An electric bulb rated for 500, W at 100, V i | vedclass

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(B) The rated power of the bulb is $P = 500\, W$ and the rated voltage is $V_{rated} = 100\, V$.The current $i$ flowing through the bulb when it opera

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$20$

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(B) The rated power of the bulb is $P = 500\, W$ and the rated voltage is $V_{rated} = 100\, V$.The current $i$ flowing through the bulb when it operates at its rated power is given by $i = \frac{P}{V_{rated}} = \frac{500}{100} = 5\, A$.Since the bulb is connected in series with a resistor $R$ to a $200\, V$ supply,the total voltage is divided between the bulb and the resistor.The voltage across the bulb is $100\, V$,so the voltage across the resistor $R$ must be $V_R = 200\, V - 100\, V = 100\, V$.Using Ohm's law for the resistor $R$,we have $V_R = i 	imes R$.Substituting the values,$100 = 5 	imes R$.Therefore,$R = \frac{100}{5} = 20\, \Omega$.

An electric bulb rated for $500\, W$ at $100\, V$ is used in a circuit having a $200\, V$ supply. The resistance $R$ that must be put in series with the bulb,so that the bulb draws $500\, W$,is ................. $\Omega$.

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