Chapter 2 Additional Problems
X2.1 Three circuit elements, R  10  , C  50  F , and L  140  H , are connected in series
and excited by a 60 Hz sinusoidal source. Identify and find the value of two seriesconnected elements that could be substituted without change in the source current.
The network impedance is
Z  R  j L  j
1
C


Z  10  j 2  60  140  106  j

1
2  60  50  106

Z  10  j53 
The network can be modeled by a series-connected R and C where
R  10 
C
1
1

 50  F
X
2  60  53
The contribution of the inductor is negligible for the frequency of this problem.
X2.2 Three circuit elements, R  10  , C  50  F , and L  140  H , are connected in parallel
and excited by a 60 Hz sinusoidal source. Identify and find the value of two parallelconnected elements that could be substituted without change in the source current.
The network admittance is
Y
1
1
j
 jC
R
L
Y
1
1
j
 j 2  60  50  106
6
10
2  60  140  10




Y  0.1  j18.928 
The network can be modeled by a parallel-connected R and L where
R
1
1

 10 
G 0.1
L
1
1

 140.141  H
B
2  60  18.928 
1
X2.3 Three circuit elements, R  10  , C  50  F , and L  140  H , are connected in series
and excited by a 60 Hz sinusoidal source. Identify and find the value of two parallelconnected elements that could be substituted without change in source current.
The network impedance is
Z  R  j L  j
1
C


Z  10  j 2  60  140  106  j

1
2  60  50  106

Z  10  j53  53.935  79.32 
Whence, the network admittance is
Y
1
1

 0.00344  j 0.01822 S
Z 53.935  79.32
The network can be modeled by a parallel-connected R and C where
R
1
1

 290.7 
G 0.00344
C
B


0.01822
 48.33  F
2  60 
X2.4 For the circuit of Fig. 2.20, let V1  V2  1200 V , Z x  Z y  10  , and Z z  j10  .
(a) Determine the value of current I1 and (b) the values of P1 and Q1 (average and
reactive power supplied to the network by source V1 ).
(a)
Ix 
V1 1200

 120 A
Zx
10
Iz 
V1  V2 2400

 24  90 A
Zz
1090
By KCL,
I1  I x  I z  12  j 24  26.83  63.43 A
(b)
SV 1  P1  jQ1  V1I1*  1200  26.83  63.43 
SV 1  1200  26.8363.43   3219.663.43 VA
2
So,
P1  SV 1 cos1  3219.6cos  63.43  1440.1 W
Q1  SV 1 sin1  3219.6sin  63.43  2879.6 VARs
X2.5 For the circuit of Fig. 2.20, let V1  2V2  2400 V and Z x  Z y  Z z  10  .
(a) Determine the value of current I g and (b) the value of average power supplied by
source V2 .
(a) By Ohm's law,
Ix 
V1 2400

 240 A
Zx
10
Iy 
V2 1200

 120 A
Zy
10
Iz 
V1  V2 3600

 360 A
Zz
10
By KCL,
I g  I x  I y  240  120  120 A
(b) By KCL,
I 2  I y  I z  120  360  480 A
The average power flow from source V2 is
P2  V2 I 2 2  120  48 cos  0   5760 W
X2.6 The instantaneous voltage and current for the network of Fig. 2.1 (shown in the phasor
domain are v  t   100sin 120 t  60 V and i  t   10cos 120 t  30  A . (a)
Calculate the value of average power (P) flowing into the network and (b) determine
the values of a series-connected R-L that model the network.
(a) The phasor terminal voltage and current are
V
100
100
  150 
30 V
2
2
I 
10
  30
2
3
Thus,
 100  10 
P  VI cos  

 cos  60   250 W
 2  2 
(b)
R
P
250

2
I
10 / 2
Z
V 100 / 2

 10 
I 10 / 2

X  Z 2  R2 
L
X



2
5 
102   52  8.66 
8.66
 22.97 mH
120
X2.7 For the one-port network of Fig. 2.1, V  V  100 V , I  100 A , and the average
power flowing into the port is P  500 W . Determine the values of a parallelconnected R and L that model the network if the source frequency is 60 Hz.
V 2 100 
R

 20 
P
500
2
The reactive power to the network is
VI 2  P 2
Q  S 2  P2 

10002   5002  866.03 VARs
V 2 100 
X

 11.547 
Q 866.03
2
L
X


11.547
 30.63 mH
2  60 
X2.8 Determine the phase angle of voltage V in Problem X2.7.
500 
P
1 
  60  V  I
  cos 
 VI 
 100 10  
  cos 1 
Since
I  0 ,
V  60 , or
V  10060 V
4
X2.9 For the network of Fig. X2.1, the two loads are
Load 1: 20 kW @ 0.8 PF lagging
Load 2: 10 kW @ 0.707 PF leading
The loads are fed from the 60 Hz source VS through a feeder line with impedance
Z  j 0.2  . Determine currents I1 , I S , and source voltage VS if the load voltage is
VL  4800 V .
With load average powers known and VL on the reference,
I1 
P1
20,000

 52.083 A
VL PF1 480  0.8 
I1  I1  cos 1  PF1   52.083  36.87 A
I2 
P2
10,000
 cos 1  PF2  
 cos 1  0.707   29.46745 A
VL PF2
480
0.707
 

