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It's said in books that a circuit is a closed path and thus that electrons come back to the source. If that's the case, what would happen when there were an earth fault in a circuit? How would the electrons return to their source?

Do the electrons actually move out of their atoms or do they just vibrate and transfer the energy that way when we apply a voltage?

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    \$\begingroup\$ I discussed more in my answer, but circuits are an abstract concept. "electrons come back to the source" in the abstract means that they must reach the reference potential. For example, a grounded battery and the Earth: the mobile charges might reach Earth or the battery negative, but because they are at the same potential, they are effectively connected. \$\endgroup\$ – DrFriedParts Jun 17 '13 at 0:24
  • \$\begingroup\$ When there is an earth fault, the electrons move through the earth fault, through an earth connection, back to the source. If there was no earth connection, there would be no current, even with an earth fault. A fully isolated circuit would be safer, but that is another question. \$\endgroup\$ – david Jun 17 '13 at 2:28
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/233851/… \$\endgroup\$ – Always Confused Jul 10 '16 at 9:29
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/243060/… \$\endgroup\$ – Always Confused Jul 10 '16 at 9:30
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Thinking about current in terms of electrons moving around is the start down a path to a poor mental model of how electricity works. Here are just a few things wrong with it:

  • Electrons are but one of many charge carriers. Any ion is also a charge carrier.

  • The protons that balance the electrons are just as important. If you had just electrons, then all the electrons in the universe would be repelled from each other and shoot out into the universe.

  • Electrons have negative charge, and you will confuse yourself for no good reason at all thinking about how they flow from negative to positive. It doesn't actually matter at all.

  • Electrons are actually swarming around in all random directions all the time, and their motion due to current is minuscule, in comparison.

The important thing is this: charge carriers (electrons being one of such) can be used to transmit an electromotive force (usually called just voltage). This is a pretty ordinary concept, really. You can push one end of a rod, and transmit a mechanical force to the other end of the rod. Does the rod move, when do you this? Well, maybe, but there are two things happening here:

  1. force is transmitted through the rod, as waves propagating at the speed of sound in that material
  2. if and only if we also transmit power, the rod moves, at most cases at a much slower speed

The difference is obvious for a rod, but since we can't see electric charge, the difference is not obvious.

So, your question was: Do electrons actually flow when a voltage is applied? Strictly speaking, the answer is maybe, and it depends on what you mean by flow. It's similar to the question, does a rope move when you pull it? Well, if it's attached to a balloon, it might move a lot. If it's attached to a brick wall, it might not move at all.

The motion of charge carriers (like electrons) is current. If we have a current, then there is a net motion of charge carriers. Really they are swarming all over, much as the individual water molecules are swarming around in a pipe, even if there's no net flow. The current describes the average movement. In the case of DC current, the average motion is in a circle.

How the individual charge carriers interact to accomplish this is complicated, and it's really a physics question, not an electronics question. However, I'd suggest you check out this MIT tutorial on fields.

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  • \$\begingroup\$ But afaik a bunch of electrons will fly apart, not cuddle together in a ball. \$\endgroup\$ – Wouter van Ooijen Jun 16 '13 at 18:09
  • \$\begingroup\$ @WoutervanOoijen yes, I guess you are right :) In any case, it would be a very different world! \$\endgroup\$ – Phil Frost Jun 16 '13 at 19:05
  • \$\begingroup\$ 90% of everything I read is just plain wrong when it comes to electrons moving around and loose electrons. \$\endgroup\$ – johnny Oct 6 '18 at 15:20
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Electrons do physically move when a voltage is applied - extremely slowly.

A circuit energized at 100VDC, powering a 1A load (like a light bulb) through 2mm diameter copper wire will see electrons moving at the rate of:

\$ \dfrac{I}{Q \cdot e \cdot R^2 \cdot \pi} \$

where

  • Q is the number of electrons per cubic centimeter of copper (roughly \$8.5 \times 10^{22}\$)
  • R is the radius of the wire
  • e is the charge per electron (roughly \$1.6 \times 10^{-19} \$ coulombs)

This works out to 8.4 cm/hour. Not exactly fast.

