Boylestad Introductory Circuit Analysis [13 ed.]
Instructor Solutions Manual for Boylestad Introductory Circuit Analysis 13th, DC: Chapters(1-11). AC: Chapters(12-25)
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Chapter 1 1.
-
2.
-
3.
-
4.
a. v = d/t, where v = speed, d = distance, t = time Given t = 2h, 15 min = Time = V=
𝑑 𝑡
2×60+15 h 60
135 60
=
ℎ = 2.25h
60.5 𝑚𝑖 ℎ 2.25
b. Speed in km/h V=
𝑑 𝑡
=
135 = 60
= 26.889 mph
60.5mi×
V = 43.264 km/h 5.
a. b.
6.
100 yd �
×
1 2.25ℎ
mph = (0.6)(160 km/h) = 96 mph km/h = (1.7)(70 mph) = 119 km/h 3𝑓𝑡 1 𝑚𝑖 �� � 1𝑦𝑑 5260𝑓𝑡
80𝑚𝑖 1ℎ 1 𝑚𝑖𝑛 �60𝑚𝑖𝑛� � 60𝑠 � ℎ
t=
1.609𝑘𝑚 1 𝑚𝑖
𝑑 𝑣
=
0.0568𝑚𝑖 0.0222 𝑚𝑖/𝑠
= 0.0568mi
= 0.222 mi/s
= 2.5586s
t = 2.56s 7.
8.
a.
= 139.33 ft/s
b.
t=
c.
u=
= 0.43 s = 40.91 mph
-
Chapter 1
1
9.
-
10.
-
11.
MKS, CGS, °C =
= 20°
SI: K = 273.15 + °C = 273.15 + 20 = 293.15 12.
4000 J ×
13.
a. b. c. d.
14.
15.
16.
17.
18.
2
0.7378𝑓𝑡−𝑙𝑏 1𝐽
= 2951.2 ft-lbs
é2.2 lbs ù 70.8 kg ê ú = 155.76 lbs ë kg û é 1 kg ù 145 lbs ê ú = 65.91 kg ë2.2 lbs û é12 in. ù é2.54 cm ù 6 ft ê úê ú = 182.88 cm ë ft û ë 1 in. û é 1 in. ù é 1 ft ù 179.9 cm ê úê ú = 5.9 ft = 5 ft 10.8 in. ë2.54 cm û ë12 in. û
a.
°F = 2(°C) + 30° = 40° + 30° = 70°
b.
°F =
c.
very close
d.
30°C Æ 90°F vs. 86°F 5°C Æ 40°F vs 41°F
a.
14.6
b.
c.
1,046.1
d.
e.
3.14159 = 3.1
a.
14.60
b.
c.
1,046.06
d.
e.
3.14159 = 3.14
a.
14.603
b.
c.
1,046.060
d.
e.
3.14159 = 3.142
a.
104
b. 106
e.
100
f. 10- 1
= 68°
56.0 = 0.0625 = 0.1
56.04 = 0.0625 = 0.06
56.042 = 0.0625 = 0.063
c. 103
d. 10- 3
Chapter 1
19.
20.
21.
a.
15 ¥ 103
e.
4.02 ¥ 10- 4 f. 2 ¥ 10- 10
b.
5 ¥ 10- 3
c. 2.4 ¥ 106
a. 4.3 × 103 + 43.0 × 103 = 47.3 × 103 = 4.73 × 104 b. 8.0 × 104 + 46.0 × 104 = 54.0 × 104 = 5.40 × 105 c. 600 × 10−6 + 6 × 10−6 = 606 × 10−6 = 6.06 × 10−4 d. 2.6 × 103 + 0.06 × 103 - 0.5 × 103 = 2.16 × 103 = 2160
a. (1000)(10000) = (103 )(104 ) = 107 b. (0.001)(100) = (1 × 10–3 )(102) = 10–1 c. (102)(107) = 109 d. (100)(0.000001) = (102)(10-6) = 10-4 e. (10−8 )(107 ) = 10−1 = 1 × 10−1 f. (104 )(10−10 )(1026 ) = 1 × 1020 a. b. c. d.
(50 ¥ 103)(2 ¥ 10-3) = 100 ¥ 100 = 100 (2.2 ¥ 103)(2 ¥ 10-3) = 4.4 ¥ 100 = 4.40 (82 ¥ 10−6)(1.2 ¥ 10-6) = 98.40 (30 ¥ 10-4)(4 ¥ 10-3)(7 ¥ 108) = 840 ¥ 101 = 8.40 ¥ 103
23.
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 = 101/10-2 = 1 ¥ 103
24.
a. 8 × 10−5 = 0.5 × 108 = 5.0 × 107
22.
d. 60 ¥ 103
4 × 103
b. c. d.
6 × 10−3 6 × 106
44 × 10−5 5 × 10−5 88 × 1018 8 × 10−8
6 6
= × 10−9 = 1× 10−9 =
=
44 5
88 8
× 100 = 8.8
× 1026 = 1.1× 1027
(102)3 = 1.0 ¥ 106 (104)8 = 100.0 ¥ 1030
a. c.
26.
a. (4002 ) = (4 × 102 )2 = 16 × 104 b. (6 × 10−4 )2 = 1296 × 10−16 = 12.96 × 10−14 c. (5 × 10−3 ) (3 × 10−4 )2 = (5 × 10−3 ) (9 × 10−8 ) = 45 × 10−11 d. ((2 × 10−4 ) (0.8 × 105 ) (0.0005 × 106 )) 3 = ((2 × 10−4 ) (8 × 104 ) (5 × 102 )) 3 = (80 × 102 )3 = (8 × 103 )3 = ((8)3 × 103 )3 = 512 × 109 = 5.12 × 1011
Chapter 1
b. d.
(10-4)1/2 = 10.0 ¥ 10- 3 (10-7)9 = 1.0 ¥ 10- 63
25.
3
27.
= (9 ¥ 104)(102)/(3 ¥ 104) = (9 ¥ 106)/(3 ¥ 104) = 3 ¥ 102 = 300
a. b.
= 2 ¥ 105
c.
= 9.0 ¥ 1012 = 1.5 ¥ 10-7 = 150.0 ¥ 10- 9
d.
= 24.0 ¥ 1012
e. f.
(16 ¥ 10-6)1/2(105)5(2 ¥ 10-2) = (4 ¥ 10-3)(1025)(2 ¥ 10-2) = 8 ¥ 1020 = 800.0 ¥ 1018
g.
= 5.64 ¥ 104 = 56.40 ¥ 103 28.
29.
4
Scientific:
a. b. c. d.
2.05 × 101 5.04 × 104 6.74 × 10−4 4.60 × 10−2
Engineering:
a. b. c. d.
20.46 × 100 50.42 × 103 674.00 × 10−6 46.00 × 10−3
Scientific
a. b. c. d.
5.0 × 10−2 4.5 × 101 1/32 = 0.03125 = 3.125 × 10−2 3.14159 = 3.142 × 100
Engineering:
a. b. c. d.
50.0 × 10−3 0.045 × 103 31.25 × 10−3 3.142 × 100
Chapter 1
30.
Scientific Notation a. 3.2 × 10−3 b. b. (7 × 107 )(4 × 101 ) = 2.8 × 108 c.
(2×104 )(6×107 ) = (5×104 ) 2 8
12 5
� � 107 = 2.4 × 107
d. (6.2) × (10 ) × (82 × 10−3 ) = (38.44)(82)(4.02) × 108 = 1.267 × 1012
Engineering Notation: a. 3.20 × 103 b. 280.0 × 106 = 0.028 × 109 c. 24.0 × 106 d. 1.267 × 1012 +2
31.
a.
6 ¥ 104 = 0.06 ¥ 10+6 = 0.06 × 10+6 -3 -3
b.
0.4 ¥ 10-3 = 400 ¥ 10-6 = 400 × 10−6 +3 -2
c.
+3
50 ¥ 105 = 5000 ¥ 103 = 5 ¥ 106 = 0.005 ¥ 109 = 0.005 × 109 -3
+2
33.
-3
12 ¥ 10-7 = 0.0012 ¥ 10-3 = 1.2 ¥ 10-6 = 1200 ¥ 10-9 = 1200 × 109 -4
32.
-3 -3
+4 d.
+3
+3
+3
a. 0.06 × 100 s = 60 × 10−3 s = 60 ms b. 4000 × 10−6 s = 4 × 10−3 s = 4 ms c. 0.08 × 10−3 s = 80 × 10−5 = 80 𝜇𝑠 d. 6400 × 10−12 s = 0.0064 × 10−6 s = 0.0064 𝜇𝑠 e. 100 × 10−3 m = 1 × 10−3 m = 1 km a.
1.5 min
b.
2 × 10−2 h
Chapter 1
= 90 s
= 72 s
5
34.
35.
36.
0.05 s
d.
0.16 m
e.
1.2 ¥ 10-7 s
f.
4 ¥ 108 s
= 0.16 ¥ 103 mm = 160 mm = 1.2 ¥ 102 ns = 120 ns = 4629.6 days
a.
80 ¥ 10-3 m
b.
60 cm
c.
é 1 µm ù 12 × 10−3 m ê -6 ú = 12 ¥ 10−3 × 10+6 µm = 12 × 103 µm ë10 m û
d.
60 cm2
a.
100 in.
b.
4 ft
c.
6 lb
d.
60 ¥ 103 dynes
e.
150,000 cm
f.
0.002 mi
5280 ft, 5280 ft
6
= 0.05 ¥ 106 µs = 50 ¥ 103 µs
c.
= 8000 ¥ 10-3 cm = 8 cm
= 60 ¥ 10- 5 km
= 60 ¥ 10- 4 m2
= 2.54 m
= 1.22 m = 26.7 N
5280 ft
= 0.13 lb
= 4921.26 ft
= 3.22 m
= 1760 yds = 1609.35 m, 1.61 km
Chapter 1
37.
38.
= 26.82 m/s
d = 15km �
1000𝑚 39.37 𝑖𝑛. 1𝑓𝑡 1 𝑚𝑖 � � 1𝑚 � �12 𝑖𝑛.� �5280𝑓𝑡� 1𝑘𝑚
= 9.321 mi ≅ 9.32055 V=
1𝑚𝑖 8.5 𝑚𝑖𝑛
,t=
=
9.321𝑚𝑖 1 𝑚𝑖 8.5 𝑚𝑖𝑛
= 79.2285 min ≅ 79.23 min
= 3600 in fi 3600 quarters
39.
100 yds
40.
60 mph:
41.
d = vt = �800
d 500 mi = = 8.33 h = 8 h: 19.8 min u 60 mph d 500 mi = 7.14 h = 7 h: 8.4 min 70 mph: t = = u 70 mph difference = 1 h: 11.4 min t=
= 1382.4m 42.
𝑑 𝑣
cm � s
d = 86 stories
[0.048h] �
60𝑚𝑖𝑛 60𝑠 1𝑚 � �1 𝑚𝑖𝑛� �100𝑐𝑚� 1ℎ
= 1605 steps
u=
43.
= 13.38 minutes
= 0.228 miles
d = (86 stories) = 44.82 min/mile
,
44.
u=
Chapter 1
distance = 86 stories
= 1204 ft
= 1.14 minutes
7
45.
= 4.74 ¥ 10- 3 Btu
a.
b.
24 ounces
c.
1.4 days
d.
1 m3
46.
6(4 × 2 + 8) = 96
47.
(42 + 6/5)/3 = 14.4
48.
= 7.1 ¥ 10- 4m3
= 1.21 ¥ 105 s
= 2113.38 pints
æ2 ö2 52 + ç ÷ = 5.044 è3ø
49.
MODE = DEGREES: cos 21.87° = 0.928
50.
MODE = DEGREES: tan-1(3/4) = 36.87°
51.
= 7.071
52.
205 ¥ 10- 6
53.
1.20 ¥ 1012
54.
6.667 ¥ 106 + 0.5 ¥ 106 = 7.17 ¥ 106
8
Chapter 1
Chapter 1
9
Chapter 2 1.
-
2.
a.
F=
b.
F=k
c.
F=
d.
Exponentially,
a.
r = 1 ft:
3.
= 18 ¥ 109 N
= 2 ¥ 109 N
=
= 0.18 ¥ 109 N
= 10 while
= 100
é12 in. ù é 1 m ù 1 ft ê úê ú = 0.305 m ë 1 ft û ë39.37 in. û 9 -6 -6 -3 kQ1Q2 (9´10 )(8´10 C)(40´10 C) 2880´10 = = F= 2 2 -3 r 93´10 (0.305 m) = 30.97 N b.
r =10 ft: = 3.05 m = 0.31 N
F= c.
r = 100 yds:
é 3 ft ù é12 in. ù é 1 m ù 100 yds ê úê úê ú = 91.4 m ë1 yd û ë 1 ft û ë39.37 in. û kQ1Q2 2880 ´10-3 2880 ´10 -3 = = F= 2 2 3 r (91.4 m) 8.35´10 = 345 µN 4.
-
5.
𝑄1 = 𝑄2 = 𝑄; 𝐹1 =
10
𝑘𝑄 2 𝑟2
⇒ 𝑄2 =
𝐹1 𝑟 2 ; 𝑘
𝑘𝑄 2
𝐹2 = (2𝑟)2 =
𝑘 2𝑟 2
�
𝐹1𝑟2 𝑘
𝑟2
� and 𝐹2 = (2𝑟)2 𝐹1 =
𝐹1 4
Ans.
Chapter 2
6.
F=
7.
F=
𝑘𝑄1 𝑄2 𝑟2
=r=�
𝑘𝑄1 𝑄2 𝑟2
⇒r= �
(9 × 109 )(30×10−6 )2 (4.5 × 104 )
= 0.01342 m = 13.42 mm
= 4(1.8) = 7.2
a.
F=
b.
Q1/Q2 = 1/2 fi Q2 = 2Q1 7.2 = kQ1Q2 = (9 ¥ 109)(Q1)(2Q1) = 9 ¥ 109
= 72 mN
= 20 µC Q2 = 2Q1 = 2(2 ¥ 10-5 C) = 40 µC 𝑊 𝑄
3.4𝐽 12𝜇𝐶
8.
V=
9.
W = VQ = (60 V)(8 mC) = 0.48 J
10.
Q=
11.
Q=
12.
a.
𝑊 𝑉
=
=
= 283.33 kv
400𝜇𝐽 40𝑚𝐶
= 10 mC
W 620 mJ = = 68.9 mC 9V V W = QV = (1 × 1012 electrons)(40 V) = 40 × 1012 eV é ù 1C 40 × 1012 ê ú = 6.41 µJ 18 ë6.242 ´ 10 electrons û
b.
Q 96 mC = = 11.43 mA t 8.4 s
13.
I=
14.
I=
15.
Q = It = (40 mA)(1.2)(60 s) = 2.88 C
16.
Q = It = (250 mA)(1.2)(60 s) = 18.0 C
17.
t=
Chapter 2
𝑄 𝑡
600 𝐶
= (4)(60𝑠) = 2.50A
=3s
11
18.
21.847 ¥ 1018 electrons = 0.29 A
I= 19.
= 3.5 C
5 min = (5)(60 s) = 300 s Q = It = (4 mA)(300 s) = 1.2 C = 7.49 ¥ 1018 electrons
1.2 C
20.
I=
21.
0.92 × 1016 electrons � I=
22.
a.
= 1.194 A > 1 A (yes)
𝑄 𝑡
=
1.474𝑚𝐶 50𝑚𝑠
1𝐶 6.242×1018 electrons
= 29.48mA
� = 1.474 mC
Q = It = (2 mA)(0.01 µs) = 2 ¥ 10-11 C 2 ¥ 10-11 C = 1.25 ¥ 108 ¢ = $1.25 ¥ 106 = 1.25 million Q = It = (100 µA)(1.5 ns) = 1.5 ¥ 10-13 C
b.
1.5 ¥ 10-13 C
= 0.94 million
(a) > (b) 23.
Q = 𝐼𝑡 = (300 × 10−3A) × (40s) = 12C V=
24.
𝑊 𝑄
=
60𝐽 12C
= 5.0V
Q = It =
= 210 C = 3.53 V
V=
25.
Q= I=
12
𝑊 𝑉
𝑄 𝑡
=
=
0.8 24𝑉
= 0.0333C
0.0333𝐶 6×10−3 𝑠
= 5.55A
Chapter 2
Ah rating 180 Ah = = 4.5 A t(hours) 40 h
26.
I=
27.
Ah rating = current × hours = (0.64A)(80h) = 51.2 Ah.
28.
t (hours) =
29.
45 Ah (for 1 h):𝑊1 = VQ = V. I. t 60𝑚𝑖𝑛 = (12V) (45A) (1h)� ��
𝐴ℎ 𝑟𝑎𝑡𝑖𝑛𝑔 𝐼
=
72𝐴ℎ 1.80𝐴
= 40h
1ℎ
60𝑠 � 1 𝑚𝑖𝑛
1ℎ
60𝑠 � 1 𝑚𝑖𝑛
= 1.944 ×106 J 75 Ah (for 1 h):𝑊2 = VQ = V. I. t 60𝑚𝑖𝑛 = (12V) (75A) (1h)� �� Ratio
𝑊2 𝑊1
= 3.240 ×106 J = 1.67 or 67% more energy available with 75Ah rating for 60s discharge:
45Ah = It = I[60𝑠] �
1𝑚𝑖𝑛 1ℎ � �60 𝑚𝑖𝑛� 60𝑠
= I (16.67×10−3 h)
75Ah = It = I[60𝑠] �
1𝑚𝑖𝑛 1ℎ � �60 𝑚𝑖𝑛� 60𝑠
= I (16.67×10−3 h)
And I =
45𝐴ℎ 16.67 ×10−3 ℎ
And I = 4500A = 𝐼2 𝐼1
= 2700A
75𝐴ℎ 16.67 ×10−3 ℎ
= 4500A
= 1.67 or 67% more starting current available at 75Ah.
30.
0.75(18 Ah) = 13.5 Ah Þ @ 250 mA
31.
(18 Ah − 15.5 Ah)/18 Ah × 100% = 13.89%
32.
At 100 mA, discharge time @ 120 H; At 25 mA, discharge time @ 425 h; @ 300 h more at 25 mA
33.
I=
4𝐴ℎ 8.0ℎ
= 500 mA;
Q = It = (500mA) (8h) �
60𝑚𝑖𝑛 60𝑠 � �1 𝑚𝑖𝑛� 1ℎ
= 41.40 kC
Energy = W = QV = (14.4kC)(12V) = 172.8kJ
34.
-
35.
-
36.
-
Chapter 2
13
37.
a.
b. c.
38.
-
39.
-
40.
-
41.
-
14
é2.54 cm ù 0.5 in ê ú = 1.27 cm ë 1 in û é30 kV ù 1.27 cm ê ú = 38.1 kV ë cm û é270 kV ù 1.27 cm ê ú = 342.9 kV ë cm û 342.9 kV:38.1 kV = 9:1
Chapter 2
Chapter 3 1.
a. 0.4 inches = 400 mils b. 1⁄32 in. = 0.03125 in. [ c. 1⁄5 in. = 0.2 in. [
1000 𝑚𝑖𝑙𝑠 1 𝑖𝑛.
d. 20 mm = 20 × 10-3m [ e. 0.02 ft [
f. 2.
3.
3cm [
12 𝑖𝑛. 1 𝑓𝑡
1 𝑖𝑛. 2.54 𝑐𝑚
][
][
1000 𝑚𝑖𝑙𝑠 1 𝑖𝑛.
] = 200 mils
39.37 𝑖𝑛. 1𝑚
1000 𝑚𝑖𝑙𝑠 1 𝑖𝑛.
1000 𝑚𝑖𝑙𝑠 1 𝑖𝑛.
] = 31.25 mils
][
1000 𝑚𝑖𝑙𝑠 1 𝑖𝑛.
] = 787.4 mils
] = 240 mils
] = 1181.10 mils
a.
ACM = (30 mils)2 = 900 CM
b.
0.08 in. = 80 mils, ACM = (80 mils)2 = 6.4 ¥ 103 CM
c.
é1" ù ê ú = 0.0625 in. = 62.5 mils, ACM = (62.5 mils)2 = 3.91 ¥ 103 CM êë16 úû
d.
2 cm
e.
0.02 ft
f.
é39.37 in. ù 4 × 10−3 m ê ú ë 1m û
= 787.4 mils, ACM = (787.4 mils)2 = 620 ¥ 103 CM
= 240 mils, ACM = (240 mils)2 = 57.60 ¥ 103 CM
é1000 mils ù 2 3 ê ú = 157.48 mils, ACM = (157.48 mils) = 24.8 × 10 CM 1 in ë û
ACM = (dmils)2 Dmils = √𝐴CM a. d = √1800𝐶𝑀 = 42.43mils = 0.0424 in. b. d = √840𝐶𝑀 = 28.98 mils = 0.029 in. c. d = √42,000𝐶𝑀 = 204.94 mils = 0.205 in. d. d = √2000𝐶𝑀 = 44.72 mils = 0.045 in. e. d = √8.25𝐶𝑀 = 2.87 mils = 0.0029 in. f. d = √6 × 103 = 77.46 mils = 0.0775 in.
Chapter 3
15
4.
R=𝜌 5.
6.
𝑙 𝐴
= (10.37)
a.
A= r
b.
d=
1 16
A CM =
9.
a.
𝑙
ACM =
16
544 CM = 23.32 mils = 23.3 ¥ 10- 3 in.
larger
c.
smaller
𝜌=
𝑅𝑙 𝑙
=
𝑅𝐴 𝜌
=
(4.4Ω)(3906.25 𝐶𝑀) 600
=
= 28.65 ft.
= 942.73 CM
942.73 CM = 30.70 mils = 30.7 ¥ 10- 3 in.
b.
a.
= 4.051 Ω
in. = 0.0625 in. = 62.5 mils, ACM = (62.5)2 = 3906.25CM.
d=
8.
(400′ ) 1024 𝐶𝑀
æ 80¢ ö l = 17 ç ÷ = 544 CM R è 2.5 Ω ø
R = 𝜌𝐴 ⇒ l =
7.
ACM = (32 mils)2 = 1024 CM.
0.032 in. = 32 mils,
(600Ω)(148𝐶𝑀) (1200 𝑓𝑡)
= 74 ⇒ Iron.
1/32≤ = 0.03125≤ = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM rl RA (3.14 Ω)(976.56 CM) Þl= = R= = 295.7 ft A r 10.37
b.
(5)(295.7) 295.7´ 1000´ = Þx = = 1.48 lbs 5 lb 1000 x
c.
−40° C: F =
9 C + 32° = 5 9 105° C: F = C = 32° = 5 F° = −40° Æ 221°
9 (-40) + 32 = −40° 5 9 (105) + 32 = 221° 5
Chapter 3
10.
a.
b.
11.
5" 8
= 0.625" = 625 mils ; 5.8" = 5800 mils ∴ A = (625 mils) (5800 mils) = 4⁄𝜋 𝐶𝑀 ⇒ 3.625 × 106 sq.mils [1 𝑠𝑞.𝑚𝑖𝑙 ] = 4.62 × 106 CM. 1" 10
= 0.1" = 100 mils, ACM = (100 𝑚𝑖𝑙𝑠)2 = 1.0 × 104 = 10 × 103 𝐶𝑀
(#10)
a.
4.62×106 𝐶𝑀 10 ×103 𝐶𝑀
= 462 wires
3" = 3000 mils, 1/2" = 0.5 in. = 500 mils Area = (3 ¥ 103 mils)(5 ¥ 102 mils) = 15 ¥ 105 sq. mils
15 ¥ 105 sq mils
b.
= 19.108 ¥ 105 CM
R=
=
= 21.71 µΩ
R=
=
= 35.59 µΩ
Aluminum bus-bar has almost 64% higher resistance. c. 12.
—
l2 = 2l1, A2 = A1/4, r2 = r1 =8 and R2 = 8R1 = 8(0.2 ) = 1.6 ∆R = 1.6 - 0.2 = 1.4
13.
A=
𝜋𝑑 2 4
⇒d=
√4𝐴 𝜋
= �
4(0.06 𝑖𝑛.2 ) 𝜋
= 0.2764 in.
dmils = 276.4 mils, ACM = (276.4 𝑚𝑖𝑙𝑠)2 = 76,396.96 CM 𝑅1 𝑅2
𝜌 𝑙1 1 �𝐴
=𝜌
𝑙 2 2�𝐴 2
and 𝑅2 =
Chapter 3
1
𝜌1 𝑙1 𝐴2
=𝜌
2 𝑙2 𝐴1
𝑅1 𝑙2𝐴1 𝑙1 𝐴2
=
=
𝑙1 𝐴2
𝑙2 𝐴1
(𝜌1 = 𝜌2 )
(900𝑚Ω)(600𝑓𝑡)(50,000 𝐶𝑀) (400𝑓𝑡)(76,396.96𝐶𝑀)
= 883.54 mΩ.
17
14.
a.
#12 = 6,529.9 CM, #14 = 4,106.8 CM 6,529.9 CM - 4,106.8 CM ´100% = 59% larger 4,106.8 CM = 1.33,
b.
= 1.59
Imax ratio = 1.33 vs Area ratio = 1.59 1.59 - 1.33 ´100% = 19.55% higher ratio for area 1.33 15.
a.
= 2 yes
b.
= 16.16 yes
= 7.5
16.
a. b.
17.
a.
#20 #30 #30 #40
1021.50𝐶𝑀 100.50 𝐶𝑀
=
100.50𝐶𝑀 9.89 𝐶𝑀
=
Yes ≅ 10
= 10.16 ≅ 10 𝑌𝑒𝑠
= 10.162
= 51,850 CM fi #3
A=
but 110 A fi #2 b. 18.
= 103,700 CM fi #0
A=
a. 𝐴�𝐶𝑀 =
b.
1.192 𝑚𝐴 𝐶𝑀
200 𝐴 167,810 𝐶𝑀
�𝜋
1𝐶𝑀
�4 𝑠𝑞.𝑚𝑖𝑙𝑠
1 𝑖𝑛.2
= 1.192 𝑚𝐴�𝐶𝑀
��
1000 𝑚𝑖𝑙𝑠 1000 𝑚𝑖𝑙𝑠 � � 1 𝑖𝑛. � 1 𝑖𝑛.
c. 6KA �1.52 𝑘𝐴� = 3.95 𝑖𝑛.2
18
= 1.52 𝑘𝐴� 2 𝑖𝑛.
Chapter 3
19. 20.
234.5+10 4Ω
=
234.5+80 𝑅2
R2 =
21.
244.5
= 5.15Ω
5 5 (℉ − 32) = (32 − 32) = 0° (= 32℉) 9 9
𝐶=
5 (68 − 32) = 20° (= 68℉) 9
234.5°+20° 2.6Ω
𝑅2 =
a.
(314.5)(4Ω)
= 0.028
𝐶=
22.
, 𝑅2 =
=
234.5°+0° 𝑅2
(234.5)(2.6Ω) = 2.396Ω 254.5
°C =
5 5 (°F - 32°) = (70° - 32°) = 21.11° 9 9
°C =
= 15.56°
234.5 + 21.11 234.5 + 15.56 = 0.025 Ω R2
b.
R2 =
(250.06)(0.025 Ω) = 24.46 mΩ 255.61
°C =
= 10°
234.5 + 21.11 234.5 + 10 = 0.025 Ω R2 R2 =
(244.5) (0.025 Ω) = 23.91 mΩ 255.61
c.
Part a: 25 mΩ − 24.46 mΩ = 0.54 mΩ Part b: 24.45 mΩ − 23.91 mΩ = 0.55 mΩ Linear 40°F fi 23.91 mΩ − 0.55 mΩ = 23.36 mΩ
d.
°C =
= −34.44°
234.5 + 21.11 234.5 - 34.44 = 25 mΩ R2
(25 mΩ)(200.06) = 19.57 mΩ 255.61 Yes, 25 mΩ − 19.57 mΩ = 5.43 mΩ
R2 =
Chapter 3
19
e.
°C =
= 48.89°
234.5 + 21.11 234.5 + 48.89 = 25 mΩ R2
(25 mΩ)(283.39) = 27.72 mΩ 255.61 Yes, 2.72 mΩ
R2 =
23.
a. b.
24.
a.
234.5+4 2Ω 234.5+4 2Ω
= =
234.5+𝑡2 2.2Ω 234.5+𝑡2 0.2Ω
, 𝑡2 = 45.45℃
, 𝑡2 = −209.05℃
68°F = 20°C
234.5 + 20 234.5 + T2 = 1Ω 2Ω 2(254.5) - 234.5 = T2 1 T2 = 274.5°C b.
#10 = 0.9989 Ω/1000´
c. din = 0.102 in @ 25.
1 ˝ 10
a.
a20 =
b.
R = R20[1 + a20(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
= 0.003929 @ 0.00393
t= 26.
20
= 83.61°C
R = R20[1 + a20(t - 20°C)] = 0.4 Ω [1 + 0.00393(16 - 20)] = 0.4 Ω [1 - 0.01572] = 0.39 Ω
Chapter 3
27.
From Table 3.2, 1000′ of #12 Copper wire = 1.588Ω @20℃ ℃=
5 5 (℉ − 32) = (170 − 32) = 76.67℃ 9 9
𝑅 = 𝑅20 [1 +∝20 (𝑡 − 20℃)]
28.
= 1.588Ω [1 + 0.00393 (76.67 − 20)] = 1.942Ω (PPM)(∆T) =
∆R =
(200)(65° - 20°) = 0.198 Ω
R = Rnominal + ∆R = 22.198 Ω 29.
(PPM)(∆T) =
∆R =
(100)(50° - 20°) = 0.30 Ω
R = Rnominal + ∆R = 100 Ω + 0.30 Ω = 100.30 Ω 30.
a.
2 times larger
b.
31. 32.
20𝑘Ω − 6.5𝑘Ω = 13.5𝑘Ω
33.
-
34.
a. b. c. d. e.
35.
a. b. c. d.
36.
a.
10 Ω ± 20% fi 8 Ω − 12 Ω 15 Ω ± 20% fi 12 Ω − 18 Ω
b.
10 Ω ± 10% fi 9 Ω − 11 Ω 15 Ω ± 10% fi 13.5 Ω − 16.5 Ω
4 times larger
6.25 kΩ and 18.75 kΩ
820 Ω ± 5%, 820 Ω ± 41 Ω, 779 Ω ´ 861 Ω 220 Ω ± 10%, 220 Ω ± 22 Ω, 198 Ω ´ 242 Ω 91 kΩ ± 20%, 91 kΩ ± 18.2 kΩ, 77.8 kΩ ´ 109.2 kΩ 9.1 kΩ ± 5%, 9100 Ω ± 455 Ω, 8,645 Ω ´ 9,555 kΩ 3.9 MΩ ± 20%, 3.9 MΩ ± 0.78 MΩ, 3.12 MΩ ´ 4.68 MΩ
78 Ω = Violet, Gray, Black, Silver 0.66Ω = Blue, Blue, Silver, Silver 44 kΩ = Yellow, Yellow, Orange, Silver 6.7 MΩ = Blue, Violet, Green, Silver
37.
470 Ω ± 10% = 470 Ω ± 47 Ω = 423 Ω ´ 517 Ω Yes
38.
No change
Chapter 3
}
no overlap, continuance
}
no overlap
21
721 = 72 × 101 Ω = 720Ω = 0.72kΩ 222 = 22 × 102 Ω = 2.2kΩ 3.9 × 104 Ω = 39kΩ = 𝑄4 𝐶5 = 1.2 × 105 Ω = 0.12MΩ
39.
a. b. c. d.
40.
a. 𝐺𝑎 =
b.
41.
1 1 = = 4.55𝑚𝑆 220Ω 𝑅𝑎 1 = 0.167𝑚𝑆 𝐺𝑏 = 6𝑘Ω 1 𝐺𝑐 = 1.1𝑀Ω = 0.91𝜇𝑆
c. d. 𝐺𝑎 > 𝐺𝑏 > 𝐺𝑐 𝑣𝑠. 𝑅𝑐 > 𝑅𝑏 > 𝑅𝑎
a.
Table 3.2, Ω/1000¢ = 1.588 Ω
= 629.72 mS
G=
= 629.69 mS (Cu)
or G =
42.
43.
b.
G=
a.
G1 =
b.
G2:G1 = 50 mS: 100 mS = 1:2 whereas R2:R1 = 20 Ω:10 Ω = 2:1. The rate of change is the same although one is increasing and the other decreasing.
c.
inverse − linear
1 1 1 = 100 mS, G2 = = 50 mS, G3 = = 10 mS 10 Ω 20 Ω 100 Ω
2 3
5 3
2 3
𝐴2 = 1 𝐴1 = 𝐴1 , 𝑙2 = �1 − � 𝑙1 = 𝐺1 𝐺2
44.
-
45.
-
46.
-
22
= 384.11 mS (Al)
=
𝐴 𝜌1 1 𝑙1 𝐴2 𝜌2 𝑙2
=
𝜌1 𝑙2 𝐴1 𝜌2 𝑙1 𝐴2
=
𝑙 � 1�𝐴1 3 𝑙1 �5�3𝐴1 �
=
1 5
𝑙1 3
, 𝜌2 = 𝜌1
Chapter 3
47.
-
æ 2.54 cm ö in. = 0.083 in. ç ÷ = 0.21 cm è 1 in. ø
48.
= 0.035 cm2
A= l=
49.
a.
" 4 ft
= 40,603 cm = 406.03 m
= 1.27 cm, 3 in. = 121.92 cm
= 21.71 µΩ
R=
50.
51. 52.
= 7.62 cm
b.
R=
= 35.59 µΩ
c.
increases
d.
decreases
𝑅𝑠 =
𝜌 𝜌 250 × 10−6 = 200 ⇒ d = = = 1.25𝜇𝑐𝑚. 200 𝑑 200
𝑅𝑠 𝑙 (225Ω)�1�2 𝑖𝑛. � 𝑙 𝑅 = 𝑅𝑠 ⇒ w= = 0.1875𝑖𝑛. = 𝑚 𝑅 600Ω a.
d = 1 in. = 1000 mils ACM = (103 mils)2 = 106 CM
r1 = b.
1 in. = 2.54 cm A=
Chapter 3
= 1 CM-Ω/ft
= 5.067 cm2
l = 1000 ft
= 30,480 cm
r2 =
= 1.66 ¥ 10- 7 Ω-cm
23
c.
= 1.66 ¥ 10- 7
k=
53.
-
54.
-
55.
-
56.
-
57.
-
58.
a.
-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.
r=
a.
Log scale:
10 fc fi 3 kΩ 100 fc fi 0.4 kΩ
b.
negative
c.
no—log scales imply linearity
d.
1 kΩ fi 30 fc 10 kΩ fi 2 fc ∆R 10 kΩ - 1k = = 321.43 Ω/fc ∆ fc 30 fc - 2 fc
59.
∆Ω - cm 300 - 30 270 Ω - cm = = @ 3.6 Ω-cm/°C 125 - 50 75° C ∆T
and 60.
24
a.
@ 0.5 mA, V 195 V @ 1 mA, V 200 V @ 5 mA, V 215 V
b.
∆Vtotal = 215 V - 195 V = 20 V
c.
5 mA:0.5 mA = 10:1 compared to 215 V: 200 V = 1.08:1
= -321.43 Ω/fc
Chapter 3
Chapter 4 1.
V = IR = (5.6 mA)(220 Ω) = 1.23 V
2.
I=
3.
R=
4.
I=
5.
𝑉 = 𝐼𝑅 = (5𝜇𝐴)(0.2𝑀Ω) = 1.0𝑉
6.
24 V V = = 10.91 mA R 2.2 kΩ = 16 kΩ
= 300 A
I=
= 2.4 mA
7.
R=
= 54.55 Ω
8.
𝐼=
9.
𝑅=
10.
R=
V 4.5 V = = 56.25 Ω I 80 mA
11.
𝑅=
𝑉 𝐼
12. 13.
𝑉 𝑅
=
𝑉 𝐼
110𝑉 7𝑘Ω
= 15.71𝑚𝐴
=
150𝑉 4.6𝐴
=
36𝑚𝑉 30𝜇𝐴
= 32.61Ω
= 1.2𝑘Ω
𝑉 = 𝐼𝑅 = (14𝐴)(0.8Ω) = 9.6𝑉
a.
R=
= 12.63 Ω
b.
t=2h
= 7200 s
W = Pt = VIt = (120 V)(9.5 A)(7200 s) = 8.21 ¥ 106 J 14. 15.
𝑉 = 𝐼𝑅 = (8.4µA)(4.2 MΩ) = 35.28𝑉 -
Chapter 4
25
16.
-
17.
-
18.
-
19.
-
20.
P=
21.
𝑡=
22.
W = t
𝑊 𝑃
=
540 J 540 J = = 2.5 W é 60 s ù 216 s 3.6 min ê ú ë1 min û 840𝐽
60𝐽�𝑠
a. 12h�
= 14𝑠
60𝑚𝑖𝑛 60𝑠 � �1𝑚𝑖𝑛� 1ℎ
= 43,200𝑠.
𝑊 = 𝑃𝑡 = (3𝑊)(43,200𝑠) = 129. 𝑘𝐽
23.
b. 𝑘𝑊ℎ =
(3𝑊)(12ℎ) 1000
= 36 × 10−3 𝑘𝑊ℎ
P = VI = (3 V)(1.4 A) = 4.20 W t=
= 2.86 s
24.
P = EI = (12 V)(40 A) = 480 W
25.
P = I2R = (7.2 mA)2 4 kΩ = 207.36 mW
26.
P = I 2R fi I =
27.
𝐼=� =�
28.
𝑃 𝑅
3𝑊 69Ω
= 10.44 mA
= 208.51𝑚𝐴
𝑉 = 𝐼𝑅 = (208.51𝑚𝐴)(69Ω) = 14.39𝑉 I=
= 1.31 mA
P = I2R = (1.31 mA)2 16.8 kΩ = 28.83 mW æ é60 min ù é 60 s ù ö W = Pt = (28.83 mW) ç ç1 h ê 1 h ú ê1 min ú ÷÷ = 103.79 J ûë ûø è ë 29.
26
E=
P 10 kW = = 208.33 V I 48 A Chapter 4
30.
𝑃 𝑅
1.5𝑊 5.0𝑀Ω
𝐼=� =�
= 547.72𝜇𝐴
No. the current will not be double. 31.
V=
32.
𝑃 = 𝑉𝐼, 𝐼 = 𝑅=
33.
34.
= 9.61 V
𝑉 𝐼
=
𝑃 𝑉
240𝑉 0.417𝐴
100 240
=
= 0.417𝐴
= 575.54Ω
V=
= 120 V
R=
= 32 Ω
a. 𝑃 = 𝑉𝐼 & 𝐼 =
𝑃 𝑉
=
0.5×10−3 𝑊 4.5𝑉
= 0.11𝑚𝐴
b. Ah-rating = (0.11𝑚𝐴)(550ℎ) = 60.5𝑚𝐴ℎ
35.
I=
= 70.71 mA = 1.42 kV
V= 36.
𝑃 = 𝑉𝐼 = (240𝑉)(40𝐴) = 9.6𝑘𝑊
𝑃𝐻𝑃 = 37.
38.
9.6𝑘𝑊
746𝑊�𝐻𝑃
=
9600𝑊 746𝑊�𝐻𝑃
= 12.87𝐻𝑃
a.
æV 2 ö æ12 V ö 2 ÷ W = Pt = ç ç R ÷ t = ç10 Ω ÷ 60 s = 86.4 J è ø è ø
b.
Energy doubles, power the same
W = Pt fi t =
39.
= 260 h
kWh =
Chapter 4
3 W 12 ´10 Wh =8h = 1500 W P
= 59.80 kWh
27
40.
a.
æ 60 min ö æ 60 s ö 6 W = Pt = (60 W)(10 h) ç ÷ç ÷ = 2.16 × 10 Ws è 1 h ø è1 min ø
b.
1 Ws = 1 J \ 2.16 × 106 J
c.
W = Pt = (60 W)(10 h) = 600 Wh = 0.6 kWh
d. e. 41.
Cost = (0.6 kWh)(12 ¢/kWh) = 7.2 ¢ 𝑃𝑡 (1000)(𝑘𝑊ℎ) ⇒P= 1000 𝑡 𝑃 125×103 𝑊 = = 543.48𝐴 𝑉 230𝑉
a. 𝑘𝑊ℎ = b. 𝐼 =
=
(1000)(1500𝑘𝑊ℎ) 12ℎ
= 125𝑘𝑊
c. 𝑃𝑙𝑜𝑠𝑡 = 𝑃𝑖 − 𝑃𝑜 = 𝑃𝑖 − η𝑃𝑖 (1 − η) = 125𝑘𝑊(1 − 0.79) = 26.25kW
42.
𝑘𝑊ℎ𝑙𝑜𝑠𝑡 =
#𝑘𝑊ℎ = 𝑘𝑊ℎ =
43.
44.
=
(26.25𝑘𝑊)(12ℎ)
= 7.14𝑘𝑊ℎ
𝑃𝑡 ⇒ 1000
t=
1000
(𝑘𝑊ℎ)(1000) 𝑃
=
a.
$120 = $4/day 30 days
b.
$4 / day = 26.7¢/h 15 h/day
c.
26.7¢/h = 2.23 kW 12¢/kWh
d.
2.23 kW = 37.17 @ 37 bulbs 60 W
e.
no
$1.00 11¢� 𝑘𝑊ℎ
𝑡=
28
$1.00 14¢
𝑃𝑡 1000
= 315𝑘𝑊ℎ
(7.14)(1000) 260𝑊
= 27.46ℎ
= 9.09𝑘𝑊ℎ
9.09𝑘𝑊ℎ 198𝑊
= 45.91ℎ
Chapter 4
45.
𝑡 = 6ℎ�𝑑𝑎𝑦 (365𝑑𝑎𝑦𝑠) = 2190ℎ
𝑘𝑊ℎ =
𝑃𝑡 1000
=
(400𝑊)(2190ℎ)
= 876𝑘𝑊ℎ
𝑘𝑊ℎ =
𝑃𝑡 1000
=
(213𝑊)(2190ℎ)
= 466.47𝑘𝑊ℎ
1000
Cost = (876𝑘𝑊ℎ)(12¢/kWh) = $105.12 1000
Cost = (466.47𝑘𝑊ℎ)(12¢/kWh) = $55.98 Cost savings = $105.12 − $55.98 = $49.14
(78 W)(4 h/day)(31 days) P×t = = 9.67 kWh 1000 1000 Cost = (12¢/kWh)(9.67 kWh) = $1.16
46.
kWh =
47.
a. 𝑃 = 𝑉𝐼 − (110𝑉)(100𝐴) = 11𝐾𝑊
b. 𝑃 = 2 × 240𝑊 + 2500𝑊 + 10 × 100𝑊 + 2000𝑊 + 1500𝑊 + 1500𝑊 = 8.98𝑘𝑊 < 11𝑘𝑊 (𝑌𝑒𝑠)
c. 𝑊 = 𝑃𝑡 = (8.98𝑘𝑊)(3ℎ) = 26.94 kWh 48.
kWh =
(1600 W)(8 h) + (1200 W)(1/3 h) + (4800 W)(1 h) + (900 W)(1/4 h) + (200 W)(1.2 h) + (50 W)(3.5 h) 1000
12, 800 Wh + 400 Wh + 4800 Wh + 225 Wh + 240 Wh + 175 Wh = = 18.64 kWh 1000 (18.64 kWh)(12¢/kWh) = $2.24
49.
kWh =
æ 1 ö (200W)(4 h) + (6)(60 W)(6 h) + (1200 W)(0.5 h) + (175 W)(3.5 h) + (250 W)ç4 h ÷ + (30 W)(8 h) è 3 ø 1000
800 Wh + 2160 Wh + 600 Wh + 612.5 Wh + 1083.32 Wh + 240 Wh = = 5.496 kWh 1000 (5.496 kWh)(12¢/kWh) = 65.95¢
50.
h=
51.
η = 𝑃𝑜 ; 𝑃𝑖 =
= 90.98%
𝑃
𝑖
𝑃𝑖 = 𝑉𝐼; 𝐼 = 52.
𝑃
𝑃𝑜
η
=
Chapter 4
0.75
= 1492𝑊
1492𝑊 𝑃𝑖 = = 12.43𝐴 𝑉 120𝑉
η = 𝑃𝑜 × 100% = 𝑖
(1.5ℎ𝑝)�746𝑊�ℎ𝑝�
(0.88)�746𝑊�ℎ𝑝� × (4.5𝐴)(220𝑉)
100% =
656.48 × 990
100% = 66.31%
29
53.
a. Pi = EI = (120 V)(1.8 A) = 216 W Pi = Po + Plost, Plost = Pi - Po = 216 W - 50 W = 166 W
h% =
b.
54.
55.
¥ 100% =
¥ 100% = 23.15%
(4.6ℎ𝑝) �746𝑊�ℎ𝑝� 𝑃𝑜 = 𝑃𝑖 = 𝑉𝐼 = ⇒I = = 18.65 𝐴 (0.80)(230𝑉) η η𝑉 𝑃𝑜
a.
Pi =
= 1657.78 W
b.
Pi = EI = 1657.78 W (110 V)I = 1657.78 W 1657.78 W I= = 13.81 A 120 V
c.
Pi =
= 2131.43 W
Pi = EI = 2131.43 W (120 V)I = 2131.43 W 2131.43 W I= = 17.76 A 120 V 56.
(16ℎ𝑝) �746𝑊�ℎ𝑝� = 13,563.64𝑊 𝑃𝑖 = = η 0.88 𝑃𝑖 13,563.64𝑊 𝐼= = = 56.52𝐴 𝑉 240𝑉 𝑃𝑜
57.
hT = h1 ◊ h2 0.75 = 0.85 ¥ h2 h2 = 0.88, h2 = 88%
58.
η 𝑇 = η 1 . η 2 = (0.86)(0.76) = 0.6536⇒ 65.36%
59.
η𝑇 = η1 . η2 = (0.82) = 0.9η2 η2 =
30
0.82 0.9
= 0.9111 ⇒ 91.11%
Chapter 4
60.
a.
hT = h1 ◊ h2 ◊ h3 = (0.93)(0.87)(0.21) = 0.170 fi 17%
b.
hT = h1 ◊ h2 ◊ h3 = (0.93)(0.87)(0.80) = 0.647 fi 64.7% ¥ 100% = 280.59%
61.
η𝑇 = 𝑛12 =
𝑃𝑜 = η1 . η2 = η2 . 2η1 = 2𝑛12 𝑃𝑖
𝑃𝑜 𝑃𝑜 200𝑊 ⇒ η1 = � =� = 0.5 2𝑃𝑖 2𝑃𝑖 2(400𝑊)
η2 = 2η1 = 2(0.5) = 1.0 η1 = 50%, η2 = 100%
Chapter 4
31
Chapter 5 1.
a. b. c. d.
E and R1 R1 and R2 E1, E2, and R1 E and R1; R3, R4, and R5
2.
a. b. c. d.
E1 and R1; E2, R3, and R4 E1 and R1; R5 and R6 R2 and R3 E1 and R1
3.
a. b. c.
RT = 0.1 kΩ + 0.39 kΩ + 1.2 kΩ + 6.8 kΩ = 8.49 kΩ RT = 1.2 Ω + 2.7 Ω + 8.2 Ω = 12.1 Ω RT = 1.2 Ω + 2.2 Ω + 3.3 Ω + 4.7 Ω = 11.4 Ω
4.
a. b. c.
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 RT = 3.3 Ω + 10 kΩ = 13.3 kΩ
5.
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
6.
a. 2MΩ will have most impact. b. 200Ω, 2kΩ can be neglected. c. 𝑅𝑇 = 200Ω + 2kΩ + 2MΩ + 400kΩ = 2.4022MΩ = 2.4022𝑀Ω vs 2.4MΩ for part (b).
7.
a. b.
Reading = 10 Ω + 33 Ω + 56 Ω + 68 Ω = 167 Ω Reading = 0.82 kΩ + 1.2 kΩ + 3.3 kΩ = 5.32 kΩ
8.
a. b.
RT = 129 kΩ = R + 56 kΩ + 22 kΩ + 33 kΩ, Reading = 18 kΩ RT = 103 kΩ = 24 kΩ + R1 + 43 kΩ + 2R1 = 67 kΩ + 3R1, R1 = 12 kΩ R2 = 24 kΩ
9.
a. b. c.
1.2 kΩ + 2.2 kΩ = 3.4 kΩ 0Ω •Ω
10.
a. 𝑅𝑇 = 12Ω + 14Ω + 20Ω = 46Ω
b. 𝐼𝑠 =
𝐸 𝑅𝑇
=
76𝑉 46Ω
= 1.65𝐴
c. 𝑉1 = 𝐼1 𝑅1 = (1.65𝐴)(12Ω) = 19.8𝑉 ; 𝑉2 = 𝐼2 𝑅2 = (1.65𝐴)(14Ω) = 23.1𝑉 𝑉3 = 𝐼3 𝑅3 = (1.65𝐴)(20Ω) = 33.0𝑉 d. 𝑃𝑠 = 𝐸𝐼𝑠 = (76𝑉)(1.65𝐴) = 125.4𝑊 e. 𝑃20Ω = 𝑉3 𝐼3 = (33.0𝑉)(1.65A) = 54.45𝑊
32
Chapter 5
11.
a. The most: 𝑅3 , the least: 𝑅1 b. 𝑅3 , 𝑅𝑇 = 2.2𝑘Ω + 6.8𝑘Ω + 84𝑘Ω = 93𝑘Ω 𝐼𝑠 =
𝐸 𝑅𝑇
=
60𝑉 93𝑘Ω
= 0.65𝑚𝐴
c. 𝑉1 = 𝐼1 𝑅1 = (0.65𝑚𝐴)(2.2𝑘Ω) = 1.43𝑉 𝑉2 = 𝐼2 𝑅2 = (0.65𝑚𝐴)(6.8𝑘Ω) = 4.42𝑉 𝑉3 = 𝐼3𝑅3 = (0.65𝑚𝐴)(84𝑘Ω) = 54.6𝑉,
Result agree with part (a) 12.
a. b.
13.
I.
II.
RT = 12 kΩ + 4 kΩ + 6 kΩ = 22 kΩ E = IRT = (4 mA)(22 kΩ) = 88 V RT = 12 Ω + 22 Ω + 82 Ω + 10 Ω = 126 Ω E = IRT = (500 mA)(126 Ω) = 63 V =4A
a.
I=
b. c. d.
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=
b.
V3.3 kΩ = (3 mA)(3.3 kΩ) = 9.9 V
R=3Ω
= 3 mA
E = 6.6 V + 9 V + 9.9 V = 25.5 V
14.
c.
R=
d.
V2.2 kΩ = 6.6 V, V3 kΩ = 9 V, V3.3 kΩ = 9.9 V
= 3 kΩ
a.
Im =
b.
RT = 1 kΩ + 2.4 kΩ + 5.6 kΩ = 9 kΩ Im =
c.
Chapter 5
= 8.18 mA, Vm =
= 18 V
= 2.5 mA, Vm = 2.5 mA(2.4 kΩ + 5.6 kΩ) = 20 V
3.3 kΩ(12 V) = 8.8 V 4.5 kΩ Vm = 12 V − 8.8 V = 3.2 V 12 V Im = = 2.67 mA 4.5 kΩ
V3.3 kΩ =
33
15.
a.
I=
10 V = 0.333 A 30 Ω
V=0V
16.
80 V = 5.33 V 60 Ω
b.
I = 0 A, V = IR =
a.
RT = 3 kΩ + 1 kΩ + 2 kΩ = 6 kΩ Is =
= 20 mA = (20 mA)(3 kΩ) = 60 V = (20 mA)(1 kΩ) = 20 V = (20 mA)(2 kΩ) = 40 V
b.
17.
=
= (20 mA)2 ◊ 3 kΩ = 1.2 W
=
= (20 mA)2 ◊ 1 kΩ = 0.4 W
=
= (20 mA)2 ◊ 2 kΩ = 0.8 W
c.
PT = PR1 + PR 2 + PR 3 = 1.2 W + 0.4 W + 0.8 W = 2.4 W
d.
PT = EIs = (120 V)(20 mA) = 2.4 W
e.
the same
f.
R1 - the largest
g.
dissipated
h.
R1: 2 W, R2 : 1/2 W, R3: 1 W
𝑃 = 28𝑊 = (2𝐴)2 . 𝑅, ⇒R = 7Ω 𝑉1 = 𝐼1𝑅1 = (2A)(3Ω) = 6𝑉 𝑉2 = 𝐼2𝑅2 = (2A)(2Ω) = 4𝑉
𝑉3 = 𝐼3𝑅3 = (2A)(7Ω) = 14𝑉
18.
𝐸 = 𝑉1 + 𝑉2 + 𝑉3 = 6𝑉 + 4𝑉 + 14𝑉 = 24𝑉
12 𝑃 = 12𝑊 = 𝐼 2 . 2Ω; 𝐼 = � = 2.449𝐴 2
𝑃 = 20𝑊 = 𝐼 2 . 𝑅1 = (2.449𝐴)2 𝑅1 𝑅1 = 3.335Ω, 𝑅𝑇 = 36Ω = 3.335Ω + 𝑅2 + 2Ω = 5.225Ω + 𝑅2 ⇒𝑅2 = 30.665Ω 𝐸 = 𝐼𝑅𝑇 = (2.449𝐴)(30.665Ω) = 75.10𝑉
34
Chapter 5
19.
a.
æ 1 ö RT = NR1 = 8 ç28 Ω ÷ = 225 Ω è 8 ø = 0.53 A
I=
20.
b.
æ 8 ö 2 æ 1 ö æ 64 ö æ 225 ö P = I R = ç A ÷ ç28 Ω ÷ = ç ÷ç ÷ =8W è15 ø è 8 ø è 225 ø è 8 ø
c.
æ 8 ö æ 225 ö Ω ÷ = 15 V V = IR = ç A ÷ ç è15 ø è 8 ø
d.
All go out!
2
𝑃𝑆 = 𝑃𝑅1 + 𝑃𝑅2 + 𝑃𝑅3 E.I. = 𝐼 2 𝑅1 + 𝐼 2 𝑅2 +40 (𝑅1 + 𝑅2 )𝐼 2 − E.I. + 40 = 0 10𝐼 2 − 40𝐼 + 40 = 0 𝐼 2 − 4𝐼 + 4 = 0
−(−4)±�(−4)2 −4(1)(4) 4±√16−16 = 2(1) 2 40𝛺 40W = (2𝐴)2 𝑅, 𝑅 = = 10 Ω. 4
I=𝑥 = P=
21.
a. b. c.
22.
a. b.
23.
a.
Vab + 4 V + 9 V − 12 V = 0, Vab = −13 V + 12 V = −1 V Vab + 4 V + 8 V − 4 V = 0, Vab = 4 V − 12 V = −8 V Vab + 12 V − 5 V + 6 V − 12 V = 0, Vab = −18 V + 17 V = −1 V
4V = 388.3 mA 10.3 Ω 16 V = 1.39 A ET = −4 V + 10 V - 12 V = −6 V, I = 11.5 Ω ET = 8 V - 32 V + 20 V = −4 V, I =
P = 8 mW = I2R, R = I=
b.
4
= 2 = 2A.
I=
= 2 kΩ
E 20 V - E = 2 mA (CW), = RT 3 kΩ + 2 kΩ = 8 mA, R =
E = 10 V
= 1.5 kΩ
E E - 4 V - 10 V E - 14 V = 8 mA (CCW) = = RT 2 kΩ + 1.5 kΩ 3.5 kΩ E = 42 V
I=
24.
a.
Chapter 5
−6 V + 4 V - 12 V - V = 0, V = −18 V + 4 V = −14 V
35
b.
+30 V − 7 V - 8 V - V = 0, V = 30 V − 15 V = 15 V
c.
−14 V - 22 V - V1 + 12 V = 0, V1 = −36 V + 12 V = −24 V V1 − V2 − 12 V = 0, V2 = V1 − 12 V = −24 V − 12 V = −36 V
25.
a. b. c.
26.
a.
I= = 1.25A 6𝛺 𝑉2 = IR = (1.25A)(4Ω) = 5V 50V-10V-𝑉1-5V = 0 𝑉1 = 50𝑉 − 15𝑉 = 35𝑉
10𝑉
+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
b.
27.
28.
a.
V1.8 Ω = IR = (3 A)(1.8 Ω) = 5.4 V 24 V − V1 − 10 V − 5.4 V = 0, V1 = 24 V − 15.4 V = 8.6 V V2.7 Ω = IR = (3 A)(2.7 Ω) = 8.1 V 10 V − 8.1 V − V2 = 0 V2 = 10 V − 8.1 V = 1.9 V
b.
+ 10 V - V1 + 6 V - 2 V - 3 V = 0, V1 = 11 V +10 V - V2 - 3 V = 0, V2 = 7 V
2𝑉 4𝛺
2𝑉 4𝛺
29.
a. b. c. d.
30.
36
= =
100𝑉 𝑅2
= , 𝑅2 =
200𝑉 ; 𝑅3
𝑅3 =
(100𝑉)(4𝛺) 2𝑉
(200𝑉)(4𝛺) 2𝑉
= 200Ω.
= 400Ω.
10 kΩ V3: V2 = 10 kΩ:1 kΩ = 10:1 V3: V1 = 10 kΩ:100 Ω = 100:1 RE (10 kΩ)(60 V) = 54.05 V V3 = 3 = 0.1 kΩ + 1 kΩ + 10 kΩ RT (R + R3 )E (1 kΩ + 10 kΩ)(60 V) V¢ = 2 = 59.46 V = 11.1 kΩ RT
a.
V=
b.
V=
= 20 V
(5 kΩ)(40 V) (2 kΩ + 3 kΩ)(40 V) = 20 V = 10 kΩ 4 kΩ + 1 kΩ + 2 kΩ + 3 kΩ
Chapter 5
(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 + 0.5Ω) 6 kΩ
c.
31.
20 V V1 (1.2 Ω)(20 V) , V1 = = 12 V = 2Ω 1.2 Ω 2Ω 20 V V2 (6.8 Ω)(20 V) , V2 = = 68 V = 2Ω 6.8 Ω 2Ω
a.
E = V1 + 20 V + V2 = 12 V + 20 V + 68 V = 100 V 120 V - V1 - 80 V = 0, V1 = 40 V 80 V - 10 V - V3 = 0, V3 = 70 V
b.
32.
68 Ω(1000 V) V 1000 V = 680 V = 2 , V2 = 68 Ω 100 Ω 100 Ω V 1000 V 2 Ω(1000 V) = 20 V = 1 , V1 = 2Ω 100 Ω 100 Ω E = V1 + V2 + 1000 V = 20 V + 680 V + 1000 V = 1700 V
a.
V1 = 0 V 10 kΩ(50 V - 30 V) V2 = 10 kΩ + 3.3 kΩ + 4.7 kΩ 10 kΩ(20 V) = = 11.11 V 18 kΩ Vx = E1 − V3.3 kΩ 3.3 kΩ(20 V) V3.3 kΩ = 18 kΩ = 3.67 V Vx = 50 V − 3.67 V = 46.33 V
b.
33.
3𝑉 3 𝑘𝛺 3𝑉 3 𝑘𝛺
I=
𝑉
2 = 4 𝑘𝛺 ; 𝑉2 =
=
𝑉4 5 𝑘𝛺
3𝑉 3 𝑘𝛺
; 𝑉4 =
= 1 mA.
(3𝑉)(4𝑘𝛺) 3 𝑘Ω
(3𝑉)(5𝑘𝛺) 3 𝑘Ω
= 4V. = 5V.
E = 3 V + 4 V + 18 V + 5 V = 30 V.
Chapter 5
37
34.
a.
b.
35.
a.
R(20 V) 2.2 kΩ + 1.8 kΩ + R 4(4 kΩ + R) = 20R 16 kΩ + 4R = 20R 16R = 16 kΩ 16 R= kΩ = 1 kΩ 16
4V=
(6 MΩ + R)(140 V) 6 MΩ + R + 3 MΩ 110(9 MΩ + R) = 840 MΩ + 140R 990 MΩ + 110R = 840 MΩ + 140R 30R = 150 MΩ 150 MΩ = 5 MΩ R= 30 110 V =
Rbulb =
= 160 Ω
Vbulb = 8 V =
b. 36.
Rbulb (12 V) 160 Ω(12 V) , Rx = 80 Ω in series with the bulb = 160 Ω + Rx Rbulb + Rx
VR = 12 V - 8 V = 4 V, P =
= 0.2 W, \ 1/4 W okay
VR1 + VR2 = 92V 1 VR2 + VR2 = 92V 5
1 5
𝑉𝑅2 � + 1� = 92 V
𝑉𝑅2 = 𝑅2 =
37.
𝑅1 =
𝑉𝑅2 𝐼𝑅2
𝑉𝑅1 𝐼𝑅1
= 76.67 V
=
=
76.67 𝑉 5 𝑚𝐴
= 15.334 kΩ
92𝑉−76.67𝑉 5𝑚𝐴
= 3.066 kΩ
𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 = 2𝑅3 + 7𝑅3 + 𝑅3 = 10𝑅3
𝑉3 = 38.
92𝑉 1.2
𝑅3 (80𝑉) 10𝑅3
= 8V,
𝑉1 = 2𝑉3 = 2(8𝑉) = 16𝑉, 𝑉2 = 7𝑉3 = 7(8𝑉) = 56V.
= 4(3VR1 ) = 12VR1
a.
E = VR1 + 3VR1 + 12VR1 \RT = R1 + 3R1 + 12R1 = 16R1 = R1 = 38
= 6.4 kΩ
= 400 Ω, R2 = 3R1 = 1.2 kΩ, R3 = 12R1 = 4.8 kΩ Chapter 5
b.
39.
40.
RT =
= 400 kΩ, R2 = 1.2 MΩ, R3 = 4.8 MΩ
= 103 and
= 103 also
a.
Va = 12 V + 5 V = 17 V Vb = 5 V + 16 V = 21 V Vab = 17 V − 21 V = −4 V
b.
Va = −6 V −6 V + 6 V + 10 V − Vb = 0, Vb = 10 V Vab = Va − Vb = −6 V − 10 V = −16 V
c.
−8 V + 3 V − Va = 0, Va = −5 V Vb = −8 V Vab = Va − Vb = −5 V − (−8 V) = −5 V + 8 V = +3 V
a.
IØ =
60 V + 20 V 80 V = 0.8 A = 18 Ω+ 82 Ω 100 Ω Va = 60 V − I(18 Ω) = 60 V − (0.8 A)(18 Ω) = 60 V − 14.4 V = 45.6 V ® 100 V - 60 V 40 V = 5 mA I= = 4(2 kΩ) 8 kΩ Va − I(2 kΩ) + 100 V = 0 Va = (I)(2 kΩ) − 100 V = (5 mA)(2 kΩ) − 100 V = 10 V − 100 V = −90 V
b.
41.
= 6.4 MΩ, R1 =
I=
54𝑉−27𝑉 1𝑘𝛺+2𝑘𝛺+3𝑘𝛺
27𝑉 6 𝑘𝛺
=
= 4.5 𝑚𝐴 (CCW).
𝑉1𝑘𝛺 = 4.5V, 𝑉2𝑘𝛺 = 9V; 𝑉3𝑘𝛺 = 13.5𝑉1𝑘𝛺
a.
b. c. 42.
𝐼𝑅2 = 𝑅1 =
𝑅3 =
Chapter 5
𝑉𝑎 = 27V, 𝑉𝑏 = 27𝑉 + 4.5𝑉 = 31.5𝑉 𝑉𝑐 = 27𝑉 + 4.5𝑉 + 9𝑉 = 40.5 𝑉 𝑉𝑑 = –13.5V, 𝑉𝑒 = 0V 𝑉𝑎𝑏 = –4.5V, 𝑉𝑑𝑐 = –54V, 𝑉𝑐𝑏 = 9V 𝑉𝑎𝑐 = –13.5V, 𝑉𝑑𝑏 = −54𝑉 + 9𝑉 = −45𝑉.
6𝑉+6𝑉 12𝑉 = 10𝛺 10𝛺
𝑉𝑅1 𝐼
𝑉𝑅3 𝐼
=
=
18𝑉−6𝑉 1.2𝐴
= 1.2A
=
−6𝑉+12𝑉 1.2𝐴
12𝑉 1.2𝐴
= 10Ω
= 5Ω
39
= 48 V - 12 V = 36 V
43.
R2 =
= 2.25 kΩ
= 12 V - 0 V = 12 V R3 =
= 0.75 kΩ
= 20 V R4 =
= 1.25 kΩ
VR1 = E - VR 2 - VR 3 - VR 4 = 100 V - 36 V - 12 V - 20 V = 32 V R1 = 44.
= 2 kΩ
a.
Va = −8 V + 14 V = +6 V, Vb = 14 V Vc = +I(10 Ω) − 6 V with 20 V 14 V + 6 V I= =1A = 10 Ω + 10 Ω 20 Ω Therefore, Vc = (1 A)(10 Ω) − 6 V = 10 V − 6 V = 4 V Vd = 0 V
b.
Vab = Va − Vb = 6 V − 14 V = −8 V Vcb= Vc − Vb = 4 V − 14 V = −10 V Vcd = Vc − Vd = 4 V − 0 V = 4 V
c.
Vad = Va − Vd = 6 V − 0 V = 6 V Vca = Vc − Va = 4 V − 6 V = −2 V
45.
V0 = 0 V, V4 = (2 kΩ)(6 mA) + 3 V = 12 V + 3 V = 15 V, V7 = 4 V V10 = V1 − V0 = 12 V − 0 V = 12 V, V23 = V2 − V3 = 4 V − (−8 V) = 4 V + 8 V = 12 V V30 = V3 − V0 = −8 V − 0 V = −8 V, V67 = V6 − V7 = 4 V − 4 V = 0 V 4 V + 8 V 12 V =3A V56 = V5 − V6 = 3 V − 4 V = −1 V, I≠ = = 4Ω 4Ω
46.
V0 = 0 V, V03 = V0 − V3 = 0 V − 0 V = 0 V, V2 = (3 mA)(3.3 kΩ) = 9.9 V V23 = V2 − V3 = 9.9 V − 0 V = 9.9 V, V12 = V1 − V2 = 20 V − 9.9 V = 10.1 V S Ii = S Io Ii = 4 mA + 3 mA + 10 mA = 17 mA
40
Chapter 5
47.
a. 𝑉𝐿 = 𝐼𝐿 . 𝑅𝐿 = (3.5A)(32Ω) = 112V. 𝑉𝑖𝑛𝑡 = 122V – 112V = 10V 𝑅𝑖𝑛𝑡 =
𝑉𝑖𝑛𝑡 𝐼
10𝑉
= 3.5𝐴 = 2.86Ω
b. VR = Voltage regulation = VR = 8.93% Ans. 48.
𝑉𝑁𝐿 − 𝑉𝐹𝐿 𝑉𝐹𝐿
× 100% =
122𝑉−112𝑉 112𝑉
× 100%
39.6 V 3.3 Ω(12 V) = 11.85 V = 3.3 Ω + 43 mΩ 3.343 Ω
a.
VL =
b.
VR =
c.
Is = IL =
¥ 100% =
¥ 100% = 1.27%
= 3.59 A
Ps = EIs = (12 V)(3.59 A) = 43.08 W Pint = I2Rint = (3.59 A)2 43 Ω = 0.554 W 49.
𝐸 𝑅𝑇
30𝑉 2.2 𝑘𝛺+6.8 𝑘𝛺
30𝑉 9 𝑘𝛺
a.
I=
b.
I=
c.
Not for most applications.
Chapter 5
𝐸 𝑅𝑇
= =
30𝑉 9 𝑘𝛺+0.450 𝑘𝛺
=
= 3.33 mA.
= 3.175 mA
41
Chapter 6 1.
a. b. c. d.
R2 and R3 E and R3 R2 and R3 R2 and R3
2.
a. b. c. d.
E, R1, R2, R3, and R4 E, R1, R2, and R3 E and R1 none
3.
a. b.
R3 and R4, R5 and R6 E and R1, R6 and R7
4.
a.
RT =
b.
RT =
c.
RT =
5.
c.
1 1 1 + + 2𝑘𝛺 4𝑘𝛺 25𝑘𝛺
1
=
1 1 1 + + 1.1kΩ 110𝛺 11𝛺
1 0.5×10−3 𝑆+0.25×10−3 𝑆+40×10−6 𝑆
=
=
1 0.79×10−3 𝑆
1 0.909×10−3 𝑆+9.09×10−3 𝑆+90.909×10−3 𝑆
=
= 1.266kΩ
1 100.908×10−3 𝑆
= 9.91Ω
(4 kΩ)(4 kΩ) 12 kΩ = 4 kΩ, 𝑅𝑇 = = 2 kΩ 3 4 kΩ+4 kΩ 20𝛺 10𝛺 𝑅′ 𝑇 = = 5Ω , 𝑅" 𝑇 = = 5Ω 4 2 (5Ω)(5Ω) = 2.5Ω 𝑅𝑇 = 5𝛺+5𝛺 1 1 1 = 𝑅𝑇 = 1 1 1 = 0.5𝑆+0.5×10−3 𝑆+0.5×10−6 𝑆 500×10−3 +0.5×10−3 +0.0005×10−3 + +
=
a.
2𝛺 2𝑘𝛺 2𝑀𝛺
1 500.5×10−3
b.
= 1.998Ω
RT =
=
= 193.57 Ω
RT =
=
42
= 13.33Ω.
a. 𝑅′ 𝑇 =
b.
6.
(40Ω)(20Ω) 40𝛺+20𝛺 1
= 304.14 Ω
Chapter 6
7.
a.
= 3 Ω || 6 Ω = 2 Ω RT = 1.61 Ω =
b.
=
, R=8Ω
= 2 kΩ
RT = 1.8 kΩ =
8.
,
c.
RT = 5.08 kΩ =
a.
RT = 1.02 Ω =
R = 18 kΩ ,
R = 6.8 kΩ
1
1.02 kΩ =
563.73´10 -6 +
1 R
1.020 kΩ =1 R 1.020 kΩ R= = 2.4 kΩ 425´10 -3
575 × 10−3 +
b.
RT = 6 kΩ = R1 = 24 kΩ
c.
9.
1 1 1 1 1 = + + + 1.11 kΩ R 8.2 kΩ 10 kΩ 2 kΩ 1 900.9 ¥ 10−6S = + 121.95 ¥ 10−6S + 100 ¥ 10−6S + 500 ¥ 10−6S R 1 = 178.95 ¥ 10−6S R 1 R= = 5.588 kΩ @ 5.6 kΩ -6 178.95´10 S
a. 2.2kΩ b. About 2 kΩ c. 𝑅𝑇 =
1
1 1 1 1 + + + 2.2𝑘𝛺 33𝑘𝛺 330𝐾𝛺 3.3𝑀𝛺
=
1 488.181×10−6 𝑆
Chapter 6
1 454.545×10−6 𝑆+30.303×10−6 𝑆+3.030×10−6 𝑆+0.303×10−6 𝑆
= 2.048 kΩ
d. 330 kΩ, 3.3 MΩ; 𝑅𝑇 =
e. 𝑅𝑇 reduced.
=
(2.2𝑘𝛺)(33𝑘𝛺) 2.2𝑘𝛺+33𝑘𝛺
= 2.06 kΩ.
43
10.
𝑅𝑇 =
a.
=
1 1 = 0.33S+0.5S+0.1S 0.93S
= 1.075 Ω.
∞ Ω. 𝑅𝑇 = 3Ω // 7Ω = 2.1Ω.
b. c. 11.
1
1 1 1 + + 3Ω 2Ω 10Ω
24 Ω || 24 Ω = 12 Ω
(Two of the 24 Ω resistors “shorted” out.) 0.1 S =
+ 0.08333 S + 0.00833 S
0.1 S =
+ 0.09167 S
= 0.1 S - 0.09167 S = 0.00833 S R1 =
12.
a.
= 120 Ω
RT = = 36 V
b. c.
=6Ω
Is =
=6A = 4.5 A = 1.5 A
d.
44
Is = I1 + I2 6 A = 4.5 A + 1.5 A = 6 A (checks)
Chapter 6
13.
𝑉𝑅1 20𝑉 = = 5A, 𝑅1 4𝛺 𝑉 20𝑉 = 1.667A. 𝐼2 = 𝑅𝑅2 = 12𝛺 2 𝑉𝑅3 20𝑉 = 0.417A. 𝐼3 = 𝑅 = 48𝛺 3 1 1 𝑅𝑇 = 1 1 1 = 0.25𝑆+0.0833𝑆+0.0208𝑆 + +
a. 𝐼1 =
b.
c. 𝐼𝑆 =
=
4𝛺 12𝛺 48𝛺
20𝑉 2.83𝛺
= 7.067𝐴 = 7.07 𝐴.
1 353.38×10−3 𝑆
= 2.83Ω.
d. 𝐼𝑆 = 𝐼1 + 𝐼2 + 𝐼3 = 5A+1.667A+0.417A = 7.07A. e. They match.
14.
a.
I R1 =
VR1 R1
=
24 V = 2.4 mA, 10 kΩ
= 20 mA,
= 3.53 mA
15.
1 1 = -6 1 1 1 100´10 S + 833.333´10 -6 S + 147.06´10 -6 S + + 10 kΩ 1.2 kΩ 6.8 kΩ 1 = = 925.93 Ω 1.08´10 -3 S
b.
RT =
c.
Is =
d.
Is = I1 + I2 + I3 = 2.4 mA + 20 mA + 3.53 mA = 25.93 mA
e.
they match
a.
RT @ 900 Ω
b.
RT =
=
= 25.92 mA
1 1 1 1 1 + + + 20 kΩ 10 kΩ 1 kΩ 91 kΩ 1 -6
-6
-3
-6
50´10 S + 100´10 S + 1´10 S + 10.99´10 S 1 = = 862.07 Ω, very close 1.16´10 -3 S c.
Chapter 6
I3 the most, I4 the least
45
d.
I R3
16.
VR 60 V 60 V = 3.0 mA, I R 2 = 2 = = 6 mA R1 R2 20 kΩ 10 kΩ VR VR 60 V 60 V = 60.0 mA, I R 4 = 4 = = 0.659 mA = 3 = R3 R4 1 kΩ 91 kΩ
I R1 =
VR1
=
60 V E = = 69.6 mA RT 862.07 kΩ Is = 3 mA + 6 mA + 60 mA + 0.659 mA = 69.66 mA (checks)
e.
Is =
f.
always greater
a. 𝑅𝑇 = 8Ω =
(20Ω)(𝑅2 ) 20Ω+𝑅2
160Ω + 8𝑅2 = 20𝑅2
12𝑅2 = 160 Ω ⇒ 𝑅2 =
b. P = 64W =
17.
𝑉2 𝑅
=
𝐸2 𝑅
=
c.
40 3
= 13.33 𝛺 =
f.
2560 3
= 29.21𝑉.
�(120𝑊)(6𝛺) = √720 = 26.83 V.
d. 𝐼𝑆 = 𝐼1 + 𝐼2 + 𝐼3 = 2.24𝐴 + 3𝐴 +
e.
40 3
40 Ω. 3
and 𝐸 2 = � � (64) or 𝐸 = �
𝑉 2 𝐸2 = 𝑅 and E = √𝑃𝑅 = 𝑅 𝐸 26.83V 𝑅2 = = = 8.94Ω 𝐼2 3𝐴 𝑉 𝐸 26.83 = 2.24A 𝐼1 = 𝑅1 = = 𝑅1 12𝛺 1
a. P = b.
𝐸2
160 12
26.83 6𝛺
= 5.24+4.47 = 9.71A.
𝑃𝑆 = E𝐼𝑆=(26.83𝑉)(9.71𝐴) = 260.52 𝑊 = 260.5𝑊
𝑃𝑅1 =
𝑃𝑅2 =
𝐸2 𝑅1
𝐸2 𝑅2
=
=
(26.83𝑉)2 12𝛺
(26.83𝑉)2 8.94𝛺
= 59.99W. =80.52W.
g. 𝑃𝑠 = 𝑃1 + 𝑃2 + 𝑃3 => 260.5W = 59.99+80.52+80.52 = 260.51W 𝑃𝑠 = 260.5𝑊 = 𝑃1 + 𝑃2 + 𝑃3 = 260.5𝑊.
46
Chapter 6
18.
I3 =
=9A
E=
= 36 V = 12.3 A - 10.8 A = 1.5 A
R1 = 19.
a. b. c.
= 24 Ω V = 48 V
48 V = 2.67 mA 18 kΩ 48 V 48 V + Is = + I2 = 16 mA + 4 mA + 2.67 mA = 22.67 mA 3 kΩ 12 kΩ I2 =
2
d.
20.
a. b. c.
21.
−
22.
a.
2
𝐼𝑅2 = 4𝐴 − 1𝐴 = 3𝐴, 𝑅2=
𝑅3 = 𝐼1 ↑ =
RT = RT = RT =
𝐸 𝐼2
=
𝑉𝑅3 𝐸 15𝑉 = = = 15V. 𝐼3 𝐼3 1𝐴 15𝑉 = 3A, 𝐼𝑆 = 𝐼1 + 4𝐴 5𝛺
1
1 1 1 + + 2𝑘𝛺 4.4𝑘𝛺 8𝑘𝛺
=
𝑉𝑅2 𝐼2
=
15𝑉 3𝐴
= 5𝛺.
= 3𝐴 + 4𝐴 = 7𝐴.
1 0.5×10−3 𝑆+227.27×10−6 𝑆+125+×10−6 𝑆
1 500×10−6 𝑆+227.27×10−6 𝑆+125×10−6 𝑆 1 852.27×10−6 𝑆 = 1.173 kΩ.
𝐼𝑅1=
b.
2
E (48 V) V = 192 mW P= = = R R 12 kΩ
𝑉𝑅1 𝑅1
𝐼𝑅3 =
=
𝑉𝑅3 𝑅3
60𝑉 2𝑘𝛺
=
= 30mA, 𝐼𝑅2 =
60𝑉 8𝑘𝛺
= 7.5 mA.
𝑉𝑅2 𝑅2
=
60𝑉 4.4𝑘𝛺
= 13.64 𝑚𝐴.
𝑃𝑅1= 𝑉𝑅1 . 𝐼𝑅1 = (60V)(30mA) = 1.8W.
𝑃𝑅2 = 𝑉𝑅3 . 𝐼𝑅2 = (60V)(13.64mA) = 818.4 mW. 𝑃𝑅3 = 𝑉𝑅2 . 𝐼𝑅3 = (60V)(7.5mA) = 450 mW.
Chapter 6
47
c.
d. e. 23.
24.
a.
𝐼𝑆=
𝐸 𝑅𝑇
=
60𝑉 1.173𝑘𝛺
= 51.15mA, 𝑃𝑠 = 𝐸𝑠 . 𝐼𝑠 = (60V)(51.15mA)
𝑃𝑆 = 3.069W ≈ 3.07 W.
𝑃𝑆 = 1.8 W + 818.4 mW + 450 mW = 3.068 W ≈ 3.07 W(𝑐ℎ𝑒𝑐𝑘𝑠). 𝑅1 = The smallest parallel resistor. Ibulb =
= 66.667 mA
b.
RT =
= 225 Ω
c.
Is =
= 0.533 A
d.
P=
=8W
e.
Ps = 8(8 W) = 64 W
f.
none, Is drops by 66.667 mA
Network redrawn:
60 V
+ −
5Ω
2Ω
20 Ω
12 Ω
RT 7.5 Ω
1.429 Ω
RT = 1.429 Ω || 7.5 Ω = 1.2 Ω E 2 (60 V)2 = Ps = = 3 kW RT 1.2 Ω 25.
a.
b.
48
5 × 60 W = 300 W Ibulbs =
= 2.5 A
Imicro =
= 10 A
ITV =
= 2.67 A
IDVD =
= 208.33 mA
Is = S I = 2.5 A + 10 A + 2.67 A + 208.33 mA = 15.38 A No
Chapter 6
26.
c.
RT =
d.
Ps = E Is = (120 V)(15.38 A) = 1,845.60 W
e.
Ps = 1845.60 W = 300 W + 1200 W + 320 W + 25 W = 1845 W (checks)
a.
b.
27.
28.
29.
6𝛺 10𝛺
= 7.8 Ω
= 3.75 Ω, 3.75Ω‖4𝛺 = 1.935 Ω.
24𝑉+4𝑉 = 14.47A. 1.935𝛺 𝑉 2 (24𝑉+4𝑉)2 = 𝑃4𝛺 = 𝑅 = 4𝛺
𝐼1 =
196W.
c. 𝐼2 = 𝐼1 = 14.47A.
𝐼𝑠 = 10 mA + 5 mA = 15 mA
𝐼2 = 5 mA – 4 mA= 1 mA. a.
SIi = SIo 5A+7A+3A=9A+I 15 A = 9 A + I 6A=I
b.
S Ii = S Io 8 mA = 2 mA + I1 I1 = 8 mA − 2 m A = 6 mA S Ii = S Io I1 + 9 mA = I2 I2 = 6 mA + 9 mA = 15 mA S Ii = S Io I2 = 10 mA + I3 I3 = 15 mA − 10 mA = 5 mA
a.
S Ii = S Io 8 A = 3 A + I2 I2 = 8 A − 3 A = 5 A, I3 = 3 A S Ii = S Io I2 + I3 = I4 I4 = 5 A + 3 A = 8 A
b.
S Ii = S Io Is = 36 mA + 4 mA = 40 mA S Ii = S Io 36 mA = I3 + 20 mA I3 = 36 mA − 20 mA = 16 mA S Ii = S Io 4 mA + 20 mA = I4 I4 = 24 mA I5 = Is = 40 mA
Chapter 6
49
30.
𝐼2 = 5 mA ̶ 2 mA = 3 mA, 𝐼1 = 9 mA − 5 mA = 4 mA, 𝐼3 = 2 mA.
E = 𝐼2 . 𝑅2 = (3 mA)(5kΩ) = 15V = Source Voltage.
𝑅1 =
RT =
31.
a.
𝑉𝑅1 𝐼1
=
𝐸 𝐼1
1
=
15𝑉 4mA
1 1 1 + + 3.75kΩ 5kΩ 7.5kΩ
R1 =
= 3.75kΩ, 𝑅3 = =
𝑉𝑅3 𝐼3
=
𝐸 𝐼3
=
15𝑉 2mA
= 7.5kΩ.
1 266.67×10−6 𝑆+200×10−6 𝑆+133.33×10−6 𝑆
=
1 600×10−6 𝑆
= 1.67 kΩ.
=5Ω
I2 = I - I1 = 3 A - 2 A = 1 A = 10 Ω
R= b.
E = I1R1 = (2 A)(6 Ω) = 12 V I2 =
= 1.33 A
I3 =
=1A
R3 =
= 12 Ω
I = I1 + I2 + I3 = 2 A + 1.33A + 1 A = 4.33 A 32.
a.
I1 =
= 64 mA
I3 =
= 16 mA
Is = I1 + I2 + I3 I2 = Is - I1 - I3 = 100 mA - 64 mA - 16 mA = 20 mA R=
= 3.2 kΩ
I = I2 + I3 = 20 mA + 16 mA = 36 mA b.
P=
= 30 V
E = V1 = 30 V I1 =
=1A
Because R3 = R2, I3 = I2 , and Is = I1 + I2 + I3 = I1 + 2I2
50
Chapter 6
2 A = 1 A + 2I2 I2 =
= 0.5 A
I3 = 0.5 A R2 = R3 =
= 60 Ω = (0.5 A)2 ◊ 60 Ω = 15 W
33.
𝐼2 =
3𝛺 𝐼 15𝛺 1
3𝛺
= 0.2𝐼1 = 0.2(10𝐴) = 2𝐴.
𝐼3 = 2𝛺 𝐼1 = 1.5(10𝐴) = 15𝐴.
𝐼4 =
34.
3𝛺 𝐼 20𝛺 1
=
3 20
3 2
(10𝐴) = 𝐴 = 1.5𝐴.
𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 = 10 A + 2 A + 15 A + 1.5 A = 28.5 A.
a.
I1 =
= 16 mA
I2 = 20 mA - 16 mA = 4 mA
35.
1 kΩ(I T ) 1 kΩ(I T ) = 1 kΩ + 2.4 kΩ 3.4 kΩ 3.4 kΩ(2.5 A) = 8.5 A and IT = 1 kΩ I1 = IT − 2.5 A = 8.5 A − 2.5 A = 6 A
b.
I2.4kΩ = 2.5 A =
a.
RT =
=
= Ix =
= 2.18 Ω I,
I1 =
(6 A) = 3.27 A
I2 =
(6 A) = 1.64 A
I3 =
(6 A) = 1.09 A
I4 = 6 A b.
4 Ω || 4 Ω = 2 Ω 20 Ω(8 A) 20 Ω(8 A) = I2 = = 5.33 A 20 Ω + 2 Ω + 8 Ω 30 Ω I1 =
= 2.67 A
I3 = 8 A − I2 = 8 A − 5.33 A = 2.67 A I4 = 8 A Chapter 6
51
36.
a.
I1 @
b.
I1/I2 = 10 Ω/1 Ω = 10,
c.
I1/I3 = 1 kΩ/1 Ω = 1000, I3 = I1/1000 = 9 A/1000
d.
I1/I4 = 100 kΩ/1 Ω = 100,000,
e.
very little effect, 1/100,000
f.
RT =
(10 A) = 9 A I2 =
@ 0.9 A 9 mA
I4 = I1/100,000 = 9 A/100,000
90 µA
= =
= 0.91 Ω
Ix =
37.
I1 =
(10 A) = 9.1 A excellent (9 A)
g.
I2 =
(10 A) = 0.91 A excellent (0.9 A)
h.
I3 =
(10 A) = 9.1 mA excellent (9 mA)
i.
I4 =
(10 A) = 91 µA excellent (90 µA)
a.
b.
38.
,
3ΩI 39 Ω(1 A) = 1 A, I = = 13 A = I2 3 Ω + 36 Ω 3Ω I1 = I − 1 A = 13 A − 1 A = 12 A CDR: I36Ω =
I3 = I = 24 mA, V12kΩ = IR = (4 mA)(12 kΩ) = 48 V 48 V V I2 = = 12 mA = R 4 kΩ I1 = I− 4 mA − I2 = 24 mA − 4 mA − 12 mA = 8 mA
a. I2 =
3kΩ I R 1
b. I1 =
9kΩ(30mA) 9kΩ+3kΩ
I
=> R= �I1 � 3 kΩ.
R = 3(3 kΩ) = 9 kΩ.
I2 =
52
I1 3
2
= 22.5 mA
= 7.5mA =
22.5mA 3
Chapter 6
39.
91 mA = 𝐼1 + 𝐼2 + 𝐼3 = 𝐼1 + 2𝐼1 + 2𝐼2 = 𝐼1 + 2𝐼1 + 2(2𝐼1) 91 mA = 𝐼1 + 2𝐼1 + 4𝐼1 = 7𝐼1 91𝑚𝐴 7
𝐼1 =
= 13mA.
𝐼2 = 2𝐼1 = 2(13 mA) = 26 mA.
𝐼3 = 2𝐼2 = 2(26 mA) = 52 mA.
𝑅1 =
𝑉𝑅1 𝐼1
𝑅3 =
𝑉𝑅3 𝐼3
𝑅2 =
40.
a.
𝑉𝑅2 𝐼2
=
= =
28𝑉 13𝑚𝐴
= 2.154 kΩ.
28𝑉 52𝑚𝐴
= 0.538 kΩ
28𝑉 26𝑚𝐴
= 1.077 kΩ
P L = V LI L 72 W = 12 V ◊ IL IL =
=6A
I1 = I2 =
41.
b. c.
Psource = EI = (12 V)(3 A) = 36 W = 36 W + 36 W = 72 W (the same)
d.
Idrain = 6 A (twice as much)
𝑅𝑇 = 6Ω ║42Ω = 5.25Ω I2 = I3 =
42.
=3A
1 2
𝐸 𝑅𝑇
=
1 2
21 5.25𝛺
= 4A.
I1 = ( 𝐼2 ) = ( 4𝐴 ) = 2𝐴. I8 Ω =
= 2 A,
I=5A-2A=3A
IR = 5 A + 3 A = 8 A, 43.
a. b.
Chapter 6
V2 =
R=
=2Ω
= 16.48 V
= 11 MΩ || 22 kΩ = 21.956 kΩ
53
V2 = c.
= 16.47 V (very close to ideal)
Rm = 20 V[20,000 Ω/V] = 400 kΩ = 400 kΩ || 22 kΩ = 20.853 kΩ V2 =
d:
a.
= 16.32 V (still very close to ideal) V2 =
= 13.33 V
= 200 kΩ || 11 MΩ = 196.429 kΩ
b.
V2 = c.
= 13.25 V (very close to ideal)
Rm = 400 kΩ = 400 kΩ || 200 kΩ = 133.333 kΩ V2 =
44.
= 11.43 V (a 1.824 V drop from Rint = 11 MΩ level)
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 = 40V.
b. c.
Vab =
22MΩ(40V) 22MΩ+2MΩ
= 20V.
R M = 400V(22,000 Ω/V) = 8.8MΩ.
Vab =
(8.8MΩ)(40V) 8.8MΩ+2MΩ
= 32.59V
(Significant drop from ideal) 𝑅𝑀 = 40V[22,000 𝛺/𝑉] = 880 𝑘𝛺 Vab =
(8.8 kΩ)(40V) 8.8 kΩ+2 MΩ
= 12.22 V
(Significant error). 45.
not operating properly, 6 kΩ not connected RT =
= 1.71 kΩ
RT = 3 kΩ || 4 kΩ = 1.71 kΩ 46.
Vab = E + I4 kΩ ◊ R4 kΩ I4 kΩ =
= 1.6 mA
Vab = 4 V + (1.6 mA)(4 kΩ) = 4 V + 6.4 V = 10.4 V 4 V supply connected in reverse so that I=
= 3.2 mA
and Vab = 12 V - (3.2 mA)(1 kΩ) = 12 V - 3.2 V = 8.8 V obtained 54
Chapter 6
Chapter 7 1.
a. b. c.
R1, R2,. and E are in series; R3, R4 and R5 are in parallel E and R1 are in series; R2, R3 and R4 are in parallel. E and R1 are in series; R2, R3 and R4 are in parallel.
2.
a. b. c.
E1 and R1 in series; R2 and R3 in parallel. E and R1 in series, R2, R3, and R4 in parallel. E, R1, R4 and R6 are in parallel; R2 and R5 are in parallel.
3.
a.
RT = 4 Ω + 10 Ω || (4 Ω + 4 Ω) + 4 Ω = 4 Ω + 10 Ω || 8 Ω + 4 Ω = 4 Ω + 4.44 Ω + 4 Ω = 12.44 Ω
b.
RT = 10 Ω +
c.
RT = 6.8 Ω + 10 Ω || (8.2 Ω + 1.2 Ω) = 6.8 Ω + 10 Ω || 9.4 Ω = 6.8 Ω + 4.85 Ω = 11.65 Ω
4.
a. b. c.
= 10 Ω + 5 Ω = 15 Ω
4Ω + 10 Ω = 2 Ω + 10 Ω = 12 Ω 2 RT = 10 Ω RT = 2 Ω + 8 Ω || (4 Ω + 6 Ω || 12 Ω) = 2 Ω + 8 Ω || (4 Ω + 4 Ω) = 2 Ω + 8 Ω || 8 Ω = 2 Ω + 4 Ω =6Ω RT =
5.
3.3𝑘Ω ∥ 15𝑘Ω = 2.676𝑘Ω
15 kΩ RT
3.3 kΩ 15 kΩ
6.
𝑅𝑇 = 9.6𝑘Ω = 𝑅1 ∥ �𝑅1 +
So that 9.6𝑘Ω =
7.
And 𝑅1 =
a. b. c. d. e.
Chapter 7
𝑅𝑇 = 2 × 2.676𝑘Ω = 5.35kΩ
3.3 kΩ
(𝑅1 )(1.5𝑅1 ) 𝑅1 +1.5𝑅1
2.5(9.6𝑘Ω) 1.5
𝑅1 � 2
=
= 16𝑘Ω
= 𝑅1 ∥ 1.5𝑅1
1.5 𝑅12 2.5𝑅1
=
1.5𝑅1 2.5
yes I2 = Is - I1 = 10 A - 4 A = 6 A yes V3 = E - V2 = 14 V - 8 V = 6 V = 4 Ω || 2 Ω = 1.33 Ω , = 4 Ω || 6 Ω = 2.4 Ω = 1.33 Ω + 2.4 Ω = 3.73 Ω RT =
55
f.
=
= 10 Ω, RT =
Is =
8.
=1A
g.
Ps = EIs = Pabsorbed = (20 V)(1 A) = 20 W
a.
= R1 || R2 = 10 Ω || 15 Ω = 6 Ω RT = || (R3 + R4) = 6 Ω || (10 Ω + 2 Ω) = 6 Ω || 12 Ω = 4 Ω
b.
Is =
= 9 A,
I1 =
I2 = c. 9.
= 10 Ω + 10 Ω = 20 Ω
=
=6A
=3A
I1 = Is − I2 = 6 A − 3 A = 3 A Va = I2R4 = (3 A)(2 Ω) = 6 V
a. 𝑅𝑇 = 𝑅1 + (𝑅2 ∥ 𝑅3 ∥ 𝑅4 ) = 11Ω + 𝐸
80𝑉
𝐼𝑠 = 𝑅 = 20Ω = 4𝐴 𝑇
27Ω 3
= 20Ω
4 3
𝐼4 = 𝐴 (𝐼𝑠 = 𝐼1 = 𝐼2 + 𝐼3 + 𝐼4 ) & 𝐼2 = 𝐼3 = 𝐼4
b. 𝑉1 = 𝐼1 𝑅1 = (4𝐴)(11Ω) = 44𝑉 (∴𝐼1 = 𝐼𝑠) 10.
4 3
𝑉3 = 𝐼3 𝑅3 = � 𝐴� (27Ω) = 36𝑉
Redrawn:
a.
b.
Va = 32 V 8 Ω || 24 Ω = 6 Ω 6 Ω(32 V) = 10.67 V Vb = 6 Ω + 12 Ω
32 V 32 V = = 1.78 A 12 Ω + 6 Ω 18 Ω RT = 72 Ω || 18 Ω || 18 Ω = 8.12 Ω I1 =
9Ω E 32 V = Is = = 3.94 A RT 8.12 Ω 11.
56
a.
Va = 36 V, Vb = 60 V Vc =
5 kΩ(60 V) = 20 V 5 kΩ + 10 kΩ Chapter 7
¬ 60 V - 36 V = 24 mA, I1 = 1 kΩ 60 V 60 V I8kΩ = = 7.5 mA, I10kΩ = = 4 mA 8 kΩ 15 kΩ
b.
24 mA ¨
¨ I = 24 mA + 7.5 mA = 31.5 mA
¨ I2 = 31.5 mA + 4 mA = 35.5 mA 12.
13.
a.
= 1.2 kΩ + 6.8 kΩ = 8 kΩ, = 2 kΩ || = + 2.4 kΩ = 1.6 kΩ + 2.4 kΩ = 4 kΩ = 1 kΩ || 4 kΩ = 0.8 kΩ RT = 1 kΩ ||
b.
Is =
c.
V=
= 19.2 V
𝐸 120 𝑉 = = 10Ω 𝐼 12𝐴
∴ 10Ω =
14.
= 60 mA
𝑅𝑇 = 2𝑅 ∥ 2𝑅 ∥ (𝑅 + 𝑅) = 2𝑅 ∥ 2𝑅 ∥ 2𝑅 ∥ =
𝑅𝑇 =
2𝑅 = 30Ω
= 2 kΩ || 8 kΩ = 1.6 kΩ
2𝑅 3
2𝑅 3
3 2
𝑎𝑛𝑑 𝑅 = (10Ω) = 15Ω RT = 1 Ω || (1 Ω + 1 Ω + RT) = 1 Ω || (2 Ω + RT) 2 Ω + RT 2 Ω + RT = = 1 Ω + 2 Ω + RT 3 Ω + RT RT(3 Ω + RT) = 2 Ω + RT 3RT + = 2 Ω + RT + 2RT − 2 Ω = 0
RT = = RT = −1 ± 1.732 = 0.732 Ω or −2.732 Ω Since RT < 1 Ω and positive choose RT = 0.732 Ω
Chapter 7
57
15.
a.
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 Is =
= 16 mA,
I2 =
= 2.33 mA
= R1 || R2 || R3 = 2 kΩ = R6 + R4 || R5 = 14 kΩ I6 =
= 2 mA
b.
V1 = E = 28 V = R4 || R5 = 6 kΩ || 9 kΩ = 3.6 kΩ V5 = I6 = (2 mA)(3.6 kΩ) = 7.2 V
c.
P=
= 261.33 mW
24 V = 6 A; 24 V − 8 V = 16 V, I2 Ø = VR 2 / R2 = 16 V/2 Ω = 8 A 4Ω 8V I1 Ø = = 0.8 A, I = I1 + I2 = 6 A + 8 A = 14 A 10 Ω I1 Ø =
16.
a.
17.
𝐼1 =
30𝑉 47Ω
a.
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
18.
= 638.298 mA 21 21𝑉 = = 209.02 mA 𝐼2 = 160Ω ∥ 270Ω 100.47Ω
Is =
=5A
I1 = I3 = I4 = b.
58
=1A =4A = (since R¢ = R2) =
= 0.5 A
Va = I3R3 - I4R5 = (4 A)(5 Ω) - (0.5 A)(6 Ω) = 20 V - 3 V = 17 V æI ö = ç 1 ÷ R2 = (0.5 A)(20 Ω) = 10 V è2 ø Chapter 7
19.
𝐸 𝑅1 +𝑅4 ∥(𝑅2 +𝑅3 ∥𝑅5 )
a. 𝐼1 = =
20𝑉 6.545Ω
= 3.056𝐴
b. CDR : 𝐼2 = 𝐼2 =
𝐼3 =
12.224𝐴 8+3 𝐼2 2
=
𝑅4 (𝐼1 ) 𝑅4 +𝑅2 +𝑅3 ∥𝑅5
= 1.111𝐴
20𝑉 4Ω+4Ω∥(4Ω+6Ω∥6Ω)
=
=
20𝑉 4Ω+4Ω∥(4Ω+3Ω)
=
20𝑉 4Ω+4Ω∥(7Ω)
=
20𝑉 4Ω+2.545Ω
4Ω(3.056𝐴) 4Ω+4Ω+6Ω∥16Ω
= 0.56𝐴
𝐼4 = 𝐼1 − 𝐼2 = 3.056A-1.111A = 1.945 A
𝑉𝑎 = 𝐼4 𝑅4 = (1.945A)(4Ω) = 7.78V 20.
a.
𝑉𝑏 = 𝐼3 𝑅3 = (0.56A)(6Ω) = 3.36V IE =
= 2 mA
IC = IE = 2 mA b.
IB = = 24 µA
= 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
d.
VCE = VC - VE = 3.6 V - 2 V = 1.6 V VBC = VB - VC = 2.7 V - 3.6 V = -0.9 V
21.
a. 𝐼2 =
44Ω 4Ω+8Ω
=
44Ω 22Ω
= 2A
b. −44𝑉 + 𝑉1 − 44𝑉 =), 𝑉1 = 88𝑉 c. 𝐼1 = 𝐼2 = 2𝐴
Chapter 7
59
22.
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.
Is = I1 + I2 + I3 + I4 + I5 = =2A+4A+8A+1A+2A = 17 A RT =
23.
a. b.
c.
60
= 1.88 Ω as above
Va = −6 V, Vb = −20 V
20 V =4A 5Ω V 14 V I 2 Ω ® = ab = =7A 2Ω 2Ω 6V I3 Ω ! =2A 3Ω I3Ω = I2Ω + I6V ≠, I6V = I3Ω − I2Ω = 2 A − 7 A = −5 A I + I6V = I5Ω, I = I5Ω − I6V = 4 A − (−5A) = 9 A I5 Ω ¯ =
Vab = Va − Vb = (−6 V) − (−20 V) = −6 V + 20 V = +14 V
Chapter 7
24.
a.
Applying Kirchoff's voltage law in the CCW direction in the upper "window": +18 V + 20 V - V8Ω = 0 V8Ω = 38 V I8Ω =
= 4.75 A
I3Ω =
=2A
KCL: I18V = 4.75 A + 2 A = 6.75 A b. 25.
V = (I3Ω)(6 Ω) + 20 V = (2 A)(6 Ω) + 20 V = 12 V + 20 V = 32 V
𝐼2 𝑅2 = 𝐼3 𝑅3 and 𝐼2 =
𝐼3 𝑅3 𝑅2
=
4𝑅3 20
=
𝑅3 5
(since the voltage across parallel elements is same) 𝐼1 = 𝐼2 + 𝐼3 =
𝑅3 5
+4
KVL: 120 = 𝐼1 𝑅1 + 𝐼3 𝑅3 And 120 =
𝑅3 5
= �
12 𝑅3 5
+ 4� (12) + 4𝑅3
+ 48 + 4𝑅3 = 6.4𝑅3 + 48
6.4𝑅3 = 72Ω
26.
𝑅3 =
72 6.4
= 11.25
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.
æ1 ö V1 = ç A ÷ (10 Ω) = è3 ø æ1 ö V2 = ç A ÷ (10 Ω) = è6 ø æ1 ö V3 = ç A ÷ (10 Ω) = è3 ø Chapter 7
10 V 3 10 V 6 10 V 3 61
E = V1 + V2 + V3 =
= 8.33 V
RT = 27.
a.
= 8.33 Ω
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 I=
b.
= 24 A
I4 =
= 12 A
I7 =
28.
=8A
c.
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
d.
P= = (8 A)22 Ω = 128 W P = EI = (240 V)(24 A) = 5760 W
a.
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 = 40 V/5.53 Ω = 7.23 A
c.
I3 =
2 Ω(7.23 A) 2 Ω( I ) = 1.69 A = 2 Ω + 6.56 2 Ω + 6.56 Ω
I7 =
= 0.375 mA = (0.375 A)2 2 Ω = 0.281 W
29.
a. 𝐸 = (48𝑚𝐴)(1.6𝑘Ω) = 76.8𝑉 b. 𝑅𝐿2 =
48𝑉 12𝑚𝐴
32𝑉 8𝑚𝐴
= 4𝑘Ω, 𝑅𝐿2 =
c. 𝐼𝑅1 = 80𝑚𝐴 − 48𝑚𝐴 = 32𝑚𝐴 𝐼𝑅2 = 32𝑚𝐴 − 12𝑚𝐴 = 20𝑚𝐴 𝐼𝑅3 = 20𝑚𝐴 − 8𝑚𝐴 = 12𝑚𝐴 𝑉
𝑅1 = 𝐼 𝑅1 = 𝑅2 =
62
𝑅1
𝑉𝑅2 𝐼𝑅2 𝑉
=
𝑅2 = 𝐼 𝑅3 = 𝑅3
64𝑉−48𝑉 32𝑚𝐴
48𝑉−32𝑉 20𝑚𝐴 32𝑉 12𝑚𝐴
=
=
16𝑉 32𝑚𝐴
16𝑉 20𝑚𝐴
= 4𝑘Ω
= 0.5 𝑘𝛺
= 0.8 𝑘𝛺
= 2.67 𝑘𝛺
Chapter 7
= 40 mA
30.
= 40 mA - 10 mA = 30 mA = 30 mA - 20 mA = 10 mA = 40 mA = 40 mA - 4 mA = 36 mA R1 =
= 0.5 kΩ
R2 =
= 2 kΩ
R3 =
= 4 kΩ
R4 =
= 1 kΩ
R5 =
= 0.6 kΩ
P1 =
= (40 mA)20.5 kΩ = 0.8 W (1 watt resistor)
P2 =
= (30 mA)22 kΩ = 1.8 W (2 watt resistor)
P3 =
= (10 mA)24 kΩ = 0.4 W (1/2 watt or 1 watt resistor)
P4 =
= (36 mA)21 kΩ = 1.3 W (2 watt resistor)
P5 =
= (40 mA)20.6 kΩ = 0.96 W (1 watt resistor)
All power levels less than 2 W. Four less than 1 W.
Chapter 7
63
31.
32.
a.
yes, RL
Rmax (potentiometer)
b.
VDR:
=3V= R2 =
R1 =
= 400 Ω fi 390 Ω
R2 =
= 266.67 Ω fi 270 Ω
= 0.25 kΩ = 250 Ω
R1 = 1 kΩ - 0.25 kΩ = 0.75 kΩ = 750 Ω = E - VL = 12 V - 3 V = 9 V (Chose
c.
rather than
since numerator of VDR
equation "cleaner")
=9V= 9R1 + 9(R2 || RL) = 12R1
R1 =
fi 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 + 39 kΩ R2 - 10 kΩ2 = 0 R2 = 0.255 kΩ, -39.255 kΩ R2 = 255 Ω R1 = 1 kΩ - R2 = 745 Ω 33.
a.
Vab =
= 32 V
Vbc = 40 V - 32 V = 8 V b.
80 Ω || 1 kΩ = 74.07 Ω 20 Ω || 10 kΩ = 19.96 Ω Vab =
= 31.51 V
Vbc = 40 V - 31.51 V = 8.49 V c.
64
P=
= 12.411 W + 3.604 W = 16.02 W
Chapter 7
d.
P=
= 12.8 W + 3.2 W = 16 W
The applied loads dissipate less than 20 mW of power. 34.
35.
12 V = 1.2 mA 10 kΩ V ab = Va − Vb = 12 V − (−18 V) = 30 V I=
36 kΩ || 6 kΩ || 12 kΩ = 3.6 kΩ = 16.88 V π 27 V. Therefore, not operating properly!
V= 6 kΩ resistor "open"
R¢ = 12 kΩ || 36 kΩ = 9 kΩ, V = 36.
= 27 V
Network redrawn:
24 V =3A 8Ω P6Ω = I2R = (3 A)2 ◊ 6 Ω = 54 W
I8Ω = I6Ω =
37.
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 I3 = I4 =
d.
Chapter 7
I10 =
c.
I6 = I4 = 0.5 A
=1A = 0.5 A
=6A
65
38.
a. 𝐼𝐶𝑆 = 1.5𝑚𝐴 0.3 1 (200𝛺)(1.5𝑚𝐴) 𝑅 𝐼 ⇒ Ω = 𝛺 = 10 mΩ. b. 𝑅𝑠ℎ𝑢𝑛𝑡 = 𝑀 𝐶𝑆 = 30 10 𝐼𝑚𝑎𝑥 − 𝐼𝐶𝑆
39.
150 mA: 𝑅𝑠ℎ𝑢𝑛𝑡 =
(1.5 𝑘𝛺)(100µ𝛺)
300 mA: 𝑅𝑠ℎ𝑢𝑛𝑡 =
40.
41.
600 mA: 𝑅𝑠ℎ𝑢𝑛𝑡 =
= 1Ω.
(1.5 𝑘𝛺)(100µ𝛺) 300𝑚𝐴−0.1 𝑚𝐴
(1.5 𝑘𝛺)(100µ𝛺) 600𝑚𝐴−0.1 𝑚𝐴
𝑉𝑚𝑎𝑥 −𝑉𝐶𝑆 𝐼𝐶𝑆
2V: 𝑅𝑆 =
2𝑉−(1𝑚𝐴)(1000𝛺) 1𝑚𝐴
=
= 0.5Ω = 0.25 Ω
20𝑉−(5µ𝐴)(1.5𝑘𝛺) 50µ𝐴
𝑅𝑆 = 400 kΩ 1 𝛺 b. �𝑉 = 1�𝐼 = = 20,000. 50µ𝐴 𝑐𝑠
2𝑉−1𝑉 1𝑚𝐴
= 1kΩ
= 19kΩ
200𝑉−1𝑉 1𝑚𝐴
200V : 𝑅𝑆 =
= 199kΩ
15MΩ = (0.5V)( 𝛺�𝑉 ) ⇒ 𝛺�𝑉 = 30 × 106 1
1 30 ×106
𝐼𝐶𝑆 = 𝛺
=
a. 𝑅𝑆 =
𝐸 𝐼𝑀
�𝑉
43.
150𝑚𝐴−0.1 𝑚𝐴
a. 𝑅𝑆 =
20V : 𝑅𝑆 =
42.
30𝐴−1.5 𝑚𝐴
b. x𝐼𝑀 =
=
=
5 𝑥200µ𝐴 5 𝑥200µ𝐴
⇒
𝐸 𝑥𝐼𝑀
+
– (𝑅𝑠𝑒𝑟𝑖𝑒𝑠 + 𝑅𝑀 +
– 1000Ω – 𝑅𝑢𝑛𝑘
𝑍𝑒𝑟𝑜 𝐴𝑑𝑗𝑢𝑠𝑡 2
2𝑘𝛺 2
= 23kΩ.
)
– (25kΩ + 1 kΩ + 1kΩ)
25×103 𝑥 3 4
𝑍𝑒𝑟𝑜 𝐴𝑑𝑗𝑢𝑠𝑡 5𝑉 = 200µ𝐴 2 𝑍𝑒𝑟𝑜 𝑎𝑑𝑗𝑢𝑠𝑡 𝑅𝑀 + + 2
− 𝑅𝑀 −
𝐸 𝑅𝑠𝑒𝑟𝑖𝑒𝑠
𝑅𝑢𝑛𝑘 =
= 0.033µ𝐴.
– 25kΩ – 25 × 103
x = , 𝑅𝑢𝑛𝑘 = 8.33 𝑘𝛺 1 2
x = , 𝑅𝑢𝑛𝑘 = 25 𝑘𝛺
66
1 4
x = , 𝑅𝑢𝑛𝑘 = 75 𝑘𝛺
Chapter 7
44.
-
45.
a.
Network redrawn:
Rohmmeter = 1.2 kΩ || (3.1 kΩ + 1.2 kΩ + 1.65 kΩ) = 1.2 kΩ || 5.95 kΩ = 1 kΩ b.
All three resistors are in parallel Rohmmeter =
Chapter 7
R 18 Ω = =6Ω 3 N
67
Chapter 8 1. 2.
3.
4.
5.
6.
𝐼1 =
(8Ω)(8𝐴) 8Ω+2Ω
= 6.4𝐴, 𝐼2 = 8𝐴 − 𝐼1 = 8𝐴 − 6.4𝐴 = 1.6 𝐴.
a. 𝐼1 = 𝐼2 = 25 𝑚𝐴 b. 𝑉2 = 𝐼2 𝑅2 = (25𝑚𝐴)(3.3 𝑘Ω) = 82.5 V 𝑉3 = 𝐼𝑅𝑇 = (25 𝑚𝐴)(2.2 𝑘 Ω + 3.3 𝑘 Ω) = 25 𝑚𝐴 (5.5 𝑘Ω) = 137.5 𝑉
E + 𝑉𝑅1 − 𝑉𝑆 = 0; 𝑉𝑅1 = (8 𝑚𝐴)(3.0 𝑘Ω) = 24 V 𝑉3 = 𝐸 + 𝑉𝑅1 = 12 𝑉 + 24 𝑉 = 36 𝑉 ± a.
Vs = E = 24 V
b.
I2 =
c.
I + I s = I 2, I s = I 2 - I = 6 A - 2 A = 4 A
=6A
𝑉1 = 𝑉2 = 𝑉3 = 𝐼𝑅𝑇 = 0.8 𝐴 [8Ω ∥ 24 Ω ∥ (16 Ω + 8 Ω) ] = 0.8 𝐴 [8Ω ∥ 24 Ω ∥ (24 Ω) ] = 0.8 𝐴 [8Ω ∥ 12 Ω ] = 3.84 V 𝑉2 3.84 𝐴 𝐼2 = = = 0.16 𝐴. 𝑅2 24 Ω 𝑅3 𝑉𝑆 16Ω (3.84𝑉) 16Ω (3.84𝑉) 𝑉3 = = = = 2.56 𝑉. 𝑅3 + 𝑅4 16 Ω + 8 Ω 24 Ω a.
I1 =
= 12 A,
KCL: I + Is - I1 = b.
+
a. I = b. I =
68
=0
- I = 12 A + 3 A - 4 A = 11 A
Vs = E = 24 V VDR: V3 =
7.
=3A
𝐸 𝑅𝑆 E RS
=
=
=6V
22 𝑉 = 4.23 𝐴, 𝑅𝑃 = 𝑅𝑆 = 5.2Ω 5.2 𝑉 12 V = 3.75 mA, 𝑅𝑃 = 𝑅𝑆 = 3.2 𝑘Ω 3.2 Ω
CHAPTER 8
8.
9.
a. E = 𝐼𝑅𝑆 = 6𝐴 (15Ω) = 90 𝑉, 𝑅𝑆 = 15Ω . b. E = 𝐼𝑅𝑆 = 18 𝑚𝐴 ( 3 𝑘Ω ∥ 9 kΩ) = 18 mA (2.25 kΩ) = 40.5 V , 𝑅𝑆 = 2.25 𝑘Ω a. CDR; 𝐼𝐿 =
𝑅𝑆 𝐼(𝑆 ) 𝑅𝑆 + 𝑅𝐿
=
95 Ω (12𝑚𝐴) 95Ω + 15Ω
= 10.36 mA.
b. 𝐸𝑆 = 𝐼𝑆 . 𝑅 =(12 mA) (95Ω) = 1.14 V 𝑅𝑆 = 95Ω 𝐸𝑆 1.14𝑉 𝐼𝐿 = = = 10.36 𝑚𝐴. 𝑅𝑆 + 𝑅𝐿 95Ω + 15Ω Yes, we obtained the same result. 10.
11.
12.
a. E = I𝑅2 = (2𝐴)(6Ω) = 12𝑉; 𝑅𝑆 = 6Ω b. 𝐸𝑇 = 18 𝑉 + 12𝑉 = 30 𝑉; 𝑅𝑇 = 12 Ω + 6 Ω = 18 Ω. c. 𝐼3 =
𝐸𝑇 𝑅𝑇 +95Ω
14.
30 𝑉 18 Ω+95 Ω
= 0.265 𝑚𝐴
a. 𝐼𝑇 = 6.2 𝐴 − 1.2 𝐴 − 0.6 𝐴 = 4.4 𝐴 b. 𝑉𝑆 = 𝐼𝑇 𝑅 = (4.4𝐴)(8Ω) = 35.2 𝑉
𝐼𝑇 ↑ = 9𝐴 − 6𝐴 = 3𝐴 CDR; 𝐼1 =
13.
=
𝑅2 𝐼𝑇 𝑅1 + 𝑅2
=
6Ω (3𝐴) 4Ω + 6Ω
= 1.8𝐴
𝑉2 = I1 𝑅1 = (1.8𝐴)(4Ω) = 7.2Ω
E2 E1 20 V 9 V = R2 R1 2Ω 3Ω = 10 A - 3 A = 7 AØ
a.
IT =
b.
Vab = -IT (R1 || R2 || R3) = - 7A (3 Ω || 6 Ω || 2 Ω) = - 7 A (1 Ω) =- 7V
c.
I3 =
a.
CHAPTER 8
7V = 1.17 A ≠ 6Ω IT = 3 mA + 6 mA - 5 mA = 4 mA ≠ RI 1 kΩ (4 mA) I R1 = 3 T = = 1.25 mA R1 + R3 2.2 kΩ + 1 kΩ V1 = I R1 (R1) = (1.25 mA)(2.2 kΩ) = 2.75 V
69
15.
4 - 4I1 - 8I3 = 0 6 - 2I2 - 8I3 = 0 I1 + I2 = I3 ────────────
a.
I1 =
b. 16.
𝐼1
b.
70
𝐼2
→ ↑ 𝐼3 ← 𝐼1 = 𝐼𝑅1 ; 𝐼2 = 𝐼𝑅3 ; 𝐼3 = 𝐼𝑅2 12 + 15 – 3𝐼3 − 4𝐼1 = 0 15 – 3𝐼3 –12𝐼2 = 0 𝐼1 + 𝐼2 = 𝐼3 𝐼1 = 4.75A; 𝐼2 = 0.25 A; 𝐼3 = 5𝐴 𝐼𝑅1 = 𝐼1 = 4.75 𝐴; 𝐼𝑅2 = 𝐼3 = 5A; 𝐼𝑅2 = 𝐼2 = 0.25A a.
17.
æ4 ö Va = I3R3 = ç A ÷ (8 Ω) = 4.57 V è7 ø
𝐸
𝐼𝐸1 = 𝑅1 = 𝑆
12 𝑉 4Ω
= 3𝐴 ↓, ↑ 𝐼𝐸2 =
𝐸2 𝑅𝑆
=
15 𝑉 3Ω
= 5𝐴 ↑
𝑅𝑃 = 𝑅𝑆 = 4Ω ; 𝑅𝑃 = 𝑅𝑆 = 3Ω Combined/Single current source = 𝐼𝑆 ↑ = 𝐼𝐸2 − 𝐼𝐸1 = 5𝐴 − 3𝐴 = 2𝐴 𝑅𝑃 = 4Ω ∥ 3Ω = 1.714 Ω (1.714Ω)(2𝐴) 𝐼12Ω = = 0.25𝐴 1.714Ω + 12Ω The same current in both parts (a) and (b).
�𝐼�1⃗ ↓ 𝐼3 ⃖𝐼�2�� 15 = 𝐼1 (5.6 𝑘Ω) − 𝐼3 (2.2 𝑘Ω) + 25 = 0 –25 + 𝐼3 (2.2 𝑘Ω) + 𝐼2 (3.3 𝑘Ω) − 40 = 0 𝐼1 + 𝐼2 = 𝐼3 𝐼1 = 𝐼𝑅1 = 2.02 𝑚𝐴, 𝐼2 = 𝐼𝑅2 = 11.01 𝑚𝐴, 𝐼3 = 𝐼𝑅3 = 13.03 𝑚𝐴
CHAPTER 8
18.
a.
-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 = I1 = 2.03 mA = I3 = 0.8 mA = I2 = 1.23 mA = I9.1kΩ
b.
19.
𝐼1 = 𝐼𝑅1 (𝐶𝑊), 𝐼2 = 𝐼𝑅2 (𝑑𝑜𝑤𝑛), 𝐼3 = 𝐼𝑅3 (𝐶𝑊), 𝐼4 = 𝐼𝑅4 (𝑑𝑜𝑤𝑛), 𝐼5 = 𝐼𝑅5 (𝐶𝑊) a.
b.
𝐸1 − 𝐼1 . 𝑅1 − 𝐼2 . 𝑅2 = 0 𝐼2 . 𝑅2 − 𝐼3 . 𝑅3 − 𝐼4 . 𝑅4 = 0 𝐼4 . 𝑅4 − 𝐼5 . 𝑅5 − 𝐸2 = 0 𝐼1 = 𝐼2 + 𝐼3 𝐼3 = 𝐼4 + 𝐼5
𝐸1 − 𝐼2 (𝑅1 + 𝑅2 ) − 𝐼3 𝑅1 = 0.
𝐼2 . 𝑅2 − 𝐼3 (𝑅3 + 𝑅4 ) + 𝐼5 . 𝑅4 = 0.
c.
d.
CHAPTER 8
𝐼3 . 𝑅4 − 𝐼5 (𝑅4 + 𝑅5 ) – 𝐸2 = 0.
𝐼2 (𝑅1 + 𝑅2 ) + 𝐼3 𝑅1 + 0 = 𝐸1 𝐼2 . 𝑅2 − 𝐼3 (𝑅3 + 𝑅4 ) + 𝐼5 𝑅4 = 0 0 + 𝐼3 𝑅4 − 𝐼5 (𝑅4 + 𝑅4 ) = 𝐸2
3𝐼2 + 2𝐼3 + 0 = 15 1𝐼2 – 9𝐼3 + 5𝐼5 = 0 0+5𝐼3 − 8𝐼5 = 12 𝐼3 = 𝐼𝑅3 = −382.17 𝑚𝐴(𝐶𝑊)
71
20.
4 - 4I1 - 8(I1 - I2) = 0 -8(I2 - I1) - 2I2 - 6 = 0 ───────────────
a.
I1 =
, I2 =
= I1 = = I2 =
æ 1 ö æ 5 ö = I1 - I2 = ç- A ÷ - ç- A ÷ = è 7 ø è 7 ø b.
æ4 ö Va = I R 3 R3 = ç A ÷ (8 Ω) = 4.57 V è7 ø
21
𝐼1 ↓ 𝐼2 ↓ –12 – 4𝐼1 − 3(𝐼1 − 𝐼2 ) – 15 = 0 15 – 3(𝐼2 − 𝐼1 ) − 15𝐼2 = 0 By solving 𝐼1 = −4.75𝐴, 𝐼2 = 0.25𝐴 𝐼𝐸1 = 4.75𝐴(𝐶𝐶𝑊) 𝐼𝐸2 = 4.75𝐴 + 0.25𝐴 = 5.0A (up) 𝐼𝑅2 = 𝐼1 − 𝐼2 = (−4.75 𝐴) − (0.25𝐴) = −5.0𝐴 (b) 𝑃𝐸2 = 𝐼𝐸2 .𝐸2 = (5.0𝐴)(15𝑉) = 75𝑊 𝑃𝑅3 = 𝐼𝑅23 . 𝑅3 (0.25)2 . 12 Ω = 750 𝑚𝑊
22.
a. 𝐼1 ↓ 𝐼2 ↓ 15 – 𝐼1 (5.6 𝑘Ω) − 2.2 𝑘Ω ( 𝐼1 − 𝐼2 ) + 25 = 0 –25 – 2.2 kΩ (𝐼2 – 𝐼1 ) – 𝐼2 (3.3 𝑘Ω) − 40 = 0 , solving for I1 and I2 ∴ 𝐼1 = 2.02 𝑚𝐴 , 𝐼2 = 11.01 𝑚𝐴 𝐼𝑅1 = 2.02 mA, 𝐼𝑅2 = 11.01 𝑚𝐴 𝐼𝑅3 = 𝐼2 − 𝐼1 = 8.99 𝑚𝐴 (𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐼2) b. 𝑉3.3 𝑘Ω = 𝐼2 𝑅2 = (11.01 𝑚𝐴)(3.3 𝑘Ω) = 36.33 𝑉
23.
a.
-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 = I1 = 2.03 mA, = 1.23 mA = 2.03 mA - 1.23 mA = 0.80 mA (direction of I1)
b.
72
Va = 6 V − I2(1.1 kΩ) = 6 V − (1.23 mA)(1.1 kΩ) = 6 V − 1.35 V = 4.65 V
CHAPTER 8
24.
25.
a. 𝐼1 ⤸ 𝐼2 ⤸ 𝐼3 ⤸ 15 – 𝐼1 (2) - 1(𝐼1 − 𝐼2 ) = 0 –1(𝐼2 − 𝐼1 ) − 𝐼2 (4) – 5(𝐼2 − 𝐼3 ) = 0 –5(𝐼3 − 𝐼2 ) − 𝐼3 (3) − 12 = 0 ________________________________ 3𝐼1 − 1𝐼2 + 0 = 15 –1𝐼1 + 10𝐼2 − 5𝐼3 = 0 0 – 5𝐼2 + 8𝐼3 = −12 __________________________________________ 𝐼1 = 4.87𝐴; 𝐼2 = −382.17𝑚𝐴; 𝐼3 = −1.261𝐴
(b), (c) – Ignore. d. 𝐼15𝑉 ↑ = 𝐼1 = 4.87𝐴 𝐼12𝑉 ↑ = – 𝐼3 = −(−1.261𝐴) = 1.261𝐴
a.
with
or
−I1 2.2 kΩ − (I1 - I2)9.1 kΩ + 18 V = 0 −18 V - (I2 - I1)9.1 kΩ - 7.5 kΩ I2 − (I2 - I3)6.8 kΩ = 0 -6.8 kΩ(I3 - I2) − 3 V - 3.3 kΩ I3 = 0 ────────────────────── 11.3 kΩ I1 − 9.1 kΩI2 = 18 V 23.4 kΩ I2 − 9.1 kΩI1 − 6.8 kΩ I3 = −18 V 10.1 kΩ I3 − 6.8 kΩ I2 = −3 V ────────────────────── 11.3 kΩ I1 − 9.1 kΩ I2 = 18 V −9.1 kΩ I2 + 23.4 kΩ I2 − 6.8 kΩ I3 = −18 V −6.8 kΩ I2 + 10.1 kΩ I3 = −3 V ─────────────────────────────
b.
I1 = 1.21 mA, I2 = −0.48 mA, I3 = −0.62 mA
c.
I E1 ¯ = I1 − I2 = 1.21 mA − (−0.48 mA) = 1.69 mA I E 2 ! = −I3 = −(−0.62 mA) = 0.62 mA
26.
a.
b. c. 27.
a.
CHAPTER 8
-4 I1 - 3(I1 - I2) - 4(I1 - I3) = 0 -3(I2 - I1) - 10 I2 - 15 - 4(I2 - I3) = 0 -7 I3 - 4(I3 - I1) - 4(I3 - I2) = 0 ───────────────────────────── I1 = -430.4 mA, I2 = -1.05 A, I3 = -395.1 mA = I1 = −430.4 mA -6.8 kΩ I1 - 4.7 kΩ(I1 - I2) + 6 - 2.2 kΩ(I1 - I4) = 0 -6 - 4.7 kΩ(I2 - I1) - 8.2 kΩ (I2 - I3) = 0 -1.1 kΩ I3 - 22 kΩ(I3 - I4) - 8.2 kΩ(I3 - I2) - 9 = 0 -1.2 kΩ I4 - 2 kΩ(I4 - I1) - 22 kΩ(I4 - I3) = 0 ──────────────────────────────────── 73
28.
b.
I1 = -0.597 mA, I2 = -2.13 mA, I3 = -2.27 mA, I4 = -2.03 mA
c.
I6V = I1 − I2 = −0.597 mA − (−2.13 mA) = 1.53 mA P6V = E I6V = (6 V)(1.53 mA) = 9.18 mW
a.
Network redrawn:
b.
-2I1 - 6 - 4I1 + 4I2 = 0 -4I2 + 4I1 - 1I2 + 1I3 - 6 = 0 -1I3 + 1I2 + 6 - 8I3 = 0
c.
I1 = -3.8 A, I2 = -4.20 A, I3 = 0.20 A
d.
= (6 V)(0.2 A) = 1.2 W = (6 V)(4.2 A) = 25.2 W = 1.2 W + 25.2 W = 26.4 W
29.
74
a.
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
b.
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
c.
b
IC/IB = 4.42 mA/63.02 µA = 70.14
CHAPTER 8
30.
I24V = I6Ω = I10Ω = I12V = 3 A (CW) I4Ω = 3 A (CCW)
24 V - 6I1 - 4I2 - 10I1 + 12 V = 0 and 16I1 + 4I2 = 36 I1 - I2 = 6 A ─────────────────── I 1 = 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
31.
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)
CHAPTER 8
75
32.
a.
b. 33.
a.
b.
34.
a. b.
35.
a.
b. c.
36.
(4 + 8)I1 - 8I2 = 4 (8 + 2)I2 - 8I1 = -6 ─────────────
æ 1 ö æ 5 ö 4 I 8Ω¯ = I1 - I 2 = ç- A ÷ - ç- A ÷ = A è 7 ø è 7 ø 7
𝐼1 ⤸ 𝐼2 ⤸ (4 + 3) 𝐼1 – 3𝐼2 = −12 − 15 (3 + 12) 𝐼2 − 3𝐼1 = 15 𝐼3Ω ↑= 𝐼2 − 𝐼1 = 0.25𝐴— 4.75𝐴 = 5.0𝐴 𝐼1 ⤸ 𝐼2 ⤸ 𝐼1 (5.6 kΩ + 2.2 kΩ) – 2.2 kΩ (𝐼2 ) = 15 + 25 = 40 𝐼2 (2.2 kΩ + 3.3 kΩ) - 2.2 kΩ (𝐼1 ) = –25 -40 = -65 𝐼𝐸1 = 𝐼1 = 2.02 𝑚𝐴, 𝐼𝐸2 ↑ = 11.01 𝑚𝐴 𝐼𝐸3 ↓ = 𝐼1 − 𝐼2 = (2.02 𝑚𝐴) − (−11.01𝑚𝐴) = 13.03 mA 𝐼1 ⤸ 𝐼2 ⤸ 𝐼3 ⤸ 𝐼1 (2 + 1) − 1𝐼2 = 15 𝐼2 (1 + 4 + 5) − 1𝐼1 − 5𝐼3 = 0. 𝐼3 (5 + 3) − 5𝐼2 = −12
𝐼1 = 4.87𝐴; 𝐼2 = −382.17 𝑚𝐴, 𝐼3 = −1.261𝐴 𝐼𝑅2 = 𝐼1 − 𝐼2 = (4.84𝐴) − (−382.17𝑚𝐴) = 5.252 𝐴
a.
b. c.
(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 Ø = I1 − I2 = 1.21 mA − (−0.48 mA) = 1.69 mA ≠ = −I3 = −(−0.62 mA) = 0.62 mA
76
CHAPTER 8
37.
a.
(3 Ω + 6 Ω)I1 − 6 Ω I2 = 9 V (6 Ω + 2 Ω)I2 − 6 Ω I1 = 20 V ──────────────────── 9 Ω I 1 − 6 Ω I2 = 9 V −6 Ω I1 + 8 Ω I2 = 20 V ──────────────────── I1 = 5.33 A, I2 = 6.5 A
38.
39.
b.
Vab = 2 Ω (I2) − 20 V = 2 Ω(6.5 A) − 20 V = 13 V − 20 V = −7 V
a.
a.1
b.
I1 = -0.597 mA, I2 = -2.13 mA, I3 = -2.27 mA, I4 = -2.03 mA
c.
I22kΩ = I4 − I3 = −2.03 mA − (−2.27 mA) = 240 µA V22kΩ = IR = (240 ¥ 10−6A)(22 kΩ) = 5.28 V
a.
I1(6.8 kΩ + 4.7 kΩ - 2.2 kΩ) - 4.7 kΩ (I2) - 2.2 kΩ I4 = 6 I2(4.7 kΩ + 8.2 kΩ) - 4.7 kΩ(I1) - 8.2 kΩ(I3) = -6 I3(8.2 kΩ + 1.1 kΩ + 22 kΩ) - 8.2 kΩ I2 - 22 kΩ I4 = -9 I4(2.2 kΩ + 1.2 kΩ + 22 kΩ) - 22 kΩ(I3) - 2.2 kΩ(I1) = 0 ───────────────────────────
a.2 𝐼1 ⤸ 𝐼2 ⤸ 𝐼3 ⤸
(1Ω + 2Ω + 4Ω) 𝐼1 − 2Ω 𝐼2 − 4Ω 𝐼3 = 16 𝑉 (2Ω + 2Ω + 10Ω) 𝐼2 − 2Ω 𝐼1 − 10Ω 𝐼3 = −25 𝑉 (4Ω + 10Ω + 8Ω) 𝐼3 − 10Ω 𝐼2 − 4Ω 𝐼1 = 25 𝑉 _____________________________________ 7𝐼1 − 2𝐼2 − 4𝐼3 = 16 2𝐼1 − 14𝐼2 − 10𝐼3 = 25 4𝐼1 + 10𝐼2 − 22𝐼3 = −25 b. 𝐼1 = 3.1788𝐴; 𝐼2 = −0.1585 𝐴 ; 𝐼3 = 1.6422𝐴
CHAPTER 8
77
40.
a.
a.3
I1(2 Ω + 4 Ω) − I24 Ω = −6 V I2(4 Ω + 1 Ω) − I14 Ω − I31 Ω = −6 V I3(1 Ω + 8 Ω) − I21 Ω = 6 V ─────────────────────────── 6I1 − 4I2 = -6 V −4I1 + 5I2 − I3 = –6 V − 1I2 + 9I3 = 6 V ─────────────────
41.
b.
I1 = −3.8 A, I2 = −4.20 A, I3 = 0.2 A I R1 = I1 − I2 = −3.8 A − (−4.20 A) = −3.8 A + 4.20 A = 0.4 A
c.
I1Ω = I2 − I3 = −4.20 A − 0.2 A = −4.4 A + -V1Ω = (I1Ω)(1 Ω) = (−4.4 A)(1 Ω) = −4.4 V
a. At V1: S Ii = S Io
0=
V -V V1 +5 A + 1 2 2Ω 8Ω
At V2: S Ii = S Io V V1 - V2 = 3A+ 2 8Ω 4Ω
é1 1 ù é1 ù and V1 ê + ú - V2 ê ú = -5 ë2 8 û ë8 û é1 ù é1 1 ù -V1 ê ú + V2 ê + ú = -3 ë8 û ë8 4 û ───────────────────────────
78
b.
V1 = −10.27 V, V2 = −11.36 V
c.
V8Ω = V1 − V2 = −10.27 V − (−11.36 V) = 1.09 V
d.
I2Ω ≠ =
10.27 V V1 = = 5.14 A 2Ω 2Ω 11.36 V V I4Ω ≠ = 2 = = 2.84 A 4Ω 4Ω
CHAPTER 8
42.
a.
𝑉01
𝑉02
− 𝑁𝑜𝑑𝑒 𝑉𝑜𝑙𝑡𝑎𝑔𝑒𝑠
At 𝑉1 ∶ ∑ 𝐼𝑖 = ∑ 𝐼𝑜 . 0=
𝑉1 8Ω
+ 10𝐴 + 𝐼6Ω
And 𝑉1 − 6Ω 𝐼 − 54𝑉 − 𝑉2 = 0 Or, I =
𝑉1 −𝑉2 −54 6Ω 𝑉
=
𝑉1 𝑉 + 2 6Ω 6Ω 𝑉
− 9𝐴 𝑉
So that, 0 = 8Ω1 + 10𝐴 + 6Ω1 − 6Ω2 − 9𝐴 or, 𝑉1 �
1 1 1 + � − 𝑉2 � � 8Ω 6Ω 6Ω
= −10𝐴 + 9𝐴 = −1𝐴. 𝑉
𝑉
2 + 5Ω2 At 𝑉2 : ∑ 𝐼𝑖 = ∑ 𝐼𝑜 ⇒ 𝐼 = 20Ω
𝑉
𝑉
𝑉
𝑉
2 Or, 6Ω1 − 6Ω2 − 9𝐴=20Ω + 5Ω2
And, 𝑉2 �
1 1 1 1 + + � − 𝑉1 � � 20Ω 6Ω 5Ω 6Ω
Resulting in 𝑉1 � 1
b.
1
= −9𝐴.
1 1 1 + � − 𝑉2 � � 8Ω 6Ω 6Ω 1
1
= −1𝐴.
+ � = −9𝐴. –𝑉1 � � + 𝑉2 � + 20Ω 6Ω 6Ω 5Ω 0.292 – 0.167 𝑉2 = −1𝐴
–0.167𝑉1 + 0.417𝑉2 = −9𝐴. c.
43.
a.
𝑉1 = −20.45𝑉; 𝑉2 = −29.77𝑉. 𝐼20Ω =
𝑉2 20Ω
= 1.49𝐴.
At V1: S Ii = S Io
4A =
V1 V1 - V2 + +2A 2Ω 4Ω
At V2: S Ii = S Io V -V V V 2A+ 1 2 = 2 + 2 4Ω 20 Ω 5 Ω or
b.
CHAPTER 8
é1 1 ù é1 ù V1 ê + ú - V2 ê ú = 2 ë2 4 û ë4 û é1 ù é1 1 1 ù -V1 ê ú + V2 ê + + ú=2 ë4 û ë 4 20 5 û
V1 = 4.8 V, V2 = 6.4 V
79
c.
44.
a.
I1: P = V1I1 = (4.8 V)(4 A) = 19.2 W I 2: P =
At V1: S Ii = S Io 0=6A +
b. c.
45.
= 3.2 W
V1 V1 - V2 V1 - V2 + + 5Ω 3Ω 2Ω
At V2: S Ii = S Io V V V - V2 V1 - V2 + = 2 + 2 7A + 1 3Ω 2Ω 4Ω 8Ω é 1 é 1 1 1 ù 1 ù + + + so that V1 ê ú - V2 ê ú = −6 A ë5 Ω 3 Ω 2 Ω û ë3 Ω 2 Ω û é 1 é 1 1 1 1 ù 1 ù V2 ê + + + + ú - V1 ê ú =7A ë4 Ω 8 Ω 3 Ω 2 Ω û ë3 Ω 2 Ω û ──────────────────────────────────── or 1.03V1 − 0.833V2 = −6 −0.833V1 + 1.21V2 = 7 ─────────────────── V1 = −2.59 V, V2 = 4 V V2Ω = V3Ω = V2 − V1 = 4 V − (−2.59 V) = 6.59 V V5Ω = V1 = −2.59 V V4Ω = V8Ω = V2 = 4 V
a. Source conversion: I3 = At V1: S Ii = S Io
0=
12 V = 3 A, Rp = R3 = 4 Ω 4Ω
V1 V V - V2 + 1 +5 A + 1 +3 A 3Ω 6Ω 4Ω
At V2: S Ii = S Io
3A+
Rewritten:
b. c.
80
V1 - V2 V = 2 +4A 4Ω 8Ω é 1 1 1 ù V2 V1 ê + + = −5 A − 3 A úë3 Ω 6 Ω 4 Ω û 4 Ω é 1 ù é 1 1 ù -V1 ê + ú + V2 ê ú = −4 A + 3 A ë4 Ω û ë4 Ω 8 Ω û
V1 = −14.86 V, V2 = −12.57 V 14.86 V I6Ω ≠ = = 2.48 A 6Ω
CHAPTER 8
46.
a.
15 V = 5 A, Rp = 3 Ω 3Ω
Source conversion: Is = V1
5A
6Ω V2
3Ω
5Ω
4Ω 3A
47.
b.
V1 = 0 V (tied to ground) S Ii = S Io V V V 0= 2 + 2 + 2 +3A 6Ω 4Ω 5Ω é 1 1 1 ù + + and V2 ê ú = −3 A ë6 Ω 4 Ω 5 Ω û V2[616.67 mS] = −3 A 3A V2 = = 4.86 V 616.67 mS
c.
V2 - = −4.86 V
+
a.
20 Ω V2
V1 20 Ω
}
2A
9Ω
10 Ω 18 Ω
4A
4Ω
V 1: S I i = S I o
0=2A+
V1 V1 - V2 + 9 Ω 10 Ω
V 2: S I i = S I o V V V1 - V2 +4A = 2 + 2 10 Ω 18 Ω 4 Ω ────────────────────────────── 0.211 V1 − 0.1 V2 = −2 −0.1 V1 + 0.405 V2 = 4 ─────────────────── b.
CHAPTER 8
V1 = −5.43 V, V2 = 8.53 V
81
VR 4 = V1 − V2 + -
48.
a.
= (−5.43 V) − (8.53 V) = −13.96 V V2
V1 2Ω
2Ω
2Ω
9Ω
5A
V3
3.94 Ω
7Ω
2Ω
4Ω
c.
1.33 Ω
V 1: S I i = S I o V - V2 V 0=5A+ 1 + 1 2Ω 2Ω V 2: S I i = S I o V2 V - V3 V1 - V2 = + 2 2Ω 3.94 Ω 2Ω V 3: S I i = S I o V3 V2 - V3 = 2Ω 1.33 Ω
49.
b.
V1 = −6.556 V, V2 = −3.113 V V3 = −1.245 V
c.
I9Ω =
3.113 V V2 = = 0.346 A ≠ 9Ω 9Ω
a.
At V1: S Ii = S Io
0=5A+
V1 V1 - V3 + 2Ω 6Ω
At V2: S Ii = S Io
5A=2A+
V2 4Ω
At V3: S Ii = S Io
V V1 - V3 +2 A + 3 6Ω 5Ω
82
CHAPTER 8
Rewritten: é 1 1 1 ù + V3 = −5 A V1 ê úë2 Ω 6 Ω û 6 Ω é 1 ù V2 ê ú=5A−2A ë4 Ω û é 1 1 1 ù + V1 = 2A V3 ê úë6 Ω 5 Ω û 6 Ω ───────────────────────
50.
b.
V1 = −6.92 V, V2 = 12 V, V3 = 2.3 V
c.
I4Ω =
12 V V1 = =3A 4Ω 4Ω
a.
Node V1:
S Ii = S Io 2A=
Supernode V2, V3: 0= Independent source: V2 - V3 = 24 V or V3 = V2 - 24 V 2 eq. 2 unknowns: =2A =0 ───────────────────── 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
CHAPTER 8
83
51.
Supernode:
S Ii = S Io 3A+4A=3A+
2 eq. 2 unk.
Subt. V2 = 16 V + V1 4A= and V1 = 48 V V2 = 16 V + V1 = 64 V ────────────────── 52.
a.
é1 1 ù é1 ù V1 ê + ú - V2 ê ú = −5 ë2 8 û ë8 û é1 ù é1 1 ù −V1 ê ú + V2 ê + ú = −3 ë8 û ë8 4 û ──────────────────── V1 = −10.27 V, V2 = −11.36 V
53.
b.
= V1 = −10.27 V,
a.
(𝑉01 𝑉02 ) 1 1 1 𝑉1 � + � − 𝑉2 � � = 8 6 6
= V2 = −11.36 V
–10A + 9A = –1A
1 1 1 1 + + � − 𝑉1 � � = −9𝐴 6 20 5 6 𝑉1 = −20.45𝑉; 𝑉2 = −29.77 𝑉 𝑉1 − 𝑉6Ω − 54𝑉 − 𝑉2 = 0 𝑉6Ω = 𝑉1 − 𝑉2 − 54𝑉 = 20.45 − (−29.77) − 54 = −44.68 𝑉. 𝑉2 �
b.
84
CHAPTER 8
54.
é 1 é 1 ù 1 ù V1 ê + ú - V2 ê ú = −2 A ë9 Ω 10 Ω û ë10 Ω û é 1 é 1 ù 1 1 ù + + V2 ê ú - V1 ê ú =4A ë10 Ω 18 Ω 4 Ω û ë10 Ω û
a.
é 1 é 1 ù 1 ù + V1 ê ú - V2 ê ú = −2 A ë9 Ω 10 Ω û ë10 Ω û é 1 ù é 1 1 1 ù + + -V1 ê ú + V2 ê ú =4A ë10 Ω û ë10 Ω 18 Ω 4 Ω û ────────────────────────────── 0.211 V1 − 0.1 V2 = −2 −0.1 V1 + .405 V2 = 4 ──────────────────────────────
b.
55.
c.
V1 = −5.43 V, V2 = 8.53 V
d.
I4Ω =
8.53 V V2 = = 2.13 AØ 4Ω 4Ω
é 1 é 1 ù 1 ù + ú - V2 ê ú = −5 A ë2 Ω 2 Ω û ë2 Ω û é 1 é 1 ù é 1 ù 1 1 1 ù + + + V2 ê ú - V1 ê ú - V3 ê ú =0 ë2 Ω 9 Ω 7 Ω 2 Ω û ë2 Ω û ë2 Ω û é 1 é 1 ù 1 1 ù + + V3 ê ú - V2 ê ú =0 ë2 Ω 2 Ω 4 Ω û ë2 Ω û ──────────────────────────────
a.
V1 ê
V1 = −6.556 V, V2 = −3.113 V V3 = −1.245 V I9Ω =
b.
56.
-3.113 V V2 = = 0.346 A ≠ 9Ω 9Ω
é 1 1 1 ù + V = −5 A V1 ê ú6 Ω 3 6 Ω 2 Ω ë û é 1 ù V2 ê ú=5A−2A ë4 Ω û é 1 1 1 ù + V = 2A V3 ê ú6 Ω 1 5 Ω 6 Ω ë û ───────────────────────
a.
b.
V1 = −6.92 V, V2 = 12 V, V3 = 2.3 V
c.
I2Ω =
CHAPTER 8
6.92 V V2 = = 3.46 A 2Ω 2Ω 85
57.
V1
a. 12 A
1Ω
2Ω
10 Ω
2A
8Ω
é 1 1 1 1 ù + + + V1 ê ú = 14 A ë1 Ω 2 Ω 10 Ω 8 Ω û V1[1.725 S] = 14 A 14 V1 = V 1.725 = 8.1 V
58.
59.
b.
Va = 0 V, Vb = 8.12 V Vab = Va − Vb = 0 V − 8.12 V = −8.12 V
a.
Same figure used for problem 46. V1 = 0 V é 1 1 1 1 ù + + V2 ê V1= −3 A ú6 Ω 4 Ω 5 Ω 6 Ω ë û
b.
V1 = 0 V \ V2[0.617 S] = −3 A -3 V2 = V 0.617 = -4.86 V
c.
I5Ω =
4.86 V V2 = = .972 A 5Ω 5Ω
Mesh analysis : 𝐼1 ⤸ 𝐼2 ⤸
𝐼3 ⤸
(2𝛺 + 6𝛺 + 10𝛺)𝐼1 − 2𝛺𝐼2 − 10𝐼3 = 18𝑉 ( 2𝛺 + 2𝛺 + 5𝛺)𝐼2 − 2𝛺 𝐼1 − 5𝛺 𝐼3 = 0
(5𝛺 + 20𝛺 + 10𝛺) 𝐼3 − 10𝛺𝐼1 − 5𝛺𝐼2 = 0
_______________________________________
Rewritten :
18𝐼1 − 2𝐼2 − 10𝐼3 = 18
−2𝐼1 + 9𝐼2 − 5𝐼3 = 0
−10𝐼1 − 5𝐼2 + 35𝐼3 = 0
___________________________________________
86
CHAPTER 8
b. c. d.
𝐼1 = 1.312𝐴 , 𝐼2 = 0.543𝐴 , 𝐼3 = 0.452 𝐴
𝐼𝑅3 = 𝐼3 − 𝐼1 = 0.452𝐴 − 1.312𝐴 = −0.86 𝐴 No No ; 2𝛺�10𝛺 = 1�5 ≠ 2𝛺�20𝛺 = 1�10 0𝑉1
60.
0𝑉2
0𝑉3
Source conversion : I =
b. c. d.
𝐸 𝑅3
=
18𝑉 6Ω
= 3𝐴.
𝑅𝑠 = 𝑅𝑝 = 6Ω. 1 1 1 1 1 + �− 𝑉2 − 𝑉 = 3𝐴. 𝑉1 � + 6Ω 2Ω 2Ω 2Ω 2Ω 3 1 1 1 1 1 + 𝑉1 − 𝑉 = 0. 𝑉2 � + �− 2Ω 5Ω 10Ω 2Ω 5Ω 3 1 1 1 1 1 + 𝑉1 − 𝑉 = 0. 𝑉3 � + �− 5Ω 2Ω 20Ω 2Ω 5Ω 2 Or, 1.167 𝑉1 − 0.5𝑉2 − 0.5𝑉3 = 3 –0.5𝑉1 + 0.8𝑉2 − 0.2𝑉3 = 0 –0.5𝑉1 − 0.2𝑉2 − 0.75𝑉3 = 0 ___________________________ 𝑉1 = 10.12𝑉; 𝑉2 = 8.58𝑉; 𝑉3 = 9.04𝑉 𝑉5Ω = 𝑉2 − 𝑉3 = −0.46𝑉 No. No; 2Ω�10Ω = 1�5 ≠ 2Ω�20Ω = 1�10.
𝐼1 ⤸ 𝐼2 ⤸ 𝐼3 ⤸ Source conversation : E = IR = (20mA)(2kΩ) = 40V. 𝑅𝑠 = 2𝑘Ω 𝐼1 (2𝑘Ω + 33𝑘Ω + 3.3𝑘Ω) − 33𝑘Ω𝐼2 − 3.3𝑘Ω𝐼3 = 40𝑉 𝐼2 (33𝑘Ω + 56𝑘Ω + 36𝑘Ω) − 33𝑘Ω𝐼1 − 36𝑘Ω𝐼3 = 0𝑉 𝐼3 (36𝑘Ω + 3.3𝑘Ω + 5.6𝑘Ω) − 3.3𝑘Ω𝐼1 − 36𝑘Ω𝐼2 = 0𝑉 __________________________________________________
61.
b. c. d. CHAPTER 8
38.3𝑘Ω𝐼1 − 33𝑘Ω𝐼2 − 3.3𝑘Ω𝐼3 = 40 −33𝑘Ω𝐼1 + 125𝑘Ω𝐼2 − 36𝑘Ω𝐼3 = 0 −3.3𝑘Ω𝐼1 − 36𝑘Ω𝐼2 + 44.9𝑘Ω𝐼3 = 0 __________________________________ 𝐼1 = 1.61𝐴 ; 𝐼2 = 0.597𝐴; 𝐼3 = 0.597𝐴. 𝐼5 = 𝐼2 − 𝐼3 = 0.597𝐴 − 0.597𝐴 = 0. Yes Yes. 87
0𝑉1
62.
0𝑉2
b. c. d.
63.
1 1 1 1 1 + + 𝑉2 − 𝑉 = 20𝑚𝐴. �− 2𝑘Ω 33𝑘Ω 56𝑘Ω 33𝑘Ω 56𝑘Ω 3 1 1 1 1 1 + + 𝑉1 − 𝑉 = 0. 𝑉2 � �− 33𝑘Ω 3.3𝑘Ω 36𝑘Ω 33𝑘Ω 36𝑘Ω 3 1 1 1 1 1 𝑉3 � + + 𝑉2 − 𝑉 = 0. �− 56𝑘Ω 36𝑘Ω 5.6𝑘Ω 36𝑘Ω 56𝑘Ω 1
𝑉1 �
Rewritten: 548.16𝑉1 − 30.3𝑉2 − 17.86𝑉3 = 20 × 103 −30.3𝑉1 + 361.11𝑉1 − 27.78𝑉3 = 0 −17.86𝑉1 − 27.78𝑉2 + 224.21𝑉3 = 0 ______________________________________ 𝑉1 = 36.78𝑉 ; 𝑉2 = 3.34𝑉; 𝑉3 = 3.34𝑉 𝑉𝑅𝑆 = 𝑉2 − 𝑉3 = 3.34𝑉 − 3.34𝑉 = 0 𝑉. Yes. Yes.
Mesh Analysis: 𝐼1 ⤸ 𝐼2 ⤸ 𝐼3 ⤸ (1𝑘Ω + 2𝑘Ω + 2𝑘Ω) 𝐼1 − 2𝑘Ω𝐼2 − 2𝑘Ω𝐼3 = 15𝑉 −2𝑘Ω𝐼1 + (2𝑘Ω + 2𝑘Ω + 2𝑘Ω) 𝐼2 − 2𝑘Ω𝐼3 = 0 −2𝑘Ω𝐼1 − 2𝑘Ω𝐼2 + (2𝑘Ω + 2𝑘Ω + 2𝑘Ω)𝐼3 = 0 ________________________________________ 𝐼1 = 5𝑚𝐴; 𝐼2 = 2.5𝑚𝐴; 𝐼3 = 2.5𝑚𝐴 𝐼1 = 𝐼15𝑉 = 5𝑚𝐴 Nodal Analysis: Source conversion: I =15𝑉�1𝑘Ω = 15𝑚𝐴; 𝑅𝑃 = 1𝑘Ω 0𝑉1 0𝑉3
88
0𝑉3
0𝑉2
1 1 1 1 1 + + 𝑉2 − 𝑉 = 15𝑚𝐴 �− 1𝑘Ω 2𝑘Ω 2𝑘Ω 2𝑘Ω 2𝑘Ω 3 1 1 1 1 1 𝑉1 + � + + 𝑉 =0 − � 𝑉2 − 2𝑘Ω 2𝑘Ω 2𝑘Ω 2𝑘Ω 3 2𝑘Ω 1 1 1 1 1 𝑉1 − 𝑉2 + � + + − �𝑉 = 0 2𝑘Ω 2𝑘Ω 2𝑘Ω 2𝑘Ω 3 2𝑘Ω __________________________________________ 𝑉1 = 10𝑉; 𝑉2 = 5𝑉; 𝑉3 = 5𝑉. 𝑉1 = 10𝑉 = 𝐸 − 𝐼𝑅𝑆 = 15𝑉 − 𝐼(1𝑘Ω) 15𝑉 − 10𝑉 𝐼= = 5𝑚𝐴. 1𝑘Ω 𝑉1 �
CHAPTER 8
64.
Mesh analysis : 𝐼1 ⤸ 𝐼2 ⤸ 𝐼3 ⤸ Source conversion : E = IRs = (4A)(10Ω) = 40V, Rs = 10Ω (10 + 10 + 20)𝐼1 − 10𝐼2 − 20𝐼3 = 40 (10 + 20 + 20)𝐼2 − 10 𝐼1 − 20 𝐼3 = 0 (20 + 20 + 10) 𝐼3 − 20𝐼1 − 20𝐼2 = 0 _______________________________________ Rewritten : 40𝐼1 − 10𝐼2 − 20𝐼3 = 40 −10𝐼1 + 50𝐼2 − 20𝐼3 = 0 −20𝐼1 − 20𝐼2 + 50𝐼3 = 0 ___________________________________________ 𝐼1 = 1.647𝐴 , 𝐼2 = 0.7059𝐴 , 𝐼3 = 0.941 𝐴 𝐼1 = 𝐼40𝑉 = 1.647𝐴 ≅ 1.65𝐴 𝑉10Ω = 16.47V , ∴ 𝑉 = 40𝑉 − 16.47 𝑉 Vs = 23.53V 𝑉
𝐼𝑠 = � �= � 𝑅 𝑠
23.53𝑉 � 10Ω
= 2.353 A
Nodal analysis :
𝑉1 �
1 1 1 1 + + � − 𝑉2 10 10 20 20
𝑉2 �
𝑉1 = 23.53 𝑉 ,
IRs =
65.
𝑉1 𝑅𝑠
𝑉3 �
= 2.353 A
1 1 1 1 + + � − 𝑉1 20 20 10 20 1 1 1 1 + + � − 𝑉1 10 20 20 10
− − −
1 𝑉 10 3 1 𝑉 20 3 1 𝑉 20 3
=4 =0 =0
𝑉2 = 9.412 𝑉 , 𝑉3 = 14.12 𝑉
I=
20 V é ù é2 ù 4 2 Ω + ê Ω + 3 Ω ú || ê Ω + 4 Ω ú 5 ë5 û ë5 û
= = 7.36 A
CHAPTER 8
89
66.
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 I=
= 1.76 mA
67.
68.
(Y-Δ conversion) 80 V 80 V = I= 12 kΩ || 12 kΩ || 6 kΩ 3 kΩ = 26.67 mA
Using ∇ − ∆ 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛
a.
60𝑉
I = (18Ω‖18Ω)‖[(18Ω‖18Ω)+(18Ω‖18Ω)] 60𝑉 60𝑉 = 9Ω‖(9Ω+9Ω) 9Ω‖18Ω 60𝑉 = = 10A 6Ω
=
b. 69.
∆ − ∇ 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛.
= 0.83 mA
90
CHAPTER 8
70.
a.
b.
R¢ = R1 + 1 kΩ = 3 kΩ R≤ = R2 + 1 kΩ = 3 kΩ R¢T =
= 1.5 kΩ
RT = 1 kΩ + 1.5 kΩ + 1 kΩ = 3.5 kΩ Is = 71.
= 5.71 mA
Using two Δ - Y conversions: c - g: 27 Ω
9Ω
27 Ω = 5.4 Ω
a - h: 27 Ω
9Ω
27 Ω = 5.4 Ω
RT = 5.4 Ω = 5.4 Ω = 4.2 Ω
CHAPTER 8
(13.5 Ω + 5.4 Ω) 18.9Ω
91
Chapter 9 1.
a.
E1 : short circuit E2 𝑅𝑇 (from source) = 4Ω + 2Ω || 15Ω = (4 + 1.765) Ω = 5.765Ω
E 1:
𝐸
16𝑉
𝐼𝑠 = � 1 �= � � = 2.8A 𝑅 5.765Ω 𝑇
[𝐼1′5Ω ] =
2Ω(2.8A) 2Ω+15Ω
= 0.32A
E2 : (taking E2 and short circuit E1) 𝑅𝑇 (from source) = 2Ω + 4Ω || 15 Ω = 2 Ω + 3.158 Ω = 5.158 Ω
E 2:
10𝑉 = 1.94A 5.158 4Ω+(2A) � 4Ω+15Ω � = 0.42A
𝐼𝑠 = E2/RT =
𝐼1"5Ω =
𝐼15Ω ↑ = 0.42A – 0.33A = 0.08A
b.
𝐼𝑠1 = E1/Rs1 =
16𝑉 4Ω
= 4A↑
𝑅𝑃1 = 𝑅𝑠1 = 4Ω
𝐼𝑠2 =
E2 𝑅𝑠2
=
10𝑉 2Ω
= 5A↓
𝐼𝑠 = 𝐼𝑠1 – 𝐼𝑠2 = 4A – 5A = 1A , 𝑅𝑠1 || 𝑅𝑠2 = 4Ω || 2Ω = 1.333Ω c. 2.
𝐼15Ω =
1.333Ω(1A) 1.333Ω+15A
= 0.08A
the same
a.
24 Ω(3 A) = 2.25 A 24 Ω + 8 Ω V¢ = I¢R = (2.25)(4.7 Ω) = 10.575 V I¢ =
V¢¢ =
4.7 Ω(12 V) = 1.763 V 4.7 Ω + 3.3 Ω + 24 Ω = 8.81 V
92
CHAPTER 9
V ¢2 (10.575 V)2 = 23.79 W = R 4.7 Ω V ¢¢2 (1.763 V)2 = 0.661 W P= = R 4.7 Ω
b.
P=
c.
3.
V 2 (8.81 V)2 = 16.51 W = R 4.7 Ω
d.
P=
e.
23.79 W + 0.661 W π 16.51 W 24.45 W ≠ 16.51 W
E:
E : considering E and open circuit 10A source 𝑅𝑇 = 12Ω + 24Ω || 60Ω = 29.14Ω
𝐼𝑠 =
𝐸 𝑅𝑇
𝐼6′0Ω =
24𝑉 = 0.824A 29.14Ω 24Ω(0.824A) = 0.235A 24Ω+60Ω
=
I : considering I (10V source) and short circuit 24V 24Ω || 60Ω = 17.14Ω
I:
12Ω(10A) = 4.118A 12Ω+17.14Ω 24Ω(4.118A) = 1.177A 𝐼6′′0Ω = 24Ω+60Ω ′ ′′ 𝐼60Ω = 𝐼 + 𝐼 = 4.118 + 1.177
𝐼 ′′′ =
4.
= 5.295A↓
Considering E1(+48V) and short circuit 30V source
E 1:
𝐼𝑇 =
𝐼𝑇 =
𝐼1 =
48𝑉 18Ω+9Ω//15Ω//10Ω 48𝑉 = 18Ω+3.6Ω
9Ω(𝐼𝑇 ) 9Ω+6Ω
=
2.22A
9Ω(2.222A) 15Ω
∴ 15Ω || 10Ω = 6Ω
5Ω || 6Ω = 3.6Ω = 1.33A
Considering E2 (30V source) and short circuit E1 (48V)
E 2:
𝐸
30𝑉 (18Ω||9Ω) +(15Ω||10Ω) 30𝑉 30𝑉 = 2.5A = 6Ω+6Ω 12Ω
𝐼𝑇 = 𝑅2 = 𝑇
=
I30V = IT + I1 = 2.5A + 1.333A = 3.833A(Direction of I1) CHAPTER 9
93
5.
I: 4.7 kΩ
+
V'R 3 −
4 mA
10 kΩ
18 kΩ
4 mA
6.43 kΩ
8 kΩ
3.3 kΩ I
6.43 kΩ(4 mA) = 1.78 mA 6.43 kΩ + 8 kΩ V R¢3 = -IR3 = −(1.78 mA)(4.7 kΩ) VR¢ = −8.37 V
I=
3
E:
−
18 V
+
5.54 kΩ(18 V) = 6.42 V 5.54 kΩ + 10 kΩ 4.7 kΩ(6.42 V) VR¢¢3 = = 3.77 V 4.7 kΩ + 3.3 kΩ VR 3 = VR¢3 + VR¢¢3 = −8.37 V + 3.77 V = −4.6 V
V= 10 kΩ
18 kΩ
+
8 kΩ V −
} 5.54 kΩ
6.
E:
Considering 40V sorce and open circuit 10mA source 6.8𝐾Ω( 40V) [𝑉2′ ] = = 14.468V = 14.47V 6.8𝐾Ω+12KΩ
I:
Considering 10V source and short circuit 40V source 𝐼2 = current through 6.8𝐾Ω 12KΩ(10mA) 𝐼2 = � � = 6.383 mA 𝑉2′′
12KΩ+6.8𝐾Ω
= 𝐼2 𝑅2 = (6.383mA) (6.8KΩ) = 43.40V V2 = 𝑉2′ + 𝑉2′′ = 57.87V = 14.47V + 43.40V = 57.87V
94
CHAPTER 9
7.
E:
I'
6.8 kΩ
1.2 kΩ
2.2 kΩ
R1
} 5.5 kΩ
+ −
+ V −
4.7 kΩ
8V
2.53 kΩ
1.2 kΩ + −
8V
Consider source (5mA) and open circuit 2mA source and short circuit 8V source [(6.8𝑘Ω)||(1.2𝑘Ω+4.7𝑘Ω)](5𝑚𝐴)
[(6.8𝑘Ω)||(5.9𝑘Ω)](5𝑚𝐴)
𝐼 ′ = [(6.8𝑘Ω)||(1.2𝑘Ω+4.7𝑘Ω)]+2.2𝑘Ω = [(6.8𝑘Ω)||(5.9𝑘Ω)]+2.2𝑘Ω 𝐼′ =
(3.16𝑘Ω)(5mA) 3.16𝑘Ω+2.2𝑘Ω
= 3.004mA↓
I (5 mA): 6.8 kΩ 5 mA
R1
2.2 kΩ
1.2 kΩ
4.7 kΩ
I''
}
𝐼
6.8 kΩ + 1.2 kΩ || 4.7 kΩ = 3.088 kΩ
I (2 mA): 6.8 kΩ 2.2 kΩ
R1
4.7 kΩ
0.956 kΩ
8.
2 mA
′′
4.288𝐾Ω 4.7𝐾Ω(1.866mA) = 9𝐾Ω+4.7𝐾Ω
= 0.640mA
&onsider 2 mA source 2.2𝐾Ω+6.8𝐾Ω=9𝐾Ω 1.2𝐾Ω || 4.7𝐾Ω = 0.956𝐾Ω 9𝐾Ω(2mA) 𝐼 ′′′ = = 1.808mA 9𝐾Ω+0.956𝐾Ω
I = 𝐼′↓ + 𝐼′′↓+𝐼′′′↑
= (3.004 + 0.640 – 1.808)mA
}
I'''
1.2 kΩ
Considering 8V source 2.2𝐾Ω + 6.8𝐾Ω = 9𝐾Ω 9𝐾Ω || 4.7𝐾Ω = 3.088𝐾Ω 𝑅𝑇 = 3.088𝐾Ω + 1.2𝐾Ω = 4.288𝐾Ω 8𝑉 = 1.866mA 𝐼𝑆 =
I = 1.836 mA
E 1:
12 V 12 Ω || 4 Ω + 4 Ω + 6 Ω 12 V 12 V = = 3 Ω + 10 Ω 13 Ω = 923.1 mA V s1 = IR = (923.1 mA)(4 Ω) = 2.492 V
I¢ =
CHAPTER 9
95
I:
9 Ω(6 A) = 4.154 A 9Ω+4Ω Vs≤ = I≤4Ω = (4.154 A)(4 Ω) = 16.62 V I≤ =
E 2:
R¢T = 12 Ω || (4 Ω + 6 Ω) = 12 Ω || 10 Ω = 5.455 Ω E 8V I= 2 = = 0.846 A RT 4 Ω + 5.455 Ω 12 Ω(0.846 A) 12 Ω( I ) = I¢≤ = = 0.462 A 22 Ω 12 Ω + 10 Ω V3¢¢¢ = I¢≤(4 Ω) = 0.462 A(4 Ω) = 1.848 V Vs (polarity of Vs¢¢) = = 16.62 V - 2.492 V - 1.848 V = 12.28 V 9.
a.
b.
𝑅𝑡ℎ = 𝑅3 + 𝑅1 ||𝑅2 = 4Ω + 6Ω||9Ω = 4Ω + 3.6Ω = 7.6Ω 𝐸𝑡ℎ =
𝐼1 =
𝐸𝑡ℎ 𝑅𝑡ℎ +𝑅
𝐼2 =
𝐼3 = 96
𝑅2 𝐸 𝑅2 +𝑅1
=
=
9Ω( 18V ) 9Ω+6Ω
10.8𝑉 = 7.6Ω+5Ω
10.8𝑉 7.6Ω+40Ω
=
9Ω( 18V ) 15Ω
= 10.8 V
0.93𝐴
= 226.89 mA
10.8𝑉 7.6Ω+120Ω
= 84.64 mA
CHAPTER 9
10.
a.
RTh: ¨ RTh = 3.3 kΩ + 1.2 kΩ || 2.4 kΩ = 3.3 kΩ + 0.8 kΩ = 4.1 kΩ
ETh:
ETh = (120 mA)(2.4 kΩ || 1.2 kΩ) = 96 V b.
RTh = 4.1 kΩ
96 V = 15.74 mA 6.1 kΩ 2 P = I R = (15.74 mA)2 2 kΩ = 0.495 W R = 100 kΩ: 96 V I= = 0.922 mA 104.1 kΩ P = I2R = (0.922 mA)2 100 kΩ = 85 mW I=
11.
a.
RTh:
¨ RTh = 6 Ω + 6 Ω || 6 Ω = 9 Ω
ETh:
ETh =
b.
𝐸𝑡ℎ 2 �R 𝑡ℎ +𝑅
R = 4 Ω ; P = �𝑅
P = 2.367 W R = 90Ω ; P = �
= �
10𝑉 �2 × 9𝛺+90𝛺
CHAPTER 9
10𝑉 2 � 9Ω+4Ω
= 10 V
4Ω
90Ω = 0.918W.
97
12. RTh
R1
R2
3Ω
RTh = 3 || 6 Ω = 2Ω
6Ω
10 Ω
+
−
ETh
+
+
−
18𝑉+9𝑉
I=� � = 3A 3Ω+6Ω V3Ω = IR = (3A) (3Ω) = 9V 𝐸𝑡ℎ = 18V – 9V = 9V
6Ω − 9V +
18 V
−
13.
I
3Ω
10 Ω
RTh:
RTh = 5.6 kΩ || 2.2 kΩ = 1.58 kΩ
I:
ETh: Superposition: 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 14.
RTh: 8Ω
RTh
2Ω
For 𝐸𝑡ℎ
IT = 4A 8 I2Ω = � �A, 4A = I2Ω + I3Ω
4Ω
3Ω
RTh = (8 Ω + 2 Ω) || (3 Ω + 4 Ω) = (8 Ω + 2 Ω) || 7 Ω = 9.556 Ω
3
NOD ANALYSIS:
ETh:
+
I
8Ω I' 4A
ETh −
8
8
𝑉
𝑉
𝑉
𝑉
2 4 = � 32 � + �2+4 � = � 32 � + � 62 �
I2Ω = �3�A, I3Ω = �6�A
2Ω
98
3Ω
4Ω
CHAPTER 9
𝐸𝑡ℎ = V6Ω + V2Ω = (IT) (6A) + I2Ω (2 Ω) 8 = 4(6A) + � 𝐴� (2 Ω) 3 16
= 24V + � 3 �V = 29.333 V RTh:
4 kΩ
4 kΩ
4 kΩ = 1.333 kΩ 3
4 kΩ
20 V
RTh =
RTh
4 kΩ
+
ETh
8 mA
4 kΩ
4 kΩ = 1 kΩ 4
Source conversion: 20 V = 5 mA, Rp = 4 kΩ I= 4 kΩ
a
+
a.
𝑅𝑡ℎ = 9.556 Ω
−
15.
𝐸𝑡ℎ = 29.333 V,
−
b
+ 1.333 kΩ
4 kΩ
5 mA
8 mA
ETh = ITRT = (3 mA)(1 kΩ) = 3 V
ETh −
b. + −
a.
3 mA
1 kΩ I
16.
}
} 1 kΩ
3V
10 kΩ
I10kΩ =
3V 3V = = 0.273 mA 1 kΩ + 10 kΩ 11 kΩ
RTh: ¨RTh = 2 Ω + 8 Ω = 10 Ω
ETh: ETh = V16Ω
CHAPTER 9
99
20 V = 825.08 mA 20 Ω + 4.24 Ω 5 Ω(825.08 mA) 5 Ω( I T ) = 125.01 mA I¢ = = 33 Ω 5 Ω + 28 Ω ETh = V16Ω = (I¢)(16 Ω) = (125.01 mA)(16 Ω) = 2 V
IT =
17.
ETh 2V 2V = = = 66.67 mA RTh + R 10 Ω + 20 Ω 30 Ω 2V 2V = 50 Ω: I = = 33.33 mA 10 Ω + 50 Ω 60 Ω 2V 2V = 100 Ω: I = = 18.18 mA 10 Ω + 100 Ω 110 Ω
b.
20 Ω:
a.
RTh:
I =
¨RTh = 3. 3 kΩ + 2.2 kΩ || 1.1 kΩ = 3.3 kΩ + 0.73 kΩ = 4.03 kΩ
E 1:
ETh: Superposition:
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 b.
100
CHAPTER 9
V= 18.
1.2 kΩ(12 V) = 2.75 V 1.2 kΩ + 4.03 kΩ
RTh: RTh = 1.2 kΩ + 3.3 kΩ + 2.2 kΩ || 5.6 kΩ = 4.5 kΩ + 1.58 kΩ = 6.08 kΩ
RTh
ETh: Source conversions: ≠I1 =
= 10 mA, Rs = 2.2 kΩ
ØI2 =
= 2.14 mA, Rs = 5.6 kΩ
Combining parallel current sources: I¢T = I1 - I2 = 10 mA - 2.14 mA = 7.86 mA≠ and 2.2 kΩ || 5.6 kΩ = 1.58 kΩ 5 mA −
7.86 mA
+
7.86 mA
3.3 kΩ + 1.58 kΩ − I'
1.2 kΩ 5 mA
+ ETh −
I¢ = 7.86 mA + 5 mA = 12.86 mA V3.3kΩ = IR = (5 mA)(3.3 kΩ) = 16.5 V V1.58kΩ = IR = (12.86 mA)(1.58 kΩ) = 20.3 V ETh = 16.5 V + 20.3 V = 36.8 V 19.
a.
RTh: ¨RTh = 51 kΩ || 10 kΩ = 8.36 kΩ
ETh: ETh =
CHAPTER 9
10 kΩ(20 V) = 3.28 V 10 kΩ + 51 kΩ
101
b.
c.
IERE + VCE + ICRC = 20 V but IC = IE and IE(RC + RE) + VCE = 20 V 20 V - VCE 20 V - 8 V 12 V or IE = = 4.44 mA = 2.2 kΩ + 0.5 kΩ 2.7 kΩ RC + RE
ETh - IBRTh - VBE - VE = 0 E - V BE - V E 3.28 V - 0.7 V - (4.44 mA)(0.5 kΩ) = and IB = Th 8.36 kΩ RTh =
20.
= 43.06 µA
d.
VC = 20 V - ICRC = 20 V - (4.44 mA)(2.2 kΩ) = 20 V - 9.77 V = 10.23 V
a.
ETh = 20 V I = 1.6 mA =
b.
ETh= 60 mV, RTh = 2.72 kΩ
c.
ETh = 16 V, RTh = 2.2 kΩ
= 12.5 kΩ
21. RTh = 4 Ω || (2 Ω + 2 Ω) =
4Ω =2Ω 2
By KVL , 10 = 4IT + 2IT + 4I4Ω + 4 10 = 4IT + 2IT + 4(IT – I4Ω) 6 = 6IT + 4I4Ω 10 = 10IT – 4I4Ω
(1) (2)
IT = 1A, I4Ω = 0 𝐸𝑡ℎ = 2 × 1 + 4 × 0 = 2𝑉 102
CHAPTER 9
22.
a. From Problem 9, RN = RTh = 7.6 Ω
𝑅𝑇 = 6Ω + 9Ω || 4Ω = 6Ω + 2.769Ω
𝐼𝑠 =
𝐼𝑁 =
23.
𝐸 𝑅𝑇
=
18𝑉 8.769Ω
9Ω(2.0527A) 9Ω+4Ω
b.
RTh = 7.6 Ω, ETh = INRN = (1.421 A)(7.6 Ω) = 10.8 V
c.
same results
a.
From Problem 10, RN = RTh = 4.1 kΩ
= 2.0527 A = 1.421 A
2.4 kΩ(120 mA) = 87.80 mA 2.4 kΩ + (1.2 kΩ || 3.3 kΩ) 1.2 kΩ(87.80 mA) IN = = 23.41 mA 1.2 kΩ + 3.3 kΩ I¢ =
b. c.
From Problem 12, RN = RTh = 2 Ω 3Ω + 18 V −
8Ω IN
−
24.
RTh = 4.1 kΩ, ETh = INRN = (23.41 mA)(4.1 kΩ) = 96 V same results.
12 V
IN 10 Ω ⇒ 6 A
3Ω
8Ω
1.5 A
+ 𝐸
18𝑉
I1 = � 1 � = � � = 6A 𝑅 3Ω 1 𝐸2
9𝑉
𝐼2 = � � = � � = 1.5A 𝑅 6Ω 2
25.
𝐼𝑁 = I1 – I2 = 6A – 1.5A = 4.5A
From Problem 13, RN = RTh = 1.58 kΩ
IN = 8 mA − 7.27 mA = 0.73 mA 26.
From Problem 14, RN = RTh = 7.56 Ω
CHAPTER 9
103
2A
4 Ω(2 A) 4 Ω + 3 Ω + 6 Ω || 2Ω 4 Ω(2 A) = = 0.941 A 7 Ω + 1.5 Ω 12 Ω(0.941 A) 2 ΩI¢ = I¢¢ = = 0.235 A 8Ω 2 Ω+6 Ω I¢ =
I'' 6Ω 2A
2Ω 4Ω
I' 3Ω
IN = 2 A − I¢¢ = 2 A − 0.235 A = 1.765 A 27.
From Problem 16, RN = RTh = 10 Ω
RT = 20 Ω + 5 Ω (12 Ω + 1.778 Ω) = 23.67 Ω E 20 V Is = T = = 844.95 mA RT 23.67 Ω 5 Ω(844.95 mA) =224. 98 mA I12Ω = 5Ω + (12 Ω + 1.778 Ω) 16 Ω(224.98 mA) = 200 mA IN = 16 Ω + 2 Ω 28.
From Problem 18, RN = RTh = 6.08 kΩ IN: Starting with figure from problem 18: I'
7.86 mA
3.3 kΩ
IN
1.58 kΩ
5 mA
1.2 kΩ IN
and coverting sources:
I'
1.58 kΩ + 12.42 V −
3.3 kΩ
1.2 kΩ
+ −
16 V
12.42 V - 6 V I¢ = 1.58 kΩ + 3.3 kΩ + 1.2 kΩ = 1.06 mA 104
CHAPTER 9
IN = I¢ + 5 mA = 1.06 mA + 5 mA = 6.06 mA 29.
From Problem 21, RN = RTh = 2 Ω
IN =
30.
2V =1A 2Ω
R N: 80 Ω 40 Ω
RN =
20 Ω
RN
1 = 11.43 Ω 1 1 1 + + 40 Ω 20 Ω 80 Ω
40 Ω
20 Ω
4A
80 Ω
200 mA
Ø IN = 4 A + 200 mA = 4.2 A
}
source conversion
31.
a.
R = RTh = 7.6 Ω from Problem 9
b.
ETh = 10.8 V from Problem 9 𝐸2
32.
𝑡ℎ
a.
R = RTh = 2 Ω from Problem 12
b.
ETh = 9 V from Problem 12 𝐸2
33.
(10.8)2
Pmax = �4𝑅𝑡ℎ � = �4(7.6Ω)� = 3.84W
𝑡ℎ
a. b.
(9𝑉)2
Pmax = �4𝑅𝑡ℎ � = �4(2Ω)� = 10.125W
R = RTh = 9.556 Ω from Problem 14 ETh = 29.333 V from Problem 14 𝐸2
(29.333)2
Pmax = �4𝑅𝑡ℎ � = �4(9.556Ω)� = 22.51 W 𝑡ℎ
CHAPTER 9
105
34.
35.
a.
R = RTh = 6.08 kΩ from Problem 18
b.
ETh = 36.8 V from Problem 18 E2 (36.8 V)2 = 55.51 mW Pmax = Th = 4RTh 4(6.08 kΩ)
a.
R = RN = RTh = 2.18 Ω b.
36.
37.
Pmax =
2 (13.33 A)2 2.18 Ω I N RN = 96.84 W = 4 4
é E ù2 Th Pmax = ê ú R4 ë RTh + R4 û with R1 = 0 Ω, ETh is a maximum and RTh is a minimum. \ R1 = 0 Ω a. V, and therefore V4 wll be its largest value when R2 is as large as possible. Therefore, choose R2 = open-circuit (∞ Ω) and P4 = b.
38.
will be a maximum.
No, examine each individually.
The voltage VL will be maximum, when R = 500 Ω because the full voltage E will appear across RL. 𝑉2
𝐸2
(16𝑉)2
Pmax = �𝑅𝐿 � = �𝑅 � = � 100Ω � = 2.56 W 𝐿
39.
𝐿
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 IL =
106
= 6.12 A
CHAPTER 9
40.
41.
-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 = = 39.3 µA IL = Req + RL 1.7346 kΩ + 5.6 kΩ VL = ILRL = (39.3 µA)(5.6 kΩ) = 220 mV
Eeq =
400𝑉
𝑅𝐿
42.
20𝑉
10𝑉
𝐼𝑇 = � � – �80𝑉� – �50𝑉� = 5A – 0.25A – 0.2A 80𝑉 𝑅𝑇 = 300Ω || 80Ω || 80Ω || 50Ω = 20.69Ω 𝑉𝐿 = 𝐼𝑇 𝑅𝑇 = (4.55A) (20.69Ω) = 94.14 V 𝑉 94.4 𝑉 𝐼𝐿 = 𝐿 = � � = 0.314 A 300𝑉
(4 A)(4.7 Ω) + (1.6 A)(3.3 Ω) 18.8 V + 5.28 V = 3.01 A = 8Ω 4.7 Ω + 3.3 Ω Req = 4.7 Ω + 3.3 Ω = 8 Ω
Ieq =
IL =
= 2.25 A
VL = ILRL = (2.25 A)(2.7 Ω) = 6.08 V 43.
=
(4 mA)(8.2 kΩ) + (8 mA)(4.7 kΩ) - (10 mA)(2 kΩ) 8.2 kΩ + 4.7 kΩ + 2 kΩ
=
= 3.38 mA
Req = 8.2 kΩ + 4.7 kΩ + 2 kΩ = 14.9 kΩ Eeq I eq (14.9 kΩ)(3.38 mA) = = 2.32 mA IL = 14.9 kΩ + 6.8 kΩ Req + RL VL = ILRL = (2.32 mA)(6.8 kΩ) = 15.78 V 44.
15 KΩ || (8 KΩ + 7 KΩ ) = 15 KΩ || 15 KΩ = 7.5 KΩ 7.5KΩ(80V)
Vab = � � = 50V 7.5KΩ+4.5KΩ Iab = �
CHAPTER 9
50𝑉 � 15KΩ
= 3.333mA
107
45.
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 = 10 V - 0.25 V = 9.75 V Iba =
46.
Vab = 0 V (short) Iab = 0 A (open) R2 any resistive value
\ R2 = short-circuit, open-circuit, any value 47.
a. b.
Is = �
30𝑉 � 8KΩ+24/3KΩ
𝐼 3
I = � 𝑆 � = 0.625 mA Is = �
c. 48.
= 1.875 mA
30𝑉 � 24KΩ+8KΩ/12KΩ
= 1.042 mA, I =
12𝑘(𝐼𝑠) 12𝑘+8𝑘
= 0.625mA.
yes
(a)
10 V 4 kΩ || 8 kΩ + 4 kΩ || 4kΩ 10 V = 2.67 kΩ + 2 kΩ
IT =
=
= 2.14 mA
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 =
108
CHAPTER 9
(b)
(8 kΩ || 4 kΩ)(10 V) 8 kΩ || 4 kΩ + 4 kΩ || 4kΩ = 5.72 V
V1 =
I1 =
= 0.71 mA
V2 = E - V1 = 10 V - 5.72 V = 4.28 V I2 =
= 1.07 mA
I = I2 - I1 = 1.07 mA - 0.71 mA = 0.36 mA 49.
a.
b.
IR2 = �
𝑅1 (𝐼) � 𝑅1 +𝑅2 +𝑅3
=�
3Ω(9A) � 3Ω+2Ω+4Ω
IR1 = �
𝑅2 (𝐼) � 𝑅1 +𝑅2 +𝑅3
=�
2Ω(9A) � 3Ω+2Ω+4Ω
V = IR2. 𝑅2 = (3A) (2Ω) = 6V
= 3A
= 2A
V = IR1.R1 = (2A)(3Ω) = 6V c.
CHAPTER 9
YES.
109
Chapter 10 1.
(a)
E =
= 36 ¥ 103 N/C
(b)
E =
= 36 ¥ 109 N/C
E (1 mm): E (2 m) = 36 × 109: 36 × 103 = 1 × 106 𝑘∅
𝑘∅
2.
∈ = � 2� = � = � 𝑟 ∈
3.
C = [∅/V] = �
4. 5.
9×103 (3𝜇𝑐) 96 𝑁/𝑐
1700𝜇𝐶 �= 34𝑉
= 16.77m
50𝜇𝐹
∅ = CV = 0.25𝜇𝐹 (220𝑉) = 55𝜇𝐶 a.
é 1m ù 1≤ ê ú = 25.4 mm ë39.37¢¢ û
= 19.69 V/m
E =
b. E =
6.
∅
= 1.97 kV/m 180𝜇𝐶
V= � � = � � = 18.37 V 𝐶 9.8𝜇𝐹 𝑉
18.37𝑉
∈=� �=� � = 6.123 KV/m 𝑑 3𝑚𝑚
7.
0.2" �
1𝑚 � 39.37"
= 5.08 mm 0.2𝑚2
𝐴
C = 8.85×10−12 ∈𝑟 � � = 8.85 × 10−12 (1) �5.08𝑚𝑚� = 348.43pF 𝑑 𝐴
0.2𝑚2
8.
C = 8.85 × 10−12 ∈𝑟 �𝑑� = 8.85 × 10−12(2.5) �5.08𝑚𝑚� = 871.06pf
9.
C = 8.85 × 10−12 ∈𝑟 � � = 8.85 × 10−12 𝑑
𝐴
d=�
8.85×10−12 (4)(0.18𝑚2 ) � 2.5𝜇𝑓
d = 2.55 𝜇𝑚 110
CHAPTER 10
10. 11.
12.
𝐶
0
a.
C = 8.85 ¥ 10-12(7)
b.
E =
c.
Q = CV = (24.78 nF)(200 V) = 4.96 µC
a.
C=
b. c.
C = 2(4.7 µF) = 9.4 µF C = 20(4.7 µF) = 94 µF æ1 ö (4)ç ÷ è 3 ø (4.7 µF) = 25.1 µF C= æ1 ö ç ÷ è4 ø
d.
13.
�
= 24.78 nF
80 V V = = 400 kV/m d 0.2 mm
(4.7 µF) = 2.35 µF
8.85×10−12 ∈𝑟 𝐴 � 𝐶
=�
D = 152.57 𝜇𝑚 � 6.007 mils�
14.
7.3𝜇𝐹
C = ∈𝑟 𝐶0 = � � = � � = 5 (mica) 𝐶 1460𝜇𝐹
8.85×10−12 (5)(0.03𝑚2 ) �= 8700𝑝𝐹
10−6 𝑚 39.37" 1000𝑚 � � 1𝑚 � � 1" � 1𝜇𝑚
5000𝑉 � 𝑚𝑖𝑙
152.57𝜇𝑚
= 6.007m
= 30.035KV = 30.04KV
mica:
= 0.24 mils
= 6.10 µm
0.24 mils
(22 µF)/°C = 4400 pF/°C
15.
[80°C] = 0.35 µF 16. 17. 18. 19.
j=+ 5% , size = 60pF + 3pF , 57pF→ 63𝑝𝐹 − −
f=+ 1% , size = 67 × 10 = 670 𝜇𝐹 −
+ −
6.7pF
663.3𝜇𝐹 → 676.7𝜇𝐹
K=+ 280pF 10% , size = 28 × 102 pF = 2800pF + − − a. τ = RC = (10 ¥ 103 Ω)(10 µF) = 100 ms
CHAPTER 10
2520pF → 3080𝑝𝐹
111
b.
uC = E(1 - e-t/τ) = 20 V(1 - e- t/100 ms)
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 = 2 mAe- t/100 ms 10 kΩ uR = Ee-t/τ = 20 Ve- t/100 ms
e.
20.
21.
υC = E(1 - e-t/τ) = 20 V(1 - e- t/1s)
a.
τ = RC = (100 kΩ)(10 µF) = 1 s
b.
c.
1τ = 12.64 V, 3τ = 19 V, 5τ = 19.87 V
d.
e.
Same as problem 21 with 5τ = 5 s and Im = 200 µA
a.
τ = RC = (2.2 kΩ + 3.3 kΩ)1 µF = (5.5 kΩ)(1 µF) = 5.5 ms
b.
uC = 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 =
e-t/τ =
20 V -t/τ e = 200 µA e- t/1s 100 kΩ υR = Ee-t/τ = 20V e- t/1s iC =
e-t/τ = 18.18 mAe- t/5.5 ms
3.3 kΩ (100 V) = 60 V 3.3 kΩ + 2.2 kΩ = 60 Ve- t/5.5 ms
VR 2 =
e.
112
CHAPTER 10
22.
a.
R = 68 kΩ + 22 kΩ = 90 kΩ τ = RC = (90 kΩ)(18 µF) = 1.62 s
b.
uC = E(1 - e-t/τ) = (80 V - 30 V)(1 - e-t/τ) uC = 50 V(1 - e−t/1.62s)
c.
iC =
e-t/τ =
50 V -t/τ e = 555.56 µAe- t/1.62s 90 kΩ
d.
23.
a. 200 𝜇𝑠
−100𝜇𝑠
b. Vc = 20 (1 – �𝑒 200𝜇𝑠 � = 20V (1 – 𝑒 −0.5 ) = 20V (1 – 0.607) = 20V(0.393) = 7.86 V −2𝑚𝑠
c. Vc = 20V (1 – �𝑒 200𝜇𝑆 � 24.
a. b. c. d. e. f.
= 20V (1 – 𝑒 −10 ) = 20V ( 1 – 45.4 × 10−6) = 20(999.95 × 10−3) = 19.999 V ≈ 20V
t = 20 ms, 5t = 5(20 ms) = 100 ms t 20 ms = t = RC, R = = 2 kΩ C 10 µF uC (20 ms) = 40 mV(1 - e-20ms/20ms) = 40 mV(1 - e-1) = 40 mV(1 - .368) = 40 mV(0.632) = 25.28 mV uC = 40 mV(1 - e-10) = 40 mV(1 - 45 ¥ 10-6) @ 40 mV Q = CV = (10 µF)(40 mV) = 0.4 µC t = RC = (1000 ¥ 106 Ω)(10 µF) = 10 ¥ 103 s 5t = 50 ¥ 103 s
25.
= 13.89 h
a.
t = RC = (4.7 kΩ)(56 µF) = 263.2 ms
b.
uC = E(1- e-t/t) = 22 V(1 - e- t/263.2ms) 22 V - t / 263.2ms E - t /t e = e iC = = 4.68 mAe- t/263.2ms 4.7 kΩ R
c.
uC(1 s) = 22 V(1 - e-1s/263.2ms) = 22 V(1 - e-3.8) = 22 V(1 - 22.37 ¥ 10-3) = 21.51 V iC (1 s) = 4.68 mAe-1s/263.2ms = 4.68 mA(22.37 ¥ 10-3) = 0.105 mA
d.
uC = 21.51 Ve-t/263.2ms iC =
CHAPTER 10
= 4.58 mAe- t/263.2ms
113
e.
26.
a.
t = RC = (3 kΩ + 2 kΩ)(2 µF) = 10 ms uC = 30 V(1 - e- t/10ms) 30 V - t /10ms e iC = = 6 mA- t/10ms 5 kΩ = iC R1 = (6 mA)(3 kΩ)e-t/10ms = 18 Ve- t/10ms
b.
100ms: e-10 = 45.4 ¥ 10-6 uC = 30 V(1 - 45.4 ¥ 10-6) = 30 V iC = 6 mA(45.4 ¥ 10-6) = 0.27 µA = 18 V(45.4 ¥ 10-6) = 0.82 mV
c.
200 ms: t¢ = R2C = (2 kΩ)(2 µF) = 4 ms uC = 30 Ve- t/4ms 30 V - t / 4ms e iC = = -15 mAe- t/4ms 2 kΩ At t = 0: u R 2 = iCR2 = (6 mA)(2 kΩ)e−t/10 ms = 12 Ve−t/10 ms At t = 200 ms: u R 2 = −(15 mA)(2 kΩ)e−t/4 ms = −30 Ve−t/4 ms
d.
114
CHAPTER 10
12 V
30 V
27.
a.
t = RC = (120 kΩ)(47 pF) = 5.64 µs u C = 40 V(1 − e−t/5.64µs) 40 V - t / 5.64µs e iC = = 333.33 µAe−t/5.64µs 120 kΩ
b.
t¢ = RC = (270 kΩ)(47 pF) = 12.69 µs At 5t uC = 40 V(1 − e-5t/t) = 40 V(1 − e-5) = 40 V(1 − 6.74 × 10−3) = 39.73 V \ uC = 39.73 Ve- t/12.69µs 39.73 V - t /12.69 ms e iC = = −331 µAe−t/12.69µs 270 kΩ
c. 40
39.73 V
28.2 63.45 µs 333.33
28.2 63.45 µs 331
CHAPTER 10
115
28.
a. b. c.
29.
𝜏 = RC = (2mΩ)(2000𝜇𝐹) = 4𝜇𝑠 5 𝜏 = 20 𝜇𝑠 𝑉 18𝑉 Im = � � = � � = 9KA 𝑅 2𝑚Ω YES.
a.
uC = Vf + (Vi - Vf)e-t/t t = RC = (4.7 kΩ)(4.7 µF) = 22.1 ms, Vf = 40 V, Vi = 6 V uC = 40 V + (6 V - 40 V)e-t/22.1ms uC = 40 V - 34 Ve- t/22.1ms
b.
Initially VR = E + uC = 40 V − 6 V = 34 V 34 V - t / 22.1ms V - t /t = e iC = R e = 7.23 mA e- t/22.1ms R 4.7 kΩ
c.
30.
116
a.
t = RC = (4.7 kΩ)(4.7 µF) = 22.1 ms, Vf = 40 V, Vi = −40 V uC = Vf + (Vi − Vf)e-t/t = 40 V + (−40 V − 40 V)e-t/22.1ms uC = 40 V − 80 Ve- t/22.1ms
b.
Initially VR = E + uC = 40 V − (−40 V) = 80 V 80 V - t / 22.1ms V e and iC = R = = 17.02 mAe- t/22.1ms 4.7 kΩ R
CHAPTER 10
c.
