- #1

zenterix

- 555

- 76

- Homework Statement
- Initially, the capacitor in a series LC circuit is charged. A switch is closed, allowing the capacitor to discharge. After a time ##t_1##, the energy stored in the capacitor is one fourth of its initial value.

- Relevant Equations
- Calculate the value of the self-inductance ##L## in terms of ##t_1## and ##C##.

We seem to have the following circuit

and the differential equation ##\ddot{q}+\omega^2q=0## where ##\omega=\frac{1}{\sqrt{LC}}##.

The solution is

$$q(t)=A\cos{(\omega t+\phi)}$$

Since ##I(t)=-\dot{q}(t)## we have

$$I(t)=-\dot{q}(t)=A\omega\sin{(\omega t+\phi)}$$

At ##t=0## we have

$$q(0)=A\cos{\phi}$$

$$I(0)=A\omega\sin{\phi}=0\implies \phi=0$$

Thus,

$$q(0)=A$$

There is no battery in the circuit and no resistance in the wires. The total energy in inductor and capacitor is constant.

The energy in the capacitor is

$$U_E(t)=\frac{q(t)^2}{2C}=\frac{A^2}{2C}\cos^2{(\omega t)}$$

and the energy in the inductor is

$$U_B(t)=\frac{LI^2(t)}{2}=\frac{L}{2}A^2\omega^2\sin^2{(\omega t)}$$

$$=\frac{A^2}{2C}\sin^2{(\omega t)}$$

I am stuck trying to find an expression for ##L## in terms of ##t_1##.

and the differential equation ##\ddot{q}+\omega^2q=0## where ##\omega=\frac{1}{\sqrt{LC}}##.

The solution is

$$q(t)=A\cos{(\omega t+\phi)}$$

Since ##I(t)=-\dot{q}(t)## we have

$$I(t)=-\dot{q}(t)=A\omega\sin{(\omega t+\phi)}$$

At ##t=0## we have

$$q(0)=A\cos{\phi}$$

$$I(0)=A\omega\sin{\phi}=0\implies \phi=0$$

Thus,

$$q(0)=A$$

There is no battery in the circuit and no resistance in the wires. The total energy in inductor and capacitor is constant.

The energy in the capacitor is

$$U_E(t)=\frac{q(t)^2}{2C}=\frac{A^2}{2C}\cos^2{(\omega t)}$$

and the energy in the inductor is

$$U_B(t)=\frac{LI^2(t)}{2}=\frac{L}{2}A^2\omega^2\sin^2{(\omega t)}$$

$$=\frac{A^2}{2C}\sin^2{(\omega t)}$$

I am stuck trying to find an expression for ##L## in terms of ##t_1##.