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Laplace Circuit Problem

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the capacitor voltage in the laplace domain.

    2. Relevant equations


    3. The attempt at a solution


    Using nodal analysis:
    [itex]\frac{Vc-20/s}{3+s}[/itex] + Vc0.5s -4=0
    => Vc= [itex]\frac{8s^2+24s+40}{s(s+1)(s+2)}[/itex]

    but the answer should be:
    Vc= [itex]\frac{24}{s(s+1)(s+2)}[/itex] which is my answer minus 8/s
    After studying the laplace equivalents of capacitors and inductors, i still feel i am doing something wrong in deciding what the polarity of the added sources is, or which voltage in the laplace circuit (the one across the capacitor or the one across the capacitor and the addded source) i should treat as the actual voltage of the capacitor.

    I hope someone could help me out. Thank you.
  2. jcsd
  3. Apr 10, 2012 #2


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    Staff: Mentor

    Hi kkasper, Welcome to Physics Forums.

    It looks to me like you've got a pretty good handle on it. The result that you found is the expression for the node voltage where the inductor and capacitor meet, which in the "real world" will be the true voltage across the capacitor. The "answer" that you're comparing it to would be the potential across the capacitor less the initial potential that it started with.

    There are two equivalent circuits that you can apply for a capacitor with an initial charge. One has a current source added in parallel as you've done, and the other places a voltage source in series with an uncharged capacitor. In this case that voltage source would be 8/s.
  4. Apr 10, 2012 #3
    Hey gneill,

    When transforming the correct answer back to the time domain Vc(0)= 8 V (the initial potential) as expected. and Vc(infinity)=20 V as expected, contradicting what you said. Did i misunderstand u ?
  5. Apr 10, 2012 #4


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    Staff: Mentor

    Maybe :smile: If I take the inverse Laplace transform of the "answer" I see:
    $$ Vc(s) = \frac{24}{s (s + 2)(s + 1)} \Rightarrow Vc(t) = 12V(1 - 2 exp(-t) + exp(-2 t))$$
    which, for t = 0, is zero volts; No 8V initial capacitor voltage there!

    On the other hand, when I take your expression and find the time domain expression, it gives 8V at time t = 0 as expected.
  6. Apr 10, 2012 #5
    OK. Now I see : ). In the solution they actually add the initial voltage afterwards. But then wouldn't my answer be correct? The original question was find Vc(s) for t≥0. This is an actual university exam question. Maybe the capacitor voltage in the laplace domain is defined, or is by convention amongst electric engineers defined to be the voltage across the capacitor element in the laplace circuit?

    Thanks a lot for your time, im studying for my exams and this helped me a lot. :)
  7. Apr 10, 2012 #6


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    Staff: Mentor

    I'm not aware of any such definition. I'd say that your answer is correct for the actual time domain voltage across the circuit element.

    If the question had asked for an expression for the change in potential across the capacitor after t=0, that would be another story.
    No problem, glad to help.
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