Correct Answer - Option 3 : 80.0% of the rated current

__Concept:__

The efficiency of Transformer \((\eta ) = \dfrac{{x\ S\cos \phi }}{{x\ S\cos \phi \ +\ {P_i} \ +\ {X^2}\ {P_{cu FL}}}}\)

Where,

x = Fraction of load

S = Apparent power in kVA

Pi = Iron losses

PcuFL = Full load copper losses

Maximum efficiency of transformer occurred at a fraction of load at

\(x = \sqrt {\dfrac{{{P_i}}}{{{P_{cuFL}}}}}\)

**A short circuit test is used in a transformer to find copper losses and the open circuit test is used to find core losses.**

__Calculation:__

Given: core losses P_{i }= 64 W,

copper losses P_{cu }at 90% of load = 81 W

P_{cu} = x^{2} P_{cuFL}

P_{cuFL} = P_{cu }/ x^{2}

PcuFL = 81 / 0.9^{2 }= 100 W

For maximum Efficiency to Occur,

The Transformer must be operated at load \(x = \sqrt {\dfrac{{{P_i}}}{{{P_{cuFL}}}}}\)

\(x = \sqrt {\dfrac{{{64}}}{{{100}}}}=0.8\)

= **80 % of Rated current**