R. Allan Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
CHAPTER 5
Exercises
E5.1 (a) We are given v (t ) = 150 cos(200t - 30 o ) . The angular frequency is the coefficient of t so we have = 200 radian/s . Then f = / 2 = 100 Hz T = 1 / f = 10 ms Vrms = Vm / 2 = 150 / 2 = 106.1 V Furthermore, v(t) attains a positive peak when the argument of the cosine function is zero. Thus keeping in mind that t has units of radians, the positive peak occurs when
tmax = 30
180
tmax = 0.8333 ms
2 (b) Pavg = Vrms / R = 225 W
(c) A plot of v(t) is shown in Figure 5.3 in the book.
E5.2
We use the trigonometric identity sin(z ) = cos(z - 90 o ). Thus 100 sin(300t + 60 o ) = 100 cos(300t - 30 o )
E5.3
= 2f 377 radian/s
T = 1 / f 16.67 ms
o
Vm = Vrms 2 155.6 V
The period corresponds to 360 therefore 5 ms corresponds to a phase angle of (5 / 16.67) 360 o = 108o . Thus the voltage is
v (t ) = 155.6 cos(377t - 108o )
E5.4
(a)
10 cos(t ) + 10 sin(t ) = 14.14 cos(t - 45 o ) (b)
V1 = 100 o + 10 - 90 o = 10 - j 10 14.14 - 45 o
I1 = 1030 o + 5 - 60 o 8.660 + j 5 + 2.5 - j 4.330 11.16 + j 0.670 11.183.44 o 10 cos(t + 30 o ) + 5 sin(t + 30 o ) = 11.18 cos(t + 3.44 o )
(c)
I2 = 200 o + 15 - 60 o 20 + j 0 + 7.5 - j 12.99 27.5 - j 12.99 30.41 - 25.28o 20 sin(t + 90 o ) + 15 cos(t - 60 o ) = 30.41 cos(t - 25.28o )
120
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
E5.5
The phasors are V1 = 10 - 30 o V2 = 10 + 30 o and V3 = 10 - 45 o
v1 lags v2 by 60 o (or we could say v2 leads v1 by 60 o ) v1 leads v3 by 15 o (or we could say v3 lags v1 by 15 o ) v2 leads v3 by 75 o (or we could say v3 lags v2 by 75 o )
E5.6
(a) Z L = jL = j 50 = 5090 o
IL = VL / Z L = 100 / j 50 = 2 - 90 o
VL = 1000 o
(b) The phasor diagram is shown in Figure 5.10a in the book. E5.7 (a) Z C = 1 / jC = - j 50 = 50 - 90 o V = 1000 o C
IC = V / Z C = 100 /( - j 50) = 290 o C
(b) The phasor diagram is shown in Figure 5.10b in the book. E5.8 (a) Z R = R = 50 = 500 o VR = 1000 o IR = VR / R = 100 /(50) = 20 o (b) The phasor diagram is shown in Figure 5.10c in the book. E5.9 (a) The transformed network is:
10 - 90 o = 28.28 - 135 o mA I= = Z 250 + j 250
Vs
121
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
i (t ) = 28.28 cos(500t - 135 o ) mA VR = RI = 7.07 - 135 o VL = jLI = 7.07 - 45 o
(b) The phasor diagram is shown in Figure 5.16b in the book. (c) i(t) lags vs(t) by 45 o. E5.10 The transformed network is:
Z =
1 = 55.47 - 56.31 o 1 / 100 + 1 /( - j 50) + 1 /( + j 200)
V = ZI = 277.4 - 56.31 o V
IL = V /( j 200) = 1.387 - 146.31 o A IR = V /(100) = 2.774 - 56.31 o A E5.11
The transformed network is:
IC = V /( - j 50) = 5.547 33.69 o A
We write KVL equations for each of the meshes: j 100I1 + 100( I1 - I2 ) = 100 - j 200 I2 + j 100I2 + 100( I2 - I1 ) = 0 Simplifying, we have (100 + j 100) I1 - 100I2 = 100 - 100 I1 + (100 - j 100) I2 = 0 Solving we find I1 = 1.414 - 45 o A and I2 = 10 o A. Thus we have
i1 (t ) = 1.414 cos(1000t - 45 o ) A and i2 (t ) = cos(1000t ).
122
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
E5.12
(a) For a power factor of 100%, we have cos( ) = 1, which implies that Also I rms = P /[ rms cos( )] = 5000 /[500 cos(0)] = 10 A. Thus we have V the current and voltage are in phase and = 0. Thus, Q = P tan( ) = 0.
I m = I rms 2 = 14.14 and I = 14.1440 o.
(b) For a power factor of 20% lagging, we have cos( ) = 0.2, which implies that the current lags the voltage by = cos-1 (0.2) = 78.46o. Thus, Thus we have I m = I rms 2 = 70.71 A and I = 70.71 - 38.46o. (c) The current ratings would need to be five times higher for the load of part (b) than for that of part (a). Wiring costs would be lower for the load of part (a).
Q = P tan( ) = 24.49 kVAR. Also, we have I rms = P /[Vrms cos( )] = 50.0 A.
