Electric Circuits, 9th Edition P45 doc

416 Balanced Three-Phase Circuits
Applying this general observation,
we can see
that
for a
three-
conductor circuit, whether balanced
or not, we
need only
two
wattmeters
to measure
the
total power.
For a
four-conductor circuit,
we
need three
wattmeters
if the
three-phase circuit
is
unbalanced,
but
only
two
wattmeters
if it is
balanced, because
in the
latter case there

is no
current
in
the neutral line. Thus, only
two
wattmeters
are
needed
to
measure
the
total average power
in any
balanced three-phase system.
The two-wattmeter method reduces
to
determining
the
magnitude
and algebraic sign
of the
average power indicated
by
each wattmeter.
We
can describe
the
basic problem
in
terms

of the
circuit shown
in
Fig. 11.20,
where
the two
wattmeters
are
indicated
by the
shaded boxes
and
labeled
W\ and W
2
. The
coil notations
cc and pc
stand
for
current coil
and
poten-
tial coil, respectively.
We
have elected
to
insert
the
current coils

of the
wattmeters
in
lines
aA and
cC. Thus, line
bB is the
reference line
for the
two potential
coils.
The load
is
connected
as a
wye,
and the
per-phase load
impedance
is
designated
as Z^ = \Z\ /jh
This
is a
general representation,
as
any
A-connected load
can be
represented

by its Y
equivalent; further-
more,
for the
balanced case,
the
impedance angle
0 is
unaffected
by the
A-to-Y transformation.
We
now
develop general equations
for the
readings
of the two
wattmeters. We assume that
the
current drawn
by the
potential coil
of the
wattmeter
is
negligible compared with
the
line current measured
by the
cur-

rent coil. We further assume that
the
loads
can be
modeled
by
passive circuit
elements
so
that
the
phase angle
of
the load impedance
(0 in
Fig. 11.20) lies
between
-90°
(pure capacitance)
and +90"
(pure inductance). Finally,
we
assume
a
positive phase sequence.
From
our
introductory discussion
of the
average deflection

of the
wattmeter,
we can see
that wattmeter
1
will respond
to the
product
of
|V
A
JJ|,
|IaA^
an
d the
cosine
of
the angle between
V
A
B
an
d
IaA-
M
we
denote
this wattmeter reading
as W
h

we can
write
Wi -IVABIPUCOS*,
= VIA cos
0
X
.
(11.54)
It follows that
W2 =
|VCBI|ICC|COS02
= VJ
L
COS0
2
. (11.55)
In
Eq.
11.54,0]
is the
phase angle between
V
AB
and I
aA
, and in Eq.
11.55,
0
2
is the

phase angle between
V
CB
and I
c(
>
To calculate
Wj and W
2
, we
express
0]
and 0
2
in
terms
of the
imped-
ance angle
0,
which
is
also
the
same
as the
phase angle between
the
phase
voltage

and
current.
For a
positive phase sequence,
0
}
= 0 + 30° = 0^ + 30°,
(11.56)
0
2
= 0 - 30° = 0$ - 30°.
(11.57)
The derivation
of Eqs. 11.56 and 11.57 is
left
as an
exercise
(see
Problem 11.34). When
we
substitute
Eqs. 11.56 and 11.57
into
Eqs. 11.54
and
11.55,
respectively,
we get
Wi
=

K
L
/
L
cos(^
+ 30°),
(11.58)
W
2
=
K
L
/
L
COS(0A
- 30°).
(11.59)
h
t
+
v
i
h
+
v
2
/3
+
v
3

m
1
•
1
•
Z
*
General
network
Figure 11.19
•
A
general circuit whose power is
supplied by
n
conductors.
IaA
cc
A
Wi
pc4 Z^ = |Z|Z0 V
AN
b«-
'CN
c+
-•—
-¾
c(
N
Figure 11.20

•
A
circuit used
to
analyze
the
two-wattmeter method
of
measuring average power
delivered to a balanced load.
11.6 Measuring Average Power in Three-Phase Circuits 417
(11.60)
To find the total power, we add W
]
and W
2
; thus
p
T
= W
]
+ W
2
= 2VL/LCOS^COS30"
= V3VL/
L
COS^,
which is the expression for the total power in a three-phase circuit.
Therefore we have confirmed that the sum of the two wattmeter readings
yields the total average power.

