Question

1.A 0.230 H inductor is wired in series with a LaTeX: 9.20\:\mu F9.20 μ F capacitor. The capacitor is fully charged when the circuit is closed at time t = 0. Calculate the rate of change in current through the inductor at t = 0. Assume that the initial charge (t = 0) on the capacitor is LaTeX: 5.30\mu C5.30 μ C

2.On the previous question at what time will the charge on the capacitor fall to half its initial value? State the first time that this occurs

Answer 1

1. Resonance frequency of the LC circuit will be

w = 1/(LC) = 1/(0.230*9.20*10^-6)

= 689.66 rad/s

If L be the inductance, C be the capacitance, Q_o be the initial charge on the capacitor, then,

L(di/dt) = Q_o/C

So, rate of change of corrent = di/dt = Q_o/LC

= (5.30*10^-6/0.230*9.20*10^-6)

= 2.504 A/s

(b) At any time t, charge on the capacitor is,

Q = Q_o cos(wt+)

Where, is the phase at t= 0.

Now, at t = 0, Q = Q_o, so, from above equation,

Q_o = Q_o cos(w*0+)

Or, cos = 1 = cos 0

So, = 0.

Let t' be the time when charge reduces half, then,

Q_o/2= Q_o cos(wt')

Or, 1/2 = cos(wt')

Or,   /3 = wt'

So, required time = /3w = (3.14/3*689.66) sec

= 1.52*10^-3 sec

= 1.52 millisecond