# Analyzing electric circuit with capacitor, inductor and resistor 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?

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}

• Thanks for your comment! I am confused though, because there isn't a battery here right? There is only a conductor, resistor and an inductor, so where does the E(t) term come from exactly? Also is my statement correct that the book is incorrect by saying that KVL applies here? May 14, 2022 at 10:18
• @Stallmp well the picture gives a battery/generator term right? The circle with the $E$? And yes, most of these books are wrong if they apply KVL. It's just not applicable here. IIRC Griffiths presents the LRC circuit the way I have written it down here using Faraday's law. Btw take a look at the lecture I linked to, it's explained very well in the first few minutes. When in doubt, ALWAYS stick to Maxwell's equations. They're literally what E&M is based on; you won't go wrong with them. May 14, 2022 at 10:23
• Wow hahah I am actually blind, thanks! Thank you very much! So basically applying Faradays law gives the same result as in the book, but then the book says that the end result is because of KVL which is basically incorrect. Thanks for the recommendation as well, I will take a look at it! May 14, 2022 at 10:28
• having said this, I think sometimes people redefine their terminology about what exactly constitutes a voltage drop or emf so that their application of KVL is consistent with Faraday's law. However, I think playing with such linguistic matters is more confusing than simply applying Faraday's law (where the definitions are crystal clear and standard for what the $E$ and $B$-fields are), because like I said, all of classical E&M is based on Maxwell's equations. May 14, 2022 at 10:36
• @peek-a-boo The KVL method is consistent with Faraday's law, because potential drop on inductor $LdI/dt$ that it uses is derived using Faraday's law. Also, "simply applying the Faraday law" is not all that your method does; the method starting with the Faraday law then needs to assume the generalized Ohm's law for resistor and null integral for inductor as well. In other words, both methods use the same assumptions and they are equivalent. May 15, 2022 at 0:17

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.

• I see now that you use a specific definition for KVL in order to make it consistent with Faradays law. The formulation that I learned for KVL is $\oint \vec(E} \ cdot d\vec{l} = 0$, which in that case wouldn't apply here. Also, why does my Latex not work for some reason? May 15, 2022 at 7:41
• That is not the KVL. Some people are very confused about this, including some teachers. The KVL is modern reformulation of the original Kirchhoff's second circuital law (which was originally stated in terms of emfs and RI's) in terms of potential drops. The integral you mention could be one possible expression of the KVL, if $\mathbf E$ was only the conservative part of total electric field. But $\mathbf E$ usually means total electric field, so the conservative part should be denoted differently, e.g. $\mathbf E_C$. The KVL then can be written $\oint \mathbf E_C \cdot d\mathbf s = 0$. May 15, 2022 at 14:08