Chapter 12, Problem 1.
If Vab = 400 V in a balanced Y-connected three-phase generator, find the phase voltages,
assuming the phase sequence is:
(a) abc
(b) acb
Chapter 12, Solution 1.
(a)
If Vab = 400 , then
400
Van =
∠ - 30° = 231∠ - 30° V
3
Vbn = 231∠ - 150° V
Vcn = 231∠ - 270° V
(b)
For the acb sequence,
Vab = Van − Vbn = Vp ∠0° − Vp ∠120°
⎛ 1
3⎞
Vab = Vp ⎜⎜1 + − j ⎟⎟ = Vp 3∠ - 30°
2 ⎠
⎝ 2
i.e. in the acb sequence, Vab lags Van by 30°.
Hence, if Vab = 400 , then
400
Van =
∠30° = 231∠30° V
3
Vbn = 231∠150° V
Vcn = 231∠ - 90° V
Chapter 12, Problem 2.
What is the phase sequence of a balanced three-phase circuit for which Van = 160 ∠30° V
and Vcn = 160 ∠ − 90° V? Find Vbn.
Chapter 12, Solution 2.
Since phase c lags phase a by 120°, this is an acb sequence.
Vbn = 160∠(30° + 120°) = 160∠150° V
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Chapter 12, Problem 3.
Determine the phase sequence of a balanced three-phase circuit in which
Vbn = 208 ∠130° V and Vcn = 208 ∠10° V. Obtain Van .
Chapter 12, Solution 3.
Since Vbn leads Vcn by 120°, this is an abc sequence.
Van = 208∠(130° + 120°) = 208∠ 250° V
Chapter 12, Problem 4.
A three-phase system with abc sequence and VL = 200 V feeds a Y-connected load with
ZL = 40 ∠30°Ω . Find the line currents.
Chapter 12, Solution 4.
VL = 200 = 3V p
200
3
V
200 < 0o
I a = an =
= 2.887 < −30o A
o
ZY
3x 40 < 30
I b = I a < −120o = 2.887 < −150o A
⎯⎯
→ Vp =
I c = I a < +120o = 2.887 < 90o A
Chapter 12, Problem 5.
For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at
the terminals are:
vAN = 150 cos ( ω t + 32º) V
vBN = 150 cos ( ω t – 88º) V
vCN = 150 cos ( ω t + 152º) V
Write the time-domain expressions for the line-to-line voltages vAN, vBC, and vCA .
Chapter 12, Solution 5.
VAB = 3V p < 30o = 3x150 < 32o + 30o = 260 < 62o
Thus,
v AB = 260 cos(ωt + 62o ) V
Using abc sequence,
vBC = 260 cos(ωt − 58o ) V
vCA = 260 cos(ωt + 182o ) V
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Chapter 12, Problem 6.
For the Y-Y circuit of Fig. 12.41, find the line currents, the line voltages, and the load
voltages.
Figure 12.41
For Prob. 12.6.
Chapter 12, Solution 6.
Z Y = 10 + j5 = 11.18∠26.56°
The line currents are
Van
220 ∠0°
Ia =
=
= 19.68∠ - 26.56° A
Z Y 11.18∠26.56°
I b = I a ∠ - 120° = 19.68∠ - 146.56° A
I c = I a ∠120° = 19.68∠93.44° A
The line voltages are
Vab = 220 3 ∠30° = 381∠30° V
Vbc = 381∠ - 90° V
Vca = 381∠ - 210° V
The load voltages are
VAN = I a Z Y = Van = 220∠0° V
VBN = Vbn = 220∠ - 120° V
VCN = Vcn = 220∠120° V
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Chapter 12, Problem 7.
Obtain the line currents in the three-phase circuit of Fig. 12.42 on the next page.
Figure 12.42
For Prob. 12.7.
Chapter 12, Solution 7.
This is a balanced Y-Y system.
440∠0° V
+
−
ZY = 6 − j8 Ω
Using the per-phase circuit shown above,
440∠0°
Ia =
= 44∠53.13° A
6 − j8
I b = I a ∠ - 120° = 44∠ - 66.87° A
I c = I a ∠120° = 44∠173.13° A
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Chapter 12, Problem 8.
In a balanced three-phase Y-Y system, the source is an abc sequence of voltages and Van
= 100 ∠20° V rms. The line impedance per phase is 0.6 + j1.2 Ω , while the per-phase
impedance of the load is 10 + j14 Ω . Calculate the line currents and the load voltages.
Chapter 12, Solution 8.
Consider the per phase equivalent circuit shown below.
Zl
Van
Ia =
+
_
ZL
Van
100 < 20o
5.396∠
=
= 5.3958
< –35.1˚
−35.1o AA
Z L + Z l 10.6 + j15.2
I b = I a < −120o = 5.3958
−155.1oAA
5.396∠<–155.1˚
5.396∠<84.9˚
I c = I a < +120o = 5.3958
84.9oAA
o
VLa = I a Z L = (4.4141 − j 3.1033)(10 + j14) = 92.83
V
92.83<∠19.35
19.35˚
A
92.83<∠−–100.65˚
VLb = VLa < −120o = 92.83
100.65o A
V
o
VLc = VLa < +120o = 92.83
92.83<∠139.35
139.35˚ V
A
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Chapter 12, Problem 9.
A balanced Y-Y four-wire system has phase voltages
Van = 120∠0°
Vbn = 120∠ − 120°
Vcn = 120∠120° V
The load impedance per phase is 19 + j13 Ω , and the line impedance per phase is
1 + j2 Ω . Solve for the line currents and neutral current.
Chapter 12, Solution 9.
Ia =
Van
120 ∠0°
=
= 4.8∠ - 36.87° A
Z L + Z Y 20 + j15
I b = I a ∠ - 120° = 4.8∠ - 156.87° A
I c = I a ∠120° = 4.8∠83.13° A
As a balanced system, I n = 0 A
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Chapter 12, Problem 10.
For the circuit in Fig. 12.43, determine the current in the neutral line.
Figure 12.43
For Prob. 12.10.
Chapter 12, Solution 10.
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Ia =
Van
220∠0°
220
=
= 7.642∠20.32°
=
Z A + 2 27 − j10 28.79∠ − 20.32°
Ib =
Vbn
220 ∠ - 120°
=
= 10 ∠ - 120°
ZB + 2
22
Ic =
Vcn
220∠120° 220∠120°
=
=
= 16.923∠97.38°
12 + j5
13∠22.62°
ZC + 2
For phase b,
For phase c,
The current in the neutral line is
I n = -(I a + I b + I c ) or - I n = I a + I b + I c
- I n = (7.166 + j2.654) + (-5 − j8.667) + (-2.173 + j16.783)
I n = 0.007 − j10.77 = 10.77∠90°A
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Chapter 12, Problem 11.
In the Y- ∆ system shown in Fig. 12.44, the source is a positive sequence with
V an = 120 ∠0° V and phase impedance Z p = 2 – j3 Ω . Calculate the line voltage V L and
the line current I L.
Figure 12.44
For Prob. 12.11.
Chapter 12, Solution 11.
VAB = Vab = 3V p < 30o = 3(120) < 30o
VL =| Vab |= 3 x120 = 207.85 V
I AB
3V p < 30o
VAB
=
=
ZA
2 − j3
I a = I AB 3 < −30o =
3V p < 0o
2 − j3
=
3 x120
= 55.385 + j83.07
2 − j3
I L =| I a |= 99.846 A
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Chapter 12, Problem 12.
Solve for the line currents in the Y-∆ circuit of Fig. 12.45. Take Z ∆ = 60∠45°Ω .
Figure 12.45
For Prob. 12.12.
Chapter 12, Solution 12.
Convert the delta-load to a wye-load and apply per-phase analysis.
Ia
110∠0° V
ZY =
+
−
ZY
Z∆
= 20 ∠45° Ω
3
110∠0°
= 5.5∠ - 45° A
20∠45°
I b = I a ∠ - 120° = 5.5∠ - 165° A
Ia =
I c = I a ∠120° = 5.5∠75° A
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Chapter 12, Problem 13.
In the balanced three-phase Y-∆ system in Fig. 12.46, find the line current I L
and the average power delivered to the load.
Figure 12.46
For Prob. 12.13.
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Chapter 12, Solution 13.
Convert the delta load to wye as shown below.
A
110∠0o V rms
2Ω
–+
110∠–120o V rms
ZY
N
2Ω
ZY
–+
110∠120o V rms
ZY
2Ω
–+
1
ZY = Z = 3 − j 2 Ω
3
We consider the single phase equivalent shown below.
2Ω
110∠0˚ V rms
+
_
3 – j2 Ω
110
= 20.4265 < 21.8o
2 + 3 − j2
I L =| I a |= 20.43 A
Ia =
S = 3|Ia|2ZY = 3(20.43)2(3–j2) = 4514∠–33.96˚ = 3744 – j2522
P = Re(S) = 3744 W.
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Chapter 12, Problem 14.
Obtain the line currents in the three-phase circuit of Fig. 12.47 on the next page.
100 –120°
Figure 12.47
For Prob. 12.14.
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Chapter 12, Solution 14.
We apply mesh analysis.
