Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A bulb is rated at 100 V, 100 W, it can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200 V and 50 Hz.
PhysicsAlternating CurrentJEE Main
Options:
  • A π3 H
  • B 100 H
  • C 2π H
  • D 3π H
Solution:
1722 Upvotes Verified Answer
The correct answer is: 3π H
From the rating of the bulb, the resistance of the bulb can be calculated.

R=Vrms2P=(100)2100=100Ω



For the bulb to be operated at its rated value the rms current through it should be 1 A

Also,

I rms=VrmsZ

1=2001002+(2π50L)2L=3π H

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.