A closer look at Eqs. 11.58 and 11.59 reveals the following about the
readings of the two wattmeters:
1.
If the power factor is greater than 0.5, both wattmeters read positive.
2.
If the power factor equals 0.5, one wattmeter reads zero.
3.
If the power factor is less than 0.5, one wattmeter reads negative.
4.
Reversing the phase sequence will interchange the readings on the
two wattmeters.
These observations are illustrated in the following example and in
Problems 11.41-11.51.
| Computing Wattmeter Readi
Calculate the reading of each wattmeter in
circuit in Fig. 11.20 if the phase voltage at
load is 120 V and (a) Z+ = 8• +;
(b) Z
0
= 8 - ;6 ft; (c) Z^ = 5 + ;5 V3 12;
(d) Z
s
= 10 /-75° ft. (e) Verify for (a)-(d)
the sum of the wattmeter readings equals the
power delivered to the load.
Solution
a) Z
(b
= 10 /36.87° ft, V
L

= 120 V3V, and
/
L
= 120/10 = 12 A.
W
]
= (120V3~)(12) cos (36.87° + 30°)
= 979.75 W,
W
2
= (120 V3)(12) cos (36.87° - 30°)
= 2476.25 W.
b) Z
&
= 10 /-36.87° ft, V
L
= 120V^ V, and
/
L
= 120/10 = 12 A.
Wi = (120V3)(12) cos (-36.87° + 30°)
= 2476.25 W,
W
2
= (120V3)(12) cos (-36.87° - 30°)
= 979.75 W.
ngs in
the
the
6 ft;

and
that
total
Three-Phase Circuits
c) Z^ = 5(1 + jV3) = 10 /60° O, V
L
= 120V3 V,
and/
L
= 12 A.
Wi = (120V5)(12) cos (60° + 30°) = 0,
W
2
= (120 V3)(12) cos (60° - 30°)
= 2160 W.
d) Z
4>
= 10 /-75° ft, V
L
=
12QV3"
V, and
/
L
= 12 A.
Wi = (120V3)(12)cos(-75° + 30°) = 1763.63 W,
W
2
= (120V5)(12) cos (-75° - 30°) = -645.53 W.
e) P

r
(a) = 3(12)
2
(8) - 3456 W,
W
}
+W
2
= 979.75 + 2476.25
= 3456 W,
P
r
(b) = P
r
(a) = 3456 W,
Wi + W
2
= 2476.25 + 979.75
= 3456 W,
/V(c) = 3(12)
2
(5) = 2160 W,
W
t
+ W
2
= 0 + 2160
= 2160W,
P
r

(d) = 3(12)
2
(2.5882) = 1118.10 W,
Wi + W
2
= 1763.63 - 645.53
= 1118.10W.
NOTE: Assess your understanding of the two-wattmeter method by trying Chapter Problems 11.43 and 11.44.
418 Balanced Three-Phase Circuits
3-0 line
-)
Generator
Plant
4
*)
/
Sub
Station
Figure 11.21 A A substation connected to a power
plant via a three-phase line.
0.6 O
-AW-
/4.8 n
^YYYV
A
v.„
13,800
/r\0
y
^3 *~

1.2
MW
1.2
MVAR
n N
Figure 11.22 • A single phase equivalent circuit for
the system in Fig.
11.21.
Practical Perspective
Transmission and Distribution of Electric Power
At the start of this chapter we pointed out the obligation utilities have to
maintain the rms voltage level at their customer's premises. Although the
acceptable deviation from a nominal level may vary among different utilities
we will assume for purposes of discussion that an allowable tolerance is
± 5.8%. Thus a nominal rms voltage of 120 V could range from 113 to
127 V. We also pointed out that capacitors strategically located on the sys-
tem could be used to support voltage levels.
The circuit shown in Fig. 11.21 represents a substation on a Midwestern
municipal system. We will assume the system is balanced, the line-to-line
voltage at the substation is 13.8 kV, the phase impedance of the distribu-
tion line is 0.6 +
/4.811,
and the load at the substation at 3 PM on a hot,
humid day in July is 3.6 MW and 3.6 magnetizing MVAR.
Using the line-to-neutral voltage at the substation as a reference, the
single-phase equivalent circuit for the system in Fig. 11.21 is shown in
Fig.
11.22. The line current can be calculated from the expression for the
complex power at the substation. Thus,
13,800

