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

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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,

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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

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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.,…