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Change of flux in an L-R circuit

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=44283&stc=1&d=1329996280.jpg

    2. Relevant equations

    afaik, flux in inductor = Li.
    at t=0, current through L is 0. so change of flux = LΔi = L(i-0)=Li

    3. The attempt at a solution


    at t=∞, i through cap. is 0.

    so, attachment.php?attachmentid=44282&stc=1&d=1329996286.jpg

    i get the following equations:
    5(i1+i2)=20
    5(i1+i2)+ 5i2=10

    i get i2=-2 and i1 = 6
    i through indctor = i1+i2 = 4

    so, flux=Li= 0.5*4=2


    book's ans. is different.
     

    Attached Files:

  2. jcsd
  3. Feb 23, 2012 #2
    t-o current is zero in the inductor lol
     
  4. Feb 23, 2012 #3
    but capacitor was charged at t-0
     
  5. Feb 23, 2012 #4

    vela

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    You did everything right except for assuming the current through the inductor is 0 right before the switch is closed.
     
  6. Feb 23, 2012 #5
    Well, um.... Current through inductor is i=i0(1-e^(-t/tau)). Putting t=0 in the equation, I get i=0.
     
  7. Feb 23, 2012 #6

    vela

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    That equation only applies to a simple LR circuit with the initial condition i(0-)=0. That's not what you have here.
     
  8. Feb 23, 2012 #7
    edit: oh I'm sorry I didn't see the battery there before the switch is closed.

    I got the ans. 1.5.
     
  9. Feb 23, 2012 #8

    vela

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    That's not entirely accurate. At steady state, a capacitor acts like an open circuit, and an inductor, like a short circuit. I wouldn't say, however, the opposite is true at the beginning of a transient.

    So suppose the switch has been open a long time before t=0. You replace the inductor with a short. What's the current flowing through the upper-left resistor? That will be the initial current through the inductor.
     
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