As the title. Between when voltage is higher; before the electrons reach the component, and afterwards.

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    \$\begingroup\$ The first thing to realize is that electrons don't move very much. What you think of as "current" is actually a neighbor-bumping-neighbor effect that propagates at a fair fraction of the speed of light, and while electrons bounce around quite rapidly in place, their actual net movement is a snail's pace. \$\endgroup\$ Aug 9, 2020 at 14:39
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    \$\begingroup\$ What do you mean by "work is done"? Let us take an example: the electrons flowing in a circuit with a boiling kettle's "heater" element. In this case the electrons find that the heater element has a "friction" (resistance) which makes it hard for the electrons to pass through. Long story short, friction causes heat, electrical energy transforms to heat energy, heat boils water, and you make your coffee, and work is done. Cheers. \$\endgroup\$
    – tlfong01
    Aug 9, 2020 at 14:42
  • \$\begingroup\$ Electrons are still electrons prior to feeding into the load as they are when leaving the load. I think you are thinking about maybe what aspect of an electron changes after it has passed through the load? \$\endgroup\$
    – Andy aka
    Aug 9, 2020 at 14:45
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    \$\begingroup\$ I'm going to try to add some context to what I think the OP is asking: what does it mean when an electron has more voltage and energy? The electron's charge doesn't change, doesn't really move down the wire any faster, its temperature/vibrations don't really increase (I think), so how what form is this extra energy in the electron taking? Perhaps the energy isn't really in the electron at all, but in the electric field trying to shove the electron? \$\endgroup\$
    – DKNguyen
    Aug 9, 2020 at 17:57

4 Answers 4


What actually happens to electrons in a circuit when work is done at a component?

Between when voltage is higher; before the electrons reach the component, and afterwards.

In both cases, the electrons are still there, and are still electrons, and are still moving in much the same way, and have the same sort of density.

Consider a bicycle transmission, with the bicycle chain being the loop of conductor that goes in a closed circuit between generator and load. The links are electrons. The chainwheel provides the energy. The chain does work on the rear sprocket when it moves.

The only difference between the two sides of the chain is the tension, that's what allows its movement to do work.

The only difference between the out wire and the return wire is voltage, an expression of the potential energy per unit charge, that's what allows the flow of current to do work.

And as transistor says in comments, the result is felt at the rear wheel immediately, even though the chain links move slowly. Well, nearly immediately. The transmission occurs at the speed of sound in the chain material, just as the transmission occurs at the speed of light along the wire.

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    \$\begingroup\$ And note that the effect at the rear wheel is felt immediately pressure is applied on the pedals even though the links are relatively slow in moving from the wheel sprocket to the driving chain ring. \$\endgroup\$
    – Transistor
    Aug 9, 2020 at 15:32
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    \$\begingroup\$ @Transistor Not quite immediately, but let's not throw the transmission line stuff at the OP ;) \$\endgroup\$
    – DKNguyen
    Aug 9, 2020 at 18:21
  • \$\begingroup\$ @DKNguyen Well, I was responding to Transistor's comment, and the bike chain analogy does lend itself to the propagation question so well, as does the hydraulic analogy. \$\endgroup\$
    – Neil_UK
    Aug 10, 2020 at 5:59
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    \$\begingroup\$ I'm very much a fan of the bike chain analogy, it conveys quite well that the power isn't "inside" the links but in the forces between them. And that it has to be a circuit. \$\endgroup\$
    – pjc50
    Aug 10, 2020 at 9:23

On the negative side of the load, where the electrons enter, they are a little closer together on average (assuming similar materials) than they are on the positive side of the load where they exit. That is essentially what it means for the (electrostatic) voltage to be lower.

The "compression" of the "electron gas" holds potential energy, and as each unit of charge moves through the load and "expands", this energy is lost to the load.

Note, though, that the electrostatic force is incredibly powerful, so the difference in electron density is very small, and is determined by the volume and self-capacitance of the circuit elements.

Comparing to @Neil_UK's answer, which is also correct, it's like how the links on the top of a bicycle chain are stretched a very little bit by the tension.


Let's stay in a classical context.

At first, let us consider what happens to an electron subject to an electric field in a vacuum. the electron will experience a force F = q E that will cause it to accelerate (or decelerate, depending on its initial conditions). Potential energy is converted to kinetic energy or viceversa.
If the electron has a lower potential energy at its final position with respect to its initial position, it will have gained kinetic energy.

Now, let us consider electrons in a resistive conductor (such as a resistor, but also the copper wire connecting it to a battery) inside a DC closed circuit.
We know that when a current is flowing in the circuit, inside the conductor there is an electric field E that follows the path of the conductor and has a magnitude compliant with Ohm's law (in its local form) \$E = j/\sigma\$.

This field is created by surface charge whose density changes in correspondence of gradients of permeability and conductivity. The charge on the lateral surface of the copper conductor and at the abrupt interfaces between copper and the resistive material is responsible to make the electric field complying with Ohm's law. In the following I will consider a resistor of the same cross section of the wires and built with a material with much smaller conductivity \$\sigma\$. (Note that even when the resistor is composed of the same material as the wires but differs in cross section, surface charges will develop where the shape changes in order to 'steer' the field lines inside the smaller section. In this case the current density will increase and since the material has the same conductivity, so will the electric field inside the resistor)

When both wires and resistor share the same cross section, the current density is the same in every point and the effect of the surface charge at the boundaries is to make the electric field inside the copper and the resistive material much different in magnitude. This will result in a different value of electric potential energy and, of course, of electric potential along the circuit path.
At first it might look like we are in the same condition as in a vacuum: the electrons inside the material are subject to an electric field E, and we should expect them to accelerate. And indeed they do, but then, in the classical model of conduction, they also bump against the lattice of ions the material is made of. The 'macroscopic' effect is to hide the sudden accelerations and deceleration due to the effect of the field and the collisions, leaving the illusion of an overall constant drift velocity that is reflected in the constant current density \$j = \sigma E\$.

The electrons that 'enter' the resistor with an higher potential energy will move at the same average speed as those that exit it at a lower potential energy, so their loss of potential energy did not raise their (average) kinetic energy. So, where did their energy go?
It was transferred to the lattice of ions: instead of simply converting potential energy into kinetic energy of the electrons like it would happen in a vacuum, the potential energy is transformed into disordered kinetic energy of the lattice atoms. The resistive conductor heats up.

The surface charges at the resistor's boundary create a stronger field inside it that will make the difference in potential energy change much bigger than in the good conductor: the greater loss of potential energy along the path that goes through the resistor corresponds to a greater gain of thermal energy of the material's lattice. The resistor will heat up, the copper wires will not.

Note that in this classical model, the electrons are basically non-interacting with themselves and the magic is worked by what is basically stationary surface charges (that is distributed along the circuit with relaxation times the very moment the circuit is closed). Conduction electrons are not like hard balls in a conduit and there is no 'linkage' or 'tension' between them as in the links of a chain. Moreover, the different value of the electric field inside the resistor is caused by the distribution of surface charge (either at the lateral surface of at the boundaries between different materials) and not by a different density of the 'electron fluid'.


They have more potential energy.

an electron in a circuit before it goes through a resistor, and towards the negative terminal, will have higher potential energy, as it moves towards the negative terminal it will dissipate the potential energy and converts it to heat (in case of wire or resistance).

Consider a positive charge in an electric field, the charge will move from the positive side (the one with higher PE) to the negative side (one with lower PE).

Same with a mass in air, it will move across the gravitational field from a position of higher potential energy to a position with lower potential energy.


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