For a battery (emf epsilon, internal resistance r) driving an external load R: I = $\frac{epsilon}{R+r}$, $P_{ext}$ = $I^{2R}$ = $epsilon^2$*$\frac{R}{R+r}$^2. Maximum $P_{ext}$ occurs when dP/dR = 0: solving gives R = r. Maximum power $P_{max}$ = epsilon^$\frac{2}{4r}$. At this condition, efficiency = $\frac{P_ext}{P_total}$ = $\frac{R}{R+r}$ = 50%. For maximum efficiency (not maximum power), R >> r — but then P is small. This trade-off is important in power systems (high efficiency) vs electronics (impedance matching for max power). JEE problems may ask for the load resistance that maximizes power, or the maximum power value.