Find the charge on the capacitor in an LRC-series circuit at t = 0.03 s when L = 0.05 h, R = 6 Ω, C = 0.005 f, E(t) = 0 V, q(0) = 4 C, and i(0) = 0 A. (Round your answer to four decimal places.)
Determine the first time at which the charge on the capacitor is equal to zero. (Round your answer to four decimal places.)

The capacitor consists of two parallel plates used to store the charge when current flows through it, and provides energy when shuts off.

By the Kirchoff's second law,

Lq'' +Lq' +q/C =E(t)

Where, L =0.05h, R = 6 Ω, C = 0.005 f, E(t) = 0 V, q(0) = 4 C, and i(0) = 0 A.

Equations becomes q''+120q' +4000 =0

The auxiliary equation is m² +120m+4000 =0

Solving this quadratic equation, we have

m =-6± 20i

Then , q(t) = Ae^(-60t)cos 20t + Be^(-60t)sin 20t where, A and B are constants.

At t=0, q(0) =7 =A

and q'(t) = A [ -20e^(-60t)sin 20t - 60e^(-60t)cos 20t ] +B [ 20e^(-60t)cos 20t - 60e^(-60t)sin 20t ]

So, i(0)= q'(0) =0 =-60A +20 B

Putting the value of A=7 above, we have B =21.

So, q(t) = 7e^(-60t)cos 20t + 21e^(-60t)sin 20t

The charge on the capacitor at time t =0.015

q(0.015) = 2.8907 C

Thus the charge on capacitor is 2.8907 C.

q(t) = 7e^(-60t)cos 20t + 21e^(-60t)sin 20t =0

Solve for t, we get

e^(-60t)cos 20t + 3e^(-60t)sin 20t\

-1/3 = sin 20t /cos20t

t =1/20 arctan(-1/3)

As, arctan is defined within -90 to +90,

So, 20t = -0.3218 +π =2.8197

t =0.1409s

Thus the first time at which the charge is zero is 0.1409s.

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