An electric bulb rated 50 W - 200 V is connected across a 100 V supply. The power dissipation of the bulb is: (a) 100 W (b) 50 W (c) 12.5 W (d) 25 W

Answer: (c). Fixed resistance: P ∝ V². P' = 50 × ((100)/(200))² = 50/4 = 12.5 W. So the answer is (c) 12.5 W.

An electric bulb rated 50 W - 200 V is connected across a 100 V… | JEE Main 2024 (4 Apr, Shift 2)

Q10JEE Main 2024 · 4 APR · SHIFT 2Easy

Electrical Power & Heating Effect

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

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Answer: (c)

Fixed resistance: $P \propto V^2$ .
$P' = 50 \times \left(\frac{100}{200}\right)^2 = \frac{50}{4} = 12.5\ \text{W}$ .
So the answer is (c) 12.5 W.

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