There are four tests in the transformer.

- Polarity test
- Open circuit test
- Short circuit test
- 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.

- Find the constant losses of the transformer.
- Find shunt branch parameters of equivalent circuit R
_{0}and X_{0}. - 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 V_{1} 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 I_{0}.

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 R**_{0} and X_{0}

_{0}and X

_{0}

The reading shows by the meters is the following quantity.

- The ammeter shows no-load current I
_{0}. - Voltmeter shows primary rated voltage V
_{1}. - Wattmeter shows iron or core loss W
_{0}.

From the approximate equivalent circuit, we can find the value of no-load resistance R_{0} and no-load reactance X_{0}.

**R _{0} = V_{1}/I_{w}**

**X _{0} = V_{1}/I_{u}**

No-load current I_{0} divides into two parts; core-loss current (I_{w}) and magnetizing current (I_{u}). From the measurement, we have I_{0}, V_{1}, and W_{0}.

This quantity which we can use to find the R_{0} and X_{0}.

**R _{0} = V_{1}/I_{w}**

**X _{0} = V_{1}/I_{u}**

Where**, I _{w} = **core loss current

**= I**

_{0}cos ф_{o}**I _{u} = **magnetizing current

**= I**

_{u}sin ф_{o}**W _{0} = **No load power

**= V**

_{1}I_{0}cos ф_{o}**Cos ф _{o} = W_{0}/V_{1} I_{0}**

By these equations, we can find the value of R_{0} and X_{0}.

**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 W_{i}/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.**

**W _{i} = A f + B f^{2}**

**W _{i}/f = A + B f**

**m = **slope of straight line** = tan θ = B**

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