# Capacitor & Inductor ?

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 ?

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

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 transfered 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 transfered 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 transfered 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.