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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
2.1
(a) A1 630 6cos 30 j sin 30 5.20 j 3
(b) A2 4 j5 
16 25 tan 1
5 6.40 128.66 6.40e j128.66 4
(c) A3 5.20 j3 4 j5 1.20 j 8 8.0181.50 (d) A4 6306.40 128.66 38.414158.658 35.78 j13.98 (e) A5 630 / 6.40 128.66 0.94 158.66 0.94e j158.66
2.2
(a) I 500 30 433.01 j250 (b) i(t ) 4 sin t 30 4 cos t 30 90 4 cos t 60 I 4 60 2.83 60 1.42 j 2.45 (c) I 5 / 2 15 4 60 3.42 j0.92 2 j3.46 
2.3
5.42 j 4.38 6.964 38.94 (a) Vmax 400 V ; I max 100 A (b) V 400
2 282.84 V; I 100
(c) V 282.8430 V; I 2.4
(a) I1 100
2 70.71A
70.71 80 A
j6 6 90 10 8 j6 j6 8
7.5 90 A
I 2 I I1 100 7.3 90 10 j7.5 12.536.87 A V I 2 j6 12.536.87 6 90 75 53.13 V (b)
2.5
(a) (t) 277 2 cos t 30 391.7 cos t 30 V I i(t ) Z (c) I
(b)
V / 20 13.8530 A 19.58 cos t 30 A j L j 2 60 10 10 3 3.77190  V Z 277 30 3.77190 73.46 60 A
i(t ) 73.46 2 cos t 60 103.9 cos t 60 A
(d) Z j 25  I V Z 27730 25 90 11.08120 A i(t ) 11.08 2 cos t 120 15.67 cos t 120 A V 75 2.6
2 15 53.03 15 ; does not appear in the answer.
(a) (b) (t) 50 2 cos t 10 ; with 377, (t) 70.71cos 377t 10 (c) A A ;
B B; C A B
c(t) a(t) b(t) 
2 Re Ce jt
The resultant has the same frequency . 2.7
(a) The circuit diagram is shown below:
(b) Z 3 j8 j4 3 j 4 553.1
2.8
(c) I 1000 553.1 20 53.1 A The current lags the source voltage by 53.1 Power Factor cos 53.1 0.6 Lagging 6 Z LT j 377 30.6 10 j11.536 m  j 377 5 10 3 j1.885 Z LL ZC j V
1 j2.88  37792110 6 
120 2 30 V 2
The circuit transformed to phasor domain is shown below:
2.9
KVL : 1200 6000.1 j0.5 VLOAD VLOAD 1200 600 0.1 j0.5
114.1 j30.0 117.9 14.7 V  p(t ) (t)i(t ) 400 cos t 30 100 cos t 80 2.10 (a) 1 400100 cos110 cos 2 t 50 2 6840.4 2 104 cos 2 t 50 W (b) P VI cos 282.8470.71 cos 30 80 6840 W Absorbed +6840 W Delivered (c) Q VI sin 282.84 70.71sin110 18.79 kVAR Absorbed (d) The phasor current I 70.71 80 180 70.71 100 A leaves the positive terminal of the generator. The generator power factor is then cos 30 100 0.3420 leading 2 2.11 (a) p(t ) (t )i(t ) 391.7 19.58 cos t 30 1 0.7669 10 4 1 cos 2 t 60   2 3.834 103 3.834 103 cos 2 t 60 W P VI cos 277 13.85 cos 0 3.836 kW Q VI sin 0 VAR Source Power Factor cos cos 30 30 1.0 (b) p(t ) (t )i(t ) 391.7 103.9 cos t 30 cos t 60 1 4.07 10 4 cos 90 cos 2 t 30 2   2.035 10 4 cos 2 t 30 W P VI cos 277 73.46 cos 30 60 0 W Q VI sin 73.46 sin 90 20.35 kVAR pf cos 0 Lagging
277
p(t ) (t )i(t ) 391.7 15.67 cos t 30 cos t 120
(c)
6.138 103
1
cos 90 cos 2 t 150 3.069 103 cos 2 t 150 W  2 P VI cos 277 11.08 cos 30 120 0 W Q VI sin 277 11.08 sin 90 3.069 kVAR Absorbed 3.069 kVAR Delivered pf cos cos 90 0 Leading pR (t) 359.3 cos t 35.93 cos t  6455 6455 cos 2 t W
2.12 (a)
(b) px (t ) 359.3 cos t 14.37 cos t 90 2582 cos 2 cot 90 2582 sin 2 t W 2
(c) P V 2
R 359.3
2
10 6455 W Absorbed 2
(d) Q V 2 (e) 2.13
X 359.3
2
25 2582 VAR S Delivered
tan 1 Q / P tan 1 2582 6455 21.8
Power factor cos cos 21.8 0.9285 Leading Z R jxc 10 j25 26.93 68.2 
i(t ) 359.3 / 26.93 cos t 68.2 13.34 cos t 68.2 A (a) pR (t) 13.34 cos t 68.2
133.4 cos t 68.2
889.8 889.8 cos 2 t 68.2 W 333.5 cos t 68.2 90
(b) px (t ) 13.34 cos t 68.2 2224 sin 2 t 68.2
W
2
(c) P I 2 R 13.34
2 10 889.8 W
(d) Q I 2 X 13.34
2 25 2224 VAR S
(e)
2
pf cos tan 1 Q / P  0.3714 Leading
cos tan 1 (2224 / 889.8)
2.14 (a) I 20 kA V ZI
3 45 20 6 45 kV
(t ) 6 2 cos t 45 kV p(t ) (t)i(t ) 6 2 cos t 45 2 2 cos t 

