Verified Solution • 2025–26 Session moderate

QJEEM22Ph746e: JEE Main 2022 Physics - Class 12 JEEM22Ph746e Solution: An electrical bulb rated 220 V, 100 W, is connected in series…

An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.

Step-by-Step Verified Solution

P100=V2R100⇒R100=V2P100{P_{100}} = {{{V^2}} \over {{R_{100}}}} \Rightarrow {R_{100}} = {{{V^2}} \over {{P_{100}}}}

P60=V2R60⇒R60=V2P60{P_{60}} = {{{V^2}} \over {{R_{60}}}} \Rightarrow {R_{60}} = {{{V^2}} \over {{P_{60}}}}

Pnet=V2R60+R100=P60P100P60+P100=60×100160=37.5{P_{net}} = {{{V^2}} \over {{R_{60}} + {R_{100}}}} = {{{P_{60}}{P_{100}}} \over {{P_{60}} + {P_{100}}}} = {{60 \times 100} \over {160}} = 37.5

This power developed is proportional to resistance.

So, P60′=Pnet×60160=37.5×60160≃14 WP{'_{60}} = {P_{net}} \times {{60} \over {160}} = 37.5 \times {{60} \over {160}} \simeq 14\,W

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