How Electricity Actually Works

I made a video about a gigantic circuit with light-second long wires that connect up to a light bulb, which is just one meter away from the battery and switch, and I asked you, after I closed the switch, how long will it take for us to get light from that light bulb? And my answer was 1/c seconds.

• And his answer is wrong.
• We would be able to communicate faster than the speed of light.
• That violates causality and common sense.
• This is actually a bit misleading.
• Misleading.
• Misleading in a way.
• Extremely unconvinced.
• Naughty Mr. Veritasium has stirred up a right hornet's nest.

Clearly, I did not do a good job of explaining what was really going on in the last video. So I wanna clear up any confusion that I created. So behind me, we have a scaled down model of this circuit. It is only 10 meters in length on either side. Obviously that's a lot shorter than one light-second, but for the first 30 nanoseconds, this model should be identical to the big circuit, and Caltech has very fast scopes, so we'll be able to see what's going on in this time.

I got a ton of help on this from Richard Abbott, who works on LIGO, the gravitational wave detector. Over here, we are going to put a little resistor, which is gonna be the stand in for our light bulb, and we're going to measure it with a scope and see essentially, what is the time delay between applying a pulse on the other side, basically flicking the switch, for us to get a voltage across our resistor.

And the magnitude of that voltage is really important. A lot of people thought it would be negligible.

• The amount of energy supplied by this is so minuscule.
• A tiny, tiny effect, right?
• The amount of power you're getting to the lamp over here, it's nuff-all.
• He meant the light turns on at any current level immediately.
• That is not what I meant.

Well, actually, with that assumption, Derek's answer is wrong. The light never turns off no matter the state of the switch. Some electrons will jump the gap and result in an extremely small continuous leakage current.

• Let me be clear about what I am claiming. Okay, it is my claim that we will see voltage and current through the load that is many orders of magnitude greater than leakage current, an amount of power that would actually produce visible light if you put it through an appropriate device, and we will see that power there in roughly the time it takes the light to cross the one meter gap, but to understand why this happens, we first have to clear up some misconceptions that I saw in responses.

Misconception number one is thinking that electrons carry the energy from the battery to the bulb. Let's say we have a simple circuit with a battery and a bulb operating at steady state. If you zoom in on the light bulb filament, you'd see a lattice of positively charged cores of atoms, the nucleus and lowest shells of electrons, surrounded by a sea of negative electrons, which are free to move around the lattice.

The actual speed of these electrons is very fast, around a million meters per second, but all in random directions. The average drift velocity of an electron is less than 0.1 millimeters per second. Now frequently, an electron will bump into a metal ion, and transfer some or all of its kinetic energy to the lattice. The electron slows down and the metal lattice starts wiggling more. It heats up. And ultimately this is what causes the filament to glow and emit light.

So a lot of people will look at this and conclude the electron carried the energy from the battery to the bulb where it dissipated its kinetic energy as heat, but consider, where did the electron get its kinetic energy from before the collision? It didn't carry that energy from the battery. In fact, if the circuit has only been on for a short time, that electron hasn't been anywhere near the battery. So how was it accelerated before the collision?

The answer is, it was by an electric field in the wire. The electron repeatedly collides with the lattice, and loses energy. And after each collision, it is again accelerated by the electric field. So although it is the electron that transfers energy to the lattice, the energy came from the electric field. So where does that electric field come from? Well, a lot of animations make it look like the electrons push each other through the circuit via their mutual repulsion.

So you might think the electric field comes from the electron behind it. There is the analogy of water flowing through a hose, or marbles in a tube. This is misconception two, thinking that mobile electrons push each other through the circuit. That is not how electrons flow in circuits. The truth is if you average over a few atoms, you find the charge density everywhere inside a conductor is zero.

The negative charge of electrons and the positive cores of atoms perfectly cancel out. So for each repulsive force between electrons, there is an equal and opposite force from the positive ion next to it. These forces cancel out. So mobile electrons cannot push each other through the wire. So where does the electric field come from?

Misconception number three is that it comes entirely from the battery. This makes intuitive sense, since the battery is the active element in the circuit, it has a positive side and a negative side. So it has an electric field, but this is not the electric field that all the electrons within the wire experience. Consider that the electric field of the battery is much larger close to the battery.

So if its field were really what's pushing the electrons around, then if you brought the light bulb close to the battery, then the bulb would glow much brighter. And it doesn't. The truth is that the electric field in the wire comes both from the battery and from charges on the surface of the wires of the circuit.