E5.13
The first load is a 10 F capacitor for which we have Z C = 1 /( jC ) = 265.3 - 90 o C = -90 o I Crms = Vrms / Z C = 3.770 A
PC = Vrms I Crms cos( C ) = 0
QC =Vrms I Crms sin( C ) = -3.770 kVAR
The second load absorbs an apparent power of Vrms I rms = 10 kVA with a power factor of 80% lagging from which we have 2 = cos -1 (0.8) = 36.87 o. Notice that we select a positive angle for 2 because the load has a lagging power factor. Thus we have P2 = Vrms I 2rms cos( 2 ) = 8.0 kW and Q2 = Vrms I 2rms sin( ) = 6 kVAR .
Now for the source we have: Ps = PC + P2 = 8 kW Qs = QC + Q2 = 2.23 kVAR
Vrms I srms = Ps 2 + Qs2 = 8.305 kVA I srms = Vrms I srms /Vrms = 8.305 A
V power factor = Ps /( rms I srms ) 100% = 96.33%
E5.14
First, we zero the source and combine impedances in series and parallel to determine the Thvenin impedance.
123
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Zt = 50 - j 25 +
1 = 50 - j 25 + 50 + j 50 1 / 100 + 1 / j 100
= 100 + j25 = 103.114.04 o
Then we analyze the circuit to determine the open-circuit voltage.
V = V = 100 t oc
100 = 70.71 - 45 o 100 + j 100
In =Vt / Zt = 0.6858 - 59.04 o
E5.15
(a) For a complex load, maximum power is transferred for Z L = Zt* = 100 - j 25 = RL + jX L . The Thvenin equivalent with the load attached is:
The current is given by 70.71 - 45 o I= = 0.3536 - 45 o 100 + j 25 + 100 - j 25 The load power is 2 PL = RL I rms = 100(0.3536 / 2 ) 2 = 6.25 W
124
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(b) For a purely resistive load, maximum power is transferred for RL = Zt = 100 2 + 252 = 103.1 . The Thvenin equivalent with the load attached is:
The current is given by 70.71 - 45 o I= = 0.3456 - 37.98 o 103.1 + 100 - j 25 The load power is 2 PL = RL I rms = 103.1(0.3456 / 2 ) 2 = 6.157 W E5.16 The line-to-neutral voltage is 1000 / 3 = 577.4 V. No phase angle was specified in the problem statement, so we will assume that the phase of Van is zero. Then we have Van = 577.40 o Vbn = 577.4 - 120 o Vcn = 577.4120 o The circuit for the a phase is shown below. (We can consider a neutral connection to exist in a balanced Y-Y connection even if one is not physically present.)
The a-phase line current is Van 577.40 o IaA = = = 4.610 - 37.02o Z L 100 + j 75.40 The currents for phases b and c are the same except for phase. IbB = 4.610 - 157.02o IcC = 4.61082.98o
V I 577.4 4.610 P = 3 Y L cos( ) = 3 cos(37.02 o ) = 3.188 kW
2 2
125
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
VI 577.4 4.610 Q = 3 Y L sin( ) = 3 sin(37.02o ) = 2.404 kVAR
2 2
E5.17
The a-phase line-to-neutral voltage is Van = 1000 / 3 0 o = 577 . 4 0 o The phase impedance of the equivalent Y is ZY = Z / 3 = 50 / 3 = 16.67 . Thus the line current is V 577.40 o = 34.630 o A IaA = an = ZY 16.67 Similarly, IbB = 34.63 - 120 o A and IcC = 34.63120 o A. Finally, the power is P = 3(I aA / 2 ) 2 Ry = 30.00 kW
Problems
P5.1 The units of angular frequency are radians per second. The units of frequency f are hertz which are equivalent to inverse seconds. The relationship between them is = 2f .
P5.2*
v (t ) = 10 sin(1000t + 30 o ) = 10 cos(1000 - 60 o )
= 1000 rad/s
f = 500 Hz
phase angle = = -60 o = - 3 radians
T = 1 f = 2 ms Vrms = Vm
2 = 10 2 = 7.071 V
P = ( rms )2 R = 1 W V
First positive peak occurs for 1000t peak - 3 = 0
t peak = 0.3333 ms
126
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.3
v (t ) = 50 sin(500t + 120 o ) = 50 cos(500t + 30 o )
= 500 rad/s
f = 250 Hz
phase angle = = 30 o = 6 radians
T = 1 f = 4 ms Vrms = Vm
2 = 50 2 = 35.36 V
P = ( rms )2 R = 25 W V
Positive peak occurs for 500t1 + 6 = 0 The first positive peak after t = 0 is at t peak = t1 +T = 3.667 ms
t1 = -0.3333 ms
P5.4
p (t ) = Ri 2 (t ) = 10 4 cos 2 (2000 t ) = 5000[1 + cos(4000 t )] W Pavg = R (I rms )2 = 100 10
i (t ) = 10 cos(2000 t ) A
(
2
)
2
= 5000 W
127
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.5
v (t ) = 1000 sin(500t ) V
p (t ) = v 2 (t ) R = 2000 sin2 (500t ) = 1000[1 - sin(1000t )] W Pavg = ( rms )2 R = 1000 W V
P5.6*
Sinusoidal voltages can be expressed in the form v (t ) =Vm cos(t + ).
