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

The resistance of the bulb remains constant and can be calculated using $R = \frac{V^2}{P} = \frac{220^2}{100} = 484\ \Omega$. When operated at $V' = 110\text{ V}$, the new power consumed is $P' = \frac{(V')^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\text{ W}$.