ELECTRICAL MACHINES

AND
ELECTRICAL APPARATUS AND
DEVICES

Hanes K. Gutierrez
BSEE – 3A
Induction Motors
1. A dc test is performed on a 460-V -connected 100-hp induction motor. If VDC = 21 V and IDC = 72 A,
what is the stator resistance R1 ? Why is this so?
SOLUTION If this motor’s armature is connected in delta, then there will be two phases in parallel with one
phase between the lines tested.

VDC R1 R1

R1

Therefore, the stator resistance R1 will be

2. A 220-V three-phase six-pole 50-Hz induction motor is running at a slip of 3.5 percent. Find:
(a) The speed of the magnetic fields in revolutions per minute
(b) The speed of the rotor in revolutions per minute
(c) The slip speed of the rotor
(d) The rotor frequency in hertz

SOLUTION
(a) The speed of the magnetic fields is
120 fe 120(50 Hz)
nsync = = = 1000 r/min
P 6
(b) The speed of the rotor is

nm = (1 − s) nsync = (1 − 0.035)(1000 r/min) = 965 r/min

(c) The slip speed of the rotor is
nslip = snsync = (0.035)(1000 r/min) = 35 r/min
(d) The rotor frequency is

3. Answer the questions in Problem 7-2 for a 480-V three-phase four-pole 60-Hz induction motor running at
a slip of 0.025.
SOLUTION
 (a) The speed of the magnetic fields is

4. A three-phase 60-Hz induction motor runs at 715 r/min at no load and at 670 r/min at full load.
(a) How many poles does this motor have?
(b) What is the slip at rated load?
(c) What is the speed at one-quarter of the rated load?
(d) What is the rotor’s electrical frequency at one-quarter of the rated load?

SOLUTION
(a) This machine has 10 poles, which produces a synchronous speed of

120 fe 120(60 Hz)

nsync = = = 720 r/min
P 10
(b) The slip at rated load is
nsync − nm 720 − 670
s= 100% = 100% = 6.94%
nsync 720
(c) The motor is operating in the linear region of its torque-speed curve, so the slip at ¼ load will be
s = 0.25(0.0694) = 0.0174
The resulting speed is

nm = (1 − s) nsync = (1 − 0.0174)(720 r/min) = 707 r/min

(d) The electrical frequency at ¼ load is

fr = sfe = (0.0174)(60 Hz) = 1.04 Hz

5. A 50-kW 440-V 50-Hz two-pole induction motor has a slip of 6 percent when operating at full-load
conditions. At full-load conditions, the friction and windage losses are 520 W, and the core losses are 500
W. Find the following values for full-load conditions:
(a) The shaft speed nm

(b) The output power in watts

(c) The load torque  load in newton-meters
(d) The induced torque  ind in newton-meters
 (e) The rotor frequency in hertz

SOLUTION
(a) The synchronous speed of this machine is
120 fe 120(50 Hz)
n = = = 3000 r/min
sync
P 2
Therefore, the shaft speed is

nm = (1 − s) nsync = (1 − 0.06)(3000 r/min) = 2820 r/min

(b) The output power in watts is 50 kW (stated in the problem).
(c) The load torque is
50 kW
 load = OUT = (2820 r/min) 2 rad  1 min  = 169.3 N  m
P
m
  
 1r  60 s 
(d) The induced torque can be found as follows:
Pconv = POUT + PF&W + Pcore + Pmisc = 50 kW + 520 W + 500 W = 51.2 kW
51.2 kW
 ind = conv = (2820 r/min) 2 rad  1 min  = 173.4 N  m
P
m
  
 1r  60 s 
(e) The rotor frequency is

fr = sfe = (0.06)(50 Hz) = 3.00 Hz

6. A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load speed
of 3440 r/min. Calculate the slip and the electrical frequency of the rotor at no-load and full-load
conditions. What is the speed regulation of this motor [Equation (4-57)]?
SOLUTION The synchronous speed of this machine is 3600 r/min. The slip and electrical frequency at no-
load conditions is
nsync − nnl 3600 − 3580
s = 100% = 100% = 0.56%
nl
nsync 3600
fr,nl = sfe = (0.0056)(60 Hz) = 0.33 Hz
The slip and electrical frequency at full load conditions is
nsync − nnl 3600 − 3440
s = 100% = 100% = 4.44%
fl
nsync 3600
fr,fl = sfe = (0.0444)(60 Hz) = 2.67 Hz
The speed regulation is
nnl − nfl 3580 − 3440
SR = 100% = 100% = 4.1%
nfl 3440
7. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent
circuit components are
R1 = 0.220  R2 = 0.127  X M = 15.0 
X1 = 0.430  X 2 = 0.430 
Pmech = 300 W Pmisc  0 Pcore = 200 W
For a slip of 0.05, find
(a) The line current IL

(b) The stator copper losses PSCL

(c) The air-gap power PAG

(d) The power converted from electrical to mechanical form Pconv

(e) The induced torque ind

(f) The load torque load

(g) The overall machine efficiency

(h) The motor speed in revolutions per minute and radians per second

SOLUTION The equivalent circuit of this induction motor is shown below:

IA
R1 jX 1 jX2 R2

+
0.22  j0.43  j0.43  0.127 

1− s 
jX R
V j15  M 2 
 s 
2.413 
-

(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance ZF
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.
IA
R1 jX 1 jXF RF

0.22  j0.43 

V

jX M
The equivalent impedance of the rotor circuit in parallel with is:
1 1
=
ZF = 1 1 1 1 = 2.337 + j0.803 = 2.4719 
+ +
jX M Z2 j15  2.54 + j0.43
The phase voltage is 208/ = 120 V, so line current I L is
V 1200 V
IL = I A = =
R1 + jX1 + RF + jX F 0.22  + j0.43  + 2.337  + j0.803 
I L = I A = 42.3 − 25.7 A
(b) The stator copper losses are

PSCL = 3IA2 R1 = 3(42.3 A) (0.22 ) = 1180 W

2

R2
The air gap power is PAG = 3I 2 = 3I A RF
2 2
(c)
s
2 R2
(Note that 3I A2 RF is equal to 3I 2 , since the only resistance in the original rotor circuit was R2 / s ,
s
and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin
equivalent circuit must be the same as the power consumed by the original circuit.)
= 3I 2R = 3(42.3 A) (2.337 ) = 12.54 kW
R2 2
P = 3I 2
AG 2 A F
s
(d) The power converted from electrical to mechanical form is

Pconv = (1 − s) PAG = (1 − 0.05)(12.54 kW) = 11.92 kW

(e) The induced torque in the motor is
PAG = 12.54 kW
= 66.5 N  m
 ind = sync (1800 r/min) 2 rad  1 min 
  
 1r  60 s 
(f) The output power of this motor is
POUT = Pconv − Pmech − Pcore − Pmisc = 11.92 kW − 300 W − 200 W − 0 W = 11.42 kW
The output speed is
nm = (1 − s) nsync = (1 − 0.05) (1800 r/min) = 1710 r/min
Therefore the load torque is
11.42 kW
 load = OUT = (1710 r/min) 2 rad  1 min  = 63.8 N  m
P
m
  
 1r  60 s 
(g) The overall efficiency is
POUT POUT
= 100% = 100%
PIN 3V I A cos
11.42 kW
=  100% = 83.2%
3(120 V)(42.3 A) cos 25.7
(h) The motor speed in revolutions per minute is 1710 r/min. The motor speed in radians per second is
  2 rad  1 min
 = (1710 r/min) = 179 rad/s
  
1 r   60 s 
m

8. For the motor in Problem 7-7, what is the slip at the pullout torque? What is the pullout torque of this
motor?
SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit
from the rotor back to the power supply, and then using that with the rotor circuit model.

jXM (R1 + jX1) =

( j15 )(0.22  + j0.43 )
= 0.208 + j0.421  = 0.47063.7 
Z =
R1 + j ( X1 + X M ) 0.22  + j (0.43  + 15 )
TH

V =
( j15 ) (1200 V) = 116.70.8 V
jX M
V =
R1 + j(X 1 + X M ) 0.22  + j(0.43  + 15 )
TH 

The slip at pullout torque is

R2
smax =
2
TH TH + X2
0.127 
smax = = 0.145
(0.208 ) + (0.421  + 0.430 )
2 2

The pullout torque of the motor is

 2
3VTH
 max =
sync R TH
2
R2TH + X2
 TH

3(116.7 V)
2

max =
 (88.5 rad/s) 0.208  + (0.208 ) + (0.421  + 0.430 ) 
2 2

 
 max = 100 N  m
9. (a) Calculate and plot the torque-speed characteristic of the motor in Problem 7-7. (b) Calculate and plot
the output power versus speed curve of the motor in Problem 7-7.
SOLUTION
(a) A MATLAB program to calculate the torque-speed characteristic is shown below.

% M-file: prob7_9a.m
% M-file create a plot of the torque-speed curve of the
% induction motor of Problem 7-7.

% First, initialize the values needed in this program.

r1 = 0.220; % Stator resistance
x1 = 0.430; % Stator reactance
r2 = 0.127; % Rotor resistance
x2 = 0.430; % Rotor reactance
xm = 15.0; % Magnetization branch reactance
v_phase = 208 / sqrt(3); % Phase voltage
n_sync = 1800; % Synchronous speed (r/min)
w_sync = 188.5; % Synchronous speed (rad/s)

% Calculate the Thevenin voltage and impedance from Equations

% 7-38 and 7-41.
v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);

% Now calculate the torque-speed characteristic for many

% slips between 0 and 1. Note that the first slip value
% is set to 0.001 instead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:1:50) / 50; % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed

% Calculate torque versus speed

for ii = 1:51
t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );
end

% Plot the torque-speed curve

figure(1);
plot(nm,t_ind,'k-','LineWidth',2.0);
xlabel('\bf\itn_{m}');
ylabel('\bf\tau_{ind}');
title ('\bfInduction Motor Torque-Speed Characteristic');
grid on;
The resulting plot is shown below:

(b) A MATLAB program to calculate the output-power versus speed curve is shown below.

% M-file: prob7_9b.m
% M-file create a plot of the output pwer versus speed
% curve of the induction motor of Problem 7-7.

% First, initialize the values needed in this program.

r1 = 0.220; % Stator resistance
x1 = 0.430; % Stator reactance
r2 = 0.127; % Rotor resistance
x2 = 0.430; % Rotor reactance
xm = 15.0; % Magnetization branch reactance
v_phase = 208 / sqrt(3); % Phase voltage
n_sync = 1800; % Synchronous speed (r/min)
w_sync = 188.5; % Synchronous speed (rad/s)

% Calculate the Thevenin voltage and impedance from Equations

% 7-38 and 7-41.
v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);

% Now calculate the torque-speed characteristic for many

% slips between 0 and 1. Note that the first slip value
% is set to 0.001 instead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:1:50) / 50; % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed (r/min)
wm = (1 - s) * w_sync; % Mechanical speed (rad/s)
 % Calculate torque and output power versus speed
for ii = 1:51
t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );
p_out(ii) = t_ind(ii) * wm(ii);
end

% Plot the torque-speed curve

figure(1);
plot(nm,p_out/1000,'k-','LineWidth',2.0);
xlabel('\bf\itn_{m} \rm\bf(r/min)');
ylabel('\bf\itP_{OUT} \rm\bf(kW)');
title ('\bfInduction Motor Ouput Power versus Speed');
grid on;
The resulting plot is shown below:

7-10. For the motor of Problem 7-7, how much additional resistance (referred to the stator circuit) would it be
necessary to add to the rotor circuit to make the maximum torque occur at starting conditions (when the
shaft is not moving)? Plot the torque-speed characteristic of this motor with the additional resistance
inserted.
SOLUTION To get the maximum torque at starting, the smax must be 1.00. Therefore,