As you go along the wire from the negative end of the battery to the positive end, there is a gradient of charge built up on its surface, starting with an excess of electrons, through to roughly no charge in the middle, as we'll see the steepest charge gradient is actually across the load to a deficiency of electrons, the exposed positive cores of atoms on the surface of the positive end of the wire.

All these charges and the charges on the battery create the electric field everywhere inside the wires. They also create an electric field in the space around the wires. These surface charges were set up almost instantaneously when the battery was inserted into the circuit. You might think you'd have to move electrons a significant distance to create this charge distribution, but that is not the case.

Even a slight expansion or contraction of the electron sea, with electrons moving on average, the radius of a proton, can establish the surface charges you see. So the time for the charges to move is completely negligible. The speed of the setup process is limited only by the speed of light. Once that surface charge distribution has been established, the battery does continuous work to maintain it, by moving electrons through the battery against the Coulomb force.

In the load, the electric field created by all the surface charges, accelerates electrons, which dissipate their energy in collisions with the lattice. So the battery is putting energy into the field, which electrons take out and transfer to the load. An electrical engineer who made a response video, Ben Watson, came up with a good analogy. The battery is like a shepherd.

The surface charges are the sheep dogs responding to his orders. And the mobile electrons are the sheep, guided by those barking dogs. The surface charge description of electric circuits is omitted from most textbooks, but there is a great treatment of it in Matter and Interactions by Chabay and Sherwood. They also have a VPython simulation where you can see the positive surface charge in red, and negative surface charge in blue.

You can see how this entire charge distribution creates a net electric field shown by the orange arrow, everywhere in and around the circuit, everywhere inside the wire, the electric field has the same magnitude and its direction is along the wire. This is really showing you the electric field in the center of the wire, but it's depicted on the surface so you can see it.

In this circuit, all the conductors are made of the same material, but the segment at the bottom has a much narrower cross section. So it represents a resistor. Since the cross sectional area is smaller, the electron drift velocity through the resistor has to be higher so that it can carry the same current as everywhere else in the circuit. Now, drift velocity is directly proportional to electric field. So the electric field must be largest inside the resistor.

And this is achieved by having the steepest gradient of surface charges here. You can also see the contribution to the net electric field from the battery in magenta, and the contribution from surface charges in green. Far from the battery, most of the electric field is due to surface charges, whereas close to the battery, it has a greater contribution and the field due to surface charges is actually in the opposite direction to the field from the battery.

So to sum up, electrons don't carry the energy from battery to bulb, nor do they push each other through the wire. They are pushed along by an electric field, which is created by charges on the battery, and charges on the surface of the wires. With this view of circuits, things that might have previously seemed mysterious make a lot more sense.

Like if electrons leave a battery at the same rate, and with the same drift velocity as they return, how do they carry energy from the battery? The answer is they don't. They are accelerated by the electric field before each collision with the lattice. At a junction, how do the correct number of electrons go down each path? Well, they're guided by the electric field, which extends everywhere throughout the circuit.

The fields are the main actors, extending everywhere throughout the circuit, and the electrons are just their pawns. So how does this apply to the big circuit? When the battery is connected into the circuit, even with the switch open, charges rearrange themselves. On the negative side of the battery, there is an excess of electrons on the surface of the wires and the switch.

On the positive side, there is a deficiency of electrons. So positive charges built up on the surface of the wires. The charges rearrange themselves until the electric field is zero everywhere inside the conductor. This electric field is due to all the surface charges and the charges on the battery. There is an electric field outside the wires due to these charges, but it's zero inside the wires.

We now have the full potential difference of the battery across the switch. And no current is full flowing, except for leakage current, which I'll assume is negligible. When we close the switch, the surface charges on both sides of the switch neutralize each other on contact. And at that instant, the electric field inside the conductor is no longer zero, and current starts flowing through the switch.

Simultaneously, the new electric field from the modified surface charges radiates outwards at essentially the speed of light. And when it reaches the bulb, the electric field inside it is no longer zero. So current starts to flow here too. This is why I said the bulb lights up in 1/c seconds, because the bulb was one meter from the switch, and the change in the electric field travels out at the speed of light.

As some of you pointed out, the answer should have been one meter divided by C. And I apologize for the casual use of units. If you were to move the switch, then the bulb would take a different amount of time to emit light, which just depends on the distance between the switch and the bulb.

