AnalysisofUnbalancedThree-PhaseCircuits
8.19
Solution
(a) We solve the problem by finding out the phasor voltage at load neutral with respect to the ground
(earth) node G by using the KCL equation at load neutral. The sum of currents flowing away from that
node is set to zero.
V
V
V
V
−
∠ °
+
+
+
−
∠ −
°
+
+
−
∠
°
−
=
230 9 0
10
0 1
0 3
230 9
120
8
6
230 9 120
8
6
.
.
.
.
.
j
j
j
00
1
10
1
0 1
0 3
1
8
6
1
8
6
230 9 0
10
230 9
,
. .,
.
.
.
.
i e
j
j
j
V
+
+
+
+
+
−
=
∠ °
+
∠
∠ −
°
+
+
∠
°
−
120
8
6
230 9 120
8
6
j
j
.
.
Solving for
V
, we get,
V
=
-
2.31
-
j5.49
=
5.96
∠-
112.8
°
V rms.
Now,
we can find wire currents as
I
I
R
Y
=
∠ ° −
∠ −
°
=
∠
°
=
∠ −
°
230 9 0
5 96
112 8
10
23 33 1 35
230 9
120
.
.
.
.
.
.
A rms
−−
∠ −
°
+
=
∠ −
°
=
∠
° −
∠ −
5 96
112 8
8
6
22 5
157 06
230 9 120
5 96
.
.
.
.
.
.
j
A rms
I
Y
1112 8
8
6
23 45 155 71
18 83 175 65
.
.
.
.
.
°
+
=
∠
°
=
+
+
=
∠
°
j
A rms
A
I
I
I
I
N
R
Y
B
rms
The load phase voltages can be determined by multiplying line current phasors by phasor impedance
of each phase.
V
V
RN
YN
=
∠
° ×
−
=
∠ −
°
=
∠ −
23 33 1 35
9 9
0 3
231 04
0 39
22 5
1
.
.
.
. )
.
.
.
(
V rms
j
557 06
7 9
5 7
219 18
121 25
23 45
155 71
.
.
. )
.
.
.
.
° ×
+
=
∠ −
°
=
∠ −
(
V rms
j
V
BN
°° ×
+
=
∠ −
°
(
V rms
7 9
6 3
237
123 65
.
. )
.
j
The line voltages across the load can be determined in terms of the phase voltage phasors now.
V
= V
V
RY
RN
YN
−
=
∠ −
° −
∠ −
° =
∠
°
231 04
0 39
219 18
121 25
391 63 28 33
.
.
.
.
.
.
V rrms
V
= V
V
YB
YN
BN
−
=
∠ −
° −
∠ −
° =
∠ −
219 18
121 25
237
123 65
398 32
90 8
.
.
.
.
.
°°
−
=
∠
° −
∠ −
° =
∠
V rms
V
= V
V
BR
BN
RN
237 123 65
231 04
0 39
400 2 147 94
.
.
.
.
.
°°
V rms
The active power and reactive power delivered to the load is
determined by adding up the
corresponding power delivered to each phase of the load.
P
P
P
P
=
+
+
=
×
×
−
° −
°
+
×
×
R
Y
B
231 04 23 33
0 39
1 35
231 04 22 5
.
.
cos ( .
.
)
.
.
cos (1121 25
157 06
237 23 45
123 65
155 71
538
.
(
.
))
.
cos (
.
(
.
))
° − −
°
+
×
×
° − −
°
=
77 1
3999
4345 9
231 04 23 33
0 39
1 3
.
.
.
.
sin ( .
.
W
W
kW
R
Y
B
+
+
=
+
+
=
×
×
−
° −
Q Q
Q
Q
55
231 04 22 5
121 25
157 06
237 23 45
123
°
+
×
×
° − −
°
+
×
×
)
.
.
sin (
.
(
.
))
.
sin (
..
(
.
))
.
.
.
.
65
155 71
163 25
2885 3
3465 7
0 744
° − −
°
=
+
−
= −
VAr
VAr
VAr
kVAr
8.20
SinusoidalSteady-StateinThree-PhaseCircuits
Two different definitions of apparent power and power factor are possible in the case of unbalanced
three-phase circuits. The first one is using the total complex power,
S
, delivered to the load. In this
definition, the magnitude of
S
is taken as apparent power and ratio of active power to the magnitude
of
S
is taken as average power factor of the circuit. The apparent power in the load in the circuit in this
example then is (13.73
2
+
0.744
2
)
°
.5
=
13.75 kVA and power factor is 0.9985 lead.
