SSC JE Electrical Engineering 24 Jan 2018 Evening Paper

For maximum Power transfer
${\mathit{R}}_{\mathit{L}}={\mathit{R}}_{\mathit{S}}$
Load impedence should equal to source impedence.
Replace all source with his internal Resistance${\mathit{R}}_{\mathrm{th}}={\mathit{R}}_{\mathit{S}}=8||1$ $=\frac{8\times 1}{9}=\frac{8}{9}$
$\mathit{I}=\frac{20-10}{9}=\frac{10}{9}$${\mathrm{V}}_{\mathrm{AB}}={\mathit{V}}_{\mathrm{th}}=20-\frac{8\times 10}{9}$ $=\frac{100}{9}\mathrm{V}$
${\mathrm{P}}_{\max }=\frac{{\mathit{V}}_{\mathrm{th}}^2}{4{\mathit{R}}_{\mathit{S}}}$$=\frac{100}{9}\frac{2\times 9}{4\times 8}$ = 34.722 Watt
So maximum power transfer is approximately 35 Watt.