SSC JE Electrical Engineering 2017 Set 2 Paper

For maximum power transform load resistance is equal to source resistance
${\mathit{R}}_{\mathit{L}}={\mathit{R}}_{\mathit{S}}$${\mathit{I}}_{\max }=\frac{{\mathit{V}}_{\mathit{m}}}{\left({\mathit{R}}_{\mathit{S}}+{\mathit{R}}_{\mathit{S}}\right)}=\frac{{\mathit{V}}_{\mathit{m}}}{2{\mathit{R}}_{\mathit{S}}}$
${\mathit{P}}_{\max }={\left(\frac{{\mathit{V}}_{\mathit{m}}}{\sqrt{2}}\right)}^2$
${\mathit{P}}_{\max }={{\mathit{I}}^2}_{\mathit{m}}{\mathit{R}}_{\mathit{L}}=\frac{{{\mathit{V}}^2}_{\mathit{m}}}{4{\mathit{R}}_{\mathit{L}}}$
efficiency = $\frac{\mathrm{out}\mathrm{put}}{\mathrm{Input}}$ $=\frac{\frac{{{\mathit{V}}^2}_{\mathit{m}}}{4{\mathit{R}}_{\mathit{L}}}}{\frac{{{\mathit{V}}^2}_{\mathit{m}}}{2{\mathit{R}}_{\mathit{L}}}}=\frac{1}{2}=50\%$
So, for maximum power transfer to a load, the efficiency of the circuit is 50%