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Capacitor & Inductor ?

  • Thread starter mkbh_10
  • Start date
222
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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 ?
 

Answers and Replies

russ_watters
Mentor
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That sounds like homework - what do you think the answer is?
 
10
0
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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