Question

A 30.0?F capacitor is charged to 210?C and then connected across the ends of a 4.20mH inductor.

A.Find the maximum current in the inductor.

B.At the instant the current in the inductor is maximal, how much charge is on the capacitor plates?

C.Find the maximum potential across the capacitor.

D.At the instant the potential across the capacitor is maximum, what is the current in the inductor?

E.Find the maximum energy stored in the inductor.

F.At the instant the energy stored in the inductor is maximum, what is the current in the circuit?

Please explain all steps clearly.

Answer 1

A.

in laplace u get the solution to be

I(S) = 7/(SL) where L is the iductance.

take inverse laplace transform to get

i(t) = 7/L*u(t) so the maximum current would be 1.66 k A.

B.

the charge in the capacitor plate is when the current is maximum is 210u C.

c.

maximum voltage at the instant V = (q/C) = 210/30 = 7V.

d.

the max current is same as deirved 7/L = 1.66k

E.

the energy stored in this system is

E = 1/2*C*V^2 = 0.5*(210*!)^-6)^2/(30*!)^-^6) J = 0.0735 m J.

F..

current here also will be 1.66k A.

actually this will be a ocsillator circuit so the votage will shift from capacitor to inductor periodically. so also the current.

u can find that the governing equation for the circuit is

V = L*di/dt and i = C*dv/dt

as no externa; voltage source is given so

LC*(d2v/dt2) = V-------------------------------------which is the general ocsillation formula with solution

V = asin(wt) + b cos(wt) , where w is the ocsillation frequany = 1/sqrt(LC)