570
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CHAPTER 11
P O LY P H A S E C I R C U I T S
SUMMARY
■
An important advantage of the balanced three-phase system
is that it provides very smooth power delivery.
■
Because of the balanced condition, it is possible to analyze
a circuit on a per-phase basis, thereby providing a
significant computational shortcut to a solution.
■
■
TABLE 11.3
QUANTITY
Ia , Ib , Ic
■
In a balanced system the voltages and currents sum to zero.
Line-to-neutral voltage AVp B
and
Phase voltage AVp B
Vab , Vbc , Vca
Line-to-line, phase-to-phase, line voltage AVLB
Phase voltage AVp B
Phase current AIp B
1. If the source/load connection is not wye–wye, then
transform the system to a wye–wye connection.
2. Determine the unknown phasors in the wye–wye
connection and deal only with the phase a.
3. Convert the now-known phasors back to the corresponding
phasors in the original connection.
Vab + Vbc + Vca = 0
I ab + I bc + I ca = 0
•
Van , Vbn , Vcn
Iab , Ibc , Ica
Van + Vbn + Vcn = 0
I a + I b + I c = 0 (no current in the neutral line)
DELTA
Phase current AIp B
The relationships between wye- and delta-connected
sources are shown in Table 11.1.
The three-phase terminology is shown in Table 11.3.
WYE
Line current AIL B
A balanced three-phase voltage source has three sinusoidal
voltages of the same magnitude and frequency, and each
voltage is 120° out of phase with the others. A positivephase-sequence balanced voltage source is one in which Vbn
lags Van by 120° and Vcn lags Vbn by 120°.
■
■
Three-phase terminology
The steps recommended for solving balanced three-phase
ac circuits are as follows:
■
Power factor correction in a balanced three-phase
environment is performed in the same manner as in the
single-phase case. Three capacitors are put in parallel with
the load to reduce the lagging phase caused by the
three-phase load.
PROBLEMS
11.1 Sketch a phasor representation of an abc-sequence
balanced three-phase Y-connected source, including Van ,
Vbn , and Vcn if Van = 120 / 15° V rms.
11.6 Find the equivalent Z of the network in Fig. P11.6.
1⍀
11.2 Sketch a phasor representation of a balanced three-phase
system containing both phase voltages and line voltages
if Van = 120 / 90° V rms. Label all magnitudes and
assume an abc-phase sequence.
11.3 Sketch a phasor representation of a balanced three-phase
system containing both phase voltages and line voltages
if Van = 100 / 45° V rms. Label all magnitudes and
assume an abc-phase sequence.
2⍀
–j2 ⍀
1⍀
1⍀
2⍀
j1 ⍀
Z
11.4 Sketch a phasor representation of a balanced three-phase
system containing both phase voltages and line voltages
if Vab = 208 / 60° V rms. Label all phasors and assume
an abc-phase sequence.
11.5 A positive-sequence three-phase balanced wye voltage
source has a phase voltage of Van = 240 / 90° V rms.
Determine the line voltages of the source.
–j1 ⍀
1⍀
–j1 ⍀
Figure P11.6
–j2 ⍀
PROBLEMS
11.7 Find the equivalent impedances Z ab, Z bc, and Z ca in the
network in Fig. P11.7.
a
1⍀
1⍀
j1 ⍀
11.13 An abc-sequence set of voltages feeds a balanced
three-phase wye–wye system. If the line current in the
a phase is 16.78. / 20.98° A rms, the line impedance is
1.2 + j1.8 ⍀, and the input voltage Vab = 440 / 70° V
rms. find the load impedance.
j1 ⍀
c
b
Figure P11.7
11.8 Find the equivalent Z of the network in Fig. P11.8.
2⍀
–j1 ⍀
1⍀
Z
j1 ⍀
1⍀
j1 ⍀
–j2 ⍀
–j2 ⍀
1/2 ⍀
–j1 ⍀
–j2 ⍀
Figure P11.8
11.9 Find the equivalent Z of the network in Fig. P11.9.
2⍀
a
1⍀
–j1 ⍀
Z
1⍀
–j1 ⍀
1⍀
c
b
2⍀
2⍀
–j1 ⍀
11.11 A positive-sequence balanced three-phase
wye-connected source supplies power to a balanced
wye-connected load. The magnitude of the line voltages
is 208 V rms. If the load impedance per phase is
36 + j12 ⍀, determine the line currents if / Van = 0°.
