If we consider a single electron going around a closed loop, with a battery giving an EMF of $6\ \mathrm V$, why does the electron have to lose the energy in the loop?

If the circuit had zero resistance, wouldn't the electron just gain more and more energy as it loops around the circuit again and again, thus breaking Kirchhoff's voltage law and failing to conserve of energy, since the chemical potential of the battery gives the energy?

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    $\begingroup$ Where does the Kirchhoff's voltage law say that electrons have to lose energy? $\endgroup$ – Dmitry Grigoryev Sep 12 at 7:50
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    $\begingroup$ Why do you say that the electron would GAIN energy as it goes around the loop? $\endgroup$ – Mike Brockington Sep 12 at 12:49

I mean the basic principle is, you're right, this setup is described by the math of classical electrical circuits as one which should drive an infinite current, and if you only have one charge to move then an infinite current necessitates it moving with infinite speed. If you do not give it a way to lose energy then it will continue to gain energy indefinitely.

Kirchhoff's voltage law can still be applied to such a condition but it is somewhat less informative. The voltage law comes from the fact that there is a scalar potential field $\varphi$ defined over space, it is just one field, it has only one value and no other. It is something like wondering why I cannot gain infinite energy by sledding down a hill: well, whatever speed you are getting on the way down, I can prove that you’re in for an equal-and-opposite trek up. So if this field is well-defined then Kirchhoff’s voltage law holds good.

This field $\varphi$ in turn comes from the Maxwell equations, which in my favorite units look like,$$ \begin{align} \nabla \cdot E &= c\rho & \nabla\times E &= -\dot B\\ \nabla \cdot B &= 0 & \nabla \times B &= J + \dot E \end{align} $$ where $\dot X = \partial X/\partial w = c^{-1} \partial X/\partial t.$

First, the bottom-left one is used to rewrite $B = \nabla \times A$ for some “vector potential” $A$ and then the top-right one says $$\nabla \times (E + \dot A) = 0$$ implying that there exists a scalar potential function $\varphi$ such that $$E + \dot A = -\nabla \varphi.$$ So this is a mathematical theorem, this scalar function $\varphi$ always exists. You can really say that there is a voltage drop of 6 volts from one side of the terminal to the other, and that is fine. Kirchhoff’s voltage law holds everywhere.

But notice what is not said: the connection to a particle’s kinetic energy depends on work which depends on this electric field $E$ and its behavior over closed loops. And $\nabla\varphi$ vanishes over closed loops, but there is no reason that $\dot A$ must, and so a particle can gain/lose energy in a closed loop, if the vector potential is increasing/decreasing in time.

So, normally we can just step back from a circuit and assume that whatever the vector potential $A$ is, the circuit comes to some equilibrium value where it is not changing and $\dot A = 0$. Then $E = -\nabla \phi$ and we know that the thing must have lost all of its energy by the time that it comes back around the loop. That is how we know that the electron has lost its energy.

But, in the case that you are describing, we know that the Liénard–Wiechert potentials $\phi, A$ for a moving charge contain an $A$ which runs parallel to the charge, in rough proportion to its speed. As the charge is going faster and faster, this $A$ is increasing and increasing around the loop, and this factor of $\dot A$ is non-negligible.

Kirchhoff’s rule is not technically wrong, what is wrong is the assumption that energy only changes from a change in potential, that these go hand-in-hand. The Maxwell equations do not say this, they say instead that there is this other thing, this vector-potential $A$, which must change in that scenario.

  • $\begingroup$ While I admire you bringing Lienard-Wiechart potentials to bear in this answer, I hope you realize you did just open a very large, quantum mechanical can of worms. See here. If we start thinking too hard and too classically about how electrons move around a circuit, we run into a multitude of problems. $\endgroup$ – Charles Hudgins Sep 12 at 7:44
  • $\begingroup$ Very good answer except for the beginning, the current won't be infinite in any finite time, mainly due to self inductance of the circuit, which makes the rate of current increase finite. Even if there was merely a single mobile charge without any circuit self inductance, it would not get accelerated to infinite speed in finite time. $\endgroup$ – Ján Lalinský Sep 12 at 17:24
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    $\begingroup$ Who said anything about finite time? $\endgroup$ – CR Drost Sep 12 at 17:29
  • $\begingroup$ It is the default in all questions about what happens, it does not need to be said. What would happen in the limit of infinite time is a mathematical detail that is not the best thing to focus on in this question. $\endgroup$ – Ján Lalinský Sep 12 at 19:18

Kirchhoff's circuit laws (both the current law and the voltage law) apply in the lumped circuit approximation only.