V3
I*
A
= (1.2 + /1.2)10*
It follows that
or
I*
A
= 150.61 + /150.61 A
I
aA
= 150.61 - /150.61 A.
The line-to-neutral voltage at the generating plant is
13,800
V3
'0° + (0.6 + /4.8)(150.61 - /150.61)
= 8780.74 + /632.58
= 8803.50/4.12° V.
Therefore the magnitude of the line voltage at the generating plant is
|V
ab
| = V3(8803.50) = 15,248.11V.
We are assuming the utility is required to keep the voltage level within
± 5.8% of the nominal value. This means the magnitude of the line-to-line
voltage at the power plant should not exceed 14.6 kV nor be less than
13 kV. Therefore, the magnitude of the line voltage at the generating plant
could cause problems for customers.
When the magnetizing vars are supplied by a capacitor bank connected
to the substation bus, the line current I
aA

becomes
I
aA
= 150.61 + /0 A.
Summary 419
Therefore the voltage at the generating plant necessary to maintain a line-
to-line voltage of 13,800 V at the substation is
13,800
Van = -^- /O: + (0-6 + /4.8)(150.61 + /0)
= 8057.80 + /722.94
= 8090.17/5.13° V.
Hence
|V
ab
|
= V3(8090.17) = 14,012.58 V.
This voltage level falls within the allowable range of 13
kV
to 14.6 kV.
NOTE:
Assess
your
understanding
of this
Practical Perspective
by trying
Chapter
Problems
11.52(a)-(b) and 11.53, 11.56, and 11.57.
Summary

• When analyzing balanced three-phase circuits, the first
step is to transform any A connections into Y connections,
so that the overall circuit is of the Y-Y configuration. (See
page 402.)
• A single-phase equivalent circuit is used to calculate the
line current and the phase voltage in one phase of the
Y-Y structure. The a-phase is normally chosen for this
purpose. (See page 404.)
• Once we know the line current and phase voltage in the
a-phase equivalent circuit, we can take analytical short-
cuts to find any current or voltage in a balanced three-
phase circuit, based on the following facts:
• The b- and c-phase currents and voltages are identi-
cal to the a-phase current and voltage except for a
120° shift in phase. In a positive-sequence circuit, the
b-phase quantity lags the a-phase quantity by 120°,
and the c-phase quantity leads the a-phase quantity
by 120°. For a negative sequence circuit, phases b and
c are interchanged with respect to phase a.
• The set of line voltages is out of phase with the set of
phase voltages by ±30°. The plus or minus sign corre-
sponds to positive and negative sequence, respectively.
• In a Y-Y circuit the magnitude of a line voltage is
V3 times the magnitude of a phase voltage.
• The set of line currents is out of phase with the set of
phase currents in A-connected sources and loads by
T30°.
The minus or plus sign corresponds to positive
and negative sequence, respectively.
• The magnitude of a line current is V3 times the mag-

nitude of a phase current in a A-connected source
or load,
(See pages 404-405 and 407-408.)
• The techniques for calculating per-phase average
power, reactive power, and complex power are identical
to those introduced in Chapter 10. (See page 410.)
• The total real, reactive, and complex power can be deter-
mined either by multiplying the corresponding per phase
quantity by 3 or by using the expressions based on line
current and line voltage, as given by Eqs. 11.36,11.38, and
11.41.
(See pages 410 and 411.)
• The total instantaneous power in a balanced three-phase
circuit is constant and equals 1.5 times the average
power per phase. (See page 412.)
• A wattmeter measures the average power delivered to a
load by using a current coil connected in series with
the load and a potential coil connected in parallel
with the load. (See page 415.)
• The total average power in a balanced three-phase cir-
cuit can be measured by summing the readings of two
wattmeters connected in two different phases of the
circuit. (See page 415.)
420 Balanced Three-Phase Circuits
Problems
AH phasor voltages in the following Problems are stated in
terms of the rms value.
Section 11.1
11.1 Verify that Eq. 11.3 is true for either Eq. 11.1 or
Eq.11.2.

