RL circuit

  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. LeonhardEuler

    LeonhardEuler 864
    Gold Member

    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. Oh, wait. That is wrong.

    With respect to I, you get:

    R + LI = 0

    So I = -R/L?
     
  5. Deriving with respect to I. Sorry.
     
  6. LeonhardEuler

    LeonhardEuler 864
    Gold Member

    This is still not right because you differentiated one term and integrated the other.
     
  7. LeonhardEuler

    LeonhardEuler 864
    Gold Member

    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. What do you mean?

    d/di(RI) = R
    d/di(L(di/dt)) = LI
     
  9. differential equation
     
  10. LeonhardEuler

    LeonhardEuler 864
    Gold Member

    Taking a derivative of a derivative doesn't make the derivative do away.
     
  11. LeonhardEuler

    LeonhardEuler 864
    Gold Member

    Yes!
     
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