υC +40 V
5τ = 110.5 ms 0
t
−40 V iC
17.02 mA
5τ = 110.5 ms 0
31.
a.
t = RC = (4.7 kΩ)(4.7 µF) = 22.1 ms, Vf = 40 V, Vi = 40 V uC = Vf + (Vi − Vf)e-t/t = 40 V + (40 V − 40 V)e-t/t = 40 V + 0e-t/t uC = 40 V
b.
Initially VR = E − uC = 40 V − 40 V = 0 V V - t /t 0 V - t /t = e and iC = R e =0A R R
c.
40 V
0
32.
t
υC
iC
40 V
0A t
0A t
t = RC = (R1 + R2)(C) = (1 kΩ + 2.2 kΩ)(180 µF) = (3.2 kΩ)(180 µF) = 576 ms uC = 20 Ve- t/576ms Vi 20 V - t / 576ms e- t /t = e iC = = 6.25 mAe- t/576ms R1 + R2 3.2 kΩ 2.2 kΩ (20 V) = 13.75 V VDR: VR 2 = 2.2 kΩ + 1 kΩ u R 2 = 13.75 Ve- t/576ms
CHAPTER 10
117
iC
υC 6.25 mA
20 V
13.75 V
0A 0 5τ = 2.88 s
t
0 5τ = 2.88 s
t
33.
uC = Vf + (Vi -Vf)e-t/t t = RC = (820 Ω)(3300 pF) = 2.71 µs, Vf = -20 V, Vi = −10 V uC = -20 V + (−10 V - (-20 V))e-t/2.71µs uC = -20 V + 10 Ve- t/2.71µs -(20 V - 10 V) -10 V = Im = = −12.2 mA 820 Ω 820 Ω iC = iR = -12.2 mAe- t/2.71µs
34.
a.
118
υR2
5τ = 2.88 s
R = 10 kΩ + 8.2 kΩ = 18.2 kΩ τ = RC = (18.2 kΩ)(6.8 µF) = 123.76 ms uC = Vf + (Vi − Vf)e−t/τ Vf = 20 V + 40 V = 60 V Vi = −8 V uC = 60 V + (−8 V − 60 V)e−t/123.76 ms uC = 60 V − 68 Ve−t/123.76 ms 8 V + 20 V + 40 V Im = = 3.74 mA 18.2 kΩ −t/123.76 ms iC = 3.74 mAe CHAPTER 10
t
b.
35.
a.
Cτ = 20 µF + 47 µF = 67 µF τ = RC = (56 kΩ)(67 µF) = 3.75 s uC = 18 V(1 − e−t/3.75 s) υC
0
18 V
5τ = 18.75 ms
b.
uC = 18 V(1 − e−10s/5.75 s) = 18 V(1 − e−2.67) = 18 V(1 − 69.25 ¥ 10−3) = 18 V(0.931) = 16.76 V
c.
uC = 18 V(1 − e−5) = 18 V(1 − 6.74 ¥ 10−3) = 18 V(.993) = 17.88 V Q20µF = CV = (20 µF)(17.88 V) = 357.6 µC Q47µF = CV = (47 µF)(17.88 V) = 840.36 µC
CHAPTER 10
t
119
36.
a.
Network redrawn: +
321.4 µA 10 kΩ
CT = 67µF ⇒
56 kΩ
10 kΩ
67µF ⇒
}
−
56 kΩ 18 V
source conversion
8.48 kΩ
source conversion + −
8.48 kΩ 2.73 V
+ 67µFυC −
τ = RC = (8.48 kΩ)(67 µF) = 568.2 ms uC = 2.73 V(1 − e−t/568.2 ms)
37.
120
b.
uC = 2.73 V(1 − e−10s/568.2 ms) = 2.73 V(1 − e−17.6) = 2.73 V(1 − 22.72 ¥ 10−9) @ 2.73 V
c.
uC = 2.73 V(1 − e−5) = 2.73 V (0.993) = 2.72 V Q20µF = CV = (20 µF)(2.72 V) = 54.4 µC Q47µF = CV = (47 µF)(2.72 V) = 127.84 µC
a.
uC = 140 mV(1 - e-1ms/2 ms) = 140 mV(1 - e-0.5) = 140 mV(1 - 0.6065) = 140 mV(0.3935) = 55.59 mV
b.
uC = 140 mV(1 - e-10) = 140 mV(1 - 45.4 ¥ 10-6) @ 139.99 mV
c.
100 mV = 140 mV(1 - e-t/2 ms) 0.714 = 1 - e-t/2 ms 0.286 = e-t/2 ms loge 0.286 = loge e-t/2 ms 1.252 = -t/2 ms t = 1.252 (2 ms) = 2.5 ms
d.
uC = 138 mV = 140 mV(1 - et/2 ms) 0.986 = 1 - e-t/2 ms -14 × 10−3 = -e-t/2 ms loge 14 × 10−3 = -t/2 ms -4.268 = -t/2 ms t = (4.268)(2 ms) = 8.54 µs
CHAPTER 10
38.
𝜏 = RC = (44 K Ω)(30 𝜇𝐹) = 1.32s −𝑡
−𝑡
Vc = 15V (1 – �𝑒 1.32𝑠 �) = 15V – 15V 𝑒 1.32𝑠 −𝑡
10V = 15V – 15V𝑒 1.32𝑠 −𝑡
–5V = –15V𝑒 1.32𝑠 −𝑡
1
�3� = 𝑒 1.32𝑠
−𝑡
0.333 = 𝑒 1.32𝑠 loge0.333 = �
−𝑡 � 1.32𝑠 −𝑡 �1.32𝑠�
–1.0996 = t = 1.0996(1.32s) = 1.45s
39.
𝑉
T= –𝜏 loge�1 − � 𝐸𝑐 ��
12s = –𝜏
12S = –𝜏 12S =
loge�1 − �
15𝑉 �� 20𝑉
loge[1 − 0.25]
287.68 × 10−3 𝜏
𝜏=�
12𝑆 � 0.28768
𝜏 = RC = 𝜏
= 41.67 s
41.67𝑆
R = �𝐶 � = �800𝜇𝐹 � = 52.09KΩ 40.
a.
τ = RC = (12 kΩ + 8.2 kΩ)(6.8 µF) = 137.36 ms uC = 60 V(1 − e−t/τ) 48 V = 60 V(1 − e−t/τ) 0.8 = 1 − e−t/τ 0.2 = 1 − e−t/τ loge0.2 = logee−t/τ −1.61 = −t/τ t = (1.61)τ = (1.61)(137.36 ms) = 221.15 ms
b.
iC =
c.
E - t /t 60 V - t / t e = e R 20.2 kΩ
= 2.97 mAe−t/137.36 ms iC(221.15 ms) = 2.97 mAe−221.15 ms/137.36 ms = 2.97 mAe−1.61 = 2.97 mA (199.89 × 10−3) = 0.594 mA t = 2τ iC = 2.97 mAe−2τ/τ = 2.97 mAe−2 = 0.4 mA 0.135 P = EI = (60 V)(0.4 mA) = 24 mW
CHAPTER 10
121
41.
a.
um = uR = Ee-t/t = 60 Ve-1t/t = 60 Ve-1 = 60 V(0.3679) = 22.07 V
b.
iC =
= 6 µAe-2
= 6 µA(0.1353) = 0.81 µA
42.
c.
uC = E(1 - e-t/t) 50 V = 60 V(1 - e-t/2 s) 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 = 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
t = RC = (10 kΩ)(15 µF) = 0.15 s uC = E(1 - e-t/t) = −15 V(1 - e- t/0.15 s)
iC =
122
15 V - t / 0.15 E - t /t e =e = −1.5 mAe- t/0.15 s 10 kΩ R
CHAPTER 10
b.
43.
a.
Source conversion and combining series resistors: E = −(4 mA)(6.8 kΩ) = −27.2 V RT = 6.8 kΩ + 1.5 kΩ = 8.3 kΩ Vf = −27.2 V, Vi = 10 V t = RC = (8.3 kΩ)(2.2 µF) = 18.26 ms uC = Vf + (Vi - Vf)e-t/t = −27.2 V + (10 V - (−27.2 V))e-t/18.26 ms uC = −27.2 V + 37.2 Ve- t/18.26 ms uR(0+) = −27.2 V - (−27.2 V))e-t/18.26 ms = −37.2 V 32.7 V - t /18.26ms e iC = 8.3 kΩ iC = −4.48 mAe- t/18.26 ms
CHAPTER 10
123
b.
44.
a.
RTh = 3.9 kΩ + 0 Ω || 1.8 kΩ = 3.9 kΩ ETh = 36 V
+ 18 µF 12 V −
t = RC = (3.9 kΩ)(18 µF) = 70.2 ms uC = Vf + (Vi - Vf)e-t/t = 36 V + (+12 V - 36 V)e-t/70.2 ms uC = 36 V - 24 Ve- t/70.2 ms uR(0+) = 24 V − 12 V = 24 V 24 V - t / 70.2ms e iC = 3.9 kΩ iC = 6.15 mAe- t/70.2 ms
b.
351 ms
124
351 ms
CHAPTER 10
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 I=
= 0.107 mA
ETh = 4 V - (0.107 mA)(6.8 kΩ) = 4 V - 0.727 V = 3.27 V
uC = 3.27 V(1 - e-t/t) t = RC = (2.69 kΩ)(20 µF) = 53.80 ms uC = 3.27 V(1 - e- t/53.80 ms) iC = = 1.22 mA e- t/53.80 ms
46.
a.
CHAPTER 10
t = RC = (6.8 kΩ)(39 µF) = 265.2 ms uC = Vf + (Vi - Vf)e-t/t = −20 V + (−8 V - (−20 V))e-t/265.2 ms uC = −20 V + 12 Ve- t/265.2 ms uR(0 +) = +8 V - 20 V = −12 V 12 V - t / 265.2ms e iC = 6.8 kΩ iC = −1.76 mAe- t/265.2 ms
125
b.
47.
𝑅𝑡ℎ = 3 M Ω // 12 M Ω = 2.4 M Ω 12M Ω (24V) 𝐸𝑡ℎ = � � = 19.2 V
a.
12M Ω+3M Ω −𝑡
b.
𝜏 = 𝑅𝑡ℎ C = (2.4 M Ω) (1 𝜇𝐹)= 2.4s
VC = 𝐸𝑡ℎ (1 – �𝑒 𝜏 � = 19.2V (1 – [𝑒−4] = 19.2V(1 – 0.0183) = 18.85v
−𝑡
𝐸
Ic = � � 𝑒 𝜏 𝑅 4𝜇𝐴 =
19.2𝑉 −𝑡 𝑒𝜏 2.4M Ω −𝑡
c.
0.5 = 𝑒 2.4𝑠 −𝑡 loge0.5 = 2.4𝑠 t = –(2.4s) (–0.693) = 1.66s Vmeter = VC
−𝑡
VC = 𝐸𝑡ℎ (1 – �𝑒 𝜏 �
−𝑡
15V = 19.2V((1 – �𝑒 2.4𝑠 �
−𝑡
15V = 19.2 – 19.2 �𝑒 2.4𝑠 � −𝑡
– 4.2= – 19.2V �𝑒 2.4𝑠 � −𝑡
0.218 = �𝑒 2.4𝑠 �
−𝑡
loge(0.218) = 2.4𝑠 t = –(2.4s) (–1.52) t = 3.65s 126
CHAPTER 10
48. 0 Æ 1 ms: iC = 2 ¥ 10-6
(20 V) = 40 mA 1 ms
1 Æ 3 ms: iC = 2 ¥ 10-6
= 0 mA
3 Æ 6 ms: iC = −2 ¥ 10-6
= −6.67 mA
6Æ 12 ms: iC = -2 ¥ 10-6
(10 V) = −3.33 mA 6 ms
iC(mA)
80 60
40
40 20 0
1
2
3
−20
4 5 −6.67
6
7
8
9
10 −3.33
11
12
t (ms)
−40 −60 −80
49.
(5 V) = -1.18 A 20 µs (10 V) = 4.7 A 20 Æ 30 µs: iC = 4.7 µF 10 µs (10 V) = −1.57 A 30 Æ 60 µs: iC = −4.7 µF 30 µs 0 Æ 20 µs: iC = −4.7 µF
(0 V) =0A 10 µs (10 V) = 4.7 A 70 Æ 80 µs: iC = 4.7 µF 10 µs 60 Æ 70 µs: iC = 4.7 µF
80 µs Æ 100 µs: iC = −4.7 µF
CHAPTER 10
(5 V) = −1.175 A 20 µs
127
iC(A) 10 8 6
4.7
4.7
4 2
0A
0
10
−2
20
30
−1.18 A
40
50
60
0 70
−1.57
80
90
100
t (ms)
−1.175
−4 −6 −8
50.
iC =
fi 0 Æ 4 ms: iC = 0 mA
=0V
4 Æ 6 ms: iC = -40 mA 6 Æ 16 ms: iC = +40 mA 16 Æ 20 ms: iC = -80 mA 20 Æ 25 ms: iC = 0 mA
=
(-40 mA) = −4 V
=
(40 mA) = +20 V
(4 ms) (-80 mA) = −16 V 20 µs =0V =
0V
51.
52.
6 𝜇𝐹 + 5 𝜇𝐹 = 11 𝜇𝐹 , 8 𝜇𝐹 + 15 𝜇𝐹= 23 𝜇𝐹 CT = �
(11𝜇𝐹)(23𝜇𝐹) 11𝜇𝐹+23𝜇𝐹
[𝐶𝑇′ ] = �
� = 7.44 𝜇𝐹
(8𝜇𝐹)(12𝜇𝐹) 8𝜇𝐹+12𝜇𝐹
�= 4.8𝜇𝐹
�𝐶𝑇" � = [𝐶𝑇′ ] + 12 𝜇𝐹 = 4.8 𝜇𝐹+ 12 𝜇𝐹 = 16.8 𝜇𝐹 [𝐶𝑇′′′ ] = �
128
CT = �
�𝐶𝑇" �(8𝜇𝐹) �𝐶𝑇" �+8𝜇𝐹
(8𝜇𝐹)(1𝜇𝐹) 8𝜇𝐹+1𝜇𝐹
�= �
(16.8𝜇𝐹)(8𝜇𝐹) 16.8𝜇𝐹+8𝜇𝐹
� = 0.889𝜇𝐹
� = 1𝜇𝐹
CHAPTER 10
53.
10 µF || 100 µF = 9.09 µF 20 µF + 9.09 µF = 29.09 µF QT = CTE = (29.09 µF)(20 V) = 581.8 µC Q20µF = CV = (20 µF)(20 V) = 400 µC V20µF = 20 V QT¢ = CT¢ E = (9.09 µF)(20 V) = 181.8 µC QT¢ = Q10µF = Q100µF = 181.8 µC Q 181.8 µC = = 18.18 V V10µF = C 10 µF Q 181.8 µC = = 1.818 V V100µF = C 100 µF
54.
360 𝜇𝐹+200 𝜇𝐹= 560 𝜇𝐹 CT �
(470𝜇𝐹)(560𝜇𝐹) 470𝜇𝐹+560𝜇𝐹
� = 255.53 𝜇𝐹
QT = CTE = 255.53 𝜇𝐹 (70V) = 17.89mC 𝑄
𝑉3 = � 𝐶3 � = � 3
55.
17.89𝑚𝐶 � 470𝜇𝐹
= 38.06 𝑉
V1 = V2 = E – V3 = 70V – 38.06 V = 31.94V Q1= V1 C1 = 11.498 mC Q2 = V2C2= (31.94v)(200 𝜇𝐹)= 6.39 mC
Steady state so ignore 10 KΩ resistor [𝐶𝑇′ ] = 330 𝜇𝐹 + 100 𝜇𝐹= 430 𝜇𝐹 CT = �
�𝐶𝑇′ � (220𝜇𝐹) �𝐶𝑇′ �+220𝜇𝐹
�=�
(430𝜇𝐹)(220𝜇𝐹) 430𝜇𝐹+220𝜇𝐹
�
CT = 145.54 𝜇𝐹 (𝑄𝑇 ) = 𝑄1 = CTE = (145.45𝜇𝐹) (30V) = 4.37mC ∅
4.37mC
𝑉1= � 𝐶1 � = � 220𝜇𝐹 � = 19.86V
56.
1
V3 = V2 = E – V1 = 30V – 19.86 V = 10.14 V 𝑄2 = C2 V2 = 330 𝜇𝐹(10.14 V) = 3.346 mC 𝑄3 = C3 V3 = (100 𝜇𝐹) (10.14 v) = 1.014 mC
V4kΩ =
= 32 V = V0.08µF
Q0.08µF = (0.08 µF)(32 V) = 2.56 µC V0.04µF = 48 V Q0.04µF = (0.04 µF)(48 V) = 1.92 µC 57. 58.
1
1
WC = � �C𝑉 2 = � �(140 pF) (20 V)2 = 28𝜇𝐽 2 2
C = 8𝜇𝐹, W = 1500J 𝑄2
W = �2𝐶 �
Q = √2𝐶𝑊 = �2(8𝜇𝐹)(1500𝐽) Q = 0.155C
CHAPTER 10
129
59.
a.
(220 kΩ + 3.3 kΩ)(12 V) = 9.85 V 2.2 kΩ + 3.3 kΩ + 1.2 kΩ (3.3 kΩ)(12 V) V100µF = = 5.91 V 2.2 kΩ + 3.3 kΩ + 1.2 kΩ
V200µF =
1 (200 µF)(9.85 V)2 = 970 mJ 2 1 W100µF = (100 µF)(5.91 V)2 = 1.75 mJ 2
W200µF =
60.
a. b. c. d. e.
130
1
WC = � �C𝑉 2 2 1
= � �(1500𝜇𝐹 ) (120V)2 = 10.8J 2 Q = CV = (1500𝜇𝐹)(120 V) = 0.18 C 𝑄
I = � �=� 𝑡
0.18𝑐 1 2500𝑠
𝑊
� = 450 A 10.8𝐽
P = Vav Iav = � � = � 𝑡 𝑄
0.18𝑐
1 𝑠 2500
t=� �=� � = 15s. 𝐼 12𝑚𝐴
� = 27 KW
CHAPTER 10
Chapter 11 1.
2.
B=
b. c.
0.04 T F = NI = (40 t)(2.2 A) = 88 At
d.
0.04 T
0.02 m2
= 0.4 ¥ 103 gauss
é2.54 cm ù é 1 m ù d = 0.15≤ ê ú = 3.81 mm úê ë 1¢¢ û ë100 cm û é2.54 cm ù é 1 m ù úê ú = 25.4 mm l = 1≤ ê ë 1¢¢ û ë100 cm û A=
= L 3.
pd 2 p(3.81 mm)2 = 11.40 ¥ 10-6 m2 = 4 4 N 2 m A (250) 2 (4π × 10−7 )(11.4 × 10−6 m 2 ) = = 35.25 μH 0.00254 m
2 −7 −6 2 2 m r m0 A (250) (600)(4π ×10 )(11.4 × 10 m ) a. L N= = 21.15 mH = 25.4 mm
b.
increase = change in µr Lnew = µrLo
m m
2 r 0 4.= L N=
5.
6.
= 4/2 ¥ 10-2 Wb/m2 = 0.02 Wb/m2
a.
(200t)2 (1000)(4π × 10−7 )(1.5 ×10−4 m 2 ) = 37.699 mH 0.20 mm
L= a.
L¢ = (3)2Lo = 9Lo = 9(4.7 mH) = 42.3 mH
b.
L¢ =
c.
L¢ =
d.
æ1 ö2 1 ç ÷ (1500)Lo 2 2 = 375(4.7 mH) = 1.76 mH L¢ = è ø 1 2
a.
39 ¥ 102 µH ± 10% fi 3900 µH ± 10% fi 3.9 mH ± 10%
CHAPTER 11
Lo =
(4.7 mH) = 1.57 mH = 16 (4.7 mH) = 75.2 mH
131
b.
68 × 100 µH ± 5% = 68 µF ± 5%
c.
47 µH ± 10%
d.
15 × 102 µH ± 10% = 1500 µH ± 10% = 15 mH ± 10% = (60 t)(140 mWb/s) = 8.4 V
7.
e=
8.
e=
25 V = 62.5 mWb/s 400 t
9.
e=
æ ö ç 1 ÷ æ ö 1 fi N = e ç ÷ = 50 mV ç ÷ = 10 turns ç df ÷ è5 m Wb/s ø ç ÷ è dt ø
10.
a.
e=
= (22 mH)(1 A/s) = 22 mV
b.
e=
= (22 mH)(20 mA/ms) = 440 mV
e=
æ 6 mA ö = (22 mH) ç ÷ = 1.32 V è100 ms ø
11.
L 470 mH = = 23.5 µs 20 kΩ R
a.
t=
b.
iL =
c.
uL = Ee-t/t = 40 Ve- t/23.5 µs uR = iRR = iLR = E(1 - e-t/t) = 40 V(1 - e- t/23.5 µs)
d.
iL: 1t = 1.264 mA, , 3t = 1.9 mA, 5t = 1.986 mA uL: 1t = 14.72 V, 3t = 1.96 V, 5t = 280 mV
40 V E (1 - e - t / t ) = (1 - e-t/t) 20 kΩ R = 2 mA(1 − e−t/23.5 µs)
e.
132
CHAPTER 11
12.
L 4.7 mH = = 2.14 µs R 2.2 kΩ
a.
t=
b.
iL =
c.
uL = Ee-t/t = 12 Ve- t/2.14 µs uR = iRR = iLR = E(1 - e-t/t) = 12 V(1 - e- t/2.14 µs)
d.
iL: 1t = 3.45 mA, , 3t = 5.18 mA, 5t = 5.41 mA uL: 1t = 4.42 V, 3t = 0.60 V, 5t = 0.08 V
(1 - e-t/t) = 5.45 mA(1 - e- t/2.14 µs)
e.
13.
14.
L 12 mH = = 0.6 µs R 20 kΩ uL = (28 V − 12 V)e−t/0.6µs = 16 Ve−t/0.6µs
a.
t=
b.
is = iR = iL =
c.
5t = 5(0.6 µs) = 3 µs
E 16 V (1 − e−t/t) = (1 − e−t/0.6µs) = 0.8 mA(1 − e−t/0.6µs) R 20 kΩ
iL = RT =
= 1.2 kΩ
æL ö R(15 ms) 1.2 kΩ(15 ms) = 5t = 5 ç ÷ = 15 µs: L = 5 5 èRø = 3.6 mH 15.
a.
iL = If + (Ii - If)e-t/t Ii = 8 mA, If =
CHAPTER 11
= 9.23 mA, t =
= 30.77 µs
133
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 = 0 and = iRR = iLR = (8 mA)(3.9 kΩ) = 31.2 V =E= 36 V - 31.2 V = 4.8 V uL = 4.8 Ve- t/30.77 µs b.
16.
a.
Ii = -8 mA, If = 9.23 mA, t =
= 30.77 µs
iL = If + (Ii - If)e-t/t = 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 = 0 (at t = 0-) but, = iRR = -iLR = (-8 mA)(3.9 kΩ) = -31.2 V =E= 36 V - (-31.2 V) = 67.2 V - t/30.77 µs uL = 67.2 V e b.
17.
134
c.
Final levels are the same. Transition period defined by 5t is also the same.
a.
t=
b.
iL =
L 100 mH 100 mH = = = 29.41 µs R R1 + R2 3.4 kΩ 2.2 kΩ(20 V) At 0+ u R 2 = = 12.94 V 2.2 kΩ + 1.2 kΩ u R 2 = 12.94Ve- t/29.41 µs E 20 V (1 - e - t / t ) = (1 - e - t / 29.41 ms ) R1 + R2 3.4 kΩ iL = 5.88 mA(1 − e- t/29.41 µs)
CHAPTER 11
c.
18.
a.
Source conversion:
t=
= 588.2 µs
iL = If + (Ii - If)e-t/t If =
= 1.76 mA
iL = 1.76 mA + (4 mA - 1.76 mA)e-t/588.2µs iL = 1.76 mA + 2.24 mA e- t/588.2µs
uR(0 +) = 4 mA(3.4 kΩ) = 13.6 V KVL: +6 V − 13.6 V - uL(0+) = 0 uL(0+) = −7.6 V - t/588.2µs uL = −7.6 Ve b.
19.
a. If = τ=
= 2 mA =
= 19.23 µs
iL = If + (Ii - If )e-t/τ = 2 mA + (6 mA - 2 mA)e-t/19.23 µs iL = 2 mA + 4 mAe- t/19.23 µs KVL: 20.8 V − 62.4 V - υL(0+) = 0 υL(0+) = −41.60 V υL = −41.6 Ve-t/19.23 µs
CHAPTER 11
135
b.
20.
a.
t=
uL = 8 Ve- t/0.278µs, iL =
21.
=
10 mH = 0.278 µs 36 kΩ
= 0.222 mA(1 - e- t/0.278µs)
b.
5t fi steady state L 10 mH = t¢ = = 0.208 µs R 12 kΩ + 36 kΩ iL = Ime-t/t¢ = 0.222 mAe- t/0.208µs uL = -(0.222 mA)(48 kΩ)e-t/t = -10 .66Ve- t/0.308µs
a.
t=
L 4.7 mH = = 2.35 µs R 2 kΩ
iL =
= 6 mA(1 - e- t/2.35µs)
uL = Ee-t/t = 12 V e- t/2.35µs b.
136
iL = 6 mA(1 - e-t/2.35µs) = 6 mA(1 - e-1µs/2.35µs) = 6 mA(1 - e-0.426) = 6 mA(1 - .653) = 6 mA(.347) = 2.08 mA L 4.7 mH = t¢ = = 392 ns R1 + R2 12 kΩ iL = 2.08 mAe- t/392ns uL = 12 Ve-t/2.35µs = 12 Ve-1µs/2.35µs = 12 Ve-0.426 = 12 V(.653) = 7.84 V CHAPTER 11
At t = 0+ after switch moved uL = −(2.08 mA)(12 kΩ) = −24.96 V and uL = −24.96 Ve- t/392ns 5t¢ = 1.96 µs c.
6 mA
iL
2.08 mA 0 12 V
1 µs τ = 2.35 µs
t
1.96 µs
7.84 V
0
1.96 µs τ 1 µs
t
−24.96 V
22.
a.
RTh = 6.8 kΩ ETh = 6 V
t=
iL =
= 0.74 µs
= 0.88 mA(1 - e- t/0.74µs)
uL = Ee-t/t = 6 Ve- t/0.74µs
CHAPTER 11
137
b.
Assume steady state and IL = 0.88 mA
t¢ =
= 0.33 µs
iL = Ime-t/t¢ = 0.88 mA e- t/0.33µs uL = -Vme-t/t¢ Vm = ImR = (0.88 mA)(15 kΩ) = 13.23 V uL = -13.23 Ve- t/0.33µs c.
d.
23.
138
a.
= ImR2 = (0.88 mA)(8.2 kΩ) = 7.22 V
RTh = 2 kΩ + 2.2 kΩ + 6.2 kΩ || 3 kΩ = 6.22 kΩ 6.2 kΩ(12 V) ETh = = 8.09 V 6.2 kΩ + 3 kΩ 8.09 V L 47 mH = If = = 1.3 mA , t = = 7.56 µs 6.22 kΩ R 6.22 kΩ CHAPTER 11
iL = 1.3 mA(1 - e- t/7.56µs) u L = 8.09 Ve- t/7.56µs
24.
b.
0.632(1.3 mA) = 0.822 mA 0.368(8.09 V) = 2.98 V
a.
Source conversion: E = IR = (4 mA)(12 kΩ) = 48 V, ENet = 48 V − 20 V = 28 V
t=
= 55.56 ns
= 0.778 mA(1 - e- t/55.56ns)
iL =
uL = Ee-t/t = 28 Ve- t/55.56ns b.
t = 100 ns: iL = 0.778 mA(1 - e-100ns/55.56ns) = 0.778 mA(1 - e-1.8) = 0.65 mA 0.165
uL = 28 Ve 25.
-1.8
= 4.62 V RTh = 2.2 kΩ || 4.7 kΩ = 1.50 kΩ 4.7 kΩ(10 V) ETh = = −6.81 V 4.7 kΩ + 2.2 kΩ
a.
t=
= 6.67 µs
= −4.54 mA(1 - e- t/6.67µs)
iL =
uL = Ee-t/t = −EThe−t/τ = −6.81 Ve- t /6.67µs b.
t = 10 µs: iL = −4.54 mA(1 - e-10µs/6.67µs) = −4.54 mA(1 - e-1.5) 0.223 = −3.53 mA uL = −6.81 V(0.223) = -1.52 V
c.
t¢ =
= 2.13 µs
iL = −3.53 mAe- t/2.13µs At t = 10 µs VL = (3.53 mA)(4.7 kΩ) = 16.59 V uL = 16.59 Ve- t/2.13µs
CHAPTER 11
139
d.
–1.52 V
26.
a.
Finding the Thevenin circuit for the inductor: RTh = R4 || (R3 + R1 || R2) = 1 kΩ || (2.7 kΩ + 8.2 kΩ || 2.2 kΩ) R1 R2 R4 RTh = 1 kΩ || 4.43 kΩ = 0.816 kΩ R3 Is E R1
R'T
+
I' R2
R4
ETh −
R3
RT ¢ = R1 + R2 || (R3 + R4) = 8.2 kΩ + 2.2 kΩ || (2.7 kΩ + 1 kΩ) = 8.2 kΩ + 2.2 kΩ || 3.7 kΩ = 8.2 kΩ + 1.38 kΩ = 9.58 kΩ
36 V = = 3.76 mA ¢ 9.58 kΩ RT R2 ( I s ) 2.2 kΩ(3.76 mA) = and I¢ = = 1.4 mA R2 + R3 + R4 2.2 kΩ + 2.7 kΩ + 1 kΩ and finally ETh = I¢R4 = (1.4 mA)(1 kΩ) = 1.4 V Then Is =
E
L 10 mH = = 12.25 µs R 0.816 kΩ t/12.25µs uL = 1.4 Ve= 1.4 Ve-25µs/12.25µs = 1.4 Ve-2.05 = 1.4 V(128.73 ¥ 10-3) = 180 mV t=
b.
140
uL = 1.4 Ve-t/12.25µs uL = 1.4 Ve-1µs/12.25µs = (1.4 V)(e-0.082) = (1.4 V)(.0921) = 1.29 V
CHAPTER 11
c.
Finding the Thevenin equivalent for R1 at 1t At 1t: uL = 1.4 Ve-t/t = 1.4 Ve-t/t = 1.4 Ve-1 = 1.4 V(0.368) = 515.2 mV +
+
36 V
515.2 mV = V'L (at 1τ)
2.2 kΩ R1
−
−
8.2 kΩ
2.7 kΩ
RTh¢ = 2.2 kΩ || 2.7 kΩ = 1.21 kΩ 2.2 kΩ
R'Th
2.7 kΩ
+
+
36 V
+ 515.2 mV
2.2 kΩ −
− − E' + Th
− 2.7 kΩ
− ETh¢ + 36 V − u R 2 = 0
ETh¢ = 36 − u R 2
u R2 =
2.2 kΩ(515.2 mV) = 0.231 V 2.2 kΩ + 2.7 kΩ
\ ETh¢ = 36 V − 0.231 V = 35.77 V −
35.77 V
+
E'Th
R'Th 1.21 kΩ
8.2 kΩ
u R1 =
8.2 kΩ(35.77 V) = 31.34 V 8.2 kΩ + 1.21 kΩ
+ V − R1
d.
27.
a.
ETh (1 − e-t/t) RTh 1.4 V (1 - e - t /12.25 ms ) = 0.81 kΩ = 1.72 mA(1 − e-t/12.25µs) 1 mA = 1.72 mA(1 − e-t/12.25µs) 0.581 = 1 − e-t/12.25µs 0.419 = e-t/12.25µs t = 12.25 µsloge0.419 = 12.25 µs(0.87) = 10.66 µs iL =
Ii =
= 2 mA
t = 0 s: Thevenin: RTh = 3.3 kΩ + 1 kΩ || 4.7 kΩ = 3.3 kΩ + 0.82 kΩ = 4.12 kΩ ETh =
= 2.81 V
iL = If + (Ii - If)e-t/t CHAPTER 11
141
If =
= 0.68 mA, t =
= 0.49 ms
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 uR(0+) = 2 mA(4.12 kΩ) = 8.24 V KVL(0+): 2.81 V - 8.24 V - uL = 0 uL = -5.43 V uL = -5.43 Ve- t/0.49 ms b.
28.
a.
8V = 5.33 mA, VL = 0 V 1.5 kΩ RTh = (3 kΩ || 12 kΩ) || 4 kΩ = 1.5 kΩ 2.4 kΩ(20 V) ETh = = 7.5 V 2.4 kΩ + 4 kΩ Steady-state: IL =
¢ = 1.5 kΩ || 1.5 kΩ = 0.75 kΩ RTh ¢ = 8 V + 7.5 V = 15.5 V ETh
t=
L 3 mH = = 4 µs R 0.75 kΩ
15.5 V = 20.67 mA Ii = 5.33 mA 0.75 kΩ iL = If + (Ii − If)e−t/τ = 20.67 mA + (5.33 mA − 20.67 mA)e−t/4 µs iL = 20.67 mA − 15.34 mAe−t/4µs υL = 15.5 Ve−t/4µs If =
142
CHAPTER 11
b.
iL(2t) = 20.67 mA − 15.34 mA e−2
} 0.135
= 18.6 mA
uL (2t) = 15.5 Ve-2 = 15.5 V(0.135) = 2.09 V c.
Ii = 18.6 mA υL + υR − 8 V = 0 υL = 8 V − υR = 8 V − (18.6 mA)(1.5 kΩ) = −19.9 V L 3 mH = t¢ = = 2 µs R 1.5 kΩ
8V = 5.33 mA 1.5 kΩ iL = If + (Ii − If)e−t/τ = 5.33 mA + (18.6 mA − 5.33 mA)e−t/2µs = 5.33 mA + 13.27 mAe−t/2µs υL = −19.9 Ve−t/2µs Ii = 18.6 mA
29.
If =
a.
RTh = 2 MΩ || 10 MΩ = 1.67 MΩ 10 MΩ(24 V) ETh = = 20 V 10 MΩ + 2 MΩ = 12 µA
iL = 12 µAe-t/5 µs 10 µA = 12 µAe-t/5 µs
CHAPTER 11
143
0.833 = e-t/5 µs loge 0.833 = -t/5 µs 0.183 = t/5 µs t = 0.183(5 µs) = 0.92 µs
30.
b.
uL (0+) = iL(0+)Rm = (12 µA)(10 MΩ) = 120 V uL = 120 Ve-t/5µs = 120 Ve-10µs/5µs = 120 Ve-2 = 120 V(0.135) = 16.2 V
c.
uL = 120 Ve-5t/t = 120 Ve-5 = 120 V(6.74 ¥ 10-3) = 0.81 V
a.
Closed Switch: RTh = 1.2 kΩ || 2.2 kΩ = 0.776 kΩ 1.2 kΩ(24 V) ETh = = 8.47 V 1.2 kΩ + 2.2 kΩ
Open Switch: ¢ = 6.9 kΩ || 1.2 kΩ = 1.02 kΩ RTh ¢ = 1.2 kΩ(24 V) = −3.56 V ETh 8.1 kΩ
−3.56 V + υR − υL = 0 υL = −3.56 V + (10.91 mA)(1.02 kΩ) = 7.57 V υL = 7.57Ve−t/1.18 ms
L 1.24 H = = 1.18 ms R 1.02 kΩ 3.56 V Iss = = 3.49 mA = If 1.02 kΩ iL = If + (Ii - If)e-t/t = -3.49 mA + ((-10.91 mA - (-3.49 mA))e-t/1.18 ms iL = −3.49 mA − 7.42 mAe−t/1.18 ms t=
144
CHAPTER 11
b.
31.
a.
iL = 150 mA(1 - e-1.5ms/15ms) = 150 mA(1 - e-1/10) = 150 mA(1 - e-0.01) = 150 mA(1 - 0.90484) = 150 mA(0.0952) = 14.28 mA
b.
iL = 150 mA(1 - e-150ms/15ms) = 150 mA(1 - e-10) = 149.99 mA
c.
75 mA = 150 mA(1 - e-t/t) 0.5 = 1 - e-t/t 0.5 = e-t/t 0.5 = e-t/t loge 0.5 = -t/t t = -(t)(loge 0.5) = -(15 ms)(loge 0.5) = -(15 ms)(-693.15 ¥ 10-3) = 10.397 ms
d.
149 mA = 150 mA(1 - e-t/15 ms) 0.99 = 1 - e-t/15ms -0.01 = -e-t/15ms 0.01 = e-t/15ms loge 0.01 = -t/15 ms t = -(15 ms)(loge 0.01) = -(15 ms)(-4.605) = 69.08 ms
CHAPTER 11
145
32.
a.
IL (1τ) = 0.632Imax = 126.4 µA Imax =
= 200 µA
iL = I m (1 - e - t / t ) -64.4 ms ö æ ç 160 mA = 200 mA 1 - e t ÷ ç ÷ è ø 0.8 = 1 - e
b.
0.2 = e
-64.4 ms t
-64.4 ms t
log e 0.2 = -1.61 =
t=
33.
-64.4 ms t
64.4 ms = 40 µs 1.61
L L = 40 ms = , L = (500 kΩ)(40 µs) = 20 mH R 500 kΩ
c.
t=
d.
Im =
a.
L fi open circuit equivalent 10 MΩ(16 V) VL = = 13.33 V 10 MΩ + 2 MΩ
E fi E = (200 µA)(500 Ω) = 100 mV R
b.
RTh = 2 MΩ || 10 MΩ = 1.67 MΩ 10 MΩ(16 V) ETh = = 13.33 V 10 MΩ + 2 MΩ = 7.98 µA c.
iL = 7.98 µA(1 - e-t/3 µs)
t=
= 3 µs
10 µA = 7.98 µA(1 - e-t/3 µs) 1.253 = 1 - e-t/3 µs 0.253 = e-t/3 µs loge(0.253) = -t/3µs 1.374 = t/3µs t = 1.374(3 µs) = 4.12 µs 146
CHAPTER 11
uL = 13.33 V e-t/3 µs = 13.33 V e-12 µs/3 µs = 13.33 V e-4 = 13.33 V(0.0183) = 0.244 V
d.
34.
eL =
é15 mA ù 0 - 4 ms, eL = (200mH) ê ú = 750 mV ë 4 ms û é0 mA ù 4 - 10 ms, eL = (200 mH) ê ú= 0 V ë 6 ms û é15 mA ù 10 - 14 ms, eL = (200 mH) ê ú = 750 mV ë 4 ms û é15 mA ù 14 - 18 ms, eL = −(200 mH) ê ú = −750 mV ë 4 ms û
:
18 - 19 ms, eL = 0 V
eL
é15 mA ù 19 - 22 ms, eL = −(200 mH) ê ú = −1 V ë 3 ms û 22 ms Æ, eL = 0 V
1 0.75 V
18 19 4
5
10
14 15
22 20
25
t (ms)
−0.75 V −1
35.
uL = 0 Æ 2 ms: uL = 0 V
æ 30 mA ö 2 Æ 6 ms: uL = −(5 mH) ç ÷ = −37.5 mV è 4 ms ø 6 Æ 10 ms: uL = 0 V æ 20 mA ö 10 Æ 14 ms: uL = (5 mH) ç ÷ = 25 mV è 4 ms ø 14 Æ 17 ms: uL = 0 V æ 5 mA ö 17 Æ 19 ms: uL = −(5 mH) ç ÷ = −12.5 mV è 2 ms ø 19 Æ, uL = 0 V
CHAPTER 11
147
ʋL(mV) 40 30
25
20 10 0
2
6 5
17 10
0
19
14 15
20
25
t (ms)
−10 –12.5
−20 −30 −40
36.
–37.5
L = 10 mH, 4 mA at t = 0 s
uL = 0 - 5 µs: uL = 0 V, ∆iL = 0 mA and iL = 4 mA 5 - 10 µs: ∆iL = 10 - 12 µs: ∆iL =
(-10 V) = -5 mA (-25 V) = -5 mA
12 - 16 µs: uL = 0 V, ∆iL = 0 mA and iL = -6 mA 16 - 24 µs: ∆iL =
10 V = 8 mA
iL(mA) 10 5 0 −5 −10
4 mA
+2 mA 5
−1 mA10
15 16
24 25
t (µs)
−6 mA
L4 + L5 = 5.6 mH + 3.4 mH = 9.0 mH L3 || (L4 + L5) = 4 mH || 9 mH = 2.77 mH = L" L' = L1 || (L2 + L") = 2.4 mH || (3.3 mH + 2.77 mH) = 2.4 mH || 6.07 mH L' = (2.4 mH)(6.07 mH)/(2.4 mH + 6.07 mH) = 1.72 mH LT = L6 + L' = 10 mH + 1.72 mH = 11.72 mH
37.
a.
38.
L2 || L4 = 20 mH || 60 mH = 15 mH L' = L2 + L3 || L4 = 55 mH + 15 mH = 70 mH L" = L5 || L' = 22 mH || 70 mH = 16.74 mH LT = L1 + L' = 18 mH + 16.74 mH = 34.74 mH
148
20
CHAPTER 11
39.
33 mH + 1.8 mH = 5.1 mH 4.7 mH || 5.1 mH = 2.45 mH
40.
= 6.2 mH + 12 mH || 36 mH + 24 mH = 39.2 mH = 9.1 µ F + 10 µF || 91 µF = 9.1 µF + 9.01 µF = 18.11 µF || 3.3 µF = 18.11 µF || 3.3 µF = 2.79 µF CT = 39.2 mH in series with 2.79 µF
41.
7 µF || 42 µF = 6 µF 12 µF + 6 µF = 18 µF 5 mH + 20 mH = 25 mH Series combination of 2.2 kΩ resistor, 25 mH coil, 18 µF capacitor
42.
a.
= 2 kΩ || 8.2 kΩ = 1.61 kΩ, 1.2 mH L¢ t= T = = 745.3 µs ¢ RT 1.61 kΩ
= 3 mH || 2 mH = 1.2 mH
iL =
36 V (1 - e - t / 745.3 ms ) = 22.36 mA(1 - e- t/745.3µs) 1.61 kΩ uL = Ee-t/t = 36 Ve -t/745.3µs =
b.
CHAPTER 11
149
43.
a.
Source conversion: E = 16 V, Rs = 2 kΩ RTh = 2 kΩ + 2 kΩ || 8.2 kΩ = 2 kΩ + 1.61 kΩ = 3.61 kΩ 8.2 kΩ(16 V) ETh = = 12.86 V 8.2 kΩ + 2 kΩ E 12.86 V 30 mH L = Im = Th = = 3.56 mA, t = = 8.31 µs RTh 3.61 kΩ R 3.61 kΩ iL = 3.56 mA(1 - e- t/8.31µs) u L1 + u L2 = 12.86 V initially (t = 0+)
10 mH of total = 10 mH + 20 mH uL = 4.29 Ve−t/8.31µs
uL =
= 4.29 V
b.
44.
a. ¨RTh = 10 kΩ || 20 kΩ = 6.67 kΩ ETh =
= 13.33 V
LT = 3 H + 4.7 H || 10 H = 3 H + 3.197 H = 6.197 H τ=
= 0.93 ms
υL = 13.33Ve- t/0.93 ms iL =
150
(1 - e-t/τ) = 2 mA(1 - e- t/0.93 ms)
CHAPTER 11
b.
c.
u L3 = (0.52)(13.33e−t/0.93 ms) = 6.93 Ve−t/0.93 ms
25 V
45.
= I 2 I= R2
= 6.25 A
E 25 V 25 V = = = 1.79 A R2 + R3 10 Ω + 4 Ω 14 Ω
+ I2 = 6.25 A + 1.79 A = 8.04 A
I1 = 46.
I1 = I2 = 0 A V1 = V2 = E = 100 V
47.
I1 =
= 3 A, I2 = 0 A
V1 = 12 V, V2 = 0 V 48.
V2 =
6 Ω + (75 V) = 15 V 6 Ω + 4 Ω + 20 Ω
V1 = ∞ V (open ckt)
75 V = 2.5 A 20 Ω + 4 Ω + 6 Ω I2 = 0 A I1 =
CHAPTER 11
151
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=
b.
R=
c.
R=
=
= 0.04 T
=
from the above R (c) > R (a) > R (b) 5.
R=
6.
R=
7.
9 in. H=
8.
µ=
9.
B=
F Φ
F Φ
=
=
500 At = 609.76 ¥ 103 At/Wb 8.2´10-4 Wb 150 gilberts 66,000 max wells
= 2.273 ¥ 10-3 rels (CGS)
= 0.2286 m
500 At F = = 2187.23 At/m 0.2286 m l = 4 ¥ 10- 4 Wb/Am
= 0.33 T
Fig. 12.7: H 800 At/m NI = Hl fi I = Hl/N = (800 At/m)(0.2 m)/75 t = 2.13 A
152
CHAPTER 12
10.
B=
= 0.6 T
Fig. 12.7, Hiron = 2500 At/m Fig. 12.8, Hsteel = 70 At/m NI = Hl(iron) + Hl(steel) (200 t)I = (Hiron + Hsteel)l (200 t)I = (2500 At/m + 70 At/m)0.3 m I= 11.
a.
200
= 3.86 A
N1I1 + N2I2 = Hl B=
=1T
Fig. 12.7: H 750 At/m N1(2 A) + (40 t)(3 A) = (750 At/m)(0.2 m) N1 = 15 t
12.
= 13.34 ¥ 10- 4 Wb/Am
b.
µ=
a.
80,000 lines l(cast steel) = 5.5 in. l(sheet steel) = 0.5 in. Area = 1 in.2 B=
= 8 ¥ 104 ¥ 10-8 Wb = 8 ¥ 10-4 Wb = 0.14 m = 0.013 m = 6.45 ¥ 10-4 m2 = 1.24 T
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.
Cast steel: µ =
= 9.73 ¥ 10- 4 Wb/Am
Sheet steel: µ =
= 26.96 ¥ 10- 4 Wb/Am
CHAPTER 12
153
13.
N 1I + N 2 =
+
(20 t)I + (30 t)I = (50 t)I = B=
" "
with 0.25 in.2
B=
= 1.6 ¥ 10-4 m2
= 0.5 T
Fig. 12.8: Hcast steel Fig. 12.7: Hcast iron
280 At/m 1500 At/m
lcast steel = 5.5 in.
= 0.14 m
lcast iron = 2.5 in.
= 0.064 m
(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.
a.
lab = lef = 0.05 m, laf = 0.02 m, lbc = lde = 0.0085 m NI = 2Hablab + 2Hbclbc + Hfalfa + Hglg = 1.2 T fi H
B=
360 At/m (Fig. 12.8)
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 = 3.33 ¥ 10- 3 Wb/Am
µsheet steel =
µair = 4π ¥ 10- 7 Wb/Am µsheet steel: µair = 3.33 ¥ 10-3 Wb/Am:4p ¥ 10-7 15.
4 cm f=
2627:1
= 0.04 m NI
(80 t)(0.9 A)
= 1.35 N
154
CHAPTER 12
16.
C = 2πr = (6.28)(0.4 m) = 2.512 m B=
= 1.54 T
Fig. 12.7: Hsteel = 2100 At/m Hg = 7.97 ¥ 105; Bg = (7.97 ¥ 105)(1.54 T) = 1.23 ¥ 106 At/m; lg = 0.002 m; lsteel = 2.512 m; N1I1 + N2I2 = Hglg + Hl(sheet steel) (250 t)(I1) + (50 t)(0.3 A) = (1.23 ¥ 106 At/m)(0.002 mm) + (2100 At/m)(2.512 m) I1 = 30.88 A 17.
a.
= 2 ¥ 10-3 m
0.2 cm
= 0.79 ¥ 10-4 m2
A=
NI = Hglg, Hg = 7.96 ¥ 105 Bg é æ 0.2´10 -4 Wb ö ù 5 ê ÷ ú 2 ¥ 10-3 m (200 t)I = (7.96´10 ) ç -4 2 ÷ ç ê è 0.79´10 m ø úû ë I = 2.02 A b.
Bg =
= 0.25 T
F@ 2N 18.
Table: Section
Φ(Wb)
-4
a-b, g-h b-c, f-g
2 ¥ 10
c-d, e-f
2 ¥ 10-4
-4
CHAPTER 12
2 ¥ 10-4
B(T)
H
l(m)
5 ¥ 10 5 ¥ 10-4
0.2 0.1
5 ¥ 10-4
0.099
-4
a-h b-g d-e
A(m2)
5 ¥ 10 2 ¥ 10-4
0.2 0.2
5 ¥ 10-4
0.002
Hl
155
Bbc = Bcd = Bg = Bef = Bfg =
= 0.4 T
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 For loop 2: SF = 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 and Hgb =
= 3300 At/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 Bab = Bha = Bgh =
= 1.02 T
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: SF = 0 NI = Hablab + Hbglbg + Hghlgh + Hahlah (200 t)I = 36 At + 659.49 At + 36 At + 36 At (200 t)I = 767.49 At 3.84 A I 19.
156
NI = Hl l = 2πr = (6.8)(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)(0.012 m2) Φ = 8.16 mWb
CHAPTER 12
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 fi H 270 At/m (Fig. 12.8) 300 At = (270 At/m)(0.8 m) + 637.6(0.47 T) 300 At π 515.67 At \ Poor approximation! ¥ 100% @ 58% Reduce Φ to 58% 0.58(0.94 ¥ 10-4 Wb) = 0.55 ¥ 10-4 Wb B=
= 0.28 T fi H
190 At/m (Fig. 12.8)
300 At = (190 At/m)(0.8 m) + 637.6(0.28 T) 300 At π 330.53 At Reduce Φ another 10% = 0.55 ¥ 10-4 Wb - 0.1(0.55 ¥ 10-4 Wb) = 0.495 ¥ 10-4 Wb B=
= 0.25 T fi H
175 At/m (Fig. 12.7)
300 At = (175 At/m)(0.8) + 637.6(0.28 T) 300 At π 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: 900 At/m ö æ ç B = 1.5 T 1 - e 700 At/m ÷ = 1.09 T ç ÷ è ø Graph: 1.1 T H = 1800 At/m: 1800 At/m ö æ B = 1.5 T ç1 - e 700 At/m ÷ = 1.39 T ç ÷ è ø Graph: 1.38 T H = 2700 At/m: 2700 At/m ö æ ç B = 1.5 1 - e 700 At/m ÷ = 1.47 T ç ÷ è ø Graph: 1.47 T
CHAPTER 12
157
Excellent comparison! 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 = e-H/700 At/m
æ B ö loge ç1 ÷= è 1.5 T ø æ B ö and H = -700 loge ç1 ÷ è 1.5 T ø d.
B = 1 T:
æ 1T ö H = -700 loge ç1 ÷ = 769.03 At/m è 1.5 T ø Graph: B = 1.4 T:
750 At/m
æ 1.4 T ö H = -700 loge ç1 ÷ = 1895.64 At/m è 1.5 T ø Graph: e.
1920 At/m
æ B ö H = -700 loge ç1 ÷ è 1.5 T ø æ 0.2 T ö = -700 loge ç1 ÷ è 1.5 T ø = 100.2 At/m I=
= 40.1 mA vs 44 mA for Ex. 12.1
158
CHAPTER 12
Chapter 13 1.
a. b. c. d. e.
10 V 15 ms: -10 V, 20 ms: 0 V 20 V 20 ms 2 cycles
2.
a. b. c. d. e.
200 µA 1 µs: 200 µA, 7 µs: -200 µA 400 µA 4 µs 2.5 cycles
3.
a. b. c. d. e.
40 mV 1.5 ms: -40 mV, 5:1 ms: -40 mV 80 mV 2 ms 3.5 cycles
4.
a. b. c.
e. f.
high 5 cycles T = 5 µs 1 1 = = 200 kHz f= T 5 ms 16 mV 32 mV
a.
T=
b.
T=
c.
T=
d.
T=
a.
f=
b.
f=
c.
f=
d.
f=
d.
5.
6.
CHAPTER 13
250 Hz
= 4 ms
50 MHz 28 kHz 2 Hz
= 20 ns
= 35.71 µs
= 0.5 s
= 1 Hz 1/36 s 75 ms 40
= 36 Hz = 13.33 Hz = 25 kHz
159
7.
8. 9. 10.
T=
2 kHz
= 0.5 ms, 5(0.5 ms) = 2.5 ms
25 ms = 0.25 ms 100 cycles 72 cycles = 9 Hz f= 8s T=
a.
Vpeak = (2.5 div.)(50 mV/div) = 125 mV
b.
T = (3.2 div.)(10 µs/div.) = 32 µs
c.
f=
11.
a. b. c. d.
Peak = 2.8 div.(10 mV/div.) = 28 mV Peak-to-peak = 2(28 mV) = 56 mV T = 2 div.(5 µs/div.) = 10 µs 5 cycles
12.
a. b. c. d.
13.
a. b. c. d.
14.
a. b.
160
1 1 = = 31.25 kHz T 32 ms
æ p ö Radians = ç ÷ 40° = 0.22π rad è180° ø æ p ö Radians = ç ÷ 60° = rad è180° ø æ p ö Radians = ç ÷ 135° = 0.75p rad è180° ø æ p ö Radians = ç ÷ 170° = 0.94p rad è180° ø æ180° ö æp ö Degrees = ç ÷ ç ÷ = 60° è p øè 3 ø æ180° ö Degrees = ç ÷ 1.2p = 216° è p ø æ180° ö 1 Degrees = ç ÷ p = 18° è p ø 10 æ180° ö Degrees = ç ÷ 0.6 p = 108° è p ø
2p 2p = = 3.93 rad/s T 1.6 s 2p 2p w= = = T 0.5 ms 0.5
w=
= 12.5 ¥ 103 rad/s
CHAPTER 13
2p 2p 2p = 897.597 ¥ 103 rad/s = = 7 µs 7´10-6 s T 2p 2p w= = 2.09 × 106 rad/s = T 3×10−6 s
w=
c. d. 15.
a. b. c. d.
w = 2p f = 2p (150 Hz) = 942.48 rad/s w = 2p f = 2p (0.5 kHz) = 3.142 × 103 rad/s w = 2p f = 2p (4 kHz) = 25.13 ¥ 103 rad/s w = 2p f = 2p (0.008 MHz) = 50.27 ¥ 103 rad/s
16.
a.
w = 2p f =
17.
f=
c.
f=
d.
f=
= 104.09 Hz
= 9.61 ms
T=
b.
654 rad/s
fif=
18 rad/s
w 2p
= 2.86 Hz, T = 349.07 ms
6600 rad/s
w 2p
=
0.19 rad/s
2p
= 1050.42 Hz, T = 0.952 ms
= 30.24 ¥ 10-3 Hz, T = 33.07 ms
æ p ö 2p radians (120°) ç ÷= è180° ø 3 t=
q
w
=
1 2p/ 3 rad 2p/ 3 rad 2 = 5.56 ms = = = 2p f 2p (60 Hz) (6)(60) 180
18.
æ p ö p (45°) ç ÷ = , a = wt fi w = è180° ø 4
19.
a.
Amplitude = 20, f =
b.
Amplitude = 12, f = 120 Hz
c.
Amplitude = 106, f =
d.
Amplitude = 8, f =
20.
-
21.
-
CHAPTER 13
4 9
= 87.27 rad/s
= 60 Hz
= 1591.55 Hz
w 2p
=
10, 058 rad/s = 1.6 kHz 2p
161
= 35.699 ms,
cycle = 17.85 ms
22.
T=
23.
i = 0.3 sin 60° = 0.3(0.866) = 0.26 A
24.
æ180° ö 1.4p ç ÷ = 252° è p ø
176
u = 25 sin 252° = 25(-0.9511) = -23.78 V 25.
26.
8 ¥ 10-3 = 40 ¥ 10-3 sin a 0.2 = sin a a = sin-1 0.2 = 11.54° and 180° - 11.54° = 168.46°
u = Vm sin a 60 = Vm sin 30° = Vm (0.5) 60 \Vm = = 120 V
1.5 ms 160°
æ 360 ö T = 1.5 ms ç ÷ = 18 ms è 30 ø = 55.56 Hz 18 w = 2p f = (2p)(55.56 Hz) = 349.07 rad/s f=
and u = 120 sin 349.07t 27.
-
28.
-
29.
a.
u = 6 × 10- 3 sin (2π 2000t + 30°)
b.
i = 20 ¥ 10- 3 sin(2π 60t - 60°)
30.
a.
u = 120 ¥ 10- 6 sin(2π 1000t - 80°)
31.
u = 12 ¥ 10- 3 sin(2π 2000t + 135°)
32.
u = 8 ¥ 10- 3 sin(2π 500t +π/6)
33.
u leads i by 90°
34.
u leads i by 10°
35.
u = 5 sin (wt - 30° + 90°) = 5 sin (wt + 60°) and i = 8 sin(wt + 50°) u leads i by 10°
36.
u = −5 cos(wt + 90°) = 5 sin(ωt + 90° + 90° + 180°) = 5 sinwt i = −3 sin(wt + 20°) = 3 sin(ωt + 20° + 180°) = 3 sin(wt + 200°) For υ = 5 sin ωt i = 3 sin(ωt + 200°) i leads u by 200°
162
CHAPTER 13
37.
T= t1 =
38.
= 1 ms
120° æ T ö 2 æ1 ms ö ç ÷= ç ÷= 180° è 2 ø 3 è 2 ø
2p T
w = 2p f = T=
2p
w
=
2p = 125.66 µs 50, 000 rad/s
æ 40° ö t1 = ç ÷ (T) = è 360° ø = 13.96 µs 39.
æ 40° ö ç ÷ (125.66 µs) è 360° ø
T = 1 ms tpeak @ 30° tpeak =
40.
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.
= 125 Hz (both)
q=
¥ 360° = 207° i leads e
or e leads i by 153° 41.
42.
G=
æ1 ö 1 1 (1 mV)(10 ms) + (5 mV)(10 ms) - (2 mV)(10 ms) + 2ç (4 mV)(5 ms)÷ - (2 mV)(10 ms) 2 è2 ø 2
40 ms 5 mV + 50 mV - 20 mV + 20 mV - 10 mV 45 mV = = 40 40 = 1.125 µV
CHAPTER 13
163
43.
1 1 ms) + (6 mV)(3 ms) - (2 mV)(5 ms) 2 2 G= 14 ms -6 mV + 4 mV + 18 mV - 5 mV +22 mV - 11 mV 11 mV = = = 14 14 14 = 0.786 mV -(6 mV)(1 ms) + (8 mV)(
44.
45.
a.
0V
b.
c. 46.
The same
1 1 (pr2) = p(20 mA)2 = 628.32 µA 2 2 628.32 mA 628.32 mA = G= = 15.71 mA d 40 mA (15.71 mA)(/ p ) - (5 mA)(/ p) G= 2/ p
Area =
= 5.36 mA 47.
48.
a.
T = ( 2 div.)(0.2 ms/div) = 0.4 ms
b.
f=
c.
Average = (-2.5 div.)(10 mV/div.) = -25 mV
a.
T = (4 div.)(10 µs/div.) = 40 µs
b.
f=
c.
G=
= 2.5 kHz
= 25 kHz
=
164
CHAPTER 13
=
= 1.713 div.
1.713 div.(10 mV/div.) = 17.13 mV Vrms = 0.7071(130 V) = 91.92 V Irms = 0.7071(5 ¥ 10−3 A) = 3.54 mA Vrms = 0.7071(9 ¥ 10−6 V) = 6.36 µV
49.
a. b. c.
50.
ω = 2πf = 2π(60 Hz) = 377 rad/s a. u = 6.8( 2)sin 377t b. c.
51.
i = 60( 2)sin 3.77t = 84.85 sin 377t u = 5 ¥ 103 ( 2 ) sin 3.77t = 7.07 ¥ 103 sin 377t
Vrms =
(2 V)2 (4 s) + (-2 V)2 (1 s)+ (-1 V)2 4s
( )=
12 s
2 2 2 16 V s + 4 V s + 4 V s 12 s
s 24 V 2/ 2 = 2V 12 / s = 1.414 V =
52.
Vrms = =
53.
(3 V)2 (2 s) + (2 V)2 (2 s) + (1 V) 2 2s + (-1 V) 2 (2 s) + (-3 V) 2 (2 s) + (1 V) 2 2s
( )
12 s
50 2 V = 12
2 4.167 V = 2.04 V
=0V
G= Vrms =
54.
a.
( )
=8V
T = (4 div.)(10 µs/div.) = 40 µs f=
= 25 kHz
Av. = (1 div.)(20 mV/div.) = 20 mV Peak = (2 div.)(20 mV/div.) = 40 mV = 34.64 mV
rms = b.
T = (2 div.)(50 µs) = 100 µs f=
= 10 kHz
Av. = (-1.5 div.)(0.2 V/div.) = -0.3 V Peak = (1.5 div.)(0.2 V/div.) = 0.3 mV rms = CHAPTER 13
= 367.42 mV 165
55.
a.
b.
A1 = Area = 96 + (4)(64) + (2)(4) = 96 + 256 + 8 = 360
56.
166
c.
rms =
= 5.48
d.
G=
e.
rms @ 1.5 (average value)
a.
Vdc = IR = (4 mA)(2 k ) = 8 V Meter indication = 2.22(8 V) = 17.76 V
b.
Vrms = 0.707(16 V) = 11.31 V
= 3.67
CHAPTER 13
Chapter 14 1.
-
2.
-
3.
a.
(377)(10) cos 377t = 3770 cos 377t
b.
(400)(20) cos(400t + 60°) = 8 ¥ 103 cos(400t + 60°)
c.
(
d.
(-200)(1) cos(t + 180°) = -200 cos(t + 180°) = 200 cos t
a.
Im = Vm/R = 160 V/20 Ω = 8 A, i = 8 sin 100t
b.
Im = Vm/R = 60 V/20 Ω = 3 A, i = 3 sin(2000t + 45°)
c.
Im = Vm/R = 6 V/3 Ω = 2 A, i = 2 sin(wt + 100°)
d.
Im = Vm/R = 12 V/3 Ω = 4 A, i = 4 sin(wt + 220°)
a.
Vm = ImR = (0.2 A)(7.8 ¥ 103 Ω) = 1.56 kV Vm = 1.56 ¥ 103 V υ = 1.56 ¥ 103 sin 500t
b.
Vm = ImR = (5 ¥ 10-3 A)(7.8 ¥ 103 Ω) = 39 V υ = 39 sin(600t - 120°)
a.
0Ω
b.
XL = 2πfL = 2π(3 mH)(60 Hz) = 1.13 Ω
c.
XL = 2π(3 mH)(8 kHz) = 150.79 Ω
d.
XL = 2π(3 mH)(1.4 MHz) = 26.39 kΩ
a.
L=
b.
L=
a.
XL = 2πfL fi f =
4.
5.
6.
7.
8.
f=
CHAPTER 14
20)(157) cos(157t - 20°) = 4440.63 cos(157t - 20°)
2.5 kΩ (12.47 kHz) 45 kΩ (5.8 kHz)
= 1.23 H
XL XL XL = = 2pL (6.28)(47 mH) 295.16 ´10 -3 H
10 Ω 295.16´10
= 31.91 mH
-3
H
= 33.88 Hz
167
9.
10.
11.
12.
13.
168
XL
4 kΩ
b.
f=
c.
f=
a.
Vm = ImXL = (25 mA)(20 Ω) = 500 mV υ = 0.5 sin(wt + 90°)
b.
Vm = ImXL = (40 ¥ 10-3 A)(20 Ω) = 0.8 V υ = 0.8 sin(wt + 150°)
c.
i = 6 sin(wt + 150°), Vm = ImXL = (6 A)(20 Ω) = 120 V υ = 120 sin(wt + 240°) = 120 sin(wt - 120°)
a.
XL = wL = (150 rad/s)(0.15 H) = 22.5 Ω Vm = ImXL = (15 A)(22.5 Ω) = 337.5 V υ = 337.5 sin(150t + 90°)
b.
XL = wL = (400 rad/s)(0.15 H) = 60 Ω -6 Vm = ImXL = (6 ¥ 10-6 A)(60 Ω) = 360 ´10 V -6 υ = 360 ¥ 10 sin(400t + 120°)
a.
Im =
Vm 120 V = = 3 A, i = 3 sin(wt - 90°) XL 40 Ω
b.
Im =
Vm 30 V = = 0.75 A, i = 0.75 sin(wt - 70°) X L 40 Ω
a.
XL = wL = (90 rad/s)(0.25 H) = 22.5 Ω Im = Vm/XL = 2.5 V/22.5 Ω = 0.111 A i = 0.111 sin(90t - 90°)
b.
XL = wL = (20 rad/s)(0.25 H) = 5 Ω Im = Vm/XL = 16 mV/5 Ω = 3.2 mA i = 3.2 ¥ 10-3 sin(20t - 85°)
a.
XC =
1 1 = =• Ω 2p fC 2p (0 Hz)(0.4´10-6 F)
b.
XC =
1 1 = = 4973.6 kΩ 2p fC 2p (80 Hz)(0.4´10-6 F)
c.
XC =
1 1 = = 159.15 Ω 2p fC 2p (2.5 kHz)(0.4´10-6 F)
295.16´10
-3
H
XL 295.16 ´10
-3
H
=
=
295.16´10 -3 H
= 13.55 kHz
12 kΩ 295.16 ´10 -3 H
= 40.66 kHz
CHAPTER 14
d.
14.
15.
16.
17.
XC =
XC =
1 1 ÞC = 2p fC 2p fX C
a.
C=
b.
C=
a.
f=
b.
f=
1 1 = = 0.68 Hz 2p CX C 2p (3.9´10 -6 F)(60 kΩ)
c.
f=
= 408.1 kHz
d.
f=
a.
Im = Vm/XC = 120 V/2.5 Ω = 48 A i = 48 sin(wt + 90°)
b.
Im = Vm/XC = 4 × 10−3 V/2.5 Ω = 0.16 A i = 1.6 × 10−3 sin(wt + 130°)
a.
υ = 30 sin 250t, XC =
1
2p (250 Hz)(75 Ω) = 8.49 µF
1
2p(36 kHz)(2.2 kΩ) = 2009.5 pF = 4.08 kHz
Im =
b.
a.
= 20.40 Hz
4 kΩ
= 4 kΩ
(250)
= 7.5 mA, i = 7.5 ¥ 10-3 sin(250t + 90°)
υ = 90 ¥ 10-3 sin 377t, XC = Im =
18.
1 1 = = 0.159 Ω 2p fC 2p (2.5 mHz)(0.4´10-6 F)
90 2.65
90
= 2.65 kΩ = 33.96 µF, i = 33.96 ¥ 10-6 sin(377t + 90°)
Vm = ImXC = (50 ¥ 10-3 A)(2 kΩ) = 100 V υ = 100 sin(wt - 90°)
CHAPTER 14
169
19.
b.
Vm = ImXC = (2 ¥ 10-6)(2 kΩ) = 4 mV υ = 4 ¥ 10-3 sin(wt - 30°)
a.
i = 0.2 sin 500t, XC =
= 4 kΩ
(500)(0.5
Vm = ImXC = (0.2 A)(4 kΩ) = 800 V, υ = 800 sin(500t - 90°) b.
i = 5 ¥ 10-3 sin (377t − 45°), XC =
(377)(0.5
= 5.305 kΩ
Vm = ImXC = (5 ¥ 10-3 A)(5.305 kΩ) = 26.53 V υ = 26.53 sin(377t - 135°) 20.
a.
υ leads i by 90° fi L, XL = Vm/Im = 550 V/11 A = 50 Ω L=
b.
= 132.63 mH
υ leads i by 90° fi L, XL = Vm/Im = 36 V/4 A = 9 Ω = 147.36 µH
L= c.
υ and i are in phase fi R R=
=7Ω
i = 5sin(wt + 90° )üï ý i leads u by 90° Þ C u = 2000 sin wt ïþ
21.
XC =
= 400 Ω, C =
I 1 = = 15.92 µF wX C (157 rad/s)(400 Ω) fiL
XL = c.
-
23.
-
170
= 254.78 mH
u = 35 sin(wt - 20°)üï ý in phase Þ R i = 7 sin(wt - 20°) ïþ R=
22.
= 40 Ω, L =
=5Ω
CHAPTER 14
24.
fi
XC =
2π(2kΩ)(1.5 µF) (18.85
10−3)
@ 53.05 Hz 25.
XL = 2πfL = R L=
26.
= 318.47 mH
XL = 2πfL 1 XC = 2pfC f2 = and f =
27.
2
29.
10−6)(80
10−3)
= 2.5 kHz
XC = XL
1 = 2πfL fi 2pfC 28.
(2
a.
P=
b.
P=
c.
P=
d.
P=
R=
(3600
= 3.58 nF
= 389.7 W, Fp = 0.866
= 50 W, Fp = 1.0
= 73.86 W, Fp = 0.985
= 2.30 W, Fp = 0.766
æ8 A ö 2 = 7 Ω, P = I2R = ç ç ÷÷ 7 Ω = 224 W è 2ø (56 V) = 224 W P=
56 V
æ56 V ö æ 8 A ö P = VI cos θ = ç ÷÷ cos 0° = 224 W ÷÷ çç ç è 2 øè 2 ø All the same!
CHAPTER 14
171
30.
P = 150 W: Fp = cos θ = P/VI = 150 W/(200 V)(2.5 A) = 0.3 P = 0 W: Fp = cos θ = 0 500 500 W P = 500 W: Fp = = =1 (200 V)(2.5 A) 500
31.
P= 600 W =
(60 V)
(0.5) fi Im = 40 A
i = 40 sin(ωt + 20° − 60°) = 40 sin(ωt - 40°) 32.
a.
Im = Em/R = 120 V/6.8 kΩ = 17.65 mA, i = 17.65 ¥ 10- 3 sin(2π60t + 20°)
b.
æ17.65 mA ö 2 P=IR= ç ÷÷ 6.8 kΩ = 1.06 W ç 2 è ø
c.
T =
2
= 16.67 ms
6(16.67 ms) = 100.02 ms @ 0.1 s 33.
34.
a.
XL = ωL = (1500 rad/s)(3 mH) = 4.5 kΩ 240 V V 240 V Im = m = = 53.33 mA, i = 53.33 sin(1500t - 45°) = X L (1500 rad/s)(3 mH) 4.5 kΩ
b.
Power loss = 0 W
a.
XC =
1 1 1 = = = 294.88 kΩ w C (2p 600 rad/s)(900 pF) (3.768)(900 ¥ 10-12)
Em = ImXC = (20 ¥ 10-3 A)(294.88 kΩ) = 5857.62 V e = 5.897 ¥ 103 sin(2π600t - 120°)
35.
b.
P=0W
a.
X C1 =
1 1 1 = = = 50 Ω 4 2p f C1 wC1 (10 rad/s)(2 µF) = 10 Ω
E = 84.85 V –60°
I1 =
= 1.697 A –150°
I2 =
= 8.485 A –150°
i1 = 2.4 sin(104t + 150°) i2 = 12 sin(104t + 150°)
172
CHAPTER 14
b.
CT = 2 µF || 10 µF = 12 µF 1 1 = XC = 4 wC (10 rad/s)(12 µF)
84.85Ð 60° E = 8.33 Ω Ð -90° XCT = 10.19 A–150° is = 14.4 sin (104t + 150°)
Is =
36.
a.
L1 || L2 = 60 mH || 120 mH = 40 mH = 2πfLT = 2π(103 Hz)(40 mH) = 251.33 Ω Vm = Im X LT = (80 A)(251.33 Ω) = 20.11 kV
and υs = 20.11 kV sin(103t + 30° + 90°)
so that υs = 20.11 ¥ 103 sin(103t + 120°) ,
b.
= 2pfL1 = 2p(103 Hz)(60 mH) = 376.99 Ω = 16 A
and i1 = 16 sin(103t + 30°) = 2πfL2 = 2π(103 Hz)(120 mH) = 753.98 Ω =8A and i2 = 8 A sin(103t + 30°) 37.
a. c. e.
7.21 –56.31° 15.81 –71.57° 2236.07 –−63.43°
b. d. f.
4.24 –45° 500.05 –5.71° 0.45 –−63.43°
38.
a. c. e.
17.89 –−116.57° 20.22 × 10−3 –−8.53° @ 200 –0°
b. d. f.
8.94 –−26.57° 8.49 × 10−3 –−135° 1000 –−178.85°
39.
a. c. e.
4.6 + j3.86 −j2000 47.97 + j1.68
b. d. f.
−6.0 + j10.39 −6 × 10−3 − j2.2 × 10−3 4.7 × 10−4 −j1.71 × 10−4
40.
a. c. e.
42 + j0.11 −3 × 10−3 − j5.20 × 10−3 −15
b. d. f.
1 × 103 − j1.73 × 103 −6.13 × 10−3 + j5.14 × 10−3 2.09 × 10−3 − j1.20
41.
a.
9.4 + j8.4
b.
246.2 + j51.7
c.
5.74 × 10−6 + j84
CHAPTER 14
173
42.
43.
44.
45.
46.
47.
48.
a.
3.2 + j0.6
b.
239.3 + j301
c.
−20.5 + j4
a.
12.17 –54.70°
b.
98.37 –13.38°
c.
28.07 –−115.91°
a.
−12.0 + j34.0
b.
86.80 + j312.40
c.
−283.90 − j637.65
a.
8.00 –20°
b.
49.68 –−64.0°
c.
40 × 10−3–40°
a.
6.0 –−50°
b.
200 × 10−6 –60°
c.
109 –−170°
a.
4
b.
−4.15 − j4.23
c.
6.69 − j6.46
a.
b.
c.
174
= 10.0 - j5.0
= 19.38 ¥ 10-3 –-15.69°
= 3.07 ¥ 103 –79.44°
CHAPTER 14
49
= 5.06 –88.44°
a.
b.
æ öæ 8 öæ ö 1 1 ÷ ç ÷ç ÷ -4 2 ÷ç ç è 4´10 Ð 20° ø è j( j ) ø è 36 - j30 ø
æ 8 öæ ö 1 (2500– −20°) ç ÷ ç ÷ è - j ø è 46.861Ð -39.81° ø (2500 –-20°)(8j)(0.0213 –39.81°) = 426 –109.81° 50.
a. (x + j5) + (3x + jy) - j6 = 16 (x + 3x) + j(5 + y - 6) = 16 + j0 x + 3x = 16 5+y-6=0 y = +6 - 5 4x = 16 y=1 x=4 b.
51.
a.
(18 –20°)(x –-60°) = 30.64 - j25.72 (18x –-40°) = 40 –(-40°) 18 x = 40° x = 2.22 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-yfi
20 - 5xy = -70 20 - 5(9 - y)y = -70 5y(9 - y) = 90 y2 - 9y + 18 = 0 y= y=
= 6, 3
For y = 6, x = 3 y = 3, x = 6 (x = 3, y = 6) or (x = 6, y = 3) = 4 –-θ = 3.464 - j2 = 4 –-30°
b.
θ = 30° 52.
a.
180.0 –40°
b.
25 ¥ 10-3 –-60°
c.
212.13 –-120°
CHAPTER 14
175
53.
54.
a.
21.21 –−180°
b.
4.24 ¥ 10-6 –90°
c.
3.96 × 10−6–50°
a.
56.57 sin(377t + 20°)
b.
169.68 sin (377t + 10°)
c.
11.31 ¥ 10- 3 sin(377t − 110°)
d.
6000 sin(377t - 180°)
55.
(Using peak values) ein = υa + υb fi υa = ein - υb = 60 V –90° - 20 V –-45° = j60 V − (14.142 V − j14.142 V) = j60 V - 14.142 V + j14.142 V = −14.142 V + j74.142 V = 75.479 V –100.8° and va = 75.48 sin(377t + 100.8°)
56.
is = i1 + i2 fi i1 = is - i2 (Using peak values) = (30 ¥ 10-6 A –80°) - (4 ¥ 10-6 A –-30°) = 5.21 ¥ 10-6 + j0.98 ¥ 10−6 − 3.46 ¥10−6 + j2 ¥10−6 = 1.86 ¥10−6 A –59.58° -6 i1 = 1.86 ¥ 10 sin (wt + 59.58°)
57.
(Using peak values) ein = υa + υb + υc υa = ein − υb − υc = 120 V –30° − 30 V –60° − 40 V –−90° = (103.92 V + j60 V) − (15 V + j25.981 V) − (−j40 V) = 88.92 V + j74.02 V = 115.70 V –39.775° va = 115.70 sin(377t + 39.78°)
58.
(Using effective values) Is = I1 + I2 + I3 I1 = Is − I2 − I3 = 12.73 A –180° − 5.66 A –90° − 2[5.66 A –90°] = 12.73 A –180° − 5.66 A –90° − 11.32 A –90° = −12.73 A − (j5.66 A) − (j11.32 A) = −12.73 A − j16.98 A = 21.22 A –−126.86° = (1.414)(21.22 A)sin 377t i1 = 30 sin 377t
176
CHAPTER 14
Chapter 15 1.
2.
a.
I = (0.7071)(20 mA – 30°) = 14.14 mA – 30°
b.
V = IR = (14.14 mA – 30°)(2 kΩ – 0°) = 28.28 V – 30°
c.
−
d.
u = (1.414)(28.28 V) sin (1000t + 30°) = 40 sin (1000t + 30°)
e.
−
a.
VR = (0.7071)(24 V) – 20° = 16.97 V – 20°
b.
I=
c.
−
d.
3.
4.
VR 16.97 V Ð 20° = = 2.5 A – 20° R 6.8 Ω Ð 0°
i = (1.414)(2.5 A) sin (300t + 20°) = 3.53 sin (300t + 20°)
e.
−
a.
IL = (0.7071)(10 mA) – 40° = 7.071 mA/– 40°
b.
VL = ILXL = (0.7071 mA – 40°)(2 kΩ – 90°) = 14.14 V – 130°
c.
−
d.
uL = (1.414)(14.14 V) sin (250t + 130°) = 20 sin (250t + 130°)
e.
−
a.
XL = wL = (750 rad/s)(40 mH) = 30 Ω
b.
VL = (0.7071)(200 µV) – 90° = 141.42 µV – 90°
c.
IL =
CHAPTER 15
VL 141.42 mV Ð 90° = = 4.71 µA – 0° XL 30 Ω Ð 90°
177
5.
d.
−
e.
iL = (1.414)(4.71 µA) sin (750t + 0°) = 6.66 ¥ 10−6sin 750t
f.
−
a.
IL = (0.7071)(6 mA) – 20° = 4.243 mA – 20° VL = (0.7071)(16 V) – 110° = 11.314 V – 110°
6.
7.
178
VL 11.314 V Ð 110° = = 2.67 kΩ – 90° = XL IL 4.243 mA Ð 20°
b.
ZL =
c.
XL = wL fi L =
d.
−
e.
−
a.
VC = (0.7071)(60 V) – 60° = 42.43 V – 60°
b.
IC =
c.
−
d.
iC = (1.414)(1.061 A) sin (400t + 150°) = 1.5 sin (400t + 150°)
e.
−
a.
XC =
b.
IC = (0.7071)(5 µA) – −80° = 3.54 µA – −80°
c.
VC = ICXC = (3.54 µA – −80°)(5 kΩ – −90°) = 17.7 mV – −170°
d.
−
e.
uC = (1.414)(17.7 mV) sin (2000t − 170°) = 25.03 ¥ 10−3 sin (2000t − 170°)
XL 2.67 kΩ = = 2.23 H w 1200 rad/s
VC 42.43 V Ð 60° = = 1.061 A – 150° XC 40 Ω Ð -90°
1 1 = = 5 kΩ wC (20, 000 rad/s)(0.01 µF)
CHAPTER 15
8.
f.
−
a.
IC = (0.7071)(60 µA) – 80° = 42.43 µA – 80° VC = (0.7071)(24 mV) – −10° = 16.97 mV – −10°
b
XC = ZC =
c.
XC =
d.
−
e.
−
VC 16.97 mV Ð -10° = = 400 Ω – −90° 42.43 µA Ð 80° IC
1 1 1 ÞC = = = 1.25 µF wC wX C (2000 rad/s)(400 Ω)
9.
−
10.
XL = 2πfL = 2π(1.2 kHz)(5 mH) = 37.7 Ω
11.
XC =
1 1 = = 79.58 Ω 2pfC 2p(100 kHz)(0.02 µF)
12.
a.
ZT = 7.8 Ω + j8.2 Ω = 11.32 Ω – 46.43°
b.
ZT = 2 Ω - j8 Ω + 20 Ω = 22 Ω - j8 Ω = 23.41 Ω – -19.98°
c.
ZT = 3 kΩ + j3.2 kΩ + 5.6 kΩ + j6.8 kΩ = 8.6 kΩ + j10 kΩ = 13.19 kΩ – 49.30°
a.
ZT = 3 Ω + j5 Ω = 5.83 Ω – 59.1°
b.
ZT = 1 kΩ + j8 kΩ + j6 kΩ - j4 kΩ = 1 kΩ + j10 kΩ = 10.05 Ω – 84.29°
c.
ZT = R + jω L1 −
13.
j + jω L2 = R + jX L1 − jX C + jX L2 ωC
Now, XL1 = w L1 = 2p fL1 = 2p (103 Hz)(47 ¥ 10-3 H) = 295.31 Ω = 0.295 kΩ XC =
= 1.592 kΩ
XL2 = ωL2 = 2πfL2 = 2π(103 Hz)(200 ¥ 10−3 H) = 1.257 kΩ ZT = 0.5 kΩ + j0.295 kΩ - j1.59 kΩ + j1.257 kΩ ZT = 0.5 kΩ + j0.038 kΩ = 500 Ω - j38 Ω 14.
a.
ZT =
E 120 V Ð 0° = = 20 Ω – -45° = 14.142 Ω - j14.142 Ω = R - jXC 6 A Ð 45° I
b.
ZT =
80 V Ð 130° E = = 4 kΩ – 90° = j4 kΩ = jXL I 20 mA Ð 40°
CHAPTER 15
179
15.
ZT =
a.
ZT = 8 Ω + j6 Ω = 10 Ω – 36.87°
c.
I = E/ZT = 100 V – 0°/10 Ω – 36.87° = 10 A – -36.87° VR = (I – q)(R – 0°) = (10 A – -36.87°)(8 Ω – 0°) = 80 V – -36.87° VL = (I – q)(XL – 90°) = (10 A – -36.87°)(6 Ω – 90°) = 60 V – 53.13°
f.
P = I2R = (10 A)2 8 Ω = 800 W
g.
Fp = cos θT = R/ZT = 8 Ω/10 Ω = 0.8 lagging
a.
uR = 113.12 sin(w t - 36.87°) uL = 84.84 sin(w t + 53.13°) i = 14.14 sin (w t - 36.87°) ZT = 18 Ω - j29.15 Ω = 34.26 Ω –-58.30° 1 1 = XC = = 29.15 Ω 2p fC 2p (60 Hz)(91 µF)
c.
I=
h. 16.
8 kV Ð 0° E = = 666.67 Ω – 30° = 577.35 Ω + j333.34 Ω = R + jXL I 12 A Ð -30°
c.
= 3.50 A –78.30°
VR = (I –θ)(R –0°) = (3.50 A –78.30°)(18 Ω –0°) = 63.0 V –78.30° VC = (I –θ)(XC –-90°) = (3.50 A –78.30°)(29.15 Ω –-90°) = 102.03 V –−11.70°
17.
f.
P = I2R = (3.50 A)2 18 Ω = 220.5 W
g.
Fp = R/ZT = 18 Ω/34.26 Ω = 0.525 leading
h.
i = 4.95 sin(377t + 78.30°) υR = 89.1 sin(377t + 78.30°) υC = 144.27 sin(377t − 11.70°)
a.
ZT = 4 Ω + j6 Ω - j10 Ω = 4 Ω - j4 Ω = 5.66 Ω –-45°
c.
XL = wL fi L = XC =
d.
I=
= 16 mH
fi
= 265 µF
= 8.83 A –45°
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°
180
CHAPTER 15
18.
f.
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.
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.
XL = wL = (20 × 103 rad/s)(0.1H) = 2 kΩ 1 1 = XC = = 6.1 kΩ 3 wC (20´10 rad/s)(8200 pF)
= 4 Ω/5.66 Ω = 0.707 leading
ZT = 1.2 kΩ + j2 kΩ − j6.1 kΩ = 1.2 kΩ − j4.1 kΩ = 4.27 kΩ –−73.69° b.
−
c.
−
d.
4.24 V Ð 60° E = = 0.993 mA –133.69° ZT 4.27 kΩ Ð -73.69° VR = IR = (0.993 mA –133.69°)(1.2 kΩ –0°) = 1.19 V –133.69° VL = IXL = (0.993 mA –133.69°)(2 kΩ –90°) = 1.99 V –223.69° VC = IXC = (0.993 mA –133.69°)(6.1 kΩ –−90°) = 6.06 V –43.69° I=
e.
−
f.
E = VR + VL + VC 4.24 V –60° = 1.19 V –133.69° + 1.99 V –223.69° + 6.06 V –43.69° = (−0.822 V + j0.80 V) + (−1.44 V − j1.37 V) + (4.38 V + j 4.19 V) = 2.12 V+ j3.62 V 4.24 V –60° @ 4.20 V –59.65°
g.
P = I2R = (0.993 mA)2(1.2 kΩ) = 1.18 mW
h.
Fp =
i.
i = 1.4 × 10−3 sin (20,000t + 133.69°) υR = 1.68 sin (20,000t + 133.69°) υL = 2.81 sin (20,000t + 223.69°) υC = 8.57 sin (20,000t + 43.69°)
CHAPTER 15
1.2 kΩ R = = 0.281 leading 4.27 kΩ ZT
181
19.
20.
a.
ZT = 30 Ω + j100 Ω − j20 Ω = 30 Ω + j80 Ω = 85.44 Ω – 69.44°
b.
Is =
c.
VR = IRR = IsR = (468.16 mA – −9.44°)(30 Ω – 0°) = 14.04 V – −9.44°
d.
Fp = cos qT =
a.
ZT = 8 Ω + j34 Ω - j16 Ω = 8 Ω + j18 Ω = 19.7 Ω – 66.04° E 48 V Ð 0° - 32 V Ð 45° Is = IL = T = 19.7 Ω Ð 66.04° ZT
E 40 V Ð 60° = = 468.16 mA – −9.44° ZT 85.44 Ω Ð 69.44°
R 30 Ω = = 0.351 lagging ZT 85.44 Ω
48 V - (22.63 V + j22.63 V) 19.7 Ω Ð 66.04° 25.37 V - j22.63 V 33.99 V Ð -41.729° = = 19.7 Ω Ð 66.04° 19.7 Ω Ð 66.04° IL = 1.726 A – -107.77° =
21.
b.
V C = I C X C = I sX C = (1.726 A – −107.77°)(16 Ω – -90°) VC = 27.62 V – -197.77°
a.
IL1 = I = 8 mA – 30°
b.
ZT = R − jXC + jXL1 + R2 + jXL2 = 2 kΩ − j4 kΩ + j8 kΩ + 5 kΩ + j4 kΩ = 7 kΩ + j8 kΩ = 10.63 kΩ – 48.81° Vs = IZT = (8 mA – 30°)(10.63 kΩ – 48.81°) = 85.04 V – 78.81°
c.
22.
VR1 = IR1 = (8 mA – 30°)(2 kΩ – 0°) = 16 V – 30°
24 V (rms) fi 33.94 V (peak) 43.20 V(p − p) fi 21.60 V (peak)
22 Ω(33.94 V) 22 Ω +R 475.20 + 21.60R = 746.68 Vscope = 21.60 V =
182
CHAPTER 15
R=
23.
a.
271.48 Ω = 12.57 Ω ≈ 12.6 Ω 21.60 æ 22.8 V ö VL (rms) = 0.7071 ç ÷ = 8.06 V è 2 ø XL =
2.4 mA
= 3.36 kΩ
XL = w L = (1000 rad/s)L = 3.36 kΩ fi L =
3.36 kΩ = 3.36 H 1000 rad/s
b. (26 V)2 = 2
VR = 611.04 611.04 = 24.72 V
VR = R=
24.
V R (rms) 24.72 V = = 10.3 kΩ I (rms) 2.4 mA
c.
6.2 H
a.
æ10.37 Vö VR(rms) = 0.7071 ç ÷ = 3.666 V è 2 ø IR(rms) =
VR (rms) 3.666 V = = 366.6 µA 10 kΩ R
b. (15 V)
(3.666 V) 211.56 211.56
14.55 V
14.55 V VC (rms) = = 39.69 kΩ I (rms) 366.6 µA 1 1 1 ÞC = = XC = = 100.25 pF ≈ 100 pF 2p fC 2p fX C 2p (40 kHz)(39.69 kΩ) XC =
25.
P = VI cos θ fi 9000 W = (200 V)(I)(0.9) 9000 W I= = 50 A 180 0.9 = cos θ
CHAPTER 15
183
θ = 25.84° V = 200 V –0°, I = 50 A –-25.84° ZT =
50 A –-25.84°
= 4 Ω –−25.84° = 3.6 Ω + j1.74 Ω
26.
P = VI cos q fi 400 W = (240 V)(3 A) cos θ cos θ = 0.5556 fi θ = 56.25° V = 240 V –0°, I = 3 A –-56.25° 240 V –0° = 80 Ω –56.25° = 44.44 Ω + j66.52 Ω ZT = 3 A –−56.25° RT = 44.44 Ω = 4 Ω + R fi R = 40.44 Ω So, the series elements that must be in the enclosed contained in Fig. 15.102 are R = 40.44 Ω and XL = 66.52 Ω.
27.
a.
b.
28.
a.
b.
(6 kΩ –0°) (200 V –60°) (6 kΩ + j8 kΩ)
V2 =
(8 kΩ –90°)(200 V –60°) = 160 V –96.87° 10 –53.13°
(40 Ω –90°)(110 V –15°) 4400 V–105° = 89.27 V –50.75° 49.29 Ω–54.25° (6.8 Ω + j40 Ω + 22 Ω) (22 ΩÐ0°)(110 V Ð15°) 2420 kV Ð15° = = 49.097 V –-39.25° V2 = 49.29 ΩÐ54.25° 49.29 Ω Ð 54.25° V1 =
(20 Ω –90°) (30 V –60°) = 21.225 V –195° (20 Ω + j20 Ω − j40 Ω) (40 Ω –−90°)(30 V –60°) = 42.43 V –15° V2 = V1 =
ZT = 4.7 kΩ + j30 kΩ + 3.3 kΩ - j10 kΩ = 8 kΩ + j20 kΩ = 21.541 kΩ –68.199° = 3.3 kΩ + j30 kΩ - j10 kΩ = 3.3 kΩ + j20 kΩ = 20.27 kΩ –80.631° (20.27 kΩ –80.631°) (150 V –0°) V1 = = 141.15 V –12.432° V2 = =
29.
a.
1200 V–60° = 120 V –6.87° 10 –53.13°
V1 =
= 3.3 kΩ - j10 kΩ = 10.53 kΩ –-71.737° (10.53 kΩ –−71.737°) (150 V –0°)
= 73.33 V –-139.94°
XL = wL = (1000 rad/s)(20 mH) = 20 Ω XC =
= 25.64 Ω
ZT = 30 Ω + j20 Ω - j25.64 Ω = 30 Ω - j5.64 Ω = 30.53 Ω –-10.65°
184
CHAPTER 15
= 655.1 mA –50.65°
I=
VR = (I – θ)(R – 0°) = (655.1 mA –50.65°)(30 Ω –0°) = 19.65 V –50.65° VC = (655.1 mA –50.65°)(25.64 Ω –-90°) = 16.80 V –−39.35° b.
cos θT =
= 0.983 leading
c.
P = I2R = (655.1 mA)2 30 Ω = 12.87 W
f.
VR =
= 19.66 V –50.65° = 16.80 V –−39.35°
VC =
30.
g.
ZT = 30 Ω - j5.64 Ω = R - jXC
a.
ZT =
–tan-1XL/R
f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
b.
ZT 1.0 kΩ 1.008 kΩ 1.181 kΩ 1.606 kΩ 2.134 kΩ 2.705 kΩ
θT 0.0° 7.16° 32.14° 51.49° 62.05° 68.3°
VL = f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
c. f 0 Hz 1 kHz
CHAPTER 15
VL 0.0 V 0.623 V 2.66 V 3.888 V 4.416 V 4.646 V
θL = 90° - tan-1 XL/R 90.0° 82.84°
185
5 kHz 10 kHz 15 kHz 20 kHz
57.85° 38.5° 27.96° 21.7°
f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
VR = RE/ZT 5.0 V 4.96 V 4.23 V 3.11 V 2.34 V 1.848 V
d.
31.
a.
ZT = │ZT│ = f 0 kHz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz
186
–-tan-1XC/R , θT = -tan-1XC/R │ZT│ •Ω 353.1 Ω 150.80 Ω 120.78 Ω 111.09 Ω 105.58 Ω
θT -90.0° -73.55° -48.46° -34.11° -25.82° -18.71°
CHAPTER 15
b.
–-90° + tan-1XC/R
VC =
│VC│ =
f 0 Hz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz
c.
θC = -90° + tan-1 XC/R f 0 Hz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz
d.
│VC│ 10.0 V 9.59 V 7.49 V 5.61 V 4.36 V 3.21 V
θC 0.0° -16.45° -41.54° -55.89° -64.18° -71.29°
│VR│ =
f 0 Hz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz
CHAPTER 15
│VR│ 0.0 V 2.83 V 6.63 V 8.28 V 9.00 V 9.47 V
187
32.
a.
ZT = f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
b.
ZT •Ω 19.31 × 103 Ω 3.40 × 103 Ω 1.21 × 103 Ω 1.16 × 103 Ω 1.84 × 103 Ω
θT -90.0° -87.03° -72.91° -34.33° +30.75° +56.99°
│VC│ = f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
│VC│ 120 V 120.62 V 136.94 V 192.4 V 133.45 V 63.29 V
c. f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
188
I 0.0 mA 6.21 mA 35.29 mA 99.17 mA 103.45 mA 65.22 mA
CHAPTER 15
33.
=Rfif=
a.
XC =
= 1.54 kHz
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 2pfC 2p (100 Hz)(0.47 mF)
ZT @ XC f = 10 kHz: XC =
1 1 = = 33.86 Ω 2pfC 2p (10 kHz)(0.47 mF)
ZT @ R d.
-
e.
f = 40 kHz: XC =
1 1 = = 8.47 kΩ 2pfC 2p (40 kHz)(0.47 mF)
q = -tan-1 34.
= -2.2°
a.
b.
c.
35.
(I):
(a)
θdiv. = 0.8 div., θT = 4 div. θ=
¥ 360° = 72°
u 1 leads u 2 by 72° (b)
CHAPTER 15
u1: peak-to-peak = (5 div.)(0.5 V/div.) = 2.5 V
189
æ 2.5 V ö V1(rms) = 0.7071 ç ÷ = 0.88 V è 2 ø u2: peak-to-peak = (2.4 div.)(0.5 V/div.) = 1.2 V æ1.2 V ö V2(rms) = 0.7071 ç ÷ = 0.42 V è 2 ø (c)
T = (4 div.)(0.2 ms/div.) = 0.8 ms f=
(II):
(a)
=
= 1.25 kHz (both)
θdiv. = 2.2 div., θT = 6 div. θ=
¥ 360° = 132°
u 1 leads u 2 by 132°
190
(b)
u1: peak-to-peak = (2.8 div.)(2 V/div.) = 5.6 V æ 5.6 V ö V1(rms) = 0.7071 ç ÷ = 1.98 V è 2 ø u2: peak-to-peak = (4 div.)(2 V/div.) = 8 V æ8 V ö V2(rms) = 0.7071 ç ÷ = 2.83 V è 2 ø
(c)
T = (6 div.)(10 µs/div.) = 60 µs 1 f= = = 16.67 kHz 60 ms
CHAPTER 15
Chapter 16 1.
a.
b.
12 kΩÐ90° (2 kΩÐ0°)(6 kΩÐ90°) = 2 kΩ + j6 kΩ 6.325Ð71.57° = 1.897 kΩ – 18.43° = 1.799 kΩ + j0.599 kΩ
ZT =
72 kΩ∠ − 90° (12 kΩ∠ − 90°)(6 kΩ∠0°) = = ZT ′ = 5.37 kΩ∠ − 26.57° n 6 kΩ − j12 kΩ 13.42∠ − 63.43°
ZT ′ X L (5.37 kΩ∠ − 26.57°)(20 kΩ∠90°) 107.4∠63.43° = = = 5.885 kΩ∠ − 11.32° ZT + X L 5.37 kΩ∠ − 26.57° + 20 kΩ∠90° 18.25∠74.75°
= ZT
Rectangular form ZT = 5.77 kΩ – j1.155kΩ c.
XL1 = ωL = 2πfL1 = 2π(10 kHz)(40 mH) = 2.513 kΩ XC =
1 1 1 = = = 2.653 kΩ wC 2p fC 2p (10 kHz)(6 nF)
XL2 = ωL = 2πfL2 = 2π(10 kHz)(20 mH) = 1.257 kΩ
(1.257 kΩ∠90°)(2.653 kΩ∠ − 90°) 3.335 kΩ∠0° ZT ′ = = = 2.389 kΩ∠90° j1.257 kΩ − j 2.653 kΩ 1.396∠ − 90° (2.513 kΩ∠90°)(2.389 kΩ∠90°) 6.004 kΩ∠180° ZT = = 1.225 kΩ∠90° = j1.225 kΩ = 4.902∠90° j 2.513 kΩ + j 2.389 kΩ 2.
a.
b. c.
1 1 1 1 1 1 1 = + + = + + ZT X L XC R 8 Ω Ð 90° 4 Ω Ð -90° 12 Ω Ð 0° = 0.125 – -90° + 0.25 – 90° + 0.083 – 0° = -j0.125 + j0.25 + 0.083 = 0.083 + j0.125 = 150.05 ¥ 10-3 – 56.416° 1 ZT = = 6.66 Ω – -56.42° = 3.68 Ω - j5.55 Ω -3 150.05´10 Ð 56.416° ZT = -j4 kΩ = 4 kΩ – -90° 1 ZT
=
1 R1
+
1 XL
+
1 XC
1 R2
=
1 1.2 kΩ Ð 0°
I. II.
-6
+
1 3.2 kΩ Ð 90°
+
1 4.6 kΩ Ð -90° -6
+
1 3.6 kΩ Ð 0°
= 833.33 ¥ 10 – 0° + 312.5 ¥ 10 – -90° + 217.39 ¥ 10 – 90° + 277.78 ¥ 10-6 – 0° = 1.111 ¥ 10-3 - j312.5 ¥ 10-6 + j217.39 ¥ 10-6 = 1.111 ¥ 10-3 - j95.11 ¥ 10-6 = 1.115 ¥ 10-3 – -4.89° 1 ZT = = 896.86 Ω – 4.89° = 893.6 Ω + j76.45 Ω -3 1.115´10 Ð -4.89° 3.
-6
+
1 1 = 0.1136 s = 0.1136 S – 0° = R 8.8 Ω 1 1 = 3.333 mS – -90° = −j3.333 mS = −j3.333´10-3 S = YL = j300 Ω 300 ΩÐ−90°
YR =
CHAPTER 16
191
III.
4.
a.
b.
YC =
1 1 = 0.333 mS – 90° = j0.333´10-3 S = 3 kΩ Ð-90° j3 kΩ
(15 Ω∠0°)(60 Ω∠90°) 900∠90° = = 14.55∠14.04° 15 Ω + j 60 Ω 61.85∠75.96° 1 1 YT= = 0.069∠ − 14.04°= 0.067S − j0.017S= G − jB = ZT 14.55 Ω∠14.04° ZT =
22 Ω || 2.2 Ω= 2 Ω (2 Ω∠0°)(8 Ω∠90°) = ZT = 1.94 Ω∠ − 14.03° 2 Ω − j8 Ω 1 1 Y = 0.515 Ω∠14.03° = = T ZT 1.94 Ω∠ − 14.03° YT = G + jB 0.49S + j0.12 =
c.
1 1 1 YT = + + 4 kΩ∠0° 6 kΩ∠90° 9 kΩ∠ − 90° = 0.25 ×10−3 S∠0° + 0.167 × 10−3 S∠ − 90° + 0.111×10−3 S 3 = 0.25 × 10−3 S − j0.056 ×10−= S 0.256 mS∠ − 12.63° YT= G − jB
= ZT 3.91 kΩ∠ − 12.63° d.
G
Y 5.
a.
-jB
ZT = 4 Ω + j8 Ω = 8.94 Ω – 63.43° YT = 0.112 S – −63.43° YT = 50.09 mS −j100.17 mS = G − jBL
b.
ZT = 33 Ω + 20 Ω − j60 Ω = 53 Ω − j60 Ω = 80.06 Ω – 48.54° YT = 12.49 mS – 48.54° = 8.27 mS + j9.36 mS = G + jBC
c.
ZT = 400 Ω − j500 Ω − j600 Ω = 400 Ω − j100 Ω = 412.31 Ω – −14.03° YT = 2.43 mS – 14.03° = 2.36 mS + j0.59 mS = G − jBC
192
CHAPTER 16
6.
I.
a.
1 kΩ = 500 Ω 2 2 kΩ Ð -90° X C1 || X C 2 = = 1 kΩ – -90° 2 1 1 + YT = 500 Ω Ð 0° 1 kΩ Ð -90° = 2 ¥ 10-3 S + 1 ¥ 10-3 S – 90° = 2 mS + j1 ms
R1 || R2 =
b. 500 Ω
II.
a.
1 kΩ
1 1 1 + + 10 kΩ Ð 0° 4 kΩ Ð 90° 8 kΩ Ð 90° = 100 ¥ 10-6 + 250 ¥ 10-6 – -90° + 125 ¥ 10-6 – -90° = 100 ¥ 10-6 + 375 ¥ 10-6 – -90° = 100 µS - j375 µs
YT =
b. 10 kΩ
7.
a.
2.67 kΩ
XL = 2π fL = 2π(2 kHz)(470 mH) = 5.91 kΩ ZT = 4.7 kΩ + j5.91 kΩ = 7.55 kΩ – 51.51°
CHAPTER 16
193
1 1 = = 132.45 µS – -51.51° ZT 7.55 kΩ Ð 51.51° = 82.43 µS - j103.67 µS = G - jBL
YT =
b.
1 1 fiR= = 12.13 kΩ R 82.43 mS 1 1 BL = 103.67 µS = fi XL = = 9.65 kΩ XL 103.67 mS G = 82.43 mS =
12.13 kΩ
c.
R
XL = 9.65 kΩ
R = 12.13 kΩ XL = 2π fL = 9.65 kΩ, L =
d. 8.
YT=
9.65 kΩ = 767.92 mH 2p (2 kHz)
Both have resistance and inductive components.
I 6 × 10−3 ∠103 t + 10° + 90° 6 × 10−3 ∠70° = = 24 V 24 ∠103 t − 60° + 90°
= GL + jBL 0.25×10−3 S∠70° = 0.086 mS + j0.235 mS Y= T G= = L 0.086 mS X= L
1 1 = R= = 11.627 kΩ L RL 0.086 mS
1 1 = = 4.255 kΩ BL 0.235 mS
X L =ω L =103 L =4.255 × 10−3 Ω; ω =10000 rad/s L=
9.
4.255 ×10−3 103
1 1 + = 0.1 S - j0.05 S = 111.8 mS –-26.57° 10 Ω Ð 0° 20 Ω Ð 90°
a.
YT =
b.
-
c.
E = Is/YT = 2 A –0°/111.8 mS –-26.57°= 17.89 V –26.57° IR =
194
H = 4.255 H
= 17.89 V –26.57°/10 Ω –0° = 1.79 A –26.57°
CHAPTER 16
IL =
10.
11.
= 17.89 V –26.57°/20 Ω –90° = 0.89 A –-63.43°
f.
P = I2R = (1.79 A)2 10 Ω = 32.04 W
g.
Fp =
h.
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
a.
XC =
b.
YT =
c.
E=
= 18.02 V – -6.1°
IR =
= 1.80 mA – -6.1°
IC =
= 0.883 mA – 83.90°
= 0.894 lagging
1 1 = = 20.4 kΩ 2p fC 2p (60 Hz)(0.13 mF) 1 1 + = 0.1 mS –0° + 0.049 mS – -90° 10 kΩ Ð 0° 20.4 kΩ Ð -90° = 0.111 mS – 26.10°
d.
-
e.
Is = IR + IC 2 mA – 20° = 1.80 mA – -6.1° + 0.883 mA – 83.90° = (1.79 mA - j0.191 mA) + (0.094 mA + j0.878 mA) = 1.88 mA + j0.687 mA 2 mA – 20° @ 2 mA – 20.07°
f.
P = I2R = (1.80 mA)2 10 kΩ = 32.4 mW
g.
Fp =
h.
ω = 2pf = 377 rad/s is = 2.83 ¥ 10- 3 sin(ωt + 20°) iR = 2.55 ¥ 10- 3 sin(ωt - 6.57°) iC = 1.25 ¥ 10- 3 sin(ωt + 83.44°) e = 25.48 sin(ωt - 6.57°)
a.
R1 || R2 = 220 Ω || 120 Ω = 77.65 Ω
CHAPTER 16
= 0.9 leading
195
XC__XL ȍ–- __ȍ– Ð Ð Ð = = ȍ– - j Ð - - j + j ZT R__R __ X&__X/ ȍ–__ȍ–
E
Y T
F
-
G
I s
Ð ´ Ð 65.21 Ω – 32.91° = j Ð
= 15.34 mS – -32.91° ZT Ð
E 9 Ð = 184.02 mA – -32.91° ZT Ð E 9 Ð = 200 mA – 90° XC Ð -
H
I C
I
H VLQwt 16.97 sin wt is ¥- VLQ wt- 260.02 ¥ 10- 3 sin (wt - 32.91°)
J
Fp FRVq7 FRV 0.84 lagging
D
+ + Ð Ð Ð - 6–°6–- °6– ° 6-j6 0.89 S –-19.81° ZT 1.12 Ω – 19.81° Y T
E
-
F
X C
fi
XL ȦLfiL
G
5.31 mH
2.40 V –79.81°
E I R I L
196
531 µF
2.00 A –79.81° 1.20 A –-10.19°
CHAPTER 16
= 0.48 A –169.81°
IC =
13.
f.
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° 3 = 2.13 A –60.01°
g.
P = I2R = (2.00 A)2 1.2 Ω = 4.8 W
h.
Fp =
i.
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°)
a.
XL = wL = (1000 rad/s)(3.9 H) = 3.9 kΩ, 1 1 = XC = = 8.33 kΩ wC (1000 rad/s)(0.12 mF) 1 1 1 + + YT = 3 kΩ Ð 0° 3.9 kΩ Ð 90° 8.33 kΩ Ð -90° = 0.333 mS –0° + 0.256 mS –-90° + 0.120 mS –90° = 0.333 mS - j0.136 mS = 0.36 mS –-22.22°
d.
E = I/YT = 3.54 mA –-20°/0.36 mS –-22.22° = 9.83 V –2.22° IR = IL = IC =
14.
= 0.941 lagging
= 9.83 V–2.22°/3 kΩ –0° = 3.28 mA –2.22° = 9.83 V–2.22°/3.9 kΩ –90° = 2.52 mA –-87.78° = 9.83 V–2.22°/8.33 kΩ –-90° = 1.18 mA –92.22°
g.
P = I2R = (3.28 mA)23 kΩ = 32.28 mW
h.
Fp = G/YT = 0.333 mS/0.36 mS = 0.925 leading
i.
e = 13.9 sin(1000t + 2.22°) iR @ 4.64 ¥ 10-3 sin(1000t + 2.22°) iL @ 3.56 ¥ 10-3 sin(1000t - 87.78°) iC = 1.67 ¥ 10-3 sin(1000t + 92.22°)
a.
I= I1 + I 2 , (22 Ω∠0°)(60 Ω∠90°) = 22 Ω + j 60 Ω (60 Ω∠90°)(30 A∠40°) I1 = = 20.66 Ω∠20° (22 Ω∠0°)(30 A∠40°) I2 = = 20.66 Ω∠20°
ZT =
CHAPTER 16
1320∠90° = 22.66 Ω∠20° 63.906∠70° 1800∠130° = 87.12 A∠110° 20.66∠20° 31.95 A∠20° 197
b.
′ 12 Ω − j 6 Ω ZT= = 13.42 Ω∠ − 26.57° (4 Ω∠90°)(13.42 Ω∠ − 26.57°) 53.68∠63.43° = = 3.141 Ω∠83.99° 16 Ω − j 6 Ω 17.09∠ − 20.56° (13.42 Ω∠ − 26.57°)(8 A∠45°) I1 = 34.19 A∠ − 65.56° = 3.141 Ω∠83.99° (4 Ω∠0°)(8 A∠45°) I2 = = 10.19 A∠ − 38.99° 3.141 Ω∠83.99° ZT =
c. (20 Ω∠90°)(30 Ω∠90°) 600∠180° = ZT ′ = j 20 Ω + j 40 Ω − j10 Ω 50∠90°
′ 12 Ω∠90°; Z= ZT= j 40 − j10 = j 30 3 kΩ + j12 Ω; and Z= T 2 (30 Ω∠90°)(5 A∠0°) (30 Ω∠90°)(5 A∠0°) = 3 A∠0° = j 20 Ω + j 40 Ω − j10 Ω 50 Ω∠90° (20 Ω∠90°)(5 A∠0°) (20 Ω∠90°)(5 A∠0°) I2 = = = 2 A∠0° j 50 Ω 50 Ω∠90°
I1=
15.
a.
–-90° + tan-1XC/R
ZT =
θT = -90° + tan-1XC/R
│ZT│ =
f 0 Hz 1 kHz 2 kHz 3 kHz 4 kHz 5 kHz 10 kHz 20 kHz
b.
198
│VC│ =
│ZT│ 40.0 35.74 28.22 22.11 17.82 14.79 7.81 3.959
θT 0.0° -26.67° -45.14° -56.44° -63.55° -68.30° -78.75° -89.86°
= I[ZT(f)]
CHAPTER 16
f 0 kHz 1 kHz 2 kHz 3 kHz 4 kHz 5 kHz 10 kHz 20 kHz c.
│IR│ = f 0 kHz 1 kHz 2 kHz 3 kHz 4 kHz 5 kHz 10 kHz 20 kHz
16.
a.
│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
│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 –90° - tan-1XL/R
ZT =
θT = 90° - tan-1XL/R
│ZT│ =
f
θT
0 Hz 1 kHz 5 kHz 7 kHz 10 kHz
b.
0.0 k 1.22 k 3.91 k 4.35 k 4.65 k
90.0° 75.86° 38.53° 29.6° 21.69°
│IL│ = f
CHAPTER 16
199
• 31.75 mA 6.37 mA 4.55 mA 3.18 mA
0 Hz 1 kHz 5 kHz 7 kHz 10 kHz
c.
17.
IR =
= 8 mA (constant)
–90° - tan-1XC/R
YT =
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
18.
YT =
(use data of Prob. 36),
f 0 Hz 1 kHz 5 kHz 7 kHz 10 kHz
200
θT 0.0° 26.67° 45.14° 56.44° 63.55° 68.30° 78.75° 89.86°
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 16
19.
a.
YT = G –0° + BL –-90° + BC –90° –tan-1
= f 0 Hz
│YT│ XL fi 0 , ZT = 0 , YT = • 1.86 mS 1.02 mS 1.00 mS 1.02 mS 1.04 mS
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 = f 0 kHz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
c.
│θT│ -90.0° -57.51° -12.63° +1.66° +9.98° +16.54°
ZT 0.0 537.63 980.39 1k 980.39 961.54
θT 90.0° 57.52° 12.63° -1.66° -9.98° -16.54°
VC(f) = I[ZT(f)] f 0 kHz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
CHAPTER 16
│VC│ 0.0 V 5.38 V 9.80 V 10 V 9.80 V 9.62 V
201
d.
IL = f 0 kHz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz
20.
a.
Rp =
Rs2 + X s2 (20 Ω)2 + (70 Ω)2 = = 265 Ω (R) Rs 20 Ω 5300 Ω = 75.71 Ω (C) 70 Ω
Xp =
b.
(2 kΩ)2 + (14 kΩ − 8 kΩ)2
Rp =
a.
= 20 kΩ (R)
(2 kΩ)2 + (6 kΩ)2 = 6.67 kΩ (L) 6 kΩ
Xp =
21.
IL 10.0 mA 8.56 mA 3.12 mA 1.59 mA 1.04 mA 0.765 mA
Rs =
= 7.02 kΩ
Xs =
= 2.88 kΩ
ZT = 7.02 kΩ - j2.88 kΩ b.
Rs =
= 17.48 Ω
Xs =
= 29.72 Ω
ZT = 17.48 Ω + j29.72 Ω 22.
a.
CT = 2 µF XC =
= 79.62 Ω
XL = ωL = 2π(103 Hz)(10 mH) = 62.8 Ω YT =
1 1 1 + + 220 Ω Ð 0° 79.62 Ω Ð -90° 62.8 Ω Ð 90°
= 4.55 mS –0° + 12.56 mS –90° + 15.92 mS –-90°
202
CHAPTER 16
= 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° = 176.68 V –36.44°/220 –0° = 0.803 A –36.44°
IR =
= 176.68 V –36.44°/62.80 –90° = 2.813 A –-53.56°
IL = b.
Fp = G/YT = 4.55 mS/5.66 mS = 0.804 lagging
c.
P = I2R = (0.803 A)2 220 Ω = 141.86 W
f.
Is = IR + 2IC + IL and IC = = = IC = -0.657 + j0.893 = 1.11 A –126.43° = 176.7 Ω –36.44°
ZT =
g.
= 142.15 Ω + j104.96 Ω = R + jXL 23.
P = VI cos θ = 8000 W = (120 V)(80 A)cos θ cos θ = 0.833 (lagging) θ = 33.56° 40 A –-33.56° 120 V –0° = 0.333 S –-33.56° = 0.277 S - j0.184 S
YT =
YT = 0.277 S - j0.184 S = GT - jBL GT = 0.277 S =
= 0.05 S +
and R¢ = =
XL
0.277 S
XL=
= 4.405 Ω
= 0.184 S = 5.435 Ω
4.405Ω 20Ω
CHAPTER 16
𝑅𝑅 ′
𝑋𝑋𝐿𝐿
5.435Ω
203
Chapter 17 1.
2.
(16 Ω∠0°)(8 Ω∠ − 90°) 16 Ω − j8 Ω = 6 Ω∠90° + 7.155 Ω∠ − 63.43°= 3.225∠ − 7.13°
a.
ZT= X L + X C || R= 6 Ω∠90° +
b.
IS =
c.
I1 =I S =6.20 A∠7.13° (from Fig. 17.38)
d.
I2 =
e.
V= X L ⋅ I= (6 Ω∠90°)(6.20 A∠7.13°) = 37.2 V∠97.13° L S
a.
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 = =
3.
jX C I S (8 Ω∠ − 90°)(6.20 A∠7.13°) 49.6∠82.87° = = = 2.77 A∠109.44° R − jX C 16 Ω − j8 Ω 17.89∠ − 26.57°
= 4.07 A –-48.57°
Z R2 I S Z R 2 + ZC
=
(2 Ω Ð 0° )(4.07 A Ð -48.57° ) 2 Ω - j8 Ω = 0.987 A –27.39°
Z LE (6 Ω Ð 90° )(30 V Ð 0° ) 180 V Ð 90° = = 7.38 Ω Ð 48.57° 7.38 Ω Ð 48.57° ZT = 24.39 V –41.43°
d.
VL =
a.
ZT ( jX L ) || (R2 − jX= = C ) ( j12 Ω) || (8 Ω − j12 Ω) =
204
E (20 V∠0°) = = 6.20 A∠7.13° ZT 3.225∠ − 7.13°
b.
I= S
c.
I2 =
(12 Ω∠90°)(14.42 ∠ − 56.3°) = 2.163 Ω∠33.7° j12 Ω + 8 Ω − j12 Ω
(50 V∠0°) E = = 23.12 A∠ − 33.7° ZT 2.163 Ω∠33.7°
(12 Ω∠90°)I S (12 Ω∠90°)(23.12 A∠ − 33°) = = 128.27 A∠22.6° 2.163 Ω∠33.7° 2.163 Ω∠33.7 °
CHAPTER 17
4.
d.
VC= I 2 ⋅ X C= (128.27 A∠22.6°)(12 Ω∠ − 90°)= 1.539 KV∠ − 67.4°
e.
P = EI cos θ = (50 V)(23.12 A)cos(33.7°) = 961.74 W
a.
ZT = R1 + Z 2 + Z3 , where R1 = 5 KΩ (4 KΩ∠ − 90°)(6 KΩ ∠90°) = (12 KΩ∠ − 90°) − j 4 KΩ + j 6 KΩ Z = (12 KΩ ||12 KΩ)||(8 KΩ ∠90°) = (6 KΩ∠0°) || (8 KΩ ∠90°) 3 R2 || R3 || X = L2
Z 2 X C || = X L1 =
=
(48 ∠90°) = (4.8 KΩ ∠36.87°) = 3.8 KΩ + j2.88 KΩ 6 + j8
Z= 5 KΩ − j12 KΩ + 3.8 KΩ + j 2.88 KΩ T = 8.8 KΩ − j9.12 = KΩ 12.67 KΩ∠ − 46° b.
V2 = I R2 ⋅ R2 = I R3 ⋅ R3 = I L ⋅ X L2 , IS =
(240 V∠60°) E = = 18.94 mA∠106°) ZT 12.67 KΩ ∠ − 46°
= IL
(6 KΩ∠0°)I S (6 KΩ∠0°)18.94 ×10−3 A∠106° = 23.68 mA∠69.13° = Z3 4.8 KΩ ∠36.87°
= V2 (23.68 mA∠69.13°)(8 KΩ∠90°) = 189.44∠159.13°
5.
c.
Fp = power factor = cos 46° = 0.695 (lagging)
a.
400 Ω –-90° || 400 Ω –-90° =
= 200 Ω –-90°
Z¢ = 100 Ω - j200 Ω = 223.61 Ω –-63.43° = -j200 Ω + j600 Ω = +j400 Ω = 400 Ω –90° (223.61 Ω Ð -63.43° )(400 Ω Ð 90° ) 89, 444.00 Ω Ð 26.57° = ZT = Z¢ || = (100 Ω - j200 Ω) + j400 Ω 223.61Ð 63.43° = 400 Ω –-36.86° I=
= 0.25 A –36.86°
(200 Ω Ð -90° )(100 Ω Ð 0° ) 20, 000 V Ð -90° = = 89.44 V –-26.57° 100 Ω - j200 Ω 223.61Ð -63.43°
b.
VC =
c.
P = EI cos θ = (100 V)(0.25 A) cos 36.86° = (25)(0.8) = 20 W
CHAPTER 17
205
6.
Z2 I (9 Ω∠ − 90°)(2 A∠30°) 18∠ − 60° = = = 2.55 A∠ − 30° Z1 + Z2 5 Ω + j 4 Ω − j9 Ω 7.071∠ − 45°
I1 a. =
b.
Z2 I 6.40∠39° = (2 A∠30 = °) 1.81 A∠114° Z1 + Z 2 7.071∠ − 45°
= I2
VC= I 2 ⋅ X C= (1.81 A∠114°)(16∠ − 90°)= 28.96 V∠24° c.
Vab = I1 ⋅ X L1 − I 2 ⋅ X L2 = (2.55 A∠ −15°)(4 Ω∠90°) − (1.81 A∠114°)(7 Ω∠90°) = 10.2 V∠75° −12.67 V∠204° = 2.64 V + j 9.85 +11.57 + j 5.15 = 14.21 + j14 = 19.95 V∠44.57°
7.
Z1 = 10 Ω –0° Z2 = 80 Ω –90° || 20 Ω –0° 1600 Ω Ð90° 1600 Ω Ð 90° = = 20 + j80 82.462 Ð75.964° = 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° = 29.206 Ω –9.273° || 6 Ω –-90° = =
= 1.42 A –18.26°
I1 =
b.
= 28.103 Ω –-18.259°
V1 = = 26.57 V –4.76°
c.
206
P = EI cos θ = (40 V)(1.423 A)cos 18.259° = 54.07 W
CHAPTER 17
8.
!! R2 = 1 kΩ a. R2
+ VL2 XL2
Is
+ E −
−
X L1 = 2π fL1 = 2π(5 kHz)(68 mH) = 2.14 kΩ
R1
X L2 = 2π fL2 = 2π(5 kHz)(100 mH) = 3.14 kΩ
XL1 + VC −
XC
XC =
1 1 = = 10.61 kΩ 2p fC 2p (5 kHz)(3 nF)
ZT = R2 + X L2 || (R1 + X L1 + XC)
kΩ + j2.14 kΩ - j10.61 kΩ ) = 1 kΩ + (3.14 kΩ –90°) || (6.8 !#####"#####$ 6.8 kΩ - j8.47 kΩ !###"### $ 10.86 kΩ Ð -51.24° = 1 kΩ + (3.14 kΩ –90°) || (10.86 kΩ –-51.24°) (3.14 kΩ Ð 90° )(10.86 kΩ Ð -51.24° ) = 1 kΩ + + j3.14 kΩ + 6.8 kΩ - j8.47 kΩ 34.10 kΩ Ð 38.76° 34.10 kΩ Ð 38.76° = 1 kΩ + = 1 kΩ + 6.8 - j5.33 8.64 Ð -38° = 1 kΩ + 3.95 kΩ –76.76° = 1 kΩ + 0.904 kΩ + j3.85 kΩ = 1.904 kΩ + j3.85 kΩ ZT = 4.3 kΩ –63.69° E 10 V Ð 0° = Is = = 2.33 mA –-63.69° ZT 4.3 kΩ Ð 63.69° b.
IC =
X L2 I s X L2 + (R 1 + X L1 + X C )
=
(3.14 kΩ Ð 90° )(2.33 mA Ð -63.69° ) + j3.14 kΩ + 6.8 kΩ - j8.47 kΩ
7.32 mA Ð26.3° 7.32 mA Ð 26.3° = = 0.847 mA –64.4° 6.8 - j5.33 8.64 Ð -38.1° VC = ICXC = (0.847 mA –64.4°)(10.61 kΩ –-90°) = 8.99 V –-25.60° =
c.
9.
VL2 = E - IsR2 = 10 V –0° - (2.33 mA –-63.69°)(1 kΩ –0°) = 10 V - 2.33 V –-63.69° = 10 V - (1.03 V - j2.09 V) = 8.97 V + j2.09 V = 9.21 V –13.12°
a. R1
+
+
VL L
VS
−
−
R2
+ C
VC
−
XL = 2π fL = 2π(10 kHz)(47 mH) = 2.95 kΩ 1 1 = XC = 2p fC 2p (10 kHz)(0.01 mF) = 1.59 kΩ
CHAPTER 17
207
1 1 1 1 = + + ZT R1 X L R 2 + XC 1 1 1 1 = + + ZT 4.7 kΩ 2.95 kΩ Ð 90° 1.8 kΩ - j1.59 kΩ 1 2.40 kΩ Ð -41.46° = 212.77 µS - j338.98 µS + 312.26 µS + j275.88 µS = 525.03 µS - j63.1 µS = 528.81 µS –-6.85° 1 = 1.89 kΩ –6.85° and ZT = 528.81 mS Ð -6.85° Vs = (I)(ZT) = (8 mA –0°)(1.89 kΩ –6.85°) = 15.12 V –6.85° = 212.77 µS –0° + 338.98 µS –-90° +
10.
X C Vs (1.59 kΩ Ð -90° )(15.12 V Ð 6.85° ) = 1.8 kΩ - j1.59 kΩ R + XC 24 V Ð -83.15° = 10 V –-41.69° = 2.40 Ð -41.46°
b.
VC =
c.
VL = Vs = 15.12 V –6.85°)
a.
Z1= 5 Ω + j1 Ω= 5.099 Ω∠13.309° Z 2= 4 Ω Z3= (6 Ω + j15 Ω − j 7 Ω= ) (6 Ω + j8 Ω= ) 17.889 Ω∠26.565° YT =
1 1 1 1 1 1 + + = + + Z1 Z 2 Z3 5.099 Ω∠13.309° 4 Ω∠0° 17.889 Ω∠26.565°
YT 0.196 S ∠ − 13.309° + 0.25 S ∠0° + 0.056 S ∠ − 26.565° = YT (0.191 S − j 0.045 S + (0.25 S ) + (0.05 S − j 0.025 S ) = S 0.496 S ∠ − 8.11° = 0.491 S − j 0.07 = 1 1 Z= = = 2.02 Ω∠8.11° T YT 0.496 S ∠ − 8.11° b.
c.
I= 1
120 V∠0° E = = 23.53 A∠ −13.309° Z1 5.099 Ω∠13.309°
I 2=
E 120 V∠0° = 30 A∠0° = 4 Ω∠ ° Z2
I= 3
120 V∠0° E = = 6.70 A∠ − 26.57° Z3 17.889 Ω∠26.565°
I= S
E 120 V∠0° = = 59.406 A∠ − 8.11° ZT 2.02 Ω∠8.11°
I S= I1 + I 2 + I 3= 23.53 A∠ − 13.309° + 30 A∠0° + 6.70 A∠ − 26.57°
59.4 A∠= − 8.11° (22.89 A − j 5.417 A) + (30 A) + (6 A − j 3 A) = 58.989 A − j8.417 A = 58.9 A − j8.4 A = 59.4 A∠ − 8.11° (checks) 208
CHAPTER 17
d. 11.
F= P
G 0.491 S = = 0.989 (lagging) YT 0.496 S
a.
X= L1 2π (103 )(0.4= H) 2513.27 Ω L1 ω= X= L2 2π (103 )(0.8H) = 5026.55 Ω L2 ω= X= C
1 1 = = 159.15 Ω 3 ωC 2π (10 )(1 µ F )
ZT = R∠0° + X L1 ∠90° + ( X C ∠ − 90°) || ( X L2 ∠90°) = 400 Ω + j 2513.27 Ω − j164.35= Ω 400 Ω + j 2348.9 Ω = 2.383 kΩ∠80.34°
IS b.=
50 V∠0° = 20.98 mA∠ − 80.34° 2.383 kΩ∠80.34°
c.= I1
X L2 I S (5.026∠90°)(20.98 mA∠ − 80.34°) = = 21.67 mA∠ − 80.34° X L2 + X C 5.026 kΩ − j0.159 kΩ
= I2
XC IS (0.159 kΩ∠ − 90°)(20.98 mA∠ − 80.34°) = X L2 + X C j 5.026 kΩ − j0.159 kΩ = 0.69 mA∠ − 260.34 = ° 0.69 mA∠99.66°
V1 I = (0.69 mA∠99.66°)(5.026 kΩ∠90°) = 3.47 V∠189.66° d. = 2 X L2 Vab = E − I S R1 = 50∠0° − (20.98 mA∠ − 80.34°)(400 Ω∠0°) = 49.07 V∠8.04°
12.
e.
= P I S2= R (20.98 mA) 2 (400 Ω= ) 0.176 W
f.
F = P
a.
ZT= 2 kΩ + Z= T = = =
b. c.
