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RC Circuit with two resistors

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a circuit with two resistors, R and r, and a capacitor of C, and EMF of V0 as shown in the diagram, find the voltage across the capacitor during charging. Prove that this voltage, V is given by V = V0 (r/(R+r)) (1-e-((R+r)t)/(RrC))

    2. Relevant equations


    3. The attempt at a solution

    This is what I have

    Loop A (with C): V0 + RI + q/c = 0 ==> I = (V0 - VC)/R
    Loop B (with r): V0 + RI + rIr = 0
    I = Ir + Ic ==> I = V/r + C(dv/dt)

    From first and third,
    (V0 - VC)/R = V/r + C(dv/dt)

    Simplify to get,
    V0 - RC(dv/dt) = V(1+R/r)
    V = (r/[R+r])(V0 - RC(dV/dt))

    The shape of the equation is getting there (I hope), but what do I do next? To get the given equation, RC(dV/dt) must be V0e-((R+r)t)/(RrC).

    RC (dV/dt) = V, solving this differential equation to get ln (V) = -t/(RC) + k, hence, RC (dV/dt) = e-t/(RC) + k. And I am totally stuck.

    Did I do something wrong somewhere? I can't think of anyway to get the RrC term in e-((R+r)t)/(RrC), not to mention the V0 and the (R+r) terms, unless k is like Rt/r. But that still does not give me a V0?

    Attached Files:

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  3. Feb 13, 2012 #2


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    The differential equation is correct. You need not solve it to prove that the given V is solution, just plug in V and dV/dt. What is the derivative of the given V(t) function?

  4. Feb 13, 2012 #3
    Hi ehild,

    Thanks for your reply.

    For the differential equation, V = RC (dV/dt), solved for V = e-t/(RC) + k,

    e-t/(RC) + k = RC (dV/dt)

    dV/dt = e-t/(RC) + k/RC

    Is this what you meant?
    Last edited: Feb 13, 2012
  5. Feb 13, 2012 #4


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    Your differential equation is not V = RC (dV/dt), but V = (r/[R+r])(V0 - RC(dV/dt)).

  6. Feb 13, 2012 #5
    Sorry for the mistake,

    That will get me

    (V - (r/[R+r])V0)-1dV = [-(r/[R+r])RC)]-1dt

    Which gets me ln [V - (r/[R+r])V0)] = t/(RrC/[R+r])
    V - (r/[R+r])V0) = e(t/(RrC/[R+r])]
    And a little rearranging gets me V = e(t/(RrC/[R+r])] + (r/[R+r])V0)

    Hmm, did i do something wrong? Even if I subsitute V = RC(dV/dt) into the equation, i will get

    V = (r/[R+r])(V0 - e(t/(RrC/[R+r])] - (r/[R+r])V0))

    which is not equal to the answer..
    Last edited: Feb 13, 2012
  7. Feb 14, 2012 #6


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    You missed to add the arbitrary constant. Your result will be is the same as the given function with the condition that the initial voltage on the capacitor is zero: V(0)=0.

    Why do you substitute V=RC(dV/dt)? It is valid for a simple RC circuit only.

  8. Feb 14, 2012 #7
    Hi again ehild, I added in the arbitrary constant and got the equation! Thank you so much!

    Regarding the V = RC (dV/dt), I have misunderstood the derivation, and assumed that it is a general formula for all RC circuits. Thanks for pointing that out, I'll read up on it tonight.

    Once again, thanks for your help!
  9. Feb 14, 2012 #8


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    You are welcome:smile:

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