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Chapter 3 : Transformers
3-1. The secondary winding of a transformer has a terminal voltage of v (^) s ( ) t = 282 8 sin 377 V. t. The turns
ratio of the transformer is 50:200 ( a = 0.25). If the secondary current of the transformer is
i s ( ) t = 7 07. sin (^377 t − 3687. °)A , what is the primary current of this transformer? What are its voltage
regulation and efficiency? The impedances of this transformer referred to the primary side are
R eq = 0 05. Ω RC = 75 Ω
X eq = 0 225. Ω X (^) M = 20 Ω
S OLUTION The equivalent circuit of this transformer is shown below. (Since no particular equivalent circuit was specified, we are using the approximate equivalent circuit referred to the primary side.)
The secondary voltage and current are
0 V 200 0 V
V S = ∠° = ∠ °
36.87 A 5 36.87 A
I S = ∠ − ° = ∠ − °
The secondary voltage referred to the primary side is
= = 50 ∠ 0 °V
V S a V S
The secondary current referred to the primary side is
= = 20 ∠- 36. 87 °A
a
S S
I
I
The primary circuit voltage is given by
P S S^ (^ R eq + jX eq)
V = V I
V P = 50 ∠ 0 °V+ ( 20 ∠− 36. 87 °A)( 0. 05 Ω+ j 0. 225 Ω) = 53. 6 ∠ 3. 2 °V
The excitation current of this transformer is
53. 6 3. 2 V
53. 6 3. 2 V
EX j
I I C I M
I EX= 2. 77 ∠− 71. 9 °
Therefore, the total primary current of this transformer is
+ EX = 20 ∠− 36. 87 °+ 2. 77 ∠− 71. 9 °= 22. 3 ∠− 41. 0 °A
I P = I S I
The voltage regulation of the transformer at this load is
VR 100 % × =
× =
S
P S aV
V aV
The input power to this transformer is
P IN = V IP P cos θ= ( 53.6 V ) ( 22.3 A cos) ª¬ 3.2 ° − −( 41.0 °) º¼
P IN =^ ( 53. 6 V)(^22. 3 A)^ cos44.2°= 857 W
The output power from this transformer is
P OUT = VS IS cosθ= ( 2 00 V)( 5 A) cos( 36. 87 °) = 800 W
Therefore, the transformer’s efficiency is
857 W
800 W
IN
= OUT^ × = × =
P
P
3-2. A 20-kVA 8000/277-V distribution transformer has the following resistances and reactances:
R P = 32 Ω RS = 0. 05 Ω
X P = 45 Ω X S = 0.06Ω
RC = 250 k Ω XM = 30 kΩ
The excitation branch impedances are given referred to the high-voltage side of the transformer.
(a) Find the equivalent circuit of this transformer referred to the high-voltage side.
(b) Find the per-unit equivalent circuit of this transformer.
(c) Assume that this transformer is supplying rated load at 277 V and 0.8 PF lagging. What is this transformer’s input voltage? What is its voltage regulation?
(d) What is the transformer’s efficiency under the conditions of part (c)?
S OLUTION
(a) The turns ratio of this transformer is a = 8000/277 = 28.89. Therefore, the secondary impedances referred to the primary side are
2 2 RS aR S
2 2 XS a X S
Therefore, the primary voltage on the transformer is
( )
V P (^) = V S R (^) EQ jX EQ I S
V P = 8000 ∠ 0 °V+ ( 73. 7 + j 95. 1 ) ( 2.50∠-36.87°A) = 8290 ∠ 0. 55 °V
The voltage regulation of the transformer under these conditions is
VR = × =
(d) Under the conditions of part (c) , the transformer’s output power copper losses and core losses are:
P OUT = S cosθ= ( 20 kVA)( 0. 8 ) = 16 kW
( 2. 5 ) ( 73. 7 ) 461 W
2 EQ
2
CU ¸ = = ¹
P = §^ I ′ R
S
275 W
2
2
core = =
C
S R
V
P
The efficiency of this transformer is
OUT CU core
OUT × =
× =
P P P
P
3-3. A 2000-VA 230/115-V transformer has been tested to determine its equivalent circuit. The results of the
tests are shown below.
Open-circuit test Short-circuit test V (^) OC = 230 V V (^) SC = 13.2 V I (^) OC = 0.45 A I (^) SC = 6.0 A P (^) OC = 30 W P (^) SC = 20.1 W
All data given were taken from the primary side of the transformer.
(a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer.
(b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0. PF leading.
(c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging.