What's key is the fact that it's the energy that races through the circuit almost instantaneously - not the electrons themselves. (The electrons make a convenient 'highway' to allow the energy to flow quickly.)

It's unfortunate that the slow drift of electrons under a voltage ended up with the same name as the energy flow that actually does work in a circuit.

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  • \$\begingroup\$ Without that slow drift of electrons, there is no current, so we can't have an energy flow. An energy flow is called power, and as we know, \$P=IE\$. If \$I=0\$, then there can be no energy flow. So maybe they only half have the same name :) \$\endgroup\$ – Phil Frost Jun 17 '13 at 2:03
  • \$\begingroup\$ True enough. Mind you, in AC, they just wiggle and don't really circulate per se. \$\endgroup\$ – Adam Lawrence Jun 17 '13 at 2:49
  • \$\begingroup\$ Q = 8.5×10^22 Elektrons/cm^3 is the total number of electrons per volume of Cu. Only a fraction of those electrons are free electrons that take part in conduction (en.wikipedia.org/wiki/Free_electron_model). So this formula is wrong. \$\endgroup\$ – Curd Sep 16 '13 at 15:29
  • \$\begingroup\$ @Curd your number is wrong, where did you get it? >"Q = 8.5×10^22 Elektrons/cm^3 is the total number of electrons per volume of Cu." No, the total number of electrons/cm^3 for copper is 2.46x10^24. Therefore, if each atom contributes just one mobile electron to the metal's electron-sea, then the free electron density = 2.46e24 / N, where N=29 for copper. Their above equation is correct. See this same calc in Halliday/Resnick physics, or wikipedia, Drift_velocity \$\endgroup\$ – wbeaty Aug 11 '15 at 2:28
  • \$\begingroup\$ @wbeaty: yup, you are right (I don't have Halliday but) I recalculated and get about rho / Mm * Na * 29 = 2.44E24 as total number of electrons per cm^3 (rho densisty, Mm molar mass, Na = AVogadro's number). I don't remember my calculation 2 years ago... \$\endgroup\$ – Curd Aug 11 '15 at 6:58
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Don't confuse convenient abstraction with physical reality

  • "Circuits" are an abstract concept designed to help us better reason about the world.
  • electrons are a physical entity.

A note about "closed" paths

Closed path circuits does not imply electrons returning to the source. Furthermore, the electrons that leave the source are extremely rarely the same electrons that return to the source's other pole (see @madmanguruman's answer for the velocity explanation).

Mechanical analogies

It's like dominoes that fall. The energy wave propagates through the falling dominoes, but the dominoes don't translate much.

Remember energy is the charge of the electron times the force applied to it (voltage). It is (overwhelmingly) the forces that are moving through the metal lattice, not the charges (electrons).

Just like in this picture:

enter image description here

The forces transfer across the balls, but the balls stay largely in place. Unlike the mechanical balls, which are balanced by gravity, with electrons in metal wires from galvanic cells (batteries), there is a slow overall drift of the electrons (like cars stuck in traffic) to the other end.

Further reading

You might consider this answer I gave to a similar related physics question.

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  • \$\begingroup\$ Heh, circuits are common macro objects, while electrons are theoretical beasts with strong QM behavior. But I do agree: we can eliminate much abstraction by using circuits built from charged sand through hoses, or charged metal balls on a rotating plastic wheel. In any case, charge drift (current) is required in any circuit. Analogy: with a mechanical drive belt, employ higher and higher force/tension at lower speed, until the belt moves at meters/hr yet transfers kilowatts. It only seems as if the force is more important than the motion. Stop the sloooow belt, and the energy stops too. \$\endgroup\$ – wbeaty Aug 11 '15 at 2:53
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We're talking about metals here. Typically, an object of metal does not consist of molecules. It consists of metal atoms, all grouped together. This is shown in the picture below:

enter image description here

The red circles are electrons. As you can see, you can't really say to what atom an electron 'belongs'. These electrons form the connections between the atoms - so they belong to two atoms.

