If power in external resistance $R$ is maximum then
$(i) R = r$ $(ii)$ Power in $R$ is $\frac{{{E^2}}}{{4R}}$
$(iii)$ Input power $\frac{{{E^2}}}{{2R}}$ $(iv)$Efficiency is $50\%$
Diffcult
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d
If power in $\mathrm{R}$ is maximum then
If power in $\mathrm{R}$ is maximum then
$\boxed{R = r\,}\,i = \frac{E}{{2R}} = \frac{E}{{2r}}$
Power in $\mathrm{R}$
$P=\left(\frac{E}{2 R}\right)^{2} \cdot R=\frac{E^{2}}{4 R}$
input power.
$P_{\text {in }}=\left(\frac{E}{2 R}\right)^{2} \cdot 2 R=\frac{E^{2}}{2 R}$
Efficiency $=\frac{P_{\text {out }}}{P_{\text {in }}} \times 100=50 \%$
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