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
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$
