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Derivation of Source-Free RC circuit

  1. Mar 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi,
    If there is a series combination of a resistor and an initially charged capacitor, I know that the decay of the voltage is given by the equation v(t) = Ae^(-t/RC) where V(0) = A = V0. But i am unsure of how to get to this equations.

    2. Relevant equations
    If I assume Ir = current flowing through resistor
    and Ic = current flowing through Capacitor and assume there both flowing out of the node.


    3. The attempt at a solution
    Then:
    Ic + Ir = 0
    C(dv/dt) + V/R = 0
    V/R = -C(dv/dt)
    1/V(dv) = -1/RCdt
    integrate both sides
    ln v = -t/RC + A
    A= integration constant

    Here is where i cant go no more I saw somewhere they got: ln v = -t/RC + ln A, from there I know how to get the solution but how to arrive at ln A?, can we just do this automatically because it is a constant to make life easier or is there some logic behind it?
     
  2. jcsd
  3. Mar 28, 2014 #2
    Two things you can do here. You're answer is already correct even though it says "+A" and not "+ln(A)".

    Let me explain.

    When you get a constant of integration, it is representative of a single value determined from initial conditions. If no conditions are given, it remains a letter; in your case, "A". Now think what ln(A) is. The natural log of a constant is what? A constant! So leaving it as "A" is totally fine, because "A" just represents a constant anyway.

    The reason the solution has "+ln(A)" is because when they evaluated the left hand integral of dV/V, then wrote it as ln(V) - ln(A). Personally, I write mine like this so I can evaluate the integral as ln(V/A). It's all preference. What really matters is how you implement the initial conditions.
     
    Last edited: Mar 28, 2014
  4. Mar 28, 2014 #3
    Now, to elaborate on the above (sorry for separate post) equation of v(t) = Ae(-t/RC) where V(0) = A = V0.

    Your equation is ln(V) = -t/RC + A.
    So let's check this out with the info I gave you in my previous post:

    Remove the natural log: eln(V) = e-t/RC + A
    Now rewrite: V = e-t/RCeA
    Rearrange: V = eAe-t/RC

    Now above I told you ln(A) is a constant, as is A..so it's literally the same thing because it just represents a constant. So why don't we look at eA. That's a constant raised to a constant power, right? So it's still just a constant. Just call it A (or C, or whatever, it doesn't matter, as long as you know it's just a single value).

    Now we have: V = Ae-t/RC.

    Now just implement your initial conditions!
     
  5. Mar 28, 2014 #4

    rude man

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    Gold Member

    ln v = -t/RC + A
    ln v(0+) = -t/RC|t=0 + A
    So A = ln v(0+)
    and ln v = ln v(0+) - t/RC
    so take exp of both sides to get your answer. Remember exp(a + b) = exp(a)exp(b).
     
  6. Mar 28, 2014 #5
    I would like to thank both of you and taking the time out to answer my question. It has made me understand the derivation thoroughly.
     
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