2025 · Q28
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:
General KnowledgePhysics
- A30 W
- B45 W
- C15 W
- D2 W
Explanation
The filament's resistance is R = V^2/P = 220^2/60. At 110 V the power is V^2/R = 110^2 x 60/220^2 = 60/4 = 15 W. Halving the applied voltage quarters the power dissipated, so the bulb delivers 15 W.