In response to my original video, Ben Watson simulated a model of the circuit using software from Ansys called HFSS. It provides a complete solution to Maxwell's equations in three dimensions. Now have worked with Ben and Ansys to make these simulations. When the switch is closed, you can see the electric field radiate out, and as it hits the far wire, it generates current.

The electric field is to the right. So the electrons flow to the left. This simulation shows the magnitude of the magnetic field, which falls off pretty rapidly as it crosses the gap. But then a magnetic field appears around the far wire, and this magnetic field is created by the current in that wire. To me, this suggests that it really is the electric field, and not the changing magnetic field that creates the current through the load.

Some commenters on the original video claimed my answer of three or four nanoseconds violates causality. I guess they were thinking that the bulb would only go on if the circuit were complete. And it wouldn't if the circuit were broken somewhere, which could be up to half a light second away. So it seemed like I was saying, we could get information about the status of the whole circuit, even half a light second away, in just nanoseconds.

But that is not what I was saying. What I should have stated explicitly, is that the bulb goes on regardless of whether the circuit is complete or not. The current flows through the load due to the electric field it experiences. To illustrate this, Ben added a wire below the circuit that is completely disconnected from it. You can see is that its response to the changing electric field is virtually identical to that of the top wire, at least up until the signal reaches the far end and reflects back.

This is why my answer doesn't break causality. At least initially, connected and disconnected wires behave exactly the same. Using this software, you can also simulate the Poynting vector that is the cross product of electric and magnetic fields. In the last video, I showed how the Poynting vector indicates the direction of energy flow.

And after the switch is closed, the Poynting vector points out of the battery, across the gap to the other wire, whether connected or not, because energy is carried by fields and not electrons, it can go straight across the gap. So you might ask, why do we need wires at all? Well, we don't. I mean, phones and toothbrushes charge without wires connecting them to a stream of electrons, and researchers have demonstrated remote charging using the energy from WiFi signals.

Wires are more efficient because they channel the fields and hence the energy from source to load. Here's another angle on the Poynting vector. And you can see once there is current in the top wire, the fields around it carry energy in both directions. Now, of course, the Poynting vector also points parallel to the first wire, carrying the energy around the circuit as most people would expect.

But again, note how the energy is carried outside the wires, not in the wires. Now admittedly, thinking about circuits this way is complicated. And since nobody wants to solve Maxwell's equations in three dimensions just to analyze a basic circuit, scientists and engineers have worked out shortcuts.

For example, Ohm's law, voltage equals current times resistance, is just the macroscopic result of all the surface charges, their electric fields, and zillions of electrons bumping into zillions of metal ions. You can simplify all that physics into a single circuit element, a resistor, and the basic quantities of current and voltage. This is called the lumped element model, lump all the spread-out multi-particle and field interactions into a few discrete circuit elements.

And we use this technique every time we draw a circuit diagram. So our original diagram of the big circuit is flawed because fields between the wires are important to the problem, but there are no circuit elements to indicate these interactions. To fix it, we need to add capacitors all down the wires. These capture the effect charges on one wire have on the other.

If there are negative charges on the surface of the bottom wire, for example, they'll induce positive charges on the surface of the top wire. Also, since these wires are long, they're gonna create significant magnetic fields around them, which resist changes in current. So we model this with inductors all the way down the wires. We could also add resistors, making what electrical engineers would recognize as the distributed element model for a transmission line.

But we're assuming that these wires are superconducting. So this is how we could model a superconducting transmission line. This diagram offers another way of understanding why current flows through the load almost immediately. When you first apply a voltage across a capacitor, current flows as opposite charge builds up on the two plates.

In the short time limit, a capacitor is a short circuit. It acts just like an ordinary wire. Once it's charged up, no more current flows, but by this point, the next capacitor is charging up. And then the next one, and then the next one. And so we have a loop of current that is expanding out at roughly the speed of light. This is of course, just another way of talking about the effect the electric field that the bottom wire has on the top wire.

One reason it's useful to look at the circuit this way, is because you can use the values of inductance and capacitance to calculate the characteristic impedance of the transmission lines. You can think of this as the resistance to alternating current that a source would see when sending a signal down the wires. The characteristic impedance is equal to the square root of inductance divided by capacitance.

And for our circuit, I measured the capacitance and the inductance of the lines, - 11.85, call it, micro Henry's. - So we got a characteristic impedance of about 550 Ohms. To maximize the power delivered to our load, we want its resistance to equal the sum of the other impedances in the circuit. So that's why we picked a 1.1 kilo-Ohm resistor.