R2
smax =
R 2
+(X +X )2
TH TH 2

R2
1.00 =
(0.208 ) + (0.421  + 0.430 )
2 2

R2 = 0.876 
Therefore, an additional 0.749  must be added to the rotor circuit. The resulting torque-speed
characteristic is:
11. If the motor in Problem 7-7 is to be operated on a 50-Hz power system, what must be done to its supply
voltage? Why? What will the equivalent circuit component values be at 50 Hz? Answer the questions in
Problem 7-7 for operation at 50 Hz with a slip of 0.05 and the proper voltage for this machine.
SOLUTION If the input frequency is decreased to 50 Hz, then the applied voltage must be decreased by 5/6
also. If this were not done, the flux in the motor would go into saturation, since
1
=
N
 v dt
T

and the period T would be increased. At 50 Hz, the resistances will be unchanged, but the reactances will
be reduced to 5/6 of their previous values. The equivalent circuit of the induction motor at 50 Hz is
shown below:
IA
R1 jX 1 jX2 R2

+ j0.358  0.127 
0.22  j0.358 
1− s 
jX R
V j12.5  M 2 
 s 
2.413 
-

(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance ZF
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.
 IA
R1 jX 1 jXF RF

0.22  j0.358 

V

The equivalent impedance of the rotor circuit in parallel with jX M is:

1 1
=
ZF = 1 1 1 1 = 2.310 + j0.804 = 2.4519.2 
+ +
jX M Z2 j12.5  2.54 + j0.358
The line voltage must be derated by 5/6, so the new line voltage is VT = 173.3 V . The phase voltage is
173.3 / = 100 V, so line current I L is
V 1000 V
IL = I A = =
R1 + jX1 + RF + jX F 0.22  + j0.358  + 2.310  + j0.804 
I L = I A = 35.9 − 24.7 A
(b) The stator copper losses are

PSCL = 3I A2 R1 = 3(35.9 A) (0.22 ) = 851 W

2

R2
The air gap power is PAG = 3I 2 = 3I A RF
2 2
(c)
s
2 R2
(Note that 3I A2 RF is equal to 3I 2 , since the only resistance in the original rotor circuit was R2 / s ,
s
and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin
equivalent circuit must be the same as the power consumed by the original circuit.)
R2
PAG = 3I 2 = 3I A RF = 3(35.9 A) (2.310 ) = 8.93 kW
2 2 2

s
(d) The power converted from electrical to mechanical form is
Pconv = (1 − s)PAG = (1 − 0.05) (8.93 kW) = 8.48 kW
(e) The induced torque in the motor is
PAG = 8.48 kW
= 54.0 N  m
 ind = sync (1500 r/min) 2 rad  1 min 
  
 1r  60 s 
(f) In the absence of better information, we will treat the mechanical and core losses as constant despite
the change in speed. This is not true, but we don’t have reason for a better guess. Therefore, the output
power of this motor is
POUT = Pconv − Pmech − Pcore − Pmisc = 8.48 kW − 300 W − 200 W − 0 W = 7.98 kW
The output speed is
 nm = (1 − s) nsync = (1 − 0.05)(1500 r/min) = 1425 r/min
Therefore the load torque is
7.98 kW
 load = OUT = (1425 r/min) 2 rad  1 min  = 53.5 N  m
P
m
  
 1r  60 s 
(g) The overall efficiency is
POUT POUT
= 100% = 100%
PIN 3V I A cos
7.98 kW
=  100% = 81.6%
3(100 V)(35.9 A) cos 24.7
(h) The motor speed in revolutions per minute is 1425 r/min. The motor speed in radians per second is
 2 rad  1 min 
m = (1425 r/min) = 149.2 rad/s
  
 1r  60 s 
12. Figure 7-16a shows a simple circuit consisting of a voltage source, a resistor, and two reactances. Find the
Thevenin equivalent voltage and impedance of this circuit at the terminals. Then derive the expressions
for the magnitude of VTH and for RTH given in Equations (7-38) and (7-42).

SOLUTION The Thevenin voltage of this circuit is

jX M
V = V
R1 + j(X 1 + X M )
TH 

The magnitude of this voltage is

XM
VTH = V
R1 + ( X 1 + X M )
2 2

If X M  X1 , then R12 + (X 1 + X M )  (X 1 + X M ) , so
2 2

XM
X1 + X M
TH

The Thevenin impedance of this circuit is

 jX M (R1 + jX1 )
Z =
TH
R1 + j(X 1 + X M )
ZTH = jX M (R1 + jX1 )R1 − j ( X 1 + X M )

R1 + j ( X 1 + X M )R1 − j ( X 1 + X M )

Z =
− R X X 1 1 M
+RX X
1 1 M 1 M
 
+ R X 2 + j R 2X + X 2X + X X 2
1 M 1 M 1 M

R 2 + (X + X )
TH 2
1 1 M

+ j R1 X M2 + X 1 X M + X) 1 X M
2
ZTH = RTH + jX TH = R1 X M 2 2 2

R1 + X 1 + X M )
2
( 2
(
R1 + X 1 + X M
2

2
The Thevenin resistance is RTH = R1 X M . If X M  R1 , then
2
(
R1 + X 1 + X M )
2

R12 + (X 1 + X M )  (X 1 + X M ) , so
2 2

2
 X 
 R1
 X1 + X M 
TH

+ X1X M 2
The Thevenin reactance is X TH = R1 X M2 + X 1 X M
2 2
.
(
R1 + X 1 + X M )
2

If X M  R1 and X M  X1 then X1 X M 2  R12 X M + X 12 X M and (X1 + X M )  X M 2  R12 , so

2

X1X M
TH
 =X

13. Figure P7-1 shows a simple circuit consisting of a voltage source, two resistors, and two reactances in series
with each other. If the resistor RL is allowed to vary but all the other components are constant, at what
value of RL will the maximum possible power be supplied to it? Prove your answer. (Hint: Derive
an expression for load power in terms of V, RS , X S , RL and X L and take the partial derivative of that
expression with respect to RL .) Use this result to derive the expression for the pullout torque [Equation
(7-52)].

SOLUTION The current flowing in this circuit is given by the equation

V
IL =
RS + jX S + RL + jX L
 V
IL =
(RS + RL ) 2
+ (X S + X L )
2

The power supplied to the load is

P = I L RL =
2

(RS + RL
(R + R )2 + ( X + X )2  V 2 − V 2R 2 ( R + R )
P  S L S L  L 2 S L 
= (R + R )2 + ( X + X )2 
RL
 S L S L 
To find the point of maximum power supplied to the load, set P / RL = 0 and solve for RL .
(R + R )2 + ( X + X )2  V 2 − V 2R 2 ( R + R ) = 0
 S L S L  L  S L 

(R + R ) + ( X + X )  = 2R ( R + R )
2 2

 S L S L  L S L

RS2 + 2RSR L + R L2 + ( X S + X L ) = 2RS RL + 2RL 2

2

RS 2 + R L2 + ( X S + X L ) = 2R L2
2

R 2 +(X + X )2 = R 2
S S L L

Therefore, for maximum power transfer, the load resistor should be

RL = RS 2 + (X S + X L )
2

14. A 440-V 50-Hz six-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit
parameters are
R1 = 0.082  R2 = 0.070  X M = 7.2 
X1 = 0.19  X 2 = 0.18 
PF&W = 1.3 kW Pmisc = 150 W Pcore = 1.4 kW
For a slip of 0.04, find
(a) The line current IL
(b) The stator power factor
(c) The rotor power factor
(d) The stator copper losses PSCL

(e) The air-gap power PAG

(f) The power converted from electrical to mechanical form Pconv

(g) The induced torque ind

(h) The load torque load
(i) The overall machine efficiency 
(j) The motor speed in revolutions per minute and radians per second

SOLUTION The equivalent circuit of this induction motor is shown below:

IA
R1 jX 1 jX2 R2

+ j0.18  0.07  I2
0.082  j0.19 
1− s 
jX R
V j7.2  M 2 
 s 
1.68 
-

(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance ZF
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.
IA
R1 jX 1 jXF RF

0.082  j0.19 

V

jX M
The equivalent impedance of the rotor circuit in parallel with is:
1 1
=
ZF = 1 1 1 1 = 1.557 + j0.550 = 1.6719.2 
+ +
jX M Z2 j7.2  1.75 + j0.18

The phase voltage is 440/ = 254 V, so line current I L is

V 2540 V
IL = I A = =
R1 + jX1 + RF + jX F 0.082  + j0.19  + 1.557  + j0.550 
I L = I A = 141 − 24.3 A
(b) The stator power factor is
PF = cos 24.3 = 0.911 lagging
(c) To find the rotor power factor, we must find the impedance angle of the rotor
X2 0.18
 = tan−1 = tan−1 = 5.87
R
R2 / s 1.75
Therefore the rotor power factor is
PFR = cos 5.87 = 0.995 lagging
(d) The stator copper losses are

PSCL = 3IA2 R1 = 3(141 A) (0.082 ) = 4890 W

2
 R2
The air gap power is PAG = 3I 2 = 3I A RF
2 2
(e)
s
2 R2
(Note that 3I A2 RF is equal to 3I 2 , since the only resistance in the original rotor circuit was R2 / s ,
s
and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin
equivalent circuit must be the same as the power consumed by the original circuit.)
= 3I 2R = 3(141 A) (1.557 ) = 92.6 kW
R2 2
P = 3I 2
AG 2 A F
s
(f) The power converted from electrical to mechanical form is

Pconv = (1 − s) PAG = (1 − 0.04)(92.6 kW) = 88.9 kW

(g) The synchronous speed of this motor is
120 fe 120(50 Hz)
nsync = = = 1000 r/min
P 62 rad  1 min
 = (1000 r/min) = 104.7 rad/s
  
1 r   60 s 
sync

Therefore the induced torque in the motor is

PAG = 92.6 kW
= 884 N  m
 ind = sync (1000 r/min) 2 rad  1 min 
  
 1r  60 s 
(h) The output power of this motor is
POUT = Pconv − Pmech − Pcore − Pmisc = 88.9 kW − 1.3 kW − 1.4 kW − 300 W = 85.9 kW
The output speed is
nm = (1 − s) nsync = (1 − 0.04) (1000 r/min) = 960 r/min
Therefore the load torque is
85.9 kW
 load = OUT = (960 r/min) 2 rad  1 min  = 854 N  m
P
m
  
 1r  60 s 
(i) The overall efficiency is
POUT POUT
=  100% =  100%
PIN 3V IA cos
85.9 kW
=  100% = 87.7%
3(254 V)(141 A) cos 24.3
(j) The motor speed in revolutions per minute is 960 r/min. The motor speed in radians per second is
2 rad  1 min 
m = (960 r/min) = 100.5 rad/s
  
 1r  60 s 
15. For the motor in Problem 7-14, what is the pullout torque? What is the slip at the pullout torque? What is
the rotor speed at the pullout torque?
SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit
from the rotor back to the power supply, and then using that with the rotor circuit model.
jX M (R1 + jX1 ) =
( j7.2 )(0.082  + j0.19 )
= 0.0778 + j0.1860  = 0.20267.3 
Z =
TH
R1 + j(X 1 + X M ) 0.082  + j(0.19  + 7.2 )
V =
( j7.2 ) (2540 V) = 247.50.6 V
jX M
V =
R1 + j(X 1 + X M ) 0.082  + j(0.19  + 7.2 )
TH 

The slip at pullout torque is

R2
smax =
RTH + (X TH + X 2 )
2 2

0.070 
smax = = 0.187
(0.0778 ) + (0.186  + 0.180 )
2 2

The pullout torque of the motor is

3V 2
 max = TH

sync RTH R2TH + (X TH + X 2 )


3(247.5 V)
2
 max =
(104.7 rad/s)0.0778  + (0.0778 ) + (0.186  + 0.180 )
 max = 1941 N  m
16. If the motor in Problem 7-14 is to be driven from a 440-V 60-Hz power supply, what will the pullout
torque be? What will the slip be at pullout?
SOLUTION If this motor is driven from a 60 Hz source, the resistances will be unchanged and the
reactances will be increased by a ratio of 6/5. The resulting equivalent circuit is shown below.
IA
R1 jX 1 jX2 R2

+
0.082  j0.228  j0.216  0.07  I2
1− s 
jX R
V j8.64  2  
 s 
M

1.68 
-

The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the
rotor back to the power supply, and then using that with the rotor circuit model.

jX M ( R1 + jX1) =
( j8.64 )(0.082  + j0.228 ) = 0.0778 + j0.223  = 0.23670.7 
Z TH = R + j X + X
( ) 0.082  + j (0.228  + 8.64 )
1 1 M
j8.64 
VTH =
jXM V = (2540 V) = 247.50.5 V
R1 + j ( X1 + X M ) 
0.082  + j (0.228  + 8.64 )
 The slip at pullout torque is
R2
smax =
RTH + (X TH + X 2 )
2 2

0.070 
smax = = 0.157
(0.0778 ) 2
+ (0.223  + 0.216 )
2

The synchronous speed of this motor is

120 fe 120(60 Hz)
nsync = = = 1200 r/min
P  62 rad  1 min 
sync = (1200 r/min) = 125.7 rad/s
  
 1 r  60 s 
Therefore the pullout torque of the motor is
3V 2
 max = TH

sync RTH R2
+ (X TH + X 2 )
 TH

3(247.5 V)
2

 max =
(125.7 rad/s)0.0778  + (0.0778 ) + (0.223  + 0.216 )

 max = 1396 N  m
17. Plot the following quantities for the motor in Problem 7-14 as slip varies from 0% to 10%: (a) ind (b)
Pconv (c) Pout (d) Efficiency . At what slip does Pout equal the rated power of the machine?
SOLUTION This problem is ideally suited to solution with a MATLAB program. An appropriate program
is shown below. It follows the calculations performed for Problem 7-14, but repeats them at many values
of slip, and then plots the results. Note that it plots all the specified values versus nm , which varies from
900 to 1000 r/min, corresponding to a range of 0 to 10% slip.

