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A single phase transformer has no load of 64 W, as obtained from an open circuit test. When a short circuit test is performed on it with 90% of the rated currents flowing in its both LV and HV winding. The measured loss is 81 W. The transformer has maximum efficiency when operated at:
1. 50.0% of the rated current
2. 64.0% of the rated current
3. 80.0% of the rated current
4. 88.8% of the rated current

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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 P= 64 W, 

copper losses Pcu at 90% of load = 81 W

Pcu = x2 PcuFL

PcuFL = Pcu / x2

PcuFL = 81 / 0.9= 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

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