1 + j 2Ω
A
a
+
100∠0 o V
-
ZL
ZL
I3
n
100∠120 o V
+
c
I1
-
100∠120 o V
+
b
I2
B
C
Z L = 12 + j12Ω
1 + j 2Ω
1 + j 2Ω
For mesh,
− 100 + 100∠120 o + I 1 (14 + j16) − (1 + j 2) I 2 − (12 + j12) I 3 = 0
or
(14 + j16) I 1 − (1 + j 2) I 2 − (12 + j12) I 3 = 100 + 50 − j86.6 = 150 − j86.6 (1)
For mesh 2,
100∠120 o − 100∠ − 120 o − I 1 (1 + j 2) − (12 + j12) I 3 + (14 + j16) I 2 = 0
or
− (1 + j 2) I 1 + (14 + j16) I 2 − (12 + j12) I 3 = −50 − j86.6 + 50 − j86.6 = − j173.2 (2)
For mesh 3,
− (12 + j12) I 1 − (12 + j12) I 2 + (36 + j 36) I 3 = 0
(3)
Solving (1) to (3) gives
I 1 = −3.161 − j19.3,
I 2 = −10.098 − j16.749,
I 3 = −4.4197 − j12.016
I aA = I 1 = 19.58∠ − 99.3 A
o
I bB = I 2 − I 1 = 7.392∠159.8 o A
I cC = − I 2 = 19.56∠58.91o A
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Chapter 12, Problem 15.
The circuit in Fig. 12.48 is excited by a balanced three-phase source with a line
voltage of 210 V. If Z l = 1 + j1 Ω , Z ∆ = 24 − j 30Ω , and ZY = 12 + j5 Ω , determine the
magnitude of the line current of the combined loads.
Figure 12.48
For Prob. 12.15.
Chapter 12, Solution 15.
Convert the delta load, Z ∆ , to its equivalent wye load.
Z∆
= 8 − j10
Z Ye =
3
(12 + j5)(8 − j10)
= 8.076 ∠ - 14.68°
20 − j5
Z p = 7.812 − j2.047
Z p = Z Y || Z Ye =
Z T = Z p + Z L = 8.812 − j1.047
Z T = 8.874 ∠ - 6.78°
We now use the per-phase equivalent circuit.
Vp
210
Ia =
,
where Vp =
Zp + ZL
3
Ia =
210
3 (8.874 ∠ - 6.78°)
= 13.66 ∠6.78°
I L = I a = 13.66 A
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Chapter 12, Problem 16.
A balanced delta-connected load has a phase current I AC = 10 ∠ − 30° A.
(a) Determine the three line currents assuming that the circuit operates in the positive
phase sequence.
(b) Calculate the load impedance if the line voltage is V AB = 110 ∠0° V.
Chapter 12, Solution 16.
(a)
I CA = - I AC = 10∠(-30° + 180°) = 10∠150°
This implies that
I AB = 10 ∠30°
I BC = 10∠ - 90°
I a = I AB 3 ∠ - 30° = 17.32∠0° A
I b = 17.32∠ - 120° A
I c = 17.32∠120° A
(b)
Z∆ =
VAB 110 ∠0°
=
= 11∠ - 30° Ω
I AB 10 ∠30°
Chapter 12, Problem 17.
A balanced delta-connected load has line current I a = 10 ∠ − 25° A. Find the phase
currents I AB , I BC , and I CA.
Chapter 12, Solution 17.
I a = I AB 3 < −30o
⎯⎯
→ I AB =
Ia
10
=
< −25o + 30o = 5.773 < 5o A
o
3 < −30
3
I BC = I AB < −120o = 5.775 < −115o A
I CA = I AB < +120o = 5.775 < 125o A
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Chapter 12, Problem 18.
If V an = 440 ∠60° V in the network of Fig. 12.49, find the load phase currents I AB , I BC,
and I CA .
Figure 12.49
For Prob. 12.18.
Chapter 12, Solution 18.
VAB = Van 3 ∠30° = (440 ∠60°)( 3 ∠30°) = 762.1∠90°
Z ∆ = 12 + j9 = 15∠36.87°
I AB =
VAB 762.1∠90°
=
= 50.81∠53.13° A
Z ∆ 15∠36.87°
I BC = I AB ∠ - 120° = 50.81∠ - 66.87° A
I CA = I AB ∠120° = 50.81∠173.13° A
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Chapter 12, Problem 19.
For the ∆ - ∆ circuit of Fig. 12.50, calculate the phase and line currents.
Figure 12.50
For Prob. 12.19.
Chapter 12, Solution 19.
Z ∆ = 30 + j10 = 31.62 ∠18.43°
The phase currents are
Vab
173∠0°
=
= 5.47 ∠ - 18.43° A
I AB =
Z ∆ 31.62 ∠18.43°
I BC = I AB ∠ - 120° = 5.47 ∠ - 138.43° A
I CA = I AB ∠120° = 5.47 ∠101.57° A
The line currents are
I a = I AB − I CA = I AB 3 ∠ - 30°
I a = 5.47 3 ∠ - 48.43° = 9.474∠ - 48.43° A
I b = I a ∠ - 120° = 9.474∠ - 168.43° A
I c = I a ∠120° = 9.474∠71.57° A
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Chapter 12, Problem 20.
Refer to the ∆ - ∆ circuit in Fig. 12.51. Find the line and phase currents. Assume that the
load impedance is ZL = 12 + j9 Ω per phase.
Figure 12.51
For Prob. 12.20.
Chapter 12, Solution 20.
Z ∆ = 12 + j9 = 15∠36.87°
The phase currents are
210∠0°
= 14∠ - 36.87° A
15∠36.87°
= I AB ∠ - 120° = 14∠ - 156.87° A
I AB =
I BC
I CA = I AB ∠120° = 14∠83.13° A
The line currents are
I a = I AB 3 ∠ - 30° = 24.25∠ - 66.87° A
I b = I a ∠ - 120° = 24.25∠ - 186.87° A
I c = I a ∠120° = 24.25∠53.13° A
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Chapter 12, Problem 21.
Three 230-V generators form a delta-connected source that is connected to a balanced
delta-connected load of ZL = 10 + j8 Ω per phase as shown in Fig. 12.52.
(a) Determine the value of IAC.
(b) What is the value of Ib?
Figure 12.52
For Prob. 12.21.
Chapter 12, Solution 21.
(a)
− 230∠120°
− 230∠120°
=
= 17.96∠ − 98.66° A(rms)
10 + j8
12.806∠38.66°
17.96∠–98.66˚ A rms
I AC =
230∠ − 120 230∠0°
−
10 + j8
10 + j8
= 17.96∠ − 158.66° − 17.96∠ − 38.66°
= −16.729 − j6.536 − 14.024 + j11.220 = −30.75 + j4.684
= 31.10∠171.34° A
I bB = I BC + I BA = I BC − I AB =
(b)
31.1∠171.34˚ A rms
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Chapter 12, Problem 22.
Find the line currents Ia, Ib, and Ic in the three-phase network of Fig. 12.53 below.
Take Z ∆ = 12 − j15Ω , ZY = 4 + j6 Ω , and Zl = 2 Ω .
208 0° V
Figure 12.53
For Prob. 12.22.
Chapter 12, Solution 22.
Convert the ∆-connected source to a Y-connected source.
Vp
208
Van =
∠ - 30° =
∠ - 30° = 120 ∠ - 30°
3
3
Convert the ∆-connected load to a Y-connected load.
Z
(4 + j6)(4 − j5)
Z = Z Y || ∆ = (4 + j6) || (4 − j5) =
3
8+ j
Z = 5.723 − j0.2153
ZL
Van
Ia
+
−
Z
Van
120∠ − 30°
= 15.53∠ - 28.4° A
=
Z L + Z 7.723 − j0.2153
I b = I a ∠ - 120° = 15.53∠ - 148.4° A
Ia =
I c = I a ∠120° = 15.53∠91.6° A
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Chapter 12, Problem 23.
A three-phase balanced system with a line voltage of 202 V rms feeds a delta-connected
load with Zp = 25 ∠60°Ω .
(a) Find the line current.
(b) Determine the total power supplied to the load using two wattmeters connected to the
A and C lines.
Chapter 12, Solution 23.
(a)
I AB =
VAB
202
=
Z∆
25∠60 o
o
I a = I AB 3∠ − 30 =
202 3∠ − 30 o
25∠60
o
= 13.995∠ − 90 o
I L =| I a |= 13.995A
(b)
⎛ 202 3 ⎞
⎟ cos 60 o = 2.448 kW
P = P1 + P2 = 3VL I L cos θ = 3 (202)⎜⎜
⎟
25
⎝
⎠
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Chapter 12, Problem 24.
A balanced delta-connected source has phase voltage Vab = 416 ∠30° V and a positive
phase sequence. If this is connected to a balanced delta-connected load, find the line and
phase currents. Take the load impedance per phase as 60 ∠30°Ω and line impedance per
phase as 1 + j1 Ω .
Chapter 12, Solution 24.
Convert both the source and the load to their wye equivalents.
Z∆
= 20 ∠30° = 17.32 + j10
ZY =
3
Vab
∠ - 30° = 240.2∠0°
Van =
3
We now use per-phase analysis.
1+jΩ
Van
Ia =
+
−
Ia
20∠30° Ω
Van
240.2
=
= 11.24∠ - 31° A
(1 + j) + (17.32 + j10) 21.37 ∠31°
I b = I a ∠ - 120° = 11.24∠ - 151° A
I c = I a ∠120° = 11.24∠89° A
But
I AB =
I a = I AB 3 ∠ - 30°
11.24 ∠ - 31°
3 ∠ - 30°
= 6.489∠ - 1° A
I BC = I AB ∠ - 120° = 6.489∠ - 121° A
I CA = I AB ∠120° = 6.489∠119° A
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Chapter 12, Problem 25.