By KCL,
I S  I1  I 2  52.083  36.87  29.46745  63.36  9.46 A
By KVL,
VS  Z I S  VL   0.290  63.36  9.46   4800  482.251.49 V
X2.10 For the network of Fig. X2.1, Load 1 is described by 10 kw @ 0.866 PF lagging and
Load 2 is a 6  resistor. Load voltage VL  2400 V . The feeder line has an
impedance Z  0.5  90  . Determine (a) the source current I S , (b) the average
power PS supplied by the source, and (c) the value of source voltage VS .
5
(a)
I1 
P1
10,000
  cos 1  PF1  
  cos 1  0.866   48.11  30 A
VL PF1
 240 0.866
I2 
VL 2400

 400 A
6
6
By KCL,
I S  I1  I 2  48.11  30  400  85.13  16.4 A
(b)
 240   19.6 kW
V2
PS  P1  L  10,000 
6
6
2
(c) By KVL,
VS  Z I S  VL   0.5  90  85.13  16.4   2400
VS  231.6  10.15 V
X2.11 Describe a single equivalent load for the two loads of Problem X2.9.
Using the value of I S from Problem X2.9,
Seq  VL I S*   4800 63.369.46
Seq  5068.89.46  4999.9  j833.1 VA
The PF of the equivalent load is
PFeq  cos  9.46   0.986 lagging
Thus,
Loadeq : Peq @ PFeq  4999.9 W @ 0.986 PF lagging
X2.12 Determine the value of a capacitor C that if placed across the parallel-connected loads
of Problem X2.9 would correct the combined load PF to unity.
From the solution of Problem X2.11, the total lagging VARs for the combined
loads is Qeq  833.1 VARs . Thus,
V 2  480 
XC  L 
 276.56 
Qeq
833.1
2
C
1
1

 9.59  F
 X C 2  60  276.56 
6
X2.13 The three-phase network of Fig. X2.2 is balanced with a  b  c phase sequence. If
I AB  1040 , determine the three line currents.
Based on [2.26],
I aA  I AB 3 e j 30  1040


3  30  17.3210 A
By three-phase symmetry,
IbB  17.32  110 A
IcC  17.32130 A
X2.14 Rework Problem X2.13 if the phase sequence is a  c  b and all else is unchanged.
Since there is no applicable formula developed in the text for a  c  b phase
sequence, I aA will be determined by use of KCL. For a  c  b sequence,
I BC  10160 A
ICA  10  80 A
But,
I aA  I AB  ICA  1040  10  80  17.3270 A
By three-phase symmetry,
IbB  17.32  170 A
IcC  17.32  50 A
X2.15 The three-phase network of Fig. X2.2 is balanced with a  b  c phase sequence. Let
Z  0 and Z L  3  j 4  . If I AB  4820 A , determine the value of source voltage
Van .
Since Z  0 ,
Vab  VAB  I AB Z L   4820  553.13  240  73.13  V
7
X2.16 Rework Problem X2.15 except let Z  j 0.1  .
Based on [2.26],
I aA  I AB 3 e j 30   4820


3  30  83.14  10 A
By three-phase symmetry,
IbB  83.14  130
By KVL and using VAB as determined in Problem X2.15,
Vab  I aA Z  VAB  IbB Z
Vab   83.14  10  0.190   24073.13  83.14  130  0.190   251.6775.1 V
Based on [2.16],
Van 
Vab e j 30
3
 145.3145.1 V
X2.17 The three-phase network of Fig. X2.2 is balanced with a  b  c phase sequence. Let
Z  j 0.1  and Z L  3  j 4  . If Van  1200 V , determine (a) the value of load
current I AB and (b) load voltage VBC .
(a) By delta-wye conversion, phase a of the network can be modeled as shown in Fig.
X2.3. Whence,
I aA 
Van
1200

 68.66  55.1 A
Z  Z L / 3 j 0.1   3  j 4  / 3
Based on [2.26], the required load current is
I AB 
I aAe j 30
3

 68.67  55.1 130  39.64  25.1 A
3
8
(b) By Ohm's law,
VAB  I AB Z L   39.64  25.1  553.13   198.228.03 V
From three-phase symmetry,
VBC  198.2  91.97 V
2.18 For the network of Problem X2.16, calculate (a) the total complex power supplied by
the three-phase source and (b) the total complex power delivered to the load.
(a) Based on [2.66], the source-supplied complex power is
STS  3 Vab I aA  3  251.67  83.14  45.1  10
STS  36.2455.1 kVA
(b) Using the result for VAB from Problem X2.15 and [2.55], the load complex power
is
*
STL  3VAB I AB
 3  24073.13  48  20   34.5653.13 kVA
X2.19 Calculate the total average power supplied to the delta-connected load for Problem
X2.17.
Since the line impedance is pure reactive, the average power may be calculated
at either the source or load. The average power delivered to the load using the results
of Problem X2.17 is
PTL  3 VAB I aA cos  Z L   3 198.2  68.66  cos  53.13   14.142 kW
X2.20 The three-phase network of Fig. X2.4 is balanced with a  b  c phase sequence. Let
R  X  10  . If the load voltage VAB  4800 , find (a) the value of source current
Iba and (b) the total average power supplied to the load.
9
(a) Based on [2.16],
VAN 
VAB
330

4800
 277.14  30 V
330
Then,
I aA 
VAN
277.14  30

 19.6  75 A
R  jX
10  j10
Based on [2.29],
Iba 
I aA
3  30

19.6  75
 11.32  45 A
3  30
(b) The load power is
2
PTL  3I aA
R  3 19.6 10   588 W
10