% M-file: prob7_17.m
% M-file create a plot of the induced torque, power
% converted, power out, and efficiency of the induction
% motor of Problem 7-14 as a function of slip.

% First, initialize the values needed in this program.

r1 = 0.082; % Stator resistance
x1 = 0.190; % Stator reactance
r2 = 0.070; % Rotor resistance
x2 = 0.180; % Rotor reactance
xm = 7.2; % Magnetization branch reactance
v_phase = 440 / sqrt(3); % Phase voltage
n_sync = 1000; % Synchronous speed (r/min)
w_sync = 104.7; % Synchronous speed (rad/s)
p_mech = 1300; % Mechanical losses (W)
p_core = 1400; % Core losses (W)
p_misc = 150; % Miscellaneous losses (W)

% Calculate the Thevenin voltage and impedance from Equations

% 7-38 and 7-41.
v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);

% Now calculate the torque-speed characteristic for many

% slips between 0 and 0.1. Note that the first slip value
% is set to 0.001 instead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:0.001:0.1); % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed
wm = nm * 2*pi/60; % Mechanical speed

% Calculate torque, P_conv, P_out, and efficiency

% versus speed
for ii = 1:length(s)

% Induced torque
t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );

% Power converted
p_conv(ii) = t_ind(ii) * wm(ii);

% Power output
p_out(ii) = p_conv(ii) - p_mech - p_core - p_misc;

% Power input
zf = 1 / ( 1/(j*xm) + 1/(r2/s(ii)+j*x2) );
ia = v_phase / ( r1 + j*x1 + zf );
p_in(ii) = 3 * v_phase * abs(ia) * cos(atan(imag(ia)/real(ia)));

% Efficiency
eff(ii) = p_out(ii) / p_in(ii) * 100;

end

% Plot the torque-speed curve

figure(1);
plot(nm,t_ind,'b-','LineWidth',2.0);
xlabel('\bf\itn_{m} \rm\bf(r/min)');
ylabel('\bf\tau_{ind} \rm\bf(N-m)');
title ('\bfInduced Torque versus Speed');
grid on;

% Plot power converted versus speed

figure(2);
plot(nm,p_conv/1000,'b-','LineWidth',2.0);
xlabel('\bf\itn_{m} \rm\bf(r/min)');
ylabel('\bf\itP\rm\bf_{conv} (kW)');
title ('\bfPower Converted versus Speed');
grid on;

% Plot output power versus speed

figure(3);
plot(nm,p_out/1000,'b-','LineWidth',2.0);
xlabel('\bf\itn_{m} \rm\bf(r/min)');
ylabel('\bf\itP\rm\bf_{out} (kW)');
title ('\bfOutput Power versus Speed');
axis([900 1000 0 160]);
grid on;

% Plot the efficiency

figure(4);
plot(nm,eff,'b-','LineWidth',2.0);
xlabel('\bf\itn_{m} \rm\bf(r/min)');
ylabel('\bf\eta (%)');
title ('\bfEfficiency versus Speed');
grid on;
The four plots are shown below:
 This machine is rated at 75 kW. It produces an output power of 75 kW at 3.4% slip, or a speed of 966
r/min.
18. A 208-V, 60 Hz, six-pole Y-connected 25-hp design class B induction motor is tested in the laboratory,
with the following results:
No load: 208 V, 22.0 A, 1200 W, 60 Hz
Locked rotor: 24.6 V, 64.5 A, 2200 W, 15 Hz
DC test: 13.5 V, 64 A
Find the equivalent circuit of this motor, and plot its torque-speed characteristic curve.
SOLUTION From the DC test,
13.5 V
2R1 =  R1 = 0.105 
64 A
IDC

R1

VD

R1 R1

In the no-load test, the line voltage is 208 V, so the phase voltage is 120 V. Therefore,

X +X =
V
=
120 V = 5.455  @ 60 Hz
1 M
I A,nl 22.0 A
In the locked-rotor test, the line voltage is 24.6 V, so the phase voltage is 14.2 V. From the locked-rotor
test at 15 Hz,
V 14.2 V
Z LR = RLR + jX LR = = = 0.2202 
I A,LR 64.5 A
−1 
−1 PLR 2200 W
LR = cos = cos   = 36.82
SLR  3(24.6 V)(64.5 A)
Therefore,

RLR = Z LR cos LR = (0.2202 )cos 36.82 = 0.176 

 R1 + R2 = 0.176 
 R2 = 0.071 

X LR = ZLR sinLR = (0.2202 )sin 36.82 = 0.132 

At a frequency of 60 Hz,
 60 Hz 
X LR =   = 0.528 
 X LR
15 Hz
 
For a Design Class B motor, the split is X 1 = 0.211  and X 2 = 0.317  . Therefore,

X M = 5.455  - 0.211  = 5.244 

The resulting equivalent circuit is shown below:
IA
R1 jX 1 jX2 R2

+ j0.317  0.071  I2
0.105  j0.211 
1− s 
jX R
V j5.244  2 
 s 
M

A MATLAB program to calculate the torque-speed characteristic of this motor is shown below:

% M-file: prob7_18.m
% M-file create a plot of the torque-speed curve of the
% induction motor of Problem 7-18.

% First, initialize the values needed in this program.

r1 = 0.105; % Stator resistance
x1 = 0.211; % Stator reactance
r2 = 0.071; % Rotor resistance
x2 = 0.317; % Rotor reactance
xm = 5.244; % Magnetization branch reactance
v_phase = 208 / sqrt(3); % Phase voltage
n_sync = 1200; % Synchronous speed (r/min)
w_sync = 125.7; % Synchronous speed (rad/s)

% Calculate the Thevenin voltage and impedance from Equations

% 7-38 and 7-41.
 v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);

% Now calculate the torque-speed characteristic for many

% slips between 0 and 1. Note that the first slip value
% is set to 0.001 instead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:1:50) / 50; % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed

% Calculate torque versus speed

for ii = 1:51
t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );
end

% Plot the torque-speed curve

figure(1);
plot(nm,t_ind,'b-','LineWidth',2.0);
xlabel('\bf\itn_{m}');
ylabel('\bf\tau_{ind}');
title ('\bfInduction Motor Torque-Speed Characteristic');
grid on;
The resulting plot is shown below:

19. A 208-V four-pole 10-hp 60-Hz Y-connected three-phase induction motor develops its full-load induced
torque at 3.8 percent slip when operating at 60 Hz and 208 V. The per-phase circuit model impedances of
the motor are
R1 = 0.33  X M = 16 
X1 = 0.42  X 2 = 0.42 
Mechanical, core, and stray losses may be neglected in this problem.
(a) Find the value of the rotor resistance R2 .

(b) Find max , smax , and the rotor speed at maximum torque for this motor.
(c) Find the starting torque of this motor.
(d) What code letter factor should be assigned to this motor?

SOLUTION The equivalent circuit for this motor is

IA
R1 jX 1 jX2 R2

+ j0.42  ???  I2
0.33  j0.42 
1− s 
jX R
V j16  M 2 
 s 

The Thevenin equivalent of the input circuit is:

jXM (R1 + jX1) =
( j16 )(0.33  + j0.42 )
= 0.313 + j0.416  = 0.52053 
Z =
TH
R1 + j ( X1 + X M ) 0.33  + j (0.42  + 16 )
V =
( j16 ) (1200 V) = 116.91.2 V
jX M
V =
R1 + j(X 1 + X M ) 0.33  + j(0.42  + 16 )
TH 

(a) If losses are neglected, the induced torque in a motor is equal to its load torque. At full load, the
output power of this motor is 10 hp and its slip is 3.8%, so the induced torque is

nm = (1 − 0.038)(1800 r/min) = 1732 r/min

ind = load =
(10 hp)(746 W/hp) = 41.1 N  m
(1732 r/min) 2 rad  60 s 
 1r  1 min 
The induced torque is given by the equation
ind = 2
3VTH R2 / s
 ( R + R / s)2 + ( X + X )2 
sync  TH 2 TH 2 
Substituting known values and solving for R2 / s yields

3(116.9 V) R22 / s
2
41.1 N  m =
(188.5 rad/s) (0.313 + R / s) + (0.416 + 0.42) 
2

 2 
7, 747 = 40, 997 R2 / s
(0.313 + R / s)2 + 0.699
 2 
(0.313 + R / s) + 0.699 = 5.292 R / s
2

 2  2
 0.098 + 0.626R / s + ( R / s) + 0.699 = 5.292 R / s
2

 2 2  2
2
 R2  R 
  − 4.666 2  + 0.797 = 0
 s   s 
 R2 
= 0.178, 4.488
 
 s 
R2 = 0.0067 , 0.17 
These two solutions represent two situations in which the torque-speed curve would go through this
specific torque-speed point. The two curves are plotted below. As you can see, only the 0.17  solution
is realistic, since the 0.0067  solution passes through this torque-speed point at an unstable location on
the back side of the torque-speed curve.

(b) The slip at pullout torque can be found by calculating the Thevenin equivalent of the input circuit
from the rotor back to the power supply, and then using that with the rotor circuit model. The Thevenin
equivalent of the input circuit was calculate in part (a). The slip at pullout torque is
R2
smax =
2
TH TH + X2
0.17 
smax = = 0.190
(0.313 ) + (0.416  + 0.420 )
2 2

The rotor speed a maximum torque is

npullout = (1 − s) nsync = (1 − 0.190)(1800 r/min) = 1457 r/min

and the pullout torque of the motor is
3V 2
 max = TH

sync RTH R2TH + (X TH + X 2 )


 3 (116.9 V)
2
 max =
(188.5 rad/s)0.313  + (0.313 ) + (0.416  + 0.420 )

 max = 90.2 N  m
(c) The starting torque of this motor is the torque at slip s = 1. It is
ind = 2
3VTH R22 / s
 ( R + R / s) + ( X +X )2 
sync  TH 2 TH 2 
3(116.9 V) (0.17 )
2

ind = = 38.3 N  m
(188.5 rad/s)(0.313 + 0.17 ) + (0.416 + 0.420) 
2 2

(d) To determine the starting code letter, we must find the locked-rotor kVA per horsepower, which is
equivalent to finding the starting kVA per horsepower. The easiest way to find the line current (or
armature current) at starting is to get the equivalent impedance Z F of the rotor circuit in parallel with
jX M at starting conditions, and then calculate the starting current as the phase voltage divided by the
sum of the series impedances, as shown below.
IA,start
R1 jX 1 jXF RF

0.33  j0.42 

V

jX M
The equivalent impedance of the rotor circuit in parallel with at starting conditions (s = 1.0) is:
1 1
Z F ,start = =
1 1 1 1 = 0.161 + j0.411 = 0.44268.6 
+ +
jXM Z2 j16  0.17 + j0.42

The phase voltage is 208/ = 120 V, so line current I L,start is

V 1200 V
IL,start = I A = =
R1 + jX1 + RF + jX F 0.33  + j0.42  + 0.161  + j0.411 
IL,start = I A = 124 − 59.4 A
Therefore, the locked-rotor kVA of this motor is

S = 3 VT I L,rated = 3(208 V) (124 A) = 44.7 kVA

and the kVA per horsepower is

44.7 kVA
kVA/hp = = 4.47 kVA/hp
10 hp
This motor would have starting code letter D, since letter D covers the range 4.00-4.50.
20. Answer the following questions about the motor in Problem 7-19.
(a) If this motor is started from a 208-V infinite bus, how much current will flow in the motor at starting?
(b) If transmission line with an impedance of 0.50 + j0.35  per phase is used to connect the induction
motor to the infinite bus, what will the starting current of the motor be? What will the motor’s
terminal voltage be on starting?
(c) If an ideal 1.2:1 step-down autotransformer is connected between the transmission line and the motor,
what will the current be in the transmission line during starting? What will the voltage be at the motor
end of the transmission line during starting?