S OLUTION
(a) OPEN CIRCUIT TEST:
230 V
0. 45 A
Y EX (^) = GC − jBM = = S
− −
- 15 230 V 0. 45 A
30 W
cos cos
1
OC OC
(^1) OC
V I
P
Y EX (^) = GC − jBM = 0.001957 ∠ − 73.15 ° S = 0.000567- 0.001873 S j
C
C G
R
M
M B
X
SHORT CIRCUIT TEST:
6.0A
13.2V
Z EQ = R EQ+ jX EQ = = Ω
− −
- 3 13.2V 6 A
2 0.1W
cos cos
1
SCSC
(^1) SC
V I
P
Z EQ = R EQ+ jX EQ= 2. 20 ∠ 75. 3 °Ω= 0. 558 + j 2. 128 Ω
R EQ= 0. 558 Ω
X EQ = j 2. 128 Ω
To convert the equivalent circuit to the secondary side, divide each impedance by the square of the turns
ratio ( a = 230/115 = 2). The resulting equivalent circuit is shown below:
R EQ, (^) S = 0.140Ω X (^) EQ, S = j 0.532Ω
RC S (^) , = 441 Ω X (^) M S , = 134 Ω
(b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred
to the secondary side. The rated secondary current is
8. 70 A
115 V
1000 VA
IS = =
We will now calculate the primary voltage referred to the secondary side and use the voltage regulation
equation for each power factor.
(1) 0.8 PF Lagging:
= + EQ = 115 ∠ 0 °V+ ( 0.140+ 0.532Ω)(8.7 ∠ 3687 °A)
V P V S Z I S j -.
= 118. 8 ∠ 1. 4 ° V
V P
VR = × =
(2) 1.0 PF:
= + EQ = 115 ∠ 0 °V+ ( 0.140+ 0.532Ω)(8.7 ∠ 0 °A)
V P V S Z I S j
= 116. 3 ∠ 2. 28 ° V
V P
The secondary current I (^) S is given by
43. 48 A
2300 V 0. 9
90 kW IS = =
I S = 43. 48 ∠− 25. 8 ° A
(a) The voltage at the power source of this system (referred to the secondary side) is
V source (^) V S I SZ (^) line + I SZ EQ
source 2300 0 V^ (^ 43.48^ 25.8^ A) (^ 1.12^ j 4.11^^ )^ (^ 43.48^ 25.8^ A) (^ 0.12^ j 0.5 )
V = ∠ ° + ∠ − ° + Ω + ∠ − ° + Ω
source =^2441 ∠^3.^7 °^ V
V
Therefore, the voltage at the power source is
( ) 14. 24 3. 7 kV
2.4kV
14 kV V source = 2441 ∠ 3. 7 °V = ∠ °
(b) To find the voltage regulation of the transformer, we must find the voltage at the primary side of the transformer (referred to the secondary side) under full load conditions:
V P = V S + I SZ EQ
= 2300 ∠ 0 °V+( 43. 48 ∠− 25. 8 °A)( 0. 12 + 0. 5 Ω) = 2314 ∠ 0. 43 °V
V P j
There is a voltage drop of 14 V under these load conditions. Therefore the voltage regulation of the transformer is
VR = × =
(c) The power supplied to the load is P OUT = 90 kW. The power supplied by the source is
IN source cos =^ (^2441 V)(^43.^48 A)^ cos29.5°=^92.^37 kW
P = V IS θ
Therefore, the efficiency of the power system is
92.37kW
90 kW 100 % IN
= OUT^ × = × =
P
P
3-5. When travelers from the USA and Canada visit Europe, they encounter a different power distribution
system. Wall voltages in North America are 120 V rms at 60 Hz, while typical wall voltages in Europe are 220-240 V at 50 Hz. Many travelers carry small step-up / step-down transformers so that they can use their appliances in the countries that they are visiting. A typical transformer might be rated at 1-kVA and 120/240 V. It has 500
1 turns of wire on the 120-V side and 1000 turns of wire on the 240-V side. The magnetization curve for this transformer is shown in Figure P3-2, and can be found in file p32.mag at this book’s Web site.
1 Note that this turns ratio was backwards in the first printing of the text. This error should be corrected in all
subsequent printings.
(a) Suppose that this transformer is connected to a 120-V, 60 Hz power source with no load connected to the 240-V side. Sketch the magnetization current that would flow in the transformer. (Use MATLAB to plot the current accurately, if it is available.) What is the rms amplitude of the magnetization current? What percentage of full-load current is the magnetization current?
(b) Now suppose that this transformer is connected to a 240-V, 50 Hz power source with no load connected to the 120-V side. Sketch the magnetization current that would flow in the transformer. (Use MATLAB to plot the current accurately, if it is available.) What is the rms amplitude of the magnetization current? What percentage of full-load current is the magnetization current?
