Analyzing electric circuit with capacitor, inductor and resistor

Question

In my Calculus book, they state the following for this circuit (image above): $L \frac{dI}{dt} + RI + \frac{Q}{C} = E(t)$, where $E(t)$ is the electromotive force (also known as $\epsilon$), due to Kirchoff's loop rule. However, I thought that you can't use Kirchoff's loop rule here, since there is a change in magnetic flux when turning the switch on due to the great self-inductance by the inductor $L$?

This is what I thought: Applying Faraday's law (not KVL), we get $$E(t) = \oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi}{dt} = -L\frac{dI}{dt}$$ $$E(t) = IR - \frac{Q}{C} = -L\frac{dI}{dt}$$ $$IR - \frac{Q}{C} + L\frac{dI}{dt} = 0$$

(The electric field in the resistor is opposite to the electric field in the capacitor) (We assume that the wires are perfect conducting wires, so that there is no electric field in the wires).

Can someone explain who is correct and what the mistake is?

Answer 1

Faraday is your best friend, especially since you're dealing with changing magnetic fluxes here. If you require some more authority to convince you that you're right, take a look at the wonderful lectures by Walter Lewin on electromagnetism.

Now, let us apply Faraday's law $\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_B}{dt}$. In calculating this contour integral, I'll start at the top left corner and assume the current is flowing clockwise. So, we have: \begin{align} IR +0 +\frac{Q}{C}+ (-E(t))&=-L\frac{dI}{dt} \end{align} Here, the first $IR$ is what you get by integrating $\vec{E}\cdot d\vec{l}$ across the resistor (left to right in the direction of the current). The $0$ is what you get by integrating across the inductor (there's no/negligible electric field in the inductors). The $\frac{Q}{C}$ is what you get by integrating across the capacitor, and the $-E(t)$ is what you get by integrating across the source battery (the minus sign is because I have assumed the current flows clockwise, so at the battery it's going from the bottom to the top, i.e from the "negative" terminal to the "positive" terminal, so the electric field is actually pointing the opposite direction, hence the minus sign). As always the $L\frac{dI}{dt}$ is the effect of the inductor. Rearranging, you get \begin{align} L\frac{dI}{dt}+IR+\frac{Q}{C}&=E(t), \end{align} or writing it all in terms of charges, \begin{align} L\frac{d^2Q}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}&=E(t). \end{align}

Answer 2

You can use the so-called KVL (the Kirchhoff Voltage Law) in AC circuits.

KVL is always valid, for trivial reason: it works intentionally with electric potential. Potential is always well-defined based on the conservative part of total electric field, and sum of its drops in closed path is always zero, whatever the induced field is.

It is true that KVL is sometimes not directly applicable (while still being true); this happens when potential drops on various elements cannot be easily expressed, such as when there is external induced electric field (due to other bodies such as moving magnets or other circuits) acting on current in the circuit. But this is not the case here. But it would be the case in transformer, for example. Then, one must go to the original Kirchhoff's second circuital law (the original formulation of KVL), which does not use potential drops, but complete loop EMFs.

So back to your example, KVL says that sum of all potential drops in a closed path equals zero. Since there is no external source of EMF (such as moving magnets or other circuit) in your question, KVL is applicable. Drop of potential on perfect inductor is $LdI/dt$ (follows from the Faraday law), drop on resistor is $RI$ (follows from absence of non-conservative field at the resistor), drop on capacitor is $Q/C$ and drop on the power source is $-E$ (the minus is because under normal circumstances, potential drop on a battery acts against its EMF). All this leads to the correct equation

$$ L\frac{dI}{dt} + RI + \frac{Q}{C} - E = 0. $$ which your textbook gives.

The same result can be obtained starting from the Faraday law and then using the generalized Ohm's law (which is just another way of stating the Kirchhoff's second circuital law) to find that integral of total electric field over the resistor is $RI$. This method is arguably easier to explain and justify.

The KVL method requires some explaining about when it is applicable (no external emfs acting on the circuit) and that it works with potential based on the conservative part of field only. But it is easier and more tidy to use this method in practice when writing down equations for complicated circuits.