INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY
POWER SYSTEM ANALYSIS AND DESIGN FIFTH EDITION
J. DUNCAN GLOVER MULUKUTLA S. SARMA THOMAS J. OVERBYE
Contents Chapter 2
1
Chapter 3
27
Chapter 4
71
Chapter 5
95
Chapter 6
137
Chapter 7
175
Chapter 8
195
Chapter 9
231
Chapter 10
303
Chapter 11
323
Chapter 12
339
Chapter 13
353
Chapter 14
379
Chapter 2 Fundamentals ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS 2.1 b 2.19 a 2.2 a 2.20 A. c 2.3 c B. a 2.4 a C. b 2.5 b 2.21 a 2.6 c 2.22 a 2.7 a 2.23 b 2.8 c 2.24 a 2.9 a 2.25 a 2.10 c 2.26 b 2.11 a 2.27 a 2.12 b 2.28 b 2.13 b 2.29 a 2.14 c 2.30 (i) c (ii) b 2.15 a (iii) a 2.16 b (iv) d 2.17 A. a 2.31 a B. b 2.32 a C. a 2.18 c
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2.1
(a) A1 = 5∠30° = 5 [ cos30° + j sin 30°] = 4.33 + j 2.5 4 = 5 ∠126.87° = 5e j126.87° −3 (c) A3 = ( 4.33 + j 2.5 ) + ( −3 + j 4 ) = 1.33 + j 6.5 = 6.635∠78.44°
(b) A2 = −3 + j 4 = 9 + 16 ∠ tan −1
(d) A4 = ( 5∠30° )( 5 ∠126.87° ) = 25 ∠156.87° = −22.99 + j 9.821 (e) A5 = ( 5∠30° ) / ( 5∠ − 126.87° ) = 1∠156.87° = 1 e j156.87° 2.2
(a) I = 400∠ − 30° = 346.4 − j 200 (b) i(t ) = 5sin (ω t + 15° ) = 5cos (ω t + 15° − 90° ) = 5cos (ω t − 75° )
( 2 ) ∠ − 75° = 3.536∠ − 75° = 0.9151− j3.415 (c) I = ( 4 2 ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 ) I = 5
= 3.743 − j 6.244 = 7.28∠ − 59.06°
2.3
(a) Vmax = 359.3V; I max = 100 A (b) V = 359.3
2 = 254.1V; I = 100
2 = 70.71A
(c) V = 254.1∠15° V; I = 70.71 ∠ − 85° A 2.4
(a) I1 = 10∠0°
− j6 6∠ − 90° = 10 = 7.5∠ − 90° A 8 + j6 − j6 8
I 2 = I − I1 = 10∠0° − 7.3∠ − 90° = 10 + j 7.5 = 12.5∠36.87° A V = I 2 ( − j 6 ) = (12.5∠36.87° ) ( 6∠ − 90° ) = 75∠ − 53.13° V
(b)
2.5
(a) υ (t ) = 277 2 cos (ω t + 30° ) = 391.7cos (ω t + 30° ) V (b)
I = V / 20 = 13.85∠30° A i(t ) = 19.58cos (ω t + 30° ) A
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(c) Z = jω L = j ( 2π 60 ) (10 × 10 −3 ) = 3.771∠90° Ω
I = V Z = ( 277 ∠30° ) ( 3.771 ∠90° ) = 73.46 ∠ − 60° A
i(t ) = 73.46 2 cos (ω t − 60° ) =103.9cos (ω t − 60° ) A
(d) Z = − j 25 Ω I = V Z = ( 277∠30° ) ( 25∠ − 90° ) = 11.08∠120° A i(t ) = 11.08 2 cos (ω t + 120° ) = 15.67cos (ω t + 120° ) A
2.6
(
(a) V = 100
)
2 ∠ − 30°= 70.7∠ − 30° ; ω does not appear in the answer.
(b) υ (t ) = 100 2 cos (ω t + 20° ) ; with ω = 377,
υ (t ) = 141.4 cos ( 377t + 20° ) (c) A = A∠α ; B = B∠β ; C = A + B c(t ) = a(t ) + b(t ) = 2 Re Ce jωt
The resultant has the same frequency ω. 2.7
(a) The circuit diagram is shown below:
(b) Z = 3 + j8 − j 4 = 3 + j 4 = 5∠53.1° Ω (c) I = (100∠0° ) ( 5∠53.1° ) = 20∠ − 53.1° A The current lags the source voltage by 53.1° Power Factor = cos53.1° = 0.6 Lagging 2.8
Z LT = j ( 377 ) ( 30.6 × 10 −6 ) = j11.536 m Ω Z LL = j ( 377 ) ( 5 × 10 −3 ) = j1.885 Ω ZC = − j V=
1 = − j 2.88 Ω ( 377 ) ( 921 × 10−6 )
120 2 2
∠ − 30° V
3 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.