V The peak voltage is Vm = 2 rms = 2 20 = 28.28 V. The frequency is
f = 1 /T = 10 kHz and the angular frequency is = 2f = 210 4 radians/s. The phase corresponding to a time interval of t = 20 s is = (t /T ) 360 o = 72o . Thus we have v (t ) = 28.28 cos(210 4t - 72o ) V.
P5.7
I m = I rms 2 = 5 2 = 7.071 A
f =
T
1
= 83.33 Hz
128
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
tmax = -90 o T o i (t ) = 7.071 cos(523.6t - 90 ) = 7.071 sin(523.6t )
= 2f = 523.6 rad/s = -360
P5.8
A sequence of MATLAB instructions to accomplish the desired plot for part (a) is Wx = 2*pi; Wy = 2*pi; Theta = 90*pi/180; t = 0:0.01:20; x = cos(Wx*t); y = cos(Wy*t + Theta); plot(x,y) By changing the parameters, we can obtain the plots for parts b, c, and d. The resulting plots are
(a)
(b)
(c)
(d)
129
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.9*
Vrms
i 2 = v (t )dt T 0
= 1 25dt 2 0
1
T
= 3.536 V
P5.10
Vrms
i T 2 v (t )dt = T 0
= = = = = 1 2 [5 + 10 cos(20 t )] dt 0 .1 0 1 2 [25 + 100 cos(20 t ) + 100 cos (20 t )]dt 0 .1 0
0.1 0.1 0.1 1 25dt + 100 cos(20 t )dt + 100 cos 2 (20 t )]dt 0 .1 0 0 0 0.1 0.1 0.1 0.1 1 25dt + 100 cos(20 t )dt + 50dt + 50 cos( 40 t )]dt 0 .1 0 0 0 0 0.1 0.1
1 [2.5 + 0 + 5 + 0] 0 .1 = 8.660 V
P5.11*
Vrms =
=
T
1
T
v
0
2
(t )dt
2
0.5 0
[5 sin(2t )] dt
0.5 0
[12.5 + 12.5 sin( 4t )]dt
t = 0.5
12.5 = 12.5t - cos( 4 t ) 4 t =0 = 6.25 = 2.5 V
130
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.12
Vrms =
=
T
1
1
T
v
0
2
(t )dt
2
[A cos(2t ) + B sin(2t )] dt
0 0.5 0
[A
2
cos 2 (2t ) + 2AB cos(2t ) sin(2t ) + B 2 sin 2 (2t ) dt
t =1
]
A2 B2 = t +0+ t 2 2 t =0 A2 + B 2 = 2
P5.13
Vrms =
=
1
T
1
1
T
v
0
2
(t )dt
2
[3 exp( -t )] dt
0
[9 exp( -2t )]dt
0
=
[- 4.5 exp( -2t )]tt =10 =
= 4.5[1 - exp( -2)] = 1.973 V
P5.14
Vrms =
= =
T
1
T
v (t )dt
2 0
1 2 (2.5t ) dt 20 1 t3 6.25 2 3 t =0
t =2
2
= 2.887 V
131
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.15*
I rms =
=
T
1
T
i (t )dt
2 0
2 4 1 4dt + 1dt 4 0 2
= 1.581 A
P5.16
The rms values of all periodic waveforms are not equal to their peak values divided by the square root of two. However, they are for sinusoids, which are important special cases.
P5.17*
v 2 (t ) = 100 sin(t ) = 100 cos (t - 90 o )
V1 = 100 0 o = 100 V2 = 100 - 90 o = - j 100 Vs = V1 + V2 = 100 - j 100 = 141 .4 - 45 o
v 1 (t ) = 100 cos (t )
v s (t ) = 141.4 cos (t - 45 o )
V lags V by 90 o 2 1 Vs lags V by 45o 1 Vs leads V by 45o 2
P5.18
v 2 (t ) = 150 sin (t + 60 o ) = 150 cos (t - 30 o )
V1 = 100 45 o = 70.71 + j 70.71 V2 = 150 - 30 o = 129 .9 - j 75 Vs = V1 + V2 = 200 .6 - j 4.29 = 200 .6 - 1.23 o
v 1 (t ) = 100 cos (t + 45 o )
v s (t ) = 200 .6 cos (t - 1.23o )
132
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
V2 lags V1 by 75 o Vs lags V1 by 46 .23 o Vs leads V2 by 28 .77 o
P5.19
The magnitudes of the phasors for the two voltages are 8 2 and 3 2 V. The phase angles are not known. If the phase angles are the same, the phasor sum would have its maximum magnitude which is 11 2 . On the other hand, if the phase angles differ by 180 o , the phasor sum would have its minimum magnitude which is 5 2 . Thus, the maximum rms value of the sum is 11 V and the minimum is 5 V.
