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
Answer: Option D
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Related Questions on Transformers
The purpose of providing an iron core in a transformer is to
A. provide support to windings
B. reduce hysteresis loss
C. decrease the reluctance of the magnetic path
D. reduce eddy current losses
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
If the percentage impedances of the two transformers working in parallel are different, then
A. transformers will be overheated
B. power factors of both the transformers will be same
C. parallel operation will be not possible
D. parallel operation will still be possible, but the power factors at which the two transformers operate will be different from the power factor of the common load
An ideal transformer will have maximum efficiency at a load such that
A. copper loss = iron loss
B. copper loss < iron loss
C. copper loss > iron loss
D. none of the above
Cu loss directly proportional to I^2. So if I become half loss will be reduced to one-fourth.
copper loss is proportional to KVA^2 or I^2 so for half load its is = (1/2)^2*1600 = 400 W
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?
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
We know that,
Pcu= x^2* Pcu(full load)
Pcu= (1/2)^2 *1600w
= 400w