Q 12-03-237JEE MainJEE Main 2020 (4 Sep, Shift 1)Easy
A battery of $3.0\ \text{V}$ is connected to a resistor dissipating $0.5\ \text{W}$ of power. If the terminal voltage of the battery is $2.5\ \text{V}$, the power dissipated within the internal resistance is:
Answer: (C) $0.10\ \text{W}$
Current: $I = \dfrac{0.5}{2.5} = 0.2$ A. Voltage across the internal resistance $= 3.0 - 2.5 = 0.5$ V.
Power in the internal resistance $= 0.5\times0.2 = 0.10$ W.
Solution by Sreeraj P, M.Sc Physics