Now, when a current starts flowing, these electrons indeed move. When a current flows, energy is transferred. Since the atoms can't move easily, the electrons have to move.

You can see this in the unit Ampere of current as well: 1 ampere equals 1 Coulomb per second. The Coulomb (C) is the unit of charge (Q). 1 Ampere means 1 Coulomb of charge passes a certain point in 1 second. This charge is produced by the electrons that actually flow from object one to object two.

When we're talking about DC current (normal battery-powered application, for example), these electrons will not return to their source. Consider this circuit:

enter image description here

At the beginning, there's a difference in charge between the negative and the positive pole: the negative pole has a surplus of electrons. This creates a force (voltage), and since there's a link between the two poles (the wire and the bulb), the electrons start to flow. The electrons move from the negative pole through the bulb to the positive pole, until there's no difference in charge anymore (or it's that little that it won't cause a current to flow).

You can now see that these electrons did not return to their source: they started at the negative pole and ended at the positive pole.

We call this a closed path because there's a circle: current starts at the battery and ends at the battery. There's confusion because the battery actually exists of two objects: the positive and the negative pole.

Look at this circuit (which is basically the same, but with a capacitor instead of a battery and a resistor instead of a bulb):

enter image description here

Current flows from the right side of the capacitor (negatively charged, electrons surplus) through the resistor to the left side of the capacitor (positively charged, electrons shortage). Here, the capacitor plates are separated, so you can easily see that it actually isn't a closed path.

We just call it a closed path, because current starts and ends at the capacitor.

Since the electrons don't really have to return to their base, you can now understand that electrons can flow into the earth as well. This is also what happens with lightning. Electrons flow from the clouds to the earth (or the other way around, I wouldn't know), just to neutralize the difference in charge.

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    \$\begingroup\$ Regarding lightning: Both directions. "On average worldwide, negative lightning flashes make up the vast majority, about 90 percent of all strikes. ... By the way, positive lightning strikes are believed to be the most dangerous, since they can produce very large currents, up to 300,000 amps!" (source) \$\endgroup\$ – Anindo Ghosh Jun 16 '13 at 13:21
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    \$\begingroup\$ I like your energy @Camil (pun intended), but you should be aware that there a number of subtle inaccuracies with this answer. The confusion is not that a battery has two poles, the confusion is that circuits don't describe the movement of any single electron -- they describe aggregate behavior and energy transfer... your answer continues to make the same confused assumptions that led the OP to ask the question. Either discuss in the abstract, in which case, the current must return to the source - or - discuss the physical with electrons and their any-equipotential-surface-will-do attitude. \$\endgroup\$ – DrFriedParts Jun 17 '13 at 0:20
  • \$\begingroup\$ p.s. - I did not down vote. Just for the record in case someone else does. -- "not me!" ;) \$\endgroup\$ – DrFriedParts Jun 17 '13 at 0:21
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    \$\begingroup\$ It would also be worth pointing that although electrons do not travel through batteries, current does. This is why a battery must have an electrolyte, and it works precisely because electrons can't travel through it, but positive ions can. The positive ions, travelling in the opposite direction of the electrons, prevent the electrons moving through the circuit from creating an equilibrium until the chemical energy is depleted. Though the ions and electrons move in opposite directions, they have opposite charges, and together make a complete circuit of current in one direction. \$\endgroup\$ – Phil Frost Jun 17 '13 at 2:11
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    \$\begingroup\$ @CamilStaps an individual electron will take a random path anywhere it can. Probably most of this motion is attributable to thermal noise, and not the electric machine in which it happens to be a part. Only if you take the average motion of many (more than billions) of electrons will you notice that they are moving in one direction more than another. And, circuits don't describe electron flow: they describe current flow. \$\endgroup\$ – Phil Frost Jun 17 '13 at 13:04

protected by Dave Tweed Jan 10 '16 at 21:31

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