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Homework Help: Capacitor & Inductor ?

  1. Mar 18, 2008 #1
    Can a capacitor discharge itself through an ideal inductor ? If yes state the reason , if No how will the charge on the capacitor will behave ?
  2. jcsd
  3. Mar 18, 2008 #2


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    Staff: Mentor

    That sounds like homework - what do you think the answer is?
  4. Mar 19, 2008 #3
    So the capacitor equation is:

    [tex]C \frac{du(t)}{dt} = i(t)[/tex]

    where u - voltage, i - current, C capacitance

    And for inductor:

    [tex]L \frac{di(t)}{dt} = u(t)[/tex]

    L - inductance

    The energy transfered from capacitor to the circuit is given by:

    [tex]W_C = - \int_{t_1}^{t_2} u(t)i(t)dt = C \int_{t_2}^{t_1} u(t)du = \left.\frac{1}{2} C u^2(t)\right|^{t_1}_{t_2} = \frac{1}{2} Cu^2(t_1) - \frac{1}{2} Cu^2(t_2)[/tex]

    The energy transfered to inductor is given by (similarly):

    [tex]W_L = \frac{1}{2} Li^2(t_2) - \frac{1}{2} Li^2(t_1)[/tex]

    Obviously the energy transfered from capacitor is accumulated in inductor so:

    [tex] W_C = W_L [/tex]

    If we assume that inductor is discharged in the instant [tex]t_1[/tex] then [tex]i(t_1) = 0[/tex], and capacitor is charged to the volgate [tex]v(t_1) = V[/tex]. Assume that the capacitor is discharged in in [tex]t_2[/tex] instant. From the energy ballance we get the current:

    [tex]i(t_2) = \sqrt{\frac{C}{L}} V[/tex]

    Which is obviously different than 0. From this observation you get the answer to your question.
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