1
24 cos 45 cos 2t 45   2 8.49 12 cos 2t 45 MW (b) P VI cos 6 2 cos 45 0 8.49 MW Delivered (c) Q VI sin 6 2 sin 45 0 8.49 MVAR Delivered 8.49 MVAR Absorbed (d) pf cos 2.15 (a) I
4
cos 45 0 0.707 Leading

2 60
 230 
2 30 A
i(t ) 2 cos t 30 A with 377 rad/s p(t ) (t )i(t ) 4 cos 30 cos 2 t 90 3.46 4 cos 2 t 90 W (b) (t), i(t), and p(t) are plotted below. (c) The instantaneous power has an average value of 3.46 W, and the frequency is twice that of the voltage or current.
2.16 (a) Z 10 j120 0.04 10 j15.1 18.156.4  pf cos 56.4 0.553 Lagging (b) V 120 0 V The current supplied by the source is I 120 0 18.156.4 6.63 56.4 A The real power absorbed by the load is given by P 120 6.63 cos 56.4 440 W 2
which can be checked by I 2 R 6.63 10 440 W The reactive power absorbed by the load is Q 120 6.63 sin 36.4 663 VAR 2
(c) Peak Magnetic Energy W LI 2 0.04 6.63 1.76 J Q W 377 1.76 663 VAR is satisfied. 2.17 (a) S V I
*
ZII
*
Z I
2
j LI
2
Q Im[S ] LI 2 (b) (t) L
di  dt
2 L I sin t

p(t ) (t) i(t ) 2 L I 2 sin t cos t 
L I 2 sin 2 t Q sin 2 t

Average real power P supplied to the inductor 0  Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a maximum value of Q.  * * 2 2.18 (a) S V I Z I I Re ZI j Im Z I 2 P jQ 2
P Z I cos Z ; Q Z I 2 sin Z (b) Choosing i(t )  Then (t) 
2 I cost, 2 Z I cos t Z
p(t ) (t ) i(t ) Z I 2 cos t Z cost Z I 2 cos Z cos 2t Z  Z I 2 cos Z cos 2t cos Z sin 2t sin Z  P(1 cos 2t) Q sin 2t
1
(c) Z R j L
jC
From part (a), P RI 2 and Q Q QL where QL LI 2 and Q
1 C
C
I2 C
which are the reactive powers into L and C, respectively. Thus p(t ) P 1 cos 2t QL sin 2t QC sin 2t If 2 LC 1,
Q Q L
Then
2.19 (a) S V I
*
150
Q 0 C
p(t ) P 1 cos 2t
10
 50 375 60 *
5
2 2 187.5 j324.8
 