11.12 In a three-phase balanced wye–wye system, the source is
an abc-sequence set of voltages with Van = 120 / 60° rms.
The per phase impedance of the load is 12 + j16 ⍀. If
the line impedance per phase is 0.8 + j1.4 ⍀, find the
line currents and the voltages.
j1 ⍀
1⍀
571
–j1 ⍀
d
Figure P11.9
11.10 A positive-sequence balanced three-phase wye-connected source with a phase voltage of 120 V rms supplies
power to a balanced wye-connected load. The per-phase
load impedance is 40 + j10 ⍀ . Determine the line currents in the circuit if / Van = 0° .
11.14 An abc-phase-sequence three-phase balanced
wye-connected source supplies a balanced deltaconnected load. The impedance per phase in the delta
load is 12 ⫹ j9 ⍀. The line voltage at the source is
Vab ⫽ 120 23 / 40° V rms. If the line impedance is
zero, find the line currents in the balanced wye–delta
system.
11.15 An abc-phase-sequence three-phase balanced
wye-connected 60-Hz source supplies a balanced deltaconnected load. The phase impedance in the load consists
of a 20 ⍀ resistor in series with a 50-mH inductor, and
the phase voltage at the source is Van = 120 / 20° V rms
V rms. If the line impedance is zero, find the line currents in the system.
11.16 An abc-phase-sequence three-phase balanced wyeconnected source supplies power to a balanced
wye-connected load. The impedance per phase in the
load is 14 + j12 ⍀. If the source voltage for the a phase
is Van = 120 / 80° V rms, and the line impedance is
zero, find the phase currents in the wye-connected
source.
11.17 An abc-sequence balanced three-phase wye-connected
source supplies power to a balanced wye-connected load.
The line impedance per phase is 1 + j5 ⍀, and the load
impedance per phase is 25 ⫹ j25 ⍀. If the source line
voltage Vab ⫽ 208 / 0° V rms find the line currents.
11.18 An abc-sequence balanced three-phase wye-connected
source supplies power to a balanced wye-connected
load. The line impedance per phase is 1 + j0 ⍀, and
the load impedance per phase is 20 + j20 ⍀. If the
source line voltage Vab is 100 / 0° V rms, find the line
currents.
11.19 An abc-sequence set of voltages feeds a balanced
three-phase wye–wye system. The line and load
impedances are 1 + j1 ⍀ and 10 + j10 ⍀,
respectively. If the load voltage on the ␣ phase is
VAN = 110 / 30° V rms, determine the line voltages of
the input.
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CHAPTER 11
P O LY P H A S E C I R C U I T S
11.20 In a balanced three-phase wye–wye system, the
source is an abc-sequence set of voltages. The load
voltage on the a phase is VAN = 108.58 / 79.81° V rms,
Zline = 1 + j1.4 ⍀, and Zload = 10 + j13 ⍀. Determine the
input sequence of voltages.
11.30 In a balanced three-phase wye–wye system, the load
impedance is 10 + j1 ⍀. The source has phase sequence
abc and the line voltage Vab = 220 / 30° V rms. If the
load voltage VAN = 120 / 0° V rms, determine the line
impedance.
11.21 A balanced abc-sequence of voltages feeds a balanced
three-phase wye–wye system. The line and load impedance are 0.6 + j0.9 ⍀ and 8 + j12 ⍀, respectively. The
load voltage on the a phase is VAN = 116.63 / 10° V rms.
Find the line voltage Vab .
11.31 In a balanced three-phase wye–wye system, the load
impedance is 20 + j12 ⍀. The source has an abc-phase
sequence and Van = 120 / 0° V rms. If the load voltage
is VAN = 111.49 /- 0.2° V rms, determine the
magnitude of the line current if the load is suddenly
short-circuited.
11.22 In a balanced three-phase wye–wye system, the source is
an abc-sequence set of voltages. The load voltage on the
a phase is VAN = 110 / 80° V rms, Z line = 1 + j1.4 ⍀,
and Z load = 10 + j13 ⍀. Determine the input sequence
of the line-to-neutral voltages.
11.32 In a balanced three-phase wye–wye system, the
source is an abc-sequence set of voltages and
Van = 120 / 40° V rms. If the a-phase line current and
line impedance are known to be 7.10 / -10.28° A rms
and 0.8 + j1 ⍀, respectively, find the load impedance.
11.23 In a balanced three-phase wye–wye system, the source
is an abc-sequence set of voltages. The load voltage
on the a phase is VAN = 120 / 60° V rms,
Z line = 2 + j1.4 ⍀, and Z load = 10 + j10 ⍀.
Determine the input voltages.
11.24 In a balanced three-phase wye–wye system, the source
is an abc-sequence set of voltages. Z line = 1 + j1 ⍀,
Z load = 14 + j12 ⍀, and the load voltage on the a
phase is VAN = 440 / 30° V rms. Find the line
voltage Vab.
11.25 A balanced abc-sequence of voltages feeds a balanced
three-phase wye–wye system. The line and load impedance are 0.6 + j 0. ⍀ and 8 + j12 ⍀ respectively. The load
voltage on the a phase is VAN = 116.63 / 10° V rms.
Find the line voltage Vab .
11.26 In a balanced three-phase wye–wye system, the source
is an abc-sequence set of voltages. Zline = 1 + j1.8 ⍀,
Zload = 14 + j12 ⍀, and the load voltage on the a phase
is VAN = 398.1 / 17.99° V rms. Find the line voltage Vab.
11.27 In a balanced three-phase wye–delta system the source has
an abc phase sequence and Van = 120 / 40° V rms. The
line and load impedances are 0.5 + j0.4 ⍀ and 24 + j18 ⍀,
respectively. Find the delta currents in the load.
11.28 In a balanced three-phase wye–wye system, the load
impedance is 8 + j4 ⍀. The source has phase
sequence abc and Van = 120 / 0° V rms. If the load
voltage is VAN = 111.62 / -1.33° V rms, determine the
line impedance.
11.29 In a balanced three-phase wye–wye system, the
total power loss in the lines is 400 W.
VAN = 105.28 / 31.56° V rms and the power factor
of the load is 0.77 lagging. If the line impedance is
2 + j1 ⍀, determine the load impedance.
11.33 In a balanced three-phase wye–wye system, the
source is an abc-sequence set of voltages and
Van = 120 / 50° V rms. The load voltage on the a phase
is 110.65 / 29.03° V rms and the load impedance is
16 + j20 ⍀. Find the line impedance.
11.34 An abc-sequence set of voltages feeds a balanced
three-phase wye–wye system. If the input voltage
Van = 440 / 10° V rms, the load voltage on the a phase
is VAN = 398.32 / 8.72° V rms, and Zload is 20 + j24 ⍀,
find the line impedance.
11.35 An abc-phase-sequence balanced three-phase source
feeds a balanced load. The system is wye–wye connected. The load impedance is 10 + j6 ⍀, the line impedance is 1 + j0.5 V, and / VAN = 60°. The total power
loss in the lines is 470.44 W. Find VAN and the magnitude of the source voltage.
11.36 An abc-sequence balanced three-phase source feeds a
balanced wye–wye system. Zline = 0.8 + j0.2 ⍀,
Zload = 12 + j6 ⍀, / VAN = 30°. The total power absorbed
by the load is 2 kW. Determine the total power loss in the
lines.
11.37 An abc-phase-sequence three-phase balanced
wye-connected 60-Hz source supplies a balanced
delta-connected load. The phase impedance in the
load consists of a 20-⍀ resistor series with a 20-mH
inductor, and the phase voltage at the source is
Van = 120 / 30° V rms. If the line impedance is zero,
find the line currents in the system.
11.38 In a three-phase balanced system, a delta-connected
source supplies power to a wye-connected load.
