Question

An electric bulb, rated as $200\text{W}$ at $100\text{V}$, is used in a circuit having a $200\text{V}$ supply. The resistance $R$, that must be put in series with the bulb, so that the
bulb delivers the same power is $\Omega$)

Answer 1

Power, $P=\frac{V^2}{R}$
$R_B=\frac{V_1^2}{P}=\frac{100\times 100}{200}=50\Omega$

For the $200\text{V}$ supply, power is still equal to $200\text{W}$
To deliver same power, the potential difference across the bulb should be equal to $100\text{V}$. This will happen by connecting an equal resistance in series with the bulb, so that the potential difference across both, the bulb and the resistance, is $100\text{V}$, each.
$\therefore R=50\Omega$