Now, I hope you're convinced that current will flow as soon as the electric field reaches the far wire. The question is, how much? Are we gonna see an appreciable voltage even with these lines a meter apart? That's what it seemed like a lot of people were doubting in the last video. So that's really what we want to find out here.

Okay, so now we're putting a pulse in there.

• Yep.

• Well looky, looky, Derek.

• So what do we got that yellow one is our-

• Got a fraction of the applied voltage overshoot. And then-

• So it looks to me like the initial voltage that we're getting is about-

• Five volts per division. So it looks like about five volts, roughly four or five volts.

• The green curve rising up to around 18 volts is the source voltage.

And the yellow line is the voltage across the resistor. So after just a few nanoseconds, this voltage rises to around four volts. Since the resistor was a kilo-Ohm, that means four milliamps of current are flowing in the resistor, before the signal goes all the way around the circuit. So we were transferring about 14 milliwatts of power. This is what 14 milliwatts of light actually looks like.

So, yeah, it's not a fully on bulb, but it is visible light and way more than you would get from just leakage current. Now, some of you may argue, it's unfair to use a little LED when I showed a bulb and car battery in the original video, but those items were for illustrative purposes only. The clue that this is actually a thought experiment is the two light-seconds of super conducting wire that stretch out into space.

This is not an engineering question about how best to wire up a light bulb in your bedroom. The question was intentionally vague. And if you want to choose circuit components such that the bulb never goes on, you are welcome to do that and I support your conclusion. Just to me, the most interesting way to approach this problem is to ask, how could you make the light go on fastest?

I was worried that those long wires would pick up all the radio waves passing through, and we wouldn't even be able to see the signal for that noise, but you can see clearly on the graph that the signal is way above the noise level. Alpha Phoenix set up a kilometer of wire and got a very similar result.

• So the light bulb turns on a little bit, and then after one light-speed delay, the light bulb turns on the rest of the way.

• YouTuber, ZY, simulated the transmission line circuit, and found that even with realistic assumptions, he transferred 12 milliwatts to the load straight away.

• Derek is actually more correct than we give him credit for.

So, I believe that he's correct on all counts. And the question is neither deceptive or requires like technicalities.

• So everyone agrees that a steady, small, but way-bigger-than-leakage-current signal flows through the load in the first second after the switch is closed. Is it enough to emit light? Yes, if you use an LED. But the point of the thought experiment was to reveal something that's normally hidden by the way that we think about and teach electric circuits.

You know, we use voltage and current and lumped elements because they're more convenient than working with Maxwell's equations, but we shouldn't forget that the main actors are actually the fields. They are what carry the energy, and you don't have to take my word for it. This is Rick Hartley, a veteran printed circuit board designer.

• I used to think in terms of voltage and current.

And I used to think that the energy in the circuit was in the voltage and current, but it's not. The energy in the circuit is in the fields. The most important thing you need to know is that when you route a trace, you better define the other side of that transmission line, because if you don't, those fields are gonna spread and they're gonna leave you an unhappy individual.

• I think one of the things that I'm most excited about the circuit's video was the response videos I saw by so many people, especially people with far better credentials in electrical engineering than me. I really enjoyed watching those videos.

So I feel like my circuits video was kind of like a mistake on my part in certain ways that I didn't delve deep enough into this part of the problem. I honestly didn't think that this was the focus of the video, but clearly everyone who watched it did, so that's on me, but by making that mistake, and by not going deep into my explanation, I invited seemingly a whole bunch of other people to make explanations, which I thought were great.

And some people like Alpha Phoenix even took up the challenge and set up his own version of the experiment. So, frankly, I'm just really excited at what came about, even though I do acknowledge that this was my fault in the first place. Like I should have done a better explanation, but by not doing so, you know, there are a lot of great explanations out there.

And that's what I love. So I'm gonna recommend a whole bunch of electrical engineering YouTubers to you in case you wanna check those out because they're a lot of great channels, and you should really see how they think about electronics, and how they explain this circuit.

Hey, this video was sponsored by Brilliant, the website and app that gets you thinking deeply about concepts in math, science, and computer science. Brilliant is sponsoring a lot of our videos this year, because they know someone who makes it to the end of a Veritasium video is exactly the sort of person who would love to learn with Brilliant.

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