CUET Preparation Today
The maximum power drawn out of cell is given by : (E is emf, r is internal resistance) |
$\frac{E^2}{2r}$ $\frac{E^2}{4r}$ $\frac{E^2}{r}$ $\frac{E^2}{3r}$ |
$\frac{E^2}{4r}$ |
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 $\left|\frac{dP}{dR}\right|_{R=r}=0$ $∴P_{max}=\left(\frac{E}{r+r}\right)^2r$ $=\frac{E^2}{4r}$ |
