tilt =
× 100% ≅ 6.9%
29 mV
CHAPTER 24
333
4.
tr ≅ (0.2 div.)(2 ms/div.) = 0.4 ms
tf ≅ (0.4 div.)(2 ms/div.) = 0.8 ms
5.
tilt =
V +V
V1 − V2
= 0.1 with V = 1 2
V
2
Substituting V into top equation,
0.95 V1
V1 − V2
or V2 = 0.905(15 mV) = 13.58 mV
= 0.1 leading to V2 =
V1 + V2
1.05
2
6.
7.
a.
tr = 80% of straight line segment
= 0.8(2 μs) = 1.6 μs
b.
tf = 80% of 4 μs interval
= 0.8(4 μs) = 3.2 μs
c.
At 50% level (10 mV)
tp = (8 − 1)μs = 7 μs
d.
prf =
a.
T = (4.8 − 2.4)div. [50 μ s/div.] = 120 μs
c.
8.
334
1
1
= 50 kHz
=
T 20 μ s
b.
f=
1
1
= 8.33 kHz
=
T 120 μs
Maximum Amplitude: (2.2 div.)(0.2 V/div.) = 0.44 V = 440 mV
Minimum Amplitude: (0.4 div.)(0.2 V/div.) = 0.08 V = 80 mV
T = (3.6 − 2.0)ms = 1.6 ms
1
1
= 625 Hz
prf = =
T 1.6 ms
tp
0.2 ms
× 100% = 12.5%
Duty cycle = × 100% =
1.6 ms
T
CHAPTER 24
9.
T = (15 − 7)μs = 8 μs
1
1
prf = =
= 125 kHz
T 8 μs
tp
(20 − 15) μ s
5
× 100% = × 100% = 62.5%
Duty cycle = × 100% =
T
8 μs
8
10.
T = (3.6 div.)(2 ms/div.) = 7.2 ms
1
1
= 138.89 Hz
prf = =
T 7.2 ms
tp
1.6 div.
× 100% = 44.4%
Duty cycle = × 100% =
T
3.6 div.
11.
a.
T = (9 − 1)μs = 8 μs
c.
prf =
d.
e.
12.
b.
tp = (3 − 1)μs = 2 μs
1
1
=
= 125 kHz
T 8 μs
Vav = (Duty cycle)(Peak value) + (1 − Duty cycle)(Vb)
tp
2 μs
× 100% = 25%
Duty cycle = × 100% =
8 μs
T
Vav = (0.25)(6 mV) + (1 − 0.25)(−2 mV)
= 1.5 mV − 1.5 mV = 0 V
or
(2 μ s)(6 mV) − (2 μ s)(6 mV)
=0V
Vav =
8μs
Veff =
(36 × 10−6 )(2 μs) + (4 × 10−6 )(6 μs)
= 3.46 mV
8 μs
Eq. 24.5 cannot be applied due to tilt in the waveform.
(Method of Section 13.6)
Between 2 and 3.6 ms
1
(3.4 ms − 2 ms)(2 V) + (3.6 ms − 3.4 ms)(7.5 V) + (3.6 ms − 3.4 ms)(0.5 V)
2
Vav =
3.6 ms − 2 ms
1
(1.4 ms)(2 V) + (0.2 ms)(7.5 V) + (0.2 ms)(0.5 V)
2
=
1.6 ms
2.8 V + 1.5 V + 0.05 V
= 2.719 V
=
1.6
CHAPTER 24
335
13.
Ignoring tilt and using 20 mV level to define tp
tp = (2.8 div. − 1.2 div.)(2 ms/div.) = 3.2 ms
T = (at 10 mV level) = (4.6 div. − 1 div.)(2 ms/div.) = 7.2 ms
3.2 ms
tp
Duty cycle = × 100% =
× 100% = 44.4%
7.2 ms
T
Vav = (Duty cycle)(peak value) + (1 − Duty cycle)(Vb)
= (0.444)(30 mV) + (1 − 0.444)(10 mV)
= 13.320 mV + 5.560 mV
= 18.88 mV
14.
Vav = (Duty cycle)(Peak value) + (1 − Duty cycle)(Vb)
tp
(decimal form)
Duty cycle =
T
(8 − 1) μ s
=
= 0.35
20 μ s
Vav = (0.35)(20 mV) + (1 − 0.35)(0)
= 7 mV + 0
= 7 mV
15.
