SSC JE Electrical Engineering 2010 Paper

∵ At maximum efficiency Iron loss = copper loss = 1000 W
∵ It occurs at full load, so full load
Copper loss = ${\mathit{I}}_{\mathrm{fl}}^2\mathit{R}$ = 1000 watt
At half load,
Copper loss $={\left(\frac{{\mathit{I}}_{\mathrm{fl}}}{2}\right)}^2\times \mathit{R}=\frac{1}{4}{\mathit{I}}_{\mathrm{fl}}^2\times \mathit{R}$
$=\frac{1000}{4}=250\mathrm{watt}$
So Copper loss at half of full load will be 250W. (Here ${\mathit{I}}_{\mathrm{fl}}$ = full load current)