If copper losses of a single-phase transformer at 50% of rated current are 1500 W, then copper losses at full load is equal to:

Question

Question

This question was previously asked in

MPPGCL JE Electrical 28 April 2023 Shift 1 Official Paper

Answer 1

Concept

The copper loss of a 1ϕ transformer is given by:

\(P_{cu}=x^2P_{cufl}\)

where, P_cu = Copper loss

P_cufl = Copper loss at full load

x = Fraction of loading

Calculation

Given, Pcu = 1500 W

x = 50% = 0.5

\(1500=(0.5)^2P_{cufl}\)

Pcufl = 6000 W