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LC Circuit

  1. Feb 25, 2008 #1
    Hello guys. I have a simple question regarding an LC circuit.

    Imagine a voltage source [itex]V_0[/itex], a capacitor [itex]C[/itex] and an inductor [itex]L[/itex], all hooked up in series. I know that the equation governing the behvior of the system is


    and hence

    [tex]q(t)=A\cos \omega t + B\sin \omega t + CV_0.[/tex]

    What I'm having trouble with is the initial conditions. Is it fair to assume that in [itex]t=0[/itex] there is no charge nor current in the system?

    If I put a switch in the system, how would the initial conditions change (assuming is open in [itex]t=0[/itex] and closed in [itex]t>0[/itex])?
  2. jcsd
  3. Feb 25, 2008 #2
    Yes because the voltage across a capacitor cannot change instantaneously and then current though an inductor cannot change instantaneously.
  4. Feb 25, 2008 #3
    So the answer is

    [tex]q(t)=CV_0 (1-\cos \omega t)[/tex]

    no matter if I have a switch or not?
  5. Feb 26, 2008 #4
    Well, in case you have been wondering, It's all wrong!

    What I have to do is imagine a RLC ciruit, solve it with conditions [itex]i(0)=V_0/R, \, i'(0)=0[/itex], integrate in [itex]t[/itex], divide by [itex]C[/itex] and then take the limit as [itex]R \rightarrow 0[/itex]. Then I'll know what's the voltage passing trough the capacitor on my original LC circuit!

  6. Feb 26, 2008 #5
    Well, you didn't give us that initial set of conditions.
  7. Feb 27, 2008 #6
    I know, I know. It was exactly that what made me realize my minstake. Thanks for the input tough, you really got me thinking.
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