Q 12-03-249JEE MainJEE Main 2019 (10 Jan, Shift 2)Easy
A current of $2\ \text{mA}$ was passed through an unknown resistor which dissipated a power of $4.4\ \text{W}$. Dissipated power when an ideal power supply of $11\ \text{V}$ is connected across it is:
Answer: (C) $11\times10^{-5}\ \text{W}$
$R = \dfrac{P}{I^2} = \dfrac{4.4}{(2\times10^{-3})^2} = 1.1\times10^6\ \Omega$.
$$P' = \frac{V^2}{R} = \frac{121}{1.1\times10^6} = 11\times10^{-5}\ \text{W}$$
Solution by Sreeraj P, M.Sc Physics