Chapter 10 Solutions Essay

10
Sinusoidal Steady State Power
Calculations

Assessment Problems
AP 10.1 [a] V = 100/ − 45◦ V,
I = 20/15◦ A
Therefore
1
P = (100)(20) cos[−45 − (15)] = 500 W,
2
Q = 1000 sin −60◦ = −866.03 VAR,
[b] V = 100/ − 45◦ ,

B→A

I = 20/165◦

P = 1000 cos(−210◦ ) = −866.03 W,

B→A

Q = 1000 sin(−210◦ ) = 500 VAR,
[c] V = 100/ − 45◦ ,

A→B

I = 20/ − 105◦

P = 1000 cos(60◦ ) = 500 W,

A→B

Q = 1000 sin(60◦ ) = 866.03 VAR,
[d] V = 100/0◦ ,

A→B

A→B

I = 20/120◦

P = 1000 cos(−120◦ ) = −500 W,

B→A

Q = 1000 sin(−120◦ ) = −866.03 VAR,

B→A

AP 10.2 pf = cos(θv − θi ) = cos[15 − (75)] = cos(−60◦ ) = 0.5 leading rf = sin(θv − θi) = sin(−60◦ ) = −0.866
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
10–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

10–2
AP 10.3

CHAPTER 10. Sinusoidal Steady State Power Calculations

Iρ
0.18
From Ex. 9.4 Ieff = √ = √ A
3
3
2
P = Ieff
R=

0.0324
(5000) = 54 W
3

AP 10.4 [a] Z = (39 + j26) (−j52) = 48 − j20 = 52/ − 22.62◦ Ω
Therefore I =

250/0◦
= 4.85/18.08◦ A (rms)
48 − j20 + 1 + j4

VL = ZI = (52/ − 22.62◦ )(4.85/18.08◦ ) = 252.20/ − 4.54◦ V (rms)
IL =

VL
= 5.38/ − 38.23◦ A (rms)
39 + j26

[b] SL = VL I∗L = (252.20/ − 4.54◦ )(5.38/ + 38.23◦ ) = 1357/33.69◦
= (1129.09 + j752.73) VA
PL = 1129.09 W;

QL = 752.73 VAR

[c] P = |I |2 1 = (4.85)2 · 1 = 23.52 W;

Q = |I |24 = 94.09 VAR

[d] Sg (delivering) = 250I∗ = (1152.62 − j376.36) VA
Therefore the source is delivering 1152.62 W and absorbing 376.36 magnetizing VAR.
|VL |2
(252.20)2
[e] Qcap =
=
= −1223.18 VAR
−52
−52
Therefore the capacitor is delivering 1223.18 magnetizing VAR.
Check:

94.09 + 752.73 + 376.36 = 1223.18 VAR and
1129.09 + 23.52 = 1152.62 W

AP 10.5 Series circuit derivation:
S = 250I∗ = (40,000 − j30,000)
Therefore I∗ = 160 − j120 = 200/ − 36.87◦ A (rms)
I = 200/36.87◦ A (rms)
Z=

V
250
=
= 1.25/ − 36.87◦ = (1 − j0.75) Ω
I
200/36.87◦

Therefore R = 1 Ω,

XC = −0.75 Ω

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problems

10–3

Parallel circuit derivation
P =

(250)2
;
R

therefore R =

Q=

(250)2
;
XC

therefore XC =

(250)2
= 1.5625 Ω
40,000
(250)2
= −2.083 Ω
−30,000

AP 10.6
S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA
S2 = 6000(0.8) − j6000(0.6) = 4800 − j3600 VA
ST = S1 + S2 = 13,800 + j8400 VA
ST = 200I∗ ;

therefore I∗ = 69 + j42

I = 69 − j42 A

Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91◦ V (rms)
AP 10.7 [a] The phasor domain equivalent circuit and the Th´evenin equivalent are shown below:
Phasor domain equivalent circuit:

Th´evenin equivalent:

VTh = 3

−j800
= 48 − j24 = 53.67/ − 26.57◦ V
20 − j40

ZTh = 4 + j18 +

−j800
= 20 + j10 = 22.36/26.57◦ Ω
20 − j40

For maximum power transfer, ZL = (20 − j10) Ω
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc.,

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