# Transformer Efficiency

The **Efficiency** of the transformer is defined as the ratio of useful power output to the input power. The input and the output power are measured in the same unit. Its unit is either in Watts (W) or KW. Transformer efficiency is denoted by Ƞ.

- V
_{2}– Secondary terminal voltage - I
_{2}– Full load secondary current - Cosϕ
_{2}– power factor of the load - P
_{i}– Iron losses = hysteresis losses + eddy current losses - Pc – Full load copper losses = I
_{2}^{2}Res

Consider, the x is the fraction of the full load. The efficiency of the transformer regarding x is expressed as

The copper losses vary according to the fraction of the load.

## Maximum Efficiency Condition of a Transformer

The efficiency of the transformer along with the load and the power factor is expressed by the given relation.

The value of the terminal voltage V_{2 }is approximately constant. Thus, for a given power factor the Transformer efficiency depends upon the load current I_{2}. In the equation (1) shown above the numerator is constant and the transformer efficiency will be maximum if the denominator with respect to the variable I_{2} is equated to zero.

i.e Copper losses = Iron losses

Thus, the transformer will give the maximum efficiency when their copper loss is equal to the iron loss.

From equation (2) the value of output current I_{2} at which the transformer efficiency will be maximum is given as

If x is the fraction of full load KVA at which the efficiency of the transformer is maximum

Then, copper losses = x^{2}P_{c } (where P_{c} is the full load copper losses)

Iron losses = P_{i}

For maximum efficiency x^{2} P_{c} = P_{i}

Therefore, Output KVA corresponding to maximum efficiency

Putting the value of x from the above equation (3) in equation (4) we will get

The above equation (5) is the maximum efficiency condition of a transformer.

## 4 Comments

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