P5.20
Vm = 3 V
T = 0.5 s
= 2f = 4 rad/s
f =
tmax = -45 o T v (t ) = 3 cos( 4t - 45 o ) V
= -360 o
T
1
= 2 Hz
V = 3 - 45 o V 3 Vrms = = 2.121 V 2 P5.21*
= 2f = 400
v 1 (t ) = 10 cos (400 t + 30 o ) v 2 (t ) = 5 cos( 400 t + 150 o ) v 3 (t ) = 10 cos (400 t + 90 o ) v 1 (t ) lags v 2 (t ) by 120 o v 1 (t ) lags v 3 (t ) by 60 o v 2 (t ) leads v 3 (t ) by 60 o
133
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.22
v 1 (t ) = 10 cos(t + 30 o ) I 1m = 2 I 1rms = 7.071
V1 = 1030 o I1 = 7.07150 o
i1 (t ) = 7.071 cos(t + 50 o )
P5.23*
We are given the expression 5 cos(t + 75 o ) - 3 cos(t - 75 o ) + 4 sin(t ) Converting to phasors we obtain 575 o - 3 - 75 o + 4 - 90 o = 1.2941 + j 4.8296 - (0.7765 - j 2.8978) - j 4 = 0.5176 + j 3.7274 = 3.76382.09 o Thus, we have 5 cos(t + 75 o ) - 3 cos(t - 75 o ) + 4 sin(t ) =
3.763 cos(t + 82.09 o )
P5.24
We are given the expression 5 sin(t ) + 5 cos(t + 30 o ) + 5 cos(t + 150 o ) Converting to phasors we obtain 5 - 90 o + 530 o + 5150 o = - j 5 + 4.3301 + j 2.500 - 4.3301 + j 2.5 = 0 Thus, we have 5 sin(t ) + 5 cos(t + 30 o ) + 5 cos(t + 150 o ) = 0
P5.25
A sequence of MATLAB commands to generate the desired plot is: t = 0:0.01:2; v = cos(19*pi*t) + cos(21*pi*t); plot(v,t) The resulting plot is
134
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Notice that the first term cos(19t) has a frequency of 9.5 Hz while the second term cos(21t) has a frequency of 10.5 Hz. At t = 0, the rotating vectors are in phase and add constructively. At t = 0.5 the vector for the second term has rotated one half turn more than the vector for the first term and the vectors cancel. At t = 1, the first vector has rotated 9.5 turns and the second vector has rotated 10.5 turns so they both point in the same direction and add constructively. P5.26 For an inductance, we have VL = jLIL . I For a capacitance, we have V = C . C jC For a pure resistance, current and voltage are in phase. For a pure inductance, current lags voltage by 90 o. For a pure capacitance, current leads voltage by 90 o.
P5.27
P5.28*
v L (t ) = 10 cos(2000 t )
= 2000
Z L = jL = j 200 = 200 90 o
IL = VL Z L = (1 20 ) - 90 o VL = 100 o
i L (t ) = (1 20 ) cos (2000 t - 90 o ) = (1 20 ) sin(2000 t )
135
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
iL (t ) lags v L (t ) by 90 o
P5.29*
v C (t ) = 10 cos(2000 t )
= 2000
ZC =
-j = - j 15.92 = 15.92 - 90 o C VC = 100 o
iC (t ) = 0.6283 cos(2000 t + 90 o ) = -0.6283 sin(2000 t )
IC = VC Z C = 0.628390 o
iC (t ) leads v C (t ) by 90 o
P5.30
(a) Notice that the voltage is a sine rather than a cosine. V V = 100 - 60 o I = 130 o Z = = 100 - 90 o = - j 100 I Because Z is pure imaginary and negative, the element is a capacitance. 1 = 200 C = = 50 F Z
V = 2500 o = 250 + j 0 I Because Z is pure real, the element is a resistance of 250 .
(b) V = 50050 o
I = 250 o
Z =
136
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(c) Notice that the current is a sine rather than a cosine. V V = 10030 o I = 1 - 60 o Z = = 10090 o = j 100 I Because Z is pure imaginary and positive, the element is an inductance. = 400
L=
Z
= 0.25 H 1 C
P5.31*
Z = jL + R - j
= 500 :
Z = j 50 + 50 - j 200
= 50 - j 150 = 158.1 - 71.57 o
= 1000 : Z = j 100 + 50 - j 100
= 50 = 500 o
= 2000 : Z = j 200 + 50 - j 50 = 50 + j 150 = 158.171.57 o
P5.32
Z =
1 1 = 1 Z L + 1 Z c 1 jL + jC
= 500 :
Z =
= j 66.67 = 1000 : Z = 1
1 1 ( j 50 ) + j 0.005
1 ( j 100 ) + j 0.01 = 1 = 2000 : Z = 1 ( j 200 ) + j 0.02 = - j 66.67 =
P5.33
Z =
1 / R + 1 Z L + 1 Zc
1
1 1 / R + 1 jL + jC
137
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
= 500 :
Z =
=
1 / 100 + 1 ( j 5) + j 0.05 1 0.01 - j 0.15
1
= 6.65286.19 o = 0.4425 + j 6.637 Because the imaginary part is positive, the impedance is inductive. 1 = 1000 : Z = 1 / 100 + 1 ( j 10 ) + j 0.1 = 1 0.01 + j 0
= 100 Because the imaginary part is zero, the impedance is purely resistive. 1 = 2000 : Z = 1 / 100 + 1 ( j 20 ) + j 0.20 = 1 0.01 + j 0.15
= 6.652 - 86.19 o = 0.4425 - j 6.637 Because the imaginary part is negative, the impedance is capacitive.
