# Can Kirchhoff's Voltage Law be applied here?

First, consider a simple circuit comprising of an ideal battery of EMF: E and a single resistor of resistance R.

Imagine that the switch is opened in the circuit. In the battery, some internal mechanism drives (by applying a force) negative charges to one terminal, leaving positive charge on other terminal, at equilibrium an opposing electric field is generated inside battery. No further accumulation of charges takes place.

As soon as the switch is closed, electrons get another 'path' to go from terminal having higher potential to that having lower relative potential through the resistor. An electric field (conservative in nature) is set up across resistor.

Now, the work done by conservative fields (present inside the battery and across the resistor also) on any charge over a closed (or cyclic loop) path is ZERO. That's why I think Kirchhoff's Voltage Law works.

My problem is that why do we apply Kirchhoff's Law for Inductors as well?

In an inductor, an EMF is induced by changing magnetic flux. The EMF is simply the work done by the NON CONSERVATIVE electric field,(produced by changing magnetic flux in the inductor coil) on moving across the a particular path i.e across the inductor coil.

Since this field is non-conservative, according to my analogy given at the starting of the question, this work done across closed path on the charge shouldn't be zero. But in books I see that Kirchhoff's Law is applied here! How can this happen?

Any help will be appreciated.

• Hey i think electric field is always conservative Sep 29, 2018 at 16:22
• Well, electrostatic field is always conservative, electric fields may or may not be conservative. Sep 29, 2018 at 16:23
• Can you give the example where you have used kirchoff's laws in inductor Sep 29, 2018 at 16:29
• "Now, the work done by conservative fields (present inside the battery and across the resistor also) on any charge over a closed (or cyclic loop) path is ZERO. That's why I think Kirchhoff's Voltage Law works." --- Your reasoning here isn't clear to me; a conservative electric field cannot drive a steady current through a closed circuit. Yet, there is a steady current through the battery and resistor and KVL works. Sep 29, 2018 at 16:30
• Google: L-R Circuits, everywhere you will see that they're applying Kirchhoff's law to an inductor and saying that net potential drop across a circuit containg inductor, a battery and a resistance is zero. Sep 29, 2018 at 16:32

It is not an issue of the field being conservative or not. Ultimately, Kirchhoff's laws are about the relationship between branch currents and node voltages in a network of lumped circuit elements. If you define three kinds of branch elements denoted by $$R,C,L$$ using the relationships $$v=Ri$$, $$i=C\frac{dv}{dt}$$, and $$v=L\frac{di}{dt}$$, respectively, then you may freely use Kirchhoff's current and voltage laws. These defining relationships between voltage and current are idealization and simplification not just for an inductor but also for a capacitor and resistor, as well. In the case of the inductor we ignore all fields outside the coil, and if we cannot because we have an inductive transformer then we include that part explicitly by defining a two-port with a pair of equations, such as $$v_1=L_{11}\frac{di_1}{dt}+L_{12}\frac{di_2}{dt}$$ and $$v_2=L_{12}\frac{di_1}{dt}+L_{22}\frac{di_2}{dt}$$, and a similar set of equations if you need more ports than two. If the capacitor is physically large then we may encounter problems with the current continuity law and will not be able to neglect the displacement current.

Note too that in no sense one could claim that the fields of a voltage or current generator are "conservative", not even for a battery: electrochemistry is not electrostatics. Somewhere, somehow you must impose a phenomenon that is outside of electricity or magnetism. Instead we postulate that certain node pairs have a predefined voltage history, and a given branch has a predefined current history independently of the rest of the circuit and thus represent a voltage or a current source, resp. In other words sources are time dependent boundary conditions. This way as you go around in a loop you must always get 0 voltage, no conservative field is needed. At the next level of abstraction you only need that in an arbitrary loop at any instant every connecting wire the current must be the same. And assuming linear superposition you can derive that the sum of branch currents at any node must be zero. So then the only questions is whether a loop is physically small enough so that the current uniformity holds. Once you have picked the defining lumped element equations between $$v$$ and $$i$$ you may say that KVL and KIL have more to do with network topology than actual physics.

• I've been working on and off on an answer but, after reading this, I decided that I there wasn't anything significant I could add beyond this answer and so I simply upvoted this one. Sep 30, 2018 at 22:29

the work done by conservative fields (present inside the battery and across the resistor also) on any charge over a closed (or cyclic loop) path is ZERO. That's why I think Kirchhoff's Voltage Law works.

Correct. Modern formulation of KVL (as opposed to the original Kirchhoff's statement about electromotive forces in a circuit) is really about drops of the Coulomb potential when passing across all elements in a circuit (in the designated positive direction of the loop). These drops have to sum up to zero because when full closed path is mentally travelled in space (assuming the Coulomb potential not changing during that process, although in reality it changes with the time, so the traveling process is only imagined and takes no time at all), the potential at the position of traveling particle acquires its starting value.

