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I Current flow across inductors in Series

  1. Mar 4, 2016 #1
    I can't articulately ask the question so I drew a diagram. In the diagram, both capacitors are equal in capacitance. The bottom inductors are both 10000 nh and the top are 100 nh.
    A. Will the capacitors charge at the same time?
    B If we switch the large inductors to the top and the small to the bottom of the diagram, will the capacitors charge at the same rate?
     

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  3. Mar 4, 2016 #2
    Can we assume that although it is not depicted, there exists some mechanism (a switch, possibly?) by which the circuit can be energized or de-energized at will?
     
  4. Mar 5, 2016 #3

    NascentOxygen

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    Staff: Mentor

    While ever there is a changing current in an inductor there will be a voltage loss across that inductor. So, after switch-on, the upper capacitor won't charge at a rate exactly identical with that of the lower one because of the additional impedance in the upper ones path (the pair of tiny 100nH inductors).
     
  5. Mar 5, 2016 #4
    Yes, I forgot to add it.
     
  6. Mar 5, 2016 #5
    Okay, now this is what I thought. But, will both inductors store energy in their magnetic fields before either capacitor begins to charge? Or will it be a first come first serve deal with the current flow?
     
  7. Mar 5, 2016 #6

    NascentOxygen

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    The top capacitor charges as current flows to it via the top pair of inductors. Inductors store energy as current flows in them. The top pair of inductors are in series, so they share the same identical current, and their current is the current charging the top capacitor. So it all happens starting together.
     
  8. Mar 5, 2016 #7
    Okay, now what determines the time it takes for a steady current to flow through an inductor? Is there a formula I can look up?
     
  9. Mar 5, 2016 #8

    NascentOxygen

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    There is no steady current in the circuit depicted. Were you hoping there'd be a steady current somewhere? If so, why?
     
  10. Mar 5, 2016 #9
    Oh i know that. I meant in general, knowing the parameters of an inductor, how can we calculate the time it takes before a steady current will flow through it.
     
  11. Mar 6, 2016 #10

    NascentOxygen

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    Analysis is generally not a simple matter, even where the inductor is assumed ideal. The simplest case would be where you have an ideal inductor connected across an ideal voltage source---and here the current never reaches a steady value, it just goes on increasing at a fixed rate forever. If you now insert a resistor in series with the inductor, to create a 3 element circuit, then after switch-on the current increases but with a decaying exponential. You might consider that after 5 or more time-constants have elapsed the current is getting close to a steady value.

    If you are unfamiliar with these terms, google for RL circuit time constant.
     
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