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ESE Electrical 2015 Paper 2: Official Paper

Option 1 : 90.9%

CT 1: Basic Concepts

18461

10 Questions
10 Marks
6 Mins

Concept:

The efficiency of the 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

Maximum efficiency of transformer occurred at a fraction of load, \(X = \sqrt {\frac{{{P_i}}}{{{P_{cu}}}}}\)

In a transformer, a short circuit test is used to find copper losses and the open circuit is used to find core losses.

Calculation:

Pi = 200 W

Pcu = 200 W

For UPF, cosϕ = 1,

For maximum efficiency(ηmax) of the transformer, \(X = \sqrt {\frac{{{P_i}}}{{{P_e}}}} = \sqrt {\frac{{200}}{200}} = 1\)

\(\% {\eta _{max}} = \frac{{\left( {1 \times 4 \times 1000 \times 1} \right) \times 100}}{{\left( {1 \times 4 \times 1000 \times 1} \right) +200 + 200}} = {\rm{\;}}90.9{\rm{\% }}\)