P Re S 187.5 W Absorbed Q Im S 324.8 VAR S Absorbed (b) pf cos 60 0.5 Lagging 187.5 tan cos 1 0.9 90.81VAR S  QC QL QS 324.8 90.81 234 VAR S
(c) Q P tan Q S
S
1 1 Y1  Z1 2030
2.20
Y 2
1
0.02 j0.03464 S G
2
1
100 0.0433 433 W Absorbed 2
Q V2B
100 0.025 250 VAR S Absorbed 1 2
P V 2G 2
100 0.02 200 W Absorbed 2 2
2
Q V B 2
0.04 60
2560
Z2
1

1
V
P1 V 2 G
0.05 30 0.0433 j0.025S G1 jB1
100 0.03464 346.4 VAR S Absorbed 2
jB 2
2
2.21 (a)
co 1 0s 6 53 . 13 . QL P tan L 500 tan 53.13 666.7 kVAR cos 1 0.9 25.84 S L
QS P tan S 500 tan 25.84 242.2 kVAR QC QL QS 666.7 242.2 424.5 kVAR SC QC 424.5 kVA 500 0.746 414.4 kW and Q 0mkVAR 0.9
(b) The Synchronous motor absorbs Pm
Source PF cos tan 1 666.7 914.4  1
2.22 (a) Y 1
1
1
4 j5 
Z1
0.808 Lagging
0.16 51.34 6.451.34 0.1 j0.12S
1 1 0.1S Y2  Z 2 10 P
2
P V G1 G2 V  P1 V 2 G
G1 G
1000  2
0.1 0.1
2
1
2
P V 2G 2
70.71 0.1 500 W 70.71 0.1 500 W
2
(b) Yeq Y1 Y2 0.1 j0.12 0.1 0.2 j0.12 I S V Yeq
0.233 30.96 S 70.710.233 16.48 A
70.71 V
2.23
S V I * 1200 15 30 1800 30 1558.85 j900 P Re S 1558.85 W Delivered
2.24
Q Im S 900 VAR S Delivered 900 VAR S Absorbed S P jQ 10 j0; S 10 cos 1 0.9 9 j4.359 1
1
1
2
10 0.746 cos 1 0.95 9.238 18.19 8.776 j2.885 0.85 0.95 SS S1 S2 S3 27.78 j1.474 27.82 3.04 S3
PS Re(SS ) 27.78 kW QS Im(SS ) 1.474 kVAR SS SS
2.25
27.82 kVA
SR V IR * RI I * I 2 R (20)2 3 1200 j0 * SL V IL * ( jX I )I jX I 2 Lj8(20)2 0 j3200 L
S V I * ( jIX C
C
)I * jX I 2 j 4(20)2 0 j1600 C
C
Complex power absorbed by the total load SLOAD SR SL SC 200053.1 Power Triangle:
Complex power delivered by the source is * SSOURCE V I * 100 020 53.1 200053.1 The complex power delivered by the source is equal to the total complex power absorbed by the load. 2.26 (a) The problem is modeled as shown in figure below: PL 120 kW pf L 0.85 Lagging cos 1 0.85 31.79 L Power triangle for the load:
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Q P tan 31.79 SL PL jQL 141.1831.79 kVA
L
L
74.364 kVAR
I SL / V 141,180 / 480 294.13 A
Real power loss in the line is zero. 2
I2X
Reactive power loss in the line is Q LINE
294.13 1 LINE
