Chapter 9 Solutions
9
Sinusoidal Steady State Analysis
Assessment Problems
AP 9.1 [a] V = 170/−40◦ V
[b] 10 sin(1000t + 20◦ ) = 10 cos(1000t − 70◦ )
. ·.
I = 10/−70◦ A
[c] I = 5/36.87◦ + 10/−53.13◦
= 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A
[d] sin(20,000πt + 30◦ ) = cos(20,000πt − 60◦ )
Thus,
V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60)
= 162.13 + j298.73 = 339.90/61.51◦ mV
AP 9.2 [a] v = 18.6 cos(ωt − 54◦ ) V
[b] I = 20/45◦ − 50/ − 30◦ = 14.14 + j14.14 − 43.3 + j25
= −29.16 + j39.14 = 48.81/126.68◦
Therefore i = 48.81 cos(ωt + 126.68◦ ) mA
[c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76
= −8.98 + j72.24 = 72.79/97.08◦ v = 72.79 cos(ωt + 97.08◦ ) V
AP 9.3 [a] ωL = (104 )(20 × 10−3 ) = 200 Ω
[b] ZL = jωL = j200 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
9–1 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.
9–2
CHAPTER 9. Sinusoidal Steady State Analysis
[c] VL = IZL = (10/30◦ )(200/90◦ ) × 10−3 = 2/120◦ V
[d] vL = 2 cos(10,000t + 120◦ ) V
−1
−1
=
= −50 Ω ωC 4000(5 × 10−6 )
[b] ZC = jXC = −j50 Ω
30/25◦
V
[c] I =
=
= 0.6/115◦ A
ZC
50/−90◦
[d] i = 0.6 cos(4000t + 115◦ ) A
AP 9.4 [a] XC =
AP 9.5 I1 = 100/25◦ = 90.63 + j42.26
I2 = 100/145◦ = −81.92 + j57.36
I3 = 100/−95◦ = −8.72 − j99.62
I4 = −(I1 + I2 + I3 ) = (0 + j0) A,
AP 9.6 [a] I =
therefore i4 = 0 A
125/−60◦
125
/(−60 − θZ )◦
=
|Z|/θz
|Z|
But −60 − θZ = −105◦
.·. θZ = 45◦
Z = 90 + j160 + jXC
.·. XC = −70 Ω;
. ·. C =
[b] I =
XC = −
1
= −70 ωC 1
= 2.86 µF
(70)(5000)
Vs
125/−60◦
=
= 0.982/−105◦ A;
Z
(90 + j90)
.·. |I| = 0.982 A
AP 9.7 [a]
ω = 2000 rad/s ωL = 10 Ω,
−1
= −20 Ω ωC © 2010 Pearson
Sinusoidal Steady State Analysis
Assessment Problems
AP 9.1 [a] V = 170/−40◦ V
[b] 10 sin(1000t + 20◦ ) = 10 cos(1000t − 70◦ )
. ·.
I = 10/−70◦ A
[c] I = 5/36.87◦ + 10/−53.13◦
= 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A
[d] sin(20,000πt + 30◦ ) = cos(20,000πt − 60◦ )
Thus,
V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60)
= 162.13 + j298.73 = 339.90/61.51◦ mV
AP 9.2 [a] v = 18.6 cos(ωt − 54◦ ) V
[b] I = 20/45◦ − 50/ − 30◦ = 14.14 + j14.14 − 43.3 + j25
= −29.16 + j39.14 = 48.81/126.68◦
Therefore i = 48.81 cos(ωt + 126.68◦ ) mA
[c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76
= −8.98 + j72.24 = 72.79/97.08◦ v = 72.79 cos(ωt + 97.08◦ ) V
AP 9.3 [a] ωL = (104 )(20 × 10−3 ) = 200 Ω
[b] ZL = jωL = j200 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
9–1 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.
9–2
CHAPTER 9. Sinusoidal Steady State Analysis
[c] VL = IZL = (10/30◦ )(200/90◦ ) × 10−3 = 2/120◦ V
[d] vL = 2 cos(10,000t + 120◦ ) V
−1
−1
=
= −50 Ω ωC 4000(5 × 10−6 )
[b] ZC = jXC = −j50 Ω
30/25◦
V
[c] I =
=
= 0.6/115◦ A
ZC
50/−90◦
[d] i = 0.6 cos(4000t + 115◦ ) A
AP 9.4 [a] XC =
AP 9.5 I1 = 100/25◦ = 90.63 + j42.26
I2 = 100/145◦ = −81.92 + j57.36
I3 = 100/−95◦ = −8.72 − j99.62
I4 = −(I1 + I2 + I3 ) = (0 + j0) A,
AP 9.6 [a] I =
therefore i4 = 0 A
125/−60◦
125
/(−60 − θZ )◦
=
|Z|/θz
|Z|
But −60 − θZ = −105◦
.·. θZ = 45◦
Z = 90 + j160 + jXC
.·. XC = −70 Ω;
. ·. C =
[b] I =
XC = −
1
= −70 ωC 1
= 2.86 µF
(70)(5000)
Vs
125/−60◦
=
= 0.982/−105◦ A;
Z
(90 + j90)
.·. |I| = 0.982 A
AP 9.7 [a]
ω = 2000 rad/s ωL = 10 Ω,
−1
= −20 Ω ωC © 2010 Pearson