Two electric bulbs A and B are rated as 60, W | vedclass

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(C) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage.Since both bulbs are connected in parallel to the same s

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$B$ draws more current than $A$

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(C) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage.Since both bulbs are connected in parallel to the same source,the voltage $V$ across each bulb is the same.From the relation $P = \frac{V^2}{R}$,we have $R = \frac{V^2}{P}$.For bulb $A$,$P_A = 60\, W$,so $R_A = \frac{V^2}{60}$.For bulb $B$,$P_B = 100\, W$,so $R_B = \frac{V^2}{100}$.The current drawn by a bulb is given by $I = \frac{V}{R}$.Substituting $R$,we get $I = \frac{V}{V^2/P} = \frac{P}{V}$.Since $V$ is constant,$I \propto P$.Because $P_B > P_A$ $(100\, W > 60\, W)$,the current drawn by bulb $B$ is greater than the current drawn by bulb $A$ $(I_B > I_A)$.

Two electric bulbs $A$ and $B$ are rated as $60\, W$ and $100\, W$. They are connected in parallel to the same source. Then,

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