The set up for **short circuit test of transformer ** is shown in the figure. An auto transformer is used in the circuit to adjust the input voltage precisely to the rated voltage.

The ammeter, voltmeter and watt meter are connected in the circuit to measure the current, voltage and power respectively.

**In this test generally, the high voltage side is connected to the AC supply and the low voltage side is short circuited with the help of thick copper wire.**

# Procedure for Short Circuit Test of Transformer

- Connect the circuit as shown in the figure.
- Keep the auto transformer output at its minimum voltage position and switch on the AC supply.
- Increase the applied voltage very slowly, and adjust it to get the current equal to the rated value of the winding. Do not increase the applied voltage further.
- Note down the watt meter, voltmeter and ammeter readings. Let these are W
_{sc}, V_{sc}and*I*_{sc}

## Parameter Calculations

The short circuit test on transformer is performed at the primary and secondary rated currents. Therefore the total copper loss is the full load copper loss.

As the iron losses are supply voltage dependent and supply voltage in this test is small, the iron losses will be negligibly small.

Hence the reading (W_{sc}) shown by the wattmeter is almost entirely corresponding to the full load copper loss.

We know that W_{sc} = V_{sc}*I*_{sc}cos φ_{sc}

or **cos φ _{sc} = W_{sc} / (V_{sc}I_{sc})**

The wattmeter reading W

_{sc}indicates the full load copper loss.

Therefore, W

_{sc}= Copper loss =

*I*^{2}

_{sc}R

_{01}

or

**R**

_{01}= W_{sc}/*I*^{2}_{sc}Where R

_{01 }is the total equivalent resistance of the transformer referred to the primary.

Similarly Z

_{01 }= V

_{sc}/

*I*_{sc}= (R

^{2}

_{01}+ X

^{2}

_{01})

^{1/2}

or

**X**

_{01}= (Z^{2}_{01}– R^{2}_{01})^{1/2}Where X

_{01}is the total equivalent reactance of the transformer referred to primary.

**In this way parameters R**

_{01}, X_{01}, Z_{01}and full load copper losses can be calculated from the short circuit test on transformer.If the transformation ratio K is known, it is possible to obtain these parameters referred to the secondary side.

## Efficiency Calculation from Short Circuit Test of Transformer

**Example**: A 250/500V transformer gave the following test results:

- Short circuit test with low voltage winding short circuited; 20V, 12A 100W.
- Open circuit test on low voltage side; 250V, 1A, 80W.

Determine the efficiency of the transformer when the output is 10A, 500V at 0.8 p.f. lagging.

**Solution**: From the short circuit test (on H.V. side)

Voltmeter reading, V_{2sc} = 20V

Ammeter reading, **I**_{2sc} = 12A

Wattmeter reading, W_{c} = 100W

In the short circuit test of transformer, the wattmeter measures the copper losses of the transformer on short circuit secondary current(**I**_{2sc}).

Output secondary current, **I**_{2} = 10A

Copper losses at this current, P_{c} = (**I**_{2}/**I**_{2sc})^{2}W_{c} = (10/12)^{2} x 100 = 69.44W

**From open circuit test:**

In the open circuit test of transformer, the wattmeter measures the iron losses occurred in the transformer.

Therefore, iron losses in the transformer (P_{i}) = 80W

Output power at given load = V_{2}**I**_{2}cosɸ_{2} = 500 x 10 x 0.8 = 4000W

Efficiency of transformer = [output/(output + P_{i} + P_{c})] x 100

= [4000/(4000 + 80 + 69.44)] x 100 = 96.4%

Thanks for reading about “short circuit test of transformer “.

## Transformer | All Posts

- Single Phase Transformer Working Principle
- Ideal Transformer
- Construction of Three Phase Transformer
- Types of Transformers
- Equivalent Resistance and Reactance of Transformer
- Equivalent Circuit of Single Phase Transformer
- Power Loss in a Transformer
- Open Circuit Test of Single Phase Transformer
- Short Circuit Test on Single Phase Transformer
- Transformer Efficiency
- Regulation of Transformer
- Autotransformer
- Instrument Transformers
- Polarity of Transformer Windings
- Significance of Vector Group of Transformer
- Buchholz Relay Construction | Working
- Why current transformer secondary should not be opened
- Dielectric Strength Test of Transformer Oil
- Transformer Moisture Removal Process

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JollyCopper loss at load current 10 A in HV winding should be [(20/12)2 x 100 ] A = 277.78 A ; because 10 A in HV winding corresponds to 20 A in LV winding.

JollyCorrection :

Copper loss at load current 10 A in HV winding should be [(20/12)2 x 100 ] W = 277.78 W

JollyCorrection in my previous comment:

Copper loss should be [(20/12)2 x 100 ] W = 277.78 W