11.2 What is the phase sequence of each of the following
sets of voltages?
a) y
a
= 208 cos (OJ? + 27°) V,
^ = 208 cos
(cot
+ 147°) V,
v
c
= 208 cos
(cot
- 93°) V.
b) v.
a
= 4160 cos
(cot
- 18°) V,
v
b
= 4160 cos
(cot
- 138°) V,
?;
c
= 4160 cos
(cot
+ 102°) V.
11.3 For each set of voltages, state whether or not the volt-
PSPICE

a
ges form a balanced three-phase set. If the set is bal-
anced, state whether the phase sequence is positive or
negative. If the set is not balanced, explain why.
a) v
a
= 139 cos 377? V,
v
b
=
139 cos (377?
+
120°)
V,
v
c
=
139 cos (377?
-
120°)
V.
b)
v.
d
=
381 cos 377?
V,
v
b
= 381 cos (377? + 240°) V,

v
c
= 381 cos (377? + 120°) V.
c) v
a
= 2771 sin (377? - 30°) V,
v
b
= 2771 cos 377? V,
v
c
= 2771 sin (377? + 210°) V.
d) v
A
= 170 sin
(cot
+ 30°) V,
v
b
= -170 cos
cot
V,
v
c
= 170 cos
(cot
+ 60°) V.
e) v
a
= 339 cos

cot
V,
% = 339 cos
(cot
+ 120) V,
v
c
= 393 cos
(cot
- 120°) V.
f) v.
A
= 3983 sin
(cot
+ 50°) V,
v
b
= 3983 cos
(cot
- 160°) V,
v
c
= 3983 cos
(cot
+ 70°) V.
Section 11.2
11.4 Refer to the circuit in Fig.
11.5(b).
Assume that there
are no external connections to the terminals a,b,c.

Assume further that the three windings are from a
balanced three-phase generator. How much current
will circulate in the A-connected generator?
Section 11.3
11.5 A balanced three-phase circuit has the following
characteristics:
• Y-Y connected;
• The line voltage at the source, V
ab
, is
240 V3~/90° V;
• The phase sequence is negative;
• The line impedance is 4 -I- /5
£l/<f>',
• The load impedance is 76 + /55
ft/cf).
a) Draw the single phase equivalent circuit for the
a-phase.
b) Calculated the line current in the a-phase.
c) Calculated the line voltage at the load in the
a-phase.
11.6 Find the rms value of I„ in the unbalanced three-
phase circuit seen in Fig. PI 1.6.
Figure PI 1.6
0.1 n /0.8ft a 0.4 ft /3.2 ft A 59.5 ft /76 ft
->WV
»****"¥">
0 /yy*^ (VYY> 0 'WW
0.1ft /0.8 ft b 0.4 ft /3.2 ft B 39.5 ft /26 ft
-• "VS/V rrvY>

9
VW—
M^ 240/-
240/-120°
V
0.1ft /0.8 ft
c 0.4 ft /3.2 ft C 19.5 ft /lift
-• /vW «^w> » VW rvw>_
N
Problems
421
11.7 The time-domain expressions
for
three line-to-neutral
voltages
at the
terminals
of
a Y-connected load
are
v
AN
=
7620 cos
(W +
30°)
V,
v
m
=

7620 cos
(a)t
+
150°)
V,
'»
CN
=
7620 cos
(cot
-
90")
V.
What
are the
time-domain expressions
for the
three
line-to-line voltages
v
Aii
,
I>BO
and
VQA?
11.8
a) Is the
circuit
in Fig.
PI 1.8

a
balanced
or
unbal-
PSPICE
anced three-phase system? Explain.
MULTISIM
b) Find
I„.
Figure PI
1.8
^2()0 /0!V
11.9
The
magnitude
of the
line voltage
at the
terminals
of
a
balanced Y-connected load
is
6600
V.
The load
impedance
is
240
- /70

0/<£. The load
is fed
from
a
line that
has an
impedance
of
0.5
+ /4 0/(£.
a) What
is the
magnitude
of the
line current?
b) What
is the
magnitude
of the
line voltage
at
the source?
11.10
a)
Find
I„ in the
circuit
in
Fig.
PI

