In an oscillating LC circuit, L = 8.07 mH and C = 1.52 F. At time t = 0, the cur

Question

Explanation / Answer

given that

L = 8.07 mH

C = 1.52 uF

I = 14.2 mA

(a)

we know that

energy of system

E = (1/2)*L*I^2

it is equal to maximum energy stored on the capacitor

E = (1/2)*C*Vmax^2

so (1/2)*L*I^2 = (1/2)*C*Vmax^2

Vmax = sqrt (L*I^2/C)

we know that

Qmax = C*Vmax

so Qmax = C*sqrt (L*I^2/C) = sqrt(L*C*I^2)

Qmax = sqrt [ 8.07*10-3 * 1.52*10-6 * (14.2*10-3)2 ]

Qmax = 15.72*10-7 C

(b)

rate of change of energy in the capacitor will be maximum when current is maximum

so that time is when t = 0 and every 360 deg of the oscillation frequency thereafter.

(c)

rate of change of energy in the capacitor maximum

we know that

I = C*dV/dt

so dV/dt = I/C = 14.2*10-3/1.52*10-6

dV/dt = 9342.10 J/s

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