If the line impedance is 0.2 + j 0.4 ⍀, the load
impedance 3 + j2 ⍀, and the source phase voltage
Vab = 208 / 10° V rms, find the magnitude of the line
voltage at the load.
11.39 In a balanced three-phase wye–wye system, the
source is an abc-sequence set of voltages and
Van = 120 / 50° V rms. The load voltage on the a
phase is 110 / 50° V rms, and the load impedance is
16 ⫹ j20 ⍀. Find the line impedance.
PROBLEMS
573
11.40 In a balanced three-phase wye–wye system, the
source is an abc-sequence set of voltages and
Van ⫽ 120 / 40° V rms. If the a-phase line current
and line impedance are known to be 6 / 15° A rms
and 1 + j1 ⍀, respectively, find the load impedance.
11.50 A three-phase load impedance consists of a balanced
wye in parallel with a balanced delta. What is the
equivalent wye load and what is the equivalent delta
load if the phase impedances of the wye and delta are
6 + j3 ⍀ and 15 + j10 ⍀ , respectively?
11.41 An abc-phase-sequence three-phase balanced
wye-connected source supplies a balanced deltaconnected load. The impedance per phase in the delta
load is 12 + j6 ⍀. The line voltage at the source is
Vab ⫽ 120 23 / 40° V rms. If the line impedance is
zero, find the line currents in the balanced wye–delta
system.
11.51 In a balanced three-phase system, the abc-phasesequence source is wye connected and
Van = 120 / 20° V rms. The load consists of two balanced wyes with phase impedances of 8 + j2 ⍀ and
12 + j3 ⍀. If the line impedance is zero, find the line
currents and the phase current in each load.
11.42 In a balanced three-phase delta–wye system, the source
has an abc-phase sequence. The line and load impedances are 0.6 + j0.3 ⍀ and 12 + j7 ⍀, respectively.
If the line current I aA ⫽ 9.6 /- 20° A rms, determine
the phase voltages of the source.
11.43 In a three-phase balanced system, a delta-connected
source supplies power to a wye-connected load. If the
line impedance is 0.2 + j0.4 ⍀, the load impedance
6 + j4 ⍀, and the source phase voltage Vab = 210 / 40°
V rms, find the magnitude of the line voltage at the load.
11.44 An abc-sequence set of voltages feeds a balanced threephase wye–wye system. If Van = 440 / 30° V rms,
VAN = 413.28 / 29.78° V rms, and Z line = 2 + j1.5 ⍀,
find the load impedance.
11.45 An abc-phase-sequence three-phase balanced wyeconnected source supplies a balanced delta-connected
load. The impedance per phase of the delta load is
20 + j4 ⍀. If VAB = 115 / 35° V rms, find the line
current.
11.46 An abc-phase-sequence three-phase balanced wyeconnected source supplies power to a balanced deltaconnected load. The impedance per phase in the load is
14 + j7 ⍀. If the source voltage for the a phase is
Van = 120 / 80° V rms and the line impedance is zero,
find the phase currents in the wye-connected source.
11.47 In a three-phase balanced delta–delta system, the
source has an abc-phase sequence. The line and load
impedances are 0.3 + j0.2 ⍀ and 9 + j6 ⍀,
respectively. If the load current in the delta is
I AB = 15 / 40° A rms, find the phase voltages of
the source.
11.48 An abc-phase-sequence three-phase balanced wyeconnected source supplies a balanced delta-connected
load. The impedance per phase of the delta load is
10 + j8 ⍀. If the line impedance is zero and
the line current in the a phase is known to be
I aA = 28.10 / -28.66° A rms, find the load voltage VAB .
11.49 In a balanced three-phase wye–delta system, the source
has an abc-phase sequence and Van = 120 / 0° V rms.
If the line impedance is zero and the line current
I aA = 5 / 20° A rms , find the load impedance per phase
in the delta.
11.52 In a balanced three-phase delta–delta system, the source
has an abc-phase sequence. The phase angle for the
source voltage is / Vab = 40° and I ab = 4 / 15° A rms.
