In the circuit shown, the switch has been open for a long time. It is now closed

Question

In the circuit shown, the switch has been open for a long time. It is now closed quickly.

1. Calculate the frequency of the resulting oscillating current.

2. What is the amplitude of the current oscillations?

3. What is the current when the energy stored in the inductor is exactly the same as the energy stored in the capacitor?

In the circuit shown, the switch has been open for a long time. It is now closed quickly. 1. Calculate the frequency of the resulting oscillating current. 2. What is the amplitude of the current oscillations? 3. What is the current when the energy stored in the inductor is exactly the same as the energy stored in the capacitor?

Explanation / Answer

L = 54 mH = 0.054 H

C = 6.2 uF = 6.2*10^-6 F

R = 14 ohm

V = 34 V

1)

Frequency of oscillating current , fr = (1/(2*pi))/sqrt(L*C)

So, fr = (1/(2*pi))/sqrt(0.054*6.2*10^-6)

So, fr = 275.1 Hz <--------answer

2)

For, peak current ,

the energy stored in Capacitor must be converted to energy stored in Inductor,

So, 0.5*CV^2 = 0.5*L*Io^2

where Io = peak current

V = 34 V

So, 6.2*10^-6*34^2 = 0.054*Io^2

So, Io = 0.36 A <--------answer

3)

energy stored in inductor,Ul = 0.5*L*I^2

energy stored in capacitor, Uc = 0.5*CV^2

Now, for Ul = Uc

So, LI^2 = CV^2

Now, total energy stored in capacitor, Uc = 0.5*C*V^2

So, V = sqrt(2*Uc/C)

Uc is maximum V = 35 V ,

when Uc = half the value, V = sqrt(0.5)*35 = 24.8 V

So, L*I^2 = C*24.8^2

So, now, I = sqrt(6.2*10^-6*24.8^2/0.054) = 0.266 A <-------answer

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