The maximum power drawn out of cell is given by : (E is emf, r is internal resistance)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{E^2}{4r}$
The power (P) delivered to the external load resistance (R)
$P=I^2R$
$=\left(\frac{E}{R+r}\right)^2R$
Maximum power at $R=r$ as $\left|\frac{dP}{dR}\right|_{R=r}=0$
$∴P_{max}=\left(\frac{E}{r+r}\right)^2r$
$=\frac{E^2}{4r}$