Q 12-03-032NEETJEE MainMedium
A battery of emf $E$ and internal resistance $r$ delivers maximum power to an external resistor $R$ when
Answer: (A) $R = r$
$P = \dfrac{E^2R}{(R + r)^2}$ is maximum when $\dfrac{dP}{dR} = 0$, which gives $R = r$. Then $P_{max} = \dfrac{E^2}{4r}$.
Solution by Sreeraj P, M.Sc Physics