P5.34*
I=
Vs R + j L
100 o = 100 + j 100 = 70.71 - 45 o mA VR = RI = 7.071 - 45 o V VL = jLI = 7.07145 o V
I lags Vs by 45 o P5.35
I= =
Vs R + j L 100 o 100 + j 50
= 89.44 - 26.56o mA VR = RI = 8.944 - 26.56o V VL = jLI = 4.47263.44 o V
138
I lags Vs by 26.56o
Allan R. Hambley, Electrical Principles Engineering: and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.36
I= =
R - j C
100 o 1000 - j 1000
Vs
= 7.07145 o mA VC = (- j C )I = 7.071 - 45 o V VR = RI = 7.07145 o V
I leads Vs by 45 o P5.37*
I= = Vs
R - j C
100 o 1000 - j 2000
= 4.47263.43o mA VC = (- j C )I = 8.944 - 26.57 o V VR = RI = 4.47263.43o V
I leads Vs by 63.43o P5.38
Is = 0.50 o V = Is 1 1 200 + 1 j 100
= 44.7263.44 o IR = V R = 0.223663.44 o IL = V jL = 0.4472 - 26.56o
V leads Is by 63.44 o P5.39*
Is = 1000 o mA V = Is 1 1 100 + 1 (- j 200 )
= 8.944 - 26.56o V IC = V ( jC ) = 44.7263.44 o mA IR = V R = 89.44 - 26.56o mA
V lags Is by 26.56o 139
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.40
Vs = 100 o I= = Vs jL + R - j C 10 j 500 + 100 - j 500
= 0.10 o A VL = j 500 I = 5090 o VR = 100 I = 100 o VC = - j 500 I = 50 - 90 o
The peak value of v L (t ) is five times larger than the source voltage! This is possible because the impedance of the capacitor cancels the impedance of the inductance. P5.41*
Is = 100 o mA V = Is 1 1 R + 1 jL + jC 1 1 1000 - j 0.005 + j 0.005
= 10 -2
= 100 o V IR = V R = 100 o mA IC = V ( jC ) = 5090 o mA IL = V jL = 50 - 90 o mA
The peak value of iL (t ) is five times larger than the source current! This is possible because current in the capacitance balances the current in the inductance (i.e., IL + IC = 0 ). P5.42
V1 = 100 - 90 o = - j 100 I= V2 = 10030 o =
R + jL
V1 - V2
=
- j 100 - (86.60 + j 50 ) 100 + j 50
o
- 86.60 - j 150 100 + j 50
= 1.549 - 146.6
VR = 100 I = 154.9 - 146.6o VL = j 50 I = 77.45 - 56.6o
140
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
I lags V by 56.6o 1 I lags VL by 90 o
P5.43
Ztotal = jL +
1 1 R + jC 1 0.01 + j 0.01
= j 100 +
= j 100 + 50 - j 50 = 50 + j 50 = 70.7145o I= 1000o
Ztotal
= 1.414 - 45o
IR = I
- j 100 ZC = (1.414 - 45o ) 100 - j 100 R + ZC 100 R = (1.414 - 45o ) 100 - j 100 R + ZC
= 1 - 90o IC = I
= 10o
P5.44 Writing KCL equations at nodes 1 and 2 we obtain V1 V - V2 + 1 = 10 o 10 5 + j 15 V2 V - V1 + 2 = 130 o - j 10 5 + j 15 Solving these equations, we obtain V1 = 6.735 - 38.54 o V2 = 16.25 - 55.52o
141
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.45*
First we write the KVL equation: V1 - V2 = 100 o Then we enclose nodes 1 and 2 in a closed surface to form a supernode and write a KCL equation: V1 V V V + 1 + 2 + 2 =0 10 j 20 15 - j 5 The solution to these equations is: V1 = 9.40229.58o V2 = 4.986111.45 o
P5.46*
Writing KVL equations around the meshes, we obtain 5I1 + j 15( I1 - I2 ) = 20 - j 10I2 + j 15( I2 - I1 ) = 10 Solving, we obtain: I1 = 1.64480.54 o I2 = 2.977 74.20 o
P5.47
The current through the current source is I1 - I2 = 2 Writing KVL around the perimeter of the circuit, we have 5I1 + (10 + j 5) I2 = 10 Solving, we obtain: I1 = 20 o and I2 = 0
P5.48
Writing KVL equations around the meshes, we obtain 10I1 + j 20( I1 - I2 ) = 0 j 20( I2 - I1 ) + 15( I2 - I3 ) = -10 - j 5I3 + 15( I3 - I2 ) = 0 Solving, we obtain: I1 = 0.9402 - 150.4 o I2 = 1.051 - 177.0 o I3 = 0.9972 - 158.6o
142
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.49
The KCL equation is
V1 - 20 V V + 10 + 1 + 1 = 0. Solving, we find j 15 - j 10 5
V1 = 20.33 - 23.50 o V. P5.50 P5.51 The KCL equation is
V1 - 10 V 1 + + 2 = 0. Solving, we find V1 = 0 V. 5 10 + j 5
(a) For a pure resistance, the power is positive and the reactive power is zero. (b) For a pure inductance, the power is zero and the reactive power is positive. (c) For a pure capacitance, the power is zero, and the reactive power is negative.