Using methods of Section 13.8:
A1 = b1h1 = [(0.2 div.)(50 μs/div.)][(2 div.)(0.2 V/div.)] = 4 μsV
A2 = b2h2 = [(0.2 div.)(50 μs/div.)][(2.2 div.)(0.2 V/div.)] = 4.4 μsV
A3 = b3h3 = [(0.2 div.)(50 μs/div.)][(1.4 div.)(0.2 V/div.)] = 2.8 μsV
A4 = b4h4 = [(0.2 div.)(50 μs/div.)][(1 div.)(0.2 V/div.)] = 2.0 μsV
A5 = b5h5 = [(0.2 div.)(50 μs/div.)][(0.4 div.)(0.2 V/div.)] = 0.8 μsV
Vav =
16.
Using the defined polarity of Fig. 24.57 for υC, Vi = −5 V, Vf = +20 V
and τ = RC = (10 kΩ)(0.02 μF) = 0.2 ms
a.
336
(4 + 4.4 + 2.8 + 2.0 + 0.8) μ sV
= 117 mV
120 μ s
υC = Vi + (Vf − Vi)(1 − e−t/τ)
= −5 + (20 − (−5))(1 − e−t/0.2 ms)
= −5 + 25(1 − e−t/0.2 ms)
= −5 + 25 − 25e−t/0.2 ms
υC = 20 V − 25 Ve−t/0.2 ms
CHAPTER 24
b.
c.
Ii = 0
iC =
d.
17.
E − υC 20 V − ⎡⎣ 20 V − 25 V e
=
R
10 k Ω
υC = Vi + (Vf − Vi)(1 − e−t/RC)
= 8 + (4 − 8)(1 − e−t/20 ms)
= 8 − 4(1 − e−t/20 ms)
= 8 − 4 + 4e−t/20 ms
= 4 + 4e−t/20 ms
υC = 4 V(1 + e−t/20 ms)
− t / 0.2 ms
⎤
⎦ = 2.5 mAe−t/0.2 ms
τ = RC = (2 kΩ)(10 μF)
= 20 ms
18.
Vi = 10 V, Vf = 2 V, τ = RC = (1 kΩ)(1000 μF) = 1 s
υC = Vi + (Vf − Vi)(1 − e−t/τ)
= 10 V + (2 V − 10 V)(1 − e−t)
= 10 − 8(1 − e−t)
= 10 − 8 + 8e−t
υC = 2 V+ 8 Ve−t
19.
Vi = 10 V, Ii = 0 A
CHAPTER 24
Using the defined direction of iC
−(10 V − 2 V) −t/τ
e
iC =
1kΩ
τ = RC = (1 kΩ)(1000 μF) = 1 s
8 V −t
iC = −
e
1kΩ
and iC = −8mAe−t
337
20.
τ = RC = (5 kΩ)(0.04 μF) = 0.2 ms (throughout)
υC = E(1 − e−t/τ) = 20 V(1 − e−t/0.2 ms)
(Starting at t = 0 for each plot)
a.
T=
1
1
=
= 2 ms
f 500 Hz
T
= 1 ms
2
5τ = 1 ms =
b.
T=
1
1
= 10 ms
=
f 100 Hz
T
= 5 ms
2
5τ = 1 ms =
c.
T=
T
2
1⎛T ⎞
⎜ ⎟
5⎝ 2 ⎠
1
1
= 0.2 ms
=
f 5 Hz
T
= 0.1 ms
2
⎛T ⎞
5τ = 1 ms = 10 ⎜ ⎟
⎝2⎠
21.
The mathematical expression for iC is the same for each frequency!
τ = RC = (5 kΩ)(0.04 μF) = 0.2 ms
20 V −t / 0.2 ms
= 4 mAe−t/0.2 ms
and iC =
e
5 kΩ
a.
T=
1
T
= 2 ms, = 1 ms
500 Hz
2
5τ = 5(0.2 ms) = 1 ms =
338
b.
T=
c.
T=
T
2
1
T
= 10 ms,
= 5 ms
100 Hz
2
1⎛T ⎞
5τ = 1 ms = ⎜ ⎟
5⎝ 2 ⎠
T
1
= 0.1 ms
= 0.2 ms,
2
5000 Hz
⎛T ⎞
5τ = 1 ms = 10 ⎜ ⎟
⎝2⎠
CHAPTER 24
22.
