An electric bulb rated 50 W–200 V is connected across a 100 V supply. The power dissipation of the bulb is:
- 100 W
- 50 W
- 12.5 W
- 25 W
Answer
(3) 12.5 W
Solution
R = Vrated2Prated = (200)250 = 800
R is fixed, P = V2R
P = (100)2800 = 12.5 W
Shortcut: P V2 half voltage 504 = 12.5 W
Answer: (3)