400 Ω R = = 0.168 (lagging) ZT 2383 Ω (1.2 kΩ∠0°)(1.8 kΩ∠ − 90°) 3.6 kΩ∠90° + 1.2 kΩ − j1.8 kΩ 2 2.16 kΩ∠ − 90° 2 kΩ + + 1.8 kΩ∠90° 2.16∠ − 56.31° 2 kΩ +1 kΩ∠ − 33.69° +1.8 kΩ∠90° 2 kΩ + 832.05 Ω − j 554.70 Ω + j1.8 kΩ 2.83 kΩ + j1.25= kΩ 3.09 kΩ∠23.83°
V= = 60 V∠0° 1 IR= 1 (30 mA∠0°)(2 kΩ∠0°)
= I1
(1.2 kΩ∠0°)(30 mA∠0°) 36 A∠0° = = 16.67 mA∠56.31° 1.2 kΩ − j1.8 kΩ 2.16 × 103 ∠ − 56.31°
(
)
d.
3.6 kΩ V2= I X L1 || X L2 = (30 mA∠0°) ∠90° = 54 V∠90° 2
e.
VS =IZT =(30 mA∠0°)(3.09 kΩ∠23.83°) =92.70 V∠23.83°
CHAPTER 17
209
13.
R3 + R4 = 2.7 kΩ + 4.3 kΩ = 7 kΩ R¢ = 3 kΩ || 7 kΩ = 2.1 kΩ Z¢ = 2.1 kΩ - j10 Ω
(40 kΩ Ð 0° )(20 mA Ð 0° ) 40 kΩ + 2.1 kΩ - j10 Ω = 19 mA –+0.014° as expected since R1
(CDR)
I¢ (of 10 Ω cap.) =
(3 kΩ Ð 0°)(19 mA Ð 0.014°) 57 mA Ð 0.014° = 10 3 kΩ + 7 kΩ = 5.7 mA –0.014° P = I2R = (5.7 mA)2 4.3 kΩ = 139.71 mW I4 =
(CDR)
14.
Z¢
Z=′ 15 Ω − j 20 Ω = 25 Ω∠ − 53.13° R4 ∠0° || Z=′ 20 Ω∠0° || 25 Ω∠ − 53.13= ° 12.40 Ω∠ − 23.36° Z ′′ = R3∠0° + R4 ∠0° || Z ′ = 15 Ω + 12.40 Ω∠ − 23.36° = 15 Ω + (11.38 Ω − j 4.92 Ω)= 26.38 Ω − j 4.92 Ω = 26.83 Ω∠ − 10.5° 536.6∠ −10.56° = 11.51 Ω∠ − 4.5° 46.64∠ − 6.06° ′′ 15 Ω + 11.51∠ − 4.5°= 15 Ω + (11.47 Ω − j 0.90) ZT= R1∠0° + R2 ∠0° || Z =
′′ 20 Ω∠0° || 26.83∠ − 10.56= R2 ∠0° || Z= °
= (26.47 − j0.90) = Ω 26.49 Ω∠ − 1.95° 220 V∠0° E IS = = = 8.31 Ω∠1.95° ZT 26.49 Ω∠ − 1.95° I R1 = I , I 5=
R2 ∠0° I S (20 Ω∠0°)(8.31 A∠1.95°) 166.2 A∠1.95° = 3.56 A∠8.01° = = ′′ 46.64∠ − 6.06° R2 ∠0° + Z 46.64∠ − 6.06°
= I R5
R4 ∠0° I R3 73∠8.01° (20 Ω∠0°)(3.56 A∠8.01°) = 1.81 A∠37.75° = = 40.31∠ − 29.74° 20 Ω + (15 − j 20) R4 ∠0° + Z ′
15.
!! XC = 10 kΩ
a.
+
E = 120 V ∠ 0°
10 kΩ
10 kΩ 40 kΩ
2.7 kΩ 3 kΩ R5
4.3 kΩ
−
40 kΩ + 10 kΩ
ZT′
!##"##$ (40 kΩ Ð 0° )(120 V Ð 0° )
(40 kΩ Ð 0° )(120 V Ð 0° ) = 96 V –0° 40 kΩ + 10 kΩ RTh = 10 kΩ || 40 kΩ = 8 kΩ ZT¢ = 3 kΩ || (2.7 kΩ + 4.3 kΩ) = 3 kΩ || 7 kΩ = 2.1 kΩ
ETh =
210
CHAPTER 17
ZTh
+
8 kΩ
10 kΩ ZT′
ETh = 96 V ∠ 0°
2.1 kΩ
−
ETh 96 V Ð 0° 96 V Ð 0° = = ZT 8 kΩ + 2.1 kΩ - j10 kΩ 10.1 kΩ - j10 kΩ 96 V Ð 0° = 6.76 mA –44.72° = 14.21 kΩ Ð -44.72° (3 kΩ Ð 0° )(IC ) (3 kΩ Ð 0° )(6.76 mA Ð 44.72° ) = I R5 = 10 kΩ Ð 0° 3 kΩ + 2.7 kΩ + 4.3 kΩ 20.28 mA Ð 44.72° = 10 = 2.03 mA –44.72° PR 5 = I2R = (2.03 mA)24.3 kΩ = 17.72 mW
IC = Is =
16.
2.21 kΩ
a.
ZT
I
2.21 kΩ
2.2 kΩ
ZT′′
2.21 kΩ
2.2 kΩ
XC = 2.2 kΩ
1 1 = 2p fC 2p (40 kHz)(1.8 nF)
= 2.21 kΩ
ZT′
ZT¢ = (2.2 kΩ –0°) || (2.2 kΩ - j2.21 kΩ) !###"###$ 3.12 kΩ Ð -45.13° 6.86 kΩ Ð -45.13° 6.86 kΩ Ð -45.13° = = 2.2 + 2.2 - j2.21 4.4 - j2.21 6.86 kΩ Ð -45.13° = 1.36 kΩ –-18.46° = 4.92 Ð -26.67° ZT¢¢ = 2.2 kΩ –0° || (2.21 kΩ –-90° + 1.36 kΩ –-18.46°) = 2.2 kΩ –0° || (-j2.21 kΩ + 1.29 kΩ - j0.43 kΩ) = 2.2 kΩ –0° || (1.29 kΩ - j2.64 kΩ) ! #"# $ 2.94 kΩ Ð -63.96°
(2.2 kΩ Ð 0° )(2.94 kΩ Ð -63.96° ) 2.2 kΩ + 1.29 kΩ - j2.64 kΩ 6.47 kΩ Ð -63.96° 6.47 kΩ Ð -63.96° = = 3.49 - j2.64 4.38 Ð -37.11° ZT¢¢ = 1.48 kΩ –-26.85° = 1.32 kΩ - j0.668 kΩ ZT = 2.21 kΩ –-90° + ZT¢¢ = -j2.21 kΩ + 1.32 kΩ - j0.668 kΩ = 1.32 kΩ - j2.89 kΩ = R - jXC = 3.18 kΩ –-65.45° =
CHAPTER 17
211
b.
Source conversion:
2.2 kΩ 2.21 kΩ
2.21 kΩ
+ 4 mA
8.8 V ∠ 0°
2.2 kΩ
2.2 kΩ
2.2 kΩ
−
Second source conversion:
2.21 kΩ IR3
Z′
I
2.2 kΩ
2.2 kΩ
+ VR3
−
} Z′′
ì Z¢ = 2.2 kΩ - j2.21 kΩ = 3.12 kΩ Ð -45.13 °ü ï ï í ý source conversion 8.8 V Ð 0° = 2.82 mA Ð 45.13° ï ïI = 3.12 kΩ Ð -45.13° î þ ZT¢¢ = 2.2 kΩ –0° || 3.12 kΩ –-45.13° (2.2 kΩ Ð 0° )(3.12 kΩ Ð -45.13° ) 6.86 kΩ Ð -45.13° = = 2.2 kΩ + 2.2 kΩ - j2.21 kΩ 4.4 - j2.21 6.86 kΩ Ð -45.13° = 1.39 kΩ –-18.46° = 1.32 kΩ - j0.44 kΩ = 4.92 Ð -26.67° ( Z¢¢)(I) (1.39 kΩ Ð -18.46° )(2.82 mA Ð 45.13° ) IR3 = = ¢ ¢ Z + 2.2 kΩ - j2.21 kΩ 1.32 kΩ - j0.44 kΩ + 2.2 kΩ - j2.21 kΩ 3.92 mA Ð 26.67° 3.92 mA Ð 26.67° = = 3.52 - j2.65 4.41Ð -36.97° = 0.89 mA –63.64° VR 3 = I R 3 R 3 = (0.89 mA –63.64°)(2.2 kΩ –0°) = 1.96 V –63.64° 17.
Source conversion: E = IZ = (0.5 A –0°)(2Ω –-90°) = 1 V –-90° Is
+
2Ω
E
−
I1
8Ω 2Ω
1 kΩ
ZT
1# Ω# Ð$ 0° ZT = -j2 Ω + j8 Ω + 2!Ω#Ð#-90° #"||# 2 Ω Ð -90° (2 Ω Ð -90° )(1 Ω Ð 0° ) = 1 - j2 2.24 Ð -63.44° = -j2 Ω + j8 Ω + 0.89 Ω –-26.56° = +j6 Ω + 0.796 Ω - j0.398 Ω
212
CHAPTER 17
= 0.796 Ω + j5.6 Ω = 5.66 Ω –81.91° E 1 V Ð -90° = Is = = 0.177 A –-171.91° ZT 5.66 Ω Ð 81.91°
(2 Ð -90° )(I s ) (2 Ð -90°)(0.177 A Ð -171.91°) = (2 Ð -90° ) + 1Ð 0° 1 - j2 354 mA Ð -261.91° = 2.24 Ð -63.44° = 158 mA –-198.47° V1 = I1R1 = (158 mA –-198.47°)(1 Ω –0°) = 158 mV –-198.47°
I1 =
CHAPTER 17
213
Chapter 18 1.
-
2.
Z = − j 5 Ω + 3 Ω∠0° || 5 Ω∠90° = − j 5 Ω +
15 Ω∠90° 3 Ω + j5 Ω
15 Ω∠90° = − j5 Ω + 2.57∠30.96° 5.83∠59.04° = − j 5 Ω + 2.20 Ω + j1.32= Ω 2.20 Ω − j 3.68= Ω 4.29 Ω∠ − 59.13° = − j5 Ω +
E 90 V∠30° I== = 20.98 A∠89.13° Z 4.29 Ω∠ − 59.13° 3.
4.
180 Ω∠90° = 14.06 Ω∠51.34° 10 Ω + j8 Ω = E IZ = (3 A∠120°)(14.06 Ω∠51.34°) = 42.18 V∠171.34° Z = 10 Ω∠0° || 8 Ω∠90° =
a.
I =
µV
16 V = 3.2 ×10−3 V; = Z 5 kΩ∠0° = R 5 × 103 Ω
I= b. 5.
𝜇𝜇𝜇𝜇 𝑅𝑅
=
16𝜇𝜇
5×103 Ω
= 3.2 × 10−3 𝑉𝑉 ; Z = 5kΩ ∠0°
V = (hI)(R) = (40I)(40 kΩ) =1.6×106 I; Z = 40 kΩ∠0°
Clockwise mesh currents: E - I 1Z 1 - I 1Z 2 + I 2Z 2 = 0 -I2Z2 + I1Z2 - I2Z3 - E2 = 0 ──────────────────── [Z1 + Z2]I1 - Z2I2 = E1 -Z2I1 + [Z2 + Z3]I2 = -E2 ──────────────────
=
6.
= 5.15 A –-24.5°
=
Clockwise mesh currents: E1 – I1 Z1 – I1 Z2 + I2 Z2 = 0 – I2 Z2 + I1 Z2 – I2 Z3 – E2 = 0
Where
[𝑍𝑍1 + 𝑍𝑍2 ] I1 – Z2 I2 = E1
Z= 1 R= 1 50 Ω∠0°
– Z2 I1 + [Z2 + Z3] I2 = –E2 I 50Ω= I= 1
214
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°
[ Z 2 + Z3 ]E1 − z2 E2 z1Z 2 + z1Z 3 + z2 Z3
Z= X C ∠ − 90°; = 40 Ω∠ − 90° 2 Z= X C ∠ − 90°; = 60 Ω∠ − 90°; 3 E= 1 6 V∠45°; E2= 30 V∠0° CHAPTER 18
7.
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: = 2.55 A –132.72° 8.
Clockwise mesh currents: E 1 - I 1Z 1 - I 1Z 2 + I 2Z 2 = 0 -I2Z2 + I1Z2 - I2Z3 - I2Z4 + I3Z4 = 0 -I3Z4 + I2Z4 - I3Z5 - E2 = 0 ─────────────────────────
X L1 = wL1 = (2π)(2 kHz)(110 µH) = 1.38 Ω X L 2 = wL2 = (2π)(2 kHz)(220 µH) = 2.76 Ω 1 1 X C1 = = wC1 2p (2 kHz)(39 mF) = 3.62 Ω 1 1 X C2 = = wC2 2p (2 kHz)(39 mF) = 2.04 Ω Z1 = 4 Ω + j1.38 Ω Z2 = -j3.62 Ω Z3 = j2.76 Ω Z4 = -j2.04 Ω Z5 = 8 Ω –0° E1 = 6 V –0° E2 = 120 V –120°
───────────────────────────────────
I R1 = I3 =
[Z2 Z 4 ]E1 + [Z2 - [Z1 + Z2 ][Z2 + Z3 + Z 4 ]]E2 2
2
2
[Z1 + Z2 ][Z2 + Z3 + Z 4 ][Z 4 + Z5 ] - [Z1 + Z2 ]Z 4 - [Z 4 + Z5 ]Z2 1000 V Ð -64.5° = 8.04 A –3.84° = 124.4 Ω Ð -68.34°
CHAPTER 18
215
9. 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 - I 3Z 1 = E1 + I2(Z2 + Z3) - I3Z2 = -E2 -I1Z3 - I 2Z 2 + I3(Z1 + Z2 + Z4) = 0 -I1Z1 ─────────────────────────────────── Applying determinants: I3 = = 48.33 A –-77.57° or I3 =
if one carefully examines the network!
10. Z1 = 7 Ω –0°, Z2 = 5 Ω –−90° Z3 = 4 Ω –0°, Z4 = 6 Ω –-90° Z5 = 4 Ω –0°, Z6 = 6 Ω + j8 Ω E1 = 25 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 ───────────────────────── Z2I2 - Z 4I 3 = E 1 (Z1 + Z2 + Z4) I1 Z5I3 = -E2 -Z2I1 + (Z2 + Z3 + Z5)I2 - Z5I2 + (Z4 + Z5 + Z6)I3 = 0 Z4I1 ───────────────────────────────────────── = Z4 + Z5 + Z6 and determinants: Using Z¢ = Z1 + Z2 + Z4, Z≤ = Z2 + Z3 + Z5,
=
216
5229.24∠74.86° = 1.45 A∠26.77° 3609.675∠48.09°
CHAPTER 18
11. 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 ───────────────────────── - Z 4 I2 +0 - Z 6I 4 = E 1 (Z4 + Z6 + Z7) I1 - Z 2I 3 +0 =0 -Z4I1 + (Z1 + Z2 + Z4)I2 - Z5I4 = -E2 0 - Z2 I2 + (Z2 + Z3 + Z5)I3 +0 - Z5I3 + (Z5 + Z6 + Z7)I4 = 0 -Z6I1 ──────────────────────────────────────────────────── Applying determinants: = 0.68 A –-162.9° 12.
Z1 = 6 kΩ –0° Z2 = 10 kΩ –0° Z3 = 2 kΩ + j4 kΩ = 4.472 kΩ –63.43°
Mesh equations I1(Z1 + Z2) - Z2I2 = -28 V I2(Z2 + Z3) - Z2I1 = 0 ────────────────── (Z1 + Z2)I1 - Z2I2 = -28 V -Z2I1 + (Z2 + Z3)I2 = 0 ─────────────────── I L = I2 =
=
−(10 kΩ∠0°)(28 V) 60∠0° + 26.83∠63.43° + 44.72∠63.43°
−280 V −280 V = (60 + 12 + j 23.99 + 20 + j 39.99)kΩ (92 + j 63.98)kΩ −280 V = = −2.49 mA∠34.82° 112.06 kΩ∠34.82° 13.
CHAPTER 18
Source Conversion: E = (I –0°)(R –0°) = (60 I)(40 kΩ –0°) = 2.4 ¥ 106 I –0° Z1 = Rs = Rp = 40 kΩ –0° Z2 = -j0.2 kΩ Z3 = R1 = 8 kΩ –0° Z4 = j4 kΩ; = 4 kΩ –90° 217
Mesh Analysis Equation I1(Z1 + Z2 + Z3) - Z3I2 = -E I2(Z3 + Z4) - Z3I1 = 0 ──────────────────── (Z1 + Z2 + Z3)I1 - Z3I2 = -E -Z3I1 + (Z3 + Z4)I2 = 0 ──────────────────── I L = I2 =
−(8 kΩ∠0°)24 ×106 I∠0° (40 kΩ − j 0.2 kΩ + 8 kΩ)(8 kΩ + 4 kΩ) − (8 kΩ∠0°)(8 kΩ∠0°) = 51.49I∠149.31° =
14.
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° ────────────────────── I2 =
=
= 1.67 mA –0° = I2kΩ
-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° I1 =
=
= 10 mA –0° 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°
15.
E1 - I1Z1 - Z2(I1 - I2) = 0 -Z2(I2 - I1) + E2 - I3Z3 = 0 ─────────────────── I 3 - I2 = I 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 = 2.69 mA –-174.8° 218
CHAPTER 18
Z1 = 1 kΩ –0° Z2 = 4 kΩ + j6 kΩ E = 10 V –0°
16.
-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° I3 = 0.6 V –0° (99 Ω)I2 + ──────────────────────────── Determinants: I3 = I6 kΩ = 1.38 mA –-56.31° 17.
Z1 = 4 kΩ –0° Z2 = 2 kΩ –90° Z3 = 8 kΩ –-90° I1 = 2 mA –0° I2 = 6 mA –30° Y1 = 0.25 mS –0° Y2 = 0.5 mS –−90° Y3 = 0.125 mS–90°
I= 1 I3 + I 4 1 1 V V −V 1 I1 = 1 + 1 2 =V1 + − V2 =I1 Z1 Z2 Z1 Z 2 Z2 I1 Or, V1[Y1 + Y2 ] −V2 [Y2 ] = I 4 = I5 + I 2 =
V1 − V2 V2 = + I2 Z2 Z3
1 1 1 V2 + − V1 = −I2 Z2 Z 2 Z3 −I2 Or, V2 [Y2 + Y3 ] − V1[Y2 ] =
Therefore, [Y1 + Y2 ]V1 − [Y2 ]V2 = I1 −I2 [Y2 ]V1 + [Y2 + Y3 ]V2 = = V1
[Y2 + Y3 ]I1 − Y2 I2 −[Y1 + Y2 ]I 2 + Y2 I1 2579∠ − 45.77° = = = 22.89 V∠ −10.54° Y1Y2 + Y1Y3 + Y2Y3 Y1Y2 + Y1Y3 + Y2Y3 112.67∠ − 56.31°
CHAPTER 18
219
18. Z1 = 3 Ω + j4 Ω = 5 Ω –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 0 = I1 +
V1[Y1 + Y2] - V2[Y2] = -I1 - I2 ────────────────────── I 2 + I4 = I5 or
I2 +
or and
220
V2[Y2 + Y3] - V1[Y2] = I2 ────────────────── [Y1 + Y2]V1 - Y2V2 = -I1 - I2 -Y2V1 + [Y2 + Y3]V2 = I2
CHAPTER 18
Applying determinants: V1 =
= 5.12 V –-79.36°
V2 =
= 2.71 V –39.96° Z1 = 5 Ω –0° Z4 = 2 Ω –0° E = 30 V –50° I = 4 A –90° XL = 2πfL = 2π(10 kHz)(0.1 mH) = 6.28Ω 1 1 = XC = 2p fC 2p (10 kHz)(4.7 mF) = 3.39 Ω Z2 = 6.28 Ω –90° Z3 = 3.39 Ω –-90°
19.
V1[Y1 + Y2 + Y3] - V2Y3 = E1Y1 -V1[Y3] + V2[Y3 + Y4] = +I ────────────────────── Using determinants:
üï ý after source conversion. ï þ
V1 = 17.92 V –59.25° and V2 = 13.95 V –93.64° 20. 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 fi or V1[Y1 + Y2 + Y3 + Y5] - V2[Y3 + Y5] = E1Y1 I3 + I5 = I4 + I fi or V2[Y3 + Y4 + Y5] - V1[Y3 + Y5] = -I resulting in
CHAPTER 18
221
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° 21.
Z1 = 15 Ω –0° Z2 = 10 Ω –-90° Z3 = 15 Ω –0° Z4 = 3 Ω + j4 Ω
é 1 1 1 1 ù 1 + + 220 V Ð 0° 100 V Ð 90° = 0 V2 ê ú15 Ω ë15 Ω - j10 Ω 15 Ω û 15 Ω V2 [133.34 ¥ 10-3 + j100 ¥ 10-3] = 14.67 + j6.67
[
]
[
]
= 96.30 V –-12.32° V1 = E1 = 220 V –0°, V3 = E2 = 100 V –90° 22.
E1 = 25 V –0° E2 = 75 V –20°
V 1:
=0
V 2: V 3: V 4:
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
=0 =0 =0
───────────────────────
222
CHAPTER 18
Rearranging: æ1 V1 ç è Z1 æ1 V2 ç è Z1 æ1 V3 ç è Z6 æ1 V4 ç è Z2
1 1 ö + ÷Z2 Z3 ø 1 1 ö + + ÷ Z 4 Z6 ø 1 1 ö + + ÷ Z 7 Z8 ø +
+
= = =
1 1 1 ö + + ÷ Z 4 Z 7 Z5 ø
=0
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° 23.
Y1 = = 0.25 S –0° Y2 = = 1 S –-90° Y3 = = 0.2 S –0° Y4 = = 0.25 S –90° Y5 =
V1[Y1 + Y2] - Y2V2 = I1 V2[Y2 + Y3 + Y4] - Y2V1 - Y4V3 = -I2 V3[Y4 + Y5] - Y4V2 = I2 ───────────────────────────
= 0.125 S –-90° I1 = 2 A –30° I2 = 3 A –150°
- Y2 V2 + 0 = I1 [Y1 + Y2]V1 - Y4 V3 = -I2 -Y2V1 + [Y2 + Y3 + Y4]V2 0 - Y4 V2 + [Y4 + Y5]V3 = I2 ──────────────────────────────────── V1 = = 5.74 V –122.76°
CHAPTER 18
223
V2 =
= 4.04 V –145.03° = 25.94 V –78.07°
V3 = 24.
Y1 = = 0.25 S –0° Y2 = = 0.167 S –0° Y3 = = 0.125 S –0° Y4 = V1[Y1 + Y2 + Y3] - Y2V2 - Y3V3 = I1 V2[Y2 + Y4 + Y5] - Y2V1 - Y4V3 = 0 V3[Y3 + Y4 + Y6] - Y3V1 - Y4V2 = -I2 ─────────────────────────── [Y1 + Y2 + Y3]V1 - Y 2V 2 - Y 3V 3 = I 1 -Y2V1 + [Y2 + Y4 + Y5]V2 - Y 4V 3 = 0 -Y3V1 - Y4V2 + [Y3 + Y4 + Y6]V3 = -I2 ──────────────────────────────────────────
= 0.5 S –90° Y5 = = 0.2 S –-90° Y6 = = 0.25 S –-90° I1 = 4 A –0° I2 = 6 A –90°
V1 = = 15.13 V –1.29° V2 =
= 17.24 V –3.73°
V3 = = 10.59 V –-0.11° 25.
Left node:
V1 4Ix = Ix + 5 mA –0° +
Right node: V2 8 mA –0° = Insert Ix =
224
CHAPTER 18
Rearrange, reduce and 2 equations with 2 unknowns result: V1[1.803 –123.69°] + V2 = 10 V1[2.236 –116.57°] + 3 V2 = 16 ────────────────────── Determinants: V1 = 4.37 V –-128.66° V2 = V1kΩ = 2.25 V –17.63° Z1 = 1 kΩ –0° Z2 = 2 kΩ –90° Z3 = 3 kΩ –-90° I1 = 12 mA –0° I2 = 4 mA –0° E = 10 V –0°
26.
0 = I1 + and
= -I1 - I2
with V2 - V1 = E Substituting and rearranging: = -I1 - I2 and solving for V1: V1 = 15.4 V – 178.2° with V2 = VC = 5.41 V – 174.87° 27.
Left node: V1 2 mA –0° = 12 mA –0° + and 1.5 V1 - V2 = -10 Right node: V2 0 = 2 mA –0° + and 2.7 V1 - 3.7 V2 = -6.6 Using determinants:
CHAPTER 18
V1 = V2kΩ = -10.67 V –0° = 10.67 V –180° V2 = -6 V –0° = 6 V –180° 225
Z1 = 2 kΩ –0° Z2 = 1 kΩ–0° Z3 = 1 kΩ –0° I = 5 mA –0°
28.
V 1: I = with I1 = and
V2 - V1 = 2Vx = 2V1 or V2 = 3V1
Substituting will result in: =I or and with
=I V1 = Vx = -2 V –0° V2 = -6 V –0°
29.
I1 =
= 1 ¥ 10-3 Ei
Y1 =
= 0.02 mS –0°
Y2 =
= 1 mS –0°
Y3 = 0.02 mS –0° I2 = (V1 - V2)Y2 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 ─────────────────────── V L = V2 =
226
= -2451.92 Ei
CHAPTER 18
30.
a.
yes
= = 2 –-90° = 2 –-90° (balanced)
b.
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 ───────────────────────
- Z 1I 2 - Z 3I 3 = E [Z1 + Z3 + Z6]I1 - Z 5I 3 = 0 -Z1I1 + [Z1 + Z2 + Z5]I2 - Z5I2 + [Z3 + Z4 + Z5]I3 = 0 -Z3I1 ─────────────────────────────────────── I2 = I3 = = I2 - I3 =
CHAPTER 18
E[Z 1 Z 4 - Z 3 Z 2 ] E[20´10 6 Ð 90° -20´10 6 Ð 90°] = =0A ZD ZD
227
c.
I=
= = 10 mA –0°
Y1 = = 0.2 mS –0° Y2 = = 0.125 mS –0° Y3 = V1[Y1 + Y2 + Y6] - Y1V2 - Y2V3 = I V2[Y1 + Y3 + Y5] - Y1V1 - Y5V3 = 0 V3[Y2 + Y4 + Y5] - Y2V1 - Y5V2 = 0 ───────────────────────── [Y1 + Y2 + Y6]V1 - Y 1V 2 - Y 2V 3 = I -Y1V1 + [Y1 + Y3 + Y5]V2 - Y 5V 3 = 0 -Y2V1 - Y5V2 + [Y2 + Y4 + Y5]V3 = 0 ─────────────────────────────────────────
= 0.4 mS –-90° Y4 = = 0.25 mS –-90° Y5 = = 0.2 mS –90° Y6 = V2 = 1 mS –0°
V2 = V3 = = V2 - V3 = =0V 31.
a.
=
1 –-90° π 1 –90° (not balanced) b.
The solution to 26(b) resulted in E(Z 1 Z 5 + Z 3 (Z 1 + Z 2 + Z 5 )) I3 = = ZD where ZΔ = (Z1 + Z3 + Z6)[(Z1 + Z2 + Z5)(Z3 + Z4 + Z5) and and
228
]
- 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° = 1.76 mA –-71.54°
CHAPTER 18
c.
The solution to 26(c) resulted in V3 =
= YΔ = (Y1 + Y2 + Y6)[(Y1 + Y3 + Y5)(Y2 + Y4 + Y5) -
where
- 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°
with
Y6 = 1 mS –0°, I = 10 mA –0° V3 = = 7.03 V –-18.46°
Source conversion: and 32.
Z 1Z 4 = Z 3Z 2 (R1 - jXC)
= R 3R 2
(1 kΩ - j1 kΩ) and
X C1 =
XC =
= 1 kΩ
= (0.1 kΩ)(0.1 kΩ) = 10 kΩ =
\ Rx = 5 Ω, Lx =
33.
]
= 5 Ω + j5 Ω
= 5 mH
1 1 1 = = kΩ = 0.25 kΩ wC1 (2000 rad/s)(2 mF) 4
Z1 = R1 ||
–90° = (2 kΩ –0°) || (0.25 kΩ –-90°) = 248.07 Ω –-82.88°
Z2 = R2 –0° = 0.5 kΩ –0°, Z3 = R3 –0° = 4 kΩ –0° Z4 = Rx + j = 1 kΩ + j12 kΩ,
X= L (2000 rad/s)(6 H) = 12 kΩ∅ Lx ω= Z1 Z 2 248.07 Ω∠ − 82.88° 0.5∠0° = = = Z3 Z 4 4 kΩ∠0° 12.04∠85.236° 62.017 × 10−3 ∠ − 82.88° ≠ 0.4152∠ − 85.236° (Unbalanced) 34.
Apply Eq. 18.6.
CHAPTER 18
229
35.
For balance: R1(Rx + j R1Rx + jR1
) = R2(R3 + j
)
= R2R3 + jR2
\ R1Rx = R2R3 and R1
= R2
and R1wLx = R2wL3
so that 36.
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.
Z6 =
= 5 Ω –38.66°
Z7 =
= 6.25 Ω –-51.34°
Z8 =
= 3.125 Ω –128.66°
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° I=
37.
230
ZY =
= 12.02 A –38.61°
ZD 12 Ω - j9Ω = = 4 Ω - j3 Ω 3 3
CHAPTER 18
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° I=
= 7.02 A –20.56°
Z4 = 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°
38.
[12.96 Ω –67.13°] = 8.64 Ω –67.13°
ZT = Z || 2Z = I= 39.
220 V–0°
= 25.46 A –-63.13°
ZΔ = 3ZY = 3(5 Ω) = 15 Ω Z1 = 15 Ω –0° || 5 Ω –-90° = 4.74 Ω –-71.57° Z2 = 15 Ω –0° || 5 Ω –-90° = 4.74 Ω –-71.57° Z3 = Z1 = 4.74 Ω –-71.57° = 1.5 Ω - j4.5 Ω ZT = Z1 || (Z2 + Z3) = (4.74 Ω –-71.57°) || (3 Ω - j9 Ω) = (4.74 Ω –-71.57°) || (9.489 Ω –-71.57°) =
44.97 Ð−143.14° 14.23 ΩÐ-71.23°
ZT = (3.16 Ω –-71.57°) = 0.999 I=
E 200 V Ð30° = = 63.29 A –101.57° ZT 3.16 Ω Ð -71.57°
CHAPTER 18
231
Chapter 19 1. = 3 Ω –0°, Z2 = 8 Ω –90°, Z3 = 6 Ω –-90° Z2 || Z3 = 8 Ω –90° || 6 Ω –-90° = 24 Ω –-90°
I=
E1 30 VÐ30° = = 1.24 A –112.875° Z1 + Z2 || Z3 3 Ω - j24 Ω = 3.72 A –-67.125°
I¢ =
Z1|| Z2 = 3 Ω –0° || 8 Ω –90° = 2.809 Ω –20.556° I= = 10.597 A –72.322° I≤ =
= 3.721 A –2.878° = 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°
2.
232
Considering Source I = 0.3 A Ð90° and short circuiting the voltage source E = 0.3 A ∠0°. Let Z1 = 10 Ω –90°, Z2 = 5 Ω –-90° I = 0.3 A –60°, E = 15 V –0° Z1I (10 Ω Ð90° )(0.3 A Ð60°) 3.0 A Ð150° = = I¢ = +j10 Ω -j5 Ω Z1 + Z2 5 Ω Ð90° = 0.6A –60° Consider E and o/p ckt I: E 15 V Ð0° 15 AÐ0° = I≤ = = +j10 Ω j5 Ω Z1 + Z 2 5 ΩÐ90° = 3 A –-90° Therefore, IC = I¢ - I≤ = 0.6 A –60° - 3 A –-90° = (0.3 A + j0.52 A) + j3 A = 0.3 A + j3.52 A = 3.53 A –85.13°
CHAPTER 19
3.
Z1 = 3 Ω –90°, Z2 = 7 Ω –-90° E = 12 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°
E:
V1 = =
(4.33 Ω Ð70.56° )(12 V Ð90° ) (1.44 Ω + j4.08 Ω) - j7 Ω
= 15.94 V –224.31° Therefore, I¢ =
=
15.94 V –224.31°
= 5.31 A –134.31° I:
Consider I and sort ckt E 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° CDR:
I3 =
=
I≤ =
=
(4 Ω Ð0°)(0.8 A Ð120°) = 0.79 A –130.62° 4 Ω - j0.75 Ω (0.79 A
= 1.38 A –130.62°
IL = I¢ - I≤ (direction of I¢) = 5.31 A –134.31° - 1.38 A –130.62° = (-3.71 A + j3.79 A) - (-0.89 A + j1.05 A) = 3.92 A –135.82°
CHAPTER 19
233
4.
I:
IC
2.41 kΩ VC1
2 kΩ
1.17 kΩ
1 = 1.17 kΩ 2p (20 kHz)(6.8 nF) 1 = = 2.41 kΩ 2p (20 kHz)(3.3 nF)
X C1 =
+−
3.9 kΩ
6 mA ∠ 180°
X C2
} (2 kΩ ∠ 0°) || (1.17 kΩ ∠ −90°) = 1.01 kΩ ∠ −59.67°
Z¢ = 2.41 kΩ –-90° + 1.01 kΩ –-59.67° = -j2.41 kΩ + 0.510 kΩ - j0.871 kΩ = 0.510 kΩ - j3.28 kΩ IC = =
R 2I (3.9 kΩ Ð 0°)(6 mA Ð 180°) = R 2 + Z¢ 3.9 kΩ + 0.510 kΩ - j3.28 kΩ 23.4 mA Ð 180° 23.4 mA Ð 180° = 4.41Ð - j3.28 5.5Ð -36.64°
= 4.25 mA –216.64°
VC1 = IC X C 2 = (4.25 mA –216.64°)(2.41 kΩ –-90°) = 10.24 V –126.64° 2 kΩ
E:
2.41 kΩ
+
+
VC2
−
1.17 kΩ
14 V ∠ 0°
3.9 kΩ
− Z′
Z¢ = (1.17 kΩ –-90°) || (3.9 kΩ - j2.41 kΩ) = (1.17 kΩ –-90°) || (4.58 kΩ –-31.71°) =
(1.17 kΩ Ð - 90°)(4.58 kΩ Ð -31.71°) 5.36 kΩ Ð -121.71° = - j1.17 kΩ + j3.9 kΩ - j2.41 kΩ 3.9 - j3.58
5.36 kΩ Ð -121.71° = 1.01 kΩ –-79.16° 5.29 Ð -42.55° (1.01 kΩ Ð - 79.16°)(14 V Ð 0°) Z¢(E) = V Z¢ = ¢ 2 kΩ + 1.01 kΩ Ð -79.16° Z + R1 =
=
14.14 V Ð -79.16° 14.14 V Ð -79.16° 14.14 V Ð -79.16° = = 2 + 0.189 - j0.991 2.189 - j0.991 2.4 Ð -24.36°
= 5.89 V –-54.8° X C2 VZ¢ (2.41 kΩ Ð -90°)(5.89 V Ð -54.8°) 14.195 V Ð -144.8° VC 2 = = = 3.9 kΩ - j2.41 kΩ X C2 + R 2 - j2.41 kΩ + 3.9 kΩ =
234
14.195 V Ð -144.8° = 3.1 V –-113° 4.58 Ð -31.71° CHAPTER 19
VC = VC1 + VC 2 = 10.24 V –126.64° + 3.1 V –-113° = (-6.11 V + j8.21 V) + (-1.21 V - j2.85 V) = -7.31 V + j5.36 V = 9.06 V –143.75° 5.
E:
R1
R2 C
−
L
E
IL1
+
Is
XL = 2π fL = 2π(10 kHz)(20 mH) = 1.26 kΩ 1 1 = XC = 2p fC 2p (10 kHz)(0.01 mF) = 1.59 kΩ
ZT = R1 –0° + XC –-90°) || (R2 –0° + XL – 90°) kΩ = 1 kΩ + (1.59 kΩ –-90°) || (2.2 !#kΩ # #+"j1.26 ### $) 2.54 kΩ Ð 29.8° (1.59 kΩ Ð -90°)(2.54 kΩ Ð 29.8° ) = 1 kΩ + - j1.59 kΩ + 2.2 kΩ + j1.26 kΩ 4.04 kΩ Ð -60.2° 4.04 kΩ Ð -60.2° = 1 kΩ + = 1 kΩ + 2.2 - j0.33 2.23Ð -8.46° = 1 kΩ + 1.81 kΩ –-51.74° = 1 kΩ + (1.12 kΩ - j1.42 kΩ) = 2.12 kΩ - j1.42 kΩ = 2.55 kΩ –-33.82° E 16 V Ð 60° = Is = = 6.27 mA –93.82° ZT 2.55 kΩ Ð -33.82° ( X C Ð -90°)(I s ) I L1 = X C Ð -90° + R2 Ð 0° + X L Ð 90° (1.59 kΩ Ð -90° )(6.27 mA Ð 93.82° ) = - j1.59 kΩ + 2.2 kΩ + j1.26 kΩ 9.97 mA Ð 3.82° 9.97 mA Ð 3.82° = = 2.2 - j0.33 2.22 Ð -8.53° ≠ I L1 = 4.49 mA –12.35° I:
2.2 kΩ
8 mA ∠ 30° 1 kΩ
IL2
(1 kΩ –0°) || (1.59 kΩ –-90°) = 847.74 Ω –-32.17° Z¢ = XL + 847.74 Ω –-32.17° = j1.26 kΩ + 717.59 Ω - j451.36 Ω = 0.72 kΩ + j0.81 kΩ = 1.08 kΩ –48.36°
1.26 kΩ 1.59 kΩ
} Z′
CHAPTER 19
235
R 2 (I) (2.2 kΩ Ð 0°)(8 mA Ð 30°) = R 2 + Z¢ 2.2 kΩ + 0.72 kΩ + j0.81 kΩ 17.6 mA Ð 30° 17.6 mA Ð 30° = = 2.92 + j0.81 3.03Ð 15.50° = 5.81 mA –14.5° IL = I L1 - I L 2 (direction of I L1 ) = (4.49 mA –12.35°) - (5.81 mA –14.5°) = (4.39 mA + j0.96 mA) - (5.62 mA + j1.45 mA) = -1.23 mA - j0.49 mA = 1.32 mA –-158.2° I L2 =
6.
AC:
1 1 1 = = 2pfC wC (1000)(4.7 mF) = 212.77 Ω XL = 2pfL = wL = (1000)(47 mH) = 47 Ω XC =
Z1 = 212.77 Ω –-90°, Z2 = 47 Ω – 0°, Z3 = 22 Ω + j47 Ω = 51.89 Ω –64.92° Z2|| Z3 = 29.23 Ω –30.66° ZT = Z1 + Z2|| Z3 = -j212.77 Ω + 25.14 Ω + j14.91 Ω = 25.14 Ω - j197.86 Ω = 199.45 Ω –-82.76° Is =
E 20 V Ð 60° = = 0.1 A –142.76° ZT 199.45 Ω Ð -82.76°
Z 3I s (51.89 Ω Ð 64.92°)(0.1 A Ð 142.76°) 5.19 A Ð 207.68° = = Z3 + Z2 22 Ω + j47 Ω + 47 Ω 83.49 Ð 34.26° I = 62.16 mA – 173.42° 3 and i = 62.16 ¥ 10- sin (1000t + 173.42°) I=
DC:
I= = 72.46 mA
3
i = -72.46 mA + 62.16 ¥ 10- sin (1000t + 173.42°)
236
CHAPTER 19
7.
DC:
DC: Consider 15V (dc) source & open ckt I (AC) VC = 15V AC: Consider I = 3 A∠0°source and short ckt. 15 V dc source
AC:
(9 ΩÐ−0°)(I) 9 Ω + 3 Ω -j1 Ω = 2.24 A –4.76°
IC =
VC = ICXC = (2.24 A –4.76°)(1 Ω –-90°) = 2.24 V –-85.24° υC = 15 V + sin(ωt − 85.24°) = 15 V + 3.17 sin(ωt - 85.24°)
8. 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° Z¢ = Z1 || Z2 = 10 kΩ –0° || 7.071 kΩ –-45° = 4.472 kΩ –-26.57° (CDR)
(4.472 kΩ Ð -26.57°)(5 mA Ð 0°) 22.36 mA Ð -26.57° Z¢I = = 4 kΩ - j2 kΩ + j5 kΩ 5Ð 36.87° Z¢ + Z3 = 4.472 mA –-63.44°
I¢ =
Z≤ = Z2 || Z3 = 7.071 kΩ –-45° || 5 kΩ –90° = 7.071 kΩ –45°
(VDR)
V¢ = = 8.945 V –26.565°
CHAPTER 19
237
= 1.789 mA –-63.435° = 0.8 mA - j1.6 mA
I≤ =
I = I¢ + I≤ = (2 mA - j4 mA) + (0.8 mA - j1.6 mA) = 2.8 mA - j5.6 mA = 6.26 mA –-63.43° 9.
Considering hI(Current ) source and short ckt. E = 10 V∠0° Let Z1 = 1Ω –0° Z2 = 10 kΩ –90° I = 2 mA –0° E = 10 V –0° H = 200 Therefore, I¢ =
=
(15 kΩ –0°)(200)(2 mA –0°)
= 0.333 A –-33.69° 15 kΩ + j10 kΩ E: Considering e and open ckt. hI I≤ =
=
18.03 kΩ –33.69° = 0.555 mA –-33.69° IL = I¢ - I≤ (direction of I¢) = 332.78 mA –-33.69° - 0.555 mA –-33.69° = 332.225 mA –-33.69° 10.
µV: Z1 = 5 kΩ –0°, Z2 = 1 kΩ –-90° Z3 = 4 kΩ –0° V = 2 V –0°, µ = 20
V¢ L =
-Z3 ( mV) -(4 kΩ Ð 0°)(20)(2 V Ð 0°) = = -17.67 V –6.34° 5 kΩ - j1 kΩ + 4 kΩ Z1 + Z2 + Z3
I: CDR: I¢ = = = 1.104 mA –6.34° 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°
238
CHAPTER 19
11.
R1 = Z1 = 20 kΩ –0° R2 + jXL = Z2 = 7 kΩ + j7 kΩ I = 2 mA –0° Η = 150
I¢ =
=
(20 kΩ –0°)(150)(2 mA –0°) = 215.13 mA –-14.53° 20 kΩ + 7 kΩ + j10 kΩ
Given V = 10 V –0°; µ = 25 (25)(10 V –0°) 20 kΩ + 7 kΩ + j7 kΩ = 8.964 mA –-14.53°
I≤ =
=
IL = I¢ - I≤ (direction of I¢) = 215.13 mA –-14.53° - 8.964 mA –-14.53° = 206.17 mA –-14.53° R1 = 2 kΩ, R2 = 2 kΩ, h = 60, VL = (hI + I)RL Let IL = hI + I = (h + 1)I And by KVL, VL = IR1 + E = −ILRL
12.
so that I = Therefore, VL = -(h + 1)IRL = -(h + 1)
R1
RL
R1
Subt. for R1, RL VL = -(h + 1)IRL = -(h + 1)
2 kΩ –0°
2 kΩ –0°
VL = -(h + 1)(VL - E) VL(2 + h) = E(h + 1) 61 (30 V –47°) = 29.52 V –47° VL = 62 13.
I 1: I1 = 1 mA –0° Z1 = 2 kΩ –0° Z2 = 5 kΩ –0°
KVL: V1 - 20 V - V = 0
I¢ =
\ I¢ =
or V =
V1 = 21 V
CHAPTER 19
239
V = I5kΩZ2 = [I1 - I¢]Z2 = I1Z2 - I¢Z2
éZ ù I¢ ê 1 + Z2 ú = I1Z2 ë 21 û [I1] =
and I¢ =
(5 kΩ Ð 0°)(1 mA Ð 0° ) = 0.981 mA –0° æ 2 kΩ Ð 0° ö ç ÷ + 5 kΩ Ð 0° 21 è ø
I 2: V1 = 20 V + V = 21 V I≤ =
fiV=
I5kΩ =
I≤ = I2 - I5kΩ = I2 -
é Z ù I≤ ê1 + 1 ú = I2 ë 21Z2 û =
I≤ =
= 1.963 mA –0°
I = I¢ + I≤ = 0.981 mA –0° + 1.963 mA –0° = 2.94 mA –0° 14.
E 1:
10 V –0° - I 10 Ω - I 2 Ω - 4 Vx = 0 with Vx = I 10 Ω Solving for I: I=
= 192.31 mA –0°
= 10 V –0° - I(10 Ω) = 10 V - (192.31 mA –0°)(10 Ω –0°) = 8.08 V –0°
240
CHAPTER 19
I:
SIi = SIo 5 A –0° +
=0
5 A + 0.1 Vx + 2.5 Vx = 0 2.6 Vx = -5 A Vx =
= -1.923 V
= -Vx = -(-1.923 V) = 1.923 V –0° = 8.08 V –0° + 1.923 V –0° = 10 V –0° Vs = 15.
For ZTh: Short ckt E and open ckt Xc terminal a and b Z1 = 3 Ω –0°, Z2 = 4 Ω –90° E = 100 V –0° ZTh = R1 || XL = (2 Ω –0° || 5 Ω –90°) =
10 –0°
5.385 –68.199° = 1.86 Ω –21.80° = 1.73 Ω + j0.69 Ω
For ETh: Open ckt voltage between terminal a and b ETh =
16.
ZTh:
ETh:
(5 Ω –90°) jXLE jXL + R1 (j5 Ω + 2 Ω) 500 V –90° = 92.85 V –21.80° 5.385 –68.199°
Z1 = 2 kΩ –0°, Z2 = 3 kΩ –−90°, Z3 = 6 kΩ –90° E = 25 V –30 ° 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 Ð-90°)(25 V Ð30°) = 2 kΩ -j3 kΩ Z2 + Z1 75 V Ð−60° = 20.79 V –-3.69° =
ETh =
CHAPTER 19
241
17.
ZTh: Z1 = 20 Ω + j20 Ω = 28.284 Ω–45° Z2 = 70 Ω –0°
ZTh = Z1 || Z2 = (28.284 Ω –45°) || (70 Ω –0°) =
1979.88 Ω Ð45° = 21.47 Ω –32.47° 92.195 Ð12.53°
ETh = IZ¢ = IZTh = (0.2 A –0°)(21.47 Ω –32.47°) = 4.29 V –32.47° 18.
ZTh
120 Ω
XL = 2π fL = 2π(1 kHz)(12 mH) = 75.4 Ω
75.4 Ω ZTh
470 Ω
Ω) ZTh = (120 Ω –0°) || 470 !#Ω #+"j75.4 ##$ 476.01Ð 9.11° (120 Ω Ð 0°)(476.01 Ω Ð 9.11° ) = 120 Ω + 470 Ω + j75.4 Ω 57.12 kΩ Ð 9.11° = 590 +" j75.4 Ω ## $ !## 594.8Ð7.28° ZTh = 96.03 Ω –1.83° ETh
Source conversion: E2 = IZ = (2 mA –-90°)(470 Ω –0°) = 0.94 V –-90° Rs = 470 Ω 120 Ω
+
8 V ∠ 0°
−
I
75.4 Ω
+ −
ETh
450 Ω
+
0.94 V ∠ −90°
−
8 V Ð 0° - 0.94 V Ð -90° 120 Ω + 470 Ω + j75.4 Ω 8 V Ð 0° - (- j0.94 V) 8 V + j0.94 V = = 590 Ω + j75.4 Ω 594.8 Ω Ð 7.28° 8.055 V Ð 6.70° = 594.8 Ω Ð 7.28° = 13.54 mA –-0.58° ETh = 8 V –0° - IR = 8 V –0° - (13.54 mA –-0.58°)(120 Ω) = 8 V –0° - 1.62 V –-0.58° = 8 V - (1.62 V - j16.4 ¥ 10-3 V) = 8 V - 1.62 V + j16.4 ¥ 10-3 V = 8 V - 1.62 V + j0.016 V = 6.38 V + j16.4 ¥ 10-3 V I=
242
CHAPTER 19
19.
X C1 = X C2 =
1 1 = = 3.39 kΩ 2p fC 2p (1 kHz)(0.047 mF)
Source conversions: Z¢ = R1 + XC1 = 2 kΩ - j3.39 kΩ = 3.94 kΩ –-59.46°
E1 10 V Ð 0° = = 2.53 mA –59.46° Z¢ 3.94 kΩ Ð -59.46° E 10 V Ð 0° I2 = 2 = = 5 mA –0°, E = IZ = (5 mA –0°)(2 kΩ –0°) = 10 V –0° R 2 2 kΩ Ð 0° I1 =
C2
3.39 kΩ I1
I2
2 kΩ
R3
−
Z′
R2
E = 10 V ∠ 0°
+
IT = I1 - I2 = 2.53 mA –59.46° - 5 mA –0° = 1.29 mA + j2.18 mA - 5 mA = -2.47 mA + j2.18 mA = 3.29 mA –138.57° Z¢ || R2 = (3.94 kΩ –-59.46°) || (2 kΩ –0°) (3.94 kΩ Ð -59.46° )(2 kΩ Ð 0° ) = 2 kΩ - j3.39 kΩ + 2 kΩ 7.88 kΩ Ð -59.46° 7.88 kΩ Ð -59.46° = = 4 - j3.39 5.24 Ð -40.28° = 1.51 kΩ –-19.18° Source conversion: E¢ = IZ = (3.29 mA –138.57°)(1.51 kΩ –-19.18°) = 4.97 V –119.39° 1.51 kΩ ∠ −19.18°
+ −
3.39 kΩ
+ ETh
−
2 kΩ
−
E′
I′
E
+
ZTh = (2 kΩ –0°) || (1.51 kΩ –-19.18° - j3.39 kΩ) = (2 kΩ –0°) || (1.43 kΩ - j0.496 kΩ - j3.39 kΩ) = (2 kΩ –0°) || (1.43 kΩ - j3.89 kΩ) = (2 kΩ –0°) || (4.14 kΩ –-69.82°) (2 kΩ Ð 0° )(4.14 kΩ Ð -69.82°) = 2 kΩ + 1.43 kΩ - j3.89 kΩ 8.28 kΩ Ð -69.82° = 3.43 - j3.89
CHAPTER 19
243
8.28 kΩ Ð -69.82° 5.19 Ð -48.6° ZTh = 1.6 kΩ –-21.22° E¢ + E I¢ = 2 kΩ - j3.39 kΩ + 1.51 kΩ Ð -19.18° 4.97 V Ð 119.39° +10 V Ð 0° = 2 kΩ - j3.39 kΩ + 1.43 kΩ Ð - j0.496 kΩ -2.44 V + j4.33 V + 10 V = 3.43 kΩ - j3.89 kΩ 7.56 V + j4.33 V 8.71 V Ð 29.8° = = 5.19 kΩ Ð -48.6° 5.19 kΩ Ð -48.6° = 1.68 mA –78.4° ETh - V2kΩ + E = 0 ETh = V2kΩ - E = (I¢)(2 kΩ) - 10 V –0° = (1.68 mA –78.4°)(2 kΩ –0°) - 10 V = 3.36 V –78.40° - 10 V = 0.68 V + j3.29 V - 10 V = -9.32 V + j3.29 V ETh = 9.88 V –160.56° =
10 Ω
20. 2Ω
4Ω
ZTh
8Ω
ZTh = (8 Ω –-90°) || (10 Ω + 2 Ω || 4 Ω –90°) æ 8 Ω Ð 90° ö = (8 Ω –-90°) || ç10 Ω + ÷ + j4 ø è!###"2# ## $ 8 Ω Ð 90° 10 Ω + 4.47 Ð# 63.44° !### "# #$
10 Ω# + 1.79 26.56° !# # #"Ω#Ð# ## $ 10 Ω + 1.6 Ω + j0.8 Ω !###"###$ 11.6 +# j0.8 Ω !#Ω #" #$ 11.63 Ω Ð 3.95° = (8 Ω –-90°) || (11.63 Ω –3.95°) 92.8 Ω Ð -86.05° 92.8 Ω Ð -86.05° = = - j8 + 11.6 + j0.8 11.6 - j7.2 92.8 Ω Ð -86.05° = 13.65Ð -31.83° ZTh = ZN = 6.79 Ω –-54.22°
244
CHAPTER 19
Z1 = 2 Ω –0°, Z3 = 8 Ω –-90° Z2 = 4 Ω –90°, Z4 = 10 Ω –0° E = 50 V –0° ETh = V2 + V4 V2 =
(4 Ω Ð 90° )(50 V Ð 0°) + j4 kΩ + 2 Ω Ð 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° (10 Ω Ð 0° )(20.972 V Ð -70.285°) V4 = = = 16.377 V –-31.625° 10 Ω - j8 Ω 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° 21.
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° ETh: Superposition: (E1) E¢ Th = = = 74.965 V –51.34° (I)
CHAPTER 19
245
E≤ Th = = 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° 22.
ZTh:
ZTh = Z = 15 Ω - j15 = 21.21 Ω –-45° ETh:
23.
a.
ETh = E - VZ = E − IZ = 20 V –40° - IZ = 20 V –40° - (0.5 A –90°)(21.21 Ω –-45°) = 20 V –40° - 10.605 V –45° = (15.321 V + j12.856 V) - (7.499 V + j7.944 V) = 7.822 V + j5.357 V = 9.48 V –34.41° AC: ETh:
Z1 = Z3 = 22 Ω + wL –90° = 22 Ω + j47 Ω = 51.89 Ω – 64°
ETh =
Z 3E (51.89 Ω Ð 64.92° )(20 V Ð 60°) = = 6.21 V – 207.36° 22 Ω + j47 kΩ - j212.77 Ω Z3 + Z1
ZTh:
246
CHAPTER 19
(212.77 Ω Ð -90° )(51.89 Ω Ð 64.92°) - j212.77 Ω + 22 Ω + j47 kΩ = 66.04 Ω –57.36 ° = 35.62 Ω + j55.61 Ω
ZTh = Z1 || Z2 =
DC:
ETh:
ETh = -5 V
RTh: RTh = 22 Ω
b.
AC: I=
6.21 V Ð 207.36° 35.62 Ω + j55.61 Ω + 47 Ω 6.21 V Ð 207.36° = 82.62 Ω + j55.61 Ω 6.21 V Ð 207.36° = 99.59 Ω Ð 33.94° = 62.36 mA –173.42° =
DC:
I= = 72.46 mA 3
i = -72.46 mA + 62.36 ¥ 10- sin (1000t + 173.42°) matching the results of Problem 4.
CHAPTER 19
247
24.
a.
ZTh:
= 9 Ω + 3 Ω = 12 Ω = RTh
¨ZTh =
DC: Considering the 15 V dc
E¢Th = 15 V
AC: Considering the 3 A∠0° AC source
¨E≤Th =
= (3 A – 0°)(9 Ω –0°) = 27 V – 0°
ETh = ETh' + ETh'' = 15 V + 27 V –0° DC: VC = 15 V
b.
AC: VC = =
(27 V –0°) 12 Ω
= 2.24 V –-85.24°
υC = 15 V+ 2.24 V –-85.24° = 15 V + 3.17 sin(wt - 85.24°) 25.
a.
ZTh:
Z1 = Z2 =
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: 248
CHAPTER 19
E1 = (I–θ)(R1–0°) = (5 mA –0°)(10 kΩ –0°) = 50 V –0° ETh =
(7.071 kΩ Ð -45° )(20 V Ð 0° + 50 V Ð 0°) (5 kΩ - j5 kΩ) + (10 kΩ) (7.071 kΩ Ð -45° )(70 V Ð 0°) = (15 kΩ - j5 kΩ) =
= = 31.31 V –-26.57° b.
ETh 31.31 V Ð -26.565° = ZTh + Z L 4.472 kΩ Ð -26.565° +5 kΩ Ð 90° 31.31 V Ð -26.565° 31.31 V Ð -26.565° = = 4 kΩ - j2 kΩ + j5 kΩ 4 kΩ + j3 kΩ
I=
= 6.26 mA –63.44°
= 26.
Z1 = 10 kΩ – 0° Z2 = 10 kΩ – 0° Z3 = 1 kΩ – -90°
ZTh = Z3 + Z1 || Z2 = 5 kΩ - j1 kΩ @ 5.1 kΩ –-11.31° ETh: (VDR) 27.
ETh =
= 10 V
ZTh: Z1 = 40 Z2 = 0.2 Z3 = 5
= 40 kΩ –0° = 0.2 kΩ –-90° = 5 kΩ –0°
ZTh = Z3 || (Z1 + Z2) = 5 kΩ –0° || (40 kΩ - j0.2 kΩ) = 4.44 kΩ –-0.03°
CHAPTER 19
249
I¢ = = = 88.89 I –0.255° ETh = -I¢Z3 = -(88.89 I –0.255°)(5 kΩ –0°) = -444.45 ¥ 103 I –0.26° 28.
ZTh: ¨
ZTh = Z1 = 15 kΩ –0°
ETh: E¢ Th = -(hI)(Z1) = -(200)(2 mA – 0°)(15 kΩ – 0°) = -6 kV –0°
E: ETh = E¢ Th + E≤ Th = -6 kV –0° + 10 V –0° = -5999 V –0°
29.
ZTh:
= 5 kΩ –0°
Z1 = 5 k
250
Z2 = -j1
¨ZTh = Z1 + Z2 = 5 kΩ - j1 kΩ = 5.10 kΩ –-11.31°
CHAPTER 19
ETh:
ETh = 1 ETh = [µV + VZ1 ]
= -µV - IZ1 = -(20)(2 V –0°) − (2 mA –0°)(5 kΩ –0°) = −50 V –0°
30.
Z1 = 20 kΩ – 0°
ZTh:
Z'2 = 7 kΩ – 0° ¨ZTh = Z1 + Z'2 = 27 kΩ – 0°
ETh: ETh = µV - (hI)(Z1) = (25)(10 V –0°) - (150)(2 mA –0°)(20 kΩ –0°) = 250 V –0° - 6000 V –0° = -5750 V –0° 31.
ETh: (Eoc) hI = -I R1 = 2 kΩ –0° \I = 0 and hI = 0 with Eoc = ETh = E = 30 V –47° Isc: Isc = -(h + 1)I =
Ε 30 V –47° = −(60 + 1) R1 2 kΩ –0°
= -915 mA –47°
ZTh =
CHAPTER 19
30 V –47° = -32.79 Ω –0° (negative resistance) -915 mA –47°
251
32.
ETh: E¢ oc = 21 V Z1 = 5 kΩ –0° V = I1Z1 = (1 mA –0°)(5 kΩ –0°) = 5 V –0° E¢ oc = E¢ Th = 21 V = 21(5 V –0°) = 105 V –0° V = I 2Z 1 = (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 = 105 –0° + 210 V –0° = 315 V –0° = ETh = 105 kΩ –0°
ZTh = 33.
Eoc: (ETh)
KVL: -6 Ix(2 kΩ) - Ix(1 kΩ) + 8 V –0° - Ix(3.3 kΩ) = 0 Ix =
= 0.491 mA –0°
Eoc = ETh = Ix(3.3 kΩ) = 1.62 V –0°
252
CHAPTER 19
Isc:
34.
Isc =
= 2.667 mA –0°
ZTh =
=
= 607.42 Ω –0°
From Problem ZN = ZTh = 1.73 Ω + j0.69 Ω IN:
R = Z1 = 2 Ω –0°, Z2 = XL = j5 Ω = 5 Ω –90° Isc = IN =
2 Ω –0°
= 50 Ω –0° 35.
From Problem 17: ZN = ZTh = 21.47 Ω –32.47° Z1 = 20 Ω + j20 Ω Z2 = 70 Ω –0° ¨ZN = Z1 || Z2 I = 0.2 A –0° For Isc: As ZN = ZTh by passed by short circuit ¨Isc = I = IN = 0.2 A –0°
CHAPTER 19
253
36.
From Problem 21: ZN = ZTh = 5.00 Ω –-38.66° I N:
ZT = Z1 + Z2 || Z3 = 10 Ω + 8 Ω –90° || 8 Ω –-90°
Superposition:
(E1)
= 10 Ω + = very large impedance Is = and
=0A =0V = E1 = 120 V –0°
with so that I¢ sc =
=
= 15 A –90° (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° 37.
a.
Z N: E = 20 V –0°, I2 = 0.4 A –20° Z1 = 6 Ω + j8 Ω = 10 Ω –53.13° = 9 Z2 = 9 Ω - j12 Ω = 15 Ω –-53.13° ZN = Z1 || Z2 = (10 Ω –53.13°) || (15 Ω –-53.13°) = 9.66 Ω –14.93° I N: (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°
254
CHAPTER 19
38.
Z N: 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 I= =
120 V Ð 30° 3 Ω + (8 Ω - j8Ω) || 4Ω Ð 90°
= = 18.05 A –-16.22° Isc = IN = 39.
a.
(8 Ω - j8Ω)(18.05 A Ð -16.22°) Z2 (I) = = 22.83 A –-34.65° 8 Ω - j8 Ω + j4 Ω Z2 + Z3
AC: I N:
Z1 = Z3 = 22 Ω + j47 Ω = 51.89 Ω – 64°
E 20 V Ð60° = = 94 mA – 150° Z1 212.77 Ω Ð - 90° ZN = ZTh (problem 23) = 66.04 –57.36° = 35.62 Ω + j55.61 Ω IN =
DC: I N:
IN =
= 227.27 mA
RN = RTh = (problem 23) = 22 Ω
CHAPTER 19
255
b.
AC: I N:
Z N (I N ) (66.04 Ω Ð 57.36°)(94 mA Ð 150°) = 35.62 Ω + j55.61 Ω + 47 Ω Z N + 47 Ω 6.21 A Ð 207.36 ° = = 62.68 mA –173.22° 99.08 Ð 34.14 °
I=
DC:
I=
= 72.46 mA
3
and i = -72.46 mA + 62.68 ¥ 10- sin (1000t + 173.22°) Same as Problem 6 and 23. 40. a.
F rom problem 24: Z Th = RTh = ZN = 12 Ω –0° DC: Considering DC source (15V) and open ckt. I(AC source) Then Norton’s Current is
I¢ N =
15 V = 1.25 A 12 Ω
AC: Considering AC source 3 A∠0 °and short ckt. 15V dc then the norton’s current is
I≤ N =
−(3 A –0°)(9 Ω ∠0°) 12 Ω ∠0° = 2.25 A –0°
IN = I'N + I"N = 1.25 A + 2.25 A –0°
256
CHAPTER 19
b.
DC: VC = IR = (1.25 A)(12 Ω) = 15 V AC: Z ¢ = 12 Ω –0° || 1 Ω –-90° =
12 –-90° 12 - j1
= 0.997 Ω –-85.25°
Therefore, VC = IZ¢ = (2.25 A –0°)(0.997 Ω –-85.24°) = 2.24 V –-85.24° Hence, VC = 15 V + 2.24 V –-85.24° = 15 V + 3.17 sin(ω t –85.24°) 41.
a.
Note Problem 25(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 =
= 7 mA –0°
IN = Isc = 7 mA –0° b.
I=
Z N (I N ) (4.472 kΩ Ð -26.565°)(7 mA Ð 0°) = Z N + Z L 4.472 kΩ Ð -26.565° + 5 kΩ Ð 90°
= =
= 6.26 mA –63.44° as obtained in Problem 25.
42. Z N:
CHAPTER 19
Z1 = 10 kΩ –0°, Z2 = 10 kΩ –0 Z3 = -j1 kΩ ZN = Z3 + Z1 || Z2 = 5 kΩ - j1 kΩ = 5.1 kΩ –-11.31°
257
I N:
-(Z 2 || Z 3 )20 V (Z 2 || Z 3 ) + Z1 -(0.995 kΩ Ð -84.29°)(20 V) = 0.1 kΩ - j0.99 kΩ + 10 kΩ V2 = -1.961 V –-78.69° V2 =
IN = Isc = 43.
