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Asked in GSEB Board July 2021 · Rated bulbs and appliances
Given: rated power P=100 W at V=220 V.
Idea: at its rated voltage the bulb dissipates its rated power, and P=(V²)/R.
R=(V²)/P=(220²)/(100).
R=(48400)/(100)=484 Ω.
Check: I=V/R=(220)/(484)=5/(11) A, and VI=220×5/(11)=100 W.
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