Q 12-03-055JEE MainJEE Main 2026 (2 Apr, Shift 2)Easy
Two resistors of $200\ \Omega$ and $400\ \Omega$ are connected in series with a battery of $100$ V. A bulb rated at $200$ V, $100$ W is connected across the $400\ \Omega$ resistance. The potential drop across the bulb is ______ V.
Answer: (B) $50$
Resistance of the bulb: $R_b=\dfrac{V^2}{P}=\dfrac{200^2}{100}=400\ \Omega$.
Bulb in parallel with $400\ \Omega$: $\dfrac{400\times400}{800}=200\ \Omega$.
This $200\ \Omega$ is in series with the other $200\ \Omega$, so the $100$ V divides equally: the bulb gets $50$ V.
Solution by Sreeraj P, M.Sc Physics