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Solving 2nd ODE for RLC circuit

  1. Mar 31, 2013 #1
    This is really more of a mathematical question than physics.

    Given a RLC circuit, I will arrive at the following DE:

    [itex]\ddot{Q}+\frac{R}{L}\dot{Q}+\frac{1}{LC}Q-\frac{\epsilon}{L}=0[/itex]

    How do I solve for [itex]Q(t)[/itex]??
     
  2. jcsd
  3. Apr 1, 2013 #2
    A good way is the Laplace transform. Given our equation [itex]\ddot{q}(t) + \frac{R}{L}\dot{q}(t) + \frac{1}{LC}q(t) = \frac{\epsilon}{L}[/itex], we can take the Laplace transform of the equation (denoted by [itex]\ell[/itex]):

    [itex]\ell \{ \ddot{q}(t) \} + \frac{R}{L} \ell \{ \dot{q}(t) \} + \frac{1}{LC} \ell \{ q(t) \} = \ell \{ \frac{\epsilon}{L} \}[/itex]

    [itex][ s^2 Q(s) - sq(0) - \dot{q}(0) ] + \frac{R}{L} [ sQ(s) - q(0) ]+ \frac{1}{LC} Q(s) = \frac{\epsilon}{Ls}[/itex]

    [itex][LCs^2 + RCs+ 1] Q(s) = \frac{C\epsilon}{s} + [LCs + RC]q(0) + LC\dot{q}(0)[/itex]

    [itex]Q(s) = \frac{C\epsilon}{s[LCs^2 + RCs+ 1]} + \frac{[LCs + RC]q(0) + LC\dot{q}(0)}{[LCs^2 + RCs+ 1]}[/itex]

    This is as far as I wanted to go without numbers :smile: If I had numbers, I would substitute them at this point and put things in terms of simpler Laplace transforms so I can do inverse Laplace transforms on each part. You can a table of selected ones here:

    http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms
     
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