V2 -1.961 V Ð -78.69° = = -1.96 ¥ 10- 3 V –11.31° 1 kΩ Ð -90° Z3
Z N:
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°
I N:
IN = Isc = = = 100 I –0.29° 44.
Z N: Z1 = 5 kΩ –0°, Z2 = 1 kΩ –-90° ¨ ZN = Z1 + Z2 = 5 kΩ - j1 kΩ = 5.1 kΩ –-11.31°
. I N: I¢ sc = = 7.843 mA –11.31° (I):
I≤ sc = = = 1.96 mA –11.31°
258
CHAPTER 19
IN = I¢ sc + I≤ sc = 7.843 mA –11.31° + 1.96 mA –11.31° = 9.81 mA –11.31° 45.
Z N:
Z1 = 20 kΩ –0°, Z'2 = 7 kΩ –0° V = 10 V –0°, µ = 25, h = 100 I = 2 mA –0° ZN = Z1 + Z'2 = 27 kΩ –0°
IN: (Due to hI) I¢ sc =
(20 kΩ Ð0°)(150)(2 mA ∠0°) 20 kΩÐ0°+ 7 kΩ ∠0° = 222.22 mA –0° =
(Due to µV)
(25)(10 V –0°) (27 kΩ –0°) = 9.26 mA –0°
I≤sc =
Therefore, IN (in direction of I¢sc) = I¢sc - I≤sc = 222.22 mA –0° - 9.26 mA –0° = 212.96 mA –0° 46.
Z1 = 2 kΩ –0° Z2 = 5 kΩ –0°
I2 = I3 + I5kΩ V = I5kΩZ2 = (I2 - I3)Z2 Eoc = ETh = 21 V = 21(I2 - I3)Z2 æ E ö = 21 çI 2 - oc ÷ Z2 Z1 ø è é Z ù Eoc ê1 + 21 2 ú = 21 Z2I2 Z1 û ë 21(5 kΩ Ð 0°)(2 mA Ð 0°) 21Z2I2 = Eoc = æ 5 kΩ Ð 0° ö Z 1 + 21 2 1 + 21ç ÷ Z1 è 2 kΩ Ð 0° ø ETh = Eoc = 3.925 V –0°
CHAPTER 19
259
20 V π -V \ V = 0 and IN = Isc = I2 = 2 mA –0°
ZN =
= 1.96 kΩ
47. Z1 = 1 kΩ –0° Z2 = 3 kΩ –0° Z3 = 4 kΩ –0° V2 = 21 V = Eoc fi V = I = I 1 + I 2, I 1 = I2 =
, I = I1 + I2 =
+
=
=
I= and
(21)(1 kΩ Ð 0°)(3 kΩ Ð 0°)(2 mA Ð 0°) 21Z1Z2I = 3 kΩ + 21(1 kΩ Ð 0°) Z2 + 21Z1 ETh = Eoc = 5.25 V –0° Eoc =
Isc =
fi V=
Isc
V = I 1Z 1 I = I1 + I¢ Isc = I = I1 + I¢ =
æZ + Z ö 3 fi I¢ = ç 2 ÷ Isc Z 2 è ø é Z æZ + Z ö Z + Z3 ù 3 + ç 2 ú I sc ÷ I sc = ê 3 + 2 êë21Z1 Z 2 úû è Z2 ø = 0.79 mA –0°
Isc =
\
IN = 0.79 mA –0° ZN =
260
= 6.65 kΩ –0°
CHAPTER 19
48. 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 ETh = = = 199.45 V –-56.31° Pmax =
= 1198.2 W
Z1 = 3 Ω + j4 Ω = 5 Ω –53.13° Z2 = 2 Ω –0° ¨ ZN = ZTh = Z1 || Z2 = 5 Ω –53.13° || 2 Ω –0°
49.
=
3 + j4 + 2
= = = 1.56 Ω –14.47° ZTh = 1.56 Ω –14.47° ⇒ ZL = 1.56 Ω ∠−14.47° ZL = 1.51 Ω - j0.39 Ω (Conjugate of ZTh) ETh = I(Z1|| Z2) = (3 A –30°)(1.562 Ω –14.47°) = 4.69 V –74.47° (4.69 V)2 Pmax = = 3.64 W 4(1.51 Ω) 50.
ZTh: Z1 = 4 Ω –90°, Z2 = 10 Ω –0° Z3 = 5 Ω –-90°, Z4 = 6 Ω –-90° E = 60 V –60°
CHAPTER 19
261
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° ETh: ETh = = = 29.85 V –-24.29° Pmax =
= (29.85 V)2/4(2.475 Ω) = 90 W
51. 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 +
- E2 = 0
ETh = E2 = = 168.97 V –112.53° ETh = E2 -
= 200 V –90° - 168.97 V –112.53° = 78.24 V –34.16° = (78.24 V)2/4(2.475 Ω) = 618.33 W
Pmax = 52.
I=
3 V –0° = 1 mA –0° 3 kΩ –0°
ZTh = 60 kΩ –0° ETh = (50 I)(60 kΩ –0°) = (50)(1 mA –0°)(60 kΩ –0°) = 3000 V –0° (3 kV)2 Pmax = = 37.5 W 4(60 kΩ)
262
CHAPTER 19
53.
ETh:
Z1 = 2 kΩ –0° Z2 = 3 kΩ –-90° Z3 = 6 kΩ –90°
Z 2E (3 kΩ Ð-90°)(25 V Ð30°) = 2 kΩ -j3 kΩ Z2 + Z1 75 V Ð−60° = 20.79 V –-3.69° =
ETh =
ZTh: ¨ ZTh = Z3 + Z1 || Z2
(2 kΩ Ð 0°)(3 kΩ Ð -90°) 2 kΩ - j3 kΩ = + j6 kΩ + 1.66 kΩ –-33.69° = + j6 kΩ + 1.38 kΩ - j920.8 Ω = 1.38 kΩ + j5.08 kΩ = 5.26 kΩ – +74.80° ZTh = ZL* for maxm . power
ZTh = +j6 kΩ +
\ZL = 5.26 kΩ –-74.80° = 1.38kΩ - j5.08 kΩ b. 54.
Pmax =
(20.79 V)2 = 78.30 mW 4(1.38 kΩ)
From Problem 24, ZTh = RTh = 12 Ω, ETh = 15 V + 27 V –0° a.
\ ZL = 12 Ω
b.
Pmax = or ETh = and Pmax =
CHAPTER 19
(15 V)2 4(12 Ω)
(27 V)2 = 4.69 W + 15.19 W = 19.88 W 4(12 Ω)
= 30.887 V (30.887 V)2 = 19.88 W 4(12 Ω)
263
55.
56.
a.
Problem 25(a): ZTh = 4.47 kΩ –-26.57° = 4 kΩ - j2 kΩ ZL = 4 kΩ + j2 kΩ ETh = 31.31 V –-26.57°
b.
Pmax =
a.
ZTh = 2 kΩ –0° || 2 kΩ –-90° = 1 kΩ - j1 kΩ For maximum power ZL = ZTH ⇒ RL = RTH − jXTH − jXL RL = RTH − j(XTH − XL)
RL =
= (31.31 V)2/4(4 kΩ) = 61.27 mW
2
RTh + ( X Th + X Load ) 2
=
2 (1 kΩ)2 + (-1 kΩ + 3 kΩ)
=
(1 kΩ)2 + (2 kΩ)
2
= 2.24 kΩ
57.
b.
Rav = (RTh + RLoad)/2 = (1 kΩ + 2.24 kΩ)/2 = 1.62 kΩ (54 V)2 Pmax = = 450 mW 4(1.62 Ω)
a.
ZTh:
1 1 = 2pfC 2p (10 kHz)(4 nF) @ 3978.87 Ω XL = 2pfL = 2p(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 = C= = 7.31 nF 2pfX C 2p (10 kHz)(2177.27 Ω) \
b.
RL = RTh = 2940.27 Ω
c.
= Pmax =
58.
264
=
(3978.87 Ω Ð -90°)(2 V Ð 0°) = 3.43 V–−25.53°) 1 kΩ + j1884.96 Ω - j3978.87 Ω = 1 mW
(5 kΩ –0°) = 1.25 mA –0° 4 kΩ + 11 kΩ Vab = (Iab)(11 kΩ –0°) = (1.25 mA∠0°)(11 kΩ∠0°) = 13.75 V –0°
Iab =
CHAPTER 19
59.
a. V=
=
= 5 V –0° = 0.83 mA –0°
I= b. V=
=
= 10 V –0° = 0.83 mA –0°
I=
60. I1 =
= 50 mA –0°
I2 = = 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 kΩ Ð 26.57°)(51.54 mA Ð 14.04°) = IC = Z¢ + Z3 1.6 kΩ + j0.8 kΩ - j4 kΩ = 25.77 mA –104.04°
CHAPTER 19
265
Chapter 20 1.
a.
PT = 60 W + 45 W + 30 W = 135 W
b.
QT = 0 VARS, ST = PT = 135 VA
c.
ST = EIs, Is =
135 VA = 0.675 A 200 V P=
d.
R, R =
(0.675 A)2
= 131.69 Ω
V = IsR = (0.675 A)(131.69 Ω) = 88.89 V V1 = V2 = E - V = 200 V - 88.89 V = 111.11 V (111.11 V)2 P1 = , R1 = = 274.34 Ω P2 =
2.
111.11 V = 0.405 A, I2 = 274.34 W
I1 =
a.
ZT = 3 Ω - j5 Ω + j9 Ω = 3 Ω + j4 Ω = 5 Ω –53.13° = 10 A –-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 = Fp =
e.
(111.11 V)2 = 411.51 Ω 30 W
111.11 V = 0.270 A 411.51 W
e.
I=
266
, R2 =
= EI = (50 V)(10 A) = 500 VA = 0.6 lagging
-
CHAPTER 20
f.
WR =
éVI ù é VI ù VI VI : WR = 2 ê ú = 2 ê ú = f1 f1 ë f2 û ë2 f1 û V = IR = (10 A)(3 Ω) = 30 V
WR = g.
=5J
VC = IXC = (10 A)(5 Ω) = 50 V WC =
= 1.33 J
VL = IXL = (10 A)(9 Ω) = 90 V WL = 3.
a.
= 2.39 J
R 1:
E 120 VÐ0° = = 60 mA –0° R1 2 kΩÐ0° PR1 = I2R = (60 mA)22 kΩ = 7.2 W QT = 0 VAR S R1 = PR1 = 7.2 VA I R1 =
C:
1 1 = 3.98 kΩ = 2p fC 2p (5 kHz)(0.02 mF) PT = 0 W E 120 V Ð 0° = Q T: I C = = 30.15 mA –90° XC 3.98 kΩ Ð -90° QC = I2XC = (30.15 mA)23.98 kΩ = 3.62 VAR ST = QC = 3.62 VA
XC =
R2-L: XL = 2π fL = 2π(2 kHz)(80 mH) = 1 kΩ Z¢ = 200 Ω + j1 kΩ = 1.02 kΩ –78.69° 120 V Ð 0° E = 117.65 mA –-78.69° IL = = Z¢ 1.02 kΩ Ð 78.69° QL = I2XL = (117.6 mA)2(1 kΩ) = 13.84 VAR QT = QL = 13.84 VAR PT = I2R = (117.65 mA)2(200 Ω) = 2.77 W ST = 2.77 W - j13.84 VAR fi 1.41 VA
QT = 13.84 VAR (L) − 3.62 VAR (C) = 10.22 VAR (L)
ST
=
14
.28
VA
b.
PT = 7.2 W + 2.77 W = 9.97 W
CHAPTER 20
267
ST = 9.97 + j10.22 = 14.28 VA –45.71° c.
(2 kΩ Ð 0°)(3.98 kΩ Ð -90°) 7.96 kΩ Ð -90° = 4.45Ð -63.32° 2 kΩ - j3.98 kΩ = 1.79 kΩ –-26.68°
Z¢ = R1 || XC =
(1.79 kΩ Ð -26.68°)(1.02 kΩ Ð 78.69°) (1.6 kΩ - j0.8 kΩ) + (0.2 kΩ + j1 kΩ) 1.83 kΩ Ð 52.06° 1.83 kΩ Ð 52.06° = = 1.8 kΩ + j0.2 kΩ 1.81Ð 6.34° = 1.01 kΩ –45.72° P 9.97 W = 0.698 lagging FP = T = ST 14.28 VA ZT = Z¢ || (R2 + jXL) =
4.
E 120 V Ð 0° = 118.81 mA –-45.72° = ZT 1.01 kΩ Ð 45.72°
d.
Is =
a.
PT = 0 + 100 W + 300 W = 400 W QT = 300 VAR(L) - 600 VAR(C) + 0 = -300 VAR(C) ST =
= 500 VA
Fp = b.
500 VA
= 0.8 (leading)
power triangle PT = EIs cos θT 400 W = (180 V)Is(0.8)
c.
= 2.778 A 180 V(0.8) Is = 2.778 A –126.86°
Is =
5.
a.
PT = 600 W + 400 W + 100 W = 1100 W QT = 1200 VAR(L) + 800 VAR(L) - 1800(C) = 200 VAR(L)
ST=
11002 + 2002= 1118.03 VA 1100 W
b.
Fp =
c.
power triangle
d.
Is =
1118.03 VA
= 0.984 (lagging)
1118.03 VA = 5.59 A FP = 0.984 fi θ = 10.30° Is = 5.59 A –10.30°
268
CHAPTER 20
6.
a.
PT = 200 W + 100 W + 0 + 50 W = 350 W QT = 50 VAR(L) + 100 VAR(L) - 200 VAR(C) - 400 VAR(C) = -450 VAR(C) ST =
= 570.09 VA
b.
Fp =
= 0.614 (leading)
c.
-
d.
PT = EIs cos θT 350 W = (50 V)Is(0.614) Is =
= 11.4 A
Is = 11.4 A –52.12° 7.
a.
200 W: Resistive: P = I 2R fi R =
P I
200 W
=
2
(2 A)
2
=
200 W 4 A2
= 50 Ω
400 VAR(L): Inductive: Q L = I 2X L fi X L =
QL
200 VAR(L)
= 100 Ω 4 A2 I XL 100 Ω = 3.18 mH XL = 2π fL fi L = = 2p f 2p (5 kHz) 2
=
600 VAR(C): Capacitive: Q C = I 2X C fi X C = XC =
8.
QC I
2
=
600 VAR(C) 4 A2
= 150 Ω
1 1 1 fiC= = 212.2 nF = 2p fC 2p fX C 2p (5 kHz)(150 Ω)
b.
ZT = R + jXL - jXC = 50 Ω + j100 Ω - j150 Ω = 50 Ω - j50 Ω = 70.71 Ω –-45°
c.
Vs = IZT = (2 A –0°)(70.71 Ω –-45°) = 141.42 V –-45°
d.
Fp = cos(45°) = 0.707 leading
e.
VC = IXC = (2 A –0°)(150 Ω –-90°) = 300 V –-90°
a.
IR =
90 V–30° = 3.6 A –30° 25 Ω–0° P = I2R = (3 A)2 25 Ω = 225 W QR = 0 VAR S = P = 225 VA
CHAPTER 20
269
b.
c.
90 V–30° = 9 A –-60° 10 Ω–30° PL = 0 W QL = I2XL = (9 A)2 10 Ω = 810 VAR(L) S = QL = 810 VA IL =
PT = 225 W + 400 W = 625 W QT = 600 VAR(L) + 810 VAR(L) = 1410 VAR(L)
ST =
PT2 + QT2 =
625 W = 0.405 (lagging) 1542.31 VA
Fp = 9.
(625 W) 2 + (1410 VAR) 2 = 1542.31 VA
a.
R3 + jXL = 4 Ω + j4 Ω = 5.66 Ω –45° (2 Ω Ð 0°)(5 Ω Ð -90°) 10 Ω Ð -90° = 1.86 Ω –-21.8° R2 || XC –-90° = = 5.39 Ð -68.2° 2 Ω - j5 Ω (5.66 Ω Ð 45°)(1.86 Ω Ð -21.8°) (4 Ω - j4 Ω) + (1.73 Ω - j0.69 Ω) 10.53 Ω Ð 23.2° 10.53 Ω Ð 23.2° = = = 1.59 Ω –-6.81° = 1.58 Ω - j0.19 Ω 6.62 Ð 30.01° 5.73 + j3.31 ZT = R1 + 1.59 Ω –-6.81° = 2 Ω + 1.58 Ω - j0.19 Ω = 3.58 Ω - j0.19 Ω = 3.59 Ω –-3.03° E 20 V Ð 0° = 5.57 A –3.03° Is = = ZT 3.59 Ω Ð -3.03°
b.
R1: PR1 = I s R1 = (5.57 A)22 Ω = 62.05 W
2
VR1 = IsR1 = (5.57 A –3.03°)(2 Ω –0°) = 11.14 V –3.03° VR 2 = E - VR1 = 20 V –0° - 11.14 V –3.03° = 20 V - (11.12 V + j0.5 V) = 8.88 V - j0.59 V = 8.89 V –-3.8° 2
R2: PR 2 =
I R3 =
VR 2
R2
=
(8.89 V)2 = 39.52 W 2Ω
VR 2 R3 + jX L
=
8.89 V Ð -3.8° 8.89 V Ð -3.8° = = 1.57 A –-48.8° 4 Ω + j4 Ω 5.66 Ω Ð 45°
2
R3: PR 3 = I R 3 R3 = (1.57 A)24 Ω = 9.86 W
270
CHAPTER 20
2
c.
d.
(8.89 V)2 X C: Q X C = = = 15.81 VAR(C) XC 5Ω XL: QL = I2XL = (1.57 A)24 Ω = 9.86 VAR(L) VR 3
R1: ST = PR1 = 62.05 VA R2: ST = PR 2 = 39.52 VA R3: ST = PR 3 = 9.86 VA C: ST = Q X C = 15.81 VA L: ST = QL = 9.86 VA
e.
PT: PR1 + PR 2 + PR 3 = 62.05 W + 39.52 W + 9.86 W = 111.43 W QT: Q X C - Q X L = 15.81 VAR(C) - 9.86 VAR(L) = 5.95 VAR(L) 2
Fp =
10.
2
PT + QT = 111.59 VA
ST =
PT 111.43 W = 0.998 (leadering) = ST 111.59 VA
f.
-
a.
Z1 = 3 Ω + j4 Ω = 5 Ω –53.13° Z2 = 3 Ω - j4 Ω = 5 Ω –-53.13° (5 Ω Ð 53.13°)(5 Ω Ð -53.13°) 25 Ω Ð 0° ZT = Z1 || Z2 = = 4.17 Ω –0° = (3 Ω + j4 Ω) + (3 Ω - j4 Ω) 6 E 50 V Ð 60° Is = = 11.99 A –60° = ZT 4.17 Ω Ð 0°
b.
IR–L =
c.
L: IR–L = 10 A –16.87° QL = I2XL = (10 A)2·4 Ω = 400 VAR(L)
E
50 V Ð 60° = 10 A –16.87° Z R - L 5 Ω Ð 53.13° P3Ω = I2R = (10 A)2·3 Ω = 300 W E 50 V Ð 60° IR–C = = 10 A –113.13° = Z R -C 5 Ω Ð -53.13° P3Ω = I2R = (10 A)2·3 Ω = 300 W =
C: IR–C = 10 A –113.13° QC = I2XC = (10 A)2·4 Ω = 400 VAR(C) d.
3 Ω: ST = P3Ω = 300 VA for each resistor L: ST = QL = 400 VA C: ST = QC = 400 VA
e.
PT = 300 W + 300 W = 600 W
CHAPTER 20
271
QT = 400 VAR(L) + 400 VAR(C) = 0 VAR ST = PT = 600 VA P 600 W Fp = T = =1 ST 600 VA 11.
a-c.
XL = wL = (400 rad/s)(0.1 H) = 40 Ω XC = = 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° = 1.78 A –-64.02°
Is =
V2 = Is(Z2 || Z3) = (1.78 A –-64.02°)(19.21 Ω –-50.19°) = 34.19 V –-114.21°
d.
= 1.37 A –-24.21°
I3 =
= 1.14 A –-114.21°
Z 1:
P = 0 W, QL =
= (1.78 A)2 40 Ω = 126.74 VAR(L), S = 126.74 VA
Z 2:
P = 0 W, QC =
= (1.37 A)2 25 Ω = 46.92 VAR(C), S = 46.92 VA
Z 3:
P=
= (1.14 A)2 30 Ω = 38.99 W, QR = 0 VAR, S = 38.99 VA
PT = 0 + 0 + 38.99 W = 38.99 W QT = +126.74 VAR(L) - 46.92 VAR(C) + 0 = 79.82 VAR(L) ST = Fp =
e.
-
f.
WR = f1 =
272
I2 =
= 88.83 VA = 0.439 (lagging)
= 0.31 J = 63.69 Hz
CHAPTER 20
g.
WL =
= 0.32 J
WC = 12.
a.
= 0.12 J
15000 VA = 68.18 A 220 V FP = 0.6 fi 53° leading Is leads E by 53° 220 V –0° ZT = = 3.228 Ω –-53° 68.18 A –53° Is =
ZT = R − JXC = 1.937 Ω - J2.582 Ω b.
13.
14.
a.
Fp =
fi PT = FpST = (0.6)(15,000 VA) = 9000 W
75000 VA = 50 A 150 V Fp = 0.9 fi 25.84° (lagging) Is leads E by 25.84° E = 150 V –0°, Is = 50 A –-25.84° 150 V –0° = 3 Ω –25.84° = 2.7 Ω + j1.038 Ω = R + jXL ZT = 50 A –−25.84° I=
b.
P = S cos θ = (75000 VA)(0.9) = 6750 W
a.
PT = 0 + 300 W = 300 W QT = 600 VAR(C) + 200(L) = 400 VAR(C) ST =
= 500 VA
Fp = b.
= 0.6 (leading)
Is =
= 16.67 A
Fp = 0.6 fi 53.13° Is = 16.67 A –53.13° c.
-
d.
Load: 600 VAR(C), 0 W R = 0, L = 0, QC = I2XC fi XC =
= 2.159 Ω
Load: 200 VAR(L), 300 W C = 0, R = P/I2 = 300 W/(16.67 A)2 = 1.079 Ω XL =
= 0.7197 Ω
ZT = -j2.159 Ω + 1.0796 Ω + j0.7197 Ω = 1.08 Ω - j1.44 Ω CHAPTER 20
273
15.
a.
PT = 0 + 300 W + 600 W = 900 W QT = 500 VAR(C) + 0 + 500 VAR(L) = 0 VAR ST = PT = 900 VA Fp =
b.
Is =
c.
-
=1
= 9 A, Is = 9 A –0°
d.
Z 1:
QC =
= 20 Ω = 5A –90°
I1 =
I2 = Is - I1 = 9 A - j5 A = 10.296 A –-29.05° Z 2:
R=
= 2.83 Ω
XL,C = 0 Ω Z 3:
R=
= 5.66 Ω
XL = 16.
a.
= 4.72 Ω, XC = 0 Ω
PT = 200 W + 30 W + 0 = 230 W QT = 0 + 40 VAR(L) + 100 VAR(L) = 140 VAR(L) ST =
= 269.26 VA
Fp =
b.
Is =
= 0.854 (lagging)
31.35°
= 2.6926 A
Is = 2.69 A –-31.35°
274
CHAPTER 20
c.
Z 1:
R=
= 50 Ω
XL,XC = 0 Ω I1 =
= 2 A –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° Z 2:
R=
= 14.63 Ω, XL =
= 19.50 Ω
XC = 0 Ω Z 3: 17.
a.
XL =
= 48.76 Ω, R = 0 Ω, XC = 0 Ω
PT = 200 W + 1000 W = 1200 W
Load 1. = ∅1 36.87° FP 0.8 (lagging); ⇒=
sin ∅1 =0.6 ∅ = = 1 P1 tan ∅ 1 200 W(tan 36.87°) = 150 VAR(C) Load 2. = FP 0.4 (leading); ⇒ = ∅ 2 66.42°
P2 tan ∅ ∅ = = 2 2 1000 W(tan 66.42°) = 2291.28 VAR(C ) ∅T = ∅1 + ∅ 2 = 150 VAR(C ) + 2291.28 VAR(C) = 2441.28 VAR(C) 2 S= PT2 + ∅= T T
F= P
2 (1200 W) 2 + (2441.28 VAR)= 2720.27 VA
PT 1200 W = = 0.441 (leading) ST 2720.27 VA
⇒= ∅ 63.83° b.
2720.27 V = 453.38 V 6A E = 453.38 V –-68.83°
ST = EI fi E =
c.
CHAPTER 20
275
I1(load1) =
2 2 S1 S1 P1 + ∅1 = = = V1 E E
(200) 2 + (150 VA) 2 = 0.5514 A 453.38 V
2 2 (1000) 2 + (2291.28 VA) 2 S2 S2 P2 + ∅ 2 = = = = 5.514 A 453.38 V V2 E E Q 150 VAR 1 2 Q = = = = 493.35 W 1 I1 QC1 ; X C 1 2 (0.5514 A) 2 I
I 2(load= 2)
Z1:
1
P1 200 W R = = = 657.80 W 1 2 I1 (0.5514 A) 2 Z1= R1 − jX C1= 657.80 W − j 493.35 W P2 1000 W 32.89 W Z 2 =jX R2 − C 2 ; R2 = = = 2 I1 (5.514 A) 2 X C= 2
Q2 2291.28 VAR = = 75.36 W (5.514 A) 2 I12
Z 2 = 32.89 W − j 75.36 W 18.
a.
b.
0.7 fi 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 = XC = XC =
c.
= 6.059 Ω
1 2p fC
C=
1 1 = 438 µF = 2p fX C (2p )(60 Hz)(6.059 Ω)
Uncompensated: Is =
= 48.08 A
Compensated: Is = d.
= 33.65 A cos θ = 0.9 θ = cos-10.9 = 25.842° tan θ = x = (7 kW)(tan 25.842°) = (7 kW)(0.484) = 3.39 kVAR y = (7.14 - 3.39) kVAR = 3.75 kVAR
276
CHAPTER 20
QC = 3.75 kVAR = XC = C=
= 11.537 Ω
1 1 = 230 µF = 2p fX C (2p )(60 Hz)(11.537 Ω)
Uncompensated: Is = 48.08 A Compensated: ST =
= 7.778 kVA
Is =
= 37.39 A
∆Is = 48.08 A - 37.39 A = 10.69 A 19.
a.
PT = 8 kW, QT = 9 kVAR(L) − 3 kVAR(C) = 6 kVAR(L) = 10 kVA
ST = b.
Fp =
c.
Is =
d.
XC =
8 kW = 0.8 (lagging) 10 kVA 10,000 kW = 50 A 200 V
1 , Q C = I 2X C = 2pfC and
XC =
(200 V)2
C= e.
(200 V)2 40,000
= 6.667 Ω (6.667 Ω)
= 397.867 µF
ST = EIs = PT Is =
CHAPTER 20
8000 W = 40 A 200 V
277
20.
a.
Load 1: Load 2:
P = 20,000 W, Q = 0 VAR θ = cos-10.7 = 45.573° tan θ =
Load 3:
θ = cos-10.85 = 31.788°
x = (10 kW)tan 45.573° = (10 kW)(1.02) = 10,202 VAR(L)
tan θ = x = (5 kW)tan 31.788° = (5 kW)(0.62) = 3098.7 VAR(L) 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.
= 37.442 kVA
QC = QL = 13,300.7 VAR XC = C=
c.
= 75.184 Ω
1 1 = 35.28 µF = 2p fX C (2p )(60 Hz)(75.184 Ω)
Uncompensated: Is =
= 37.44 A
Compensated: ST = PT = 35 kW Is =
= 35 A
DIs = 37.44 A - 35 A = 2.44 A 21.
278
a.
ZT = R1 + R2 + R3 + jXL - jXC = 4 Ω + 3 Ω + 2 Ω + j3 Ω - j12 Ω = 9 Ω - j9 Ω = 12.728 Ω –-45° 60 V –0° = 4.714 A –+45° I= 12.728 Ω –−45° P = VI cos θ = (60 V)(4.71 A) cos 45° = 199.83 W CHAPTER 20
b.
22.
23.
a-b: P = I2R = (4.71 A)2 4 Ω = 88.74 W b-c: P = I2R = (4.71 A)2 3 Ω = 66.55 W a-c: 88.74 W + 66.55 W = 155.29 W a-d: 155.29 W c-d: 0 W d-e: 0 W f-e: P = I2R = (4.71 A)2 2 Ω = 44.368 W = 44.37 W
a.
ST = 600 VA = EIs 600 VA = 5.455 A Is = 110 V θ = cos-10.85 = 31.79° E = 110 V –0°, Is = 5.455 A –-31.79° P = EI cos θ = (110 V)(5.455 A)(0.85) = 510.04 W Wattmeter = 510.04 W, Ammeter = 5.455 A, Voltmeter = 110 V
b.
ZT =
a.
R= XL = L=
110 V –0° = 20.16 Ω –31.79° = 17.14 Ω + j10.62 Ω = R + jXL 5.455 A –−31.79° = 5 Ω, ZT = 2
ZT - R2 = (50 Ω)2 - (5 Ω)2 = 49.75 Ω XL 49.57 Ω = 132.03 mH = (2p )(60 Hz) 2p f
b.
R=
= 10 Ω
c.
R=
= 15 Ω, ZT =
XL = L= 24.
a.
= 50 Ω
= 100 Ω = 98.87 Ω
= 262.39 mH
XL = 2πfL = (6.28)(50 Hz)(0.08 H) = 25.12 Ω ZT = I=
= 25.44 Ω = 2.358 A
P = I2R = (2.358 A)2 4 Ω = 22.24 W b.
I=
= 2.07 A
ZT =
= 28.99 Ω
CHAPTER 20
279
XL = L= c.
XL 28.13 Ω = 89.54 mH = 2p f (2p )(50 Hz)
P = I2R = (1.7 A)2 10 Ω = 28.9 W ZT = XL = L=
280
= 28.13 Ω
= 35.29 Ω = 33.84 Ω = 107.77 mH
CHAPTER 20
Chapter 21 1.
a.
ws =
(1.5 H)(10 F) 258.19 rad/s
fs = b.
ws =
c.
1851.85 rad/s
ws =
a.
= 294.73 Hz = 20,203.05 rad/s
20,203.05 rad/s
= 3215.42 Hz
XC = 30 Ω, XL = Xc at resonance =5Ω
b.
3.
= 1851.85 rad/s
(0.35 mH)(7.0 F)
fs = 2.
= 41.09 Hz
(0.81 H)(0.36 F)
fs =
= 258.19 rad/s
c.
d.
VR = IR = (12 mA)(5 Ω) = 60 mV = E VL = IXL = (12 mA)(30 Ω) = 360 mV VC = IXC = (12 mA)(30 Ω) = 360 mV VL = VC
e.
Qs =
a.
XL = 2 kΩ
b.
I=
c.
VR = IR = (120 mA)(100 Ω) = 12 V = E VL = IXL = (120 mA)(2 kΩ) = 240 V VC = IXC = (120 mA)(2 kΩ) = 240 V VL = VC = 20 VR
d.
Qs =
e.
XL = 2πfL, L = XC =
CHAPTER 21
5Ω
= 6 (low Q)
f.
I=
60 mV = 12 mA 5Ω
P = I2R = (12 mA)2 5 Ω = 0.72 mW
= 120 mA
= 20 (high Q)
,C=
= 63.7 mH = 15,920 pF
281
4.
f.
BW =
g.
f2 = fs +
= 5 kHz +
= 5.13 kHz
f1 = fs -
= 5 kHz -
= 4.88 kHz
a.
fs =
fi
b.
XL = 2πfL = 2π(1.8 kHz)(3.91 mH) = 44.2 Ω
= 250 Hz
= 3.91 mH
XC =
= 44.2 Ω
XL = XC c.
Erms = (0.707)(20 mV) = 14.14 mV Irms =
= 3.01 mA
d.
P = I2R = (3.01 mA)2 4.7 Ω = 42.58 µW
e.
ST = PT = 42.58 µVA
g.
Qs =
= 191.49 Hz
é ù 2 4 ú 1 ê R 1 æRö + f2 = ç ÷ + LC ú 2p ê2L 2 è L ø êë úû é ù 2 1 æ 4.7 Ω ö 4 1 ê 4.7 Ω ú + = ç ÷ + 2p ê2(3.91 mH) 2 è 3.91 mH ø (3.91 mH)(2 mF) ú êë úû 1 601.02 + 11.324´103 = 2p = 1897.93 Hz é ù 2 4 ú 1 ê R 1 æRö + f1 = ç ÷ + LC ú 2p ê 2L 2 è L ø êë úû 1 -601.02 + 11.324´103 = 2p = 1.71 kHz
[
]
[
PHPF =
282
Fp = 1
= 9.4
BW =
h.
f.
]
Pmax =
(42.58 µW) = 21.29 µW
CHAPTER 21
5.
6.
a.
BW = fs/Qs = 4500 Hz/15 = 300 Hz
b.
f2 = fs +
= 4500 Hz + 150 Hz = 4650 Hz
f1 = fs -
= 4500 Hz - 150 Hz = 4350 Hz
c.
Qs =
fi XL = QsR = (15)(4 Ω) = 60 Ω = XC
d.
PHPF =
Pmax =
a.
L=
(I2R) = ( 0.6 A)2 4Ω = 720 mW
= 3.185 mH
@ 250 Hz
BW =
= 40, BW =
or Qs = b.
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.
Qs =
d.
I=
= 250 Hz
= 40
= 6 A –0°, VL = (I –0°)(XL –90°)
= (6 A –0°)(200 Ω –90°) = 1200 V –90° VC = (I –0°)(XC –-90°) = 1200 V –-90°
7.
e.
P = I2R = (6 A)2 5 Ω = 180 W
a.
BW =
b.
Qs =
c.
L= C=
CHAPTER 21
fi Qs = fs/BW = 2500 Hz/250 Hz = 10
fi XL = QsR = (10)(5 Ω) = 50 Ω 50 = 3.185 mH (2500 Hz) (2500 Hz)(50 Ω)
= 1.274 µF
283
d. 8.
f2 = fs + BW/2 = 2500 Hz + (250 Hz/2) = 2625 Hz f1 = fs - BW/2 = 2500 Hz - (250 Hz/2) = 2375 Hz
a.
BW = 6000 Hz - 5600 Hz = 400 Hz
b.
BW = fs/Qs fi fs = QsBW = (14.5)(400 Hz) = 5800 Hz
c.
Qs =
d.
L=
fi XL = XC = QsR = (14.5)(2.5 Ω) = 36.25 Ω 36.25 Ω = 9.952 ´ 10−4 H = 0.995 mH (5800 Hz)
C=
9.
IM =
(5800 Hz)(36.25 Ω)
fiR=
=
= 0.757 µF
= 10 Ω
BW = fs/Qs fi Qs = fs/BW = 8400 Hz/120 Hz = 70 Qs =
fi XL = QsR = (70)(10 Ω) = 700 Ω
XC = XL = 700 Ω X 700 Ω L= L = = 13.26 mH (2p )(8.4 kHz) 2pf 1 1 = = 27.07 nF C= 2pfX C (2p )(8.4 kHz)(0.7 kΩ) f2 = fs + BW/2 = 8400 Hz + 120 Hz/2 = 8.46 kHz f1 = fs - BW/2 = 8400 Hz - 60 Hz = 8.34 kHz 10.
Qs =
fi XL = QsR = 20(3 Ω) = 60 Ω = XC
BW =
fi fs = QsBW = (20)(600 Hz) = 12 kHz
L= C=
60 Ω = 796.18 µH (6.28)(12 kHz) (12 kHz)(60 Ω)
= 221.16 nF
f2 = fs + BW/2 = 12,000 Hz + 600 Hz/2 = 12.3 kHz f1 = fs - BW/2 = 12,000 Hz - 600 Hz/2 = 11.7 kHz
284
CHAPTER 21
11.
a.
fs =
= 1 MHz
= 0.16 fi BW = f2 - f1 = 0.16 fs = 0.16(1 MHz) = 160 kHz
b. c.
fiR=
P= BW =
fi
= 0.716 mH
fiC=
fs =
12.
= 720 Ω
d.
=
a.
=
= 80 fi
= 35.38 pF
=
= 56.23 Ω
=
= 50.27 Ω = 0.2 Qs = R=
= 125.66 Ω
R = Rd + 125.66 Ω = Rd + 50.27 Ω and Rd = 125.66 Ω - 50.27 Ω = 75.39 Ω c.
XC =
= XL C=
13.
a.
fp =
1 1 = = 253.3 pF 2pfX C 2p (1 MHz)(628.32 Ω)
(0.2 mH)(10 nF)
= 112.597 kHz
b.
Vc at Resonance (Z = ∞Ω) = (4 kΩ)(2 A) = 8 V CHAPTER 21
285
c.
8V
IL =
8V
IC =
14.
d.
Qp =
a.
fs =
4 kΩ
8V = 56.57 mA 141.42 Ω 8V = 56.57 mA 141.42 Ω 4 kΩ = 28.28 141.42 Ω = 13.4 kHz
= 49.46 ≥ 10 (yes)
b.
=
c.
Since
d.
XL = 2πfpL = 2π(13.4 kHz)(4.7 mH) = 395.72 Ω
≥ 10, fp @ fs = 13.4 kHz
XC =
= 395.91 Ω
XL = XC e. f. g.
= VC =
286
= (10 mA)(19.57 kΩ) = 195.7 V
≥ 10, Qp = BW =
h.
= (49.46)2 8 Ω = 19.57 kΩ
IL = IC =
= 49.46 = 270.9 Hz = (49.46)(10 mA) = 494.6 mA
CHAPTER 21
15.
a.
fs =
b.
= 1.027 MHz
=
= 86.04 Ω
Z p=
c.
= 19.36 kΩ
2
2
P = I R = (120 mA) (950.9 Ω) = 13.69 W
d.
XL = 2πfL = 2π(1.027 MHz)(200 µH) = 1.291 kΩ (86.04 Ω Ð 0°)(114.1 V) = = 7.587 V 86.04 Ω + j1.291 kΩ P=
= 669 mW
13.69 W: 669 mW @ 20:1 16.
= 5 £ 10
=
a.
fi XC =
\
b.
ZT = Rs || Rp = Rs ||
c.
E= IC =
= 104 Ω
2000 Ω
= 412.69 Ω
= (8 mA –0°)(412.69 Ω –0°) = 3.302 V –0° 3.302 V –0°
= 31.75 mA –90°
ZL = 20 Ω + j100 Ω = 101.98 Ω –78.69° 3.302 V –0° IL = = 32.38 mA –-78.69°
CHAPTER 21
287
d.
L= C=
e.
1 2pfX C
Qp =
= 636.94 µH 2p(25000 Hz) 1 = = 61.24 nF 2p(25 kHz)(104 Ω)
412.69 Ω
= 3.97
BW = fp/Qp = 25,000 Hz/3.97 = 6300.13 Hz
17.
æ 2´106 ö Hz ÷÷ (1 mH) 2p çç X 2000 Ω X è 2p ø = Qt = 35 = L Þ Rℓ = L = = 57.14 Ω 35 Rℓ 35 35 fp 2´10 6 / 2p Hz Q ℓ ³10 : Q p = = = 20 100, 000 Hz BW
And 40,000 = So R = 93.33 kΩ fi use R = 91 kΩ (standard value) 1 1 = Qp ≥ 10, XC = XL = 2000 Ω = 6 æ ö 2p fC 2´10 2p çç Hz ÷÷ C è 2p ø C = 250 pF fi use C = 240 pF (standard value) 18.
a.
fs = fp = fs
= 102.73 kHz = 102.73 kHz
= 102.73 kHz(.99958) = 102.69 kHz
fm = fs
= 102.73 kHz(0.99989) = 102.72 kHz
Since fs @ fp @ fm fi high Qp b.
XL = 2πfpL = 2π(102.69 kHz)(80 µH) = 51.62 Ω XC =
= 51.66 Ω
XL @ XC
288
CHAPTER 21
= Rs ||
c.
=
= 34.41 = 1.51 kΩ
d.
Qp =
= 29.25
BW = e.
= 3.51 kHz
Converting the voltage source to a current source: Is =
= 10 mA
And Rs = Rp = 10 kΩ Then IT =
= 8.49 mA
IC = IL @
19.
f.
VC =
a.
fs =
= (34.41)(8.49 mA) = 292.14 mA = (10 mA)(1.51 kΩ) = 15.1 V = 7.12 kHz
fp = fs
= 7.12 kHz
fm = fs
= 7.12 kHz(0.9338) = 6.65 kHz
= 7.12 kHz
1-
1 é(8 Ω)2 (1 mF) ù ê ú = 7.12 kHz (0.9839) 4 êë 0.5 mH úû = 7.01 kHz
Low Qp b.
XL = 2πfpL = 2π(6.647 kHz)(0.5 mH) = 20.88 Ω 1 1 = XC = = 23.94 Ω 2pfC 2p (6.647 kHz)(1 mF) XC > XL (low Q) = Rs || Rp = Rs ||
c.
= 500 Ω ||
= 500 Ω || 62.5 Ω
= 55.56 Ω d.
Qp =
CHAPTER 21
= 2.32
289
BW =
e.
= 2.87 kHz
One method: VC =
= (40 mA)(55.56 Ω) = 2.22 V
IC =
= 92.73 mA
IL = f. 20.
= 99.28 mA
VC = 2.22 V = 40 kΩ
a. (40 Ω)2 +
X L= b.
= (40 kΩ)(40 Ω)
160 × 104 − 1.6 ×103=
1600 × 103 − 1.6 ×103= 1264.28 Ω
1264.28 Ω = 31.607 ≥ 10 40
Q=
\ XC = XL = 1264.28 Ω
21.
c.
XL = 2πfpL fi fp =
d.
XL =
X L 1264.28 Ω = = 11.18 kHz 2pL 2p(18 mH)
fi
a.
= 20 > 10
b.
=
(11.18 kHZ)(1264.28 Ω) fp = fs =
= 11.27 nF
= 3558.81 Hz
X L 2pfL 2pfL 2p (3558.81 Hz)(0.2 H) = = fi Rℓ = = 223.61 Ω 20 Rℓ Rℓ Qℓ
= Rs ||
= Rs ||
= 20 kΩ || (20)2 223.61 Ω
= 16.345 kΩ Converting the voltage source to a current source: 120 V = 6 mA 20 kΩ Rp = Rs = 20 kΩ = (6 mA)(16.345 kΩ) = 98.07 V VC =
290
CHAPTER 21
c.
P = I2R = (6 mA)216.345 kΩ = 588.42 mW
d.
Qp = BW =
22.
a.
16.345 kΩ fp Qp
=
c.
Qp =
= 50
= 25 fi fp = QpBW = (25)(1000 Hz) = 25 kHz
BW = =
d.
23.
a.
3558.81 Hz = 975.02 Hz 3.65
Ratio of XC to suggests high Q system. \ XL = 400 Ω = XC
b.
e.
=
= 3.65
= (0.1 mA)(10 kΩ) = 1 V
f2 = fp + BW/2 = 25 kHz +
= 25.5 kHz
f1 = fp - BW/2 = 25 kHz -
= 24.5 kHz
XC =
fi
- X LX C +
=0
- 100 XL + 144 = 0 XL = = 50 Ω ±
= 50 Ω ± 48.54 Ω
XL = 98.54 Ω or 1.46 Ω b.
=
c.
Qp = =
= 8.21
= 8.05
BW = fp/Qp fi fp = QpBW = (8.05)(1 kHz) = 8.05 kHz
CHAPTER 21
291
=
d. e.
24.
a.
= (6 mA)(804.66 Ω) = 4.83 V
f2 = fp + BW/2 = 8.05 kHz +
= 8.55 kHz
f1 = fp - BW/2 = 8.05 kHz -
= 7.55 kHz
fs =
= 41.09 kHz
fp = fs
= 41.09 kHz
fm = fs
= 41.09 kHz(0.9978) = 41 kHz
= 41.09 kHz
= 41.09 kHz(0.0995) = 41.07 kHz
High Qp b.
= 4 mA –0°, Rs = 20 kΩ
I= =
X L 2pfL 2p (41 kHz)(0.5 mH) = = = 21.47 (high Q coil) 6Ω Rℓ Rℓ
Qp =
= 18.86 (high Qp)
= c.
= Rs || Rp = 20 kΩ || 2.771 kΩ = 2.43 kΩ
d.
VC =
e.
BW =
= 2.17 kHz
f.
XC =
1 1 = = 129.39 Ω 2pfC 2p (41 kHz)(30 nF)
= (4 mA)(2.43 kΩ) = 9.74 V
IC = IL =
292
= 75.25 mA = 75.50 mA
CHAPTER 21
25.
fi
=
(25 kHz)(2.5 mH) = 4.36 Ω 90
=
BW = fp/Qp fi Qp = fp/BW = 25 kHz/1.84 kHz = 13.59 f p @ fs =
High Q\
fiC=
= 16.21 nF
13.59 = 5.377 kΩ = 5.34 kΩ (25 kHz)(16.21 nF)
fi R = Q pX C =
Qp =
(25 kHz)2(2.5 mH)
= (90)2 4.36 Ω = 35.32 kΩ
Rp =
R = Rs || Rp =
35.32 kΩ − 5.34 kΩ 2.2 V
fi
26.
(35.32 kΩ)(5.34 kΩ)
fi Rs =
Qp =
= 6.29 kΩ
= 11 kΩ 11 kΩ = 343.75 Ω 32
fi
XL = XC = 343.75 Ω fi fp = QpBW = (32)(600 Hz) = 19.2 kHz
BW =
343.75 Ω = 2.85 mH (19.2 kHz)
L=
1 1 = = 24.11 nF 2pfX C 2p(19.2 kHz)(343.75 Ω)
C=
(Rs= • Ω) =
Qp =
27.
a.
343.75 Ω = 10.74 Ω 32
fi
fs =
= 251.65 kHz = 15.81 ≥ 10
= fp = fs = 251.65 kHz = Rs ||
b.
c.
Qp =
d.
BW =
CHAPTER 21
= 40 kΩ || (15.81)2 20 Ω = 4.44 kΩ
= 14.05
= 17.91 kHz
293
e.
20 µH, 20 nF fs the same since product LC the same fs = 251.65 kHz =
= 1.581
Low
: fp = fs = (251.65 kHz)(0.775) = 194.93 kHz XL = 2πfpL = 2π(194.93 kHz)(20 µH) = 24.496 Ω Rp =
= 50 Ω
= Rs || Rp = 40 kΩ || 50 Ω = 49.94 Ω Qp =
= 2.04
BW =
f.
= 95.55 kHz
0.4 mH, 1 nF fs = 251.65 kHz since LC product the same = 31.62 ≥ 10
=
\ fp = fs = 251.65 kHz = Rs || Qp = BW =
g.
h.
Network
= 40 kΩ || (31.62)2 20 Ω = 40 kΩ || (@ 20 kΩ) @ 13.33 kΩ = 21.08 = 11.94 kHz
= 100 ¥ 103
part (e)
= 1 ¥ 103
part (f)
= 400 ¥ 103
Yes, as Also, Vp =
ratio increased BW decreased. and for a fixed I,
and therefore Vp will increase with increase in the
L/C ratio.
294
CHAPTER 21
Chapter 22 1.
a.
left:
d1 =
≤ = 0.1875≤, d2 = 1≤
Value = 103 ¥ = 103 ¥ 1.54 = 1.54 kHz 1 center: d1 = ≤ = 0.5≤, d2 = 1≤ 2 Value = 103 ¥ 100.5/1 = 103 ¥ 10.5 = 103 ¥ 3.16 = 3.16 kHz right:
d1 =
≤ = 0.75≤, d2 = 1≤ Value = 103 ¥ = 103 ¥ 5.623 = 5.62 kHz
bottom: d1 =
b.
≤ = 0.3125≤, d2 =
≤ = 0.9375≤
Value = 10-1 ¥ = 10-1 ¥ 2.153 = 0.22 V center: d1 =
top:
= 10-1 ¥ 100.333
7.5 ≤ = 0.469≤, d2 = 0.9375≤ 16 Value = 10-1 ¥ 100.469/0.9375 = 10-1 ¥ 100.5 = 10-1 ¥ 3.16 = 0.316 V d1 =
≤ = 0.6875≤, d2 = 0.9375≤ Value = 10-1 ¥ = 10-1 ¥ 5.248 = 0.52 V
= 10-1 ¥ 100.720
a.
5
b.
-4
c.
8
d.
-6
e.
1.30
f.
3.94
g.
4.75
h.
-0.498
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 54 = 1.732
2.
3.
CHAPTER 22
2.996
d.
9.07
295
log10 89 + log10 6 = 0.954 + 0.778 = 1.732 6.
log10 0.4 = -0.398 log10 18 - log10 45 = 1.255 - 1.653 = -0.398
7.
log10 0.25 = -0.602 -log10 4 = -(0.602) = -0.602 −log10 1/0.25 = −(0.602) = −(0.602) -log10 3125 = 3.495 5 log10 5 = 5(0.699) = 3.495
8.
log10 3125 = 3.495 5 log10 5 = 5(0.699) = 3.495
9.
a.
bels = log10
b.
dB = 10 log10
10.
= log10
320 MW = log10 64 = 1.806 5 MW
= 10(log10 64) = 10(1.806) = 18.06
dB = 10 log10 8 dB = 10 log10
150 W 150 W
0.8 = log10 x, where x = 150 W x = 6.309 = P1 =
150 W = 23.77 W 6.309
11.
dB = 10 log10
12.
dBm = 10 log10 dBm = 10 log10
13.
dBv = 20 log10
14.
dBυ = 20 log10 26 = 20 log10
= 10 log10
220 mW
= 10 log10 20 = 13.01
= 10 log10 220 = 23.42
= 20 log10
18.8 V 0.2 V
= 20 log 10 94 = 39.46
15 mV
1.3 = log10 x; where x =
15 mV
x = 19.953 =
15 mV Vo = 299.29 mV 296
CHAPTER 22
15.
dBs = 20 log10 dBs = 20 log10
0.002
= 20
0.032
= 44.08
dBs = 20 log10
Increase = 24.08 dBs 16.
60 dBs fi 90 dBs quiet loud 60 dBs = 20 log10
= 20 log10x
3 = log10x x = 1000 90 dBs = 20 log10
= 20 log10y
4.5 = log10y y = 31.623 ¥ 103 =
=
=
and P2 = 31.62 P1 17.
–
18.
a.
8 dB = 20 log10 0.4 = log10 = 2.512 V2 = (2.512)(0.775 V) = 1.947 V P=
b.
=
= 6.32 mW
-5 dB = 20 log10 -0.25 = log10 = 0.562 V2 = (0.562)(0.775 V) = 0.436 V P=
CHAPTER 22
=
= 0.32 mW
297
19.
a.
1
–-90° + tan-1 XC/R =
Aυ =
–-tan-1 R/XC
2
æ R ö ç ÷ +1 è XC ø fc =
= 3617.16 Hz
f = f c:
Aυ =
f = 0.1fc:
At fc, XC = R = 2.2 kΩ 1 1 é 1 ù 1 = = XC = ê ú = 10[2.2 kΩ] = 22 kΩ 2pfC 2p 0.1 f c C 0.1 êë2pf c C úû 1 1 1 Aυ = = 0.995 = = æ R ö2 æ 2.2 kΩ ö 2 (.1)2 + 1 ç ÷ +1 ç ÷ +1 è XC ø è 22 kΩ ø
f = 0.5fc =
:
= 0.707
XC =
Aυ =
f = 2fc:
1 = 2p fC
1 æf ö 2p ç c ÷ C è2ø
1 æ 2.2 kΩ ö 2 ç ÷ +1 è 4.4 kΩ ø
=
298
= 0.894
1 2
=
1 (2)2 + 1
= 0.447
XC = Aυ =
θ = -tan-1 R/XC f = f c:
(0.5)2 + 1
= 1.1 kΩ
æ 2.2 kΩ ö ç ÷ +1 è1.1 kΩ ø
b.
1
XC = Aυ =
f = 10fc:
é 1 ù = 2ê ú = 2[2.2 kΩ] = 4.4 kΩ êë2p f c C úû
= 0.22 kΩ
1 æ 2.2 kΩ ö 2 ç ÷ +1 è 0.22 kΩ ø
=
1 (10)2 + 1
= 0.0995
θ = -tan-1 = -45°
f = 0.1fc:
θ = -tan-1 2.2 kΩ/22 kΩ = -tan-1
= -5.71°
f = 0.5fc:
θ = -tan-1 2.2 kΩ/4.4 kΩ = -tan-1
= -26.57°
f = 2fc:
θ = -tan-1 2.2 kΩ/1.1 kΩ = -tan-1 2 = -63.43° CHAPTER 22
f = 10fc: 20.
a.
θ = -tan-1 2.2 kΩ/0.22 kΩ = -tan-1 10 = -84.29°
1 1 = = 248.68 Hz 2p RC 2p(3.2 kΩ)(0.2 mF) f = 2fc = 497.36 Hz 1 1 = XC = = 1.5999 kΩ 2p fC 2p(497.36 Hz)(0.2 mF) V 1.5999 kΩ XC Aυ = o = = 0.4469 = 2 2 Vi R +X (3.2 kΩ)2 + (1.5999 k Ω)2 fc =
C
Vo = 0.4469 V; Vi = 0.4469(10 mV) = 4.47 mV b.
(248.68 Hz) = 24.868 Hz
f=
1 1 = = 31.99 kΩ 2p fC 2p(248.68 Hz)(0.2 mF) V 31.99 kΩ XC = Aυ = o = = 0.995 2 2 2 2 Vi (3.2 kΩ) + (31.99 kΩ) R + XC Vo = 0.995 V; Vi = 0.995(10 mV) = 9.95 mV XC =
c.
Yes, at f = fc, Vo = 7.07 mV at f =
, Vo = 9.95 mV (much higher)
at f = 2fc, Vo = 4.47 mV (much lower) 21.
fc = 500 Hz = C= Aυ =
= 0.265 µF
Vo = Vi
1 æ R ö2 ç ÷ +1 è XC ø
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 At f = 250 Hz =
fc, θ = -26.54°
At f = 1 kHz = 2fc, θ = -63.41°
CHAPTER 22
299
22.
= 70.74 kHz
a.
fc =
b.
f = 0.1 fc = 0.1(70.74 kHz) 7.074 kHz 1 1 = = 49.997 kΩ XC = 2pfC 2p(7.074 kHz)(450 pF) 49.997 kΩ = 0.995 Aυ = 2 2 (5 kΩ) + (49.997 kΩ)
c.
d.
2p(5 kΩ)(450 pF)
f = 10fc = 707.4 kHz 1 XC = = 499.97 Ω 2pfC 2p (7.074 kHz)(450 pF) 499.97 Ω = 0.0995 Aυ = 2 2 (5 kΩ) + (49.997 kΩ) Aυ =
1
0.1
= 0.01 = = 100 XC
R2 + R2 = 104 XC = XC =
300
= 104 -
= 9.999 5 kΩ 50 Ω
fif=
(50 Ω)(450 pF)
= 7.07 MHz
CHAPTER 22
23.
a.
1
–tan-1 XC/R =
Aυ =
æX ö 1+ ç C ÷ è R ø fc =
–tan-1 XC/R
= 3.62 kHz
f = f c:
Aυ =
f = 2fc:
At fc, XC = R = 2.2 kΩ ù 1 1 1é 1 1 = = ê XC = ú = [2.2 kΩ] = 1.1 kΩ 2pfC 2p (2 f c )C 2 êë2p (2 f c )C úû 2 1 Aυ = = 0.894 æ1.1 kΩ ö 2 1+ ç ÷ è 2.2 kΩ ø
f=
f c:
XC =
Aυ =
f = 10fc:
= 0.707
1 æf ö 2p ç c ÷ C è2ø 1
f c:
f=
é 1 ù = 2ê ú = 2[2.2 kΩ] = 4.4 kΩ êë2pf c C úû
æ 4.4 kΩ ö 2 1+ ç ÷ è 2.2 kΩ ø
= 0.447
XC =
= 0.22 kΩ
1
Aυ =
= 0.995 æ 0.22 kΩ ö 2 1+ ç ÷ è 2.2 kΩ ø é 1 ù 1 = 10 ê XC = ú = 10[2.2 kΩ] = 22 kΩ æf ö êë2pf c C úû c 2p ç ÷ C è 10 ø Aυ =
b.
2
1 æ 22 kΩ ö 2 1+ ç ÷ è 2.2 kΩ ø
= 0.0995
f = f c,
θ = 45°
f = 2fc,
θ = tan-1 (XC/R) = tan-1 1.1 kΩ/2.2 kΩ = tan-1
f=
f c,
θ = tan-1
f = 10fc,
θ = tan-1
CHAPTER 22
= 26.57°
= tan-1 2 = 63.43° = 5.71°
301
f=
24.
θ = tan-1
f c,
a.
f = f c: A υ =
b.
fc =
= 84.29°
= 0.707
1 1 = = 23.544 kHz 2p RC 2p(130 kΩ)(52 pF)
f = 4fc = 4(23.544 kHz) = 94.176 kHz XC =
1 1 = = 32.499 kΩ 2p fC 2p (94.176 kHz)(52 pF)
Aυ =
Vo 130 kΩ R = = = 0.970 (significant rise) 2 2 2 Vi R + XC (130 kΩ)2 + (32.499 kΩ)
c.
f = 100fc = 100(23.544 kHz) = 23544 kHz = 23.544 MHz 1 1 = 129.998 kΩ = XC = 2 p (23.544 MHz)(52 pF) 2p fC R 130 kΩ Aυ = = 0.99999 1 = 2 2 2 2 + (130 kΩ) + (129.998 kΩ) R XC
d.
At f = fc, Vo= 0.707Vi = 0.707(15 mV) = 10.605 mV 2
2 (10.605 mV) Po = V o = = 0.865 nW R 130 kΩ
25.
Aυ =
Vo = Vi
fc =
1 æX ö 1+ ç C ÷ è R ø
2
0.9 nW
–tan-1 XC/R
fi
= 795.77 Ω
R = 795.77 Ω fi 750 Ω" +## 47$ Ω = 797 Ω !# # nominal values
\ fc = At
= 1996.93 Hz using nominal values f = 1 kHz, Aυ = 0.458 f = 4 kHz, Aυ 0.9 θ = tan-1 f = 1 kHz, θ = 63.4° f = 4 kHz, θ = 26.53°
302
CHAPTER 22
26.
a.
fc =
= 79.58 kHz
b.
f = 0.01fc = 0.01(79.577 kHz) = 0.7958 kHz 796 Hz 1 1 = XC = = 9.997 MΩ 2pfC 2p (796 Hz)(20 pF) Aυ =
c.
= 0.01
f = 100fc = 100(79.577 kHz) 7.96 MHz 1 1 = XC = = 999.72 Ω 2pfC 2p (7.96 MHz)(20 pF) Aυ =
d.
0
= 0.99995
1
Aυ = = 2R R2 +
= 4R2
= 4R2 - R2 = 3R2 XC = XC =
27.
a.
(100 kΩ) = 173.2 kΩ
1 fif= 2pfC f = 45.95 kHz
low-pass section:
= 795.77 Hz
high-pass section:
= 1.94 Hz
For the analysis to follow, it is assumed (R2 +
) || R1
R1 for all frequencies of
interest.
CHAPTER 22
303
At
= 795.77 Hz: = 0.707 Vi
X C2 =
1 = 24.39 kΩ 2pfC2
=
At
= 0.925
Vo = (0.925)(0.707 Vi) = 0.654 Vi = 1.94 kHz: Vo = 0.707
X C1 =
1 = 41 Ω 2pfC1 = 0.925 Vi
= (0.707)(0.925 Vi) = 0.64 Vi At f = 795.77 Hz + = 58.1 Ω,
= 1.37 kHz = 14.17 kΩ = 0.864 Vi
=
= 0.817
Vo = 0.817(0.864 Vi) = 0.706Vi and Aυ =
= 0.706 (@ maximum value)
After plotting the points it was determined that the gain should also be determined at f = 500 Hz and 4 kHz: f = 500 Hz:
= 159.15 Ω,
= 38.82 kΩ,
= 0.532 Vi, Vo = 0.968 f = 4 kHz:
Vo = 0.515 Vi = 19.89 Ω,
= 4.85 kΩ,
= 0.981 Vi, Vo = 0.437 Vo = 0.429 Vi b.
304
Using 0.707(.706) 0.5 to define the bandwidth BW 3.4 kHz - 0.48 kHz = 2.92 kHz and BW 2.9 kHz
CHAPTER 22
æ 2.9 kHz ö with fcenter = 480 Hz + ç ÷ = 1930 Hz è 2 ø
28.
f1 =
= 4 kHz
Choose R1 = 1 kΩ = 39.8 nF \ Use 39 nF
C1 = f2 =
= 80 kHz
Choose R2 = 20 kΩ = 99.47 pF \ Use 100 pF
C2 =
Center frequency = 4 kHz + At f = 42 kHz, Assuming Z2
= 42 kHz
= 97.16 Ω,
= 37.89 kΩ
Z1 = 0.995Vi
= Vo = 0.884 as f = f1:
CHAPTER 22
= 0.884Vi = 0.884(0.995Vi) = 0.88 Vi = 0.707Vi,
= 221.05 kΩ
305
and Vo = 0.996 so that Vo = 0.996
= 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. 29.
a.
fs =
b.
Qs =
= 98.1 kHz
= 16.84
BW =
c.
= 5.83 kHz
At f = fs:
= 0.93 V and Aυ =
Since Qs ≥ 10,
f1 = fs -
= 98.1 kHz -
f2 = fs +
= 101.02 kHz
= 0.93
= 95.19 kHz
At f = 95.19 kHz: XL = 2πfL = 2π(95.19 kHz)(4.7 mH) = 2.81 kΩ XC = Vo =
= 2.99 kΩ
160 Ω(1 V Ð 0°) 160 V Ð 0° = 172 + j2.81 kΩ - j2.99 kΩ 172 - j180
=
= 0.643 V–46.30°
At f = 101.02 kHz: XL = 2πfL = 2π(101.02 kHz)(4.7 mH) = 2.98 kΩ 1 1 = XC = = 2.81 kΩ 2pfC 2p (101.02 kHz)(560 pF) 160 Ω(1 V Ð 0°) 160 V Ð 0° = Vo = 172 + j2.98 kΩ - j2.81 kΩ 172 + j170 = d.
f = f s:
= 0.66 V–-44.66°
= 0.93 V
f = f1 = 95.19 kHz, Vo = 0.707(0.93 V) = 0.66 V f = f2 = 101.02 kHz, Vo = 0.707(0.93 V) = 0.66 V 30.
306
a.
fp =
159.15 kHz
CHAPTER 22
=
= 62.5 = (62.5)2 16 Ω = 62.5 kΩ and Vo
10
4 kΩ
Vi at resonance.
However, R = 3.3 kΩ affects the shape of the resonance curve and BW = fp/ applied. For Aυ =
= 0.707,
cannot be
= R for the following configuration
For frequencies near fp, XL and X = XL || XC.
and ZL =
+ jXL
XL
For frequencies near fp but less than fp X= and for Aυ = 0.707 =R
Substituting XC =
and XL = 2πf1L
the following equation can be derived: =0 For this situation: = 48.23 ¥ 103 = 2.53 ¥ 1010 and solving the quadratic equation, f1 = 135.83 kHz and
= fp - f1 = 159.15 kHz - 135.83 kHz = 22.32 kHz
so that f2 = fp +
CHAPTER 22
= 159.15 kHz + 18.75 kHz = 177.9 kHz
307
b.
Qp =
= 3.57
BW = 2(18.75 kHz) = 37.5 kHz 31.
a.
Qs =
b.
BW =
7000 Ω 7000 Ω = 15.56 440 Ω + 10 Ω 450 Ω 15.56
f1 = 5000 Hz f2 = 5000 Hz +
= 321.34 Hz
321.34 Hz 321.34 Hz
= 4.84 kHz = 5.16 kHz
c.
