Question

Full load copper loss in a transformer is 1600 watts. At half load the copper loss will be:
1. 6400 W
2. 1600 W
3. 800 W
4. 400 W

Answer 1

Correct Answer - Option 4 : 400 W

Concept

Copper losses:

These losses occur due to the ohmic resistance of the transformer windings.

If I1and I2 are the primary and the secondary current. R1 and R2 are the resistance of primary and secondary winding then the copper losses occurring in the primary and secondary winding will be I12R1 and I22R2 respectively.

Therefore, the total copper losses will be

\({P_C} = I_1^2{R_1} + I_2^2{R_2}\)

At no load value of current is negligible, so that copper losses are also negligible. 

The copper loss P_cu at any load also given by

Pcu = x2 Pcf

Where, x = Fraction of load, Pcf = Full load copper loss,

Calculation:

Given that full load copper loss = 1600 W

Copper loss at x load = x2 × full load copper loss

At half of the load x = 0.5

Copper loss at half load = 0.52 × 1600 = 400 W

Therefore the copper loss at half of the load is 400 W

• At no load copper loss is negligible.
• These losses are determined with the help of the short-circuit test. 
• The short circuit test is performed on the high voltage winding of the transformer or electrical machine while the low voltage side is short-circuited i.e. performed at rated current.
• The equivalent resistance, impedance, and leakage reactance are also calculated by the short circuit test.