Problem 1
For the circuit in Fig. 1, determine the current in the neutral line.
Figure 1
Solution
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Ia =
Van
220 ∠0°
= 6.55∠36.53°
=
Z A + 2 27 − j20
Ib =
Vbn
220∠ - 120°
=
= 10 ∠ - 120°
22
ZB + 2
Ic =
Vcn
220∠120°
=
= 16.92 ∠97.38°
12 + j5
ZC + 2
For phase b,
For phase c,
The current in the neutral line is
I n = -(I a + I b + I c )
or
- In = Ia + Ib + Ic
- I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78)
I n = 1.91 − j12.02 = 12.17 ∠ - 81° A
Problem 2
Solve for the line currents in the Y-∆ circuit of Fig. 2. Take Z∆ = 60∠45°Ω.
Figure 2
Solution
Convert the delta-load to a Wye-load and apply per-phase analysis.
Ia
110∠0° V
ZY =
Ia =
+
-
Z∆
= 20 ∠45° Ω
3
110 ∠0°
= 5.5∠ - 45° A
20∠45°
I b = I a ∠ - 120° = 5.5∠ - 165° A
I c = I a ∠120° = 5.5∠75° A
ZY
Problem 3
The following three parallel-connected three-phase loads are fed by a balanced threephase source.
Load 1: 250 kVA, 0.8 pf lagging
Load 2: 300 kVA, 0.95 pf leading
Load 3: 450 kVA, unity pf
If the line voltage is 13.8 kV, calculate the line current and the power factor of the source.
Assume that the line impedance is zero.
Solution
pf = 0.8 (lagging) 
→ θ = cos -1 (0.8) = 36.87°
S1 = 250 ∠36.87° = 200 + j150 kVA
pf = 0.95 (leading) 
→ θ = cos -1 (0.95) = -18.19°
S 2 = 300 ∠ - 18.19° = 285 − j93.65 kVA
pf = 1.0 
→ θ = cos -1 (1) = 0°
S 3 = 450 kVA
S T = S1 + S 2 + S 3 = 935 + j56.35 = 936.7 ∠3.45° kVA
S T = 3 VL I L
IL =
936.7 × 10 3
3 (13.8 × 10 3 )
= 39.19 A rms
pf = cos θ = cos(3.45°) = 0.9982 (lagging)
Problem 4
A professional center is supplied by a balanced three-phase source. The center
has four plants, each a balanced three-phase load as follows:
Load 1: 150 kVA at 0.8 pf leading
Load 2: 100 kW at unity pf
Load 3: 200 kVA at 0.6 pf lagging
Load 4: 80 kW and 95 kVAR (inductive)
If the line impedance is 0.02 + j0.05 Ω per phase and the line voltage at the loads
is 480 V, find the magnitude of the line voltage at the source.
Solution
pf = 0.8 (leading) 
→ θ1 = -36.87°
S 1 = 150 ∠ - 36.87° kVA
pf = 1.0 
→ θ 2 = 0°
S 2 = 100 ∠0° kVA
pf = 0.6 (lagging) 
→ θ3 = 53.13°
S 3 = 200∠53.13° kVA
S 4 = 80 + j95 kVA
S = S1 + S 2 + S 3 + S 4
S = 420 + j165 = 451.2∠21.45° kVA
S = 3 VL I L
S
451.2 × 10 3
IL =
=
= 542.7 A
3 VL
3 × 480
For the line,
S L = 3 I 2L Z L = (3)(542.7) 2 (0.02 + j0.05)
S L = 17.67 + j44.18 kVA
At the source,
S T = S + S L = 437.7 + j209.2
S T = 485.1∠25.55° kVA
S
485.1 × 10 3
VT = T =
= 516 V
3 IL
3 × 542.7
Problem 5
Refer to the unbalanced circuit of Fig. . Calculate:
(a) the line currents
(b) the real power absorbed by the load
(c) the total complex power supplied by the source
Figure 7
Solution
(a)
Consider the circuit below.
