# Transformer Efficiency

The **Efficiency** of the transformer is defined as the ratio of useful power output to the input power, the two being 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

If *x* is the fraction of the full load, the transformer efficiency at this fraction is given by the relation shown below

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

## Maximum Efficiency Condition of a Transformer

The transformer efficiency at a given load and power factor is given by the relation shown below

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 efficiency of a transformer will be maximum when the copper or variable losses are equal to iron or constant losses.

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.

## 1 Comment

thanks for great works i found it useful .keep it up