# Fundamental Questions about Voltage and Resistance

I hope all is well.

In this post, I’ve asked a couple of very fundamental questions regarding the topic of electrical engineering that my current education vaguely answered—and unsatisfactorily so since it was mainly through the use of analogies—and that my current schedule prohibits the “deep dive” that their answers most likely require. In either case, if pointing me towards the right resources might be a better option than answering them, please also let me know.

My first question is regarding voltage. I think it might me more productive that I explain my current understanding of the topic. Through my current education, we first started with point charges in space. From there we learned about electrical force, electrical field, and then voltage. But this understanding becomes vague —at least in my mind—when we start talking about circuits. Voltage has the unit J/C. We can easily multiply the unit by the appropriate unit ratio (charge per electron, or 1.602 x 10^(-19) coulombs) to get some joules per one electron. We also can assume that this energy per electron comes from the kinetic energy of the electron, which of course depends on its mass and speed (and, for the purposes of the question, let’s also assume that the mass does not change). Now, here’s the question: When a battery reads 5V vs another battery reading 10V, does that mean that every electron that moves through the circuit connected to the second battery does more work than any electron moving in the circuit connected to the first battery? If so, does this mean that those corresponding electrons move at different speeds? Lastly, does this mean that we’d have an upper limit on how absolutely high a voltage value can get (the limit being set by the speed of light)?

My second question is about resistance. Across any component on a circuit (or even a length of wire, in the “real” world) we can observe a voltage drop. If we could get a microscope that somehow allowed us to see the path of electrons going through any of these components, what do we actually see? Do the electrons slow down? How does this voltage drop (and the dissipated energy) actually look like?

Thank you again for reading through this long post. Please let me know if anything I’m asking is unclear.

UPDATE: I’ve edited my question based on DKNguyen’s answer:

Firstly, thank you so very much for the detailed response. I’ve wrote a follow up below:

Higher voltage means the more work can be performed per electron. Bt it doesn't mean the more work can be performed by the one same one electron. Why this distinction?

Because the physical electron itself doesn't actually move through the entire circuit very quickly similar to how individual air molecules do not travel with the sound wave. Air molecules actually moving is wind, not sound and travels far slower than sound. Similarly, the electrons move through the conductor much more slowly than the electrical energy/wave does.

Got it. Then if we could actually look “inside” of a wire connected in, let’s say, a DC circuit, we’d see each electron being “pulled” towards the positive terminal while the electron immediately “behind” it gets “pulled” to the empty space (created by the first electron). Right?

There's just a general limit in physics where if you cram too much energy into a space, it becomes a black hole. So I guess yes? I think other stuff happens first though in this case...like the electron turning into something else after you pump so much energy into it. At least, that's what the math says anyways.

Thank you for your response, though the answer to one of my questions is still remains: let’s say you have 2 circuits—one powered by a 5V battery and the other by a 10V battery (the circuits are otherwise completely identical). Let’s say you can somehow peer into the wire leaving the (-) terminal of the battery for both circuits, and you can actually see the individual electrons “in action.” How is what you see in circuit 1 different from circuit 2?

You see electrons bumbling back and forth, not at all traveling down the wire as you might imagine. The energy they carry travels far and fast, but individual electrons don't travel very far or fast over long distances. Bumbling around as they are, they do travel at different speeds, and in different directions not just down the length of the wire. On average they have a "drift" velocity which is the motion of the physical electrons you are asking about but this isn't the same as the speed at which the energy in the electrical wave is propagated. For 60Hz mains AC , the electrons themselves aren't traveling all the way back and forth between your house and the power plant at 60 times per second.

Thank you again, though my question about resistance still remains. Let’s say, again, you have 2 circuits each composed of a battery, a resistor, a switch, and the circuits connecting the components. The only difference between these circuits is in the resistance value of the resistors: 50 ohms vs 100 ohms. Again, let’s see you can somehow peer into the resistors and can see the electrons “in action.” You flip the switch to “on” in the first circuit, and immediately look “into” the resistor in circuit one. Then you do the same for the second circuit. How is what you see different in both circuits?

Thank you again.

• youtu.be/ZSrzvoJcaJ8?t=24
– G36
Jan 29 '20 at 19:49
• Let me just note, that these questions are far from "fundamental" :) These are more about intuition, which is far from the true mechanics of the things. Jan 29 '20 at 19:51
• Since you've simplified the Volt into J/C, you're now saying "the amount of work to move 6.25 * 10^18 electrons in one second". It would be like you pushing a piano through ice versus you pushing a piano through a rubber floor. Voltage doesn't have to do with speed. What controls the speed is the medium that it travels through, i.e. a copper wire vs. a silver wire... (piano being pushed on ice vs. rubber floor). The voltage is basically how "easily" those electrons are being "pushed" through the medium of travel.
– user103380
Jan 29 '20 at 19:55
• @KingDuken I think it would be much easier to push a piano on ice than on a rubber floor. I don't know about you, but to me, pushing a piano through ice is about as hard as pushing it through a rubber floor. ;) Jan 29 '20 at 19:58
• @DKNguyen Depends where you are standing :) I would like to see someone trying to push piano on ice while standing on ice themselves... Jan 29 '20 at 19:59

When a battery reads 5V vs another battery reading 10V, does that mean that every electron that moves through the circuit connected to the second battery does more work than any electron moving in the circuit connected to the first battery?

Higher voltage means the more work can be performed per electron. Bt it doesn't mean the more work can be performed by the one same one electron. Why this distinction?