1.10.
PSPICE
b) FindV
AN
.
c) FindV
AB
.
d)
Is the
circuit
a
balanced
or
unbalanced three-
phase system?
11.11
The
magnitude
of the
phase voltage
of an
ideal
balanced three-phase Y-connected source
is
125
V. The
source
is
connected

to a
balanced
Y-connected load
by a
distribution line that
has an
impedance
of
0.1
+ /0.8
Cl/<j).
The
load impedance
is
19.9 +
/14.2
Cl/4>.
The
phase sequence
of the
source
is acb. Use the
a-phase voltage
of the
source
as the
reference. Specify
the
magnitude
and

phase angle
of the
following quantities:
(a) the
three line currents,
(b) the
three line voltages
at
the source, (c)
the
three phase voltages
at the
load,
and
(d) the
three line voltages
at the
load.
Section
11.4
11.12
A
balanced, three-phase circuit
is
characterized
as
follows:
•
Y-A
connected;

• Source voltage
in the
c-phase is 20/-90°
V;
• Source phase sequence
is abc;
• Line impedance
is
1
+ /3
CL/<f>;
• Load impedance
is 117
—
/99
12/(/).
a) Draw
the
single phase equivalent
for the
a-phase.
b) Calculate
the
a-phase line current.
c) Calculate
the
a-phase line voltage
for the
three-
phase load.

11.13
An acb
sequence balanced three-phase Y-connected
source supplies power
to a
balanced, three-phase
A-
connected load with
an
impedance
of 12 + /9
fl/4>.
The source voltage
in the
b-phase
is
240/-50°
V.
The line impedance
is 1 + /1
fl/4>.
Draw
the
single
phase equivalent circuit
for the
a-phase
and use it
to find
the

current
in the
a-phase
of
the load.
11.14 A balanced A-connected load
has an
impedance
of
864
—
/252
Q,/(f).
The load
is fed
through
a
line hav-
ing
an
impedance
of
0.5
4- /4
Q/<f).
The
phase volt-
age
at the
terminals

of the
load
is 69
kV.
The phase
sequence
is
positive.
Use V
AB
as the
reference.
a) Calculate
the
three phase currents
of the
load.
b) Calculate
the
three line currents.
c) Calculate
the
three line voltages
at the
sending
end
of the
line.
Figure P11.10
0.4

ft /212 a
1.6
n
-vw-
/4
ft
7811
-AlVir
/54
a
O
Till (FV
/1.6
b
2.6
0 /2.4 ft B 77 fl
/56
O
—•
WV
ooor>
1 w^
0.2
ft /1.2 ft c 0.8 ft /3.8 ft C 79 ft /55 ft
—•
VA/
orvv>
0 VvV
rwYV
N

422 Balanced Three-Phase Circuits
11.15 A balanced Y-connected load having an imped-
ance of 72 + /21
Cl/(f)
is connected in parallel with
a balanced A-connected load having an imped-
ance of 150/0°
Q,/<f>.
The paralleled loads are fed
from a line having an impedance of /1 0/4>. The
magnitude of the line-to-neutral voltage of the Y-
load is 7650 V.
a) Calculate the magnitude of the current in the
line feeding the loads.
b) Calculate the magnitude of the phase current in
the A-connected load.
c) Calculate the magnitude of the phase current in
the Y-connected load.
d) Calculate the magnitude of the line voltage at
the sending end of the line.
11.16 In a balanced three-phase system, the source is a bal-
anced Y with an abc phase sequence and a line voltage
V
ab
= 208/50° V. The load is a balanced Y in paral-
lel with a balanced A.The phase impedance of the Y is
4 + /3 il/cf) and the phase impedance of the A is
3 - /9 0./4). The line impedance is 1.4 -I- /0.8
Oj4>.
Draw the single phase equivalent circuit and use it to

calculate the line voltage at the load in the a-phase.
11.17 The impedance Z in the balanced three-phase cir-
cuit in Fig. PI 1.17 is 100 - /75 O. Find
a) IAB> IBO and I
CA
,
b)
IaA*
\^ and I
cC
,
c) I
ba
, I
cb
, and I
ac
.
Figure P11.17
13.2/-120° kV
11.18 For the circuit shown in Fig. PI 1.18, find
PSPKE
a) the phase currents I
AB
, I
RC
, and I
CA
MULTISIM
/ r