If the total power absorbed by the load is 1400 W, find
the load impedance.
11.53 In a balanced three-phase system, the source is a
balanced wye with an abc-phase sequence and
Vab = 215 / 50° V rms. The load is a balanced wye in
parallel with a balanced delta. The phase impedance of
the wye is 5 + j3 ⍀, and the phase impedance of the
delta is 18 + j12 ⍀. If the line impedance is
1 + j0.8 ⍀, find the line currents and the phase
currents in the loads.
11.54 In a balanced three-phase system, the source is a
balanced wye with an abc-phase sequence and
Vab = 208 / 60° V rms. The load consists of a balanced
wye with a phase impedance of 8 + j5 ⍀ in parallel
with a balanced delta with a phase impedance of
21 + j12 ⍀. If the line impedance is 1.2 + j1 ⍀, find
the phase currents in the balanced wye load.
11.55 In a balanced three-phase system, the source has an
abc-phase sequence and is connected in delta. There are
two loads connected in parallel. The line connecting the
source to the loads has an impedance of 0.2 + j0.1 ⍀.
Load 1 is connected in wye, and the phase impedance is
4 + j2 ⍀. Load 2 is connected in delta, and the phase
impedance is 12 + j9 ⍀. The current I AB in the delta
load is 16 / 45°A rms. Find the phase voltage of the
source.
11.56 In a balanced three-phase system, the source has an
abc-phase sequence and is connected in delta. There are
two parallel wye-connected loads. The phase impedance of load 1 and load 2 is 4 + j4 ⍀ and 10 + j4 ⍀,
respectively. The line impedance connecting the source
to the loads is 0.3 + j0.2 ⍀. If the current in the a
phase of load 1 is I AN1 = 10 / 20° A rms, find the delta
currents in the source.
11.57 An abc-phase-sequence balanced three-phase source
feeds a balanced load. The system is connected wyewye and / Van = 0°. The line impedance is
0.5 + j0.2 ⍀, the load impedance is 16 + j10 ⍀, and
the total power absorbed by the load is 2000 W.
Determine the magnitude of the source voltage Van.
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CHAPTER 11
P O LY P H A S E C I R C U I T S
11.58 A balanced three-phase delta-connected source
supplies power to a load consisting of a balanced
delta in parallel with a balanced wye. The phase
impedance of the delta is 24 + j12 ⍀, and the
phase impedance of the wye is 12 + j8 ⍀. The
abc-phase-sequence source voltages are
Vab = 440 / 60° V rms, Vbc = 440 /- 60° V rms, and
Vca = 440 /- 180° V rms, and the line impedance per
phase is 1 + j0.08 ⍀. Find the line currents and the
power absorbed by the wye-connected load.
11.59 The magnitude of the complex power (apparent power)
supplied by a three-phase balanced wye–wye system is
3600 VA. The line voltage is 208 V rms. If the line
impedance is negligible and the power factor angle of
the load is 25°, determine the load impedance.
11.60 An abc-sequence wye-connected source having a phase-a
voltage of 120 / 0° V rms is attached to a wye-connected
load having a per-phase impedance of 100 / 70° ⍀. If the
line impedance is 1 / 20° Æ , determine the total complex
power produced by the voltage sources and the real and
reactive power dissipated by the load.
11.61 A three-phase balanced wye–wye system has a line
voltage of 208 V rms. The line current is 6 A rms and
the total real power absorbed by the load is 1800 W.
Determine the load impedance per-phase, if the line
impedance is negligible.
11.62 A three-phase abc-sequence wye-connected source supplies 14 kVA with a power factor of 0.75 lagging to a
delta load. If the delta load consumes 12 kVA at a
power factor of 0.7 lagging and has a phase current of
10 /- 30° A rms, determine the per-phase impedance of
the load and the line.
11.63 A balanced three-phase source serves the following
loads:
Load 1: 60 kVA at 0.8 pf lagging
Load 2: 30 kVA at 0.75 pf lagging
The line voltage at the load is 208 V rms at 60 Hz.
Determine the line current and the combined power
factor at the load.
