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.

13 Ω

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230 Ω

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46 Ω

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26 Ω

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Solution

The correct option is B $26 Ω$
Using $P = V I$ the current through bulb is $I = P / V_{b} = 500 / 100 = 5 A$
and the resistance of the bulb is $R_{b} = V_{b}^{2} / P = 100^{2} / 500 = 20 Ω$
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 $R = 26 Ω$

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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.

The correct option is B 26 ΩUsing P=VI the current through bulb is I=P/Vb=500/100=5Aand the resistance of the bulb is Rb=V2b/P=1002/500=20ΩAs the resistance R is connected with bulb in series so same current will flow through them, i.e I=5ABy Ohm's law, 230=(R+Rb)I or 230=(R+20)(5) or R=26Ω

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