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 Pi = 64 W,
copper losses Pcu at 90% of load = 81 W
Pcu = x2 PcuFL
PcuFL = Pcu / x2
PcuFL = 81 / 0.92 = 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