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Inductance, RL Circuits

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data
    An inductor is connected in series with a 1000 ohm resistor and a battery. What mus be the inductance in millihenries of the inductor if the current in the circuit is required to reach 90% of its final value after 20 microseconds?


    2. Relevant equations
    I am not actually sure which equatins to use. All I know is E(naught) - L(dI/dt) = IR


    3. The attempt at a solution

    First I tried to just plug in the numbers but I am obviously missing some values. Another equation I was given was I = (E/R) (1- exp(-t/(L/R))
     
  2. jcsd
  3. Dec 8, 2012 #2

    gneill

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    Staff: Mentor

    Hi a_patel32. Welcome to Physics Forums.

    That second equation is in fact the solution of the first equation (which is a differential equation in variable I), so that's a good place to start.

    Have you worked out what the steady-state (final) value of the current will be?
     
  4. Dec 8, 2012 #3
    Using the second equation? I don't think I can because I need the battery voltage and inductance right?
     
  5. Dec 8, 2012 #4

    gneill

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    Staff: Mentor

    Nope.

    Suppose that you happened to have a value for E and a value for L. What would you write for the final current?
     
  6. Dec 8, 2012 #5
    It would be I(final) = (E/1000)(1 - exp(-20/(L/1000)) ?
     
  7. Dec 8, 2012 #6
    The final current will be the value of the current at a very late time, so to find it you should take the limit I(t→[itex]\infty[/itex]). Then ask your self: on what time, I(t) will be 90% of the final current. Do some algebra and get the solution!
     
  8. Dec 8, 2012 #7

    gneill

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    Where did the "20" come from in the exp argument?

    What is the behavior of exp(-x) when x is allowed to go to infinity (a very long time indeed).
     
  9. Dec 8, 2012 #8
    Sorry the 20 was the time in the equation, but understand we are talking about the final current. In the exp(-x) it would eventually go down to zero. So it is just going to be I(final) = E/1000?
     
  10. Dec 8, 2012 #9
    Also, the answer given is 8.7 millihenries
     
  11. Dec 8, 2012 #10

    gneill

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    Yes, that's right. So let's call that final current Imax. The current over time is then
    $$I(t) = I_{max}\left(1 - exp(-t/\tau)\right)~~~~~~\tau = L/R$$
    You're interested in a particular point in time when the value of I(t) will be 90% of the final value. So replace I(t) with (90/100)Imax and continue.
     
  12. Dec 8, 2012 #11
    Sorry, I'm not sure how to continue beyond this?
     
  13. Dec 8, 2012 #12

    gneill

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    Just replace I(t) with (90/100)Imax (that's 90% of Imax). At this point it's just algebra. Cancel what you can, then solve what's left for the time constant ##\tau##.
     
  14. Dec 8, 2012 #13
    Great thank you. i just got the answer. This was very helpful. I don't know why I didn't see this before. How are you able to find a solution because when I am taking physics exam I'm not exactly sure how to go about solving problems like these?
     
  15. Dec 8, 2012 #14

    gneill

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    You're welcome :smile:
    It's a matter of practice. Eventually you build up a "tool kit" of approaches to these problems, and you'll quickly recognize how to attack them. Good luck!
     
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