At resonance 10 Ω(Vi) Vo = 10 Ω + 440 Ω = 0.022 Vi d.
32.
a.
10 Ω || 4 kΩ = 9.975 Ω 9.975 ΩVi @ 0.022Vi as above in part (c) Vo = 9.975 Ω + 440 Ω
At resonance,
=
= 40 = (40)2 20 Ω = 32 kΩ
At resonance, Vo = and Aυ =
1 kΩ = 0.97Vi
= 0.97
For the low cutoff frequency note solution to Problem 30: =0 C= L=
1 1 = = 19.9 nF 2pfX C 2p (20 kHz)(400 Ω) = 3.18 mH
Substituting into the above equation and solving f1 = 16.4 kHz
308
CHAPTER 22
with
= 20 kHz - 16.4 kHz = 3.6 kHz
and BW = 2(3.6 kHz) = 7.2 kHz Qp =
= 2.78
b.
-
c.
At resonance = 32 kΩ || 100 kΩ = 24.24 kΩ with Vo = and Aυ =
= 0.96Vi = 0.96 vs 0.97 above
At frequencies to the right and left of fp, the impedance
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 = 32 kΩ || 20 kΩ = 12.31 kΩ with Vo =
= 0.925Vi vs 0.97 Vi above
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 =
= 726.44 kHz (band-stop) =0 =0 =0
=0
=0 =0
CHAPTER 22
309
=0 L sL pw 2 -
1 [Ls + Lp] = 0 C
= 2.01 MHz (pass-band)
f=
34.
a.
fi Ls =
fs =
= 12.68 mH
XL = 2πfL = 2π(30 kHz)(12.68 mH) = 2388.91 Ω 1 1 = XC = = 26.54 kΩ 2pfC 2p (30 kHz)(200 pF) XC - XL = 26.54 kΩ - 2388.91 Ω = 24.15 kΩ(C) = 24.15 kΩ Lp = 35.
= 128.19 mH
a.
At low frequencies, L1, L2 fi short circuits and C fi open circuit. The result is VL very close to Vi at low frequencies. At high frequencies, XC shorts to ground and X L2 has a high impedance so VL approaches 0 V.
b.
Determine fequency when RL = X L2 due to voltage divide action.
R 220 Ω = = 159.15 kHz 2pL 2p (0.22 mH) Since fc less than split between XL and RL let us try 140 kHz. XL = R fi 2π fL = R fi f =
IT
Z1
Z3
+ E
+
Z2
−
I4 Z4 VL
−
Z1 = X L1 = 2π fL1 = 2π(140 kHz)(0.47 mH) = 413.43 Ω
1 1 = = 227.36 Ω 2p fC 2p (140 kHz)(5 nF) Z3 = X L2 = 2π fL2 = 2π(140 kHz)(0.22 mH) = 193.52 Ω Z4 = R = 220 Ω Z2 = XC =
ZT = Z1 + Z2 || (Z3 + Z4) (227.36 Ω Ð -90°)(220 Ω + j193.52 Ω) = j413.43 Ω + (- j227.36 Ω) + (220 Ω + j193.52 Ω) (227.36 Ω Ð -90°)(293 Ω Ð 41.34°) = j413.43 Ω + 220 Ω - j33.84 Ω 66.61 kΩ Ð -48.66° = j413.43 Ω + 222.59 Ð -8.74° = j413.43 Ω + 299.25 Ω –-39.92°
310
CHAPTER 22
= j413.43 Ω + 229.51 Ω - j192.03 Ω = 229.51 Ω + j221.40 Ω ZT = 318.89 Ω –43.97°
E 10 V Ð 0° = = 31.36 mA –-43.97° ZT 318.89 Ω Ð 43.97°
IT =
Z 2I T (227.36 Ω Ð -90°)(31.36 mA Ð -43.97°) = Z2 + Z3 + Z 4 - j227.36 Ω + j193.52 Ω + 220 Ω 7.13 A Ð -133.97° 7.13 A Ð -133.97° = = 220 - j33.84° 222.59 Ð -8.74° = 32.03 mA –-125.23° VL = I4Z4 = (32.03 mA –-125.23°)(220 Ω –0°) = 7.05 V –-125.23°
I4 =
VL 10 V
0
36.
a.
7.05 V ≅ 7.07 V
140 Hz ≅ fc
f
At very low frequencies XC fi open-circuit and XL fi short-circuit resulting in VL fi 0 V. At very high frequencies XC fi short-circuit and XL fi open-circuit resulting in VL fi 20 V.
b.
Utilize X L2 = XC as a starting point to establish a frequency of application: 2π fL2 =
1 1 1 = 1.45 kHz Þ f = = 2pfC 2p LC 2p (100 mH)(0.12 mF)
Try f = 1 kHz: 1 1 = XC = = 1.33 kΩ 2p fC 2p (1 kHz)(0.12 mF) X L2 = 2π fL2 = 2π(1 kHz)(100 mH) = 628.32 Ω Z¢ = X L2 || RL = (628.32 Ω –90°) || (1.2 kΩ –0°)
(628.32 Ω Ð 90°)(1.2 kΩ Ð 0°) 753.98´103 Ω Ð 90° = 3 1.2 kΩ + j628.32 Ω 1.35´10 Ð 27.64° Z¢ = 558.50 Ω –62.36° =
VL =
(558.50 Ω Ð 62.36°)(20 V Ð 0°) Z¢×E = ¢ 558.50 Ω Ð 62.36° - j1.33 kΩ Z + XC =
CHAPTER 22
11.17´103 V Ð 62.36° 2.591 + j494.76 - j1.33 kΩ
311
11.17´10 3 V Ð 62.36° 11.17´10 3 V Ð 62.36° = 2.591 - j835 874.28 Ð -72.76° VL = 12.78 V –135.12° which is very close to the cutoff value of 0.707(20 V) = 14.14 V. The cutoff frequency will be slightly higher than 1 kHz. =
VL 20 V 14 V 12.78 V 0
37.
a.
1 kHz fc
f
At very low frequencies XL fi short-circuit and XC fi open-circuit. The result is VL fi 60 V. At very high frequencies XL fi open-circuit and XC fi short-circuit. The result is VL fi 0 V.
b.
Defining impedances: Z1
Z3
+ E
Z5
Z4
Z2
−
Source conversion: Z3 E Z1
Z1
Z2
Z4
Z5
New definition: V1 E Z1
Z3 Z′
V2
Z¢ = Z1 || Z2 Z≤ = Z4 || Z5
Z′′
Applying Nodal Analysis é1 1 ù 1 E V1 ê + V2 = úêë Z¢ Z 3 úû Z 3 Z1
312
CHAPTER 22
é1 1 ù 1 V2 ê + V =0 úêë Z¢¢ Z 3 úû Z 3 1 _______________________ é1 1 ù 1 E V1 ê + V2 = úêë Z¢ Z 3 úû Z 3 Z1 é1 ù é1 1 ù -V1 ê ú + ê + ú V2 = 0 ëê Z 3 ûú ëê Z¢¢ Z 3 ûú _______________________
VL = V2 =
=
1 1 + Z¢ Z3 1 Z3 1 1 + ¢ Z Z3 1 Z3
E Z¢ 0 -
1 Z3
1 1 + Z¢¢ Z3
é1 éE ù é 1 ù 1 ù ê ¢+ ú 0 - ê ú êú Z Z3 û Z1 û ë Z3 û ë ë = é1 1 ùé 1 1 ù é 1 ùé 1 ù ê ¢+ ú ê ¢¢ + ú - êú êú Z3 û ë Z3 û ë Z3 û ë Z Z3 û ë Z
[]
E / Z1Z3 é1 ù 1 é1 1 ù 1 ê ¢+ ú ê ¢¢ + ú- 2 Z3 û Z ë Z Z3 û ë Z 3
1 XL to bring VL down to 0.707 level 3 1 1 3 1 = 123.13 kHz = 2π fL and f = = 2p fC 3 2p LC 2p (1 mH)(5 nF) Using f = 120 kHz Z1: XL = 2π fL = 2π(120 kHz)(1 mH) = 754 Ω –90° 1 1 = Z 2: X C = = 265 Ω –-90° 2p fC 2p (120 kHz)(5 nF) Z3: 754 Ω –90° Z4: 265 Ω –-90° Z5: 2.2 kΩ –0° _______________________ Z1◊Z3 = (754 Ω –90°)(754 Ω –90°) = 568.52 ¥ 103 Ω2 –180° E 60 V Ð 0° = 105.5 ¥ 10-6 –-180° = Z1Z3 568.52´103 Ω2 Ð 180° _______________________
Choosing XC =
Z¢ = Z1 || Z2 =
(754 Ω Ð 90°)(265 Ω Ð -90°) + j754 Ω - j265 Ω
199.81´103 Ω Ð 0° = 489 Ð 90° = 408.61 Ω –-90°
CHAPTER 22
313
(265 Ω Ð -90°)(2.2 kΩ Ð 0°) 538´10 3 Ω Ð -90° = 2.2 kΩ - j265 Ω 2.2 Ð -6.87° = 262.61 Ω –-83.13° é1 ù 1 ù é 1 1 + ê ¢+ ú=ê ú ë Z Z3 û ë 408.61 Ω Ð -90° 754 Ω Ð 90° û = 2.44 ¥ 10-3 –90° + 1.33 ¥ 10-3 –-90° = j2.44 ¥ 10-3 - j1.33 ¥ 10-3 = j1.11 ¥ 10-3 é1 1 ù 1 1 + ê ¢¢ + ú= Z3 û 262.61Ð -83.13° 754 Ω Ð 90° ëZ = 3.81 ¥ 10-3 –83.13° + 1.33 ¥ 10-3 –-90° = 0.455 ¥ 10-3 + j3.78 ¥ 10-3 - j1.33 ¥ 10-3 = 0.455 ¥ 10-3 + j2.54 ¥ 10-3 = 2.49 ¥ 10-3 –79.48° [1.11 ¥ 10-3 –90°][2.49 ¥ 10-3 –79.48°] = 2.76 ¥ 10-6 –169.48° _______________________ Z≤ = Z4 || Z5 =
2
Z3 = (754 Ω –90°)(754 Ω –90°) = 5.69 ¥ 10-6 –180° 1 1 = 0.176 ¥10-6 –-180° = 2 6 5.69´10 Ð 180° Z3 _______________________ and 105.5´106 Ð -180° VL = 2.58´106 Ð 168.8° = 40.89 V –-11.2° @ 41 V which is very close to 0.707(60 V) = 42.42 V VL 60 V 42.4 V 41 V
0
38. a, b.
314
fc =
fc 120 Hz
f
= 7.2 kHz
CHAPTER 22
c.
f=
f c:
=
f = 2fc:
=
f=
=
f c:
f = 10fc:
d.
f=
f c:
f = 2fc:
=
= -7 dB
= -0.969 dB = -20.04 dB = -0.043 dB
=
Aυ =
= 0.447
Aυ =
= 0.894
e.
39.
a.
fc =
1 1 1 = 1.83 kHz = = 2p RC 2p (6.8 kΩ || 12 kΩ)0.02 mF 2p (4.34 kΩ)(0.02 mF)
æ 1 and Vo = ç ç 2 è 1 + ( fc / f )
CHAPTER 22
ö ÷V ÷ i ø
315
b.
c. & d.
e.
Remember the log scale! 1.5fc is not midway between fc and 2fc = 20 log10 Aυ -1.5 = 20 log10 Aυ -0.075 = log10 Aυ Aυ =
f.
40. a, b.
θ = tan-1 fc/f
316
–-tan-1f/fc
Aυ = fc =
c.
= 0.84
1 1 = = 13.26 kHz 2p RC 2p (12 kΩ)(1000 pF)
f = fc/2 = 6.63 kHz
CHAPTER 22
= -0.97 dB f = 2fc = 26.52 kHz = -6.99 dB f = fc/10 = 1.326 kHz = -0.04 dB f = 10fc = 132.6 kHz = -20.04 dB
d.
41.
f = fc/2:
Aυ =
f = 2fc:
Aυ =
= 0.894 = 0.447
e.
θ = tan-1 f/fc f = fc/2: θ = -tan-1 0.5 = -26.57° f = f c: θ = -tan-1 1 = -45° f = 2fc: θ = -tan-1 2 = -63.43°
a.
R2 || XC =
æ - jR X ö 2 C ç ÷ Vi R jX C ø R2 X C Vi 2 è Vo = = -j R2 X C R1 ( R2 - jX C ) - jR2 X C R1 - j R2 - jX C CHAPTER 22
317
- jR2 X C Vi - jR2 X C Vi = R1 R2 - jR1 X C - jR2 X C R1 R2 - j( R1 + R2 )X C R2 X C Vi R2 Vi = = jR1 R2 + ( R1 + R2 )X C RR j 1 2 + ( R1 + R2 ) XC æ R2 ö ç ÷V R1 + R2 ø i R2 Vi è = = æ RR ö 1 R1 R2 R1 + R2 + j 1 + jç 1 2 ÷ XC è R1 + R2 ø X C R2 V R1 + R2 and Aυ = o = æ RR ö Vi 1 + jw ç 1 2 ÷ C è R1 + R2 ø =
or Aυ = defining fc =
é ù 1 ê ú êë1 + jf / f c úû é ù R2 ê 1 -1 Ð - tan f / f c ú and Aυ = ú R1 + R2 ê 1 + ( f / f ) 2 c ë û é ù R2 ê 1 ú |V | with = i R1 + R2 ê 1 + ( f / f ) 2 ú c ë û Aυ =
for f
R2 R1 + R2
f c, V o =
= 0.852Vi
at f = fc: Vo = 0.852[0.707]Vi = 0.602Vi fc =
318
= 1.02 kHz
CHAPTER 22
b.
c. & d.
-20 log10
= -20 log10
= -20 log10 1.174 = -1.39 dB -1.39 dB - 0.5 dB = -1.89 dB
e.
= 20 log10 Aυ -1.89 = 20 log10 Aυ 0.0945 = log10 Aυ Aυ = f.
= 0.80
θ = -tan-1 f/fc
42. = 24.79 kΩ
a.
From Section 21.11,
CHAPTER 22
319
Aυ =
Vo jf / f1 = Vi 1 + jf / f c
f1 =
= 642.01 Hz
fc =
= 457.47 Hz
= = −2.94 dB b.
θ = 90° - tan-1 f = f 1: f = f c: f= f=
= + tan-1
θ = 45° θ = 54.52° f1 = 321 Hz, θ = 63.44° f1 = 64.2 Hz, θ = 84.29°
f = 2f1 = 1,284 Hz, θ = 26.57° f = 10f1 = 6420 Hz, θ = 5.71°
43.
a.
VTh =
12 kΩ V i = 0.682 Vi 12 kΩ + 5.6 kΩ
RTh = 5.6 kΩ || 12 kΩ = 3.82 kΩ
320
CHAPTER 22
f = • Hz: (C fi short circuit) Vo =
= 0.465 Vi
At fc: V0 = 0.707(0.465 Vi) = 0.329 Vi
R2 (0.682 Vi ) 0.682 R2 Vi = R1 + R2 - jX C R1 + R2 - jX C V 0.682 R2 j2pf (0.682 R2 )C = and Aυ = o = Vi R1 + R2 - jX C 1 + j2pf ( R1 + R2 )C jf / f1 so that Aυ = with f1 = 1 + jf / f c = 284.59 Hz voltage-divider rule: Vo =
and fc = = 132.41 Hz
20 log10 f/f1 = 20 log10 = 20 log10 0.465 = -6.65 dB
b.
θ = 90° - tan-1 f/fc = +tan-1 fc/f = tan-1 132.6 Hz/f or
CHAPTER 22
321
1 1 = = 19.89 kHz 2p R2 C 2p (10 kΩ)(800 pF) 1 1 = fc = 2p ( R1 + R2 )C 2p (10 kΩ + 91 kΩ)(800 pF) = 1.97 kHz f1 =
44.
a.
Aυ =
b.
θ = tan-1 f/f1 - tan-1 f/fc f = 10 kHz 10 kHz 10 kHz θ = tan-1 - tan-1 = 29.66° - 87.62° = -57.96° 19.89 kHz 1.97 kHz f = fc: (f1 = 10 fc) θ = tan-1
322
= 5.71° - 45° = -39.29°
CHAPTER 22
45.
a.
R1 no effect! Note Section 22.12. Aυ =
Vo 1 + j( f / f1 ) = Vi 1 + j( f / f c )
f1 =
= 2.84 kHz
fc =
= 904.3 Hz
Note Fig. 22.65. Asymptote at 0 dB from 0 Æ fc -6 dB/octave from fc to f1 æ ö 12 kΩ + 5.6 kΩ = -9.5 dB÷ -9.95 dB from f1 on ç-20 log 5.6 kΩ è ø (b)
Note Fig. 22.67. 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/2.84 kHz - tan-1 1500 Hz/904.3 Hz = 27.83° - 58.92° = -31.09°
46.
a.
Aυ =
1 1 = 964.58 Hz = 2p R1C 2p (3.3 kΩ)(0.05 mF) 1 1 = fc = = 7.74 kHz 2p ( R1 || R2 )C 2p (3.3 kΩ || 0.47 kΩ) (0.05 m F) !## #"### $ 0.411 kΩ f1 =
-20 log10
= -18.08 dB
b.
CHAPTER 22
323
θ = -tan-1
+ tan-1
f = f1 : q = - tan -1 = - tan -1 1 + tan -1
f1 f + tan -1 c f1 f1 7.74 kHz 964.58 Hz
= -45° + 92.1° = 47.1°
964.58 Hz 7.74 kHz + tan -1 4 kHz 4 kHz
f = 4 kHz : q = - tan -1
= -15.06° + 69.63° = 54.57°
f = f c : q = - tan -1
f 964.58 Hz + tan -1 c 7.74 kHz fc
= -tan-1 0.1246 + tan-1 1 = -7.89° + 45° = 37.11° 47.
R3V i 4.7 kΩ V i = = 0.59 Vi R3 + R1 + R2 4.7 kΩ + 2 kΩ + 1.2 kΩ R3V i 4.7 kΩ V i = f = high: Vo = = 0.7 Vi R3 + R1 4.7 kΩ + 2 kΩ f = 0 Hz: Vo =
V0 Vi R3Vi = 0.7 Vi R3+R1
R3Vi = 0.59 Vi R3+R1+R3 0
Av =
f
Vo R3 = Vi R3 + R1 + R2 || X C Ð -90° Define R¢ = R3 + R1
324
and Au =
R3 = ¢ R + R2 || X C Ð -90°
=
R3 ( R2 - jX C ) ¢ R ( R2 - jX C ) - jR2 X C
=
R2 R3 - jR3 X C R¢R2 - jR¢X C - jR2 X C
R3 R (- jX C ) R¢ + 2 R2 - jX C
CHAPTER 22
=
R2 R3 - jR3 X C R¢R2 - j( R¢ + R2 )X C
R3 X C R2 R3 = R¢R2 ( R¢ + R2 ) - j XC R2 R3 R2 R3 1- j
XC R2 = ù R¢ é ( R¢ + R2 ) R3 XC ú ê1 - j R3 ë R2 R3 R¢ û 1- j
XC R2 = é ù ¢ ¢ R R +R ê1 - j ¢ 2 X C ú R3 ë R R2 û 1- j
=
R3 [1 - jX C / R2 ] æ R¢ + R2 ö ù R¢ é ê1 - jç ÷ XC ú êë è R¢R2 ø úû
é 1 ù ê1 - j ú 2pR2C û R3 ë = ù R1 + R3 é 1 ê1 - j ú 2p ( R¢ || R2 )C û ë
and Au =
R3 [1 - j(f1 / f )] R1 + R3 [1 - j(f c / f )]
Au dB = 20 log10
1 2pR2C 1 fc = ¢ 2p ( R || R2 )C f1 =
R3 1 2 + 20 log10 1 + ( f1 / f ) + 20 log10 R1 + R3 1 + ( f c / f )2
= -20 log10
with f1 =
R1 + R3 1 2 + 20 log10 1 + ( f1 / f ) + 20 log10 2 R3 1 + ( fc / f ) 1 1 , fc = , R¢ = R1 + R3 ¢ 2pR2C 2p ( R || R2 )C
q = -tan-1(f1/f) + tan-1(fcf) 48.
a.
Au 1 = Aumax æ 100 Hz ö æ 130 Hz ö æ f öæ f ö ç1 - j ÷ ç1 - j ÷ ç1 + j ÷ ç1 + j ÷ f øè f øè 20 kHz ø è 50 kHz ø è
CHAPTER 22
325
Proximity of 100 Hz to 130 Hz will raise lower cutoff frequency above 130 Hz: Testing: f = 180 Hz: (with lower terms only) = -20 log10
æ100 ö 2 æ130 ö 2 1+ ç ÷ - 20 log10 1 + ç ÷ è f ø è f ø
æ100 ö 2 æ130 ö 2 = -20 log10 1 + ç ÷ - 20 log10 1 + ç ÷ è180 ø è180 ø = 1.17 dB - 1.82 dB = -2.99 dB -3 dB Proximity of 50 kHz to 20 kHz will lower high cutoff frequency below 20 kHz: Testing: f = 18 kHz: (with upper terms only)
æ f ö2 æ f ö2 = -20 log10 1 + ç 20 log 1 + ÷ ç ÷ 10 è 20 kHz ø è 50 kHz ø æ 18 kHz ö 2 æ13 kHz ö 2 = -20 log10 1 + ç 20 log 1 + ÷ ç ÷ 10 è 20 kHz ø è 20 kHz ø = -2.576 dB - 0.529 dB = -3.105 dB
b.
Testing:
f = 1.8 kHz: θ = tan-1 = 3.18° + 4.14° - 5.14° - 2.06° 0° = 0.12°
326
CHAPTER 22
49.
50 kHz vs 23 kHz Æ 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 and using f = 90 Hz:
æ10 Hz ö 2 æ 45 Hz ö 2 æ 68 Hz ö 2 Au¢dB = -20 log10 1 + ç ÷ - 20 log10 1 + ç ÷ - 20 log10 1 + ç ÷ è f ø è f ø è f ø = −0.0532 dB − 0.969 dB − 1.96 dB = −2.98 dB (excellent) High frequency cutoff: Try f = 20 kHz
æ f ö2 æ f ö2 Au¢dB = -20 log10 1 + ç ÷ - 20 log10 1 + ç ÷ è 23 kHz ø è 50 kHz ø = −2.445 dB − 0.6445 dB = −3.09 dB (excellent
\ BW = 20 kHz - 90 Hz = 19,910 Hz f1 = 90 Hz, f2 = 20 kHz
20 kHz
Testing: f = 100 Hz θ= = tan-1 0.1 + tan-1 0.45 + tan-1 0.68 - tan-1 0.00435 - tan-1 .002 = 5.71° + 24.23° + 34.22° - 0.249° - 0.115° = 63.8° vs about 65° on the plot
CHAPTER 22
327
50.
flow = fhigh - BW = 38 kHz - 37.5 kHz = 0.5 kHz = 500 Hz Aυ =
51.
-140 æ ö æ 500 ö æ 50 f ö çç 1 - j ç1 - j ÷ ÷ ç1 + j f øè f øè 38 kHz ø è
Aυ =
=
52.
Aυ =
AυdB = -20 log20
328
and f1 = 2000 Hz
1 æ f ö 1+ ç ÷ è 2000 ø
2
,
f = 1 and f = 2 kHz 2000
CHAPTER 22
jf /1000 (1 + jf /1000)(1 + jf /10, 000)
53.
Aυ =
54.
æ f öæ f ö ç1 + j ÷ ç1 + j ÷ 1000 ø è 2000 ø è Aυ = 2 æ f ö 1 + j ç ÷ 3000 ø è AυdB = -20 log10
55.
=
CHAPTER 22
=
æ f1 ö 2 æ f2 ö 2 1+ ç ÷ + 20 log10 1 + ç ÷ + 40 log10 è1000 ø è 2000 ø
=
,
1 æ f3 ö 2 1+ ç ÷ è 3000 ø
=
329
56.
a.
Woofer − 400 Hz: XL = 2πfL = 2π(400 Hz)(4.7 mH) = 11.81 Ω 1 = = 10.20 Ω XC = 2pfC R || XC = 8 Ω –0° || 10.20–-90° = 6.3 Ω –-38.11° Vo =
=
Vo = 0.673 –-96.11° Vi and Aυ =
= 0.673 vs desired 0.707 (off by less than 5%)
Tweeter − 5 kHz: XL = 2πfL = 2π(5 kHz)(0.39 mH) = 12.25 Ω 1 XC = = = 11.79 Ω 2pfC R || XL = 8 Ω –0° || 12.25 Ω –90° = 6.7 Ω –33.15° Vo = Vo = 0.678 –88.54° Vi and Aυ = b.
= 0.678 vs 0.707 (off by less than 5%)
Woofer − 3 kHz: XL = 2πfL = 2π(3 kHz)(4.7 mH) = 88.59 Ω 1 XC = = = 1.36 Ω 2pfC R || XC = 8 Ω –0° || 1.36 Ω –-90° = 1.341 Ω –-80.35° Vo =
=
Vo = 0.015 –-170.2° Vi
330
CHAPTER 22
and Aυ =
= 0.015 vs desired 0 (excellent)
Tweeter − 3 kHz: XL = 2πfL = 2π(3 kHz)(0.39 mH) = 7.35 Ω 1 XC = = = 19.65 Ω 2pfC R || XL = 8 Ω –0° || 7.35 Ω –90° = 5.42 Ω –47.42° Vo =
=
Vo = 0.337 –124.24° Vi and Aυ =
c.
= 0.337 (acceptable since relatively close to cut frequency for tweeter)
Mid-range speaker − 3 kHz:
and
CHAPTER 22
V1 =
=
Vo =
=
Aυ =
= 1.11 –8.83° Vi = 0.998 –-46.9° Vi
= 0.998 (excellent)
331
Chapter 23 1.
a.
M=
fi
b.
ep =
= (20)(0.08 Wb/s) = 1.6 V
c.
2.
a.
b.
3.
es = kNs
= (0.8)(80 t)(0.08 Wb/s) = 5.12 V
ep =
= (40 mH)(0.3 ¥ 103 A/s) = 12 V
es =
= (80 mH)(0.03 ¥ 103 A/s) = 24 V
k=1 (a)
Ls =
= 32 mH
(b)
ep = 1.6 V, es = kNs
(c)
ep = 15 V, es = 12 V
= 6.4 V
k = 0.2 (a)
Ls =
= 0.8 H
(b)
ep = 1.6 V, es = kNs
(c)
ep = 15 V, es = 12 V
= (0.2)(80 t)(0.08 Wb/s) = 1.28 V
a.
Ls =
= 355.56 mH
b.
ep =
= (300 t)(0.08 Wb/s) = 24 V
es = kNs
= (0.9)(25 t)(0.08 Wb/s) = 1.8 V
c.
332
= 50 mH
ep and es the same as problem 1: ep = 15 V, es = 12 V
CHAPTER 23
4.
5.
a.
Es =
b.
Φmax =
a.
Es =
b.
Φm(max) =
6.
Ep =
7.
f=
8.
Ep =
(40 V) = 240 V
= 7.51 mWb
=5V
Es =
= 625.63 µWb
(240 V) = 20 V
= 120 Hz
1 a. = = I L aI A) 0.75 A P (3 = 4 3 = VL I L = Z L A (3 = Ω) 2.25 V 4 2
1 2 RL (3 = Ω ) 0.185 Ω = b. R in a = 4 9.
10.
11.
140 V = 28 Ω 5A
Zp =
1 = V g aV = V) 100 V L (500 = 5 100 V Ip = = 20 A 5Ω IL = Is =
220 V = 11 A 20 Ω fi
Np =
11 A 0.04 A
60
(11)60 = 16,500 turns 0.04
CHAPTER 23
333
12.
a.
600 t = 1200 t 2
a=
æ1 ö 2 Zi = a ZL = ç ÷ (10 Ω + j10 Ω) = 2.5 Ω + j2.5 Ω = 3.536 Ω –45° è2 ø Ip = Vg/Zi = 120 V/3.536 Ω = 33.94 A 2
13.
(33.94 A) = 16.97 A, VL = ILZL = (16.97 A)(14.14 Ω) = 239.96 V ≅ 240V
b.
IL = aIp =
a.
Z p = a 2Z L fi a =
2
Zp = a=
=3
fi
b. P= 14.
= 36 Ω
= 2.78 W
a.
Re = Rp + a2Rs = 4 Ω + (4)2 1 Ω = 20 Ω
b.
Xe = Xp + a2Xs = 12 Ω + (4)2 2 Ω = 44 Ω
c.
d.
Ip =
e.
aVL = or
334
f.
-
g.
VL =
= 0.554 A –-11.73°
= I pa 2R L VL = aIpRL–0° = (4)(0.554 A –-11.73°)(20 Ω –0°) = 26.59 V –-11.73°
(120 V) = 30 V
CHAPTER 23
15.
a.
a=
=4
Re = Rp + a2Rs = 4 Ω + (4)2 1 Ω = 20 Ω Xe = Xp + a2Xs = 12 Ω + (4)2 2 Ω = 44 Ω Zp = + + = 20 Ω + j44 Ω + j(4)2 20 Ω = 20 Ω + j44 Ω + j320 Ω = 20 Ω + j364 Ω = 364.55 Ω –86.86° b.
= 329.17 mA –-86.86°
Ip =
= (I–θ)(Re–0°) = (329.17 mA –-86.86°)(20 Ω –0°)
c.
= 6.58 V –-86.86° = (I–θ)(Xe–90°) = (329.17 mA –-86.86°)(44 Ω –90°) = 14.48 V –3.14° = I(a2
) = (329.17 mA –-86.86°)(320 Ω –90°)
= 105.33 V –3.14° 16.
a.
a = Np/Ns = 4 t/1 t = 4, Re = Rp + a2Rs = 4 Ω + (4)2 1 Ω = 20 Ω Xe = Xp + a2Xs = 12 Ω + (4)2 2 Ω = 44 Ω Zp = Re + jXe - ja2XC = 20 Ω + j44 Ω - j(4)2 20 Ω = 20 Ω + j44 Ω - j320 Ω = 20 Ω - j276 Ω = 276.72 Ω –-85.86°
b.
Ip =
= 0.43 A –85.86° = (Ip–θ)(Re–0°) = (0.43 A –85.86°)(20 Ω –0°) = 8.6 V –85.86°
c.
= (Ip–θ)(Xe–90°) = (0.43 A –85.86°)(44 Ω –90°) = 18.92 V –175.86° = (Ip–θ)(a2XC–-90°) = (0.43 A –85.86°)(320 Ω –-90°) = 137.60 V –-4.14° 17.
-
18.
Coil 1: L1 - M12 Coil 2: L2 - M12 LT = L1 + L2 - 2M12 = 5 H + 8 H - 2(1 H) = 11 H
19. M12 =
(0.9) (300 mH)(600 mH) = 381.84 mH ≈ 382 mH
= 300 mH + 600 mH = 2(382 mH) = 1.64 H
CHAPTER 23
335
20.
M23 = Coil 1: Coil 2: Coil 3:
21.
E 1 - I 1[
1 (1.5 H)(6 H) = 3 H L1 + M12 - M13 = 3 H + 0.2 H - 0.1 H = 3.1 H L2 + M12 - M23 = 1.5 H + 0.2 H - 3 H = -1.3 H L3 - M23 - M13 = 6 H - 3 H - 0.1 H = 2.9 H LT = 3.1 H - 1.3 H + 2.9 H = 4.7 H +
] - I2[Zm] = 0
I 2[ + ] + I1[Zm] = 0 ────────────────────── + ) + I2(Zm) = E1 I 1( I1(Zm) + I2( + )=0 ─────────────────────── 22.
Z i = Zp +
= Rp + j
Xm = -wM –90°
+
Rp = 2 Ω,
= wLp = (103 rad/s)(8 H) = 8 kΩ
Rs = 1 Ω,
= wLs = (103 rad/s)(2 H) = 2 kΩ
M=
= 0.2 H
Zi = 2 Ω + j8 kΩ + = 2 Ω + j8 kΩ + = 2 Ω + j8 kΩ + 0.21 Ω - j19.99 Ω = 2.21 Ω + j7980 Ω Zi = 7980 Ω –89.98° 23.
336
3600 V
= 30
a.
a=
b.
12,000 VA = VsIs fi Is =
c.
Ip =
d.
a=
12,000 VA
12,000 VA
12,000 VA 3600 V
12,000 VA
= 100 A
= 3.33 A
= 3600 V 30 12,000 VA Is = = 3.33 A, Ip = 100 A 3600 V
CHAPTER 23
24.
Is = I1 = 2 A, Ep = VL = 48 V Es = Vg - VL = 240 V - 48 V = 192 V 240 V (2 A) = 10 A VgI1 = VLIL fi IL = Vg/VL ◊ I1 = 48 V Ip + I1 = IL fi Ip = IL - I1 = 10 A - 2 A = 8 A
25.
a.
Es = (100 V –0°) = 25 V –0° = VL
=
= 5 A –0° = IL
Is =
b.
æ N ö2 p Zi = a ZL = ç ç N ÷÷ ZL = è sø 2
= 20 Ω –0°
c.
26.
a.
æ100 t ö 2 2 ç ÷ 5 Ω –0° = (4) 5 Ω –0° = 80 Ω –0° 25 t è ø
E2 =
E1 =
(60 V –0°) = 10 V –0°
E3 =
E1 =
(60 V –0°) = 30 V –0°
I2 =
= 1.25 A –0°
I3 =
= 6 A –0°
=
b.
= = 0.05347 S R1 = 18.70 Ω 27.
a.
E2 =
æ 40 t ö E1 = ç ÷ (120 V –60°) = 40 V –60° è120 t ø
I2 =
= 3.33 A –60°
E3 =
æ 30 t ö E1 = ç ÷ (120 V –60°) = 30 V –60° è120 t ø
CHAPTER 23
337
= 3 A –60°
I3 = b.
= = =
= 0.0155 S
R1 = 28.
ZM =
= 64.52 Ω = ΩM12 –90°
E - I 1Z 1 - I 1 E - I1(Z1 +
- I1(-Zm) - I2(+Zm) - I1 - Zm +
- I1(-Zm) = 0
+ I2
- Zm) - I2(Zm -
)=0
or
I1(Z1 + + - 2 Zm) + I2(Zm )=E ────────────────────────────────────────────── (I2 - I1) - I1(+Zm) = 0 -I2Z2 -
or
I1(Zm ) + I2(Z2 + )=0 ────────────────────────── E 1 - I 1Z 1 - I 1
29.
- I2(-
) - I3(+
)=0
or
E1 - I1[Z1 + ] + I2 - I3 =0 ──────────────────────────────── ) + I3Z2 - I1()=0 -I2(Z2 + Z3 +
or
-I2(Z2 + Z3 + ) + I 3Z 2 + I 1 =0 ──────────────────────────────── ) + I2Z2 - I1(+ )=0 -I3(Z2 + Z4 +
or
-I3(Z2 + Z4 + ) + I 2Z 2 - I 1 =0 ───────────────────────────── ]I1 I2 + [Z1 +
\
I1 - [Z2 + Z3 +
]I2 +
I3 = E 1 Z 2I 3 = 0
I1 Z2I2 + [Z2 + Z4 + ]I3 = 0 ──────────────────────────────────────── 30.
338
Ip =
Ns 250 t Is = (400 mA) = 100 A 1t Np
CHAPTER 23
Chapter 24 1.
a.
= EL/
b.
=
d.
IL =
= 131.64 V
b.
=
Z f = 14 Ω - j20 Ω = 24.41 Ω –55°
d.
IL =
c.
2.
c.
131.64 V = 8.78 A 15 Ω
=
a.
= EL/
a. c.
4.
= 131.64 V
b.
= 8.78 A
= 131.64 V = 5.39 A
= 131.64 V
64 –-90° Zf = (8 Ω –0° || (8 Ω –-90°) = = 5.659 Ω –-45° 11.31 –45° 131.64 V = = 23.26 A 5.659 Ω
d.
IL = 23.26 A
a.
θ2 = -120°, θ3 = 120°
b.
Van = 120 V –0°, Vbn = 120 V –-120°, Vcn = 120 V –120°
c.
Ian =
= 6 A –0° = 6 A –-120°
Ibn =
= 6 A –120°
Icn =
5.
= 131.64 V
131.64 V = 5.39 A 24.41 Ω
= 3.
= 228 V/1.732 = 131.64 V
d.
IL =
= 6A
e.
VL =
=
(120 V) = 207.8 V
a.
θ2 = -120°, θ3 = +120°
b.
Van = 120 V –0°, Vbn = 120 V –-120°, Vcn = 120 V –120° = 9 Ω + j12 Ω = 15 Ω –53.13°
c.
Ian =
= 8 A –-53.13°, Ibn =
Icn =
= 8 A –66.87°
CHAPTER 24
= 8 A –-173.13°
339
e. 6.
IL =
a, b.
f.
EL =
= (1.732)(120 V) = 207.85 V
The same as problem 4. = 6 Ω –0° || 8 Ω –-90° = 4.8 Ω –-36.87°
c.
d. 7.
=8A
Ian =
= 25 A –36.87°
Ibn =
= 25 A –-83.13°
Icn =
= 25 A –156.87°
IL =
= 25 A
e.
= Van = Vbn = Vcn =
VL =
=
(120 V) = 207.84 V
= 127.0 V
= 10 Ω - j10 Ω = 14.42 Ω –-45° = Ian = Ibn = Icn = IL = IAa = IBb = ICc = 8.
= 8.98 A = 8.98 A
= 12 Ω + j16 Ω = 20 Ω–53.13° =
= 2.5 A = 13 Ω + j16 Ω = 20.62 Ω –50.91°
= VL =
9.
a.
= (2.5 A)(20.62 Ω) = 51.55 V =
(51.55 V) = 89.29 V
= 12.7 kV –-30°
EAN =
= 12.7 kV –-150°
EBN =
= 12.7 kV –90°
ECN =
b, c. IAa = Ian =
= = = 11.29 A –-97.54°
340
CHAPTER 24
10.
IBb = Ibn =
=
= 11.29 A –-217.54°
ICc = Icn =
=
= 11.29 A –22.46°
d.
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°
a.
Total load loss = 750 kW + 3I L R = 750 kW + 3(80 A)22 Ω = 750 kW + 38.4 kW = 788.4 kW
b.
Ef =
c.
Zline = 3 Ω + j20 Ω Vline(load) = EL -ILZline = 14.7 kV – 0° - 80 A (2 Ω + j20 Ω) = 14.7 kV - 160 V - j1600 V = 14,540 V - j1600 V = 14.63 kV –-6.28° 14.63 kV 14.63 kV = |Vf(load)| = = 8.45 kV 1.732 3 P(load)T = 3 Vf L I f L cos q 788.4 kW = 3(8.45 kV)(80 A) cos q 788.4 kW cos q = Fp = 3(8.45 kV)(80 A) Fp = 0.388
d.
0.388 fi 74.63° Vf L 8.45 kV |Zf| = = 105.63 Ω = If L 80 A
2
EL 3
=
14.7 kV = 8.49 kV 1.732
Zf = 105.63 –74.63° e.
Vf(load) =
14.7 kV
= 8.49 kV 3 P(load)T = 3 Vf L I f L cos q
788.4 kW 3(8.49 kV)(80 A) Fp = 0.387 lagging Fp = cos q =
CHAPTER 24
341
11.
a.
= E L/
c.
12.
=
a. c.
= 10.4 A
= E L/
13.
= (1.732)(10.4 A) = 18 A
b.
= EL = 208 V
b.
= 208 V
= (1.732)(13.36 A) = 23.14 A
= V L/
= 208 V/
= 120.09 V
c.
=
= 16.34 A
d.
IL =
a.
θ2 = -120°, θ3 = +120°
b.
Vab = 208 V –0°, Vbc = 208 V –-120°, Vca = 208 V –120°
c.
-
d.
Iab =
= (1.732)(16.34 A) = 28.30 A
= 9.46 A –0°
Vbc 208 V Ð -120° = 9.46 A –-120° Z bc 22 Ω Ð 0° = 9.46 A –120°
Ica = e. f.
342
IL =
Zf = 18 Ω –0° || 18 Ω –-90° = 12.728 Ω –-45°
Ibc =
15.
d.
= EL = 208 V
= 13.36 A
IL =
a.
14.
= 208 V/1.732 = 120.1 V
b.
Zf = 6.8 Ω + j14 Ω = 15.564 Ω –64.09° =
d.
= 208 V/1.732 = 120.1 V
IL = = E L/
= (1.732)(9.46 A) = 16.38 A = 208 V/1.732 = 120.1 V
a.
θ2 = -120°, θ3 = +120°
b.
Vab = 208 V –0°, Vbc = 208 V –-120°, Vca = 208 V –120°
CHAPTER 24
c.
-
d.
Zf = 100 Ω - j100 Ω = 141.42 Ω–-45°
e.
Iab =
= 1.47 A –45°
Ibc =
= 1.47 A –-75°
Ica =
= 1.47 A –165°
IL =
f. 16. a, b.
= E L/
= (1.732)(1.471 A) = 2.55 A = 208 V/1.732 = 120.1 V
The same as problem 13.
c.
-
d.
Zf = 3 Ω –0° || 4 Ω –90° = 2.4 Ω –36.87° Iab =
= 86.67 A –-36.87°
Ibc =
= 86.67 A –-156.87° = 86.67 A –83.13°
Ica = e. 17.
IL =
= (1.732)(86.67 A) = 150.11 A
= 120.1 V
Vab = Vbc = Vca = 220 V Zf = 10 Ω + j10 Ω = 14.142 Ω–45° Iab = Ibc = Ica =
18.
f.
a.
= 15.56 A
Iab = Iab = 15.33 A –-73.30°
b.
Ibc =
= 15.33 A –-193.30°
Ica =
= 15.33 A –46.7°
IAa - Iab + Ica = 0
CHAPTER 24
343
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° c.
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°
19.
a.
c.
20.
a. c.
21. a, b. c.
= EL = 208 V
=
= 4.00 A
= EL = 208 V =
344
d.
b. = 7.08 A
d.
=
= 120.1 V
IL =
4A
= EL IL =
= 120.09 V
= 7.08 A
The same as problem 19. = 15 Ω –0° || 20 Ω –-90° = 12 Ω –-36.87° =
d.
b.
IL =
10 A 10 A
CHAPTER 24
22.
Van = Vbn = Vcn =
= 69.28 V
Ian = Ibn = Icn =
= 2.89 A
IAa = IBb = ICc = 2.89 A 23.
Van = Vbn = Vcn =
= 69.28 V
= 10 Ω + j20 Ω = 22.36 Ω–63.43°
Vf
69.28 V = 3.10 A Zf 22.36 Ω IAa = IBb = ICc = = 3.10 A Ian = Ibn = Icn =
24.
=
Van = Vbn = Vcn = 69.28 V = 20 Ω –0° || 15 Ω –-90° = 12 Ω –-53.13° Ian = Ibn = Icn =
= 5.77 A
IAa = IBb = ICc = 5.77 A 25.
a.
= EL = 440 V
c. 26.
=
=2A
a.
= EL = 440 V
c.
= 12 Ω - j9 Ω = 15 Ω –-36.87° =
d. 27. a, b.
IL =
d. b.
= EL = IL =
= 440 V = (1.732)(2 A) = 3.46 A
= EL = 440 V
= 29.33 A
= (1.732)(29.33 A) = 50.8 A
The same as problem 25. = 22 Ω –0° || 22 Ω –90° = 15.56 Ω –45°
c.
=
28.
b.
= 28.28 A
d.
IL =
a.
θ2 = -120°, θ3 = +120°
CHAPTER 24
= (1.732)(28.28 A) = 48.98 A
345
b.
Vab = 100 V –0°, Vbc = 100 V –-120°, Vca = 100 V –120°
c.
-
d.
Iab =
= 5 A –0° = 5 A –-120°
Ibc =
= 5 A –120°
Ica =
29.
e.
IAa = IBb = ICc =
(5 A) = 8.66 A
a.
θ2 = -120°, θ3 = +120°
b.
Vab = 100 V –0°, Vbc = 100 V –-120°, Vca = 100 V –120°
c.
= 12 Ω + j16 Ω = 20 Ω –53.13°
d.
Iab = Ibc = Ica =
30.
= 5 A –-173.13° = 5 A –66.87°
e.
IAa = IBb = ICc =
a.
θ2 = -120°, θ3 = 120°
b.
Vab = 100 V –0°, Vbc = 100 V –-120°, Vca = 100 V –120°
c.
-
d.
346
= 5 A –-53.13°
= (1.732)(5 A) = 8.66 A
= 20 Ω –0° || 20 Ω –-90° = 14.14 Ω –-45° Iab =
= 7.07 A –45°
Ibc =
= 7.07 A –-75°
Ica =
= 7.07 A –165°
CHAPTER 24
e. 31.
IAa = IBb = ICc =
PT =
= 3(6 A)2 12 Ω = 1296 W
QT =
= 3(6 A)2 16 Ω = 1728 VAR(C)
ST =
= 2160 VA
Fp = 32.
(7.07 A) = 12.25 A
= 0.6 (leading)
= 120 V, PT =
= 120 V/20 Ω = 6 A
= 3(6 A)2 20 Ω = 2160 W
QT = 0 VAR ST = PT = 2160 VA Fp = 33.
PT =
= 3(8.98 A)2 10 Ω = 2419.21 W
QT =
= 3(8.98 A)2 10 Ω = 2419.21 VAR(C)
ST = Fp = 34.
=1
= 3421.28 VA = 0.7071 (leading)
= 208 V
æV 2 ö (208 V)2 f ç ÷ PT = 3 = 7210.67 W = 3× ç Rf ÷ 18 Ω è ø æV 2 ö (208 V)2 f ç ÷ QT = 3 = 7210.67 VAR(C) = 3× ç Xf ÷ 18 Ω è ø ST = Fp = 35.
= 10,197.42 VA = 0.707 (leading)
PT =
= 3(1.471 A)2 100 Ω = 649.15 W
QT =
= 3(1.471 A)2 100 Ω = 649.15 VAR(C)
ST = Fp =
CHAPTER 24
= 918.04 VA = 0.7071 (leading)
347
36.
PT =
= 3(15.56 A)2 10 Ω = 7.26 kW
QT =
= 3(15.56 A)2 10 Ω = 7.26 kVAR
ST =
37.
= 10.27 kVA
Fp =
= 0.7071 (lagging)
PT =
= 2884.80 W
QT =
= 2163.60 VAR(C)
ST =
= 3605.97 VA
Fp = 38.
= 10 Ω + j20 Ω = 22.36 Ω –63.43° = =
= 69.28 V = 3.098 A
PT =
= 3(3.098 A)2 10 Ω = 287.93 W
QT =
= 3(3.098 A)2 20 Ω = 575.86 VAR
ST =
39.
= 0.8 (leading)
= 643.83 VA
Fp =
= 0.447 (lagging)
PT =
= 26.4 kW
QT = PT = 26.4 kVAR(L) ST = Fp = 40.
= 0.707 (lagging)
= 12 Ω + j16 Ω = 20 Ω –53.13° =
348
= 37.34 kVA
=5A
PT =
= 3(5 A)2 12 Ω = 900 W
QT =
= 3(5 A)2 16 Ω = 1200 VAR(L)
CHAPTER 24
ST =
= 1500 VA
Fp =
41.
= 0.6 (lagging)
PT =
ELIL cos θ
4800 W = (1.732)(200 V)IL (0.8) IL = 17.32 A =
= 10 A
θ = cos-1 0.8 = 36.87° = 20 Ω –36.87° = 16 Ω + j12 Ω
=
42.
PT =
ELIL cos θ (208 V)IL(0.6) fi IL = 5.55 A
1200 W = =
= 120.1 V
θ = cos-1 0.6 = 53.13° (leading)
Ω - j17.31 Ω = 21.64 Ω –-53.13° = 12.98 ! #"# $ ! #"# $ R XC
=
43.
Δ:
= 15 Ω + j20 Ω = 25 Ω–53.13° =
=5A
PT =
= 3(5 A)2 15 Ω = 1125 W
QT =
= 3(5 A)2 20 Ω = 1500 VAR(L)
Y:
= V L/
= 125 V/1.732 = 72.17 V
= 3 Ω - j4 Ω = 5 Ω–-53.13° = PT = QT =
= 14.43 A = 3(14.43 A)2 3 Ω = 1874.02 W = 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 =
CHAPTER 24
= 3161 VA = 0.949 (leading)
349
44.
a.
=
= 9,237.6 V
c.
b.
IL =
= 80 A
= 400 kW P4Ω = (80 A)24 Ω = 25.6 kW PT = = 3(25.6 kW + 400 kW) = 1276.8 kW
d.
Fp =
, ST =
V LI L =
Fp = e.
= 0.576 lagging
θL = cos-1 0.576 = 54.83° (lagging) fi
IAa =
f.
(16 kV)(80 A) = 2,217.025 kVA
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°
g.
= 97.168 Ω –49.95°
=
Ω + j77.38 Ω = 62.52 ! #"# $ ! #"# $ R XC
45.
h.
Fp(entire system) = 0.576 (lagging) Fp(load) = 0.643 (lagging)
i.
η=
a.
Vf(load) =
b.
350
= 0.9398 fi 93.98%
12, 400 V
=
12, 400 V = 7,159.35 V 1.732
3 PT = 3VfIfcos q PT 2400 kW = If = 3Vf cos q 3(7,159.35 V)(0.6) If = 186.19 A IL = If = 186.19 A
CHAPTER 24
c.
1Ω
10 Ω
Iφ
+
+
Vφ
EAN
−
−
46.
d.
EAB =
a.
-
b.
3 EAN =
=
Ef = IfZline + Vf = (186.19 A –-53.13°)(10.05 Ω –84.29°) + 7,159.35 V –0° = 1.871 ¥ 103 V –31.16° + 7,159.35 –0° = 1.6 ¥ 103 V + j0.968 ¥ 103 V + 7,159.35 V = 8.759 ¥ 103 V + j0.968 ¥ 103 V = 8,810 V –6.31°
3 (8,810 V) = 15,259 V
= 10 Ω - j10 Ω = 14.14 Ω–-45°
= 127.02 V,
=
q = cos-1 0.6 fi 53.13° If = 186.19 A –-53.13° Zline = 1 Ω + j10 Ω = 10.05 Ω –84.29°
= 8.98 A = 3(8.98 A)2 10 Ω = 2419.2 W
PT =
Each wattmeter:
= 806.4 W
47.
b.
PT = 5899.64 W, Pmeter = 1966.55 W
48.
a.
-
b.
PT =
c.
0.2 fi
+ Ph = 85 W + 200 W = 285 W
Ph = PT = Ph 49.
-
50.
a.
Iab =
= 0.5 = 200 W = 200 W - 100 W = 100 W
= 20.8 A –0°
E BC 208 V Ð -120° 208 V Ð -120° = = = 14.708 A –-165° R + jX C 10 Ω + j10 Ω 14.142 Ω Ð 45° ECA 208 V Ð 120° 208 V Ð 120° = = Ica = = 14.708 A –165° R - jX C 10 Ω - j10 Ω 14.142 Ω Ð -45° Ibc =
CHAPTER 24
351
51.
b.
IAa + Ica - Iab = 0 IAa = Iab - Ica = 20.8 A –0° - 14.708 A –165° = 20.8 A - (-14.207A + j3.807 A) = 35.007 A - j3.807 A = 35.213 A –-6.207° IBb + Iab - Ibc = 0 IBb = Ibc - Iab = 14.708 A –-165° - 20.8 A –0° = (-14.207 A - j3.807 A) - 20.8 A = -35.007 A - j3.807 A = 35.213 A –-173.79° ICc + Ibc - Ica = 0 ICc = Ica - Ibc = 14.708 A –165° - 14.708 A –-165° = (-14.207 A + j3.807 A) - (-14.207 A - j3.807 A) = 7.614 A –90°
c.
P1 = VacIAa
d.
PT = P1 + P2 = 4.326 kW + 4.327 kW = 8.653 kW
Vca = Vca – θ - 180° = 208 V –120° - 180° = 208 V –-60° IAa = 35.213 A –-6.207° P1 = (208 V)(35.213 A) cos 53.793° = 4.326 kW P2 = VbcIBb Vbc = 208 V –-120° IBb = 35.213 A –-173.79° P2 = (208 V)(35.213 A) cos 53.79° = 4.327 kW
a.
=
b.
Ian =
= 8.49 A
Ibn =
= 7.08 A
Icn =
= 42.47 A
c.
PT =
=
= 120.09 V
10 Ω +
12 Ω +
2
2Ω
= (8.49 A) 10 Ω + (7.08 A) 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 =
352
2
= 6.97 kVA
CHAPTER 24
Fp = d.
= 0.707 (lagging)
Ean = 120.09 V–-30°, Ebn = 120.09 V–-150°, Ecn = 120.09 V–90° = 8.49 A –-75°
Ian =
= 7.08 A –-195°
Ibn =
= 42.47 A –45°
Icn = e.
52.
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° Z Δ = Z 1Z 2 + Z 1Z 3 + Z 2Z 3 = (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° E AB Z 3 - E CA Z 2 (200 V Ð 0°)(20 Ω Ð 0°) - (200 V Ð 120°)(5 Ω Ð 53.13°) = Ian = ZD ZD = 10.71 A –29.59°
= Ibn =
E BC Z1 - E AB Z 3 (200 V Ð -120°)(20 Ω Ð -53.13°) - (200 V Ð 0°)(20 Ω Ð 0°) = ZD ZD = 17.12 A –-145.61°
= Icn =
E CA Z 2 - E BC Z1 (200 V Ð 120°)(5 Ω Ð 53.13°) - (200 V Ð -120°)(20 Ω Ð -53.13°) = ZD ZD = 6.51 A –42.32°
=
PT = 12 Ω + 4Ω+ 20 Ω = 1376.45 W + 1172.38 W + 847.60 W = 3396.43 W QT = 16 Ω + 3 Ω = 1835.27 VAR(C) + 879.28 VAR(L) = 955.99 VAR(C) ST =
CHAPTER 24
= 3508.40 VA
353
Fp =
354
= 0.968 (leading)
CHAPTER 24
Chapter 25 1.
I.
II.
2.
I.
II.
a.
positive-going
d.
Amplitude = 12 V
e.
% tilt = 0%
a.
positive-going
b.
-6 V
c.
1 ms
d.
8V
e.
% tilt =
a.
positive-going
b.
4V
c.
1 ms
d.
4V
e.
% tilt =
a.
negative-going
b.
-2 V
c.
3 µs
d.
-6 V
e.
% tilt =
CHAPTER 25
b.
=0V
c.
tp = 2 µs
V1 - V2 ¥ 100% V 8 V + 6 V 14 V V= =7V = 2 2 8V-6V 2 % tilt = ´100% = ¥ 100% = 28.57% 7V 7
V1 - V2 ¥ 100% V 8 V + 7 V 15 V V= = 7.5 V = 2 2 8V-7V 1 % tilt = ¥ 100% = 13.33% ´100% = 7.5 V 7.5
V1 - V2 ¥ 100% V -8 V - 6 V -14 V V= = -7 V = 2 2
355
-8 V - (-7 V) -8 V + 7 V ¥ 100% ´100% = -7 V -7 V -1 V = ¥ 100% = 14.28% -7 V
% tilt =
3.
a.
positive-going
b.
d.
Amplitude = (30 - 10)mV = 20 mV
e.
% tilt = V=
4.
tr tf
5.
tilt =
c.
æ8 ö tp = ç ÷ 4 ms = 3.2 ms è10 ø
= 29 mV ¥ 100%
% tilt =
= 10 mV
6.9%
(0.2 div.)(2 ms/div.) = 0.4 ms (0.4 div.)(2 ms/div.) = 0.8 ms = 0.1 with V =
Substituting V into top equation, = 0.1 leading to V2 =
6.
356
or V2 = 0.905(15 mV) = 13.58 mV
a.
tr = 80% of straight line segment = 0.8(2 µs) = 1.6 µs
b.
tf = 80% of 10 µs interval = 0.8(10 µs) = 8 µs
c.
At 50% level (10 mV) mid-pt. of rise = 1 µs, mid-pt. of fall = 7 µs tp = 7 µs - 1 µs = 6 µs
CHAPTER 25
7.
8.
9.
10.
d.
prf =
= 50 kHz
a.
T = (4.8 - 2.4)div.
c.
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
I.
T = 5 µs 1 1 prf = = 200 kHz = T 5ms tp 2.5 div. Duty cycle = ´100% = ´100% = 50% T 5 div.
II.
T = 4 ms 1 1 prf = = 250 Hz = T 4 ms tp 1 div. Duty cycle = ´100% = ´100% = 25% T 4 div.
I.
T = (6 - 1)ms = 5 ms 1 1 prf = = 200 Hz = T 5 ms tp 1 div. Duty cycle = ´100% = ´100% = 20% T 5 div.
II.
T = (8 - 1)µs = 7 µs 1 1 prf = = 142.86 Hz = T 7 ms tp 3 div. Duty cycle = ´100% = ´100% = 42.86% T 7 div.
b.
f=
= 8.33 kHz
T = (3.6 div.)(2 ms/div.) = 7.2 ms prf =
= 138.89 Hz ¥ 100% = 44.4%
Duty cycle = 11.
= 120 µs
a.
T = (9 - 1)µs = 8 µs
c.
prf =
d.
Vav = (Duty cycle)(Peak value) + (1 - Duty cycle)(Vb)
CHAPTER 25
b.
tp = (3 - 1)µs = 2 µs
= 125 kHz
357
Duty cycle =
= 25%
Vav = (0.25)(6 mV) + (1 - 0.25)(-2 mV) = 1.5 mV - 1.5 mV = 0 V or Vav =
12.
e.
Veff =
I.
Vb = 0 V, Vpeak = 12 V, duty cycle = 0.5 Vav = (duty cycle)(peak value) + (1 - duty cycle)(Vb) = (0.5)(12 V) + (1 - 0.5)(0 V) =6V
14.
2 ms = 0.4 ms/div. 5 div. 10 mV On the vertical: = 2 mV/div. 5 div. T = 6 ms + 3(0.4 ms) = 7.2 ms 1 1 1 (0.8 ms)(20 mV) + (0.8 ms)(18 mV) + (2.4 ms)(18 mV) + (2.4 ms)(2 mV) 2 2 Vav = 2 7.2 ms 8´10 -6 sV + 7.2´10 -6 sV + 4.32´10 -6 sV + 2.4´10 -6 sV = 7.2 ms 60.8´10 -6 / sV = = 8.44 mV 7.2´10 -3 / s On horizontal:
1 1 (2 ms)(20 mV) + (10 ms)(20 mV) 2 2 Gav = 2 ms =
15.
= 3.46 mV
æ1 ö +ç (1 ms)(2 V) - (3 ms)(6 V)÷ 2 ø Vav = è 4 ms 1 msV - 18 msV -17 V = = -4.25 V = 4 ms 4
II.
13.
=0V
20´10 -9 / sV + 100´10 -9 / sV 20´10 -6 / s
=
120´10 -9 V 20´10 -6
= 6 mV
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
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CHAPTER 25
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.
= 117 mV
Using the defined polarity of Fig. 24.57 for υC, Vi = -6 V, Vf = +26 V and τ = RC = (8 kΩ)(0.02 µF) = 0.16 ms a.
υC = Vi + (Vf - Vi)(1 - e-t/τ) = -6 + (26 - (-6))(1 - e- t/0.16 ms) = -6 + 32(1 - e-t/0.16 ms) = 26 V - 32 Ve-t/0.16 ms υC = 26 V - 32 Ve- t/0.16 ms
b.
c.
Ii = 0 iC =
d.
20 V − [26 V − 32 Ve−t/0.16 ms] = 4 mAe- t/0.16 ms 8 kΩ
17.
τ = RC = (2 kΩ)(10 µF) = 20 ms Vi = 2 V, Vf = 10 V υC = Vi + (Vf - Vi)(1 - e-t/RC) = 2 V + (10 V - 2 V)(1 - e-t/τ) = 2 V + 8 V(1 - e-t/ t) = 2 V + 8 V - 8 Ve-t/t υC = 10 V - 8e- t/t
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
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359
19.
Vi = 10 V, Ii = 0 A
Using the defined direction of iC e-t/τ
iC =
τ = RC = (1 kΩ)(1000 µF) = 1 s iC = and iC = -8mAe- t
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=
= 10 ms
= 5 ms 5τ = 1 ms =
b.
T=
1 æT ö ç ÷ 5 è2 ø = 2 ms
= 1 ms 5τ = 1 ms =
c.
T=
= 0.2 ms
= 0.1 ms
æT ö 5τ = 1 ms = 10ç ÷ è2 ø
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CHAPTER 25
21.
The mathematical expression for iC is the same for each frequency! τ = RC = (5 kΩ)(0.04 µF) = 0.2 ms = 4 mAe- t/0.2 ms
and iC =
a.
T= 5τ = 5(0.2 ms) = 1 ms =
b.
T=
= 10 ms, 5τ = 1 ms =
c.
T=
= 5 ms
1 æT ö ç ÷ 5 è2 ø
= 0.2 ms,
= 0.1 ms
æT ö 5τ = 1 ms = 10ç ÷ è2 ø 22.
τ = 0.2 ms as above T=
= 2 ms 5τ = 1 ms =
0Æ
: υC = 20 V(1 - e-t/0.2 ms)
Æ T: Vi = 20 V, Vf = -20 V 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 TÆ
: Vi = -20 V, Vf = +20 V 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
23.
υC = Vi + (Vf - Vi)(1 - e-t/RC) Vi = 20 V, Vf = 20 V
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361
υC = 20 + (20 - 20)(1 - e-t/RC) = 20 V (for 0 Æ
For
)
Æ T, υi = 0 V and υC = 20 Ve- t/τ
τ = RC = 0.2 ms with
For T Æ
= 1 ms and 5τ =
, υi = 20 V
υC = 20 V(1 - e- t/τ) For
Æ 2T, υi = 0 V υC = 20 Ve- t/τ
24.
τ = RC = 0.2 ms 5τ = 1 ms = Vi = -10 V, Vf = +20 V 0Æ
: 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
25.
362
Æ T:
Vi = 20 V, Vf = 0 V υC = 20 Ve- t/0.2 ms
1 1 = = 5.31 MΩ 2pfC 2p (10 kHz)(3 pF) (9 MΩ Ð 0°)(5.31 MΩ Ð -90°) Zp = = 4.573 MΩ –-59.5° 9 MΩ - j5.31 MΩ
Z p:
XC =
Z s:
CT = 18 pF + 9 pF = 27 pF 1 1 = XC = = 0.589 MΩ 2pfCT 2p (10 kHz)(27 pF) (1 MΩ Ð 0°)(0.589 MΩ Ð -90°) Zs = = 0.507 MΩ –-59.5° 1 MΩ - j0.589 MΩ
CHAPTER 25
Vscope =
Z sVi (0.507 MΩ Ð -59.5°)(100 V Ð 0°) = Z s + Z p (0.257 MΩ - j0.437 MΩ) + (2.324 MΩ - j3.939 MΩ) = 10 V –0° =
=
(100 V –0°)
= -59.5° 26.
Z p: X C = Zp = Z s:
= 3.333 MΩ
(9 MΩ Ð 0°)(3.333 MΩ) = 3.126 MΩ –-69.68° 9 MΩ - j3.333 MΩ
XC = Zs =
= 0.370 MΩ
(1 MΩ Ð 0°)(0.370 MΩ Ð -90°) = 0.347 MΩ –-69.68° 1 MΩ - j0.370 MΩ
3 Vscope =
Z sVi (0.347 MΩ Ð -69.68°)(100 V Ð 0°) = Z s + Z p (0.121 MΩ - j0.325 MΩ) + (1.086 MΩ - j2.931 MΩ)
= 10 V –0° =
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(100 V –0°)
363