Examples on OC and SC
Test on Transformer
By : Dr. Atul R. Phadke
Associate Professor in Electrical Engineering
Government College of Engineering, Karad (Maharashtra)
EXAMPLE 1:
Open-circuit and short-circuit tests conducted on a 10 kVA, 500/2000 V, 50 Hz, single phase
transformer gave the following readings: Open circuit test on primary side: 500 V, 120 W, Short-circuit
test on primary side: 15 V, 20 A, 100 W. Determine (a) efficiency on full load unity power factor, (b)
secondary terminal voltage on full load 0.8 lagging and leading power factor.
Given data: kVA rating = 10, Vi = 500 V, V2 = 2000 V, OC test: Vi = 500 V, W0 = 120 W = 0.12 kW, SC
test: VSC = 15 V, ISC = 20 A, WSC = 100 W = 0.1 kW
Iron loss Wi = Wo = 0.12 kW
Full load current on primary side =
𝑘𝑉𝐴×1000
𝑉1
=
10 ×1000
500
= 20 𝐴
𝐼𝑆𝐶 = 𝐼𝐹𝐿 = 20 𝐴
 Copper loss at full load, WC = 0.1 kW
𝑥𝑘𝑉𝐴𝑐𝑜𝑠∅
% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑡 𝑥 % 𝑙𝑜𝑎𝑑, 𝑥 =
× 100
𝑥𝑘𝑉𝐴𝑐𝑜𝑠∅ + 𝑊𝑖 + 𝑥 2 𝑊𝑐
1 × 10 × 1
∴ 𝐹𝐿 =
× 100 = 97.85%
1 × 10 × 1 + 0.12 + (1)2 × 0.1
𝑅01 =
𝑊𝑆𝐶
2
𝐼𝑆𝐶
 𝑅01 =

100
202
= 0.25 
2
EXAMPLE 1:
Given data: kVA rating = 10, Vi = 500, Vi = 2000, OC test: Vi = 500 V, W0 = 120 W0 = 0.12 kW, SC
test: VSC = 15 V, ISC = 20 A, WSC = 100 W = 0.1 kW
𝑅01 =
𝑍01
100
202
𝑉𝑆𝐶 15
=
=
= 0.75 
𝐼𝑆𝐶 20
 𝑋01 =
𝐾=
𝑅02
𝑋02
= 0.25 
2
2
𝑍01
− 𝑅01
= 0.752 − 0.252 = 0.71 
𝑉2 2000
=
=4
𝑉1
500
= 𝑅01 𝐾 2 = 0.25 × 42 = 4 
= 𝑋01 𝐾 2 = 0.71 × 42 = 11.36 
Full load current on secondary side =
𝑘𝑉𝐴×1000
𝑉2
=
10 ×1000
2000
=5𝐴
Approximate voltage drop at 0.8 p.f. lagging = 𝐼2 𝑅02 𝑐𝑜𝑠∅ + 𝐼2 𝑋02 𝑠𝑖𝑛∅
= 5 × 4 × 0.8 + 5 × 11.36 × 0.6 = 50.08 𝑉
 𝑉2 = 2000 − 50.08 = 1949.92 𝑉
For 0.8 leading power factor 𝑉2 = 2000 − 𝐼2 𝑅02 𝑐𝑜𝑠∅ − 𝐼2 𝑋02 𝑠𝑖𝑛∅ = 2018.08 𝑉
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EXAMPLE 2:
Open circuit and short circuit tests were conducted on a 50 kVA, 6360/240 V, 50 Hz, single phase
transformer to find its efficiency. The observations during the test are: Open-circuit test: Voltage across
primary winding 6360 V, primary current 1.0 A, power input 2 kW. Short-circuit test: voltage across
primary 180 V, current in secondary 175 A, power input 2 kW. Calculate the efficiency of transformer at
full load 0.8 lagging power factor.
