An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, what will be the power consumed?

25 W

V and P are given. The two parameters missing are R and I. The current through the appliance may vary but the internal resistance of the appliance remains constant. Thus, we will focus on finding R first, using : P = V²/R

R = 484 \(\Omega\)

Now, we will see if the voltage is changed to 110 Volts, how will the power dissipated change, using the same expression.

$ P'= \frac{V'^2}{R} = \frac{110^2}{484} = 25W$