^°
OK
-
Kl
^
b) the line currents I
aA
, I
bB
, and F
cC
when Z, = 2.4 - /0.7 ft, Z
2
= 8 + /6 Q, and
Z
3
= 20 + /0 ft.
Figure P11.18
a A
480/-120° V
11.19 A three-phase A-connected generator has an inter-
nal impedance of 9 -(- /90 mft/<jf>. When the load is
removed from the generator, the magnitude of the
terminal voltage is 13,800
V.
The generator feeds a
A-connected load through a transmission line with
an impedance of 20 + /180 mft/$. The per-phase
impedance of the load is 7.056 + /3.417 ft.
a) Construct a single-phase equivalent circuit.

b) Calculate the magnitude of the line current.
c) Calculate the magnitude of the line voltage at
the terminals of the load.
d) Calculate the magnitude of the line voltage at
the terminals of the source.
e) Calculate the magnitude of the phase current in
the load.
f) Calculate the magnitude of the phase current in
the source.
11.20 A balanced three-phase A-connected source is
shown in Fig. PI 1.20.
a) Find the Y-connected equivalent circuit.
b) Show that the Y-connected equivalent circuit
delivers the same open-circuit voltage as the
original A-connected source.
c) Apply an external short circuit to the terminals
A, B, and C. Use the A-connected source to find
the three line currents I
aA
, I
bB
, and I
cC
.
d) Repeat (c) but use the Y-equivalent source to
find the three line currents.
Figure PI 1.20
• A
2.7 O
/13.511

4156/-120°
V
• C
11.21 The A-connected source of Problem 11.20 is con-
nected to a Y-connected load by means of a bal-
anced three-phase distribution line. The load
Problems 423
impedance is 1910-/636
fl/<f).
and the line imped-
ance is 9.1 + /71.5(2/0.
a) Construct a single-phase equivalent circuit of
the system.
b) Determine the magnitude of the line voltage at
the terminals of the load.
c) Determine the magnitude of the phase current
in the A-source.
d) Determine the magnitude of the line voltage at
the terminals of the source.
11.27 The three pieces of computer equipment described
below are installed as part of a computation center.
Each piece of equipment is a balanced three-phase
load rated at 208 V. Calculate (a) the magnitude of
the line current supplying these three devices and
(b) the power factor of the combined load.
• Hard Drive: 4.864 kW at 0.79 pf lag
• CD/DVD drive: 17.636 kVA at 0.96 pf lag
. CPU: line current 73.8 A, 13.853 kVAR
11.28 Calculate the complex power in each phase of the
unbalanced load in Problem 11.18.

Section 11.5
11.22 In a balanced three-phase system, the source
has an abc sequence, is Y-connected, and
V
an
= 120/20° V. The source feeds two loads, both
of which are Y-connected. The impedance of load 1
is 8 + /6
il/cf).
The complex power for the a-phase
of load 2 is 600/36° VA. Find the total complex
power supplied by the source.
11.23 A balanced three-phase source is supplying 60 kVA
at 0.6 lagging to two balanced Y-connected parallel
loads.
The distribution line connecting the source to
the load has negligible impedance. Load 1 is purely
resistive and absorbs 30 kW. Find the per-phase
impedance of Load 2 if the line voltage is 120 V5 V
and the impedance components are in series.
11.24 A three-phase positive sequence Y-connected
source supplies 14 kVA with a power factor of 0.75
lagging to a parallel combination of a Y-connected
load and a A-connected load. The Y-connected load
uses 9 kVA at a power factor of 0.6 lagging and has
an a-phase current of 10/-30° A.
a) Find the complex power per phase of the
A-connected load.
b) Find the magnitude of the line voltage.
11.29 A balanced three-phase distribution line has an

impedance of
1
+ /8 fl/</>. This line is used to sup-
ply three balanced three-phase loads that are con-
nected in parallel. The three loads are
L! = 120 kVA at 0.96 pf lead, L
2
= 180 kVA at
0.80 pf lag, and L
3
= 100.8 kW and 15.6 kVAR
(magnetizing). The magnitude of the line voltage at
the terminals of the loads is 2400 V3 V.
a) What is the magnitude of the line voltage at the
sending end of the line?
b) What is the percent efficiency of the distribution
line with respect to average power?
11.30 The line-to-neutral voltage at the terminals of the
balanced three-phase load in the circuit shown in
Fig. PI 1.30 is 1200 V. At this voltage, the load is
absorbing 500 kVA at 0.96 pf lag.
a) Use V
AN
as the reference and express I
na
in
polar form.
b) Calculate the complex power associated with
the ideal three-phase source.
c) Check that the total average power delivered

