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Time constant for the current through the inductor

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    wjD8Y.png

    http://i.imgur.com/wjD8Y.png

    In the circuit below, V = 6 volts, R = 10 ohms, L = 100 mH. The switch has been open for a long time. Then, at time t = 0, the switch is closed.



    What is the time constant for the current through the inductor?

    2L / R


    2. Relevant equations

    TAU=l/R

    3. The attempt at a solution

    I am not sure how do I find the equvialent resistance in this case.
     
    Last edited: Oct 25, 2011
  2. jcsd
  3. Oct 25, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Kill the source (short the voltage supply) and remove the inductor. Find the equivalent resistance at the terminals where the inductor was.

    attachment.php?attachmentid=40334&stc=1&d=1319604076.gif
     

    Attached Files:

  4. Oct 26, 2011 #3

    Equvialent resistance is 2r then time constant should be L/2R?

    instead of 2L/R?

    Thanks
     
  5. Oct 26, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    Nope. Those two resistors are not in series with respect to the open terminals. So the resistance is R/2. That makes your time constant 2L/R.
     
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