86.512 kVAR SS PS jQS 120 j 74.364 86.512 200.753.28 kVA The input voltage is given by VS SS / I 682.4 V (rms) The power factor at the input is cos 53.28 0.6 Lagging (b) Applying KVL, VS 480 0 j1.0 294.13 31.79 635 j250 682.421.5 V (rms) ( pf ) S cos 21.5 31.79 0.6 Lagging 2.27 The circuit diagram is shown below:
P
50 kW; cos 1 0.8 36.87; 
36.87; Q
old
OLD
P tan old
old
old
37.5 kVAR Sold 50, 000 j37, 500 50, 000 j50, 000 tan 18.19 cos 0.95 18.19; S new new 1
50, 000 j16, 430 Hence Scap Snew Sold j21, 070 VA C
21, 070 377 220
2
1155 F 
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2.28
S1 15 j6.667 S 2 30.96 j3 sin cos 1 0.96 
2.88 j0.84
S3 15 j0 STOTAL S1 S2 S3 32.88 j5.827 kVA (i) Let Z be the impedance of a series combination of R and X *
V2  * , it follows that  Since S V I * V  Z Z  V
240
V2
2
Z*
S 32.88 j5.827 103 Z 1.698 j0.301 
(1.698 j0.301) 
(ii) Let Z be the impedance of a parallel combination of R and X 2
Then
R
240 1.7518  32.88103
X
240 9.885  5.827 103
2
Z 1.7518
j9.885
2.29 Since complex powers satisfy KCL at each bus, it follows that S13 1 j1 1 j1 0.4 j0.2 0.4 j1.8  S S * 0.4 j1.8  31
13
Similarly, S23 0.5 j0.5 1 j1 0.4 j0.2 0.1 j0.7  * S32 S23 0.1 j0.7 
At Bus 3, SG 3 S31 S32 0.4 j1.8 0.1 j0.7 0.5 j1.1  2.30 (a) For load 1: 1 cos 1 (0.28) 73.74 Lagging S1 12573.74 35 j120 S2 10 j 40 S3 15 j0 STOTAL S1 S2 S3 60 j80 10053.13 kVA P jQ PTOTAL 60 kW; QTOTAL 80 kVAR; kVATOTAL STOTAL 100 kVA .  Supply pf cos 53.13 0.6 Lagging  (b) I
S*
100 103 53.13
100 53.13 A
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TOTAL
V*
10000
At the new pf of 0.8 lagging, PTOTAL of 60kW results in the new reactive power Q , such that
13 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
cos 1 0.8 36.87 and Q 60 tan 36.87 45 kVAR The required capacitor’s kVAR is QC 80 45 35 kVAR  It follows then X C
and
1000
V2
2
j 28.57 j35000
SC*
106 2 60 28.57
C
92.85 F 
S*
60, 000 j 45, 000 60 j45 75 36.87 A V* 10000 The supply current, in magnitude, is reduced from 100A to 75A  V  V11V2 2 V1  90 2.31 (a) I 90 2  1 2 12 X X X 90 The new current is I
Complex power S
VI* V
12
1 12

