I've read in many books and through many answers on this and other sites about this concept. Now I'm not sure what to "believe". In the subject at my university, we are taught that conducting wires are perfect. For example if in non-ideal wires electrons move randomly, but when you apply E-field they move in that direction, still somehow "hitting the nuclei" of atoms and basically they have this average velocity that's pretty small, (but that's the one that we are studying, not the actual speed when they randomly move).

Now in ideal wires, we consider that when an electron hits the nucleus, it's an elastic collision and no energy is lost... so electrons again have this average velocity and everything's the same. Now in that analogy, when you connect a battery of cells to a conducting (perfect) wire, you only need a "starting force", so that the electrons start moving (?), and since there's no friction -- they must move infinitely (although not accelerating). What's bothering me are these thoughts which I don't know if they are true:

(1) is the electric potential in the ideal short-circuit the same at all points (except in the source) because no E-field is present? While in the non-ideal short-circuit E-field is constant, meaning voltage must be present (therefore the electric potential in any 2 points must be different)?

(2) when electrons move in a non-ideal short circuit, their voltage lowers through the circuit such that they get from (for example) 20V to 0V with the distance. Of course the electron that's on half-way of the circuit already has a smaller el. potential like 10V?

(3) when you have a resistor in a non-ideal circuit, the "voltage drop across the resistor" is just a measure of energy (per charge) that a charge has lost on its way through that resistor? Also, why is the current the same before and after a resistor? I think it's because the electrons repel each other. For example the electrons that enter the resistor are slower (because they hit more nuclei in the resistor) and they exert a force on the electrons before them and so on? Therefore there aren't many electrons that exit the resistor which means smaller $I$ ($I=Q/t$), and there are many electrons before the resistor, but they move even slower because of the repulsion previously described, so their $t$ is big, again, $I$ are the same? (I don't get the water analogy.)

If someone could clarify this. (Explain where I'm wrong.) Thank you very much.


What you are missing is called conservation of charge. The fundamental fact is that electrons do not simply 'disappear' or vanish into thin air inside a resistor. This means that, if there is a certain number of electrons flowing into the resistor, these electrons must also flow out on the other side. In particular, in a steady state (i.e. the flow of electrons per second is constant, by the definition of 'steady'), we have that the number of electrons entering a resistor on one side per second must equal the number of electrons coming out per second.

This basic argument can very simply be made mathematically precise by noting that $I=\frac{dQ}{dt}$, where $Q$ is the charge flowing into (or out of) the resistor. I'll leave it up to you to transcribe my first paragraph into a mathematical equation, which will give you exactly what you're looking for.

  • $\begingroup$ If electrons don't lower their drift velocity after leaving the resistor, I don't see how it's possible for them, if the quantity/time is lower, to reach the other end (the beginning of the resistor, again) without piling up in front of the resistor or something... $\endgroup$ – user71801 Oct 8 '14 at 21:14
  • $\begingroup$ It seems that, ironically, you're now saying you don't understand the opposite of what you were asking about earlier... $\endgroup$ – Danu Oct 8 '14 at 21:27
  • $\begingroup$ It's a paradox to me, because it doesn't matter which way I look at the problem it turns out impossible. Whether you assume it's true that they pile up or they don't -- you (I) will come to the same conclusion. In this question, I think I expressed myself wrong -- I don't think that what enters the resistor doesn't leave (as in: smaller amount of charge leaves than it enters), but that it leaves with a small rate due to a lower velocity because of more collisions (eventually all charges leave the resistor). The only way I can see it happening is if the electrons themselves repel $e^-$ behind. $\endgroup$ – user71801 Oct 8 '14 at 21:36
  • $\begingroup$ @user71801 OK, let's get this straight: 1) Do you understand what is happening in a steady state, i.e. when the current is constant? $\endgroup$ – Danu Oct 8 '14 at 21:37
  • $\begingroup$ The drift velocity (average) of all the charges is the same, the charges go from - to + (or opposite, conventionally)... $\endgroup$ – user71801 Oct 8 '14 at 21:41

Idealizations are just that, approximately true in many situations. If you connect a wire across a battery, the non-ideal nature of the wire and battery will show-the wire will have some resistance and the battery will not be a perfect voltage source. In most circuits, the resistance of the other circuit elements are much higher than the resistance of the wire, so taking the wire resistance to be zero is a reasonable approximation.

So for (1) there is no answer-you need to consider the non-ideal properties. For (2) it s quite reasonable to assume the wire has a certain resistance per unit length. The voltage will drop linearly along the wire. Depending on the wire resistance, the battery voltage may be lower than you expect. For (3) the conservation of charge ensures that the same number of electrons enter and leave the resistor. The ones leaving are at a lower potential than the ones entering, but the number is the same.

  • $\begingroup$ About conservation... (3), what I'm thinking is, if less electrons enter, this means that they pile up before the resistor. But I know that cannot occur. $\endgroup$ – user71801 Oct 8 '14 at 20:40
  • $\begingroup$ As I said, the same number enter as leave. They just leave with less energy than they entered at. This energy difference is the heat dissipated by the resistor. $\endgroup$ – Ross Millikan Oct 8 '14 at 21:54
  • $\begingroup$ Do you want to say that their drift velocity decreases after they leave the resistor or what? $\endgroup$ – user71801 Oct 9 '14 at 6:11
  • $\begingroup$ No, electrons have a potential energy based on the voltage level they are at. That is what changes on the passage through the resistor. It is like a rock rolling down a hill. When it is high on the hill it has a lot of potential energy. After it stops rolling at the bottom, it has the same kinetic energy (zero) as it had before it started rolling, but less potential energy. Electrons in a potential field work the same way. $\endgroup$ – Ross Millikan Oct 9 '14 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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