Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Alternating Current
0 votes

A bulb is rated $100\;V,100\;W$ The inductance of an inductor that should be connected in series with the bulb to operate the bulb at rated power, with ac source $200 V, 50 \;Hz$

$(a)\;\large\frac{\sqrt 3}{\pi}\;H \\ (b)\;\sqrt 3 H \\(c)\;\sqrt 3 \pi H \\(d)\;3H $

Can you answer this question?

1 Answer

0 votes
$P= \large\frac{V^2}{R}$
$R= 100\; \Omega$
$V= IR$
$I= 1 \;amp$
$I= \large\frac{V}{Z}$
=> $Z= \large\frac{V}{I}$
$\qquad= \large\frac{200}{1}$$=200 \; \Omega$
$X_L=\sqrt {3} \times 100$
$X_L=wL=> L =\large\frac{\sqrt 3}{\pi}$
Hence a is the correct answer.
answered Mar 25, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App