In the circuit of Fig. 12.54, if Vab = 440 ∠10° , Vbc = 440 ∠250° , Vca = 440 ∠130°
V, find the line currents.
Figure 12.54
For Prob. 12.25.
Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and
consider the single-phase equivalent.
Ia =
where
440 ∠(10° − 30°)
3 ZY
Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78°
Ia =
440 ∠ - 20°
3 (14.32 ∠ - 24.78°)
= 17.74∠4.78° A
I b = I a ∠ - 120° = 17.74∠ - 115.22° A
I c = I a ∠120° = 17.74 ∠124.78° A
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Chapter 12, Problem 26.
For the balanced circuit in Fig. 12.55, Vab = 125 ∠0° V. Find the line currents IaA, IbB, and
IcC.
Figure 12.55
For Prob. 12.26.
Chapter 12, Solution 26.
Transform the source to its wye equivalent.
Vp
Van =
∠ - 30° = 72.17 ∠ - 30°
3
Now, use the per-phase equivalent circuit.
Van
,
Z = 24 − j15 = 28.3∠ - 32°
I aA =
Z
I aA =
72.17 ∠ - 30°
= 2.55∠2° A
28.3∠ - 32°
I bB = I aA ∠ - 120° = 2.55∠ - 118° A
I cC = I aA ∠120° = 2.55∠122° A
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Chapter 12, Problem 27.
A ∆-connected source supplies power to a Y-connected load in a three-phase
balanced system. Given that the line impedance is 2 + j1 Ω per phase while the load
impedance is 6 + j4 Ω per phase, find the magnitude of the line voltage at the load.
Assume the source phase voltage Vab = 208 ∠0° V rms.
Chapter 12, Solution 27.
Since ZL and Z l are in series, we can lump them together so that
ZY = 2 + j + 6 + j 4 = 8 + j 5
VP
< −30o
208 < −30o
Ia = 3
=
ZY
3(8 + j 5)
208(0.866 − j 0.5)(6 + j 4)
VL = (6 + j 4) I a =
= 80.81 − j 43.54
3(8 + j 5)
|VL| = 91.79 V
Chapter 12, Problem 28.
The line-to-line voltages in a Y-load have a magnitude of 440 V and are in the positive
sequence at 60 Hz. If the loads are balanced with Z1 = Z 2 = Z 3 = 25 ∠30° , find all line
currents and phase voltages.
Chapter 12, Solution 28.
VL = Vab = 440 = 3VP or VP = 440/1.7321 = 254
For reference, let VAN = 254∠0˚ V which leads to
VBN = 254∠–120˚ V and VCN = 254∠120˚ V.
The line currents are found as follows,
Ia = VAN/ZY = 254/25∠30˚ = 10.16∠–30˚ A.
This leads to, Ib = 10.16∠–150˚ A and Ic = 10.16∠90˚ A.
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Chapter 12, Problem 29.
A balanced three-phase Y-∆ system has Van = 120 ∠0° V rms and Z ∆ = 51 + j 45Ω .
If the line impedance per phase is 0.4 + j1.2 Ω , find the total complex power delivered to
the load.
Chapter 12, Solution 29.
We can replace the delta load with a wye load, ZY = Z∆/3 = 17+j15Ω.
The per-phase equivalent circuit is shown below.
Zl
Van
Ia =
+
_
ZY
Van
120
=
= 5.0475
< −42.955o
5.0475∠–42.96˚
ZY + Z l 17 + j15 + 0.4 + j1.2
S = 3S p = 3 | I a |2 ZY = 3(5.0475) 2 (17 + j15) = 1.3 + j1.1465 kVA
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Chapter 12, Problem 30.
In Fig. 12.56, the rms value of the line voltage is 208 V. Find the average power
delivered to the load.
Figure 12.56
For Prob. 12.30.
Chapter 12, Solution 30.
Since this a balanced system, we can replace it by a per-phase equivalent, as
shown below.
+
Vp
S = 3S p =
ZL
3V 2 p
,
Z*p
Vp =
VL
3
(208) 2
V 2L
=
= 1.4421∠45 o kVA
o
*
Z p 30∠ − 45
P = S cosθ = 1.02 kW
S=
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Chapter 12, Problem 31.
A balanced delta-connected load is supplied by a 60-Hz three-phase source with a line
voltage of 240 V. Each load phase draws 6 kW at a lagging power factor of 0.8. Find:
(a) the load impedance per phase
(b) the line current
(c) the value of capacitance needed to be connected in parallel with each load phase to
minimize the current from the source
Chapter 12, Solution 31.
(a)
Pp = 6,000,
cosθ = 0.8,
Q p = S P sin θ = 4.5 kVAR
Sp =
PP
= 6 / 0.8 = 7.5 kVA
cosθ
S = 3S p = 3(6 + j 4.5) = 18 + j13.5 kVA
For delta-connected load, Vp = VL= 240 (rms). But
S=
(b)
3V 2 p
Z*p
⎯
⎯→
Z*p =
Pp = 3VL I L cosθ
3V 2 p
3(240) 2
=
,
S
(18 + j13.5) x10 3
⎯
⎯→
IL =
6000
3 x 240 x0.8
Z P = 6.144 + j 4.608Ω
= 18.04 A
(c ) We find C to bring the power factor to unity
Qc = Q p = 4.5 kVA
⎯
⎯→
C=
Qc
4500
=
= 207.2 µF
2
ωV rms 2πx60 x 240 2
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Chapter 12, Problem 32.
A balanced Y-load is connected to a 60-Hz three-phase source with Vab = 240 ∠0° V.
The load has pf = 0.5 lagging and each phase draws 5 kW. (a) Determine the load
impedance ZY . (b) Find Ia, Ib, and Ic.
Chapter 12, Solution 32.
(a) | Vab |= 3V p = 240
⎯⎯
→ Vp =
240
= 138.56
3
Van = V p < −30o
pf = 0.5 = cos θ
⎯⎯
→ θ = 60o
P
5
P = S cos θ
⎯⎯
→ S=
=
= 10 kVA
cos θ 0.5
Q = S sin θ = 10sin 60 = 8.66
S p = 5 + j8.66 kVA
But
SP =
(b)
V p2
Z *p
V p2
138.562
⎯⎯
→ Z =
=
= 0.96 − j1.663
S p (5 + j8.66) x103
*
p
Zp = 0.96 + j1.663 Ω
Van 138.56 < −30o
Ia =
=
= 72.17 < −90o A
ZY 0.96 + j1.6627
I b = I a < −120o = 72.17 < −210o A
I c = I a < +120o = 72.17 < 30o A
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Chapter 12, Problem 33.
A three-phase source delivers 4800 VA to a wye-connected load with a phase voltage of
208 V and a power factor of 0.9 lagging. Calculate the source line current and the source
line voltage.
Chapter 12, Solution 33.
S = 3 VL I L ∠θ
S = S = 3 VL I L
For a Y-connected load,
IL = Ip ,
VL = 3 Vp
S = 3 Vp I p
IL = Ip =
S
4800
=
= 7.69 A
3 Vp (3)(208)
VL = 3 Vp = 3 × 208 = 360.3 V
Chapter 12, Problem 34.
A balanced wye-connected load with a phase impedance of 10 – j16 Ω is connected to a
balanced three-phase generator with a line voltage of 220 V. Determine the line current
and the complex power absorbed by the load.
Chapter 12, Solution 34.
V
220
Vp = L =
3
3
Ia =
Vp
ZY
=
220
3 (10 − j16)
=
127.02
= 6.732∠58°
18.868∠ − 58°
I L = I p = 6.732A
S = 3 VL I L ∠θ = 3 × 220 × 6.732∠ - 58° = 2565∠ − 58°
S = 1359.2–j2175 VA
Chapter 12, Problem 35.
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Three equal impedances, 60 + j30 Ω each, are delta-connected to a 230-V rms,
three-phase circuit. Another three equal impedances, 40 + j10 Ω each, are wyeconnected across the same circuit at the same points. Determine:
(a) the line current
(b) the total complex power supplied to the two loads
(c) the power factor of the two loads combined
Chapter 12, Solution 35.
(a) This is a balanced three-phase system and we can use per phase equivalent circuit.
The delta-connected load is converted to its wye-connected equivalent
Z '' y =
1
Z ∆ = (60 + j 30) / 3 = 20 + j10
3
IL
+
230 V
-
Z’y
Z’’y
Z y = Z ' y // Z '' y = (40 + j10) //( 20 + j10) = 13.5 + j 5.5
IL =
230
= 14.61 − j 5.953 A
13.5 + j 5.5
(b) S = Vs I * L = 3.361 + j1.368 kVA
(c ) pf = P/S = 0.9261
Chapter 12, Problem 36.
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A 4200-V, three-phase transmission line has an impedance of 4 + j10 Ω per phase. If it
supplies a load of 1 MVA at 0.75 power factor (lagging), find:
(a) the complex power
(b) the power loss in the line
(c) the voltage at the sending end
Chapter 12, Solution 36.
(a)
(b) S = 3V p I * p
S = 1 [0.75 + sin(cos-10.75) ] = 0.75 + j0.6614 MVA
⎯
⎯→
I*p =
S
(0.75 + j 0.6614) x10 6
=
= 59.52 + j 52.49
3V p
3x 4200
PL =| I p | 2 Rl = (79.36) 2 (4) = 25.19 kW
(c) Vs = VL + I p (4 + j ) = 4.4381 − j 0.21 kV = 4.443∠ - 2.709 o kV
Chapter 12, Problem 37.
