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# The full-load copper loss of a transformer is 1600 W. At half-load, the copper loss will be

A. 6400 W

B. 1600 W

C. 800 W

D. 400 W

This Question Belongs to Electrical Engineering >> Transformers

## Comments ( 5 )

1. Cu loss directly proportional to I^2. So if I become half loss will be reduced to one-fourth.

2. copper loss is proportional to KVA^2 or I^2 so for half load its is = (1/2)^2*1600 = 400 W

3. In the open circuit test of a 260/130 V single phase transformer, obtained data were: primary current 0.4 A, Power input 25 W. If the resistance of the primary winding is 0.5 ohm. Determine the core loss. Answer pleas?

4. for full load, copper loss = I^2 * R = 1600
For half load, I = I/2
therefore, half load copper loss = (I/2)^2 * R
= (1/4)*(I)^2 * R
= (1/4) * 1600
= 400

5. We know that,
Pcu= x^2* Pcu(full load)
Pcu= (1/2)^2 *1600w
= 400w

Related Questions on Transformers

A good voltage regulation of a transformer means

A. output voltage fluctuation from no load to full load is least

B. output voltage fluctuation with power factor is least

C. difference between primary and secondary voltage is least

D. difference between primary and secondary voltage is maximum