Cannot catch a minus sign mistake when deriving the ODE for an LC circuit

Question

It's probably a very basic question but I just cannot wrap my mind around it. I'll just try to derive the differential equation for an LC circuit:

According to the law of induction, in a solenoid we have:

$$U_L=-N\frac{\text{d}\Phi}{\text{d}t} = -NA\frac{\text{d}B}{\text{d}t}$$

where $N$ is the winding number of the solenoid, $A$ is its constant cross section area and $B$ is the magnetic flux density induced by the solenoid.

The flux density in a solenoid is known to be

$$B=\mu_0\mu_r\frac{N}{l}I_L$$

where $l$ is the solenoid's length and $I_L$ is the current that flows through it. Since only the current is time dependent we get:

$$U_L=-\mu_0\mu_r\frac{N^2A}{l}\frac{\text{d}I_L}{\text{d}t}$$

The constant prefactor is known as the solenoid's inductivity $L=\mu_0\mu_r\frac{N^2A}{l}$ which gives

$$U_L=-L\frac{\text{d}I_L}{\text{d}t}$$

Now, since the capacitor and the solenoid are the only elements in the circuit we get

$$U_C + U_L=0$$

For a (plate) capacitor with capacity $C$ we have

$$U_C = \frac{Q_C}{C}$$

where $Q_C$ is the charge on one of the capacitor's plates.

The current is constant in the whole circuit: $$I_C=I_L$$

But the current through the capacitor is given as $$I_C=\frac{\text{d}Q_C}{\text{d}t}$$

Plugging it all together, we get:

$$0=U_C+U_L = \frac{Q_C}{C} + \left(-L\frac{\text{d}I_L}{\text{d}t}\right) = \frac{Q_C}{C} - L\frac{\text{d}^2Q_C}{\text{d}t^2} $$

As you can see, this can't be correct, as we get an exponential rather than a sine/cosine solution.

So my question is: How to properly define the sign/"direction" for $Q$, $U$ and $I$ in this setup to get the signs right? Where did the mistake happen which led to the wrong result?

My thoughts so far:

• The charge $Q$ is defined by considering one of the capacitor plates. Say, the initially positively charged plate, that is, $Q_C(t=0)>0$. So, in the beginning, when the capacitor starts to discharge, the positive charge on this plate decreases and thus $I_C$, and thus $I_L$ as well, is negative.
• Then, according to the law of induction, $U_L$ is positive (Lenz's law, the minus sign).
• But we know that then the capacitor voltage $U_C=-U_L$ is negative.
• But this contradicts to the assumption that the capacitor's charge is initially positive and violates $Q_C=+CU_C$.

Where is the flaw in this consideration?

Answer 1

You're using opposite sign conventions for the potential differences across the capacitor and the inductor.

Arbitrarily designate one terminal of the capacitor as $C_{in}$ and the other one as $C_{out}$. Similarly, for the inductor, designate the terminals as $L_{in}$ and $L_{out}$. Connect $C_{in}$ to $L_{out}$ and $C_{out}$ to $L_{in}$. Define $I$ to be positive when it flows through the components from input to output.

You then have to choose a sign convention for each of $U_L$ and $U_C$. For example, you can write $U_L = V_{L_{out}} - V_{L_{in}}$ or you can write $U_L = V_{L_{in}} - V_{L_{out}}.$ Neither is inherently better, but the conventions must be consistent, or else you will not have $U_L + U_C = 0$.

Your reasoning for the inductor is correct if $U_L = V_{L_{out}} - V_{L_{in}}.$ (An increasing current through the inductor produces a voltage directed toward the input.)

Your reasoning for the capacitor is correct if $U_C = V_{C_{in}} - V_{C_{out}}$.

However, these sign conventions are inconsistent:

$$\begin{align} U_L + U_C &= V_{L_{out}} - V_{L_{in}} + V_{C_{in}} - V_{C_{out}} \\ &= 2V_{C_{in}} - 2V_{C_{out}} \\ &\ne 0 \end{align}$$

You can pick either of the sign conventions to reverse.

Answer 2

You seem to be confusing, in the law of induction, induced emf on the inductor (line integral of induced electric field in the positive direction, which has value $-L\frac{dI}{dt}$) with voltage on the inductor (drop of electric potential in the positive direction, thus integral of conservative field in the positive direction, with value $+L\frac{dI}{dt}$).

In AC circuit theory, and in particular, in KVL, voltage on an element means drop of electric potential, when going from one terminal to another, in the chosen positive direction. On a perfect inductor with zero ohmic resistance, this voltage can be expressed as

$$ V_{perfect~inductor} = L\frac{dI}{dt}, $$ where $I$ denotes current flowing through it in the positive direction.

This is so because in the conductive coils of the perfect inductor, the conservative part of net electric field completely cancels out the (non-conservative) induced part of net electric field. These two electric fields cancel each other in the conductor the inductor is made of, because net electric field vanishes there, as zero net field is needed for current to flow.

For a capacitor in a given circuit element with chosen positive direction, we define $Q$ in such a way that drop of potential on it in the positive direction is $Q/C$, and current in the positive direction is $I=\dot{Q}$. This implies that $Q$ is charge on that plate, into which the chosen positive direction of current enters:

     <---    |    |    <---
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             |    | 
                  Q

For a simple LC circuit, the KVL rule gives

$$ V_{L} + V_{C} = 0, $$ and using the above expressions, we get the equation $$ L\frac{dI}{dt} + \frac{Q}{C} = 0. $$ Taking time derivative of both sides, we get the correct equation for current $I$, which predicts harmonic oscillation of the current: $$ L\frac{d^2I}{dt^2} + \frac{I}{C} = 0. $$