Q 12-07-105JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
A $220$ V, $50$ Hz AC source is connected to a $25$ V, $5$ W lamp and an additional resistance $R$ in series (as shown in figure) to run the lamp at its peak brightness, then the value of $R$ (in ohm) will be
Numerical value type. Enter your answer.
Answer: 975
At full (rated) brightness the lamp carries
$$I = \frac{P}{V} = \frac{5}{25} = 0.2\ \text{A}, \qquad R_{\text{lamp}} = \frac{V^2}{P} = \frac{625}{5} = 125\ \Omega$$
Total resistance needed with the 220 V source:
$$R_{\text{total}} = \frac{220}{0.2} = 1100\ \Omega \Rightarrow R = 1100 - 125 = 975\ \Omega$$
Solution by Sreeraj P, M.Sc Physics