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I'm a 12th grader and recently read about inductors, here is a question I came across:

Which bulb will have more intensity if the switch is closed and then opened?

According to my understanding when the switch is closed, the bulb connected with a resistance has maximum intensity and decreases but in case of inductor I'm not able to comprehend how the current or potential varies. Any help will be appreciated

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If both bulbs have identical normal tungsten filaments which are rated to run directly from the given battery, bulb B will come on bright but quickly dim because the resistance of tungsten increases with temperature. The inductor limits the rate at which the current increases in bulb A. Its brightness will start low and come up fairly fast. Since the inductor has no resistance, bulb A will end up brighter than bulb B. When the switch is opened, the energy stored in the inductor will keep both bulbs burning briefly.

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  • $\begingroup$ Can we say anything about potential across both the branches $\endgroup$
    – imposter
    Commented Jun 22, 2020 at 16:13
  • $\begingroup$ The two branches are in parallel and will have the same voltage (same as the battery when the switch is closed). $\endgroup$
    – R.W. Bird
    Commented Jun 22, 2020 at 18:05
  • $\begingroup$ Thank you so much $\endgroup$
    – imposter
    Commented Jun 23, 2020 at 6:13

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