P5.52 P5.53*
See Figure 5.22 in the book. This is a capacitive load because the reactance is negative. 2 P = I rms R = (15) 2 100 = 22.5 kW
2 Q = I rms X = (15) 2 ( -50) = -11.25 kVAR Q = tan -1 = tan -1 ( -0.5) = 26.57 o P
power factor = cos( ) = 89.44% P5.54
= v - i = 30 o - 60 o = -30 o
power factor = cos( ) = 86.60% leading P = Vrms I rms cos( ) = 12.99 kW Q = Vrms I rms sin( ) = -7.5 kVAR
Z =
P5.55
V 1000 230 o = = 66.67 - 30 o o I 15 260
Vrms = 10 4 V
I rms = 20 A = v - i = 10 o - ( -20 o ) = 30 o power factor = cos 100% = 86.60% lagging P =Vrms I rms cos() = 173.2 kW Q =Vrms I rms sin() = 100 kVAR Apparent Power = Vrms I rms = 200 KVA
This is an inductive load. 143
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.56
I=
240 250 o - 220 230 o = 52.0352.71o 1 + j2
I rms = 36.79 A
Delivered by Source A: PA = 240I rms cos(50 - 52.71) = 8.820 kW Absorbed by Source B: PB = 220I rms cos(30 - 52.71) = 7.467 kW Absorbed by resistor: 2 PR = I rms R = 1.353 kW Absorbed by inductor: 2 QL = I rms X = 2.707 kVAR P5.57
VA = (10 + j 10)10 2170 o + 240 2 - 20 o = 278.1 - 56.10 o
QA = 240I rms sin(50 - 52.71) = -0.418 kVAR
QA = 220I rms sin(30 - 52.71) = -3.125 kVAR
VArms = 278.1 / 2 = 196.6 V rms
Delivered by Source A: PA = 196.6(10) cos(-56.10 - 170) = -1.364 kW Absorbed by Source B: PB = 240(10) cos(-20 - 170) = -2.364 kW
QA = 196.6(10) sin( -56.10 - 170) = 1.417 kVAR
QB = 240(10) sin( -20 - 170) = 0.4168 kVAR
Absorbed by resistor: 2 PR = I rms R = 1.000 kW Absorbed by inductor: 2 QL = I rms X = 1.000 kVAR P5.58
1000 20 o 1000 20 o I= + j 188.5 100 = 14.14 - j 7.502 = 16.01 - 27.95 o
P = Vrms I rms cos = 10 kW Q = Vrms I rms sin = 5.305 kVAR Apparent power =Vrms I rms = 11.32 kVA
Power factor = cos(27.95 o ) = 0.8834 = 88.34% lagging
144
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.59*
I=
1000 20 o 1000 20 o + = 14.14 + j 5.331 100 - j 265.3
= 15.1120.66o
P = Vrms I rms cos = 10 kW Q = VrmsI rms sin = -3.770 kVAR Apparent power = Vrms I rms = 10.68 kVA
Power factor = cos(20.66o ) = 0.9357 = 93.57% leading
P5.60
R Q = QL + QC
= =
P =
( rms )2 V
=
500 2 = 5000 W 50
( rms )2 ( rms )2 V V
XL
+
XC
500 2 500 2 + 188.5 ( -265.3) = 383.9 VAR
Apparent power = P 2 + Q 2 = 5015 VA Power factor =
P5.61 I= 500 20 o 500 20 o = R + jL - j C 50 + j 188.5 - j 265.3
Apparent power
P
= 99.71% lagging
= 5.458 256.93o
Q = QL + QC = X L (I rms )2 + X C (I rms )2
= 188.5(5.458) - 265.3(5.458)
2 2
P = R (I rms )2 = 1490 W
= 5611 - 7897 = -2286 VAR Apparent power = Vrms I rms = 2728 VA Power factor = cos(56.93) = 0.5457 = 54.57% leading
145
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.62*
Load A:
QA = PA tan A = 4.843 kVAR
Load B:
A = cos -1 (0.9 ) = 25 .84 o
PA = 10 kW
Vrms I Brms = 15 kVA
QB = Vrms I Brms sin (B ) = 9 kVAR PB = Vrms I Brms cos (B ) = 12 kW
Source:
B = cos -1 (0.8 ) = 36 .87 o
Ps = PA + PB = 22 kW Qs = QA + QB = 13.84 kVAR
Apparent power = Power factor =
(Ps )2 + (Qs )2
Ps
= 26 kVA = 0.8462 = 84.62% lagging
Apparent power
P5.63
Load A:
QA = PA tan A = 2.421 kVAR
Load B:
A = cos -1 (0.9 ) = 25.84 o
PA = 5 kW
B = cos -1 (0.8) = -36.86o
PB = 10 kW
QB = PB tan B = -7.5 kVAR
Source:
( is negative for a leading power factor.)
