SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Copper losses are directly proportional to the square of the load on the transformer.
${\mathrm{W}}_{\mathrm{cu}}={\mathit{x}}^2{\mathrm{W}}_{\mathrm{cufl}}$
Here, x is the percentage of a full load of the transformer.
${\mathrm{W}}_{\mathrm{cufl}}\mathit{­}$ is the copper losses at the full load.
Given that, x = 25% = 0.25 = $\frac{1}{4}$
Copper losses at full load $={\mathrm{P}}_{\mathrm{cu}}$
Copper loss at 25% of full load $={\left(0.25\right)}^2\times {\mathrm{P}}_{\mathrm{cu}}=\frac{{\mathrm{P}}_{\mathrm{cu}}}{16}$