Question

Power transferred to load by cell :
$p=I^{2}R=\frac{E^{2}R}{(r+R{)}^2}$ $\Rightarrow P=P_{max}$ $if\frac{dp}{dr}+0\Rightarrow r+R$
Power transferred by cell to load is maximum when
$r=R$ and $P_{m}ax=\frac{E^2}{4r}=\frac{E^2}{aR}$

Answer 1

Power is ‘r’ - $I^{2}R$

$P=\frac{E^2}{\left(R+r^2\right)}R$

$P=P_{max}$

$\frac{dp}{dR}=E^2\left[\frac{{\left(R+r\right)}^2.1-2R\left(R+r\right)}{{\left(R+r\right)}^4}\right]$

$\frac{dp}{dR}=E^2\left(R+r\right)\left[\frac{R+r-2R}{{\left(R+r\right)}^4}\right]$

So, $R+r-2R=0$

$r-R=0$

$r=R$

So, $P_{max}=\frac{E^2}{4r}$