That means they apply when the circuit can be accurately modeled as a collection of lumped components whose physical extent is insignificant; connected by ideal wires, with no significant magnetic fields coupled to the wires, and no significant charge storage in the wires.

As such, we don't expect these laws to apply to the situation of an isolated electron floating in space.

If you had in mind a situation where the two ends of your battery are connected by a wire, then it's unphysical to imagine there's only a single free electron present in that wire, or that the electrons don't interact with the atoms in the wire (which will transfer energy to the wire).


If the circuit had zero resistance, wouldn't it just gain more and more energy as it loops around the circuit again and again,

Assuming that the fundamental circuit-theory assumptions* hold, then if the circuit had zero resistance it would still have non-zero inductance. So the energy from the battery would go into the magnetic field of the inductance. There would be a voltage across the rest of the circuit equal and opposite the voltage across the battery. So Kirchoff’s voltage law would hold.

*The fundamental assumptions of circuit theory are that there is no net charge on any circuit element, there is no magnetic coupling between circuit elements, and the circuit is small compared to the speed of light and the time scales of interest. Once those assumptions are met Kirchoff’s current and voltage laws follow.

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    $\begingroup$ By the "usual circuit-theory assumptions", do you mean ideal circuit theory assumptions? If so, I don't think it is correct to say that zero resistance wires have non-zero inductance (in ideal circuit theory). See, for example, the answer by The Photon. $\endgroup$ – Alfred Centauri Sep 12 at 2:00
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    $\begingroup$ Dale, ideal wires in (ideal) circuit theory have zero inductance. This is well known. Do you disagree? $\endgroup$ – Alfred Centauri Sep 12 at 2:13
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    $\begingroup$ Dale, the OP didn't specify ideal wires but you invoked the assumptions of circuit theory within which wires have zero inductance. The first sentence of your answer is almost certainly false. For example: "In circuit theory, the wire form and size are assumed to have no effect on circuit behavior and they are assumed to be short-circuit interconnections". Of course, physical wires do have non-zero inductance but, in circuit theory, they don't. Circuit simulators operate under this assumption (two junctions connected by a wire are the same node) $\endgroup$ – Alfred Centauri Sep 12 at 10:15
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    $\begingroup$ Dale, the point Alfred is making is that your wording is problematic, because "the usual circuit-theory" does not assign wires a nonzero inductance. By assuming nonzero inductance, you are extending the discussion beyond that theory, to Maxwell field equations or at least to models with distributed parameters. Which is a fine thing to mention, after all, real wires do have self-inductance and it is important to take it into account in the scenario asked about. $\endgroup$ – Ján Lalinský Sep 12 at 21:36
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    $\begingroup$ Dale, I'm afraid there's been a misunderstanding so let me try a slightly different wording so that the point I attempted to make in my previous comment is clear: "... the OP didn't specify ideal wires but you invoked the assumptions of circuit theory, and within circuit theory, wires have zero inductance". I believe that this is uncontroversial. Also, I have no interest in a continued discussion in chat as I honestly don't see any value to be gained from it. $\endgroup$ – Alfred Centauri Sep 12 at 23:50

In the particular case you describe, the electric field is generated by the separated charges in the battery. A battery roughly consists of 2 chambers, one with negatively charged matter, one with positively charged one (there are different ways to achieve this setup, most of them make the charge separation happen by electrochemical processes).

For the sake of the argument it is only important that there is a mechanism which separates the charges in a way that going from terminal 1 (connected to chamber 1) to terminal 2 (connected to chamber2)), a charge of one $C$ will get the Energy of $6 J$.

Since the electric field is generated only from this separated charges, it is a conservative field. Moving in this field, a particle can't get energy from one complete roundtrip. Connecting both terminals of the battery, a testcharge of $ 1 C$ will be brought from one terminal to the other terminal by this electric field, gaining $6 J$ of Energy. It would require exactly this $6 J$ again to move it from the second terminal, through the battery, back to terminal 1.

