# Sheet 2

```Sheet 2
Example 1: A 20 kVA, 2500/250 V, 50Hz, single-phase transformer gave the
following test result:
Open-circuit test: 250V, 1.5A, and 100 watts measured on the low voltage side
Short-circuit test: 100V, 8A, and 320 watts measured on the on high voltage side
(a) Compute the parameters of the equivalent circuits and draw the equivalent circuit
of the transformer showing all the values.
Solution: From the given test results data, we have
Req1 
Psc 320
 2 =5Ω
I sc2
8
2
V 
X eq1   sc   52 = 12 Ω
 I sc 
Reh 
Voc2 2502
=12 Ω

Poc 100
2
V 
X m   oc   122 =173Ω
 I oc 
Re+h and Xm are low-voltage side values, which can be converted to the high voltage
side
2
Reh
N 
 Reh *  1   62.5 k Ω
 N1 
2
N 
X m  X m *  1   17.3 k Ω
 N1 
Example 2: A 10 kVA, 2400/240 V, single-phase transformer have the following
parameters: R1 = 3Ω, R2 = 15 Ω the leakage reactance are X1 = 0.03 Ω and X2 = 0.15
Ω. Calculate the full-load voltage regulation of the transformer when operated at rated
secondary voltage and (a) 0.8 power factor lagging and (b) 0.8 power factor leading.
Solution:
R eq2 = R2 + R1' = R2 + R1 / α 2 =
= 0.03 + 3/ (10)2 = 0.06 Ω
X eq2 = X2 + X1' = X2 + X1 / α 2 =
= 0.15 + 15/ (10)2 = 0.3 Ω
Zeq2 = Req2 + j X eq2 = 0.06 + j 0.3 Ω = 0.3 78.7 0
(a) For a 0.8 power factor, lagging
I2 
S
= 41.7 A
V2
Referring to Fig. 10.17
V1
 V2  I 2 Z eq 2  240 0  41.7  37 0 X 0.3 78.7 0

= 249.5 + j 8.5 = 249.65 1.950
V1
 V2
249.65  240

VR% =
X 100 =
X 100 = 4%
240
V2
(b) For a 0.8 power factor, leading
I 2 = 41.7   37 0 A
V1

 V2  I 2 Z eq 2  240 0  41.7  37 0 X 0.3 78.7 0
= 234.5 + j 11.5 = 235 2.80
249.65  240
VR% =
* 100 = -2.18%
240
Practical Problems
1- A 50 kVA, 4400/220 V transformer has R1 = 3.45 Ω, R2 = 0.009 Ω. The
values of reactances are X1 = 5.2 Ω and X2= 0.015 Ω. Calculate for the
transformer :
a.
b.
c.
d.
e.
Equivalent resistance as referred to primary
Equivalent resistance as referred to secondary
Equivalent reactance as referred to both primary and secondary,
Equivalent impedance as referred to both primary and secondary, and
Total copper loss, first using individual resistances of the two windings and
secondly, using equivalent resistance as referred to each side.
2. The primary and secondary winding resistance of a 40 kVA, 6600/250 V singlephase transformer are 10 Ω and 0.05 Ω respectively. The equivalent leakage reactance
as referred to the primary winding is 35 Ω. Find the full-load regulation for load
power factors of
a. unity ;
b. 0.8 lagging ; and
3. A 20 kVA, 2200/220 V 50-Hz distribution transformer is tested for efficiency
and regulation as follows;
d.
e.
f.
g.
h.
i.
j.
k.
l.
O.C. test (L.V. side) : 220 V, 4.2 A, 148 W.
S.C. test (H.V. side) : 86 V, 10.5 A, 360 W.
Determine:
(i) Core loss
(ii) Equivalent resistance referred to primary.
(iii)Equivalent resistance referred to secondary.
(iv) Equivalent reactance referred to primary.
(v) Equivalent reactance referred to secondary.
(vi) Regulation of transformer at 0.8 power factor lagging current.
```