A light bulb rated as 60 W at 220 V has a potential difference of 110 V across its ends. The power dissipated in this light bulb is :

The resistance of the bulb is calculated from its rating: R = V²/P = (220)²/60 = 48400/60 ≈ 806.67 Ω. When the applied voltage is 110 V (half of the rated voltage), the power dissipated = V²/R = (110)²/806.67 = 12100/806.67 = 15 W. Alternatively, since power varies as the square of voltage for a constant resistance, halving the voltage reduces the power to one-quarter: 60/4 = 15 W.