Ps = PA + PB = 15 kW Qs = QA + QB = -5.079 kVAR
Apparent power = Power factor =
(Ps )2 + (Qs )2
Ps
= 15.84 kVA = 0.9472 = 94.72% leading
Apparent power
146
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.64*
(a)
cos = 0.25 = 75.52 o
P = Vrms I rms cos( ) 100 kW P = = 400 A I rms = Vrms cos( ) 1 kV (0.25 )
I = 400 2 - 75.52 o
(b)
Qload = Vrms I rms sin = 387 .3 kVAR Qtotal = 0 = Qload + QC QC = -387 .3 10 =
3
( rms )2 V
XC
X C = -2.582 = - C = 1027 F
1 C
The capacitor must be rated for at least 387.3 kVAR. With the capacitor in place, we have:
P = 100 kW = Vrms I rms I rms = 100 A
I = 100 0o
(c) The line current is smaller by a factor of 4 with the capacitor in place, reducing I 2R losses in the line by a factor of 16.
P5.65
Zt.
The ac steady state Thvenin equivalent circuit for a two-terminal circuit consists of a phasor voltage source Vt in series with a complex impedance
The ac steady state equivalent circuit for a two-terminal circuit consists of a phasor current source In in parallel with a complex impedance Zt.
Vt is the open-circuit voltage of the original circuit. In is the short circuit current of the original network. The impedance can be found by zeroing the independent sources and finding the impedance looking into the terminals of the original network. Also we have V = Zt It . t
147
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.66
To attain maximum power, the load must equal (a ) the complex conjugate of the Thvenin impedance if the load can have any complex value; (b) the magnitude of the Thvenin impedance if the load must be a pure resistance. (a) Zeroing the current source, we have:
P5.67*
Thus, the Thvenin impedance is Zt = 100 + j 50 = 111.826.57 o Under open circuit conditions, there is zero voltage across the inductance, the current flows through the resistance, and the Thvenin voltage is V = Voc = 2000 o t
In = V Zt = 1.789 - 26.57 o t
Thus, the Thvenin and Norton equivalent circuits are:
(b) For maximum power transfer, the load impedance is Z load = 100 - j 50
Iload =
Pload = Rload (I rms -load )2 = 100(1 / 2 ) 2 = 50 W
V 200 t = =1 Zt + Z load 100 + j 50 + 100 - j 50
(c) In the case for which the load must be pure resistance, the load for maximum power transfer is Z load = Zt = 111.8
I load =
Pload = Rload (I rms -load )2 = 47.21 W
V 200 t = = 0.9190 - 13.28 o Zt + Z load 100 + j 50 + 111.8
148
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.68
Zeroing sources, we have:
Thus, the Thvenin impedance is 1 Zt = = 4.47263.43o = 2 + j 4 1 10 + 1 j 5 Writing a current equation for the node at the upper end of the current source under open circuit conditions, we have Voc - 10045 o Voc + =5 10 j5
V = V = 62.5693.80 o t oc In = V Zt = 13.9930.36o t
Thus, the Thvenin and Norton equivalent circuits are:
For the maximum power transfer, the load impedance is Z load = 2 - j 4
Iload = V 62.5693.80 o t = = 15.6493.80 o Zt + Z load 2 + j 4 + 2 - j 4
Pload = Rload (I rms -load )2 = 244.6 W
In the case for which the load must be pure resistance, the load for maximum power transfer is Z load = Zt = 4.472
I load =
Pload = Rload (I rms -load )2 = 151.2 W
V 62.5693.80 o t = = 8.223 - 62.08 o Zt + Z load 2 + j 4 + 4.472
149
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.69
For maximum power transfer, the impedance of the load should be the complex conjugate of the Thvenin impedance:
Z load = 10 - j 5 = Rload - j (C load )
Setting real parts equal: Rload = 10 Setting imaginary parts equal: - 5 = - 1 (C load )
C load = 1 (5) = 530.5 F
P5.70*
For maximum power transfer, the impedance of the load should be the complex conjugate of the Thvenin impedance:
Z load = 10 - j 5 Yload = 1 Z load = 0.08 + j 0.04 Yload = 1 Rload + jC load = 0.08 + j 0.04
Setting real parts equal: 1 Rload = 0.08
Rload = 12.5
Setting imaginary parts equal: C load = 0.04
C load = 106.1 F
P5.71
We are given v an (t ) = 100 cos(377t + 90 o ) (a) By inspection,
= 2f = 377
f = 60 Hz
(b) v (t ) = 100 cos(377t - 30o ) bn
vcn (t ) = 100 cos(377t - 150o ) vcn (t ) = 100 cos (377t - 30 o )
(c) v bn (t ) = 100 cos (377t - 150 o )
150
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.72
We are given:
v bn (t ) = 100 cos(t + 60 o ) v cn (t ) = -100 cos(t )
The phasor diagram is:
v an (t ) = 100 cos(t - 60 o )
