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Q. A filament bulb $(500\, W, 100 \,V)$ is to be used in $230\, V$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500\, W$. The value of $R$ is

NEETNEET 2016Current Electricity

Solution:

$P = \frac{V^{2}}{R} $
$ R_{b} = \frac{V^{2}}{P} = \frac{10000}{500} = 20\, \Omega $
$ i = \frac{100}{R_{b}} = \frac{130}{R} $
$ \frac{100}{20} = \frac{130}{R}$
$ R = 26\,\Omega$

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