Current Electricity

JEE Main level, single correct, medium

246 of 380

Question 246 of 380

An electric bulb of 500W500W at 100V100V is used in a circuit having a 200V200V supply. Calculate the resistance RR to be connected in series with the bulb so that the power delivered by the bulb is 500W500W.

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Answer: (C) 20Ω20\Omega

Resistance of bulb
P=V2rP=\dfrac{{V}^{2}}{r}
r=(100)2500=100×100500=20r=\dfrac{(100{)}^{2}}{500}=\dfrac{100\times 100}{500}=20
i=PV=500100=5Ai=\dfrac{P}{V}=\dfrac{500}{100}=5A
ΔV\Delta V in resistance R=100VR=100V
R=1005=20ΩR=\dfrac{100}{5}=20\Omega

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