Verified Solution • 2025–26 Session moderate

QJEEM21Phcabc: JEE Main 2021 Physics - Class 12 JEEM21Phcabc Solution: An electric bulb rated as 200 W at 100 V is used in a circuit…

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____________ Ω\Omega .

Step-by-Step Verified Solution

Power, P=V2RBP = {{{V^2}} \over {{R_B}}}

RB=V2P=100×100200{R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}

RB=50Ω{R_B} = 50\Omega

JEE Main 2021 (Online) 25th July Morning Shift Physics - Current Electricity Question 182 English Explanation

To produce same power, same voltage (i.e. 100 V) should be across the bulb.

Hence, R = RB

R = 50 Ω\Omega

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring