An electric bulb rated 50 W-200 V is connected across a 100 V supply. The power dissipation of the bulb is:

Solution

Physics sir · 3 steps

The bulb's resistance stays fixed, so find R from the rating first, then use it at the new voltage.

1. Resistance from rating. The rating tells you the bulb draws 50 W at 200 V. Since P = V²/R, the resistance is R = 200²/50 = 800 Ω.
2. Same filament. Connecting it to a different supply does not change the filament, so R stays 800 Ω. Only the voltage across it changes.

   P' = V'²/R = 100²/800 = 12.5 W

3. So. Halving the voltage cuts the power to (1/2)² = 1/4 of 50 W, which is 12.5 W, option (c).

The idea to keep. A rated appliance has a fixed resistance R = V_(rated)²/P_(rated). On any other supply, power scales as the square of the voltage.

Answer from the NTA final answer key. Solution by PrepWiser.