Correct Answer - Option 2 : 50.6%
Concept:
For maximum efficiency to occur in a transformer
Ironlosses = copperlosses
Wi = Wcu
Wi = x2 × WcuFL
\(x = \sqrt {\frac{{{W_i}}}{{{W_{cuFL}}}}} \)
Here,
x = percentage of loading at maximum efficiency
WcuFL = full load copper losses
Calculation:
Given:
Rprimary = 0.5 Ω, Irated = 25 A, Wi = 80 W, ηmax = 98%
WcuFL = I2rated × Rprimary = 252 × 0.5 = 312.5 W
\(x = \sqrt \frac{{80}}{{312.5}}\)
x = 0.505 = 50.5%