V1 1
90

V2
1
90
1
2
X X 2 V1 VV 90 1 2 90 1 2 X X The real and reactive power at the sending end are P
V12
12
Q 12
cos 90
V1V2
X
V12
cos 90 
X V1V2 1 sin X
sin 90
V1V2
X
1
2
2
sin 90  1
X V1 X
 
V V cos
1
2
2
 1
2

Note: If V1 leads V2 , 1 2 is positive and the real power flows from node 1 to node 2. If V1 Lags V2 , is negative and power flows from node 2 to node 1. (b) Maximum power transfer occurs when 90 PMAX
1 2
V1V2  X
2.32 4 Mvar minimizes the real power line losses, while 4.5 Mvar minimizes the MVA power flow into the feeder.
14 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2.33 Qcap
Mvar Losses
0
0.42
0.84
0.5
0.4
0.8
1
0.383
0.766
1.5
0.369
0.738
2
0.357
0.714
2.5
0.348
0.696
3
0.341
0.682
3.5
0.337
0.675
4
0.336
0.672
4.5
0.337
0.675
5
0.341
0.682
5.5
0.348
0.696
6
0.357
0.714
6.5
0.369
0.738
7
0.383
0.766
7.5
0.4
0.801
8
0.42
0.84
8.5
0.442
0.885
9
0.467
0.934
9.5
0.495
0.99
0.525
1.05
10 2.34
MW Losses
7.5 Mvars
2.35
15 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
.4950 j 4.950 
.3846 .4950 j 10 1.923 4.950  
.4950 j4.950
V10
.3846 .4950 j 10 1.923 4.95
V 20
1.961 48.69 1.961 78.69 1.961 48.69
0.8796 j3.127
0.4950 j 4.950 V 
V10
0.4950 j 4.950
0.8796 j3.127 V 
20
1.961 78.69
2.36 Note that there are two buses plus the reference bus and one line for this problem. After converting the voltage sources in Fig. 2.29 to current sources, the equivalent source impedances are: 0.1 j0.5 j0.1 0.1 j0.5 j0.1
0.1 j0.5 // j0.1
Z S1 Z S 2
0.509978.690.1 90 0.412375.96 0.005882 j0.1235 
0.1237 87.27
The rest is left as an exercise to the student. 2.37 After converting impedance values in Figure 2.30 to admittance values, the bus admittance matrix is: 1 1
2 Ybus
1 1
1 1
3 1
0
0 1 j1
1  j1 4
1
3 1
j1
3
j1 j
3
4 1
j
1
4
1
0
0
j

4
j 2
1
1
4 1
j
1
4
4
0
V10 
0.25
VV20
j0.25
V30
j
1 4

3
Writing nodal equations by inspection: 1
1
0 j1
1 2.083

0.3333 j1
0 0.3333 j10.3333 j0.25  0
0.25

j0.25
0.25 j0.08333 V


10 0 0  230 40
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2.38 The admittance diagram for the system is shown below:
Y12
Y13
Y14 
Y22
Y23
Y24
Y31 Y32
Y33
Y34
Y41 Y42
Y43 Y44
Y11 YBUS
Y21
8.5 2.5 j
5.0
2.5 8.75 5.0
0 0
S
5.0 5.0
22.5 12.5
0
12.5 12.5
0
Y11 y10 y12 y13 ; Y22 y20 y12 y23 ; Y23 y13 y23 y34
where
Y44 y34 ; Y12 Y21 y12 ; Y13 Y31 y13 ; Y23 Y32 y23 and
Y34 Y43 y34
2.39 (a) Yc Yd Y f
Yd
Yd
(b) j
Yb Yd Ye
Yc
Yb
Yf
Ye
Yc
Yf
V1
Yb
Ye 0
VV2 I 2 0
Ya Yb Yc
2.5
4
8.8
5
0 Y
0 8.3
V I ;Y 1
V3  V4 Y
I3

I4

14.5 8 4 2.5 V1 0 17 4 5 VV2 0 8 4
 Ye Y f Yg V4
0
V3
I1 0
1 90 0.62 135 V Y1 I
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BUS
BUS
BUS
BUS
18 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Y1
where
0.7187
 Z
BUS
0.6688 0.7045
0.6688 j  BUS
0.6307 0.6194
0.7045 0.6258
0.6307 0.6242 0.6840 0.5660
0.6194 0.6258  0.5660  0.6840
1 V YBUS I
V1 where
V
0 VV2 V  and I 3 V4 