SOLUTION
(a) The equivalent circuit of this induction motor is shown below:
IA
R1 jX 1 jX2 R2

+ j0.42  0.17  I2
0.33  j0.42 
1− s 
jX R
V j16  M 2 
 s 

The easiest way to find the line current (or armature current) at starting is to get the equivalent impedance
Z F of the rotor circuit in parallel with jX M at starting conditions, and then calculate the starting current
as the phase voltage divided by the sum of the series impedances, as shown below.
IA
R1 jX 1 jXF RF

0.33  j0.42 

V

jX M
The equivalent impedance of the rotor circuit in parallel with at starting conditions (s = 1.0) is:
1 1
=
ZF = 1 1 1 1 = 0.161 + j0.411 = 0.44268.6 
+ +
jX M Z2 j16  0.17 + j0.42

The phase voltage is 208/ = 120 V, so line current I L is

V 1200 V
IL = I A = =
R1 + jX1 + RF + jX F 0.33  + j0.42  + 0.161  + j0.411 
IL = I A = 124 − 59.4 A
(b) If a transmission line with an impedance of 0.50 + j0.35  per phase is used to connect the
induction motor to the infinite bus, its impedance will be in series with the motor’s impedances, and the
starting current will be
I =I = V ,bus
R line + jX line + R1 + jX1 + RF + jX F
L A
 1200 V
IL = I A =
0.50  + j0.35 + 0.33  + j0.42 + 0.161 + j0.411 
IL = I A = 77.8 − 50.0 A
The voltage at the terminals of the motor will be
V = I A (R1 + jX1 + RF + jX F )
V = (77.8 − 50.0 A)(0.33  + j0.42  + 0.161  + j0.411 )
V = 75.19.4 V

Therefore, the terminal voltage will be (75.1 V) = 130 V . Note that the terminal voltage sagged by
37.5% during motor starting, which would be unacceptable.
(c) If an ideal 1.2:1 step-down autotransformer is connected between the transmission line and the
motor, the motor’s impedances will be referred across the transformer by the square of the turns ratio a =
1.2. The referred impedances are
R1 = a2R1 = 1.44 (0.33 ) = 0.475 
X1 = a2 X1 = 1.44 (0.42 ) = 0.605 
RF = a2RF = 1.44 (0.161 ) = 0.232 
X F = a2 X F = 1.44 (0.411 ) = 0.592 
Therefore, the starting current referred to the primary side of the transformer will be
V ,bus
IL = IA = + jX + R + jX  + R + jX 
Rline line 1 1 F F

1200 V
IL = IA =
0.50 + j0.35 + 0.475 + j0.605 + 0.232 + j0.592 
IL = IA = 61.2 − 52 A
The voltage at the motor end of the transmission line would be the same as the referred voltage at the
terminals of the motor
V = IA (R1 + jX1 + RF + jX F )
V = (61.2 − 52 A)(0.475  + j0.605  + 0.232  + j0.592 )
V = 85.07.4 V

Therefore, the line voltage at t he motor end of the transmission line will be (85 V) = 147.3 V . Note
that this voltage sagged by 29.2% during motor starting, which is less than the 37.5% sag with case of
across-the-line starting. Since the sag is still large, it might be possible to use a bigger autotransformer
turns ratio on the starter.
21. In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current
drawn by an induction motor. While this technique works, an autotransformer is relatively expensive. A
much less expensive way to reduce the starting current is to use a device called Y- starter. If an
induction motor is normally -connected, it is possible to reduce its phase voltage V (and hence its
starting current) by simply re-connecting the stator windings in Y during starting, and then restoring the
connections to  when the motor comes up to speed. Answer the following questions about this type of
starter.
 (a) How would the phase voltage at starting compare with the phase voltage under normal running
conditions?
(b) How would the starting current of the Y-connected motor compare to the starting current if the motor
remained in a -connection during starting?

SOLUTION

(a) The phase voltage at starting would be 1 / = 57.7% of the phase voltage under normal running
conditions.

(b) Since the phase voltage decreases to 1 / = 57.7% of the normal voltage, the starting phase
current will also decrease to 57.7% of the normal starting current. However, since the line current for the
original delta connection was times the phase current, while the line current for the Y starter
connection is equal to its phase current, the line current is reduced by a factor of 3 in a Y- starter.

For the -connection: I L, = I ,

For the Y-connection: I L,Y = I ,Y
But I , = 3I ,Y , so IL, = 3IL,Y

22. A 460-V 50-hp six-pole -connected 60-Hz three-phase induction motor has a full-load slip of 4 percent, an
efficiency of 91 percent, and a power factor of 0.87 lagging. At start-up, the motor develops 1.75 times
the full-load torque but draws 7 times the rated current at the rated voltage. This motor is to be started
with an autotransformer reduced voltage starter.
(a) What should the output voltage of the starter circuit be to reduce the starting torque until it equals the
rated torque of the motor?
(b) What will the motor starting current and the current drawn from the supply be at this voltage?

SOLUTION
(a) The starting torque of an induction motor is proportional to the square of VTH ,
2 2
 V   VT 
 =  2

 start2
=
TH2

start1  VTH1   VT 1 
2 2
 V  V 
 =  T2 
 start2
=  TH2

start1  VTH1   VT 1 
If a torque of 1.75  rated is produced by a voltage of 460 V, then a torque of 1.00  rated would be
produced by a voltage of
1.00
2
 V 
1.75
rated
=  T2 
rated  460 V 

VT 2 =
(460 V)2
= 348 V
1.75
(b) The motor starting current is directly proportional to the starting voltage, so
 348 V 
I L2 = I = (0.756)IL1 = (0.756) (7I rated ) = 5.296 I rated
  L1
 460 V 
 The input power to this motor is

PIN =
POUT
=
(50 hp) (746 W/hp) = 40.99 kW
 0.91
The rated current is equal to

Irated =
PIN
=
(40.99 kW) = 59.1 A
3 VT PF 3 (460 V)(0.87)
Therefore, the motor starting current is
I L2 = 5.843 I rated = (5.296) (59.1 A) = 313 A
The turns ratio of the autotransformer that produces this starting voltage is
NSE + NC = 460 V = 1.32
NC 348 V
so the current drawn from the supply will be
Istart = 313 A = 237 A
Iline =
1.32 1.32
23. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and
with a load of about 25 percent of the rated value for the machine. If the rotor resistance of this machine
is doubled by inserting external resistors into the rotor circuit, explain what happens to the following:
(a) Slip s
(b) Motor speed nm

(c) The induced voltage in the rotor

(d) The rotor current
(e)  ind
(f) Pout

(g) PRCL

(h) Overall efficiency 

SOLUTION
(a) The slip s will increase.
(b) The motor speed nm will decrease.

(c) The induced voltage in the rotor will increase.

(d) The rotor current will increase.
(e) The induced torque will adjust to supply the load’s torque requirements at the new speed. This will
depend on the shape of the load’s torque-speed characteristic. For most loads, the induced torque will
decrease.
 (f) The output power will generally decrease: POUT =  ind  m 

(g) The rotor copper losses (including the external resistor) will increase.
(h) The overall efficiency  will decrease.
24. Answer the following questions about a 460-V -connected two-pole 100-hp 60-Hz starting code letter F
induction motor:
(a) What is the maximum current starting current that this machine’s controller must be designed to
handle?
(b) If the controller is designed to switch the stator windings from a  connection to a Y connection
during starting, what is the maximum starting current that the controller must be designed to handle?
(c) If a 1.25:1 step-down autotransformer starter is used during starting, what is the maximum starting
current that will be drawn from the line?

SOLUTION
(a) The maximum starting kVA of this motor is

Sstart = (100 hp)(5.60) = 560 kVA

Therefore,
S 560 kVA
I start = = = 703 A
3 VT (460 V)
(b) The line voltage will still be 460 V when the motor is switched to the Y-connection, but now the
phase voltage will be 460 / = 265.6 V.
Before (in ):
460 V
I , =
(RTH + R2 ) + j(X TH + X 2 )
 797 V
I L, = 3I , =
(RTH + R2 ) + j(X TH + X 2 )
After (in Y):
265.6 V
I L,Y = I ,Y =
(RTH + R2 ) + j(X TH + X 2 )
Therefore the line current will decrease by a factor of 3! The starting current with a -Y starter is
703 A
Istart = = 234 A
3
(c) A 1.25:1 step-down autotransformer reduces the phase voltage on the motor by a factor 0.8. This
reduces the phase current and line current in the motor (and on the secondary side of the transformer) by a
factor of 0.8. However, the current on the primary of the autotransformer will be reduced by another
factor of 0.8, so the total starting current drawn from the line will be 64% of its original value. Therefore,
the maximum starting current drawn from the line will be
Istart = (0.64) (703 A) = 450 A
25. When it is necessary to stop an induction motor very rapidly, many induction motor controllers reverse the
direction of rotation of the magnetic fields by switching any two stator leads. When the direction of
rotation of the magnetic fields is reversed, the motor develops an induced torque opposite to the current
direction of rotation, so it quickly stops and tries to start turning in the opposite direction. If power is
removed from the stator circuit at the moment when the rotor speed goes through zero, then the motor has
been stopped very rapidly. This technique for rapidly stopping an induction motor is called plugging.
The motor of Problem 7-19 is running at rated conditions and is to be stopped by plugging.
(a) What is the slip s before plugging?
(b) What is the frequency of the rotor before plugging?
(c) What is the induced torque  ind before plugging?
(d) What is the slip s immediately after switching the stator leads?
(e) What is the frequency of the rotor immediately after switching the stator leads?
(f) What is the induced torque ind immediately after switching the stator leads?
SOLUTION
(a) The slip before plugging is 0.038 (see Problem 7-19).

