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
2.15 a (ii) b
2.16 b (iii) a
2.17 A. a (iv) d
B. b 2.31 a
C. a 2.32 a
2.18 c
1
© 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.
, 2.1 (a) A1 = 5∠30° = 5 [ cos30° + j sin 30°] = 4.33 + j 2.5
4
(b) A2 = −3 + j 4 = 9 + 16 ∠ tan −1 = 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°
(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
I = 5
(c) I = ( 4 2 ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 )
= 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
− j6 6∠ − 90°
2.4 (a) I1 = 10∠0° = 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
2
© 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.
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
2.15 a (ii) b
2.16 b (iii) a
2.17 A. a (iv) d
B. b 2.31 a
C. a 2.32 a
2.18 c
1
© 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.
, 2.1 (a) A1 = 5∠30° = 5 [ cos30° + j sin 30°] = 4.33 + j 2.5
4
(b) A2 = −3 + j 4 = 9 + 16 ∠ tan −1 = 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°
(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
I = 5
(c) I = ( 4 2 ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 )
= 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
− j6 6∠ − 90°
2.4 (a) I1 = 10∠0° = 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
2
© 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.
