An L - C circuit consists of an inductor with an inductance of 61.0 mH and a cap

Question

An L-C circuit consists of an inductor with an inductance of 61.0 mH and a capacitor with a capacitance of 300 F . The initial charge on the capacitor is 6.40 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?

E)When the current in the inductor has half its maximum value, what is the energy stored in the inductor?

Explanation / Answer

here,

inductance , L = 61 mH

L = 0.061 H

capacitance , C = 300 uF

C = 0.0003 F

initial charge , Q' = 6.4 uC

Q' = 6.4 * 10^-6 C

initial current is zero

total energy , E = 0.5 Q^2/C + 0.5 LI^2

E = 0.5 *(6.4 * 10^-6)^2 / 0.0003

E = 6.83 * 10^-8 J

(a)

let thge maximum voltage be V

for maximum voltage

E = 0.5 * C * V^2

6.83 * 10^-8 = 0.5 * 0.0003 * V^2

V = 0.021 V

the maximum voltage across the capacitor is 0.021 V

(b)

let the maximum current in the inductor be I

for maximum current,

E = 0.5 * L * I^2

6.83 * 10^-8 = 0.5 * 0.061 * I^2

I = 1.5 * 10^-3 A

the maximum current in the inductor is 1.5 * 10^-3 A

(C)

using conservation of energy

the maximum energy stored in the inductor = E

the maximum energy stored in the inductor is 6.83 * 10^-8 J

(d)

current in the inductor , I' = I/2

I' = 0.75 * 10^-3 A

let the charge on the capacitor be Q'

using conservation of energy

E = 0.5 * Q'^2/C + 0.5 * L * I'^2

6.83 * 10^-8 = (0.5 * Q'^2)/(0.0003) + 0.5 * 0.061 * ( 0.75 * 10^-3)^2

Q' = 5.54 * 10^-6 C

the charge on the capacitor is 5.54 * 10^-6 C

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