(c) In which case is the magnetization current a higher percentage of full-load current? Why?
S OLUTION
(a) When this transformer is connected to a 120-V 60 Hz source, the flux in the core will be given by
the equation
( ) cos t N
V
t P
M
The magnetization current required for any given flux level can be found from Figure P3-2, or alternately
from the equivalent table in file p32.mag. The MATLAB program shown below calculates the flux
level at each time, the corresponding magnetization current, and the rms value of the magnetization
current.
% M-file: prob3_5a.m
% M-file to calculate and plot the magnetization
% current of a 120/240 transformer operating at
The rms magnetization current is 0.318 A. Since the full-load current is 1000 VA / 120 V = 8.33 A, the
magnetization current is 3.82% of the full-load current. The resulting plot is
(b) When this transformer is connected to a 240-V 50 Hz source, the flux in the core will be given by
the equation
( ) cos t N
V
t P
M ω
The magnetization current required for any given flux level can be found from Figure P3-2, or alternately
from the equivalent table in file p32.mag. The MATLAB program shown below calculates the flux
level at each time, the corresponding magnetization current, and the rms value of the magnetization
current.
% M-file: prob3_5b.m
% M-file to calculate and plot the magnetization
% current of a 120/240 transformer operating at
% 240 volts and 50 Hz. This program also
% calculates the rms value of the mag. current.
% Load the magnetization curve. It is in two
% columns, with the first column being mmf and
% the second column being flux.
load p32.mag;
mmf_data = p32(:,1);
flux_data = p32(:,2);
% Initialize values
S = 1000; % Apparent power (VA)
Vrms = 240; % Rms voltage (V)
VM = Vrms * sqrt(2); % Max voltage (V)
NP = 1000; % Primary turns
% Calculate angular velocity for 50 Hz
freq = 50; % Freq (Hz)
w = 2 * pi * freq;
% Calculate flux versus time
time = 0:1/2500:1/25; % 0 to 1/25 sec
flux = -VM/(wNP) * cos(w . time);**
% Calculate the mmf corresponding to a given flux
% using the MATLAB interpolation function.
mmf = interp1(flux_data,mmf_data,flux);
% Calculate the magnetization current
im = mmf / NP;
% Calculate the rms value of the current
irms = sqrt(sum(im.^2)/length(im));
disp(['The rms current at 50 Hz is ', num2str(irms)]);
% Calculate the full-load current
i_fl = S / Vrms;
% Calculate the percentage of full-load current
percnt = irms / i_fl * 100;
disp(['The magnetization current is ' num2str(percnt) ...
'% of full-load current.']);
% Plot the magnetization current.
figure(1);
plot(time,im);
title ('\bfMagnetization Current at 240 V and 50 Hz');
xlabel ('\bfTime (s)');
ylabel ('\bf\itI_{m} \rm(A)');
axis([0 0.04 -0.5 0.5]);
grid on;
When this program is executed, the results are
» prob3_5b
The rms current at 50 Hz is 0.
The magnetization current is 5.5134% of full-load current.
The rms magnetization current is 0.318 A. Since the full-load current is 1000 VA / 240 V = 4.17 A, the
magnetization current is 5.51% of the full-load current. The resulting plot is shown below.
The voltage regulation is
VR = × =
(b) As before, the easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is again a = 34.78. Thus the load impedance referred to the primary side is
2 Z (^) L j j
The referred secondary current is
2. 025 88. 8 A
7967 0 V
7967 0 V
j j
I S
and the referred secondary voltage is
S S Z^ L^ (^ 2.25^ 88.8^ A) (^ j^4234 )^8573 1.2^ V
V = I = ∠ ° − Ω = ∠ − °
The actual secondary voltage is thus
246. 5 1. 2 V
8573 1. 2 V
a
S S
V
V
The voltage regulation is
VR = × =−
3-7. A 5000-kVA 230/13.8-kV single-phase power transformer has a per-unit resistance of 1 percent and a
per-unit reactance of 5 percent (data taken from the transformer’s nameplate). The open-circuit test performed on the low-voltage side of the transformer yielded the following data:
V OC = 138. kV I (^) OC = 15. 1 A P OC = 44. 9 kW
(a) Find the equivalent circuit referred to the low-voltage side of this transformer.
(b) If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging, find the voltage regulation of the transformer. Find its efficiency.
S OLUTION
(a) The open-circuit test was performed on the low-voltage side of the transformer, so it can be used to directly find the components of the excitation branch relative to the low-voltage side.
