Correct Answer - Option 1 :
95.2 %
Concept:
The efficiency of Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)
Where, X = Fraction of load
S = Apparent power in kVA
Pi = Iron losses
Pcu = Copper losses
Calculation:
Given that, iron loss = 400 W
Copper loss at full load = 1600 W
Power factor = 0.8
Fraction of full load (x) = 0.5
\({\rm{\% \eta }} = \frac{{\left( {0.5 \times 40 \times 1000 \times 0.8} \right) \times 100}}{{\left( {0.5 \times 40 \times 1000 \times 0.8} \right) + \left( {400} \right) + {{0.5}^2}\left( {1600} \right)}} = 95.23{\rm{\% }}\)