A filament bulb $$\left( {500\,W,\,100\,V} \right)$$   is to be used in a $$230\,V$$  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 :
If a rated voltage and power are given, then
$${P_{{\text{rated}}}} = \frac{{V_{{\text{rated}}}^2}}{R}$$
$$\therefore $$ Current in the bulb, $$i = \frac{P}{V}\,\,\left( {\because P = Vi} \right)$$
$$i = \frac{{500}}{{100}} = 5\,A$$
$$\therefore $$ Resistance of bulb, $${R_b} = \frac{{100 \times 100}}{{500}} = 20\,\Omega $$
$$\because $$ Resistance $$R$$ is connected in series.
$$\eqalign{ & \therefore {\text{Current,}}\,\,i = \frac{E}{{{R_{{\text{net}}}}}} = \frac{{230}}{{R + {R_b}}} \cr & \Rightarrow R + 20 = \frac{{230}}{5} = 46 \cr & \therefore R = 26\,\Omega \cr} $$