For counterclockwise rotation, the sequence of phasors is acb. Thus, this is a negative sequence source. From the phasor diagram, we can determine that
Vab = Van - Vbn = 100 3 - 90 o Vbc = Vbn - V = 100 3 + 30 o cn V = Vcn - Van = 100 3 + 150 o ca
v bc (t ) = 100 3 cos(t + 30 o )
v ab (t ) = 100 3 cos(t - 90 o )
v ca (t ) = 100 3 cos(t + 150 o )
P5.73*
VL = 3 VY = 3 440 = 762.1 V rms V 440 = 14.67 A rms IL = Y = 30 R P = 3VYI L cos( ) = 3 440 14.67 cos(0 )
= 19.36 kW
151
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
P5.74*
ZY =
=
1 1 R + j C 1 1 50 + j 377 10 -4
= 10.98 - j 20.70 = 23.43 - 62.05 o
Z = 3ZY
= 70.29 - 62.05 o
P5.75
Vab = Van - Vbn = 3 VY - 30 o Vbc = Vbn - Vcn = 3 VY 90 o Vca = Vcn - Van = 3 VY - 150 o
P5.76
VY =
VL
3
=
440 = 254.0 V rms 3
Van = 254 20 o Vbn = 254 2 - 120 o Vcn = 254 2120 o
152
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vab = 440 230 o Vbc = 440 2 - 90 o Vca = 440 2150 o Ia = Van = 2.054 2 - 66.15o 50 + j 377 (0.3)
Ib = 2.054 2 - 186.15 o Ic = 2.054 253.85 o
P = 3 Yrms I Lrms cos( ) = 3 254 2.054 cos(66.15 o ) V Q = 3 Yrms I Lrsm sin( ) = 3 254 2.054 sin(66.15 o ) V
= 1431 VAR = 632.9 W
P5.77
This is a positive sequence source. The phasor diagram is shown in Figure 5.40 in the book. Thus, we have 440 2 0 o 3 The impedance of an equivalent wye-connected load is
Van =
= 3.333 - j 0.6667 3 The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:
ZY =
Z
Thus, the line current is
153
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
IaA =
Van 0. 5 + j . 5 + Z Y
VAn = Van - IaA (0.5 + j 0.5)
= 225.1 2 - 8.821 o
= 66.21 22.490 o
VAB = 389.9 221.18o IAB = VAB
Z
= 38.23 232.49 o
Pload = 3(I ABrms )2 10 = 43.85 kW
Pline = 3(I aArms )2 0.5 = 6576 kW
P5.78*
This is a positive sequence source. The phasor diagram is shown in Figure 5.40 in the book. Thus, we have:
Van =
440 2 0 o 3
The impedance of a equivalent wye-connected load is
ZY =
Z
3
= 1.667 - j 0.6667
The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:
Thus, the line current is:
154
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
IaA =
Van 0.5 + j 0.5 + ZY
VAn = Van - IaA (0.5 + j 0.5)
= 209.8 2 - 17.40 o
= 116.9 24.40 o
VAB = 363.4 212.60 o IAB = VAB
Z
= 67.49 234.40 o
Pload = 3(I ABrms )2 5 = 68.32 kW
Pline = 3(I aArms )2 0.5 = 20.50 kW
P5.79
The line-to-line voltage is 240 3 = 415.7 V rms. The impedance of each arm of the delta is 1 = 4.47263.44 o Z = 1 / 10 + 1 /( j 5) The equivalent wye has impedances of 3 Then working with one phase of the wye-wye, we have 240 = 161.0 A rms I line =
ZY =
Z
= 1.49163.44 o
ZY
power factor = cos(63.44 o ) 100% = 44.71% P = 3(240)(161.0)(0.4471) = 51.83 kW
P5.80
As suggested in the hint given in the book, the impedances of the circuits between terminals a and b with c open must be identical.
155
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Equating the impedances, we obtain: Z Z + Z B ZC 1 Za + Zb = = A C 1 / Z C + 1 /(Z A + Z B ) Z A + Z B + Z C Similarly for the other pairs of terminals, we obtain Z Z + Z B ZC 1 Z a + Zc = = A B 1 / Z B + 1 /(Z A + Z C ) Z A + Z B + Z C
(1)
(2)
Zb + Zc =
Z Z + ZAZB 1 = A C 1 / Z A + 1 /(Z C + Z B ) Z A + Z B + Z C
(3)
Then adding the respective sides of Equations 1 and 2, subtacting the corresponding sides of Equation 3, and dividing boths sides of the result by 2, we have:
Za =
Z B ZC Z A + Z B + ZC
Similarly we obtain:
Zb =
P5.81
Z AZC Z A + Z B + ZC
and
Zc =
ZAZB Z A + ZB + ZC
As suggested in the hint, consider the circuits shown below. The admittances of the circuits between terminals must be identical.
First, we will solve for the admittances of the delta in terms of the impedances of the wye. Then we will invert the results to obtain relationships between the impedances. Z b + Zc 1 YC + YB = = (1) 1 Z a Z b + Z b Zc + Z a Zc Za + 1 / Z b + 1 / Zc Similarly working with the other terminials, we obtain
156
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Za + Zb (2) Z a Z b + Z b Zc + Z a Zc Z a + Zc YA + YC = (3) Z a Zb + Z b Zc + Z a Zc YA + YB =
Then adding the respective sides of Equations 2 and 3, subtacting the corresponding sides of Equation 1, and dividing boths sides of the result by 2, we have:
YA =
Za Z a Zb + Z b Zc + Z a Zc
Inverting both sides of this result yields:
ZA =
Z a Z b + Z b Zc + Z a Zc Za
Similarly, we obtain:
ZB =
Z a Z b + Z b Zc + Z a Zc Zb
and Z C =
Z a Z b + Zb Zc + Z a Zc Zc
157