Because the physical electron itself doesn't actually move through the entire circuit very quickly similar to how individual air molecules do not travel with the sound wave. Air molecules actually moving is wind, not sound and travels far slower than sound. Similarly, the electrons move through the conductor much more slowly than the electrical energy/wave does.

If so, does this mean that those corresponding electrons move at different speeds? Lastly, does this mean that we’d have an upper limit on how absolutely high a voltage value can get (the limit being set by the speed of light)?

There's just a general limit in physics where if you cram too much energy into a space, it becomes a black hole. So I guess yes? I think other stuff happens first though in this case...like the electron turning into something else after you pump so much energy into it. At least, that's what the math says anyways.

My second question is about resistance. Across any component on a circuit (or even a length of wire, in the “real” world) we can observe a voltage drop. If we could get a microscope that somehow allowed us to see the path of electrons going through any of these components, what do we actually see? Do the electrons slow down? How does this voltage drop (and the dissipated energy) actually look like?

You see electrons bumbling back and forth, not at all traveling down the wire as you might imagine. The energy they carry travels far and fast, but the aveage motion of the individual electrons as they bumble around is neither far nor fast. Bumbling around as they are, they do travel at different speeds, and in different directions not just down the length of the wire. On average they have a "drift" velocity which is the motion of the physical electrons you are asking about but this isn't the same as the speed at which the energy in the electrical wave is propagated. For 60Hz mains AC , the electrons themselves aren't traveling all the way back and forth between your house and the power plant at 60 times per second.

UPDATE:

Got it. Then if we could actually look “inside” of a wire connected in, let’s say, a DC circuit, we’d see each electron being “pulled” towards the positive terminal while the electron immediately “behind” it gets “pulled” to the empty space (created by the first electron). Right?

Not really. This is more true in a semiconductor where there are a more limited number of locations where electrons can sit (so-called holes) where the movie theatre analog of people moving from seat to seat works and there are a more limited number of spots where someone can sit.

In a conductor electrons can move, much more freely and around each other. The electrons are get repulsed from each other so they ideally all have equal space between them and every electron. More like a gas. If there is an area where there is more space, the electrons there get repelled by the others to fill that space up and it is a effect that travels down the line.

Thank you for your response, though the answer to one of my questions is still remains: let’s say you have 2 circuits—one powered by a 5V battery and the other by a 10V battery (the circuits are otherwise completely identical). Let’s say you can somehow peer into the wire leaving the (-) terminal of the battery for both circuits, and you can actually see the individual electrons “in action.” How is what you see in circuit 1 different from circuit 2?

The electrons are repel each other harder with the 10V battery than with the 5V. But the resistance they have to push through to get away from each other is the same so the ones repelling harder cause more to move.

Thank you again, though my question about resistance still remains. Let’s say, again, you have 2 circuits each composed of a battery, a resistor, a switch, and the circuits connecting the components. The only difference between these circuits is in the resistance value of the resistors: 50 ohms vs 100 ohms. Again, let’s see you can somehow peer into the resistors and can see the electrons “in action.” You flip the switch to “on” in the first circuit, and immediately look “into” the resistor in circuit one. Then you do the same for the second circuit. How is what you see different in both circuits?

The electrons are repulsing each other twice as hard in the 10V battery as the 5V battery. But the ones with the 10V batter have to push through a resistance twice as hard to get away from each other so the same number end up getting through in both circuits.

• ... and drift velocity will be mm/s, not anywhere near c. Jan 29 '20 at 20:36
• @Transistor ..if we are not taking "quantum leaps" in account.. Jan 29 '20 at 20:44
• Firstly, thank you! Secondly I edited my question with followups. Jan 29 '20 at 21:36
• Please, do not vandalize your question with "follow-ups"! It is not a discussion, it is Q&A site. Jan 29 '20 at 21:38
• @Arian Do not remove your original post. It makes subseqeuent answers out of context. Instead, add to your post with an update. I have done this for you already. Jan 29 '20 at 21:38

No, your fundamental misunderstanding here is thinking that the energy available in a unit of charge is related to the velocity of that unit of charge. That's not the case. It is possible to have high voltage when there is no movement of charge at all.

INAP (I'm not a physicist) but I think the more appropriate way of thinking about the energy in an electron, or another charge carrier, is how much work is required to prevent that charge carrier from moving toward a charge of the opposite sign. If it's a conduction electron in a metal, how much work do we need to do to keep that electron from falling back into one of the ionized metal atoms?

Again, INAP, but my understanding is that there is a lower limit to voltage, which is represented by a single electron infinitely far from a single proton. An absolute zero voltage by definition but not very useful in engineering.

An upper bound on voltage would then be determined by how closely you can hold two charges apart without letting them combine.

We also can assume that this energy per electron comes from the kinetic energy of the electron, which of course depends on its mass and speed (and, for the purposes of the question, let’s also assume that the mass does not change).

This is where your description starts to diverge from mainstream physics.

The electrical potential energy has nothing to do with how fast the electron is moving. It only has to do with how close it is to other charged particles.

As an analogy, think about gravitation potential energy. If I lift a bowling ball over my head, it has gravitational potential energy whether I move it around or hold it still. It also doesn't depend how fast I moved the ball when I lifted it. It has potential energy simply by being farther away from the center of the Earth than the ground.

Same thing with electrons. If I try to push one electron close to another one, it will gain energy. It doesn't matter how fast its moving, and it doesn't matter how fast I moved it to its new location. The electrical potential energy comes just from being close to the other electron.

The electrical potential energy also has nothing to do with how much mass an object has, only with how much charge it carries and how close to other charged objects it is.