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Differentiating Kirchoff's voltage law expression

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Differentiate V0 - iR - q/C = 0 to prove that di/dt = -i/RC.

    2. Relevant equations

    V0 - iR - q/C = 0
    ^ derived from previous question for a circuit that had one battery with emf V0, a resistor of resistance R and a capacitor of capacitance C (all in series).
    di/dt = -i/RC

    3. The attempt at a solution

    V0 - iR - q/C = 0
    V0 - (dq/dt)R - q/C = 0
    V0C - (dq/dt)RC - q = 0
    (dq/dt)RC = V0C - q
    d/dt((dq/dt)RC) = d/dt(V0C - q)
    RC(d/dt(dq/dt) = -dq/dt
    RC(di/dt) = -i
    di/dt = -i/RC

    ^ not sure if my maths makes sense (not very good at doing dif/integration when there are multiple variables). I did get the "answer" (which is easy, since its given), but obviously the method's what matters.

    Thanks a lot!
     
  2. jcsd
  3. Nov 6, 2012 #2
    It's right but you may have gotten there quicker if you just differentiated this:

    V0 - iR - q/C = 0

    with respect to time right away. You'll notice the answer is in terms of di/dt so you don't really want to substitute i=dq/dt into that equation.
     
  4. Nov 6, 2012 #3
    Thanks for the reply.

    Do you mean like this?:

    V0 - iR - q/C = 0
    d/dt(V0 - iR - q/C) = d/dt(0)
    d/dt(V0) + d/dt(-iR) + d/dt(-q/C) = 0
    0 - R(di/dt) - (1/C)(dq/dt) = 0
    -R(di/dt) - (i/C) = 0
    di/dt = -i/(RC)

    ^ I pull out the R and C like they're constants (which I think is correct), and got rid of V0 like you would when you dif a constant.

    Edit: I realise the 3rd step isn't required, but just to step it out for myself, I included it.
     
  5. Nov 6, 2012 #4
    Yes that's right too. I wouldn't have shown so many steps but it's not too different anyway.

    V0 - iR - q/C = 0
    d/dt (V0 - iR - q/C) = d/dt (0)
    -R di/dt - i/C = 0 ;; maybe they want you to say i = dq/dt here
    di/dt = - i/(RC)
     
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