EX
15.1 A
13.8 kV
Y = G C − jBM = =
− −
- 56 13.8kV 1 5.1A
4 4.9kW cos cos
1
OC OC
(^1) OC
V I
P
Y EX (^) = GC − jBM = 0.0010942 ∠ − 77.56 ° S = 0.0002358 − j 0.0010685 S
C
C G
R
M
M B
X
The base impedance of this transformer referred to the secondary side is
5000 kVA
- 8 kV
2
base
2 base base S
V
Z
so R EQ = ( 0. 01 )( 38. 09 Ω) = 0. 38 Ω and X EQ= (^0. 05 )(^38. 09 Ω)^ = 1. 9 Ω. The resulting equivalent
circuit is shown below:
R eq,s (^) = 0.38Ω X (^) eq,s = j 1.9Ω
RC (^) , s = 4240 Ω XM , s = 936 Ω
(b) If the load on the secondary side of the transformer is 4000 kW at 0.8 PF lagging and the secondary voltage is 13.8 kV, the secondary current is
362. 3 A
13.8kV 0. 8
4000 kW
PF
= LOAD^ = =
S
S V
P
I
I S = 362. 3 ∠− 36. 87 ° A
The voltage on the primary side of the transformer (referred to the secondary side) is
V P = V S + I SZ EQ
P 13,800^0 V^ (^ 362.3^ 36.87^ A) (^ 0.38^ j 1.9^ ) 14,330^ 1.9^ V
V = ∠ ° + ∠ − ° + Ω = ∠ °
There is a voltage drop of 14 V under these load conditions. Therefore the voltage regulation of the transformer is
VR = × =
The transformer copper losses and core losses are
( 3 62.3 A) ( 0. 38 ) 49. 9 kW
2 EQ,
2 P CU = IS R S = Ω =
48. 4 W
1 4,330 V
2
2
core = Ω
C
P
R
V
P
Therefore the efficiency of this transformer at these conditions is
4000 W 49.9W 48.4W
4000 W
OUT CU core
OUT =
× =
P P P
P
3-8. A 150-MVA 15/200-kV single-phase power transformer has a per-unit resistance of 1.2 percent and a per-
unit reactance of 5 percent (data taken from the transformer’s nameplate). The magnetizing impedance is j 100 per unit.
(c) This problem is repetitive in nature, and is ideally suited for MATLAB. A program to calculate the
secondary voltage of the transformer as a function of load is shown below:
% M-file: prob3_8.m
% M-file to calculate and plot the secondary voltage
% of a transformer as a function of load for power
% factors of 0.8 lagging, 1.0, and 0.8 leading.
% These calculations are done using an equivalent
% circuit referred to the primary side.
% Define values for this transformer
VP = 15000; % Primary voltage (V)
amps = 0:125:12500; % Current values (A)
Req = 0.018; % Equivalent R (ohms)
Xeq = 0.075; % Equivalent X (ohms)
% Calculate the current values for the three
% power factors. The first row of I contains
% the lagging currents, the second row contains
% the unity currents, and the third row contains
% the leading currents.
I(1,:) = amps .* ( 0.8 - j*0.6); % Lagging
I(2,:) = amps .* ( 1.0 ); % Unity
I(3,:) = amps .* ( 0.8 + j*0.6); % Leading
% Calculate VS referred to the primary side
% for each current and power factor.
aVS = VP - (Req.I + j.Xeq.I);*
% Refer the secondary voltages back to the
% secondary side using the turns ratio.
VS = aVS * (200/15);
% Plot the secondary voltage versus load
plot(amps,abs(VS(1,:)),'b-','LineWidth',2.0);
hold on;
plot(amps,abs(VS(2,:)),'k--','LineWidth',2.0);
plot(amps,abs(VS(3,:)),'r-.','LineWidth',2.0);
title ('\bfSecondary Voltage Versus Load');
xlabel ('\bfLoad (A)');
ylabel ('\bfSecondary Voltage (%)');
legend('0.8 PF lagging','1.0 PF','0.8 PF leading');
grid on;
hold off;
The resulting plot of secondary voltage versus load is shown below:
3-9. A three-phase transformer bank is to handle 400 kVA and have a 34.5/13.8-kV voltage ratio. Find the
rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) if the transformer bank is connected to (a) Y-Y, (b) Y-Δ, (c) Δ-Y, (d) Δ-Δ.
S OLUTION For these four connections, the apparent power rating of each transformer is 1/3 of the total apparent power rating of the three-phase transformer.
