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Drive RL Circuit

  1. May 8, 2014 #1

    dwn

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    1. The problem statement, all variables and given/known data

    Image

    2. Relevant equations

    KCL
    DRIVEN RL CIRCUIT

    3. The attempt at a solution

    I got the right answer after trying a different tactic, but I don't understand why it is done this way.

    My first approach:

    KCL: 100/45 + 60/2 = iL for t < 0

    32.22 - (100/45)e^(-45*(0.00001)/0.5) = 29.999 A (NOT THE RIGHT ANSWER)

    Second approach:

    30 - (100/45)e^(-45*(0.00001)/0.5) = 27.78 A

    Why is the initial current 30 A? I know how they got it, but why didn't they use KCL? I mean what is happening to the 100 V source?
     

    Attached Files:

  2. jcsd
  3. May 9, 2014 #2

    NascentOxygen

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    Staff: Mentor

    In forming this equation, have you taken into account the polarities of the battery connections?

    In any case, you are making too much out of this question. For an inductor, in the absence of sparks http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon10.gif [Broken] current at 0+ = current at 0.
     
    Last edited by a moderator: May 6, 2017
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