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RL Circuit with current source

  1. May 8, 2014 #1

    dwn

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

    Image


    2. Relevant equations

    Current divider

    3. The attempt at a solution

    The only thing I see is that I solve for R1 and R2 for t < 0. Other than that I'm confused and not even sure I'm right about solving t<0.
     

    Attached Files:

  2. jcsd
  3. May 9, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Up until t=0, what is the equation for current through R2?

    How does the circuit change at t=0?
     
  4. May 9, 2014 #3

    dwn

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    Current Divider : 2 ( R1 + R2)/R1

    The source is gone, stuck in the closed loop (essentially), and the inductor becomes the current source (the energy within the inductor decays exponentially).
     
  5. May 9, 2014 #4

    NascentOxygen

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    Staff: Mentor

    That fraction is > 1.0 ?

    So what is the general form of the equation describing this decay?
     
  6. May 9, 2014 #5

    dwn

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    Oh no I'm sorry, it should be 2* ((Rtotal/R1))
    Rtotal = (R1*R2)/(R1 + R2)

    Ie-Rt/L (R equivalent)
     
  7. May 9, 2014 #6

    NascentOxygen

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    Staff: Mentor

    Which resistor carries that current?
     
  8. May 9, 2014 #7

    dwn

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    R2, if I'm not mistaken.
     
  9. May 9, 2014 #8

    NascentOxygen

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    Staff: Mentor

    Guessing is not advisable. Don't you have some way to work it out?
     
  10. May 9, 2014 #9

    dwn

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    Of course, the reason I put that was bc the forum requires min 4 characters...

    We choose the "unwanted" resistor to divide from the total because it has something to do with impedance (never covered this topic in class). I know the larger the resistance the less current flowing through it.
     
    Last edited: May 9, 2014
  11. May 9, 2014 #10

    dwn

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    The bread crumbs are not helping.
     
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