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RL circuit

  1. Apr 3, 2010 #1
    1. The problem statement, all variables and given/known data

    There is a Battery connected to a single loop circuit containing two resistors, R1 and R2, and one capacitor L.

    After a long time, the battery is removed, so there is a single loop circuit with just two resistors and a capacitor.

    What is the current going through R1?

    3. The attempt at a solution

    This is what I thought:

    So, the EMF is removed. Thus, using a loop rule, we have the formula:

    0 = IR + L(di/dt) where R is R1+R2

    Integrating, we have

    0 = (1/2)(I^2)(R) + (L)(I)

    Dividing everything by I, we have

    0 = (1/2)(I)(R) + L

    Thus, I = (2L) / R

    However, this is incorrect.

    Any ideas?

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 3, 2010 #2

    LeonhardEuler

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    There are a few problems here. You call this an RL circuit, and you use the equations for an RL circuit, but you say it contains a capacitor instead of an inductor. I assume you are just using the wrong word.

    The second problem is here:
    Are you sure you did that correctly? What variable are you integrating with respect to?
     
  4. Apr 3, 2010 #3
    Oh, wait. That is wrong.

    With respect to I, you get:

    R + LI = 0

    So I = -R/L?
     
  5. Apr 3, 2010 #4
    Deriving with respect to I. Sorry.
     
  6. Apr 3, 2010 #5

    LeonhardEuler

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    This is still not right because you differentiated one term and integrated the other.
     
  7. Apr 3, 2010 #6

    LeonhardEuler

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    You might want to rethink the whole strategy of trying to take an integral or a derivative.
    You have both I and dI/dt in this equation. What kind of equation does that make it?
     
  8. Apr 3, 2010 #7
    What do you mean?

    d/di(RI) = R
    d/di(L(di/dt)) = LI
     
  9. Apr 3, 2010 #8
    differential equation
     
  10. Apr 3, 2010 #9

    LeonhardEuler

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    Taking a derivative of a derivative doesn't make the derivative do away.
     
  11. Apr 3, 2010 #10

    LeonhardEuler

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    Yes!
     
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