11.64 A balanced three-phase source serves two loads:
Load 1: 36 kVA at 0.8 pf lagging
Load 2: 18 kVA at 0.6 pf lagging
The line voltage at the load is 208 V rms at 60 Hz.
Find the line current and the combined power factor at
the load.
11.65 The following loads are served by a balanced
three-phase source:
Load 1: 18 kVA at 0.8 pf lagging
Load 2: 8 kVA at 0.8 pf leading
Load 3: 12 kVA at 0.75 pf lagging
The load voltage is 208 V rms at 60 Hz. If the line
impedance is negligible, find the power factor at the
source.
11.66 A balanced three-phase source serves the following
loads:
Load 1: 20 kVA at 0.8 pf lagging
Load 2: 10 kVA at 0.7 pf leading
Load 3: 10 kW at unity pf
Load 4: 16 kVA at 0.6 pf lagging
The line voltage at the load is 208 V rms at 60 Hz, and
the line impedance is 0.02 + j0.04 ⍀. Find the line
voltage and power factor at the source.
11.67 A small shopping center contains three stores that
represent three balanced three-phase loads. The power
lines to the shopping center represent a three-phase source
with a line voltage of 13.8 kV rms. The three loads are
Load 1: 400 kVA at 0.9 pf lagging
Load 2: 200 kVA at 0.85 pf lagging
Load 3: 100 kVA at 0.90 pf lagging
Find the power line current.
11.68 The following loads are served by a balanced
three-phase source:
Load 1: 20 kVA at 0.8 pf lagging
Load 2: 4 kVA at 0.8 pf leading
Load 3: 10 kVA at 0.75 pf lagging
The load voltage is 208 V rms at 60 Hz. If the line impedance is negligible, find the power factor at the source.
11.69 A balanced three-phase source supplies power to three
loads:
Load 1: 30 kVA at 0.8 pf lagging
Load 2: 24 kW at 0.6 pf leading
Load 3: unknown
If the line voltage and total complex power at the load
are 208 V rms and 60 / 0° kVA, respectively, find the
unknown load.
11.70 A balanced three-phase source supplies power to three
loads. The loads are
Load 1: 24 kVA at 0.6 pf lagging
Load 2: 10 kW at 0.75 pf lagging
Load 3: unknown
If the line voltage at the load is 208 V rms, the magnitude of the total complex power is 35.52 kVA, and the
combined power factor at the load is 0.88 lagging, find
the unknown load.
11.71 A balanced three-phase source supplies power to three
loads:
Load 1: 18 kVA at 0.8 pf lagging
Load 2: 10 kW at 0.6 pf leading
Load 3: unknown
If the line voltage at the loads is 208 V rms, the line
current at the source is 116.39 A rms, and the combined
power factor at the load is 0.86 lagging, find the
unknown load.
PROBLEMS
11.72 A balanced three-phase source supplies power to three
loads:
Load 1: 30 kVA at 0.8 pf lagging
Load 2: 24 kW at 0.6 pf leading
Load 3: unknown
The line voltage at the load and line current at the
source are 208 V rms and 166.8 A rms, respectively. If
the combined power factor at the load is unity, find the
unknown load.
575
11.77 Find C in the network in Fig. P11.77 such that the total
load has a power factor of 0.9 lagging.
+
C
4.6 kV rms
Balanced
three-phase
source
60 Hz
-
Balanced
three-phase
load
6 MVA
0.8 pf
lagging
C
C
11.73 A balanced three-phase source supplies power to three
loads:
Load 1: 24 kW at 0.8 pf lagging
Load 2: 10 kVA at 0.7 pf leading
Figure P11.77
Load 3: unknown
If the line voltage at the load is 208 V rms, the magnitude of the total complex power is 41.93 kVA, and the
combined power factor at the load is 0.86 lagging, find
the unknown load.
11.74 A three-phase abc-sequence wye-connected source
with Van = 220 / 0° V rms supplies power to a
wye-connected load that consumes 50 kW of power in
each phase at a pf of 0.8 lagging. Three capacitors are
found that each have an impedance of -j2.0 ⍀ , and
they are connected in parallel with the load in a wye
configuration. Determine the power factor of the combined load as seen by the source.