τ = 0.2 ms as above
1
= 2 ms
T=
500 Hz
T
5τ = 1 ms =
2
T
0 → : υC = 20 V(1 − e−t/0.2 ms)
2
T
→ T: Vi = 20 V, Vf = −20 V
2
υC = Vi + (Vf − Vi)(1 − e−t/τ)
= 20 + (−20 − 20)(1 − e−t/0.2 ms)
= 20 − 40(1 − e−t/0.2 ms)
= 20 − 40 + 40e−t/0.2 ms
υC = −20 V+ 40 Ve−t/0.2 ms
T→
23.
3
T : Vi = −20 V, Vf = +20 V
2
υC = Vi + (Vf − Vi)(1 − e−t/τ)
= −20 + (20 − (−20))(1 − e−t/τ)
= −20 + 40(1 − e−t/τ)
= −20 + 40 − 40e−t/τ
υC = 20 V − 40 Ve−t/0.2 ms
υC = Vi + (Vf − Vi)(1 − e−t/RC)
Vi = 20 V, Vf = 20 V
υC = 20 + (20 − 20)(1 − e−t/RC)
T
= 20 V (for 0 → )
2
For
T
→ T, υi = 0 V and υC = 20 Ve−t/τ
2
τ = RC = 0.2 ms
with
For T →
T
T
= 1 ms and 5τ =
2
2
3
T , υi = 20 V
2
υC = 20 V(1 − e−t/τ)
For
3
T → 2 T, υ i = 0 V
2
υC = 20 Ve−t/τ
CHAPTER 24
339
24.
25.
τ = RC = 0.2 ms
T
5τ = 1 ms =
2
Vi = −10 V, Vf = +20 V
T
0→ :
2
υC = Vi + (Vf − Vi)(1 − e−t/τ)
= −10 + (20 − (−10))(1 − e−t/τ)
= −10 + 30(1 − e−t/τ)
= −10 + 30 − 30e−t/τ
υC = +20 V − 30 Ve−t/0.2 ms
XC =
Zs:
CT = 18 pF + 9 pF = 27 pF
1
1
=
= 0.589 MΩ
XC =
2π fC T 2π (10 kHz)(27 pF)
Zs =
Vscope =
26.
Vi = 20 V, Vf = 0 V
υC = 20 Ve−t/0.2 ms
1
1
= 5.31 MΩ
=
2π fC 2π (10 kHz)(3 pF)
(9 M Ω ∠0°)(5.31 M Ω ∠ − 90°)
= 4.573 MΩ ∠−59.5°
Zp =
9 M Ω − j 5.31 M Ω
Zp:
θ Zs
T
→ T:
2
(1 M Ω ∠0°)(0.589 M Ω ∠ − 90°)
= 0.507 MΩ ∠−59.5°
1 M Ω − j 0.589 M Ω
(0.507 M Ω ∠ − 59.5°)(100 V ∠0°)
ZsVi =
Z s + Z p (0.257 M Ω − j 0.437 M Ω) + (2.324 M Ω − j 3.939 M Ω)
50.7 × 106 V ∠ − 59.5°
1
(100 V ∠0°)
= 10 V ∠0° =
6
10
5.07 × 10 ∠ − 59.5°
= θ Z p = −59.5°
=
Zp: XC =
1
1
= 3.333 MΩ
=
5
ωC (10 rad/s)(3 pF)
(9 M Ω ∠0°)(3.333 M Ω)
= 3.126 MΩ ∠−69.68°
9 M Ω − j 3.333 M Ω
1
1
=
XC =
= 0.370 MΩ
5
ωC (10 rad/s)(27 pF)
(1 M Ω ∠0°)(0.370 M Ω ∠ − 90°)
= 0.347 MΩ ∠−69.68°
Zs =
1 M Ω − j 0.370 M Ω
✓ θ Z p =θ Z s
Zp =
Zs:
340
CHAPTER 24
Vscope =
(0.347 M Ω ∠ − 69.68°)(100 V ∠0°)
ZsVi =
Z s + Z p (0.121 M Ω − j 0.325 M Ω) + (1.086 M Ω − j 2.931 M Ω)
34.70 × 106 V ∠ − 69.68°
3.470 × 106 ∠ − 69.68°
1
≅ 10 V ∠0° =
(100 V ∠0°)
10
=
CHAPTER 24
341
Chapter 25
1.
2.