(b) The frequency of the rotor before plugging is fr = sfe = (0.038)(60 Hz) = 2.28 Hz

(c) The induced torque before plugging is 41.1 Nm in the direction of motion (see Problem 7-19).
(d) After switching stator leads, the synchronous speed becomes –1800 r/min, while the mechanical
speed initially remains 1732 r/min. Therefore, the slip becomes
nsync − nm − 1800 − 1732
s= = = 1.962
nsync − 1800

(e) The frequency of the rotor after plugging is fr = sfe = (1.962)(60 Hz) = 117.72 Hz

(f) The induced torque immediately after switching the stator leads is
Transformer
1. The high-voltage coil of a transformer is wound with 700 turns of wire, and the low-
voltage coil is wound with 292 turns. When used as a step-up transformer (the low-
voltage coil is used as the primary), the load current is 10.5 A. Find the load component
of the primary current.
A. 43.5 A B. 4.38 A C. 25.18 A D. 2.518 A
Solution:
𝑁𝑝 𝐼𝑠
=
𝑁𝑠 𝐼𝑝
𝐼 = 𝐼𝑠𝑁𝑠
=
(700)(10.5)
𝑝 𝑁𝑝 292

• 𝑰𝒑 = 𝟐𝟓. 𝟏𝟕 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

REE – May 2008

2. A transformer has a primary winding of 2, 000 turns and of 2, 400 Volts and current of
8.66 − 𝑗5 Ampere with an impedance 𝑍2 connected across the secondary winding. If the
secondary winding has 500 turns, what is the value of the secondary current?
A. 20 − 𝑗34.64 𝐴 B. 𝟑𝟒. 𝟔𝟒 − 𝒋𝟐𝟎 𝑨 C. 34.64 + 𝑗20 𝐴 D. 20 + 𝑗34.64 𝐴
Solution:
𝑁𝑝 𝐼𝑠
=
𝑁𝑠 𝐼𝑝
𝐼 = 𝐼𝑠𝑁𝑠
=
(8.66−𝑗5)(2,000)
𝑝 𝑁𝑝 500
• 𝑰𝒑 = 𝟑𝟒. 𝟔𝟒 − 𝒋𝟐𝟎 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

3. A 120 V to 27.5 V, 400 Hz step-down transformer is to be operated at 60 Hz. What is the

highest safe input voltage?
A. 200 V B. 400 V C. 120 V D. 18 V
Solution:
𝐸1 𝑓1
=
𝐸2 𝑓2
𝐸 = 𝐸1𝑓2 = (120)(60)
2 𝑓1 400
• 𝑬𝟐 = 𝟏𝟖 𝑽𝒐𝒍𝒕𝒔

REE – September 2011

4. When a welding transformer is used in a resistance welding, it will
A. step up voltage B. step down voltage
C. step up current D. step down current
∗ 𝑁𝑜𝑡𝑒:
𝐼𝑛 𝑎 𝑤𝑒𝑙𝑑𝑖𝑛𝑔 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟, 𝑎 ℎ𝑖𝑔ℎ 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝑛𝑒𝑒𝑑𝑒𝑑 𝑡𝑜 𝑚𝑒𝑙𝑡 𝑒𝑎𝑠𝑖𝑙𝑦 𝑡ℎ𝑒 𝑖𝑟𝑜𝑛.
 5. A 4, 600/230 V, 60 Hz step-down transformer has core dimension of 76.2 mm by
111.8 mm. A maximum flux density of 0.93 𝑊𝑏/𝑚2 is to be used. Assuming 9 percent loss of area due to
stacking factor of laminations, calculate the primary and secondary turns required.
A. 2, 395 and 120 B. 120 and 2, 395 C. 2, 180 and 109 D. 109 and 2, 180
Solution:
𝐴𝑒𝑓𝑓 = (1 − 0.09)(76.2 𝑚𝑚 × 111.8 𝑚𝑚)
1𝑚 2
𝐴𝑒𝑓𝑓 = 7, 752.436𝑚𝑚2 ( )
1,000 𝑚𝑚
• 𝐴𝑒𝑓𝑓 = 7.752 × 10−3 𝑚2
∅𝑚 = 𝛽𝑚𝐴𝑒𝑓𝑓
𝑊𝑏 −3 2
∅𝑚 = (0.93 ) (7.752 × 10 𝑚 )
𝑚2
• ∅𝑚 = 7.209 𝑚𝑊𝑏
𝐸𝑟𝑚𝑠 = 4.44𝑓∅𝑁
𝐸𝑟𝑚𝑠 4,600
𝑁𝑝 = 4.44𝑓∅ = (4.44)(60)(7.209×10−3)
• 𝑵𝒑 = 𝟐, 𝟑𝟗𝟓 𝒕𝒖𝒓𝒏𝒔
𝐸𝑟𝑚𝑠 230
𝑁𝑠 = =
4.44𝑓∅ (4.44)(60)(7.209×10−3)

• 𝑵𝒔 = 𝟏𝟐𝟎 𝒕𝒖𝒓𝒏𝒔

REE – October 1997

6. A small single-phase transformer has 10.2 watts no-load loss. The core has a volume of
750 cubic cm. The maximum flux density is 10, 000 gauss and the hysteresis constant of
the core is 5 × 10−4, using the Steinmetz law to find the hysteresis, determine the eddy
current loss.
A. 4.55 Watts B. 5.55 Watts C. 3.55 Watts D. 2.55 Watts
Solution:
∗ 𝑆𝑡𝑒𝑖𝑛𝑚𝑒𝑛𝑡𝑧 𝑙𝑎𝑤
𝑃ℎ 𝛼 𝛽𝑚1.6
∗ 𝑘ℎ = (5 × 10−4)(750)
• 𝑘ℎ = 0.375
𝑑𝑦𝑛𝑒−𝑐𝑚 1𝑁 1𝑚
𝑃 = (0.375)(60)(10, 000)1.6 ( )( )
ℎ 𝑠𝑒𝑐 105𝑑𝑦𝑛𝑒 100 𝑐𝑚
• 𝑃ℎ = 5.652 𝑊𝑎𝑡𝑡𝑠
𝑃𝑒 = 𝑃𝑖𝑛 − 𝑃ℎ
𝑃𝑒 = 10.2 − 5.652
• 𝑷𝒆 = 𝟒. 𝟓𝟒𝟖 𝑾𝒂𝒕𝒕𝒔
REE – September 2006
7. The primary of transformer has 200 turns and is excited by a 240 V, 60 Hz source. What
is the maximum value of the core flux?
A. 4.04 mWb B. 4.40 mWb C. 4.13 mWb D. 4.32 mWb
Solution:
𝐸 240
∅𝑚 = 4.44𝑓𝑁
=
(4.44)(60)(200)

• ∅𝒎 = 𝟒. 𝟓𝟎 𝒎𝑾𝒃

REE – September 2008

8. A transformer is rated 1 kVA, 220/110 V, 60 Hz. Because of an emergency this
transformer has to be used on a 50 Hz system. If the flux density in the transformer core
is to be kept the same as at 60 Hz and 220 V, what is the kilovolt-ampere rating at 50 Hz.
A. 0.890 kVA B. 0.833 kVA C. 0.909 kVA D. 0.871 kVA
Solution:
𝑆𝛼𝑓
𝑆1 𝑓1
=
𝑆2 𝑓2
𝑆 = 𝑆1𝑓2
=
(1)(50)
2 𝑓1 60
• 𝑺𝟐 = 𝟎. 𝟖𝟑𝟑 𝒌𝑽𝑨

9. A single-phase transformer has a no-load power input of 250 Watts, when supplied at 250
V, 50 Hz has a p.f of 0.25. What is the magnetizing component of the no-load current?
A. 4.00 A B. 3.87 A C. 1.00 A D. none of these
Solution:
𝑃 250
𝑆= =
𝑝.𝑓 0.25
• 𝑆 = 1, 000 𝑉𝐴
𝑝𝑓 = 𝑐𝑜𝑠−1 0.25
• 𝑝𝑓 = 75.52°
𝑆 = 𝑉𝐼
1,000∠75.52
𝐼= = 4∠75.52° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
250
• 𝐼 = 1 + 𝑗 3.873 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
• 𝑰𝒎 = 𝟑. 𝟖𝟕𝟑 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
REE – September 2011
10. A 4, 400 V, 60 Hz transformer has a core loss of 840 Watts, of which one-third is eddy
current loss. What is the core loss when the x’former is connected to a 4, 600 V, 50 Hz
source?
A. 977 Watts B. 907 Watts C. 927 Watts D. 944 Watts
Solution:
𝑃𝑐𝑜𝑟𝑒 = 𝑃 𝑒𝑑𝑑𝑦
1
3
1
𝑃𝑒1 = 3 (840)
• 𝑃𝑒1 = 280 𝑊𝑎𝑡𝑡𝑠
𝑃𝑐𝑜𝑟𝑒 = 𝑃ℎ + 𝑃𝑒
𝑃ℎ1 = 840 − 280
• 𝑃ℎ1 = 560 𝑊𝑎𝑡𝑡𝑠
𝑃𝑒2 𝑘 (𝐸 )2
= 2 2 2
𝑃𝑒1 𝑘1(𝐸1)
𝑃 = 280 4,6002
𝑒2 ( 2)
4,440
• 𝑃𝑒2 = 306.033 𝑊𝑎𝑡𝑡𝑠
𝐸 1.6
𝑘2( 2 0.6)
𝑃 ℎ2 𝑓2
= 𝐸1
1.6
𝑃ℎ1 𝑘1( 0.6)
𝑓1
4,6001.6
( 500.6 )
𝑃ℎ2 = 560 [ 4,4001.6
]
( )
600.6
• 𝑃ℎ2 = 670.788 𝑊𝑎𝑡𝑡𝑠
𝑃𝑐𝑜𝑟𝑒2 = 𝑃ℎ2 + 𝑃𝑒2 = 670.788 + 306.033
• 𝑷𝒄𝒐𝒓𝒆𝟐 = 𝟗𝟕𝟔. 𝟖𝟐𝟏 𝑾𝒂𝒕𝒕𝒔

REE – September 2004

11. In an ideal transformer, what is the efficiency?
A. 100% B. 90% C. 80% D. 70%
 12. A 100 kVA distribution transformer has a full-load copper loss of 1, 180 Watts. For what
kilowatt load, at a power factor of 0.71, will the copper losses in the transformer be 1,
500 Watts?
A. 90.25 B. 71 C. 112.75 D. 80.05
Solution:
𝑃𝑐𝑢−𝑎𝑛𝑦2 𝑆𝑎𝑛𝑦 2

2 =( )
𝑃𝑐𝑢−𝐹𝐿 𝑆 𝑟𝑎𝑡𝑒𝑑
𝑃
𝑆=
𝑝.𝑓
1,500 𝐾𝑊𝑙𝑜𝑎𝑑/0.71 2
=( )
1,180 100
𝐾𝑊𝑙𝑜𝑎𝑑 2 1,500
( ) = 1002 ( )
0.71 1,180
1,500
𝐾𝑊𝑙𝑜𝑎𝑑 = √(0.71)2(100)2 ( )
1,180

• 𝑲𝑾𝒍𝒐𝒂𝒅 = 𝟖𝟎. 𝟎𝟓𝟎 𝒌𝑾

13. Given a 10-kVA transformer with full-load losses amounting to 70 Watts in the iron and
140 Watts in the copper. Calculate the efficiency at half-load unity power factor.
A. 98.62% B. 97.97% C. 97.28% D. 97.94%
Solution:
𝑃𝐶𝑢𝐻𝐿 𝑆𝐻𝐿 2
=( )
𝑆𝐹𝐿
𝑃𝐶𝑢𝐹𝐿
140(52)
𝑃𝐶𝑢 𝐻𝐿
=
102
• 𝑃𝐶𝑢 𝐻𝐿 = 35 𝑊𝑎𝑡𝑡𝑠
𝑃𝑜𝐻𝐿 5,000
ɳ 𝐻𝐿 = × 100% = × 100%
𝑃𝑜𝐻𝐿+𝑃𝑐𝑜𝑟𝑒+𝑃𝐶𝑢𝐻𝐿 5,000+35

• ɳ 𝑯𝑳 = 𝟗𝟕. 𝟗𝟒%

REE – April 2004

14. Instrument transformers are used in indicating and metering and with protective devices,
they are used for .
A. measuring B. detecting C. relaying D. sensing

REE – September 2003

15. What type of transformer bank is used to convert 2-phase to 3-phase power?
A. open-delta B. scott-T C. wye-delta D. delta-wye
16. A 100-kVA 2, 400/240-Volt 60 cycle transformer has the following constants: 𝑟𝑝 =
0.42 Ω, 𝑋𝑝 = 0.72 Ω; 𝑟𝑠 = 0.0038 Ω, 𝑋𝑠 = 0.0068 Ω. What is the equivalent
impedance in primary terms?
A. 0.016 Ω B. 1.612 Ω C. 0.161 Ω D. 16.12 Ω
Solution:
𝑍𝑒 𝑝 = (𝑟𝑝 + 𝑎2𝑟𝑠) + 𝑗(𝑋𝑝 + 𝑎2𝑋𝑠)
𝑉𝑝 2,400
𝑎= =
𝑉𝑠 240
• 𝑎 = 10
𝑍𝑒 𝑝 = (0.42) + (102)(0.0038) + 𝑗(0.72) + (10)2(0.0068)
𝑍𝑒𝑝 = 0.8 + 𝑗1.4 = 1.61∠60.26°
• |𝒁𝒆𝒑| = 𝟏. 𝟔𝟏 𝒐𝒉𝒎𝒔

