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My approach: As they are arranged in series, the current through each of the resistors is the same, and so $P=I^2 R$ should be the most feasible relation to use. So $P$ is directly proportional to $R$. Therefore, the 40W bulb should have the minimum resistance (least $P$). But the answer is given as 100W. My friends are saying that we should use $P=V^2/R$. Can anyone please explain the correct approach?

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    $\begingroup$ The question is not fully specified, which is why you are getting conflicting answers. Your answer is correct if the light bulbs actually are using 40, 60, and 100W in the series circuit. But if the the bulbs are labelled with "40W," "60W," and "100W" (i.e., if that's how much each one is supposed to use when supplied with some rated voltage) then the 100W bulb would be the one with the lowest resistance, and the "series circuit" was just a red herring. (The resistance of the bulbs won't change because of being hooked up differently.) $\endgroup$ Commented Feb 6, 2019 at 13:48

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You may refer to this answer. That is, the definition of the power of the bulbs (here, 40W, 60W, and 100W) are all referring to the power at a specified operational voltage, and would not be such power when they are rearranged in series. Since the resistance of the bulbs are independent of how they are arranged, their resistance should be compared using the equation $P = V^2/R$, in which the operational voltage are the same.

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  • $\begingroup$ That is, the bulbs operates with power = 40W, 60W, and 100W only when they are connected in parallel with some specified operational voltage V. Even they are rearranged in series afterwards, their resistance do not change. It is only meaningful to define the power of the bulbs when using the same operational voltage. $\endgroup$
    – pinchun
    Commented Feb 6, 2019 at 4:50
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    $\begingroup$ The resistance will change depending on the current through the bulb because of the changes in temperature of the filament. $\endgroup$
    – Jasper
    Commented Feb 6, 2019 at 5:30
  • $\begingroup$ Oh yes, but I think we can ignore the effects caused by temperature here (suppose the bulbs' efficiency is very high, i.e. ideal bulbs). $\endgroup$
    – pinchun
    Commented Feb 6, 2019 at 5:33
  • $\begingroup$ Light bulbs are known to have very bad efficiency, about 5%. $\endgroup$
    – Jasper
    Commented Feb 6, 2019 at 13:28
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    $\begingroup$ Yes, I know that. But it seems that this is not the point which the problem want to discuss with. $\endgroup$
    – pinchun
    Commented Feb 6, 2019 at 15:18

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