Two resistors of 200Ω and 400Ω are connected in series with a battery of 100 V . A bulb rated at 200 V ,100 W is connected across the 400Ω resistance. The potential drop across the bulb is V.

Physics · Current Electricity

JEE Main 2026 — 2 April, Evening Shift — Question 11

Two resistors of $200\Omega$ and $400\Omega$ are connected in series with a battery of $100\mathrm{V}$ . A bulb rated at $200\mathrm{V},100\mathrm{W}$ is connected across the $400\Omega$ resistance. The potential drop across the bulb is V.

Option A:
25
Option B:
50
Correct
Option C:
66.6
Option D:
100

Answer: B

Step-by-step solution

Bulb resistance $R = V^2/P = 400\Omega$ . The two $400\Omega$ in parallel give 200Ω, then series with 200Ω gives total 400Ω, current = 100/400=0.25A, voltage across bulb = 0.25×200=50V

Answer key and solution verified before publishing.

Practise Current Electricity

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge