A bulb is marked 220 V, 40 W. Find the current it takes at 220 V…

Step 1. (a) $I = \dfrac{40\ \text{W}}{220\ \text{V}} = 0.182\ \text{A}$.

Step 2. (b) $R = \dfrac{(220\ \text{V})^2}{40\ \text{W}} = 1210\ \Omega$.

Step 3. (c) $P = \dfrac{(165\ \text{V})^2}{1210\ \Omega} = 22.5\ \text{W}$.

What do we see? Three quarters of the voltage gives $\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}$ of the power.