equals the total average power absorbed.
d) Check that the total magnetizing reactive power
delivered equals the total magnetizing reactive
power absorbed.
11.25 The total apparent power supplied in a balanced,
three-phase Y-A system is 4800 VA. The line volt-
age is 240 V. If the line impedance is negligible and
the power factor angle of the load is -50°, deter-
mine the impedance of the load.
11.26 Show that the total instantaneous power in a bal-
anced three-phase circuit is constant and equal to
1,5V
m
I
m
cos
0$,
where V
m
and I
m
represent the
maximum amplitudes of the phase voltage and
phase current, respectively.
Figure PI 1.30
0.18 fi
J
iA4n
# Vs/V ^VY-r>_
:-/180 a

0.18 0 /1.44«
—/yyV e-v-Y-v>_
0.18 a /1-440
* WV
i^r>nr\_
A
500 kVA
0.% pf
lag
-•C
424 Balanced Three-Phase Circuits
PSPICE
MULTISIM
11.31 a) Find the rms magnitude and the phase angle of
I
CA
in the circuit shown in Fig. PI 1.31.
b) What percent of the average power delivered by
the three-phase source is dissipated in the three-
phase load?
Figure PI 1.31
/2 ft
a 1.5 ft
-•—'vw-
f^J
1365/0!
6
565/0° V
/2 0
b

i.
5
r>
1365/-120° V
1365/120° V
/2 ft
:85.5 0
/114 a
B
85.5
XI
!yii4n
:85.5 ft
;/ii4 n
c 1.5 ft
11.32 At full load, a commercially available 100 hp, three-
phase induction motor operates at an efficiency of
97%
and a power factor of 0.88
lag.
The motor is sup-
plied from a three-phase outlet with a line-voltage
rating of 208
V.
a) What is the magnitude of the line current drawn
from the 208 V outlet? (1 hp = 746 W.)
b) Calculate the reactive power supplied to
the motor.
11.33 Three balanced three-phase loads are connected in
parallel. Load 1 is Y-connected with an impedance

of 400 + /300 ft/0; load 2 is A-connected with an
impedance of 2400 - /1800 ft/<£; and load 3 is
172.8 + /2203.2 kVA. The loads are fed from a dis-
tribution line with an impedance of 2 + /16
fl/cj).
The magnitude of the line-to-neutral voltage at the
load end of the line is 24 V3 kV.
a) Calculate the total complex power at the send-
ing end of the line.
b) What percentage of the average power at the
sending end of the line is delivered to the loads?
11.34 A three-phase line has an impedance of
0.1 + /0.8
Cl/(f).
The line feeds two balanced
three-phase loads connected in parallel. The first
load is absorbing a total of 630 kW and absorbing
840 kVAR magnetizing vars. The second load
is Y-connected and has an impedance of
15.36 - /4.48
0,/4>.
The line-to-neutral voltage at
the load end of the line is 4000 V. What is the
magnitude of the line voltage at the source end of
the line?
11.35 A balanced three-phase load absorbs 96 kVA at a
lagging power factor of 0.8 when the line voltage at
the terminals of the load is 480
V.
Find four equiva-

lent circuits that can be used to model this load.
11.36 The total power delivered to a balanced three-
phase load when operating at a line voltage of
2400 V3 V is 720 kW at a lagging power factor of
0.8.
The impedance of the distribution line sup-
plying the load is 0.8 + /6.4
Q,/<fi.
Under these
operating conditions, the drop in the magnitude
of the line voltage between the sending end and
the load end of the line is excessive. To compen-
sate,
a bank of A-connected capacitors is placed
in parallel with the load. The capacitor bank is
designed to furnish 576 kVAR of magnetizing
reactive power when operated at a line voltage
of 2400 V3 V.
a) What is the magnitude of the voltage at the
sending end of the line when the load is operat-
ing at a line voltage of 2400 V3 V and the capac-
itor bank is disconnected?
b) Repeat (a) with the capacitor bank connected.
c) What is the average power efficiency of the line
in (a)?
d) What is the average power efficiency in (b)?
e) If the system is operating at a frequency of 60 Hz,
what is the size of each capacitor in microfarads?
11.37 A balanced bank of delta-connected capacitors is
connected in parallel with the load described in

