Q. A cell of internal resistance r drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when
Solution:
Current $i = \frac{E}{r + R}$
Power generated in R
$P = i^2 R$
$P = \frac{E^2R}{(r + R)^2}$
for maximum power $\frac{dP}{dR} = 0$
$E^{2} \left[\frac{\left(r+R\right)^{2} \times1-R \times2\left(r+R\right)}{\left(r+R\right)^{4}}\right] = 0 $
$ \Rightarrow r =R $
