A bulb rated at ($100\,W$ - $200\,V$) is used on a $100\,V$ line. The current in the bulb is
Medium
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(a) $P = \frac{{{V^2}}}{R} \Rightarrow 100 = \frac{{{{(200)}^2}}}{R}$
$ \Rightarrow $ $R = \frac{{4 \times {{10}^4}}}{{{{10}^2}}} = 400\,\Omega $
Now, $i = \frac{V}{R} = \frac{{100}}{{400}} = \frac{1}{4}\,amp$
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