A capacitor of a given capacitance C is directly connected to a battery of volta

Question

A capacitor of a given capacitance C is directly connected to a battery of voltage Delta V and allowed to charge Once fully charged it is removed from the battery and its ends directly connected to the respective ends of an inductor with an inductance L. The circuit is finally connected once a switch is closed at time t - 0. You may assume this to be an ideal circuit with zero resistance. a) What is the charge on the capacitor at time t = 0? b) What is the current through the inductor at time t = 0? c) Al what lime docs the capacitor first reach its initial charge, but with opposite polarity? d) At what time docs the current first reach its maximum value? e) How much energy is stored in the electric field inside the capacitor's volume at time t = 0?

Explanation / Answer

Charge is q = CV where V is the voltage of battery

At t = 0 , Current in the inductor = 0. because inductor does not allow capacitor to discharge suddenly.

VOltage in circuit varies as sqrt(L/c) * I max * cos( t/sqrt(LC) + phi)

So q = CV = sqrt (LC) * I max * cos ( t/sqrt ( LC) + phi)

at t = 0 phi = 0 and CV = sqrt (LC) * I max. So Imax = sqrt ( C/L) * V

So from initial conditions

q = CV * cos ( t/sqrt ( LC)) so after t = pi * sqrt (LC) polarity of capacitor is reversed

I = - I max * sin( t /sqrt (LC)) .So at t = pi/2 * sqrt (LC) current reaches maximum.

At t =0 energy stored is 1/2 C V^2

As energy is conserved Energy is always 1/2 CV^2 = 1/2 L Imax^2

Rms current = I max * sqrt 2 = I rms

So I rms = sqrt (C/L) * v * sqrt(2)

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