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The total power measured in a three-phase system feeding a balanced wye-connected
load is 12 kW at a power factor of 0.6 leading. If the line voltage is 208 V, calculate the
line current IL and the load impedance ZY.
Chapter 12, Solution 37.
S=
P
12
=
= 20
pf 0.6
S = S∠θ = 20∠θ = 12 − j16 kVA
But
IL =
S = 3 VL I L ∠θ
20 × 10 3
3 × 208
S = 3 Ip
= 55.51 A
2
Zp
For a Y-connected load, I L = I p .
Zp =
S
3 IL
2
=
(12 − j16) × 10 3
(3)(55.51) 2
Z p = 1.298 − j1.731 Ω
Chapter 12, Problem 38.
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Given the circuit in Fig. 12.57 below, find the total complex power absorbed by the
load.
Figure 12.57
For Prob. 12.38.
Chapter 12, Solution 38.
As a balanced three-phase system, we can use the per-phase equivalent shown
below.
Ia =
110∠0°
110∠0°
=
(1 + j2) + (9 + j12) 10 + j14
Sp =
1
I
2 a
2
ZY =
1
(110) 2
⋅
⋅ (9 + j12)
2 (10 2 + 14 2 )
The complex power is
3 (110) 2
S = 3S p = ⋅
⋅ (9 + j12)
2 296
S = 551.86 + j735.81 VA
Chapter 12, Problem 39.
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Find the real power absorbed by the load in Fig. 12.58.
Figure 12.58
For Prob. 12.39.
Chapter 12, Solution 39.
Consider the system shown below.
5Ω
a
100∠120°
c
+
−
+
− +
100∠-120°
−
100∠0°
A
4Ω
-j6 Ω
I1
5Ω
b
8Ω
B
I2
j3 Ω
I3
C
10 Ω
5Ω
For mesh 1,
100 = (18 − j6) I 1 − 5 I 2 − (8 − j6) I 3
(1)
100 ∠ - 120° = 20 I 2 − 5 I 1 − 10 I 3
20∠ - 120° = - I 1 + 4 I 2 − 2 I 3
(2)
0 = - (8 − j6) I 1 − 10 I 2 + (22 − j3) I 3
(3)
For mesh 2,
For mesh 3,
To eliminate I 2 , start by multiplying (1) by 2,
200 = (36 − j12) I 1 − 10 I 2 − (16 − j12) I 3
Subtracting (3) from (4),
(4)
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200 = (44 − j18) I 1 − (38 − j15) I 3
(5)
Multiplying (2) by 5 4 ,
25∠ - 120° = -1.25 I 1 + 5 I 2 − 2.5 I 3
(6)
Adding (1) and (6),
87.5 − j21.65 = (16.75 − j6) I 1 − (10.5 − j6) I 3
(7)
In matrix form, (5) and (7) become
⎡
⎤ ⎡ 44 − j18 - 38 + j15 ⎤⎡ I 1 ⎤
200
⎢87.5 − j12.65⎥ = ⎢16.75 − j6 - 10.5 + j6 ⎥⎢ I ⎥
⎣
⎦ ⎣
⎦⎣ 3 ⎦
∆ = 192.5 − j26.25 ,
∆ 1 = 900.25 − j935.2 ,
∆ 3 = 110.3 − j1327.6
I1 =
∆ 1 1298.1∠ - 46.09°
=
= 6.682 ∠ - 38.33° = 5.242 − j4.144
∆
194.28∠ - 7.76°
I3 =
∆ 3 1332.2∠ - 85.25°
=
= 6.857∠ - 77.49° = 1.485 − j6.694
∆
194.28∠ - 7.76°
We obtain I 2 from (6),
1
1
I 2 = 5∠ - 120° + I 1 + I 3
4
2
I 2 = (-2.5 − j4.33) + (1.3104 − j1.0359) + (0.7425 − j3.347)
I 2 = -0.4471 − j8.713
The average power absorbed by the 8-Ω resistor is
2
2
P1 = I 1 − I 3 (8) = 3.756 + j2.551 (8) = 164.89 W
The average power absorbed by the 4-Ω resistor is
2
P2 = I 3 (4) = (6.8571) 2 (4) = 188.1 W
The average power absorbed by the 10-Ω resistor is
2
2
P3 = I 2 − I 3 (10) = - 1.9321 − j2.019 (10) = 78.12 W
Thus, the total real power absorbed by the load is
P = P1 + P2 + P3 = 431.1 W
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Chapter 12, Problem 40.
For the three-phase circuit in Fig. 12.59, find the average power absorbed by the
delta-connected load with Z ∆ = 21 + j 24Ω .
Figure 12.59
For Prob. 12.40.
Chapter 12, Solution 40.
Transform the delta-connected load to its wye equivalent.
Z∆
= 7 + j8
ZY =
3
Using the per-phase equivalent circuit above,
100∠0°
Ia =
= 8.567 ∠ - 46.75°
(1 + j0.5) + (7 + j8)
For a wye-connected load,
I p = I a = I a = 8.567
S = 3 Ip
2
Z p = (3)(8.567) 2 (7 + j8)
P = Re(S) = (3)(8.567) 2 (7) = 1.541 kW
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Chapter 12, Problem 41.
A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging. If the
three-phase system has an effective line voltage of 400 V, find the line current.
Chapter 12, Solution 41.
P 5 kW
=
= 6.25 kVA
S=
pf
0.8
S = 3 VL I L
But
IL =
S
3 VL
=
6.25 × 10 3
3 × 400
= 9.021 A
Chapter 12, Problem 42.
A balanced three-phase generator delivers 7.2 kW to a wye-connected load with
impedance 30 – j40 Ω per phase. Find the line current IL and the line voltage VL.
Chapter 12, Solution 42.
The load determines the power factor.
40
tan θ =
= 1.333 ⎯
⎯→ θ = 53.13°
30
pf = cos θ = 0.6 (leading)
⎛ 7.2 ⎞
S = 7.2 − j⎜ ⎟(0.8) = 7.2 − j9.6 kVA
⎝ 0.6 ⎠
S = 3 Ip
But
2
Ip
=
2
Zp
S
(7.2 − j9.6) × 10 3
=
= 80
3Zp
(3)(30 − j40)
I p = 8.944 A
I L = I p = 8.944 A
VL =
S
3 IL
=
12 × 10 3
3 (8.944)
= 774.6 V
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Chapter 12, Problem 43.
Refer to Fig. 12.48. Obtain the complex power absorbed by the combined loads.
Chapter 12, Solution 43.
S = 3 Ip
2
I p = I L for Y-connected loads
Zp ,
S = (3)(13.66) 2 (7.812 − j2.047)
S = 4.373 − j1.145 kVA
Chapter 12, Problem 44.
A three-phase line has an impedance of 1 + j3 Ω per phase. The line feeds a balanced
delta-connected load, which absorbs a total complex power of 12 + j5 k VA. If the line
voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line
voltage at the source end and the source power factor.
Chapter 12, Solution 44.
For a ∆-connected load,
Vp = VL ,
IL = 3 Ip
S = 3 VL I L
IL =
S
3 VL
=
(12 2 + 5 2 ) × 10 3
3 (240)
= 31.273
At the source,
VL' = VL + I L Z L
VL' = 240∠0° + (31.273)(1 + j3)
VL' = 271.273 + j93.819
VL' = 287.04 V
Also, at the source,
S ' = 3VL' I *L
S ' = 3 (271.273 + j93.819)(31.273)
⎛ 93.819 ⎞
⎟ = 19.078
θ = tan -1 ⎜
⎝ 271.273 ⎠
pf = cos θ = 0.9451
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Chapter 12, Problem 45.
A balanced wye-connected load is connected to the generator by a balanced transmission
line with an impedance of 0.5 + j2 Ω per phase. If the load is rated at 450 kW, 0.708
power factor lagging, 440-V line voltage, find the line voltage at the generator.
Chapter 12, Solution 45.
S = 3 VL I L ∠θ
IL =
IL =
S ∠-θ
3 VL
,
(635.6) ∠ - θ
3 × 440
P 450 × 10 3
S =
=
= 635.6 kVA
pf
0.708
= 834 ∠ - 45° A
At the source,
VL = 440 ∠0° + I L (0.5 + j2)
VL = 440 + (834 ∠ - 45°)(2.062 ∠76°)
VL = 440 + 1719.7 ∠31°
VL = 1914.1 + j885.7
VL = 2.109∠24.83° kV
Note, this is not normally experienced in practice since transformers are use which can
significantly reduce line losses.
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Chapter 12, Problem 46.
A three-phase load consists of three 100- Ω resistors that can be wye- or delta-connected.
Determine which connection will absorb the most average power from a three-phase
source with a line voltage of 110 V. Assume zero line impedance.
Chapter 12, Solution 46.
For the wye-connected load,
IL = Ip ,
VL = 3 Vp
S = 3 Vp I *p =
S=
VL
Z*
2
3 Vp
2
=
Z*
I p = Vp Z
3 VL
2
Z*
(110) 2
= 121 W
100
=
For the delta-connected load,
Vp = VL ,
IL = 3 Ip ,
S = 3 Vp I *p =
S=
3
3 Vp
Z*
2
=
3 VL
I p = Vp Z
2
Z*
2
(3)(110)
= 363 W
100
This shows that the delta-connected load will deliver three times more average
Z∆
power than the wye-connected load. This is also evident from Z Y =
.