a
A
440∠0° + −
b
440∠120°
+
−
j10 Ω
I1
B
− +
440∠-120°
I2
I3
20 Ω
c
For mesh 1,
440∠ - 120° − 440∠0° + j10 (I 1 − I 3 ) = 0
(440)(1.5 + j0.866)
I1 − I 3 =
= 76.21∠ - 60°
j10
For mesh 2,
440∠120° − 440∠ - 120° + 20 (I 2 − I 3 ) = 0
(440)( j1.732)
I3 − I2 =
= j38.1
20
-j5 Ω
C
(1)
(2)
For mesh 3,
j10 (I 3 − I 1 ) + 20 (I 3 − I 2 ) − j5 I 3 = 0
Substituting (1) and (2) into the equation for mesh 3 gives,
(440)(-1.5 + j0.866)
I3 =
= 152.42∠60°
j5
From (1),
I 1 = I 3 + 76.21∠ - 60° = 114.315 + j66 = 132∠30°
From (2),
I 2 = I 3 − j38.1 = 76.21 + j93.9 = 120.93∠50.94°
I a = I 1 = 132∠30° A
I b = I 2 − I 1 = -38.105 + j27.9 = 47.23∠143.8° A
I c = - I 2 = 120.9∠230.9° A
(b)
2
S AB = I 1 − I 3 ( j10) = j58.08 kVA
2
S BC = I 2 − I 3 (20) = 29.04 kVA
2
S CA = I 3 (-j5) = (152.42) 2 (-j5) = -j116.16 kVA
S = S AB + S BC + S CA = 29.04 − j58.08 kVA
Real power absorbed = 29.04 kW
(c)
Total complex supplied by the source is
S = 29.04 − j58.08 kVA
(3)
Problem 6
Determine the line currents for the three-phase circuit of Fig. . Let Va = 110
Vb = 110 ∠ − 120° , Vc = 110 ∠120° V.
∠0° ,
Figure 8
Solution
We apply mesh analysis to the circuit shown below.
Ia
+
Va
–
80 + j 50Ω
I1
–
20 + j 30Ω
–
Vc
+
Vb
+
60 − j 40Ω
Ib
I2
Ic
(100 + j80) I 1 − (20 + j 30) I 2 = Va − Vb = 165 + j 95.263
(1)
− (20 + j 30) I 1 + (80 − j10) I 2 = Vb − Vc = − j190.53
(2)
I 2 = 0.9088 − j1.722 .
Solving (1) and (2) gives I 1 = 1.8616 − j 0.6084,
I a = I 1 = 1.9585∠ − 18.1o A,
I c = − I 2 = 1.947∠117.8 o A
I b = I 2 − I 1 = −0.528 − j1.1136 = 1.4656∠ − 130.55 o A
Problem 7
A three-phase, four-wire system operating with a 208-V line voltage is shown in Fig. The source voltages are balanced. The power absorbed by the resistive wyeconnected load is measured by the three-wattmeter method. Calculate:
(a) the voltage to neutral
(b) the currents I1, I2, I3, and In
(c) the readings of the wattmeters
(d) the total power absorbed by the load
Figure 9
Solution
VL
Vp =
(b)
Because the load is unbalanced, we have an unbalanced three-phase
system. Assuming an abc sequence,
120 ∠0°
I1 =
= 2.5∠0° A
48
120∠ - 120°
I2 =
= 3∠ - 120° A
40
120∠120°
I3 =
= 2∠120° A
60
3
=
208
(a)
3
= 120 V
⎛
⎛
3⎞
3⎞
- I N = I 1 + I 2 + I 3 = 2.5 + (3) ⎜⎜ - 0.5 − j ⎟⎟ + (2) ⎜⎜ - 0.5 + j ⎟⎟
2 ⎠
2 ⎠
⎝
⎝
IN = j
3
= j0.866 = 0.866∠90° A
2
Hence,
I1 = 2.5 A ,
(c)
I2 = 3 A ,
P1 = I12 R 1 = (2.5) 2 (48) = 300 W
P2 = I 22 R 2 = (3) 2 (40) = 360 W
P3 = I 32 R 3 = (2) 2 (60) = 240 W
(d)
PT = P1 + P2 + P3 = 900 W
I3 = 2 A ,
I N = 0.866 A
Problem 8
Meter readings for a three-phase wye-connected alternator supplying power to a motor
indicate that the line voltages are 330 V, the line currents are 8.4 A, and the total line
power is 4.5 kW. Find:
(a) the load in VA
(b) the load pf
(c) the phase current
(d) the phase voltage
Solution
(a)
S = 3 VL I L = 3 (330)(8.4) = 4801 VA
(b)
P = S cos θ ⎯
⎯→ pf = cos θ =
pf =
4500
= 0.9372
4801.24
(c)
For a wye-connected load,
I p = I L = 8.4 A
(d)
Vp =
VL
3
=
330
3
= 190.53 V
P
S
Problem 9
The two-wattmeter method gives P1 = 1200 W and P2 = –400 W for a three-phase motor
running on a 240-V line. Assume that the motor load is wye-connected and that it draws a
line current of 6 A. Calculate the pf of the motor and its phase impedance.