Given data: kVA rating = 50, V1 = 6360 V, V2 = 240 V, I0 = 1 A, W0 = 2 kW, VSC = 180 V, WSC = 2 kW
In OC test, voltage across primary = 6360 V = rated primary voltage
 𝑊𝑖 = 𝑊0 = 2 𝑘𝑊
Full load current on secondary side =
𝑘𝑉𝐴×1000
𝑉2
=
50 ×1000
240
= 208.33 𝐴
ISC = 175 A, which is not equal to the full load current
 WSC is not equal to WC, i.e., copper loss at full load.
If I2 = 175 A, the percentage of loading 𝑥 =
175
208.33
= 0.84
𝑥 2 𝑊𝑐 = 𝑊𝑆𝐶
 Copper loss at full load, 𝑊𝑐 =
𝑊𝑆𝐶
𝑥2
=
2
0.84 2
= 2.83 𝑘𝑊
𝑥𝑘𝑉𝐴𝑐𝑜𝑠∅
1 × 50 × 0.8
% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑡 𝑥 % 𝑙𝑜𝑎𝑑, 𝑥 =
× 100 =
× 100
𝑥𝑘𝑉𝐴𝑐𝑜𝑠∅ + 𝑊𝑖 + 𝑥 2 𝑊𝑐
1 × 50 × 0.8 + 2 + 12 × 2.83
= 89.23%
EXAMPLE 3:
A 250/500 V transformer gave the following test results. Open circuit test: 250 V, 1 A, 80 W on low
voltage side, Short circuit test: 20 V, 12 A, 100 W on high voltage side. Calculate the circuit parameters
and draw the equivalent circuit of transformer referred to primary.
Given data: V1 = 250 V, V2 = 500 V, W0 = 80 W, I0 = 1 A,VSC = 20 V, WSC = 100 W, ISC = 12 A
From Open-circuit test 𝑊0
80
𝑐𝑜𝑠∅0 =
=
= 0.32
𝑉1 𝐼0 250 × 1
The working component of no-load current, 𝐼𝑤 = 𝐼0 𝑐𝑜𝑠∅0 = 1 × 0.32 = 0.32 𝐴
∅0 = 𝑐𝑜𝑠 −1 0.32 = 71.340
 𝑠𝑖𝑛∅0 = sin 71.34 = 0.947
The magnetizing component of no-load current 𝐼𝜇 = 𝐼0 𝑠𝑖𝑛∅0 = 1 × 0.947 = 0.947 𝐴
𝑉1
𝐼𝑤
=
250
0.32
 Magnetizing reactance, 𝑋0 =
𝑉1
𝐼𝜇
=
 Core loss resistance, 𝑅0 =
= 781.25 
250
0.947
= 263.99 
5
EXAMPLE 3:
A 250/500 V transformer gave the following test results. Open circuit test: 250 V, 1 A, 80 W on low
voltage side, Short circuit test: 20 V, 12 A, 100 W. Calculate the circuit parameters and draw the
equivalent circuit of transformer referred to primary.
Given data: V1 = 250 V, V2 = 500 V, W0 = 80 W, I0 = 1 A,VSC = 20 V, WSC = 100 W, ISC = 12 A
From Open-circuit test - 𝑅0 =
𝑉1
𝐼𝑤
=
250
0.32
= 781.25 , 𝑋0 =
𝑉1
𝐼𝜇
=
250
0.947
= 263.99 
From Short-circuit test –
𝑅02 =
𝑍02
𝑊𝑆𝐶
2
𝐼𝑆𝐶
=
100
122
I1
= 0.694 
Iw
𝑉𝑆𝐶 20
=
=
= 1.67 
𝐼𝑆𝐶 12
 𝑋02 =
2
𝑍02
−
2
𝑅02
=
0.174 
𝑅01
1.672
−
0.6942
= 1.52 
𝑅0
V1
781.25 
I0
0.38 
𝑋01
𝐼2′
Iµ
𝑋0
263.99 
𝑉2′
𝑉2 500
𝐾=
=
=2
𝑉1 250
𝑅01 = 𝑅02ൗ𝐾2 = 0.694Τ22 = 0.174 
𝑋01 = 𝑋02ൗ𝐾2 = 1.52ൗ22 = 0.38 
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