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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Alternating Current

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 $

1 Answer

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

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