3
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Chapter 12, Problem 47.
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.
Chapter 12, Solution 47.
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)
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Chapter 12, Problem 48.
A balanced, positive-sequence wye-connected source has Van = 240 ∠0° V rms and
supplies an unbalanced delta-connected load via a transmission line with impedance
2 + j3 Ω per phase.
(a) Calculate the line currents if ZAB = 40 + j15 Ω , ZBC = 60 Ω , ZCA = 18 – j12 Ω .
(b) Find the complex power supplied by the source.
Chapter 12, Solution 48.
(a) We first convert the delta load to its equivalent wye load, as shown below.
A
A
18-j12 Ω
ZA
40+j15 Ω
ZB
ZC
C
B
C
60 Ω
ZA =
(40 + j15)(18 − j12)
= 7.577 − j1.923
118 + j 3
ZB =
60(40 + j15).
= 20.52 + j7.105
118 + j3
ZC =
60(18 − j12)
= 8.992 − j 6.3303
118 + j 3
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B
The system becomes that shown below.
a
2+j3
A
+
240<0o
240<120o
+
c
-
-
I2
ZA
I1
240<-120o
+
b
ZC
ZB
2+j3
B
C
2+j3
We apply KVL to the loops. For mesh 1,
− 240 + 240∠ − 120 o + I 1 (2Z l + Z A + Z B ) − I 2 ( Z B + Z l ) = 0
or
(32.097 + j11.13) I 1 − (22.52 + j10.105) I 2 = 360 + j 207.85
For mesh 2,
240∠120 o − 240∠ − 120 o − I 1 ( Z B + Z l ) + I 2 (2Z l + Z B + Z C ) = 0
or
(1)
− (22.52 + j10.105) I 1 + (33.51 + j 6.775) I 2 = − j 415.69
Solving (1) and (2) gives
I 1 = 23.75 − j 5.328,
I 2 = 15.165 − j11.89
(2)
I aA = I 1 = 24.34∠ − 12.64 o A,
I bB = I 2 − I 1 = 10.81∠ − 142.6 o A
I cC = − I 2 = 19.27∠141.9 o A
(b) S a = (240∠0 o )(24.34∠12.64 o ) = 5841.6∠12.64 o
S b = (240∠ − 120 o )(10.81∠142.6 o ) = 2594.4∠22.6 o
S b = (240∠120 o )(19.27∠ − 141.9 o ) = 4624.8∠ − 21.9 o
S = S a + S b + S c = 12.386 + j 0.55 kVA = 12.4∠2.54 o kVA
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Chapter 12, Problem 49.
Each phase load consists of a 20- Ω resistor and a 10- Ω inductive reactance. With a line
voltage of 220 V rms, calculate the average power taken by the load if:
(a) the three-phase loads are delta-connected
(b) the loads are wye-connected
Chapter 12, Solution 49.
(a) For the delta-connected load, Z p = 20 + j10Ω,
S=
3V 2 p
3 x 220 2
=
= 5808 + j 2904 = 6.943∠26.56 o kVA
(20 − j10)
Z*p
or 5.808kW
(b) For the wye-connected load, Z p = 20 + j10Ω,
S=
V p = VL = 220 (rms) ,
V p = VL / 3 ,
3V 2 p
3 x 220 2
=
= 2.164∠26.56 o kVA or 1.9356 kW
*
3(20 − j10)
Z p
Chapter 12, Problem 50.
A balanced three-phase source with VL = 240 V rms is supplying 8 kVA at 0.6 power
factor lagging to two wye-connected parallel loads. If one load draws 3 kW at unity
power factor, calculate the impedance per phase of the second load.
Chapter 12, Solution 50.
S = S 1 + S 2 = 8(0.6 + j 0.8) = 4.8 + j 6.4 kVA,
Hence,
S 1 = 3 kVA
S 2 = S − S 1 = 1.8 + j 6.4 kVA
3V 2 p
But S 2 = * ,
Z p
Z*p =
Vp =
VL
3
240 2
V *L
=
(1.8 + j 6.4) x10 3
S2
⎯
⎯→
⎯
⎯→
.V 2 L
S2 = *
Z p
Z p = 2.346 + j8.34Ω
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Chapter 12, Problem 51.
Consider the ∆ - ∆ system shown in Fig. 12.60. Take Zi = 8 + j6 Ω , Z2 = 4.2 –
j2.2 Ω , Z3 = 10 + j0 Ω .
(a) Find the phase current IAB, IBC, ICA.
(b) Calculate line currents IaA, IbB, and IcC.
a
240 0° V
b
A
−
+
+
−
240 −120° V
c
+−
Z3
C
Z1
Z2
B
240 120° V
Figure 12.60
For Prob. 12.51.
Chapter 12, Solution 51.
This is an unbalanced system.
240 < 0o 240 < 0o
=
= 19.2-j14.4 A = 19.2–j14.4 A
I AB =
8 + j6
Z1
I BC =
240∠120°
240∠120°
=
= 50.62∠147.65˚ = –42.76+j27.09 A
Z2
4.7413∠ − 27.65
I CA =
240∠ − 120° 240∠ − 120°
=
= –12–j20.78 A
Z3
10
At node A,
I aA = I AB − I CA = (19.2 − j14.4) − (−12 − j 20.78) = 31.2 + j 6.38 A = 31.2+j6.38 A
IbBI b = I BC − I AB = (−42.76 + j 27.08) − (19.2 − j14.4) = −61.96 + j 41.48 A
= –61.96+j41.48 A
IcCI c = I CA − I
= (−12 − j 20.78) − (−42.76 + j 27.08) = 30.76 − j 47.86 A
= 30.76–j47.86 A
BC
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Chapter 12, Problem 52.
A four-wire wye-wye circuit has
Van = 120 ∠120° ,
Vbn = 120 ∠0°
Vcn = 120 ∠ − 120° V
If the impedances are
ZAN = 20 ∠60° ,
ZBN = 30 ∠0°
Zcn = 40 ∠30°Ω
find the current in the neutral line.
Chapter 12, Solution 52.
Since the neutral line is present, we can solve this problem on a per-phase basis.
Van 120 ∠120°
=
= 6 ∠60°
Ia =
20 ∠60°
Z AN
Vbn 120 ∠0°
=
= 4 ∠0°
Ib =
30 ∠0°
Z BN
Vcn 120 ∠ - 120°
=
= 3∠ - 150°
Ic =
40 ∠30°
Z CN
Thus,
- In
- In
- In
- In
= Ia + Ib + Ic
= 6 ∠60° + 4 ∠0° + 3∠ - 150°
= (3 + j5.196) + (4) + (-2.598 − j1.5)
= 4.405 + j3.696 = 5.75∠40°
I n = 5.75∠220° A
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Chapter 12, Problem 53.
In the Y-Y system shown in Fig. 12.61, loads connected to the source are unbalanced.
(a) Calculate Ia, Ib, and Ic. (b) Find the total power delivered to the load. Take
Vp = 240 V rms.
Figure 12.61
For Prob. 12.53.
Chapter 12, Solution 53.
Applying mesh analysis as shown below, we get.
Ia
+
_
VP∠120˚
+ –
VP∠0˚
100Ω
I1
VP∠–120˚
– +
80Ω
Ib
Ic
60 Ω
I2
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240 < −120o − 240 < 0o + I1 x160 − 60 I 2 = 0
⎯⎯
→
240 < 120o − 240 < −120o + 140 I 2 − 60 I1 = 0
In matrix form, (1) and (2) become
160 I1 − 60 I 2 = 360 + j 207.84 (1)
⎯⎯
→
140 I 2 − 60 I1 = − j 415.7
(2)
⎡160 −60 ⎤ ⎡ I1 ⎤ ⎡360 + j 207.84 ⎤
⎢ −60 140 ⎥ ⎢ I ⎥ = ⎢ − j 415.7 ⎥
⎣
⎦⎣ 2⎦ ⎣
⎦
Using MATLAB, we get,
>> Z=[160,-60;-60,140]
Z=
160 -60
-60 140
>> V=[(360+207.8i);-415.7i]
V=
1.0e+002 *
3.6000 + 2.0780i
0 - 4.1570i
>> I=inv(Z)*V
I=
2.6809 + 0.2207i
1.1489 - 2.8747i
I1 = 2.681+j0.2207 and I2 = 1.1489–j2.875
Ia = I1 = 2.69∠4.71˚ A
Ib = I2 – I1 = –1.5321–j3.096 = 3.454∠–116.33˚ A
Ic = –I2 = 3.096∠111.78˚ A
S a =| I a |2 Z a = (2.69) 2 x100 = 723.61 W
Sb =| I b |2 Z b = (3.454) 2 x60 = 715.81 W
Sc =| I c |2 Z c = (3.0957) 2 x80 = 766.67 W
P = Pa + Pb + Pc = 2.205 kW
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Chapter 12, Problem 54.
A balanced three-phase Y-source with VP = 210 V rms drives a Y-connected three-phase
load with phase impedance ZA = 80 Ω , ZB = 60 + j90 Ω , and ZC = j80 Ω . Calculate the
line currents and total complex power delivered to the load. Assume that the neutrals are
connected.
Chapter 12, Solution 54.
Consider the load as shown below.
Ia
A
A
NN
Ib
B
C
Ic
Ia =
210 < 0o
= 2.625 A
80
Ib =
210∠0°
210
=
= 1.9414∠–56.31˚ A
60 + j90 108.17∠56.31°
210 < 0o
= 2.625 < −90o A
Ic =
j80
*
S a = VI a = 210 x 2.625 = 551.25
Sb = VI b* =
| V |2
2102
=
= 226.15 + j 339.2
60 − j 90
Z b*
| V |2 2102
=
= j 551.25
− j80
Z c*
S = S a + Sb + Sc = 777.4 + j890.45 VA
Sc =
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Chapter 12, Problem 55.
A three-phase supply, with the line voltage 240 V rms positively phased, has an
unbalanced delta-connected load as shown in Fig. 12.62. Find the phase currents and the
total complex power.
Figure 12.62
For Prob. 12.55.
Chapter 12, Solution 55.
The phase currents are:
IAB = 240/j25 = 9.6∠–90˚ A
ICA = 240∠120˚/40 = 6∠120˚ A
IBC = 240∠–120˚/30∠30˚ = 8∠–150˚ A
The complex power in each phase is:
S AB =| I AB |2 Z AB = (9.6) 2 j 25 = j 2304
S AC =| I AC |2 Z AC = (6) 2 40 < 0o = 1440
S BC =| I BC |2 Z BC = (8) 2 30 < 30o = 1662.77 + j 960
The total complex power is,
S = S AB + S AC + S BC = 3102.77 + j 3264 VA = 3.103+j3.264 kVA
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Chapter 12, Problem 56.
Refer to the unbalanced circuit of Fig. 12.63. Calculate:
(a) the line currents
(b) the real power absorbed by the load
(c) the total complex power supplied by the source
Figure 12.63
For Prob. 12.56.
Chapter 12, Solution 56.
(a)
Consider the circuit below.
a
A
440∠0° + −
b
440∠120°
+
−
j10 Ω
I1
B
− +
440∠-120°
I2
I3
-j5 Ω
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
C
(1)
(2)
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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
(3)
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
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Chapter 12, Problem 57.
Determine the line currents for the three-phase circuit of Fig. 12.64. Let Va = 110 ∠0° ,
Vb = 110 ∠ − 120° , Vc = 110 ∠120° V.
Figure 12.64
For Prob. 12.57.
Chapter 12, Solution 57.
We apply mesh analysis to the circuit shown below.
Ia
+
Va
–
80 + j 50Ω
I1
–
20 + j 30Ω
–
Vb
Vc
+
+
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 b = I 2 − I 1 = −0.528 − j1.1136 = 1.4656∠ − 130.55 o A
I c = − I 2 = 1.947∠117.8 o A
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Chapter 12, Problem 58.
Solve Prob. 12.10 using PSpice.
Chapter 12, Solution 58.
The schematic is shown below. IPRINT is inserted in the neutral line to measure the
current through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq. =
0.1592, and End Freq. = 0.1592. After simulation, the output file includes
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E–01
1.078 E+01
–8.997 E+01
i.e.
In = 10.78∠–89.97° A
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Chapter 12, Problem 59.
The source in Fig. 12.65 is balanced and exhibits a positive phase sequence. If f = 60 Hz,
use PSpice to find VAN, VBN, and VCN.
Figure 12.65
For Prob. 12.59.
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Chapter 12, Solution 59.
The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq
= 60, and End Freq = 60. After simulation, we obtain an output file which includes
i.e.
FREQ
VM(1)
VP(1)
6.000 E+01
2.206 E+02
–3.456 E+01
FREQ
VM(2)
VP(2)
6.000 E+01
2.141 E+02
–8.149 E+01
FREQ
VM(3)
VP(3)
6.000 E+01
4.991 E+01
–5.059 E+01
VAN = 220.6∠–34.56°, VBN = 214.1∠–81.49°, VCN = 49.91∠–50.59° V
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Chapter 12, Problem 60.
Use PSpice to determine Io in the single-phase, three-wire circuit of Fig. 12.66. Let
Z1 = 15 – j10 Ω , Z2 = 30 + j20 Ω , and Z3 = 12 + j5 Ω .
Figure 12.66
For Prob. 12.60.
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Chapter 12, Solution 60.
The schematic is shown below. IPRINT is inserted to give Io. We select Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation,
the output file includes
from which,
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E–01
1.421 E+00
–1.355 E+02
Io = 1.421∠–135.5° A
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Chapter 12, Problem 61.
Given the circuit in Fig. 12.67, use PSpice to determine currents IaA and voltage VBN.
Figure 12.67
For Prob. 12.61.
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Chapter 12, Solution 61.
The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to
measure IaA and VBN. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592,
and End Freq = 0.1592. Once the circuit is simulated, we get an output file which
includes
FREQ
VM(2)
VP(2)
1.592 E–01
2.308 E+02
–1.334 E+02
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
1.115 E+01
3.699 E+01
from which
IaA = 11.15∠37° A, VBN = 230.8∠–133.4° V
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Chapter 12, Problem 62.
The circuit in Fig. 12.68 operates at 60 Hz. Use PSpice to find the source current Iab and
the line current IbB.
Figure 12.68
For Prob. 12.62.
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Chapter 12, Solution 62.
Because of the delta-connected source involved, we follow Example 12.12. In the AC
Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After
simulation, the output file includes
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
6.000 E+01
5.960 E+00
–9.141 E+01
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
6.000 E+01
7.333 E+07
1.200 E+02
From which
Iab = 7.333x107∠120° A, IbB = 5.96∠–91.41° A
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Chapter 12, Problem 63.
Use PSpice to find currents IaA and IAC in the unbalanced three-phase system shown in
Fig. 12.69. Let
ZI = 2 + j,
Z1 = 40 + j20 Ω ,
Z2 = 50 – j30 Ω , Z3 = 25 Ω
220 –120° V
220 120° V
Figure 12.69
For Prob. 12.63.
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Chapter 12, Solution 63.
Let ω = 1 so that L = X/ω = 20 H, and C =
1
= 0.0333 F
ωX
The schematic is shown below.
.
When the file is saved and run, we obtain an output file which includes the following:
FREQ
1.592E-01
FREQ
1.592E-01
IM(V_PRINT1)IP(V_PRINT1)
1.867E+01
1.589E+02
IM(V_PRINT2)IP(V_PRINT2)
1.238E+01
1.441E+02
From the output file, the required currents are:
I aA = 18.67∠158.9 o A, I AC = 12.38∠144.1o A
Chapter 12, Problem 64.
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For the circuit in Fig. 12.58, use PSpice to find the line currents and the phase currents.
Chapter 12, Solution 64.
We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq =
0.1592, and End Freq = 0.1592. After simulation the output file includes
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592 E–01
4.710 E+00
7.138 E+01
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592 E–01
6.781 E+07
–1.426 E+02
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592 E–01
3.898 E+00
–5.076 E+00
FREQ
IM(V_PRINT4)
IP(V_PRINT4)
1.592 E–01
3.547 E+00
6.157 E+01
FREQ
IM(V_PRINT5)
IP(V_PRINT5)
1.592 E–01
1.357 E+00
9.781 E+01
FREQ
IM(V_PRINT6)
IP(V_PRINT6)
1.592 E–01
3.831 E+00
–1.649 E+02
from this we obtain
IaA = 4.71∠71.38° A, IbB = 6.781∠–142.6° A, IcC = 3.898∠–5.08° A
IAB = 3.547∠61.57° A, IAC = 1.357∠97.81° A, IBC = 3.831∠–164.9° A
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Chapter 12, Problem 65.
A balanced three-phase circuit is shown in Fig. 12.70 on the next page. Use PSpice to
find the line currents IaA, IbB, and IcC.
Figure 12.70
For Prob. 12.65.
Chapter 12, Solution 65.
Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1,
Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it
is saved and simulated, we obtain an output file which includes
Thus,
FREQ
IM(V_PRINT1)
IP(V_PRINT1)
1.592E-01
1.140E+01
8.664E+00
FREQ
IM(V_PRINT2)
IP(V_PRINT2)
1.592E-01
1.140E+01
-1.113E+02
FREQ
IM(V_PRINT3)
IP(V_PRINT3)
1.592E-01
1.140E+01
1.287E+02
IaA = 11.02∠12° A, IbB = 11.02∠–108° A, IcC = 11.02∠132° A
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Since this is a balanced circuit, we can perform a quick check. The load resistance is
large compared to the line and source impedances so we will ignore them (although it
would not be difficult to include them).
Converting the sources to a Y configuration we get:
Van = 138.56 ∠–20˚ Vrms
and
ZY = 10 – j6.667 = 12.019∠–33.69˚
Now we can calculate,
IaA = (138.56 ∠–20˚)/(12.019∠–33.69˚) = 11.528∠13.69˚
Clearly, we have a good approximation which is very close to what we really have.
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Chapter 12, Problem 66.
A three-phase, four-wire system operating with a 208-V line voltage is shown in Fig.
12.71. 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 12.71
For Prob. 12.66.
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Chapter 12, Solution 66.
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 ,
I3 = 2 A ,
I N = 0.866 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
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Chapter 12, Problem 67.
* As shown in Fig. 12.72, a three-phase four-wire line with a phase voltage of 120
V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf
lagging. The motor load is connected to the three main lines marked a, b, and c. In
addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line c to
the neutral, 15 kW from line b to the neutral, and 9 kW from line c to the neutral.
