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 power dissipated by an electrical appliance is determined by the relationship P = V²/R, where V is the potential difference and R is the resistance. For a bulb rated at 60 W and 220 V, the resistance is constant and can be calculated as R = V_rated² / P_rated, which equals 220² / 60. When the bulb is connected to a 110 V supply, the new power dissipated (P_new) is calculated using the same resistance: P_new = V_new² / R. Substituting the values, P_new = 110² / (220² / 60). This simplifies to P_new = 60 * (110/220)², which is 60 * (1/2)² = 60 / 4 = 15 W. Thus, if the voltage is halved, the power dissipated becomes one-fourth of the original rated power.