It is easy to visualize gravitational potential energy as a function of the position of height, and a change in this potential is manifested in a change in height. Further, by the work-energy theorem any change in potential energy results in work done, which results in a corresponding change in kinetic energy.

In an electric circuit, the potential energy difference is provided by the battery. Yet it is not easy to see how a voltage drops manifests itself. When energy is lost (let's say by going through a resistor), what physically happens to the electrons.

In the case of gravitation potential energy it is easy to see that the height of the object changes. For an electron the only way I can see would have to do with its energy level. Also, by the work-KE theorem doesn't this also mean that the kinetic energy of the electron would have to change? Yet this is clearly untrue for the current before and after (assuming a simple series circuit) is the same so the drift velocity cannot have changed.

(And before this is marked as a duplicate of this question I already looked into every answer there and none of them answer my question — they merely restate the definitions of potential, voltage, and ohms law or provide an analogy to water level. In the case of a resistor I already know that the reason for energy loss can be considered at a superficial level to be due to collisions — I am instead asking for an explanation of how this loss of energy manifests itself in the electric field or electrons themselves, perhaps via a change in occupied orbital or something similar.)

  • $\begingroup$ The same things changes in the electrical system as in the gravitational system: the electron's position in the system. Mind you, the potential function isn't necessarily as simple as the gravitational potential. Of course, by considering only the intrinsic resistivity of a long uniform wire you can construct a system in which is is as simple. $\endgroup$ Jan 17, 2016 at 4:40
  • $\begingroup$ @dmckee If it were solely position, however, then the resistance of a resistor would not matter for the voltage drop. $\endgroup$
    – 1110101001
    Jan 17, 2016 at 4:46
  • 3
    $\begingroup$ What it the world makes you say that? The resistor controls how steep the energy landscape is. That's what I meant by saying that the potential function isn't as simple as that for gravity. You might look at my answer physics.stackexchange.com/a/215807/520 (and the others to that question) for another way to look at these things (specifically how the electric field is modified by the presence of a resistance). $\endgroup$ Jan 17, 2016 at 4:52
  • 1
    $\begingroup$ If the electrons travel in free space their velocity will change as they get accelerated by the potential difference (CRT, x-ray tubes, accelerators). In a wire the electrons will accelerate for a little bit, then hit the atomic lattice and release their energy in form of heat. The average electron velocity will be very low because electrons can't travel very far before these lossy collisions happen. At least that's the simplified classical model. In reality it's a little more complicated because of quantum mechanics, but not much more so. $\endgroup$
    – CuriousOne
    Jan 17, 2016 at 5:39
  • 2
    $\begingroup$ Nothing. The electron doesn't change at all. Potential energy is the ability of the system to perform work on the electron. If the electron doesn't move, no work will be performed. If it does move, it's still the same electron, just in a different state of motion. Relative motion doesn't change physics of objects in motion. $\endgroup$
    – CuriousOne
    Jan 17, 2016 at 6:44

3 Answers 3


The current in a circuit is a collective phenomenon from zillions of electrons. It appears due to conductivity, another collective phenomenon . It is a cumulative behavior of atoms and electrons in matter.

In insulators, electrons occupy energy levels and have to be actively kicked out of them, with the energy provided by an interaction. Insulators can be charged ( tribo electricity) but the potentials induced on the surface do not have sufficient energy to dislodge more electrons.

In conductors a single electron is free to move in collective energy levels of the conductor, and in metals there exist in a continuum unoccupied energy levels which an electron can move into given a very small energy. This is the band theory for solids.


A voltage drop sets up a field that individual electrons ( and ions) react to by moving according to the field. The individual motion of electrons in space is small, it is called the drift velocity.

The ensemble of motions of all outer level electrons generates the current. The drift velocity depends on the conductivity, and depends on the type of solid.

In resistors the conductivity is reduced , it takes more energy to go to the conduction band. For the same voltage drop across a resistor the drift velocity is much smaller ( depending on how resisitive it is) than for a same size conductor, which would short!


Microscopic view of current.

You say:

For an electron the only way I can see would have to do with its energy level.

In the band model, the band is wide, it will have less kinetic energy.

Also, by the work-KE theorem doesn't this also mean that the kinetic energy of the electron would have to change?


Yet this is clearly untrue for the current before and after (assuming a simple series circuit) is the same so the drift velocity cannot have changed.

It is changed from the case where that particular resistor is not added. The resistor acts as the maestro of the drift velocity. By lowering it within , conservation laws assure that it will get lower in the whole cirtuit, which is assured by the voltage drop across the resistor. The electrons in the metal see a smaller field than before the new resistor was introduced. ( I am talking of new resistor because one should always be there , otherwise one would have a short).


You seem to be asking, "what happens to an electron when its voltage changes?" In short, nothing happens. An electron has no idea what voltage it's at.

You claim this is different from an object's height, but it's actually exactly the same. If you only look at an object, by itself, you'll have no idea what height it's at. Nothing about the object changes when it moves up or down.

The height/gravitational potential energy is a global property, of the object plus the large thing it's being gravitationally attracted too. Voltage is just the same way.


If you put the two poles of the electric source close together and put a carbon filament between them, perhaps you will see a plasma arc. The carbon has gone, but for a powerful enough source the flow of electrons does not stops (until the source is not exhausted). The potential difference (the voltage) is responsible for the velocity of the flowing electrons (and the area of the electrodes is somehow responsible for the possible amount of flowing electrons and the volume of the arc). Then higher the potential difference then higher the flow velocity. The same happens in a wire. And the higher velocity led to higher frequency of collisions with the hull atoms and the other electrons.

Short answer to your question: In situations of different electric potentials it changes the velocity of the involved charges. In a high potential difference the velocity of the involved electrons is higher than in a lower potential difference, where the velocity is lower. The drift velocity may be higher / lower too, but this is not a must. It depends from the electric resistance of the wires material. So for metals - since the material heats up - it increases and for - heated up by more potential difference (higher voltage) - semiconductors it decreases.

  • $\begingroup$ You're avoiding why the voltage drop happens then. You're saying that before a resistor where there is higher potential, the speed of electrons is higher than after the resistor where voltage drop has accured. This is not true. Can you please explain voltage drop and what happens to electrons during the voltage drop. $\endgroup$
    – MaDrung
    Feb 9, 2017 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.