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(B) The resistance of a bulb is given by $R = V^2 / P$. Since $V$ is constant,$R \propto 1/P$. Thus,the $40\, W$ bulb has a higher resistance than the

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The potential drop across the $40\, W$ bulb is greater than the potential drop across the $100\, W$ bulb.

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(B) The resistance of a bulb is given by $R = V^2 / P$. Since $V$ is constant,$R \propto 1/P$. Thus,the $40\, W$ bulb has a higher resistance than the $100\, W$ bulb.When connected in series,the current $I$ flowing through both bulbs is the same.The potential drop across a bulb is given by $V = IR$. Since $I$ is constant,$V \propto R$.Because the $40\, W$ bulb has a higher resistance,the potential drop across it will be greater than the potential drop across the $100\, W$ bulb.

Two bulbs,as in the previous context ($40\, W$ and $100\, W$ bulbs rated at the same voltage),are connected in series to a $200\, V$ line. Then:

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