I:
a.
no
b.
no
c.
yes
d.
no
e.
yes
II:
a.
yes
b.
yes
c.
yes
d.
yes
e.
no
III:
a.
yes
b.
yes
c.
no
d.
yes
e.
yes
IV:
a.
no
b.
no
c.
yes
d.
yes
e.
yes
b.
i=
c.
2I m ⎛ 2
2
2
⎞
⎜1+ cos(2ωt − 90°) − cos(4ωt − 90°) + cos(6ωt − 90°) + … ⎟
π ⎝ 3
15
35
⎠
⎡ π⎤
⎢1 − 4 ⎥
π
⎣
⎦
2
2
⎡ π 2
⎤
i = 2 I m ⎢1 − + cos(2ωt − 90°) −
cos(4ωt − 90°) +
cos(6ωt − 90°) + …⎥
15
35
π ⎣ 4 3
⎦
2Im
−
Im 2Im
=
π
2
d.
2
2
2
⎡ π
⎤
i = −2 I m ⎢1 − +
cos (2ωt − 90°) −
cos (4ωt − 90°) +
cos (6ωt − 90°) + …⎥
15
35
π ⎣ 4 3
⎦
3.
342
a.
υ = −4 + 2 sin α
b.
υ = (sin α)2
CHAPTER 25
c.
4.
i = 2 − 2 cos α
a.
b.
CHAPTER 25
343
5.
a.
b.
c.
344
CHAPTER 25
6.
a.
Vav = 100 V
Veff =
Ieff =
8.
(50 V) 2 + (25 V) 2
= 107.53 V
2
Iav = 3 A
b.
7.
(100 V) 2 +
(3 A) 2 +
(2 A) 2 + (0.8 A) 2
= 3.36 A
2
a.
Veff =
(20 V) 2 + (15 V) 2 + (10 V) 2
= 19.04 V
2
b.
Ieff =
(6 A) 2 + (2 A) 2 + (1 A) 2
= 4.53 A
2
PT = V0I0 + V1I1 cos θ1 + … + VnIn cos θn
(50 V)(2 A)
(25 V)(0.8 A)
cos 53° +
cos 70°
= (100 V)(3 A) +
2
2
= 300 + (50)(0.6018) + (10)(0.3420)
= 333.52 W
(20 V)(6 A)
(15 V)(2 A)
(10 V)(1 A)
cos 20° +
cos 30° +
cos 60°
2
2
2
= 60(0.9397) + 15(0.866) + 5(0.5)
= 71.87 W
9.
P=
10.
a.
DC: E = 18 V, Io =
b.
Ieff =
E 18 V
= 1.5 A
=
R 12 Ω
ω = 400 rad/s:
XL = ωL = (400 rad/s)(0.02 H) = 8 Ω
Z = 12 Ω + j8 Ω = 14.42 Ω ∠33.69°
30 V/ 2 ∠0°
2.08 A
E
I= =
=
∠−33.69°
Z 14.42 Ω ∠33.69°
2
⎛ 2.08 ⎞
i = 1.5 + 2 ⎜
⎟ sin(400t − 33.69°)
⎝ 2 ⎠
i = 1.5 + 2.08 sin(400t − 33.69°)
CHAPTER 25
(1.5 A) 2 +
(2.08 A) 2
= 2.10 A
2
345
c.
11.
346
⎛ 2.08 A
⎞
DC: υR = E = 18 V, VR = ⎜
∠ − 33.69° ⎟ (12 Ω ∠0°)
2
⎝
⎠
24.96 V
=
∠ − 33.69°
2
⎛ 24.96 ⎞
υR = 18 + 2 ⎜
⎟ sin(400t − 33.69°)
⎝ 2 ⎠
υR = 18 + 24.96 sin(400t − 33.69°)
(24.96 V ) 2
= 25.21 V
2
d.
2
V Reff = (18 V ) +
e.
DC: VL = 0 V
f.
2
V Leff = 0 +
g.
2
P = I eff
R = (2.101 A)2 12 Ω = 52.97 W
a.
DC: IDC =
b.
Ieff =
c.
υR = iR = i(12 Ω)
= 24 + 24.96 sin(400t − 33.69°) + 6 sin(800t − 53.13°)
d.