Assessment Problem 11.9. The effect is to place a
capacitor in parallel with the load in each phase.
The line voltage at the terminals of the load thus
remains at 2450
V.
The circuit is operating at a fre-
quency of 60 Hz.The capacitors are adjusted so that
the magnitude of the line current feeding the paral-
lel combination of the load and capacitor bank is at
its minimum.
a) What is the size of each capacitor in microfarads?
b) Repeat (a) for wye-connected capacitors.
c) What is the magnitude of the line current?
11.38 A balanced three-phase source is supplying 150 kVA
at 0.8 pf lead to two balanced Y-connected parallel
loads.
The distribution line connecting the source to
the load has negligible impedance. The power associ-
ated with load
1
is 30 + /30 kVA.
a) Determine the types of components and their
impedances in each phase of load 2 if the line
voltage is 2500 V3 V and the impedance compo-
nents are in series.
b) Repeat (a) with the impedance components in
parallel.
Problems 425
11.39 The output of the balanced positive-sequence
three-phase source in Fig. PI 1.39 is 41.6 kVA at a

lagging power factor of 0.707. The line voltage at
the source is 240 V.
a) Find the magnitude of the line voltage at the load.
b) Find the total complex power at the terminals of
the load.
Figure P11.39
Balanced
three-phase'
source
0.04 n
/°-
03
n
WV
/-Y-YV^_
0.04 a
/°-
03
n
—Vs/V
/"YVY-N—
0.04 0 /0-°3
U
—VvV or-or^
Balanced
"•three-phase
load
11.43 The two wattmeters in Fig. 11.20 can be used to
compute the total reactive power of the load.
a) Prove this statement by showing that

V3(W
2
- Wj) = V3V
L
/
L
siiiV
b) Compute the total reactive power from the
wattmeter readings for each of the loads in
Example 11.6. Check your computations by cal-
culating the total reactive power directly from
the given voltage and impedance.
11.44 The two-wattmeter method is used to measure the
power at the load end of the line in Example 11.1.
Calculate the reading of each wattmeter.
11.45 The wattmeters in the circuit in Fig. 11.20 read as
follows: W
x
= 40,823.09 W, and W
2
= 103,176.91 W.
The magnitude of the line voltage is 2400 V5 V.
The phase sequence is positive. Find Z^.
Section 11.6
11.40 Derive Eqs. 11.56 and 11.57.
11.41 In the balanced three-phase circuit shown in
Fig. PI 1.41, the current coil of the wattmeter is con-
nected in line aA, and the potential coil of the
wattmeter is connected across lines b and c. Show
that the wattmeter reading multiplied by V3 equals

the total reactive power associated with the load.
The phase sequence is positive.
Figure PI 1.41
11.46 a) Calculate the reading of each wattmeter in the
circuit shown in Fig. PI 1.46. The value of Z^ is
40/-30° a.
b) Verify that the sum of the wattmeter readings
equals the total average power delivered to the
A-connected load.
Figure PI 1.46
24Q/W
V |
a •
K m
D
•
r-
C
'
±
cc
[
[
W
1
1
pcj
A
B
C

Z<b
z*
z*
N
bl±
Wi
-1-) »_| /-YVV^
240/120
o
~V~
4^
240/-120° V
-•
W,
+™~ l-\
c
11.42 The line-to-neutral voltage in the circuit in
Fig. PI 1.41 is 680 V, the phase sequence is positive,
and the load impedance is 16 - j 12 0/$.
a) Calculate the wattmeter reading.
b) Calculate the total reactive power associated
with the load.
11.47 The two-wattmeter method is used to measure
the power delivered to the unbalanced load in
Problem 11.18. The current coil of wattmeter 1 is
placed in line aA and that of wattmeter 2 is
placed in line bB.
a) Calculate the reading of wattmeter 1.
b) Calculate the reading of wattmeter 2.
c) Show that the sum of the two wattmeter read-

ings equals the total power delivered to the
unbalanced load.