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

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(A) The rated power of the bulb is $P = 500 \, W$ at a rated voltage $V_{bulb} = 100 \, V$.The resistance of the bulb is $R_{bulb} = \frac{V_{bulb}^2}

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$20$

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(A) The rated power of the bulb is $P = 500 \, W$ at a rated voltage $V_{bulb} = 100 \, V$.The resistance of the bulb is $R_{bulb} = \frac{V_{bulb}^2}{P} = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega$.Since the bulb is to operate at its rated power of $500 \, W$,it must draw its rated current $I = \frac{P}{V_{bulb}} = \frac{500}{100} = 5 \, A$.The total supply voltage is $V_{total} = 200 \, V$. The voltage across the bulb is $100 \, V$,so the voltage across the series resistor $R$ must be $V_R = V_{total} - V_{bulb} = 200 - 100 = 100 \, V$.Using Ohm's law for the series resistor $R$,$V_R = I 	imes R$.$100 = 5 	imes R$.$R = \frac{100}{5} = 20 \, \Omega$.

An electric bulb rated $500 \, W$ at $100 \, V$ is used in a circuit having a $200 \, V$ supply. Calculate the resistance $R$ to be connected in series with the bulb so that the power delivered by the bulb is $500 \, W$. (in $\Omega$)

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