(a) If three wattmeters are arranged to measure the power in each line, calculate the
reading of each meter.
(b) Find the magnitude of the current in the neutral line.
Figure 12.72
For Prob. 12.67.
* An asterisk indicates a challenging problem.
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Chapter 12, Solution 67.
(a)
The power to the motor is
PT = S cos θ = (260)(0.85) = 221 kW
The motor power per phase is
1
Pp = PT = 73.67 kW
3
Hence, the wattmeter readings are as follows:
Wa = 73.67 + 24 = 97.67 kW
Wb = 73.67 + 15 = 88.67 kW
Wc = 73.67 + 9 = 82.67 kW
(b)
The motor load is balanced so that I N = 0 .
For the lighting loads,
24,000
Ia =
= 200 A
120
15,000
Ib =
= 125 A
120
9,000
Ic =
= 75 A
120
If we let
I a = I a ∠0° = 200∠0° A
I b = 125∠ - 120° A
I c = 75∠120° A
Then,
- I N = Ia + Ib + Ic
⎛
⎛
3⎞
3⎞
- I N = 200 + (125)⎜⎜ - 0.5 − j ⎟⎟ + (75)⎜⎜ - 0.5 + j ⎟⎟
2 ⎠
2 ⎠
⎝
⎝
- I N = 100 − j43.3 A
I N = 108.97 A
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Chapter 12, Problem 68.
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
Chapter 12, Solution 68.
(a)
S = 3 VL I L = 3 (330)(8.4) = 4801 VA
(b)
P = S cos θ ⎯
⎯→ pf = cos θ =
pf =
P
S
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
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Chapter 12, Problem 69.
A certain store contains three balanced three-phase loads. The three loads are:
Load 1: 16 kVA at 0.85 pf lagging
Load 2: 12 kVA at 0.6 pf lagging
Load 3: 8 kW at unity pf
The line voltage at the load is 208 V rms at 60 Hz, and the line impedance is
0.4 + j0.8 Ω . Determine the line current and the complex power delivered to the loads.
Chapter 12, Solution 69.
For load 1,
S 1 = S1 cos θ 1 + jS1 sin θ1
pf = 0.85 = cos θ1
⎯⎯
→ θ1 = 31.79o
S 1 = 13.6 + j8.43 kVA
For load 2,
S 2 = 12 x0.6 + j12 x0.8 = 7.2 + j 9.6 kVA
For load 3,
S 3 = 8 + j 0 kVA
S = S 1 + S 2 + S3 = 28.8 + j18.03 = 28.8+j18.03 kVA
But SP = VPIP* with IP = IL
S
(28800 + j18030)
I*L = P =
VP
3x120.08
IL = 79.95 – j50.05 = 94.32∠–32.05˚ A. Note, this is relative to 120.08∠0˚ V. If we
assume a positive phase rotation and Vab = 208∠0˚, then Van = 120.08∠–30˚ which
yields Ia = 94.32∠–62.05˚ A, Ib = 94.32∠177.95˚ A, Ic = 94.32∠57.95˚ A.
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Chapter 12, Problem 70.
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.
Chapter 12, Solution 70.
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
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Chapter 12, Problem 71.
In Fig. 12.73, 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 12.73
For Prob. 12.71.
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Chapter 12, Solution 71.
(a)
If Vab = 208∠0° , Vbc = 208∠ - 120° , Vca = 208∠120° ,
Vab 208∠0°
=
= 10.4 ∠0°
I AB =
20
Z Ab
Vbc
208∠ - 120°
=
= 14.708∠ - 75°
I BC =
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
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Chapter 12, Problem 72.
If wattmeters W1 and W2 are properly connected respectively between lines a and b and
lines b and c to measure the power absorbed by the delta-connected load in Fig. 12.44,
predict their readings.
Chapter 12, Solution 72.
From Problem 12.11,
VAB = 220 ∠130° V
and
I aA = 30∠180° A
P1 = (220)(30) cos(130° − 180°) = 4242 W
VCB = -VBC = 220∠190°
I cC = 30∠ - 60°
P2 = (220)(30) cos(190° + 60°) = - 2257 W
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Chapter 12, Problem 73.
For the circuit displayed in Fig. 12.74, find the wattmeter readings.
Figure 12.74
For Prob. 12.73.
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Chapter 12, Solution 73.
Consider the circuit as shown below.
I1
Ia
240∠-60° V
+
−
Z
Z
240∠-120° V
−
+
Z
I2
Ib
Ic
Z = 10 + j30 = 31.62∠71.57°
240∠ - 60°
= 7.59∠ - 131.57°
31.62∠71.57°
240 ∠ - 120°
Ib =
= 7.59∠ - 191.57°
31.62∠71.57°
Ia =
I c Z + 240∠ - 60° − 240 ∠ - 120° = 0
- 240
Ic =
= 7.59∠108.43°
31.62∠71.57°
I 1 = I a − I c = 13.146∠ - 101.57°
I 2 = I b + I c = 13.146∠138.43°
P1 = Re [ V1 I 1* ] = Re [ (240∠ - 60°)(13.146 ∠101.57°) ] = 2360 W
P2 = Re [ V2 I *2 ] = Re [ (240 ∠ - 120°)(13.146∠ - 138.43°) ] = - 632.8 W
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Chapter 12, Problem 74.
Predict the wattmeter readings for the circuit in Fig. 12.75.
Figure 12.75
For Prob. 12.74.
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Chapter 12, Solution 74.
Consider the circuit shown below.
Z = 60 − j30 Ω
208∠0° V
+
−
I1
Z
208∠-60° V
−
+
I2
Z
For mesh 1,
208 = 2 Z I 1 − Z I 2
For mesh 2,
- 208∠ - 60° = - Z I 1 + 2 Z I 2
In matrix form,
⎡
⎤ ⎡ 2 Z - Z ⎤⎡ I 1 ⎤
208
⎢ - 208∠ - 60°⎥ = ⎢ - Z 2 Z ⎥⎢ I ⎥
⎣
⎦ ⎣
⎦⎣ 2 ⎦
∆ = 3Z2 ,
∆ 1 = (208)(1.5 + j0.866) Z ,
∆ 2 = (208)( j1.732) Z
I1 =
∆ 1 (208)(1.5 + j0.866)
=
= 1.789∠56.56°
∆
(3)(60 − j30)
I2 =
∆ 2 (208)( j1.732)
=
= 1.79∠116.56°
∆
(3)(60 − j30)
P1 = Re [ V1 I 1* ] = Re [ (208)(1.789∠ - 56.56°) ] = 208.98 W
P2 = Re [ V2 (- I 2 ) * ] = Re [ (208∠ - 60°))(1.79∠63.44°) ] = 371.65 W
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Chapter 12, Problem 75.
A man has a body resistance of 600 Ω . How much current flows through his ungrounded
body:
(a) when he touches the terminals of a 12-V autobattery?
(b) when he sticks his finger into a 120-V light socket?
Chapter 12, Solution 75.
(a)
I=
V 12
=
= 20 mA
R 600
(b)
I=
V 120
=
= 200 mA
R 600
Chapter 12, Problem 76.
Show that the I 2 R losses will be higher for a 120-V appliance than for a 240-V
appliance if both have the same power rating.
Chapter 12, Solution 76.
If both appliances have the same power rating, P,
P
I=
Vs
For the 120-V appliance,
For the 240-V appliance,
⎧ P2 R
⎪
2
Power loss = I 2 R = ⎨ 120
2
⎪P R
⎩ 240 2
Since
P
.
120
P
.
I2 =
240
I1 =
for the 120-V appliance
for the 240-V appliance
1
1
, the losses in the 120-V appliance are higher.
2 >
120
240 2
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Chapter 12, Problem 77.
A three-phase generator supplied 3.6 kVA at a power factor of 0.85 lagging. If 2500 W
are delivered to the load and line losses are 80 W per phase, what are the losses in the
generator?
Chapter 12, Solution 77.
Pg = PT − Pload − Pline ,
But
pf = 0.85
PT = 3600 cos θ = 3600 × pf = 3060
Pg = 3060 − 2500 − (3)(80) = 320 W
Chapter 12, Problem 78.
A three-phase 440-V, 51-kW, 60-kVA inductive load operates at 60 Hz and is wyeconnected. It is desired to correct the power factor to 0.95 lagging. What value of
capacitor should be placed in parallel with each load impedance?
Chapter 12, Solution 78.
51
= 0.85 ⎯
⎯→ θ1 = 31.79°
60
Q1 = S1 sin θ1 = (60)(0.5268) = 31.61 kVAR
cos θ1 =
P2 = P1 = 51 kW
cos θ 2 = 0.95 ⎯
⎯→ θ 2 = 18.19°
P2
S2 =
= 53.68 kVA
cos θ 2
Q 2 = S 2 sin θ 2 = 16.759 kVAR
Q c = Q1 − Q 2 = 3.61 − 16.759 = 14.851 kVAR
For each load,
Qc
= 4.95 kVAR
3
Q c1
4950
= 67.82 µF
C=
2 =
ωV
(2π )(60)(440) 2
Q c1 =
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Chapter 12, Problem 79.
A balanced three-phase generator has an abc phase sequence with phase voltage
Van = 255 ∠0° V. The generator feeds an induction motor which may be represented by a
balanced Y-connected load with an impedance of 12 + j5 Ω per phase. Find the line
currents and the load voltages. Assume a line impedance of 2 Ω per phase.
