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A 40 kVA, 100/400 V, 1-ϕ, 50 Hz transformer has iron loss of 400 W and copper loss of 1600 W at full load. What is the efficiency of the transformer at half load with 0.8 power factor lagging?
1. 95.2 %
2. 86.6 %
3. 99.1 %
4. 90.8 %

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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

\({\rm{\% \eta }} = \frac{{\left( {0.5 \times 40 \times 1000 \times 0.8} \right) \times 100}}{{\left( {0.5 \times 40 \times 1000 \times 0.8} \right) + \left( {400} \right) + {{0.5}^2}\left( {1600} \right)}} = 95.23{\rm{\% }}\)

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