# Three bulbs of 40W, 60W and 100W are arranged in series with 220V. Which bulb has minimum resistance? [closed]

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?

• 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.) Feb 6, 2019 at 13:48

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.