Determining the Valnes of Components in the Transformer Model
The values of the resistances and inductances in the transformer equivalent circuis can
be found by the open-circuit test and the short-circuit test.
The open-circuit test: in this test, one of the transformer winding is open-circuited,
and the other winding is connected to full rated supply voltage, see the circuit below.
As seen from the transformer equivalent circuit, the input current flows through the
excitation branch of the transformer. The series elements, R P and X P are too small
in comparison to RC and X M , so essentially all the input voltage is dropped across
the excitation branch.
The open-circuit test connections are shown in the following figure.
Usually, full line voltage is applied to one side of the transformer, and the input
voltage, input current, and input power to the transformer are measured.
From the figure, we conclude that,:
- the magnitude of conductance of the excitation branch is:
I
YM  OC
VOC
- The power factor angle of magnetization branch is:
 P

 OC  cos 1  OC 
 VOC .I OC 
Notice that the power factor is always lagging for a real transformer, so the angle of
the current always lags the angle of the voltage by  degrees.
We know that the conductance of the excitation branch is:
1
1
YM 
j
 GC  jBM
RC
XM
From this, we find:
I
1
 GC  Re{YM }  YM . cos  OC  OC cos  OC
RC
VOC
And:
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department
I
1
 BM   Im{YM }   YM . sin  OC   OC sin  OC
XM
VOC
The short-circuit test: In the short-circuit test, the low-voltage terminals of the
transformer are short-circuited, and the high-voltage tenninals are connected to a
variable voltage source, as shown below.
The input voltage is adjusted until the current in the short-circuited windings is equal
to its rated value. The input voltage, current and power are again measured.
Since the input voltage is so low during the short-circuit test, negligible current flows
through the excitation branch. If the excitation current is ignored, then all the voltage
drop in the transfonner can be attributed to the series elements in the circuit.
The magnitude of the series impedances referred to the primary side of the
transformer is:
V
Z CS  SC
I SC
- The power factor angle of circuit in the short-circuit test is:
 P

 SC  cos 1  SC 
 VSC .I SC 
Notice that the power factor is always lagging for a real transformer, so the angle of
the current always lags the angle of the voltage by  degrees.
We know that, in the short-circuit test:
Z CS  Z P  a 2 Z S  RP  a 2 RS   j X P  a 2 X S 
Thus:
V
(1)
RP  a 2 RS  Re{Z SC }  Z SC . cos  SC  SC cos  SC
I SC
And:
V
(2)
X P  a 2 X S  Im{ Z SC }  Z SC . sin  SC  SC sin  SC
I SC
For some simplification, the resistance of the primary winding , R P , can be estimated
by measuring it by an ohmmeter while the seconary winding is open-circuited. After
finding R P , the resistance of the secondary winding refered to primary,
RS'  a 2 RS can be found from (1), as the transformation ratio a is usually knowown,
or easily can be found, the resistance RS is found.
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department
As regard to the reactances of the primary and secondary windings, and for practical
transformers, X P and X S'  a 2 X S are realively equal, as each of them can easily be
found using (2).
Example:
Build the equivalent circuit of a 20kVA , 8000 / 240V , 60Hz transformer, knowing
that the open-circuit test was performed on the secondary side of the transformer (to
reduce the maximum voltage to be measured) and the short-circuit test were
performed on the primary side of the transformer (to reduce the maximum current to
be measured). The following data were taken:
Open-circuit test
(on secondary)
VOC  240V
I OC  7.133 A
POC  400W
Short-circuit test
(on primary)
VSC  489V
I SC  2.5 A
PSC  240W
Solution:
The transformation ratio of the transformer is:
V
8000
a P 
 33.33
VS
240
From the open-circuit test:
The power factor angle of magnetization branch is:
 P

400


0
 OC  cos 1  OC   cos 1 
  76.5
 240 x7.133 
 VOC .I OC 
The magnitude of conductance of the excitation branch is:
I
7.133
YM  OC 
 0.0297
VOC
240



YM  YM  OC  0.0297  76.5 0  0.0297 cos 76.5 0  j 0.0297 sin 76.5 0

 0.00693  j 0.02888
But:
YM 
1
1
j
RC
XM
Thus:
1
1
 144
 0.00693 and RC 
0.00693
RC
1
1
 34.63
 0.02888 and X M 
0.02888
XM
From the short-circuit test:
- The power factor angle is:
 240 
0
 SC  cos 1 
  78.7
 489 x 2.5 
The magnitude of the series impedances is:
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department
Z CS 
VSC 489

