A filament bulb (500 ,W,,, 100 ,V) is to be u | vedclass

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Question

Resistance of bulb, $R_{B}=\frac{V^{2}}{P}=\frac{(100)^{2}}{500}=20\, \Omega$

Power of the bulb in the circuit,

$P =V I $

$I =\frac{P}{V_{B}}

Vedclass · Accepted Answer

$26$

Vedclass · Answer

Resistance of bulb, $R_{B}=\frac{V^{2}}{P}=\frac{(100)^{2}}{500}=20\, \Omega$

Power of the bulb in the circuit,

$P =V I $

$I =\frac{P}{V_{B}} $

$=\frac{500}{100}=5\, \mathrm{A}$

$V_{R} =I R \Rightarrow(230-100)=5 	imes R $

$	herefore \quad R =26 \,\Omega$

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 .................. $\Omega$

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