Solution
PT = P1 + P2 = 1200 − 400 = 800
Q T = P2 − P1 = -400 − 1200 = -1600
tan θ =
Q T - 1600
=
= -2 ⎯
⎯→ θ = -63.43°
PT
800
pf = cos θ = 0.4472 (leading)
Zp =
VL 240
=
= 40
IL
6
Z p = 40 ∠ - 63.43° Ω
Problem 10
In Fig. 12, two wattmeters are properly connected to the unbalanced load supplied by
a balanced source such that Vab = 208 ∠0° V with positive phase sequence.
(a) Determine the reading of each wattmeter.
(b) Calculate the total apparent power absorbed by the load.
Figure 12
Solution
(a)
If Vab = 208∠0° , Vbc = 208∠ - 120° , Vca = 208∠120° ,
Vab 208∠0°
I AB =
=
= 10.4 ∠0°
20
Z Ab
Vbc
208∠ - 120°
I BC =
=
= 14.708∠ - 75°
Z BC 10 2 ∠ - 45°
I CA =
Vca
208∠120°
=
= 16 ∠97.38°
Z CA 13∠22.62°
I aA = I AB − I CA = 10.4∠0° − 16∠97.38°
I aA = 10.4 + 2.055 − j15.867
I aA = 20.171∠ - 51.87°
I cC = I CA − I BC = 16∠97.83° − 14.708∠ - 75°
I cC = 30.64 ∠101.03°
P1 = Vab I aA cos(θ Vab − θIaA )
P1 = (208)(20.171) cos(0° + 51.87°) = 2590 W
P2 = Vcb I cC cos(θ Vcb − θ IcC )
But
Vcb = -Vbc = 208∠60°
P2 = (208)(30.64) cos(60° − 101.03°) = 4808 W
(b)
PT = P1 + P2 = 7398.17 W
Q T = 3 (P2 − P1 ) = 3840.25 VAR
S T = PT + jQ T = 7398.17 + j3840.25 VA
S T = S T = 8335 VA
Problem 11
A balanced three-phase source furnishes power to the following three loads:
Load 1: 6 kVA at 0.83 pf lagging
Load 2: unknown
Load 3: 8 kW at 0.7071 pf leading
If the line current is 84.6 A rms, the line voltage at the load is 208 V rms, and the
combined load has a 0.8 pf lagging, determine the unknown load.
Solution
S = S1 + S 2 + S 3 = 6[0.83 + j sin(cos −1 0.83)] + S 2 + 8(0.7071 − j 0.7071)
S = 10.6368 − j 2.31 + S 2 kVA
(1)
But
S = 3VL I L ∠θ = 3 (208)(84.6)(0.8 + j 0.6) VA = 24.383 + j18.287 kVA
(2)
From (1) and (2),
S 2 = 13.746 + j 20.6 = 24.76∠56.28 kVA
Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.