In this case Kirchhoffs Voltage-Law is perfectly valid, because it only makes statements about the Work that is done by the electric field (There are other cases with non-conservative electric fields where the law is still valid, essentially because one looks only at a part of the electric field and makes certain further assumptions about it in those cases, but that should not matter here).

Your second observation is however not wrong: You mention the chemical potential of the battery, which is responsible for the charge separation. This chemical potential moves (by a nonconservative force) a charge from terminal 2 back to terminal 1, against the electric field in the battery. By that it ensures that the battery maintains its terminal voltage of $6 V$. It however isn't said that it will move exactly the electron that already completed one roundtrip.

Let's assume that the chemical potential would always bring the same electron that completed one roundtrip back to terminal 1. Then this electron would effectively see a forcefield which is a combination of the electric field (conservative) and the nonconservative "force"-Field by the electrochemics of the battery. This would all in all (as you observed) be a nonconservative field, and an electron would gain energy for each roundtrip. Energy conservation doesn't apply here, as chemical energy is transformed to kinetic energy of the electron.

However, Kirchhoff's Law still holds here, because energy conservation and Kirchhoff's Law are two things.

  • $\begingroup$ "Connecting both terminals of the battery, a testcharge of 1C will be brought from one terminal to the other terminal by this electric field, gaining 6J of Energy. It would require exactly this 6J again to move it from the second terminal, through the battery, back to terminal 1" So then if I then add a resistor it takes more energy to go around the circuit than is given by the battery since it still takes 6J to move through the battery thus disproving Kirchoff's law. $\endgroup$ – John Sep 12 at 13:22
  • $\begingroup$ Again: Kirchhoff's Law doesn't make a statement about the energy a testcharge gets while completing one roundtrip. It makes (in the electrostatic case) a statement about that portion of the energy which is generated by the electric field acting on the testcharge. Neither the resistance (nonconservative nonelectric "friction"-force acting on the charge, sucking out energy) nor the chemical potential (nonconservative nonelectric "chemical"-force acting on the charge, bringing in energy) do play a role here. $\endgroup$ – Quantumwhisp Sep 12 at 13:39

If we consider a single electron going around a closed loop, with a battery giving an EMF of 6 V, why does the electron have to lose the energy in the loop?

It is important to make sure we know which energy we are talking about. In real wires which pose dissipative resistance to motion of mobile charge carriers, the electron, as it moves along the wire, releases some energy into the wire. This energy comes (usually) from electromagnetic field near the electron, not from the electron's own store of energy; if electric current is constant in time, as in stationary DC circuit, the electron itself does not lose any kinetic or rest energy; it just funnels EM energy around it into the conductor where it becomes internal energy (the conductor heats up).

In situations where the EM field in conductor is mainly due to spatial variation of electric potential $\varphi$ (such as in a DC circuit with a battery or charged capacitor), we can restate the above and say the released energy is due to loss of potential energy of the electron $q\varphi$ as it moves from lower potential place to higher potential place. The conservative electric field with potential $\varphi$ is maintained in the circuit by the battery or capacitor.

But there are other situations where the EM field maintaining the current is not due to spatial variation of electric potential, but due to induced electric field.

Induced electric field has the same effect on mobile charges and conductor as conservative electric field: it may initiate and maintain electric current, while heat is being released into the conductor. This electric field cannot be described by any electric potential $\varphi$.

The electric potential $\varphi$ can always be defined based on the above conservative field, but in cases where the whole electric field is equal to induced field, the potential will be the same in the whole circuit. Because $\varphi$ is the same everywhere now, the electron does not lose any potential energy as it goes around the circuit. It however still releases heat, so energy of EM field is still being transformed into heat. Again, the electron is just a device to release the energy of the EM field into the conductor, but not the loser of energy.

A conductor with zero resistance will not allow the EM energy to change into heat. So if there is electric field in such conductor, the mobile charge carriers will be accelerated, the electric current will increase and the EM energy will be transformed partly into magnetic energy of the circuit (most of it), partly into kinetic energy of the mobile charge carriers (a very minor part).

If the circuit had zero resistance, wouldn't the electron just gain more and more energy as it loops around the circuit again and again, thus breaking Kirchhoff's voltage law and failing to conserve of energy, since the chemical potential of the battery gives the energy?

Yes, it would gain kinetic energy, but not potential energy. Potential energy would increase when passing from one terminal of the battery to another, but would decrease by the same amount when traversing the rest of the circuit.


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