An electric bulb rated as 100 W-220 V is co | vedclass

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(A) The power rating of the bulb is $P = 100 \ W$ and the voltage rating is $V_{rms} = 220 \ V$.Using the formula for power in an $ac$ circuit,$P = V_

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$0.64$

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(A) The power rating of the bulb is $P = 100 \ W$ and the voltage rating is $V_{rms} = 220 \ V$.Using the formula for power in an $ac$ circuit,$P = V_{rms} 	imes I_{rms}$.Therefore,the $rms$ current is $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{220} \approx 0.4545 \ A$.The peak current $I_0$ is related to the $rms$ current by the relation $I_0 = \sqrt{2} 	imes I_{rms}$.Substituting the values,$I_0 = 1.414 	imes 0.4545 \approx 0.64 \ A$.

An electric bulb rated as $100 \ W-220 \ V$ is connected to an $ac$ source of $rms$ voltage $220 \ V$. The peak value of current through the bulb is: (in $A$)

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