Start free

Physics · Work, Power & Energy

JEE Main 2024 — 4 April, Shift 2 — Question 41

An electric bulb rated 50 W−200 V50 \mathrm{~W}-200 \mathrm{~V} is connected across a 100 V supply. The power dissipation of the bulb is :

  1. Option A:

    12.5 W

    Correct
  2. Option B:

    25 W

  3. Option C:

    50 W

  4. Option D:

    100 W

Answer: A

Step-by-step solution

Rated power & voltage gives resistance R=V2P=(200)250=4000050\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(200)^{2}}{50}=\frac{40000}{50} R=800\mathrm{R}=800

P=(Vapplied )2R=(100)2800\mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^{2}}{\mathrm{R}}=\frac{(100)^{2}}{800}

P=12.5\mathrm{P}=12.5 watt

Answer key and solution verified before publishing.

Practise Work, Power & Energy

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 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Work, Power & Energy
Topic
Power

More Work, Power & Energy questions from this paper

An electric bulb rated 50 W -200 V is connected across a 100 V… | JEE Main 2024 PYQ with Solution · DhiX AI