Can a capacitor discharge itself through an ideal inductor?

[mkbh_10]

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 ?

 

[russ_watters]

That sounds like homework - what do you think the answer is?

 

[a-lbi]

So the capacitor equation is:

C \frac{du(t)}{dt} = i(t)

where u - voltage, i - current, C capacitance

And for inductor:

L \frac{di(t)}{dt} = u(t)

L - inductance

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

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)

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

W_L = \frac{1}{2} Li^2(t_2) - \frac{1}{2} Li^2(t_1)

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

W_C = W_L

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

i(t_2) = \sqrt{\frac{C}{L}} V

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