JEE Main 2024 Physics question
An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is:
Answer: (A)
Solution
R = 200²/50 = 800 Ω; P = 100²/800 = 12.5 W.
An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is:
R = 200²/50 = 800 Ω; P = 100²/800 = 12.5 W.