11.75 If the three capacitors in the network in Problem 11.74
are connected in a delta configuration, determine the
power factor of the combined load as seen by the
source.
11.78 Find the value of C in Fig. P11.78 such that the total
load has a power factor of 0.87 lagging.
+
34.5 kV rms
Balanced
three-phase
source
60 Hz
-
C
C
11.79 Find C in the network in Fig. P11.79 so that the total
load has a power factor of 0.9 leading.
+
C
4.6 kV rms
34.5 kV rms
Balanced
three-phase
source
60 Hz
Figure P11.76
-
C
C
C
Balanced
three-phase
load
20 MVA
0.707 pf
lagging
Balanced
three-phase
load
20 MVA
0.707 pf
lagging
Figure P11.78
11.76 Find C in the network in Fig. P11.76 such that the total
load has a power factor of 0.87 leading.
+
C
Balanced
three-phase
source
60 Hz
-
Balanced
three-phase
load
6 MVA
0.8 pf
lagging
C
C
Figure P11.79
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CHAPTER 11
P O LY P H A S E C I R C U I T S
11.80 A standard practice for utility companies is to divide customers into single-phase users and three-phase users. The utility
must provide three-phase users, typically industries, with all three phases. However, single-phase users, residential and
light commercial, are connected to only one phase. To reduce cable costs, all single-phase users in a neighborhood are
connected together. This means that even if the three-phase users present perfectly balanced loads to the power grid, the
single-phase loads will never be in balance, resulting in current flow in the neutral connection. Consider the 60-Hz,
abc-sequence network in Fig. P11.80. With a line voltage of 416 / 30° V rms, phase a supplies the single-phase users
on A Street, phase b supplies B Street, and phase c supplies C Street. Furthermore, the three-phase industrial load, which
is connected in delta, is balanced. Find the neutral current.
A
a
±
–
240 0°
V rms
Three-phase
36 kW
pf=0.5
lagging
B
b
240 –120°
V rms
±
–
C
c
IAN
±
–
240 120°
V rms
n
A Street
48 kW
pf=1
InN
ICN
IBN
B Street
30 kW
pf=1
C Street
60 kW
pf=1
N
Figure P11.80
•
TYPICAL PROBLEMS FOUND ON THE FE EXAM
11PFE-1 A wye-connected load consists of a series RL
impedance. Measurements indicate that the rms
voltage across each element is 84.85 V. If the rms
line current is 6A, find the total complex power for
the three-phase load configuration.
a. 1.25 / -45° kVA
b. 4.32 / 30° kVA
c. 3.74 / 60° kVA
d. 2.16 / 45° kVA
11PFE-2 A balanced three-phase delta-connected load consists
of an impedance of 12 + j12 ⍀. If the line voltage at
the load is measured to be 230 V rms, find the total
real power absorbed by the three-phase configuration.
a. 6.62 kW
b. 2.42 kW
c. 3.36 kW
d. 5.82 kW
11PFE-3 Two balanced three-phase loads are connected in parallel. One load with a phase impedance of 24 + j18 ⍀
is connected in delta, and the other load has a phase
impedance of 6 + j4 ⍀ and is connected in wye. If
the line-to-line voltage is 208 V rms, determine the line
current.
a. 15.84 / -60.25°A rms
b. 28.63 / -35.02°A rms
c. 40.49 / 30.27°A rms
d. 35.32 / 90.53°A rms
11PFE-4 The total complex power at the load of a three-phase
balanced system is 24 / 30° kVA. Find the real power
per phase.
a. 3.24 kW
b. 4.01 kW
c. 6.93 kW
d. 8.25 kW
11PFE-5 A balanced three-phase load operates at 90 kW with
a line voltage at the load of 480 / 0° V rms at 60 Hz.
The apparent power of the three-phase load is
100 kVA. It is known that the load has a lagging
power factor. What is the total three-phase reactive
power of the load?
a. 22.43 kvar
b. 30.51 kvar
c. 25.35 kvar
d. 43.59 kvar