In a simple battery + resistor circuit, what form of energy is lost from the electrons upon exiting the resistor?

I will give this question a little context. Firstly, as I understand it, as soon as I "close the switch" on a circuit, electric current pretty quickly establishes a steady state where, at any given cross section along a wire, the average kinetic energy of that slice is both constant and equal anywhere along the wire. (Please correct me if I am misunderstanding).

If the average kinetic energy is constant anywhere throughout the wire in the circuit (which confuses me, as I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded as much), that means that the average velocity is constant. So, in the example of a simple circuit with a battery of v Volts and a resistor of o Ohms, my question is the following:

Because a 'voltage drop' is known to occur across the resistor, what type energy is being "traded" to generate the heat that radiates from the resistor? I would think that the reflexive answer is "electrical energy"...hence the VOLTAGE drop (the sacrificed energy is clearly not kinetic, as the velocity is constant everywhere). However, I find this confusing. Does this mean that if I had a positive test charge, it would be easier for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister?

Further, if the correct answer IS "electrical energy", why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? I always here the comment Oh! It's because all the electrons are running into densely packed crap which impedes their flow but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case. So, ultimately, I guess the real question is:

**What goes on in the resistor that is literally removing electrical energy **. I am sure that this is a quantum mechanics question, but my quantum mechanics knowledge is not terribly strong. If I could get an answer that is devoid of crazy wave function equations, I would greatly appreciate it. Thanks!

• "a resistor of 0 Ohms"? So... no resistor? Aug 29 '18 at 5:47
• – rob
Aug 29 '18 at 13:30
• @Steeven o Ohms. Not zeros ohms. Sorry for the confusion. I did not realize that this website depicted the letter o and the number zero almost identically. Aug 29 '18 at 14:56
• Isn’t it conventional to use “$R$” Ohms for an unknown resistor? Sep 4 '18 at 11:36

Let's assume that the average speed of an electron should be the same everywhere in the circuit, including wires and resistors. This is generally not true (for instance, when the diameter of a wire changes), but is a reasonable simplification.

If we track an individual electron, we'll, presumably, see that it speeds up due to the electric field, then it collides with an atom (ion), loses its speed and its kinetic energy (which is turned into heat energy) and speeds up again.

If we assume that the frequency of the collisions in a resistor is greater than in a wire, then, given the same electric field (i.e., the same acceleration), the average speed of electrons in the resistor would be lower than in the wire, so the electrons will pile up at the front end of the resistor and will be at deficit at the back end, which will increase the electric field inside the resistor.

As a result, the electrons inside the resistor will accelerate faster, so that, even with the increased frequency of collisions, they will have the same average (and top) speed as the electrons in the wire. Since the frequency of collisions has increased and the kinetic energy loss (heat gain) associated with each collision is the same (due to the same top speed), the heat generated by each passing electron per unit length in the resistor will be greater than in the wire. but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case.

If there were no other forces on the charges, you would be correct. But there is also an electric field in the resistor that is providing a forward force. These forces balance. With a net force of zero, there is no acceleration of the charge and kinetic energy doesn't change.

Imagine a block sliding down a ramp with friction. At the right angle, the loss from friction is exactly offset by the gain from potential energy. Kinetic energy is unchanged, but potential energy is lost and the system dissipates heat.

The energy came from whatever process raised the block to the top of the ramp. In the circuit, the battery is the means that places the charges at the correct point with sufficient potential energy.

I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded as much

Stuff will only accelerate if a net force pushes/pulls. And there is no such net force after exiting the resistor. The electrons won't start moving just because there is "space".

what type energy is being "traded" to generate the heat that radiates from the resistor?

That would be electrical potential energy. The battery "pushes" the electrons through the "constriction" or "filter", which the resistor is, and this "pushing energy" is converted into heat in the passing. This "pushing energy" is called electric potential energy.

Does this mean that if I had a positive test charge, it would be easier for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister?

I'm not sure what you mean here. Surely it is easier to let the charge "follow the flow" as part of the current. The battery voltage "pushes" them through the resistor from one side to the other. In the process, the resistor "sucks out" some energy.

For you to move a charge backwards through the resistor, you would have to add not only another round of energy for the resistor to "suck out", but also to add enough energy to overcome the battery "push" (to regain the drop in voltage). This is a weird situation that requires external interference - I am guessing that you meant something different...

why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? [...] because all the electrons are running into densely packed crap which impedes their flow but that to me sounds like a reason for their KINETIC energy to be reduced.

But this is exactly true. In the classical model (so that we don't have to dive into quantum mechanics), electrons will bump into the atoms of the more densely packed resistor material. An impacted atom will thus be "pushed" and start moving (vibrating; this is what we at the macro-scale call thermal energy or temperature - it will in turn be radiated away as heat) - it gets some kinetic energy from the electron, which conversely looses some kinetic energy.

Now, remember that the battery "push" is still acting. The electron right behind is still moving at full speed. It pushes on the electron that has lost a bit of kinetic energy and brings it back up to full speed again. This other electron thus looses some kinetic energy, but it too has an electron behind it. This "loss of kinetic energy" therefore propagates all the way to the battery terminal. In the end the battery takes the energy loss and resupplies it to the electrons to keep the current running - which is why batteries run out of juice eventually.

So, it is indeed micro-scale kinetic energy which in the resistor is turned into heat and dissipated. But when this kinetic energy loss is replenished right away so that all electrons overall move at the same constant speed, then we call the situation a steady state.

The electrical potential difference (electric field) between the two ends of the resistor does work on the electrons in the resistor causing them to accelerate and gain kinetic energy. However, the velocity of the electrons increases only for short interval of time as each accelerated electron suffers frequent collisions with positive ions in the resistor and looses its kinetic energy, generating heat. As a result the current (a function of the average drift velocity of the electrons) going into the resistor equals the current leaving the resistor.

In a simple battery + resistor circuit, what form of energy is lost from the electrons upon exiting the resistor?

Electrons leave the resistor with less potential energy than they enter the resistor with (the electric potential at the exit terminal is more positive than than that at the entry terminal, and electrons have negative electric charge).

Yes, electrons give up kinetic energy to the lattice that makes up the resistor due to collisions. This is why the resistor is heated by the electron current through.

However, there is also an electric field through the resistor (there must if there is a potential difference across the resistor terminals). So, while the electrons give up energy (on average) due to collisions with the lattice, they also gain kinetic energy as they are accelerated by the electric field inside.

On the whole, the kinetic energy lost to the lattice is balanced by the gain in kinetic energy due to the electric field, so the energy delivered to the resistor lattice is balanced by the potential energy lost by the electrons.

(Towards the goal of consolidating answers to like questions, the above is a copy and paste of my answer to this recently asked and closed as a duplicate question here)