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RL Circuit Parallel

  1. Jun 8, 2015 #1
    1. The problem statement, all variables and given/known data
    There is a DC circuit with a resistor and inductor in parallel. Graph the current through the resistor vs time and the inductor vs. time.

    2. Relevant equations
    i=i(0)e^(-Rt/L)+V/R(1-e^(RT/L)

    3. The attempt at a solution
    My first thought was that initially current through the inductor would be 0 Amps, while current through the resistor would be V/R. At t=infinity, the inductor would become an ideal wire so all current would go through the inductor. But if that is the case then I'm not sure what the current would be since the circuit would be essentially reduced to a battery and an inductor.
     
  2. jcsd
  3. Jun 8, 2015 #2

    phinds

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    2016 Award

    Since you have not bothered to draw the circuit, I have to assume (and it's bad form to force people to make assumptions in order to answer your question) that you mean a DC voltage source with an R and an L hooked to it through a switch and you then throw the switch (closing it).

    IF that is the case, then your analysis is correct so you have to ask yourself if it is actually possible to have an ideal inductor hooked directly across an ideal power source.
     
  4. Jun 8, 2015 #3
    Sorry about that. Here is the picture. They didn't include a switch, but I don't think the question makes sense otherwise. Thank you!
     

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  5. Jun 8, 2015 #4

    phinds

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    Do you think that graph is correct if the elements are ideal elements? Do you really think the current through the resistor goes to zero?

    You still have not answered my original question.
     
  6. Jun 8, 2015 #5
    I'm guessing it's not possible to actually have an ideal inductor hooked across an ideal power source, but I'm not sure what would actually happen. From I=V/R it would seem like infinite current, which is impossible. Since it isn't possible, then some current would have to go through the resistor the entire time. Since voltage is constant and resistance is constant, then current through the resistor would be constant. Current still increases through the inductor. If voltage across the inductor is also constant, then v=Ldi/dt givesI=vt/L, so it would be linear.
     
    Last edited: Jun 8, 2015
  7. Jun 8, 2015 #6
    Are you sure that you have an DC source?
     
  8. Jun 8, 2015 #7

    NascentOxygen

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    It is okay to connect an inductor across a voltage source---with the proviso that you don't leave it connected for too long. (As a separate issue, there also needs to be provision made for safely disconnecting the inductor when the time comes to do so.)

    Your description of the currents sounds right, so the current drawn from the source will be the sum of these and (for the time under consideration you can see it contains no exponential term).

    Since you earlier attached an incorrect sketch, it would be a good idea to provide an updated sketch, to assist those others following this thread. :smile:
     
  9. Jun 9, 2015 #8
    I'm sure I have a DC circuit. Thanks for explaining what would actually happen!
     

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  10. Jun 9, 2015 #9

    NascentOxygen

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    Staff: Mentor

    Your final graph looks good.

    Would you like to have a go at explaining (using some formula) why disconnecting the inductor after some time might be dangerous?
     
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