0 1
90 0.62 135
Then solve for V1 , V2 , V3 , and V4 . 240
2.40 (a) VAN
0 138.560 V
(Assumed as Reference)
3 VAB 24030 V; VBC 240 90 V ; I A 15 90 A Z Y
(b) I AB  Z
VAN 138.56 0 9.2490 0 j9.24  IA 15 90 IA 3
30
15
90 30 8.66 60A 3
Vq AB 24030 27.7190 0 j27.71 I AB 8.66 60
Note: ZY Z / 3 2.41
1 3VLL I L cos pf
S3 
3 480 20 cos 1 0.8 16.627 103 36.87 13.3 103 j(9.976 103 ) P3 Re S3 13.3 kW
Delivered
Q3 I m S3 9.976 kVAR Delivered 2.42 (a) With Vab as reference
Van
208
30
Z
4 j3 536.87
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3
3 Ia
Van 120.130 24.02 66.87 A 536.87 Z  / 3
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S3
an a
3V I * 3 120.1 3024.02 66.87 865436.87 6923 j5192
P3 6923 W; Q3 5192 VAR ; both absorbed by the load pf cos 36.87 0.8 Lagging; S3 S3
8654 VA
(b)
Vab 2080 V
I a 24.02 66.87 A
13.87 36.87 A
2.43 (a) Transforming the -connected load into an equivalent Y, the impedance per phase of the equivalent Y is 60j 45 Z2 3 With the phase voltage V1
120 3 3
20 j15 
120 V
taken as a reference, the per-phase equivalent
circuit is shown below:
Total impedance viewed from the input terminals is Z 2 j4 I
30 j 40 20 j15  2 j 4 22 j 4 24  30 j 40 20 j15 
V1 1200 50 A Z 24
The three-phase complex power supplied S 3V1 I * 1800 W P 1800 W and Q 0 VAR
delivered by the sending-end source
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(b) Phase voltage at load terminals V2 1200 2 j4 5 0 110 j20 111.8 10.3 V The line voltage magnitude at the load terminal is VLOAD 
 3 111.8 193.64 V
L -L
(c) The current per phase in the Y-connected load and in the equiv.Y of the -load: V2 1 j2 2.236 63.4 A Z1
I
1
VV2
I 2
4 j2 4.472 26.56 A
Z2
The phase current magnitude in the original -connected load I ph 
I2
4.472 3
2.582 A
3
(d) The three-phase complex power absorbed by each load is 1
S 3V2 I1 * 430 W j600 VAR
2
S 3V2 I2 * 1200 W j900 VAR
The three-phase complex power absorbed by the line is I 2 32 j 4 (5)2 150 W j300 VAR
S 3 R jX L
L
L
The sum of load powers and line losses is equal to the power delivered from the supply: S1 S2 SL 450 j600 1200 j900 150 j300  1800 W j0 VAR 2.44 (a) The per-phase equivalent circuit for the problem is shown below:
2200 3 2200 V taken as Ref. 3 Total complex power at the load end or receiving end is Phase voltage at the load terminals is V2
SR3 560.10.707 j0.707 132 528 j396 66036.87 kVA
22 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
With phase voltage V2 as reference, * RV3
S 660, 00036.87 100 36.87 A 3V2* 3 22000
I
Phase voltage at sending end is given by V1 22000 0.4 j2.7100 36.87 2401.74.58 V The magnitude of the line to line voltage at the sending end of the line is V1 
L -L
 3V1 
3 2401.7 4160 V
(b) The three-phase complex-power loss in the line is given by 2
SL 3 3RI
2
j3 I
2
2
3 0.4 100
j3 2.7100
12 kW j81kVAR (c) The three-phase sending power is S
S 3
3V I *1 32401.74.5810036.87 540 kW j 477 kVAR
Note that SS 3 SR3 SL 3  2.45 (a)
IS
S
S 3VLL
25.001 103 30.07 A 3 480
(b) The ammeter reads zero, because in a balanced three-phase system, there is no neutral current. 2.46 (a)
Using voltage division: V AN
208
