LC circuit

From Physics wiki

Given a closed circuit with a source of constant voltage $V$ , an ideal capacitor with capacitance $C$ and an ideal inductor with inductance $L$ , assuming that the wires and circuit elements have negligible internal resistances, we analyse the circuit as follows.

Assign a direction to the unknown time-varying current $I (t)$ and use Kirchoff's voltage law. Counting voltage increases around the loop we have: $V - \frac{Q(t)}{C} - L\frac{dI(t)}{dt} = 0$

Where $Q (t)$ is the charge accumulated on the capacitor at time $t$ . Since $I(t) = \frac{dQ(t)}{dt}$ we can write this as a second-order inhomogeneous differential equation:

$\frac{d^{2}Q(t)}{dt^2} + \frac{Q(t)-VC}{LC} = 0$ We define $A(t)\equiv \frac{Q(t)-VC}{LC}$ , consequently $\frac{d^{2}A(t)}{dt^2} = \frac{1}{LC}\frac{d^{2}Q(t)}{dt^2}$ which removes the inhomogeneous term:

$\frac{d^{2}A(t)}{dt^2} + \frac{A(t)}{LC} = 0$

Multiplying by $\frac{dA(t)}{dt}$ and integrating results in a constant of motion:

$\frac{dA(t)}{dt}\frac{d^{2}A(t)}{dt^2} + \frac{dA(t)}{dt}\frac{A}{LC} = 0$

$\frac{1}{2}\frac{d}{dt}\left(\frac{dA(t)}{dt}\right)^2 + \frac{1}{2LC}\frac{d}{dt}A(t)^2 = 0$

$\frac{d}{dt}\left(\frac{dA(t)}{dt}^2 + \frac{A(t)^2}{LC}\right) = 0$

$\frac{dA(t)}{dt}^2 + \frac{A(t)^2}{LC} = B$

Where $B$ is the constant of motion in question. Replacing the substitution for $A (t)$ we find that $B$ is (up to a multiplicative factor of $\frac{L^3 C^2}{2}$ ) the total energy of the system. Thus this is a statement of energy conservation.

The equation above is now separable and we integrate it.

$\frac{dA(t)}{dt}^2 = B - \frac{A(t)^2}{LC}$

$\frac{dA(t)}{dt} = \pm\sqrt{B-\frac{A(t)^2}{LC}}$

$\pm\int^A_{A_0}\frac{dA}{\sqrt{B-\frac{A^2}{LC}}} = \int^t_{t_0}dt$

$\sqrt{LC}\int\frac{dz}{\sqrt{1-z^2}} = \pm(t-t_0)$ where we substituted $z = \frac{A}{\sqrt{BLC}}$

We can perform the $z$ integral using the derivative of inverse sine.

$\sqrt{LC}\left(\arcsin{\left(\frac{A(t)}{\sqrt{BLC}}\right)} - \arcsin{\left(\frac{A(t_0)}{\sqrt{BLC}}\right)}\right) = \pm(t-t_0)$

$A(t) = \sqrt{BLC}\sin{\left(\pm\frac{t-t_0}{\sqrt{LC}} + \arcsin{\left(\frac{A(t_0)}{\sqrt{BLC}}\right)}\right)}$

Putting back all of our substitutions we have:

$Q(t) = \sqrt{B(LC)^3}\sin{\left(\pm\frac{t-t_0}{\sqrt{LC}} + \arcsin{\left(\frac{Q(t_0)-VC}{\sqrt{B(LC)^3}}\right)}\right)} + VC$

$I(t) = \frac{dQ(t)}{dt} = \pm LC\sqrt{B}\cos{\left(\pm\frac{t-t_0}{\sqrt{LC}} + \arcsin{\left(\frac{Q(t_0)-VC}{\sqrt{B(LC)^3}}\right)}\right)}$

We can check our units by substituting these back into the equation defining $B$ and seeing if it is self-consistent. Recall that $\frac{BL^3 C^2}{2} = E_{total}$ where $E t o t a l$ is the total energy of the system and determines the amplitude of the oscillations.

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LC circuit

From Physics wiki

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