These drops of the Coulomb potential ("voltages") are a valid concept even for circuits with inductors or other elements because the Coulomb potential is always defined, even if electromagnetic induction is taking place at some regions of space. Electrostatic potential just isn't sensitive to induced electric field at all.

My problem is that why do we apply Kirchhoff's Law for Inductors as well? In an inductor, an EMF is induced by changing magnetic flux. The EMF is simply the work done by the NON CONSERVATIVE electric field,(produced by changing magnetic flux in the inductor coil) on moving across the a particular path i.e across the inductor coil. Since this field is non-conservative, according to my analogy given at the starting of the question, this work done across closed path on the charge shouldn't be zero. But in books I see that Kirchhoff's Law is applied here! How can this happen?

In the sense described above, even inductor has definite voltage between its terminals: it is difference of the Coulomb potential, or integral of electrostatic component of electric field along any path joining the terminals.

This drop of potential can be measured by an oscilloscope with proper quality probes that eliminate random (depending on cable shielding imperfection and cable configuration in space) influences of induced electric field.

This modern KVL does not include any contribution due to induced EMF or electrochemical EMF, only contributions due to potential drops. However for ideal inductors, potential drop is $$LdI/dt$$ (with plus sign) which sometimes confuses people to think EMF $$-LdI/dt$$ (with minus sign) contributes in KVL.

Potential drop on ideal inductor is $$LdI/dt$$ (not $$-LdI/dt$$) because in ideal inductor conductor, net electric field has to be zero (no resistance, zero field is needed to drive current). So electrostatic field is exactly cancelling the induced electric field so their integral over a path going inside inductor's coils is the same but has opposite sign. Thus in short - EMF is (for any inductor) $$-LdI/dt$$, but potential drop (only for ideal inductor) is $$+LdI/dt$$.

Let's consider two scenarios.

In the first scenario, we'll place an inductor in a changing magnetic field and some emf will be induced into it. If a resistor is connected between the terminals of the inductor, we'll have an LR circuit with some current flowing in it and some power dissipated on the resistor. The source of this power is the magnetic field, which was generated outside the circuit, so the conservation of energy and, therefore, KVL, is not applicable to this circuit.

In the second scenario, we'll connect an inductor to a battery. The battery voltage will be applied to the inductor and will cause the current in the circuit, formed by the battery and the inductor, to grow as $$\frac {di}{dt}=\frac V L$$ and, with it, the magnetic field around the incuctor will grow as well. We can also say that the applied voltage, $$V$$ is equal or opposed by the back emf generated by the inductor, complying with KVL. In this case, the source of the magnetic field energy comes from the battery, which is part of the circuit, and, therefore, both conservation of energy and KVL are applicable.

So, we can conclude that the applicability of the conservation of energy and KVL to a circuit with an inductor depends on whether the energy comes form the circuit or from somewhere else.

• KVL is used to solve circuits with transformers and coupled inductors (e.g.,$v_R = v_{L1} = L_1\frac{di_{L1}}{dt} + M_{12}\frac{di_2}{dt}$) so I don't think the conclusion of the 2nd paragraph is true. Sep 29, 2018 at 22:17
• @AlfredCentauri Sure, but you have to take into account the source of energy, e.g., both primary and secondary.
– V.F.
Sep 29, 2018 at 22:21
• KVL states that the (signed) sum of the voltages 'round a closed loop is zero. This is valid for the circuit in your 2nd paragraph that contains just two voltages and thus $v_R = v_L$ by KVL. The fact that $v_L$ involves a term other than the circuit current doesn't invalidate KVL does it? On the other hand, if there were a changing magnetic flux through the surface bounded by this circuit, it wouldn't be true that $v_R = v_L$ and KVL wouldn't hold. Sep 29, 2018 at 22:25
• @AlfredCentauri In the equation you've referenced, $L_1\frac{di_{L1}}{dt}$ could be interpreted as $v_{L1}$ and $M_{12}\frac{di_2}{dt}$ as emf induced by external magnetic field. By equating $v_R$ and $v_L$ you are assuming that KVL is applicable.
– V.F.
Sep 29, 2018 at 23:56
• the inductive law for a transformer does not depend on the KVL, it is rather an essentially constitutive relationship of $i,v$ for a lumped circuit element that when is made to be part of the network the latter will satisfy the KVL, the battery is boundary condition; see my answer below. Sep 30, 2018 at 19:14

You are correct that Kirchoff’s voltage law does break down when you apply it along a loop where there is an E field with nonzero curl.

However, there is an interesting result which is derived in section 11.3 in this free online textbook: http://web.mit.edu/6.013_book/www/book.html

Basically, it shows that as long as there is some surface enclosing a component and the electromagnetic induction is negligible on that surface, then the usual circuit energy conservation laws hold. That is, as you surmised, the basis of KVL.

So even if within an inductor that condition is violated, as long as the magnetic fields are well localized, so there is a surface surrounding the inductor where those fields are negligible, then KVL will hold outside that surface regardless of how it might break down inside.

That is the basis for applying KVL in circuits with inductors