Question

An *L-C* circuit consists of a 69.5-mH inductor and a
240-µF capacitor. The initial charge on the capacitor is 5.95 µC,
and the initial current in the inductor is zero.

(a) What is the maximum voltage across the capacitor?

__________ V

(b) What is the maximum current in the inductor?

__________ A

(c) What is the maximum energy stored in the inductor?

__________ J

(d) When the current in the inductor has half its maximum value,
what is the charge on the capacitor?

__________C

When the current in the inductor has half its maximum value, what
is the energy stored in the inductor?

_____________ J

Answer #1

A.

Q = C*V

Vmax = Q/C = 5.95*10^-6/(240*10^-6)

Vmax = 0.0247 V

Part B

Energy stored is given by:

U = 0.5*L*i)max^2 = 0.5*Q^2/C

i)max = Q/sqrt (LC)

i)max = 5.95*10^-6/sqrt (69.5*10^-3*240*10^-6)

i)max = 0.00146 Amp = 1.46*10^-3 Amp

Part C

Umax = 0.5*L*i)max^2

Umax = 0.5*69.5*10^-3*(1.46*10^-3)^2 = 7.41*10^-8 J

Part D

If i = i)max/2, then energy stored in inductor

U_{L} = Umax/4

U_{C} = 3*Umax/4

U_{C} = 3*Q^2/(4*2*C) = q^2/2C

q = Q*sqrt (3/4)

**q = 5.95*10^-6*sqrt (3/4) = 5.15*10^-6 C**

Part E.

energy stored in inductor will be

U_{L} = Umax/4

U_{L} = 7.41*10^-8/4 = 1.85*10^-8 J

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