Veff =
⎛ 2.08 A
⎞
ω = 400 rad/s: VL = ⎜
∠ − 33.69° ⎟ (8 Ω ∠90°)
2
⎝
⎠
16.64 A
∠56.31°
=
2
υL = 0 + 16.64 sin(400t + 56.31°)
(16.64 V) 2
= 11.77 V
2
24 V
=2A
12 Ω
ω = 400 rad/s:
Z = 12 Ω + j(400 rad/s)(0.02 H) = 12 Ω + j8 Ω = 14.422 Ω ∠33.69°
30 V ∠0°
= 2.08 A ∠−33.69° (peak values)
I=
14.422 Ω ∠33.69°
ω = 800 rad/s:
Z = 12 Ω + j(800 rad/s)(0.02 H) = 12 Ω + j16 Ω = 20 Ω ∠53.13°
10 V ∠0°
I=
= 0.5 A ∠−53.13° (peak values)
20 Ω ∠53.13°
i = 2 + 2.08 sin(400t − 33.69°) + 0.5 sin(800t − 53.13°)
(2 A) 2 +
(2.08 A) 2 + (0.5 A 2 )
= 2.51 A
2
(24 V) 2 +
(24.96 V) 2 + (6 V) 2
= 30.09 V
2
CHAPTER 25
12.
e.
DC: VL = 0 V
ω = 400 rad/s: VL = (2.08 A ∠−33.69°)(8 Ω ∠90°)
= 16.64 V ∠56.31°
ω = 800 rad/s: VL = (0.5 A ∠−53.13°)(16 Ω ∠90°)
= 8 V ∠36.87°
υL = 0 + 16.64 sin(400t + 56.31°) + 8 sin(800t + 36.87°)
f.
Veff =
g.
2
R = (2.508 A)2 12 Ω = 75.48 W
PT = I eff
a.
DC: I = −
b.
c.
d.
(0) 2 +
(16.64 V) 2 + (8 V) 2
= 13.06 V
2
60 V
= −5A
12 Ω
ω = 300 rad/s: XL = ωL = (300 rad/s)(0.02 H) = 6 Ω
Z = 12 Ω + j16 Ω = 13.42 Ω ∠26.57°
E = (0.707)(20 V) ∠0° = 14.14 V ∠0°
E
14.14 V ∠0°
I= =
= 1.054 A ∠−26.57°
Z 13.42 Ω ∠26.57°
ω = 600 rad/s: XL = ωL = (600 rad/s)(0.02 H) = 12 Ω
Z = 12 Ω + j12 Ω = 16.97 Ω ∠45°
E = −(0.707)(10 V) ∠0° = −7.07 V ∠0°
E
7.07 V ∠0°
= −0.417 A ∠−45°
I= =−
Z
16.97 Ω ∠45°
i = −5 + (1.414)(1.054)sin(300t − 26.57°) − (1.414)(0.417)sin(600t − 45°)
i = −5 + 1.49 sin(300t − 26.57°) − 0.59 sin(600t − 45°)
Ieff =
(10 A) 2 +
(1.49 A) 2 + (0.59 A) 2
= 10.06 A
2
DC: V = IR = (−5 A)(12 Ω) = −60 V
ω = 300 rad/s: VR = (1.054 A ∠−26.57°)(12 Ω ∠0°)
= 12.648 V ∠−26.57°
ω = 600 rad/s: VR = (−0.417 A ∠−45°)(12 Ω ∠0°)
= −5 V ∠−45°
υR = −60 + (1.414)(12.648)sin(300t − 26.57°) − (1.414)(5)sin(600t − 45°)
υR = −60 + 17.88 sin(300t − 26.57°) − 7.07 sin(600t − 45°)
2
V R eff = (60 V ) +
CHAPTER 25
(17.88 V ) 2 + (7.07 V ) 2
= 61.52 V
2
347
e.
f.
13.
DC: VL = 0 V
ω = 300 rad/s:
VL = (1.054 A ∠−26.57°)(6 Ω ∠90°) = 6.324 V ∠63.43°
ω = 600 rad/s:
VL = (−0.417 A ∠−45°)(6 Ω ∠90°) = −2.502 V ∠45°
υL = 0 + (1.414)(6.324)sin(300t + 63.43°) − (1.414)(2.502)sin(600t + 45°)
υL = 8.94 sin(300t + 63.43°) − 3.54 sin(600t + 45°)
(8.94 V ) 2 + (3.54 V ) 2
= 6.8 V
V Leff =
2
g.
2
R = (10.06 A)2 12 Ω = 1214.44 W
P = I eff
a.
