Three bulbs of 40, W,60, W,and 100, W are con | vedclass

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(A) In a series circuit,the current $(I)$ flowing through all components is the same.According to Ohm's Law,the potential difference $(V)$ across a re

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$40\, W$ (highest),$100\, W$ (lowest)

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(A) In a series circuit,the current $(I)$ flowing through all components is the same.According to Ohm's Law,the potential difference $(V)$ across a resistor is given by $V = I 	imes R$.Since $I$ is constant,$V$ is directly proportional to the resistance $(R)$: $V \propto R$.The power rating of a bulb is given by $P = \frac{V^2}{R}$,which implies $R = \frac{V^2}{P}$.For a fixed voltage rating,the resistance is inversely proportional to the power rating $(R \propto \frac{1}{P})$.Therefore,the bulb with the lowest wattage $(40\, W)$ has the highest resistance,and the bulb with the highest wattage $(100\, W)$ has the lowest resistance.Consequently,the potential difference is highest across the $40\, W$ bulb and lowest across the $100\, W$ bulb.

Three bulbs of $40\, W$,$60\, W$,and $100\, W$ are connected in series across the mains. Across which bulb is the potential difference highest? Across which is it lowest?

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