a single phase 100 kVA, 1000/100 V transformer gives the following test results:

Question

a single phase 100 kVA, 1000/100 V transformer gives the following test results:
Open Circuit test: 100V, 6.0 A and 400 W
Short Circuit test: 50V, 100A and 1800W

At what load will maximum efficiency occur?

Explanation / Answer

we know maximum efficiency occurs when copper losses is equal to iron losses cu losses = iron losses x^2(i2^2*R02) = iron losses------->1 where x - represent fraction of full load i2^2*R02 - represents full load copper losses from 1 x = sqrt(iron losses/full load copper losses) iron losses = 400 from open circuit test full load copper losses = 1800 from short circuit test x = sqrt(400/1800) = 0.471 means maximum efficieny occurs at 47.1 % of full load

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