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Solving a Second Order Circuit for Capacitor Voltage

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data

    upload_2014-10-10_11-20-43.png
    upload_2014-10-10_11-20-49.png
    2. Relevant equations

    Here is the technique I am using:

    upload_2014-10-10_11-21-43.png

    3. The attempt at a solution

    img001.jpg
    img002.jpg

    I understand how to solve the problem using the technique provided by the solution but I was wondering where I messed up in the technique that I used. I prefer the second technique because there is less memorization.
     
  2. jcsd
  3. Oct 10, 2014 #2

    gneill

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    Staff: Mentor

    Did you have a question? The provided solution looks pretty thorough except that they didn't explicitly state the steady-state solution.
     
  4. Oct 10, 2014 #3
    The problem is that the provided solution doesn't derive the differential equations from scratch. It uses equations that we are supposed to memorize. I would prefer to just solve the problem from scratch instead of plugging in and chugging. I am trying to derive the differential equations from scratch and I did so but my final solution is incorrect.
     
  5. Oct 10, 2014 #4

    gneill

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    Staff: Mentor

    When I do nodal analysis I see:
    $$\frac{v}{R} + \frac{1}{L} \int v\;dt + C \frac{dv}{dt} = 0 $$
    Clear the integral by differentiating the whole thing:
    $$\frac{1}{R} \frac{dv}{dt} + \frac{1}{L} v + C \frac{d^2 v}{dt^2} = 0$$
    $$\frac{d^2 v}{dt^2} + \frac{1}{R C} \frac{dv}{dt} + \frac{1}{L C} v = 0$$
    Plugging in component values and changing notation:
    RC = 1
    LC = 1/4

    thus:

    v'' + v' + 4v = 0

    So you have complex conjugate roots, not a double real root. This makes sense since the circuit is underdamped.
     
  6. Oct 10, 2014 #5
    I just did it this way and it seems to be correct. Is there anything I did wrong with my method?

    Actually, I think I figured out what I did wrong. I substituted C=0.25 instead of C=1 and I substituted L=1 instead of L=0.25. Opps.
     
    Last edited: Oct 10, 2014
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