An electric bulb is rated 220text{ V}, 100text{ W}. When it is operated on 110text{ V}, the power consumed by the bulb is:

The resistance $R$ of the bulb remains constant and is given by $R = \frac{V^2}{P} = \frac{220^2}{100} = 484\;\Omega$. When operated at $V' = 110\text{ V}$, the power consumed is $P' = \frac{(V')^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\text{ W}$. Alternatively, power is proportional to $V^2$, so halving the voltage reduces the power to $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ of its initial value, which is $100 / 4 = 25\text{ W}$.