Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

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 200V and 50 Hz.

Option: 1

\mathrm{\frac{\pi}{\sqrt{3}} \mathrm{H}}


Option: 2

\mathrm{100 \mathrm{H}}


Option: 3

\mathrm{\frac{\sqrt{2}}{\pi} \mathrm{H}}


Option: 4

\mathrm{\frac{\sqrt{3}}{\pi} \mathrm{H}}


Answers (1)

best_answer

From the rating of the bulb, the resistance of the bulb can be calculated.

\mathrm{\mathrm{R}=\frac{\mathrm{V}_{\mathrm{rms}}^2}{\mathrm{P}}=100 \Omega}

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

\mathrm{\mathrm{rms}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}} \therefore 1=\frac{200}{\sqrt{100^2+(2 \pi 50 \mathrm{~L})^2}} \Rightarrow \mathrm{L}=\frac{\sqrt{3}}{\pi} \mathrm{H}}

Posted by

Ritika Kankaria

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question