DC: I = 0 A
1
1
= 20 Ω
=
ωC (400 rad/s)(125 μ F)
Z = 15 Ω − j20 Ω = 25 Ω ∠−53.13°
E = (0.707)(30 V) ∠0° = 21.21 V ∠0°
E
21.21 V ∠0°
I= =
= 0.848 A ∠53.13°
Z 25 Ω ∠ − 53.13°
i = 0 + (1.414)(0.848)sin(400t + 53.13°)
i = 1.2 sin(400t + 53.13°)
ω = 400 rad/s:
(1.2 A) 2
= 0.85 A as above
2
b.
Ieff =
c.
DC: VR = 0 V
ω = 400 rad/s: VR = (0.848 A ∠53.13°)(15 Ω ∠0°) = 12.72 V ∠53.13°
υR = 0 + (1.414)(12.72)sin(400t + 53.13°)
υR = 18 sin(400t + 53.13°)
d.
348
XC =
(18 V ) 2
= 12.73 V
V R eff =
2
e.
DC: VC = 18 V
ω = 400 rad/s: VC = (0.848 A ∠53.13°)(20 Ω ∠−90°)
= 16.96 V ∠−36.87°
υC = 18 + (1.414)(16.96)sin(400t − 36.87°)
υC = 18 + 23.98 sin(400t − 36.87°)
f.
2
V C eff = (18 V ) +
g.
2
R = (0.848 A)2 15 Ω = 10.79 W
P = I eff
(23.98 V ) 2
= 24.73 V
2
CHAPTER 25
14.
a.
400
400
cos 2ωt −
cos 4ωt
π
3π
15 π
= 63.69 + 42.46 sin(2ωt + 90°) − 8.49 sin(4ωt + 90°)
ω = 377 rad/s:
e = 63.69 + 42.46 sin(754t + 90°) − 8.49 sin(1508t + 90°)
DC: XL = 0 ∴ VL = 0 V
1
1
ω = 754 rad/s: XC =
=
= 1330 Ω
ωC (754 rad/s)(1 μ F)
e=
200
+
XL = ωL = (754 rad/s)(0.1 H) = 75.4 Ω
Z′ = (1 kΩ ∠0°) || 75.4 Ω ∠90° = 75.19 Ω ∠85.69°
E = (0.707)(42.46 V) ∠90° = 30.02 V ∠90°
(75.19 Ω ∠85.69°)(30.02 V ∠90°)
Z′(E)
=
= 1.799 V ∠−94.57°
Vo =
Z′ + ZC 75.19 Ω ∠85.69° + 1330 Ω ∠ − 90°
ω = 1508 rad/s:
XC =
1
1
= 6631.13 Ω
=
ωC (1508 rad/s)(1 μ F)
XL = ωL = (1508 rad/s)(0.1 H) = 150.8 Ω
Z′ = (1 kΩ ∠0°) || 150.8 Ω ∠90° = 149.12 Ω ∠81.42°
E = (0.707)(8.49 V) ∠90° = 6 V ∠90°
Z′(E)
(149.12 Ω ∠81.42°)(6 V ∠90°)
=
Vo =
′
Z + ZC 149.12 Ω ∠81.42° + 6631.13 Ω ∠ − 90°
= 1.73 V ∠−101.1°
υo = 0 + 1.414(1.799)sin(754t − 94.57°) − 1.414(1.73)sin(1508t − 101.1°)
υo = 2.54 sin(754t − 94.57°) − 2.45 sin(1508t − 101.1°)
15.
b.
Voeff =
c.
P=
(2.54 V ) 2 + (2.45 V ) 2
= 2.50 V
2
( V eff ) 2 (2.50 V ) 2
= 6.25 mW
=
R
1kΩ
i = 0.318Im + 0.500 Im sin ωt − 0.212Im cos 2ωt − 0.0424Im cos 4ωt + … (Im = 10 mA)
i = 3.18 × 10−3 + 5 × 10−3 sin ωt − 2.12 × 10−3 sin(2ωt + 90°)
− 0.424 × 10−3 sin(4ωt + 90°) + …
i ≅ 3.18 × 10−3 + 5 × 10−3 sin ωt − 2.12 × 10−3 sin(2ωt + 90°)
DC: Io = 0 A, Vo = 0 V
ω = 377 rad/s;
XL = ωL = (377 rad/s)(1.2 mH) = 0.452 Ω
1
1
XC =
=
= 13.26 Ω
ωC ( 377 rad/s ) (200 μ F)
Z′ = 200 Ω − j13.26 Ω = 200.44 Ω ∠−3.79°
I = (0.707)(5 × 10−3)A ∠0° = 3.54 mA ∠0°
ZLI
(0.452 Ω ∠90°)(3.54 mA ∠0°)
=
Io =
= 7.98 μA ∠93.66°
′
j 0.452 Ω + 200 Ω − j13.26 Ω
ZL + Z
CHAPTER 25
349
Vo = (7.98 μA ∠93.66°)(200 Ω ∠0°) = 1.596 mV ∠93.66°
ω = 754 rad/s: XL = ωL = (754 rad/s)(1.2 mH) = 0.905 Ω
1
1
XC =
= 6.63 Ω
=
ωC (754 rad/s)(200 μ F)
Z′ = 200 Ω − j6.63 Ω = 200.11 Ω ∠−1.9°
I = (0.707)(2.12 mA) ∠90° = 1.5 mA ∠90°
(0.905 Ω ∠90°)(1.5 mA ∠90°)
Z LI
Io =
=
= 6.8 μA ∠181.64°
j 0.905 Ω + 200 Ω − j 6.63 Ω
Z L + Z′
Vo = (6.8 μA ∠181.64°)(200 Ω ∠0°) = 1.36 mA ∠181.64°
υo = 0 + (1.414)(1.596 × 10−3)sin(377t + 93.66°)
− (1.414)(1.360 × 10−3)sin(754t + 181.64°)
−3
υo = 2.26 × 10 sin(377t + 93.66°) + 1.92 × 10−3 sin(754t + 1.64°)
16.
a.
b.
17.
350
60 + 70 sin ωt + 20 sin(2ωt + 90°) + 10 sin(3ωt + 60°)
+20 + 30 sin ωt − 20 sin(2ωt + 90°) + 5 sin(3ωt + 90°)
DC: 60 + 20 = 80
ω:
70 + 30 = 100 ⇒ 100 sin ωt
2ω: 0
3ω: 10 ∠60° + 5∠90° = 5 + j8.66 + j5 = 5 + j13.66 = 14.55 ∠69.9°
Sum = 80 + 100 sin ωt + 14.55 sin(3ωt + 69.9°)
20 + 60 sin α + 10 sin(2α − 180°) + 5 sin(3α + 180°)
−5 + 10 sin α +
0
− 4 sin(3α − 30°)
DC: 20 − 5 = 15
α:
60 + 10 = 70 ⇒ 70 sin α
2α: 10 sin(2α − 180°)
3α: 5 ∠180° − 4 ∠−30° = −5 − [3.46 − j2] = −8.46 + j2
= 8.69 ∠166.7°
Sum = 15 + 70 sin α + 10 sin(2α − 180°) + 8.69 sin(3α + 166.7°)
iT = i1 + i2
= 10 + 30 sin 20t
− 0.5 sin(40t+ 90°)
+20 + 4 sin(20t + 90°) + 0.5 sin(40t+ 30°)
DC: 10 A + 20 A = 30 A
ω = 20 rad/s: 30 A ∠0° + 4 A ∠90° = 30 A + j4 A = 30.27 A ∠7.59°
ω = 40 rad/s: −0.5 A ∠90° + 0.5 A ∠30°
= −j0.5 A + 0.433 A + j0.25 A
= 0.433 A − j0.25 A = 0.5 A ∠−30°
iT = 30 + 30.27 sin(20t + 7.59°) + 0.5 sin(40t − 30°)
CHAPTER 25
18.
e = υ1 + υ2
= 20 − 200 sin 600t + 100 sin(1200t + 90°) + 75 sin 1800t
−10 + 150 sin(600t + 30°)
+0
+ 50 sin(1800t + 60°)
DC: 20 V − 10 V = 10 V
ω: 600 rad/s:
−200 V ∠0° + 150 V ∠30° = 102.66 V ∠133.07°
ω = 1200 rad/s: 100 sin(1200t + 90°)
ω = 1800 rad/s: 75 V ∠0° + 50 V ∠60° = 108.97 V ∠23.41°
e = 10 + 102.66 sin(600t + 133.07°) + 100 sin(1200t + 90°) + 108.97 sin(1800t + 23.41°)
CHAPTER 25
351
352
CHAPTER 25