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Circuit Question Involving Resistors and Inductors

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data

    For the circuit below, if L=2.0 H, emf=10 V, R1=10 ohms, R2= 20 ohms, what is the current through each component

    a) at the instant the switch is closed
    b)a long time after the switch is closed
    c) If after a long time the switch is reopened, what is the time constant for the current?




    2. Relevant equations

    V=IR

    3. The attempt at a solution

    For a) I think the resistors are in series so equivalent resistance would be 30 ohms. Then I used V=IR to find a current of 0.33 A.

    Could someone please check if a) is right and show me the steps to do the other parts. Thanks!
     

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    Last edited by a moderator: Aug 5, 2014
  2. jcsd
  3. Aug 5, 2014 #2

    berkeman

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    Staff: Mentor

    The current you calculated for right after the switch is closed is correct. What about the current a long time after the switch is closed? What is an inductor's impedance at DC?
     
  4. Aug 5, 2014 #3
    After a long time, the inductor will act like a piece of wire with no resistance so current would still be flowing through R1 but zero current would be in R2. The current in R1 would be I=10/10=1 A. Am I correct?

    For part c, I know that the time constant is tau=L/R so do I just use L=2.0 H and R=0 ohms?
     
  5. Aug 5, 2014 #4

    berkeman

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    Staff: Mentor

    Yes, very good. :smile:

    When the switch is opened, what is in parallel with the inductor? The inductor current cannot change instantly, so it will decay with what time constant?
     
  6. Aug 5, 2014 #5
    R2 is in parallel with the inductor. I know that current decrease is represented by
    I(final)=I(initial)*e^(-t/τ).
     
  7. Aug 5, 2014 #6

    berkeman

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    Staff: Mentor

    Looks good!
     
  8. Aug 5, 2014 #7
    So the time constant is τ=-t/ln(Ifinal/Iinitial)
    There is no value for part c?
     
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