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Question

A filament bulb $$(500\ W, 100\ V)$$ is to be used in a $$230\ V$$ main supply. When a resistance $$R$$ is connected in series, it works perfectly and the bulb consumes $$500\ W$$. The value of $$R$$ is. 


A
13 Ω
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B
230 Ω
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C
46 Ω
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D
26 Ω
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Solution

The correct option is B $$26\ \Omega $$
Using $$P=VI$$ the current through bulb is $$I=P/V_b=500/100=5 A$$
and the resistance of the bulb is $$R_b=V_b^2/P=100^2/500=20 \Omega$$
As the resistance R is connected with bulb in series so same current will flow through them, i.e $$I=5 A$$
By Ohm's law, $$230=(R+R_b)I$$ or $$230=(R+20)(5)$$ or $$R=26 \Omega$$  

Physics

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