# Open Circuit Test of Transformer There are four tests in the transformer.

1. Polarity test
2. Open circuit test
3. Short circuit test
4. Sumpner’s test or back to back test

In this article, we will explain the open circuit test for the transformer.

## Why we are performing the open circuit test transformer?

The open-circuit test is to be performed on the transformer to find the following objective.

1. Find the constant losses of the transformer.
2. Find shunt branch parameters of equivalent circuit R0 and X0.
3. To separate iron losses and find hysteresis loss and eddy current loss.

## Open Circuit Test

Open circuit test and short circuit test performed to determine the efficiency, regulation and circuit constant. In this test, the load is not connected with the transformer.

This test gives accurate results compared to the taking measurement on a full load transformer. Also, these tests consume a very low amount of power compared to the full load output of the transformer.

As we have seen in the construction of the transformer, there are two windings; primary winding and secondary winding.

Let us assume that; the primary winding is low voltage winding and the secondary winding is high voltage winding.

Voltmeter, Ammeter and Wattmeter connected with low voltage winding. And the secondary winding of the transformer is kept open. Therefore, this test named an open circuit test.

The transformer supplied with the rated voltage and frequency.

Hence, the voltmeter shows the rated voltage V1 of the primary. A very small amount of current will flow through the primary winding because the secondary is open. This current is known as no-load current I0.

There are two types of loss occurs in the primary; core loss and copper loss.

The copper depends on the current flowing through the winding. But in this case, only no-load current is flowing through the primary side which is just 2-5% of the full load primary current. Therefore, the copper loss in the primary winding is neglected.

The core loss depends on the flux. Since the primary supplied with rated voltage. So, the flux set by the rated voltage has a normal value. And this core loss is the same as the core loss occurred at full load.

Therefore, the wattmeter reads the power loss and that is the core loss or iron loss of that transformer.

### Find R0 and X0

The reading shows by the meters is the following quantity.

• The ammeter shows no-load current I0.
• Voltmeter shows primary rated voltage V1.
• Wattmeter shows iron or core loss W0.

From the approximate equivalent circuit, we can find the value of no-load resistance R0 and no-load reactance X0.

R0 = V1/Iw

X0 = V1/Iu

No-load current I0 divides into two parts; core-loss current (Iw) and magnetizing current (Iu). From the measurement, we have I0, V1, and W0.

This quantity which we can use to find the R0 and X0.

R0 = V1/Iw

X0 = V1/Iu

Where, Iw = core loss current = I0 cos фo

Iu = magnetizing current = Iu sin фo

W0 = No load power = V1 I0 cos фo

Cos фo = W0/V1 I0

By these equations, we can find the value of R0 and X0.

### Find Constant Loss

Total loss = Iron loss + Dielectric loss + Copper loss

The iron loss and dielectric loss are the constant losses which we want to find. Here as we have discussed, the copper loss is very small and which we can neglect. So, the loss shown by the wattmeter is a constant loss.

### Separation of Iron Loss

Iron loss divided into two types; hysteresis loss and eddy current loss. By this test, these losses can be separated.

For that, this test should be conducted with variable frequency and voltage. The value of voltage and frequency is chosen in such a way that, the ratio of voltage and frequency (V/F) remains constant.

In the first test, supply with rated voltage and frequency. After that, reduce the voltage and set the value of frequency constraint that the ratio of V/F remains constant.

Repeat this test four to five times and make a graph for Wi/F vs frequency. From the graph find the constant A and B. Where A is on the Y-axis and B is on the slope of the curve.

Now from this you can calculate the hysteresis loss and eddy current loss at rated frequency and voltage by below equations.

Hysteresis loss at rated frequency = A x rated frequency

Eddy current loss at rated frequency = B x square of the rated frequency.

Wi = A f + B f2

Wi/f = A + B f

m = slope of straight line = tan θ = B