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2023 · Q106

An incandescent bulb rated as 100W at 110V is connected to a 220V power supply. The power dissipated in the bulb would be:

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  1. A50 W
  2. B100 W
  3. C200 W
  4. D400 W

Explanation

The bulb's resistance is R = V^2/P = 110^2/100 = 121 ohms. At 220 V the power dissipated is P' = V^2/R = 220^2/121 = 48400/121 = 400 W. Doubling the voltage quadruples the power for a fixed resistance. Hence (d).