An L-C circuit consists of a 69.0-mH inductor and a 255-plF capacitor. The initi

Question

An L-C circuit consists of a 69.0-mH inductor and a 255-plF capacitor. The initial charge on the capacitor is 6.20 ?C, and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (d) When the current in the inductor has half its maximum value, what is the charge on the capacitor? When the current in the inductor has half its maximum value, what is the energy stored in the inductor?

Explanation / Answer

A.

Q = C*V

Vmax = Q/C = 6.20*10^-6/(255*10^-6)

Vmax = 0.024 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 = 6.20*10^-6/sqrt (69*10^-3*255*10^-6)

i)max = 0.00147 Amp = 1.47*10^-3 Amp

Part C

Umax = 0.5*L*i)max^2

Umax = 0.5*69*10^-3*(1.47*10^-3)^2 = 7.45*10^-8 J

Part D

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

UL = Umax/4

UC = 3*Umax/4

UC = 3*Q^2/(4*2*C) = q^2/2C

q = Q*sqrt (3/4)

q = 6.2*10^-6*sqrt (3/4) = 5.37*10^-6 C

Part E.

energy stored in inductor will be

UL = Umax/4

UL = 7.45*10^-8/4 = 1.86*10^-8 J

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