The ratings for each transformer in the bank for each connection are given below:
Connection Primary Voltage Secondary Voltage Apparent Power Turns Ratio Y-Y 19.9 kV 7.97 kV 133 kVA 2.50: Y-Δ 19.9 kV^ 13.8 kV^ 133 kVA^ 1.44: Δ-Y 34.5 kV^ 7.97 kV^ 133 kVA^ 4.33: Δ-Δ 34.5 kV 13.8 kV 133 kVA 2.50:
3-10. A Y-connected of three identical 100-kVA 7967/ 277 -V
2 transformers is supplied with power directly from a large constant-voltage bus. In the short-circuit test, the recorded values on the high-voltage side for one of these transformers are
V SC = 560 V I SC = 12. 6 A P SC = 3300 W
(a) If this bank delivers a rated load at 0.88 PF lagging and rated voltage, what is the line-to-line voltage on the primary of the transformer bank?
(b) What is the voltage regulation under these conditions?
(c) Assume that the primary line voltage of this transformer bank is a constant 13.8 kV, and plot the secondary line voltage as a function of load current for currents from no-load to full-load. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading.
(d) Plot the voltage regulation of this transformer as a function of load current for currents from no-load to full-load. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading.
2 This voltage was misprinted as 7967/480-V in the first printing of the text. This error should be corrected in all
subsequent printings.
Note: It is much easier to solve problems of this sort in the per-unit system, as we shall see in the next problem.
(c) This sort of repetitive operation is best performed with MATLAB. A suitable MATLAB program is
shown below:
% M-file: prob3_10c.m
% M-file to calculate and plot the secondary voltage
% of a three-phase Y-Y transformer bank as a function
% of load for power factors of 0.85 lagging, 1.0,
% and 0.85 leading. These calculations are done using
% an equivalent circuit referred to the primary side.
% Define values for this transformer
VL = 13800; % Primary line voltage (V)
VPP = VL / sqrt(3); % Primary phase voltage (V)
amps = 0:0.04184:4.184; % Phase current values (A)
Req = 6.94; % Equivalent R (ohms)
Xeq = 24.7; % Equivalent X (ohms)
% Calculate the current values for the three
% power factors. The first row of I contains
% the lagging currents, the second row contains
% the unity currents, and the third row contains
% the leading currents.
re = 0.85;
im = sin(acos(re));
I(1,:) = amps .* ( re - j*im); % Lagging
I(2,:) = amps .* ( 1.0 ); % Unity
I(3,:) = amps .* ( re + j*im); % Leading
% Calculate secondary phase voltage referred
% to the primary side for each current and
% power factor.
aVSP = VPP - (Req.I + j.Xeq.*I);
% Refer the secondary phase voltages back to
% the secondary side using the turns ratio.
% Because this is a delta-connected secondary,
% this is also the line voltage.
VSP = aVSP * (277/7967);
% Convert secondary phase voltage to line
% voltage.
VSL = sqrt(3) * VSP;
% Plot the secondary voltage versus load
plot(amps,abs(VSL(1,:)),'b-','LineWidth',2.0);
hold on;
plot(amps,abs(VSL(2,:)),'k--','LineWidth',2.0);
plot(amps,abs(VSL(3,:)),'r-.','LineWidth',2.0);
title ('\bfSecondary Voltage Versus Load');
xlabel ('\bfLoad (A)');
ylabel ('\bfSecondary Voltage (V)');
legend('0.85 PF lagging','1.0 PF','0.85 PF leading');
grid on;
hold off;
The resulting plot is shown below:
(d) This sort of repetitive operation is best performed with MATLAB. A suitable MATLAB program is
shown below:
% M-file: prob3_10d.m
% M-file to calculate and plot the voltage regulation
% of a three-phase Y-Y transformer bank as a function
% of load for power factors of 0.85 lagging, 1.0,
% and 0.85 leading. These calculations are done
% using an equivalent circuit referred to the primary side.
% Define values for this transformer
VL = 13800; % Primary line voltage (V)
VPP = VL / sqrt(3); % Primary phase voltage (V)
amps = 0:0.04184:4.184; % Phase current values (A)
Req = 6.94; % Equivalent R (ohms)
Xeq = 24.7; % Equivalent X (ohms)
% Calculate the current values for the three
% power factors. The first row of I contains
% the lagging currents, the second row contains
% the unity currents, and the third row contains
% the leading currents.
re = 0.85;
im = sin(acos(re));
I(1,:) = amps .* ( re - j*im); % Lagging
I(2,:) = amps .* ( 1.0 ); % Unity
I(3,:) = amps .* ( re + j*im); % Leading
% Calculate secondary phase voltage referred
% to the primary side for each current and
% power factor.
