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Asked in GUJCET 2022 · Rated bulbs and appliances
Given: P = 200 W at V = 220 V.
Idea: P = (V²)/R, so R = (V²)/P.
R = (220²)/(200) = (48400)/(200) = 242 Ω.
On the same supply a higher-power bulb has a lower resistance; a 100 W bulb has 484 Ω.
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