17. Calculate the all-day efficiency of a 100-kVA transformer operating under the following
conditions: 6 hours on a load of 50 kW at 0.73 power factor; 3 hours on a load of 90 kW
at 0.82 power factor; 15 hours with no load on secondary. The iron loss is 1, 000 Watts
and the full-load copper loss is 1, 060 Watts.
A. 96.31% B. 94.87% C. 95.33% D. 95.29%
Solution:
𝑃𝑜𝑢𝑡 = (50, 000)(6) + (90, 000)(3) + (0)(15)
• 𝑃𝑜𝑢𝑡 = 570 𝑘𝑊ℎ𝑟
𝑃𝑐𝑜𝑟𝑒 = (1, 000)(24)
• 𝑃𝑐𝑜𝑟𝑒 = 24 𝑘𝑊ℎ𝑟
𝑘𝑉𝐴 2
𝑃𝐶𝑢1 1
=( )
𝑃𝐶𝑢𝐹𝐿 𝑘𝑉𝐴𝐹𝐿
50/0.73 2
𝑃𝐶𝑢1 = 1, 060 ( )
100
• 𝑃𝐶𝑢1 = 497.279 𝑊𝑎𝑡𝑡𝑠
90/0.82 2
𝑃 = 1, 060 ( )
𝐶𝑢2 100
• 𝑃𝐶𝑢2 = 1, 276.919 𝑊𝑎𝑡𝑡𝑠
• 𝑃𝐶𝑢3 = 0 𝑊𝑎𝑡𝑡𝑠
𝑃𝐶𝑢𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐶𝑢1(6) + 𝑃𝐶𝑢2(3) + 𝑃𝐶𝑢3(15)
𝑃𝐶𝑢𝑡𝑜𝑡𝑎𝑙 = (487.279)(6) + (1, 276.919)(3) + (0)(15)
• 𝑃𝐶𝑢𝑡𝑜𝑡𝑎𝑙 = 6.814 𝑘𝑊ℎ𝑟
𝑃𝑜𝑢𝑡 570
ɳ = × 100% = × 100%
𝑎𝑙𝑙 𝑑𝑎𝑦 𝑃𝑜𝑢𝑡+𝑃𝑐𝑜𝑟𝑒+𝑃𝐶𝑢 570+24+6.814
 • ɳ𝒂𝒍𝒍 𝒅𝒂𝒚 = 𝟗𝟒. 𝟖𝟕%
REE – September 2005
18. A 50-kVA, single-phase transformer has 96% efficiency when it operates at full-load
unity power factor for 8 hours per day. What is the all-day efficiency of the transformer if
the copper loss is 60% of full-load losses?
A. 92% B. 90% C. 89.5% D. 93%
Solution:
50
0.96 =
50+𝑃𝑙𝑜𝑠𝑠𝑒𝑠𝐹𝐿
50
𝑃𝑙𝑜𝑠𝑠𝑒𝑠 𝐹𝐿 = 0.96 − 50
• 𝑃𝑙𝑜𝑠𝑠𝑒𝑠𝐹𝐿 = 2.083 𝑘𝑊
𝑃𝐶𝑢𝐹𝐿 = (0.6)(2, 083)
• 𝑃𝐶𝑢𝐹𝐿 = 1, 249.8 𝑊
𝑃𝑐𝑜𝑟𝑒 = 𝑃𝑙𝑜𝑠𝑠 + 𝑃𝐶𝑢𝐹𝐿 = 2, 083 − 1, 249.8
• 𝑃𝑐𝑜𝑟𝑒 = 833.2 𝑊𝑎𝑡𝑡𝑠
𝑃𝑜𝑢𝑡𝑡𝑜𝑡𝑎𝑙 = (50 𝑘𝑉𝐴)(1.0)(8)
• 𝑃𝑜𝑢𝑡𝑡𝑜𝑡𝑎𝑙 = 400𝑘𝑊ℎ𝑟
𝑃𝑐𝑜𝑟𝑒 𝑡𝑜𝑡𝑎𝑙 = (833.2)(24)
• 𝑃𝑜𝑢𝑡𝑡𝑜𝑡𝑎𝑙 = 19.997 𝑘𝑊ℎ𝑟
𝑃𝐶𝑢𝑡𝑜𝑡𝑎𝑙 = (1, 249.8)(8)
• 𝑃𝐶𝑢𝑡𝑜𝑡𝑎𝑙 = 9.998 𝑘𝑊ℎ𝑟
400
ɳ𝑎𝑙𝑙 =
400+19.997+9.998
× 100%
𝑑𝑎𝑦
• ɳ𝒂𝒍𝒍 𝒅𝒂𝒚 = 𝟗𝟑. 𝟎𝟐%

Asst. EE – October 1991

19. A 10 kVA, 2, 400/240 V, single-phase transformer has the following resistances and
leakage reactances;
𝑟𝑝 = 3 Ω 𝑟𝑠 = 0.03 Ω
𝑋𝑝 = 15 Ω 𝑋𝑠 = 0.15 Ω
Find the primary voltage required to produce 240 V at the secondary terminals at full-load, when the load
power factor is 0.8 lagging.
A. 2, 400 V B. 2, 496.5 V C. 2, 348 V D. 2, 445.5 V
Solution:
2,400
𝑎=
240
• 𝑎 = 10
 𝑎𝑉𝑠 = 10(240)
• 𝑎𝑉𝑠 = 2, 400
10,000 ∠𝑐𝑜𝑠−1(0.8)
𝐼𝑡 = 2,400
• 𝐼𝑡 = 4.17 ∠ − 36.87° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝑍𝑒 𝑝 = [3 + (10)2(0.03)] + 𝑗[15 + (10)2(0.15)]
• 𝑍𝑒𝑝 = 6 + 𝑗30 𝑜ℎ𝑚𝑠
𝑉𝑝 = 𝐼𝑡𝑍𝑒𝑝 + 𝑎𝑉𝑠
𝑉𝑝 = (4.17 ∠ − 36.87)(6 + 𝑗30 ) + 2, 400
𝑉𝑝 = 2, 496.526 ∠1.95° 𝑉𝑜𝑙𝑡𝑠
• |𝑽𝒑| = 𝟐, 𝟒𝟗𝟔. 𝟓𝟐𝟔 𝑽𝒐𝒍𝒕𝒔

20. A 500 kVA, single-phase, 13, 200/2, 400 Volts transformer has 4% reactance and 1%
resistance. The leakage reactance and resistance of the high voltage (primary) winding
are 6.34 Ω and 1.83 Ω, respectively. The core loss under rated condition is 1, 800 Watts.
Calculate the leakage reactance and resistance of the low voltage (secondary) winding.
A. 7.56 Ω and 1.66 Ω B. 13.69 Ω and 3.42 Ω
C. 0.25 Ω and 0.055 Ω D. 13.9 Ω and 3.48 Ω
Solution:
(𝑉𝑏)2 (13,200)2
𝑍𝑏 = =
𝑆𝑏 500,000
• 𝑍𝑏 = 348.48 𝑜ℎ𝑚𝑠
𝑅𝑒𝑝 = 𝑅𝑝𝑢𝑍𝑏𝑎𝑠𝑒 = (0.01)(348.48)
• 𝑅𝑒𝑝 = 3.4848 𝑜ℎ𝑚𝑠
𝑋𝑒𝑝 = 𝑋𝑝𝑢𝑍𝑏𝑎𝑠𝑒 = (0.04)(348.48)
• 𝑋𝑒𝑝 = 13.9392 𝑜ℎ𝑚𝑠
𝑟𝑒𝑓𝑒𝑟𝑟𝑒𝑑 𝑡𝑜 𝑝𝑟𝑖𝑚𝑎𝑟𝑦
𝑅𝑒 𝑝 = 𝑟𝑝 + 𝑎2𝑟𝑠
13,200
𝑎=
2,400
• 𝑎 = 5.5
𝑅𝑒𝑝−𝑟𝑝 3.4848−1.83
𝑟𝑠 = =
𝑎2 5.52
• 𝒓𝒔 = 𝟎. 𝟎𝟓𝟓
𝑋𝑒𝑝 = 𝑋𝑝 + 𝑎2𝑋𝑠
 𝑋𝑒𝑝−𝑋𝑝 13.9392−6.34
𝑋𝑠 = =
𝑎2 5.52
• 𝑿𝒔 = 𝟎. 𝟐𝟓𝟏 𝒐𝒉𝒎𝒔

21. In Problem No.20, calculate the %V.R and efficiency of the transformer at full-load,
0.85 p.f. lagging and 2, 400 Volts.
A. 4% and 97.8% B. 6% and 95.4% C. 5% and 96.8% D. 3% and 98.4%
Solution:
𝑟𝑝 1.83
=
𝑎2 5.52
• 𝑟𝑝 = 0.0605 𝑜ℎ𝑚𝑠
𝑎2
𝑋𝑝 6.34
𝑎2
=
5.52
• 𝑋𝑝 = 0.21 𝑜ℎ𝑚𝑠
𝑎2
𝜃 = 𝑐𝑜𝑠−1(0.85)
• 𝜃 = 31.79°
500,000 ∠−31.79°
𝐼𝑇 = 2,400
• 𝐼𝑟𝑇 = 208.33𝑋∠ − 31.79°
𝑍 = ( 𝑝 + 𝑟 ) + 𝑗 ( 𝑝 + 𝑋 ) = (0.0605 + 0.055) + 𝑗(0.21 + 0.25)
𝑒𝑠 𝑎2 𝑠 𝑎2 𝑠

• 𝑍𝑒𝑠 = 0.1155 + 𝑗0.46 Ω

𝑉𝑝 = 𝐼𝑇𝑍𝑒𝑠 + 𝑉𝑠
𝑉𝑝 = (208.33 ∠ − 31.79°)(0.1155 + 𝑗0.46) + 2, 400
• 𝑉𝑝 = 2, 471.89 ∠1.6° 𝑉𝑜𝑙𝑡𝑠
2,471.89−2,400
%𝑉𝑅 = × 100
2,400
• %𝑽𝑹 = 𝟑%
𝑃𝑜𝑢𝑡 = (500 𝑘𝑉𝐴)(0.85)
• 𝑃𝑜𝑢𝑡 = 425 𝑘𝑊
• 𝑃𝑐𝑜𝑟𝑒 = 1, 800 𝑊
𝑃𝑐𝑢 = 𝐼𝑇 2𝑅𝑇 = (208.33)2(0.1155)
• 𝑃𝑐𝑢 = 5, 012.86 𝑊
425,000
ɳ= × 100
425,000+1,800+5,012.86
• ɳ = 𝟗𝟖 %
22. An 11, 000/230 V, 150 kVA, single-phase, 50 Hz transformer has a core loss of 1.4 kW
and a full-load copper loss of 1.6 kW. What is the value of maximum efficiency at unity
p.f?
A. 98.17% B. 98.04% C. 97.22% D. 97.64%
Solution:
𝑃𝑐𝑢𝑚𝑎𝑥 = 𝑃𝑐𝑜𝑟𝑒
• 𝑃𝑐𝑢𝑚𝑎𝑥 = 1, 400 𝑊
𝑃𝑐𝑢𝑚𝑎𝑥 𝑘𝑉𝐴𝑚𝑎𝑥 2
=( )
𝑃𝑐𝑢𝐹𝐿 𝑘𝑉𝐴𝐹𝐿
1,400
𝑘𝑉𝐴𝑚𝑎𝑥 = √(150 ) (
2
)
1,600

• 𝑘𝑉𝐴𝑚𝑎𝑥 = 140.31 𝑘𝑉𝐴

𝑘𝑊𝑚𝑎𝑥 = 𝑘𝑉𝐴𝑚𝑎𝑥(𝑝. 𝑓) = (140.31)(1.0)
• 𝑘𝑊𝑚𝑎𝑥 = 140.13 𝑘𝑊
140.13
ɳ= × 100
140.13+2(1.4)
• ɳ = 𝟗𝟖. 𝟎𝟒 %