 195.6
I SC
2.5



Z CS  Z CS  SC  195.678.7 0  195.6 cos 78.7 0  j195.6 sin 78.7 0

 38.4  j192 
But:
Z CS  Z P  a 2 Z S  RP  a 2 RS  j X P  a 2 X S  Req  jX eq
Thus:
Req  38.4 and X eq  192

 

Therefore, the equivalent resistance and reactance referred to the high-voltage
(primary) side are:
Req  38.4 and X eq  192
The resulting simplified equivalent circuit referred to the high-voltage (primary) side
can be found by converting the excitation branch values to the high-voltage side.
Or:
2
RC , P  a 2 RC ,S  33.33 144  159k
X M , P  a 2 X M ,S  33.33 34.63  38.4k
The resulting equivalent circuit is shown in the figure below.
2
THE PER-UNIT SYSTEM OF MEASUREMENTS
Circuits containing transformers can be solved easily using the per-unit (pu) system of
measurements. This method provides the following advantages:
1- per unit data provides relative magnitude information.
2- simplifies the analysis of circuits containing transformers. The per unit
impedance of a transformer in three-phase system is the same, regardless of
the type of the winding connections (star-delta, delta-star, stat-star, or deltadelta). The per unit impedance of a transformer is the same when referred to
either the primary or the secondary side,
3- circuit parameter are in narrow ranges (around 1 p.u) and this makes error
checking easier.
4- Manufacturers usually specify the impedance of an equipment in per unit on
the base of its nameplate rating.
5- The per unit impedance values of various ratings of equipment lie in a narrow
range, (whereas the actual ohmic values may vary widely). Therefore when
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department
actual values are unknown, a good approximate (typical) value of similar
equipment can be used.
This method is suitable for systems with different voltage levels (sections connected
with transformers).
In this method base values are selected, and all the actual values are referred to their
respective base values, as:
actual value
value in p.u 
base value
normally two base values, base VA, VAB and the base voltage, VB are chosen.
then for each section VB is found from the transformers ratios. then the base current
I B and the base impedance Z B are calculated for each section. the p.u values then are
found.
For the case of single phase systems:
- select the base VA, S base
- select Vbase and calculate it for each section.
- the base current for each section is calculated as:
S
I B  base
Vbase
- the base impedance for each section is calculated as:
V 2
V
- Z B  base  base
I base
S base
Example:
A simple power system shown below contains a 480V generator connected to an
ideal 1 : 10 step-up transformer, a transmission line, an ideal 20 : 1 step-down
transformer, and a load. The impedance of the transmission line is 20  j 60  and the
impedance of the load is 1030 0  . The base values for this system are chosen to
be 480V and 10kVA at the generator.
a) Find the base voltage, current, impedance, and apparent power at every point in the
power system.
b) Convert this system to its per-unit equivalent circuit.
c) Find the power supplied to the load in this system.
d) Find the power lost in the transmission line.
Solution:
a)
In the generator region: S base  10kVA , Vbase1  480V , thus:
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department
I base1 
S base 10000

 20.83 A
Vbase1
480
Vbase1
480

 23.04
I base1 20.83
In the transmission line region: S base  10kVA , Vbase2  4800V , thus:
S
10000
I base2  base 
 2.083 A
Vbase2
4800
Z base1 
Vbase2 4800

 2304
I base2 2.083
In the load region: S base  10kVA , Vbase3  240V , thus:
S
10000
I base3  base 
 41.67 A
Vbase3
240
Z base2 
Vbase3
240

 5.76
I base3 41.67
b) the per-unit values of the circuit parameters are:
the generator voltage in p.u:
VG ,actual 480
VG. pu 

 1.0 p.u
Vbase1
480
the impedance of the transmission line in p.u:
Z Line,actual 20  j 60
Z Line. pu 

 0.0087  j 0.026 p.u
Z base2
2304
the impedance of the load in p.u:
Z Load ,actual 1030 0
Z Load . pu 

 1.73630 0 p.u  1.503  j 0.868 p.u
Z base3
5.76
The per-unit equivalent circuit of the power system is shown below.
Z base3 
c) From this circuit:
I G. pu  I Line. pu  I Load . pu
VG , pu
10 0

 0.569  30.6 0 p.u
0
0.0087  j 0.026  1.73630
Z line. pu  Z load. pu
The per-unit power of the load is:
2
2
PLoad . pu  I Load . pu  .RLoad , p.u  0.569 1.503  0.487 p.u  0.487 x10000  4870W
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department
d) The per-unit power loss in the transmission line is:
2
2
PLine. pu  I Line. pu  .RLine, p.u  0.569 0.0087  0.00282 p.u  0.00282 x10000  28.2W
Yasin A. Al Shiboul,
Al Balqa Applied University,
Faculty of Engineering Technology,
EE Department