The second definition accepts the sum of apparent powers of load phases as the three-phase apparent
power. Power factor is taken as the ratio of active power to apparent power.. The apparent power in the
load in the circuit in this example then is 15.88 kVA and power factor is 0.865. We cannot specify it
as a lead or lag power factor in this case.
The second definition is more appropriate in the sense that power factor and apparent power are
expected to give us an indication as to the utilisation efficiency of current in carrying active power. The
first method of calculating apparent power and power factor will hide the fact that underutilization of
current is taking place when one phase takes leading reactive power and another phase takes lagging
reactive power. The phases may consume large reactive power and yet the complex power may come
out with only real part or with small imaginary part. Therefore, the definition of apparent power and
power factor based on apparent power of individual phases of load is a more meaningful one.
(b) The neutral-shift voltage in this case is zero since the source neutral and load neutral are tied
together by a zero impedance link. The circuit becomes a collection of
three single-phase circuits
sharing a common point at neutral. Therefore, the line
currents can be obtained as
I
I
R
Y
=
∠ °
=
∠ °
=
∠ −
°
+
=
∠ −
230 9 0
10
23 09 0
230 9
120
8
6
23 09
156
.
.
.
.
.
A rms
j
887
230 9 120
8
6
23 09 156 87
°
=
∠
°
−
=
∠
°
=
+
+
A rms
A rms
I
I
I
I
I
B
N
R
Y
B
.
.
.
j
== −
19 38
. A rms
The load phase voltages can be determined by multiplying line current phasors by phasor impedance
of each phase.
V
V
RN
YN
=
∠ ° ×
−
=
∠ −
°
=
∠ −
23 09 0
9 9
0 3
228 7
1 74
23 09
156
.
.
. )
.
.
.
.
(
V rms
j
887
7 9
5 7
224 93
121 06
23 09
156 87
° ×
+
=
∠ −
°
=
∠ −
° ×
(
V rms
.
. )
.
.
.
.
j
V
BN
((
V rms
7 9
6 3
233 31 118 3
.
. )
.
.
−
=
∠
°
j
The line voltages across the load can be determined in terms of the phase voltage phasors now.
V
V
V
RY
RN
YN
=
−
=
∠ −
° −
∠ −
° =
∠
°
228 7
1 74
224 93
121 06
391 51 28 32
.
.
.
.
.
.
V rm
ms
V
V
V
YB
YN
BN
=
−
=
∠ −
° −
∠
° =
∠ −
224 93
121 06
233 31 118 3
398 15
90 8
.
.
.
.
.
.
°°
=
−
=
∠
° −
∠ −
° =
∠
V rms
V
V
V
BR
BN
RN
233 31 118 3
228 7
1 74
400 2 147 9
.
.
.
.
.
. 55
391 51 28 32
398 15
90 8
400 2 147
°
=
−
=
∠
° +
∠ −
° +
∠
V rms
V
V
V
RY
YB
BR
.
.
.
.
.
..
.
.
.
.
.
.
95
344 65
185 73 5 56
398 11 339 204
212 37
°
=
+
−
−
−
+
≈
j
j
j
0 withinn round-off error
The active power and reactive power delivered to the load are determined by adding up the
corresponding power delivered to each phase of the load.
AnalysisofUnbalancedThree-PhaseCircuits
8.21
P
P
P
P
=
+
+
=
×
×
° −
°
+
×
×
−
R
Y
B
231 04 23 33
0
1 74
231 04 22 5
121
.
.
cos(
.
)
.
.
cos(
..
(
.
))
.
cos(
.
.
))
.
06
156 87
237 23 45
118 3
156 87
5287 2
4
° − −
°
+
×
×
° −
°
=
+
W
2211 9
4211 9
13 7
231 04 23 33
0
1 74
.
.
.
.
.
sin(
.
W
W
kW
R
Y
B
+
=
=
+
+
=
×
×
° −
Q Q
Q
Q
°°
+
×
×
−
° − −
°
+
×
×
)
.
.
sin(
.
(
.
))
.
sin(
231 04 22 5
121 06
156 87
237 23 45
118..
.
)
.
.
.
.
3
156 87
159 94
3038 9
3358 8
0 48
° −
°
=
+
−
= −
VAr
VAr
VAr
kVAr
The three-phase apparent power is 15.86 kVA and average power factor is 0.864.
(c) The neutral points are unconnected and hence the system is a three-wire system in this case. The
neutral-shift
voltage
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