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Asked in JEE Main 28th July 2nd Shift 2022 · Rated bulbs and appliances
R₁=(220²)/(100)=484 Ω, R₂=(220²)/(60)=806.7 Ω.
Series current: I=(220)/(484+806.7)=0.1705 A.
Power in the 100 W bulb: P=I²R₁=(0.1705)²×484≈14.06 W.
About 14 W.
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