Question 3.23: Efficiency of 400/200 V, 200 kVA transformer is 98.5 per cen...

Let W_{c} be the core loss and W_{cu} be the full-load copper loss.

\eta_{F.L} = 0.985 = \frac{200  ×  0.8}{200  ×  0.8  + W_{c}  +  W_{cu}}

or,                                                          0.985 = \frac{160}{160  +  W_{c}  +  W_{cu}}

or,                                                          \frac{160}{0.985} = 160  +  W_{c}  +  W_{cu}

or,                                                          W_{c}  +  W_{cu} =  2.43  kW = 2430  W                                                                (3.50)

At half load, W_{c} will remain the same but W_{cu} will be one-fourth its full-load value. Efficiency at half load is

or,                                   0.975 = \frac{320}{320  +  4  W_{c}  +  W_{cu}}

or,                                   \frac{320}{0.975} = 320  +  4  W_{c}  +  W_{cu}

or,                                   4  W_{c} + W_{cu} = 8.2  kW = 8200  W                                                                    (3.51)

From (3.50) and (3.51),                     W_{c} = 1923  W

and                                                         W_{cu} = 507  W