23. A 300-kVA, single-phase transformer is designed to have a resistance of 1.5% and

maximum efficiency occurs at a load of 173.2 kVA. Find its efficiency when supplying
full-load at 0.8 p.f. lagging at normal voltage and frequency.
A. 97.56% B. 96.38% C. 98.76% D. 95.89%
Solution:
𝑃𝑐𝑢𝐹𝐿
𝑟𝑝.𝑢 = 𝑆𝐹𝐿
𝑃𝑐𝑢𝐹𝐿 = 𝑟𝑝.𝑢𝑆𝐹𝐿 = (0.015)(300, 000)
• 𝑃𝑐𝑢𝐹𝐿 = 4, 500 𝑊𝑎𝑡𝑡𝑠
𝑃𝑐𝑢𝑚𝑎𝑥 173.2 2
=( )
4,500 300
• 𝑃𝑐𝑢𝑚𝑎𝑥 = 1, 499.91 𝑊𝑎𝑡𝑡𝑠
𝑃𝑐𝑢𝑚𝑎𝑥 = 𝑃𝑐𝑜𝑟𝑒
𝑃𝑜𝑢𝑡 300(0.8) × 100
ɳ 𝐹𝐿 = × 100 =
𝑃𝑜𝑢𝑡+𝑃𝑐𝑜𝑟𝑒+𝑃𝑐𝑢𝐹𝐿 (300)(0.8)+1.5+4.5

• ɳ𝑭𝑳 = 𝟗𝟕. 𝟓𝟔 %
REE – September 2002
24. A 20 kV/7.87 kV autotransformer has 200 A current in the common winding. What is the
secondary line current?
A. 143.52 B. 200 C. 56.48 D. 329
Solution:
𝐼𝑐
=𝑎−1
𝐼𝑝
20,000
𝑎=
7,870
• 𝑎 = 2.54
200
= 2.54 − 1
𝐼𝑝
• 𝐼𝑝 = 129.76 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐼𝑠 = 𝐼𝑝 + 𝐼𝑐 = 129.76 + 200
• 𝑰𝒔 = 𝟑𝟐𝟗. 𝟕𝟔 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

25. An autotransformer is adjusted for an output voltage of 85.3 Volts when operated from a
117 Volts line. The variable power load draws 3.63 kW at unity power factor at this
setting. Determine the transformed power and the connected power from the source to the
load.
A. 980 Watts and 2, 650 Watts B. 1, 343 Watts and 2, 287 Watts
C. 1, 815 Watts and 1, 815 Watts D. 1, 210 Watts and 2, 420 Watts
Solution:
117
𝑎=
85.3
• 𝑎 = 1.37
1 1
𝑃 = 𝑃 (1 − ) = (3, 630) (1 − )
𝑡𝑟𝑎𝑛𝑠 𝑖𝑛 𝑎 1.37
• 𝑷𝒕𝒓𝒂𝒏𝒔 = 𝟗𝟖𝟎. 𝟑𝟔 𝑾𝒂𝒕𝒕𝒔
𝑃 3,630
𝑃𝑐𝑜𝑛 = 𝑖𝑛 =
𝑎 1.37
• 𝑷𝒄𝒐𝒏 = 𝟐, 𝟔𝟒𝟗. 𝟔𝟒 𝑾𝒂𝒕𝒕𝒔

REE – April 2006

26. What would happen if you connect a transformer to a dc circuit with a voltage of 20% of
nameplate ratings after a steady state condition is reached?
A. No voltage is registered at the secondary
B. Rated no-load current flows to the secondary
C. primary current is equal to voltage over equivalent primary impedance
D. Voltage is established at secondary
 27. A short-circuit test was performed upon a 10 kVA, 2, 300/230-Volt transformer with the
following results: 𝐸𝑠𝑐 = 137 𝑉𝑜𝑙𝑡𝑠; 𝑃𝑠𝑐 = 192 𝑊𝑎𝑡𝑡𝑠; 𝐼𝑠𝑐 = 4.34 𝐴𝑚𝑝𝑒𝑟𝑒𝑠. Calculate in
secondary terms the transformer equivalent.
A. 29.88 Ω B. 2.988 Ω C. 0.2988 Ω D. 298.8 Ω
Solution:
𝑃𝑠𝑐 192
𝑅= 2 =
𝐼𝑠𝑐 4.342
• 𝑅 = 10.19 Ω
𝐸𝑠𝑐 137
𝑍= 𝐼𝑠𝑐
=
4.34
• 𝑍 = 31.57 Ω
𝑋 = √𝑍2 − 𝑅2 = √31.572 − 10.192
• 𝑿 = 𝟐𝟗. 𝟖𝟖 Ω

REE – April 2007

28. A transformer is rated 500 kVA, 4, 800/480 V, 60 Hz when it is operated as a
conventional two winding transformer. This transformer is to be used as a 5280/4800 V
stepdown autotransformer in a power distribution system. In the autotransformer, what is
the transformer rating when used in this manner?
A. 5 MVA B. 6 MVA C. 5.5 MVA D. 6.5 MVA
Solution:
𝑆 500,000
𝐼𝑝 = =
4,800
𝑉𝑝
• 𝐼𝑝 = 104.17 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝑆 500,000
𝐼𝑠 = =
4,80
𝑉𝑠
• 𝐼𝑠 = 1, 041.7 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐼𝑝′ = 𝐼𝑠
• 𝐼𝑝′ = 1, 041.7 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝑆 = 𝑉𝑝′𝐼𝑝′ = (5, 280)(1, 041.7)
• 𝑺 = 𝟓. 𝟓 𝑴𝑽𝑨

REE – September 2008

29. Two identical transformers bank on open delta serve a balanced three-phase load of 26
kVA at 240 V, 60 Hz. What is the minimum size of each in kVA needed to serve this
load?
A. 25 B. 10 C. 30 D. 15
Solution:
𝑆∅𝑙𝑜𝑎𝑑 26,000
𝑆∅𝑟𝑎𝑡𝑒𝑑 = =
(√3) √3
• 𝑺∅𝒓𝒂𝒕𝒆𝒅 = 𝟏𝟓 𝒌𝑽𝑨
 30. Two single-phase, 100 kVA transformers are connected in V (open delta) bank supplying
a balanced three-phase load. If the balanced three-phase load is 135 kW at 0.82 p.f
lagging and 0.823 efficiency, determine the overload kVA on each transformer.
A. 10.5 B. 5.5 C. 15.5 D. 20.5
Solution:
𝑃 135,000
𝑆𝐿 = 𝑝.𝑓
=
0.82
• 𝑆𝐿 = 164.634 𝑘𝑉𝐴
𝑃𝑜𝑢𝑡
ɳ= 𝑃𝑖𝑛
135
𝑃𝑖𝑛 = 0.823
• 𝑃𝑖𝑛 = 164.034 𝑘𝑊
𝑆 = 𝑃𝑖𝑛 = 164.034 𝑘𝑊
𝑖𝑛 𝑝.𝑓 0.82

• 𝑆𝑖𝑛 = 200.041 𝑘𝑉𝐴

𝑆 200.041
𝑆∅ = 𝑖𝑛 =
√3 √3
• 𝑆∅ = 115.494 𝑘𝑉𝐴
• 𝑆∅𝑟𝑎𝑡𝑒𝑑 = 100 𝑘𝑉𝐴
𝑆𝑂.𝐿 = 𝑆𝐿∅ − 𝑆∅𝑟𝑎𝑡𝑒𝑑 = 115.494 − 100
• 𝑺𝑶.𝑳 = 𝟏𝟓. 𝟒𝟗𝟒 𝒌𝑽𝑨

31. In problem No. 30, determine the p.f of each transformer secondary.
A. 0.820 lagging and 0.820 lagging B. 0.996 lagging and 0.424 leading
C. 0.996 lagging and 0.424 lagging D. 0.410 lagging and 0.410 lagging
Solution:
𝜃 = 𝑐𝑜𝑠−1(0.82)
• 𝜃 = 34.92°
𝑝. 𝑓1 = 𝑐𝑜 𝑠(30 + 𝜃) = 𝑐𝑜 𝑠(30 + 34.92)
• 𝒑. 𝒇𝟏 = 𝟎. 𝟒𝟐𝟒 𝒍𝒂𝒈𝒈𝒊𝒏𝒈
𝑝. 𝑓2 = 𝑐𝑜 𝑠(30 − 𝜃) = 𝑐𝑜 𝑠(30 − 34.92)
• 𝒑. 𝒇𝟐 = 𝟎. 𝟗𝟗𝟔 𝒍𝒂𝒈𝒈𝒊𝒏𝒈

REE – April 2005

32. What is the normal secondary circuit current of a current transformer?
A. 15 A B. 20 A C. 5 A D. 10 A
33. In Problem No.30, what is the minimum size in kVAR of a capacitor bank to be
connected across the load so that each transformer is loaded 96% of its rated capacity?
A. 87 kVAR B. 114 kVAR C. 27 kVAR D. 66 kVAR
Solution:
𝑆∅ = (0.96)(𝑆∅𝑟𝑎𝑡𝑒𝑑 ) = (0.96)(100)
• 𝑆∅ = 96 𝑘𝑉𝐴
• 𝑃𝑇 = 164.034 𝑘𝑊
• 𝑆 = 200.041 𝑘𝑉𝐴
𝑄 = √𝑆2 − 𝑃𝑇2 = √(200.041)2 − (164.034)2
• 𝑄 = 114.5 𝑘𝑉𝐴𝑅
𝑆𝑛𝑒𝑤 = (96 𝑘𝑉𝐴)(√3)
• 𝑆𝑛𝑒𝑤 = 166.277 𝑘𝑉𝐴
𝑆 = 𝑃2 + 𝑄2
2

𝑄 = √𝑆𝑛𝑒𝑤2 − 𝑃𝑇 2 = √(166.277)2 − (164.034)2

• 𝑸 = 𝟐𝟕. 𝟐𝟐 𝒌𝑽𝑨𝑹

34. A polarity test is performed upon a 1, 150/115 V transformer. If the input voltage is 116,
calculate the voltmeter reading if the polarity is subtractive.
A. 127.6 V B. 106 V C. 126 V D. 104.4 V
Solution:
𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑣𝑒
1,150
𝑎=
115
• 𝑎 = 10
116
𝑉𝑟𝑒𝑎𝑑𝑖𝑛𝑔 = 116 −
10
• 𝑽𝒓𝒆𝒂𝒅𝒊𝒏𝒈 = 𝟏𝟎𝟒. 𝟒 𝑽𝒐𝒍𝒕𝒔

35. A 20:1 potential transformer is used with a 150-V voltmeter. If the instrument
deflection is 118 Volts, calculate the line voltage.
A. 3, 000 V B. 2, 850 V C. 2, 360 V D. 2, 242 V
Solution:
𝑉𝐿
𝑎= 𝑉𝑟𝑒𝑎𝑑𝑖𝑛𝑔
20 𝑉𝐿
=
1 118
• 𝑽𝑳 = 𝟐, 𝟑𝟔𝟎 𝑽𝒐𝒍𝒕𝒔
REE –September 2010
36. A three-phase wye-delta connected, 50 MVA, 345/34.5 kV transformer is protected by
differential protection. The current transformer on the high side for differential protection
is 150:5. What is the current on the secondary side of CT’s?
A. 3.83 A B. 2.53 A C. 4.50 A D. 4.83 A
Solution:
𝑆 50,000,000
𝐼∅𝑝 = = 345,000
3𝑉∅𝑝 3( )
√3
• 𝐼∅𝑝 = 83.67 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝑠𝑖𝑛𝑐𝑒 𝑤𝑦𝑒 − 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛
𝐼∅𝑝 = 𝐼𝐿 𝑝
• 𝐼𝐶𝑇 𝑝 = 𝐼𝐶𝑇 𝑝 = 83.67 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐿 ∅
𝐼𝑝 𝑁𝑝
𝑎= =
𝐼𝑠 𝑁𝑠
150 83.67
=
5 𝐼𝐶𝑇𝑠
∅
(83.67)(5)
𝐼𝐶𝑇 𝑠 =
∅ 150
• 𝐼𝐶𝑇 𝑠 = 2.789 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
∅

𝐼𝐶𝑇𝑠𝑒𝑐 = √3 (2.789)
• 𝑰𝑪𝑻𝒔𝒆𝒄 = 𝟒. 𝟖𝟑𝟏 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