0
Z/ 3V
V an
3
( Z  / 3) Z
1030 1030 0.8 j0.6
120.091030
Load voltage = VAB
LINE
1200.930  10.9930.62
9.46 j5.6 109.3 0.62 V  3 109.3 189.3 V Line-to-Line
(b)
Z eq 10 30 j20 11.547 0 V AN
q V an
q
Z eq
Z eq Z LINE
208
11.547 3 11.547 0.8 j0.6 
1386.7 12.3622.78
112.2 2.78 V
Load voltage Line-to-Line VAB 
3 112.2 194.3 V
2.47
(a) I G1
15 103 cos 1 0.8 23.53 36.87 A 8 460 0.8 460 0 1.4 j1.6 23.53 36.87 3 216.9 2.73 V Line to Neutral Load Voltage VL  3 216.9 375.7 V Line to line
VL VG1 Z LINE1 I G1
30 10 3 3 375.7 0.8
(b) I L
2.73
cos1 0.8 57.63  39.6 A
I G 2 I L I G1 57.63 39.6 23.53 36.87 34.14 41.49 A VG 2 VL Z LINE 2 I G 2 216.9 2.73 0.8 j134.14 41.49 259.7 0.63 V 3 259.7
Generator 2 line-to-line voltage VG 2 
449.8 V (c)
*
SG 2 3VG 2 2I G 3 259.7 0.63
34.1441.49
20.12 103 j17.4 103 PG 2 20.12 kW; QG 2 17.4 kVAR; Both delivered 2.48 (a)
(b) pf cos 31.32 0.854 Lagging L
(c) I L
26.93 10 3 32.39 A
S 3VLL
3 480 2
(d) Q Q 14 103 VAR 3 V C
L
LL 2
X (e)
3 480 49.37  14 103 I C VLL / X 480 / 49.37 9.72 A
I LINE
L
P 3 VLL
23 103 27.66 A 3 480
/X 
2.49 (a) Let ZY Z A Z B ZC
for a balanced Y-load Z Z AB Z BC Z CA
Using equations in Fig. 2.27 Z 2 qZY 2 Z Y2
Yq
Z
3Z Y
ZY and Z
Z Y
(b) Z
ZB
A
Z
2
Z
Z
Z
3
j10 j25 j 50  j10 j20 j25 j10 j20 j5
j 40 ; Z
j20 j25 C
2.50 Replace delta by the equivalent WYE:
j100 
j5
Y
Z
j
2 3
Per-phase equivalent circuit is shown below:
Noting that j1.0 j
2
j 2 , by voltage-divider law, 3 V1 
1
j2 1000 1050 j2 j0.1
(t ) 105 2 cos (t 0) 148.5 cost V 
In order to find i2 (t) in the original circuit, let us calculate VAB VAB VAN  VBN  3 e j 30 V AN  173.230
Then
I AB
173.230 86.6120 j2
i2 (t ) 86.6 2 cos t 120 122.5 cos t 120 A
1 2.51 On a per-phase basis S1  3 150 j120 50 j40 kVA 50j 40 103 I1  2000
25 j20 A Note: PF Lagging
Load 2: Convert into an equivalent Y 1 Z 2Y  150 j 48 50 j16  3 I2
20000 38.117.74 50 j16 36.29 j11.61 A Note : PF Leading
S3 per phase
1
120 0.6 j 120 sincos 1 0.6  3 I3 
24 j32 kVA
24j32 10 3 2000
12 j16 A Note :PF Leading
Total current drawn by the three parallel loads I T I1 I 2 I 3 I TOTAL 73.29 j7.61 A Note :PF Leading Voltage at the sending end: VAN 20000 73.29 j 7.610.2 j1.0  2007.05 j74.81 2008.442.13 V Line-to-line voltage magnitude at the sending end  32008.44 3478.62 V
2160
2.52 (a) Let VAN be the reference: V AN
3
0 ≃ 24000 V
Total impedance per phase Z 4.7 j9 0.3 j1 5 j10
Line Current
24000 214.7 63.4 A I 5 j10

A
With positive A-B-C phase sequence, I B 214.7 183.4 A; I C 214.7 303.4 214.756.6 A
(b) VAN  LOAD
VBN
24000 214.7 63.40.3 j1 24000 224.159.9 2179.2 j38.54

2179.5 1.01 V
LOAD

2179.5 121.01 V ; VC N□
(c) S / Phase VAN 
LOAD
2179.5 241.01 V LOAD
I A 2179.5214.7 467.94 kVA
Total apparent power dissipated in all three phases in the load S3
3 467.94 1403.82 kVA LOAD
Active power dissipated per phase in load P1  LOAD
2179.5214.7 cos 62.39 216.87 kW P3
3 216.87 650.61 kW LOAD
Reactive power dissipated per phase in load Q1  LOAD
2179.5214.7sin 62.39 414.65 kVAR Q3
3 414.65 1243.95 kVAR LOAD
(d) Line losses per phase P1

LOSS
2
214.7 0.3 13.83 kW
□
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LOSS