A capacitor with C = 3.30 times 10^-3 F has an initial charge Q_0 = 7.50 times 1

Question

A capacitor with C = 3.30 times 10^-3 F has an initial charge Q_0 = 7.50 times 10^-3 C. It is in a loop containing a switch and an inductor with L = 4.50 times 10^- 3 H and negligible resistance. At t = 0, the switch is closed. I. Calculate the energy in the capacitor at immediately after the switch is closed (t = 0). 8.52 times 10^-3 J. II. Calculate minimum time it takes for the energy in the capacitor to return to its initial value after the switch is closed. III. Calculate the magnitude of the current through the inductor at time t = 3.61 times 10^-3 s.

Explanation / Answer

B.

the minimum time it takes to reach it's initial value after the switch is closed is half of the time t

t = 2*pi/w

w = sqrt (1/LC)

t = 2*pi*sqrt (LC)

t1 = t/2

t1 = 2*pi*sqrt (4.5*10^-3*3.3*10^-3)/2

t1 = 0.0121 sec

C.

I = I0*sin wt

I0 = V/(XL) = V/(w*L)

Q = C*V

V = Q/C

we already know that

w = sqrt (1/LC)

So,

I = Q/(sqrt (LC))*sin (t/sqrt (LC))

I = 7.5*10^-3/(sqrt (3.3*4.5*10^-6))*sin (3.61*10^-3/(sqrt (3.3*4.5*10^-6)))

I = 1.568 Amp.

Let me know if you have any doubt.

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