REE – October 2000

37. The CT ratio and PT ratio used to protect a line are 240 and 2,000, respectively. If the
impedance of each line is 10 Ω, what is the relay impedance to protect the line from
fault?
A. 83.33 ohms B. 1.2 ohms C. 48, 000 ohms D. 12 ohms
Solution:
𝑃𝑇𝑟 2,000
𝑍𝑟 = =
𝐶𝑇𝑟 240
• 𝑍𝑟 = 8.33 𝑜ℎ𝑚𝑠
𝑍
𝑍𝑟 = 𝑍 𝑙𝑖𝑛𝑒
𝑟𝑒𝑙𝑎𝑦
𝑍 10
𝑟𝑒𝑙𝑎𝑦 = 8.33

• 𝒁𝒓𝒆𝒍𝒂𝒚 = 𝟏. 𝟐 𝒐𝒉𝒎𝒔
 38. Two transformers 1 and 2 are connected in parallel supplying a common load of 120 kVA.
Transformer 1 is rated 50 kVA, 7, 620/240-V single-phase and has an equivalent
impedance of 8.5 Ω while transformer 2 is rated 75 kVA, 7, 620/240-V single-phase and
has an equivalent impedance of 5.1 Ω. The two transformers operate with the same power
factors. What is the kVA load of each transformer?
A. 48 & 72 B. 45 & 75 C. 42 & 78 D. 40 & 80
Solution:
𝑍𝑒1 𝑆2
=
𝑍𝑒2 𝑆1
8.5 𝑆2
=
5.1 𝑆1
• 𝑆2 = 1.67𝑆1 → 𝑒𝑞′𝑛 1
• 𝑆𝐿 = 𝑆1 + 𝑆2 → 𝑒𝑞′𝑛 2
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑒𝑞’𝑛 1 𝑖𝑛 2
120 = 𝑆1 + 1.67𝑆1
• 𝑺𝟏 = 𝟒𝟒. 𝟗𝟒 𝒌𝑽𝑨
𝑆2 = 1.67(44.94)
• 𝑺𝟐 = 𝟕𝟓. 𝟎𝟓 𝒌𝑽𝑨

39. Two single-phase transformers are connected in parallel at no-load. One has a turns ratio
of 5, 000/440 and rating of 200 kVA, the other has a ratio of 5, 000/480 and rating of 350
kVA the leakage reactance of each is 3.5%. The no-load circulating current is .
A. 207 A B. 702 A C. 720 A D. 270 A
Solution:
2
𝑋 = (𝑉 𝑏𝑎𝑠𝑒1)
=
(440)2
𝑏𝑎𝑠𝑒1 𝑆𝑏𝑎𝑠𝑒1 200,000

• 𝑋𝑏𝑎𝑠𝑒1 = 0.968 Ω
(𝑉 )2 (480)2
𝑏𝑎𝑠𝑒2
𝑋𝑏𝑎𝑠𝑒2 = =
𝑆𝑏𝑎𝑠𝑒2 350,000

• 𝑋𝑏𝑎𝑠𝑒2 = 0.658 Ω
𝑋𝑒−𝑠1 = 𝑋𝑒𝑋𝑏𝑎𝑠𝑒1 = (0.035)(0.968)
• 𝑿𝒆−𝒔𝟏 = 𝟎. 𝟎𝟑𝟑𝟗 Ω
𝑋𝑒−𝑠2 = 𝑋𝑒𝑋𝑏𝑎𝑠𝑒2 = (0.035)(0.658)
• 𝑿𝒆−𝒔𝟐 = 𝟎. 𝟎𝟐𝟑 Ω
REE – October 1997
40. A power transformer rated 50, 000 kVA, 34.5/13.8 kV is connected Y-grounded primary
and delta on the secondary. Determine the full load phase current at the secondary side.
A. 2, 092 A B. 1, 725 A C. 1, 449 A D. 1, 208 A
Solution:
𝑆 = √3 𝑉𝐿𝐼𝐿
50,000,000
𝐼𝐿 𝑝 = (34,500)(√3)
• 𝐼𝐿 𝑝 = 836.74 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐼𝐿𝑝 = 𝐼∅𝑝
𝑉𝑝 𝐼𝑠
=
𝑉𝑠 𝐼𝑝
(34,500)(836.74)
𝐼∅𝑠 = 13,800 (√3)
• 𝑰𝒔∅ = 𝟏, 𝟐𝟎𝟕. 𝟕𝟑 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
50,000,000
𝐼𝐿𝑠 = (√3) (13,800)
• 𝐼𝐿𝑠 = 2, 091.849 𝐴𝑚𝑝𝑒𝑟𝑒𝑠
𝐼 =
𝐼𝐿 𝑠
=
2,091.849
∅𝑠 √3 √3
• 𝑰∅𝒔 = 𝟏, 𝟐𝟎𝟕. 𝟕𝟑 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

REE – April 2006

41. A 2, 000 kW, 2, 400-V, 75% p.f load is to be supplied from a 34, 5000-V, 3-phase line
through a single bank of transformers. Give the primary and secondary line currents in
amperes for the wye-wye connections.
A. 50/700 B. 48/650 C. 60/800 D. 45/642
Solution:
𝑌 − ∆ 𝐶𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛
2 𝑀𝑊 ∠𝑐𝑜𝑠−1 ∗
0.75 (0.75)
𝐼𝐿 𝑝 =[ ]
√3 (34,500)
𝐼𝐿𝑝 = 44.626∠44.41°

• 𝑰𝑳𝒑 = 𝟒𝟒. 𝟔𝟐𝟔 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

2 𝑀𝑊 ∠𝑐𝑜𝑠 −1 ∗
0.75 (0.75)
𝐼𝐿 𝑠 =[ ]
√3 (2,400)
𝐼𝐿𝑠 = 641.5∠44.41°
• 𝑰𝑳𝒔 = 𝟔𝟒𝟏. 𝟓 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
REE – April 2005
42. A 3, 000 kVA, 2, 400 V, 75% power factor load is to be supplied from a 34, 500-V,
three-phase line through a single bank of transformers. What is the voltage rating of each
transformer if the connection is wye-wye?
A. 20, 000/1, 380 B. 18, 500/1, 350 C. 18, 000/1, 850 D. 19, 000/1, 350
Solution:
34,500
𝑉∅𝑝 =
√3
• 𝑽∅𝒑 = 𝟏𝟗, 𝟗𝟏𝟖. 𝟓𝟖𝟒 𝑽𝒐𝒍𝒕𝒔
2,400
𝑉∅𝑠 = √3
• 𝑽∅𝒔 = 𝟏, 𝟑𝟖𝟓. 𝟔𝟒𝟏 𝑽𝒐𝒍𝒕𝒔

REE – March 1998

43. A 13.8 kV/480 V, 10 MVA three-phase transformer has 5% impedance. What is the
impedance in ohms referred to the primary?
A. 0.952 ohm B. 0.03 ohm C. 5.125 ohm D. 9.01 ohm
Solution:
𝑍𝑎𝑐𝑡𝑢𝑎𝑙
𝑍𝑝𝑢 = 𝑍𝑏
(𝑉𝑏)2 (13,800)2
𝑍𝑏 = =
𝑆𝑏 10,000,000
• 𝑍𝑏 = 19.044 𝑜ℎ𝑚𝑠
𝑍𝑒𝑝 = 𝑍𝑝𝑢𝑍𝑏 = (0.05)(19.044)
• 𝒁𝒆𝒑 = 𝟎. 𝟗𝟓𝟐 𝒐𝒉𝒎𝒔

REE – May 2009

44. A three-phase transformer is rated 15 MVA, 69/13.2 kV has a series impedance of 5%.
What is the new per unit impedance if the system study requires a 100 MVA base and 67
kV base?
A. 0.354 B. 0.347 C. 0.372 D. 0.333
Solution
2
𝑍 =𝑍 ( 𝑆𝑛 ) ( 𝑉𝑜 )
𝑝𝑢𝑛𝑒𝑤 𝑝𝑢𝑜𝑙𝑑 𝑆𝑜 𝑉𝑁
100 69 2
𝑍 = (0.05) ( )( )
𝑝𝑢𝑛𝑒𝑤 15 67
• 𝒁𝒑𝒖𝒏𝒆𝒘 = 𝟎. 𝟑𝟓𝟒
REE – April 2004
45. A transformer rated 2, 000 kVA, 34, 500/240 volts has 5.75% impedance. What is the per
unit impedance?
A. 0.0635 B. 0.0656 C. 0.0575 D. 34.2
Solution:
𝑍𝑝𝑢 = 57.5%
• ∴ 𝑍𝑝𝑢 = 0.0575

46. Three 5:1 transformers are connected in delta-wye to step up the voltage at the beginning
of a 13, 200-Volt three-phase transmission line. Calculate the line voltage on the high
side of the transformers.
A. 114, 300 V B. 66, 000 V C. 132, 000 V D. 198, 000 V
Solution:
5 𝑉∅2 𝑉∅2
= =
1 𝑉∅1 13,200 𝑉
𝑉∅2 = 𝑎𝑉𝑝
𝑉∅2 = (5)(13, 200)
• 𝑉∅2 = 66, 000 𝑉𝑜𝑙𝑡𝑠
𝑉𝐿 = √3 𝑉∅2 = √3 (66, 000)
• 𝑽𝑳 = 𝟏𝟏𝟒, 𝟑𝟏𝟓. 𝟑𝟓𝟑 𝑽𝒐𝒍𝒕𝒔

47. A 150 kVA, 2, 400/480-V, three-phase transformer with an equivalent impedance of 4%is
connected to an infinite bus and without load. If a three-phase fault occurs at the
secondary terminals, the fault current in amperes is .
A. 4, 512 A B. 3, 908 A C. 7, 815 A D. 1, 504 A
Solution:
3 ∅ 𝑓𝑎𝑢𝑙𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙𝑠
𝑆𝐵
𝐼𝐹 3∅ =
√3 𝑉𝐵𝑋𝑝𝑢
150,000
𝐼𝐹 3∅ = √3 (480)(0.04)
• 𝑰𝑭𝟑∅ = 𝟒, 𝟓𝟏𝟎. 𝟓𝟒𝟗 𝑨𝒎𝒑𝒆𝒓𝒆𝒔
48. Transformer 1 is in parallel with Transformer 2
Transformer 1 Transformer 2
150 kVA, single-phase 300 kVA, single-phase
6, 600/240 V 6, 600/240 V
𝑍𝑒−𝑠1 = 0.02425∠62.9° Ω 𝑍𝑒−𝑠2 = 0.01067∠62.9° Ω
Determine the maximum kVA load the bank can carry without overloading any of the two transformers,
assuming that the two transformers operate at the same power factors.
A. 450 kVA B. 432 kVA C. 420 kVA D. 412 kVA
Solution:
𝑆1 + 𝑆2 = 𝑆𝑙𝑜𝑎𝑑
𝐶𝑎𝑠𝑒 𝐼:
𝑇1 @ 150 𝑘𝑉𝐴
𝑍𝑒−𝑠1 𝑆2
=
𝑍𝑒−𝑠2 𝑆1
𝑍𝑒−𝑠1 0.02425∠62.9°
= 0.01067∠62.9°
𝑍𝑒−𝑠2
• 𝑍𝑒−𝑠1 = 2.273
𝑍𝑒−𝑠2
𝑆2
2.273 =
𝑆1
𝑆2 = 2.273 (150, 000)
• 𝑆2 = 340.95 𝑘𝑉𝐴
• ∴ 𝑜𝑣𝑒𝑟𝑙𝑜𝑎𝑑 𝑎𝑡 𝑇2
𝐶𝑎𝑠𝑒 𝐼𝐼:
𝑇2 @ 300 𝑘𝑉𝐴
300,000
𝑆1 = 2.273
• 𝑆1 = 131.984 𝑘𝑉𝐴
𝑆𝑙𝑜𝑎𝑑 = 𝑆1 + 𝑆2 = (131.984) + (300)
• 𝑺𝒍𝒐𝒂𝒅 = 𝟒𝟑𝟏. 𝟗𝟖𝟏 𝒌𝑽𝑨