Chapter 12, Solution 79.
Consider the per-phase equivalent circuit below.
Ia
2Ω
a
A
+
−
Van
ZY = 12 + j5 Ω
n
Ia =
Thus,
N
Van
255∠0°
=
= 17.15∠ - 19.65° A
Z Y + 2 14 + j5
I b = I a ∠ - 120° = 17.15∠ - 139.65° A
I c = I a ∠120° = 17.15 ∠100.35° A
VAN = I a Z Y = (17.15∠ - 19.65°)(13∠22.62°) = 223∠ 2.97° V
Thus,
VBN = VAN ∠ - 120° = 223∠ - 117.03° V
VCN = VAN ∠120° = 223∠122.97° V
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Chapter 12, Problem 80.
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.
Chapter 12, Solution 80.
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.
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Chapter 12, Problem 81.
A professional center is supplied by a balanced three-phase source. The center has four
balanced three-phase loads 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.
Chapter 12, Solution 81.
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
= 542.7 A
IL =
=
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
VT =
ST
3 IL
=
485.1 × 10 3
3 × 542.7
= 516 V
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Chapter 12, Problem 82.
A balanced three-phase system has a distribution wire with impedance 2 + j6 Ω per
phase. The system supplies two three-phase loads that are connected in parallel. The first
is a balanced wye-connected load that absorbs 400 kVA at a power factor of 0.8 lagging.
The second load is a balanced delta-connected load with impedance of 10 + j8 Ω per
phase. If the magnitude of the line voltage at the loads is 2400 V rms, calculate the
magnitude of the line voltage at the source and the total complex power supplied to the
two loads.
Chapter 12, Solution 82.
S 1 = 400(0.8 + j 0.6) = 320 + j 240 kVA,
V 2p
S2 = 3 *
Z p
For the delta-connected load, V L = V p
(2400) 2
S 2 = 3x
= 1053.7 + j842.93 kVA
10 − j8
S = S 1 + S 2 = 1.3737 + j1.0829 MVA
Let I = I1 + I2 be the total line current. For I1,
S1 = 3V p I *1 ,
I *1 =
S1
Vp =
VL
3
(320 + j 240) x10 3
=
,
3VL
3 (2400)
For I2, convert the load to wye.
I 1 = 76.98 − j 57.735
2400
3∠ − 30 o = 273.1 − j 289.76
10 + j8
I = I 1 + I 2 = 350 − j 347.5
I 2 = I p 3∠ − 30 o =
Vs = VL + Vline = 2400 + I (3 + j 6) = 5.185 + j1.405 kV
⎯
⎯→
| Vs |= 5.372 kV
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Chapter 12, Problem 83.
A commercially available three-phase inductive motor operates at a full load of 120 hp (1
hp = 746 W) at 95 percent efficiency at a lagging power factor of 0.707. The motor is
connected in parallel to a 80-kW balanced three-phase heater at unity power factor. If the
magnitude of the line voltage is 480 V rms, calculate the line current.
Chapter 12, Solution 83.
S1 = 120 x746 x0.95(0.707 + j 0.707) = 60.135 + j 60.135 kVA,
S 2 = 80 kVA
S = S1 + S 2 = 140.135 + j 60.135 kVA
But | S |= 3VL I L
⎯
⎯→
IL =
|S|
3VL
=
152.49 x10 3
3 x 480
= 183.42 A
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Chapter 12, Problem 84.
* Figure 12.76 displays a three-phase delta-connected motor load which is connected to
a line voltage of 440 V and draws 4 kVA at a power factor of 72 percent lagging. In
addition, a single 1.8 kVAR capacitor is connected between lines a and b, while a 800-W
lighting load is connected between line c and neutral. Assuming the abc sequence and
taking Van = V p ∠0° , find the magnitude and phase angle of currents Ia, Ib, Ic, and In.
Figure 12.76
For Prob. 12.84.
* An asterisk indicates a challenging problem.
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Chapter 12, Solution 84.
We first find the magnitude of the various currents.
For the motor,
S
IL =
3 VL
=
4000
440 3
= 5.248 A
For the capacitor,
IC =
Q c 1800
=
= 4.091 A
VL
440
For the lighting,
Vp =
I Li =
440
3
= 254 V
PLi 800
=
= 3.15 A
Vp 254
Consider the figure below.
Ia
a
IC
+
-jXC
Vab
b
−
I1
Ib
I2
Ic
I3
c
ILi
In
R
n
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If Van = Vp ∠0° ,
Vab = 3 Vp ∠30°
Vcn = Vp ∠120°
IC =
Vab
= 4.091∠120°
-j X C
I1 =
Vab
= 4.091∠(θ + 30°)
Z∆
where θ = cos -1 (0.72) = 43.95°
I 1 = 5.249 ∠73.95°
I 2 = 5.249 ∠ - 46.05°
I 3 = 5.249∠193.95°
I Li =
Vcn
= 3.15∠120°
R
Thus,
I a = I 1 + I C = 5.249∠73.95° + 4.091∠120°
I a = 8.608∠93.96° A
I b = I 2 − I C = 5.249∠ - 46.05° − 4.091∠120°
I b = 9.271∠ - 52.16° A
I c = I 3 + I Li = 5.249∠193.95° + 3.15∠120°
I c = 6.827 ∠167.6° A
I n = - I Li = 3.15∠ - 60° A
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Chapter 12, Problem 85.
Design a three-phase heater with suitable symmetric loads using wye-connected
pure resistance. Assume that the heater is supplied by a 240-V line voltage and is to give
27 kW of heat.
Chapter 12, Solution 85.
Let
ZY = R
Vp =
VL
3
=
240
3
= 138.56 V
Vp2
27
P = Vp I p =
= 9 kW =
2
R
R=
Thus,
Vp2
P
=
(138.56) 2
= 2.133 Ω
9000
Z Y = 2.133 Ω
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Chapter 12, Problem 86.
For the single-phase three-wire system in Fig. 12.77, find currents IaA, IbB, and InN.
Figure 12.77
For Prob. 12.86.
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Chapter 12, Solution 86.
Consider the circuit shown below.
1Ω
a
A
+
−
120∠0° V rms
I1
24 – j2 Ω
1Ω
n
N
+
−
120∠0° V rms
I2
15 + j4 Ω
1Ω
B
b
For the two meshes,
120 = (26 − j2) I 1 − I 2
120 = (17 + j4) I 2 − I 1
In matrix form,
⎡120⎤ ⎡ 26 − j2
- 1 ⎤⎡ I 1 ⎤
=
⎢120⎥ ⎢ - 1
17 + j4⎥⎦⎢⎣ I 2 ⎥⎦
⎣ ⎦ ⎣
∆ = 449 + j70 ,
∆ 1 = (120)(18 + j4) ,
(1)
(2)
∆ 2 = (120)(27 − j2)
∆ 1 120 × 18.44 ∠12.53°
=
= 4.87 ∠3.67°
∆
454.42 ∠8.86°
∆ 2 120 × 27.07 ∠ - 4.24°
=
I2 =
= 7.15∠ - 13.1°
∆
454.42 ∠8.86°
I1 =
I aA = I 1 = 4.87 ∠ 3.67° A
I bB = - I 2 = 7.15∠166.9° A
∆ 2 − ∆1
∆
(120)(9 − j6)
=
= 2.856∠ - 42.55° A
449 + j70
I nN = I 2 − I 1 =
I nN
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Chapter 12, Problem 87.
Consider the single-phase three-wire system shown in Fig. 12.78. Find the current in the
neutral wire and the complex power supplied by each source. Take Vs as a 115 ∠0° -V,
60-Hz source.
Figure 12.78
For Prob. 12.87.
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Chapter 12, Solution 87.
L = 50 mH ⎯
⎯→
Consider the circuit below.
jωL = j (2π)(60)(5010 -3 ) = j18.85
1Ω
115 V
+
−
I1
20 Ω
2Ω
15 + j18.85 Ω
115 V
+
−
I2
30 Ω
1Ω
Applying KVl to the three meshes, we obtain
23 I 1 − 2 I 2 − 20 I 3 = 115
- 2 I 1 + 33 I 2 − 30 I 3 = 115
- 20 I 1 − 30 I 2 + (65 + j18.85) I 3 = 0
In matrix form,
- 20
⎡ 23 - 2
⎤ ⎡ I 1 ⎤ ⎡115⎤
⎢ - 2 33
⎥ ⎢I ⎥ = ⎢115⎥
- 30
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣- 20 - 30 65 + j18.85⎥⎦ ⎢⎣I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
∆ = 12,775 + j14,232 ,
∆ 2 = (115)(1825 + j471.3) ,
(1)
(2)
(3)
∆ 1 = (115)(1975 + j659.8)
∆ 3 = (115)(1450)
∆ 1 115 × 2082∠18.47°
=
= 12.52∠ - 29.62°
∆
19214∠48.09°
∆ 2 115 × 1884.9 ∠14.48°
I2 =
= 11.33∠ - 33.61°
=
∆
19124 ∠48.09°
∆ − ∆ 1 (115)(-150 − j188.5)
I n = I 2 − I1 = 2
=
= 1.448∠ - 176.6° A
∆
12,775 + j14,231.75
I1 =
S 1 = V1 I *1 = (115)(12.52∠ 29.62°) = 1252 + j711.6 VA
S 2 = V2 I *2 = (115)(1.33∠33.61°) = 1085 + j721.2 VA
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