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How do electrons behave in an electric circuit? There was a question before about a voltage drop which was interesting, but I can not participate in that discussion, so I am writing this question hoping to clarify this topic.

Electrons in a wire behave like an incompressible fluid. Driven by the potential difference, they all must move at the same time so they all gain energy instantly and then they all lose this energy due to moving through a specific resistor (like a wire itself).

For example, when you turn on the light, it starts glowing instantly although electrons are moving very slowly, but like cars in a row in front of a traffic light they all feel the signal to move so they start moving.

Suppose it is a battery that is making them move. If it is 1.5 V battery, it means that each unit of 1 C charge gets about 1.5 J. Not in the battery, but anywhere in a circuit. Electrons get what we call a drift speed; now they are moving slowly through the wire (1 cm/s) so there is no way that electrons from negative terminal are those we have to wait for, oh no!

Almost instantly, the influence of the field of the battery is spread through the wire. All electrons move now. Even if there is no resistance, electrons must come to positive terminal and sink in. And when they do, we say that they spent their energy, like a ball falling from a high building. This is a short circuit and it drains battery very fast since without any resistance, the current gets very high. When you put a light-bulb or a hater in this circuit, it will converse this energy of the electrons into light or heat or something.

So electron are entering and giving their energy away, then moving further, going to the positive terminal. They are slowing those electrons behind and a constant current is established in a very short period of time.

Knowing all of this, the question is, how do we have that the sum of voltage drops is always equal to a voltage of a source? Is the energy spent in a light-bulb equal to the whole energy of charge carriers? I should say not likely?

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  • $\begingroup$ you missed one, i wrote: so electron are entering, and it should be, electrons...plural. $\endgroup$ – Žarko Tomičić Nov 11 '14 at 16:17
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You are asking for a proof of Kirchhoff's voltage law or KVL, that the algebraic sum of voltage differences around any loop in a DC circuit is zero. KVL stems from Faraday's law, which essentially says that (unless there is time varying magnetic field, which, for an ordinary circuit is not true) the E field is curl-less, therefore the integral of E around any loop, which equals the voltage, is always zero.

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  • $\begingroup$ Well, no, that much I know, but why is the entire energy spent in the bulb? As far as I know, all of its energy should be spent when it comes to the other pole, in this case, positive terminal of the battery. So it should have some of its energy left of course, coming out of the last bulb..here, bulb can be anything, of course, that causes voltage drop. I know all about conservative fields and line integrals of force... $\endgroup$ – Žarko Tomičić Nov 13 '14 at 21:31
  • $\begingroup$ Well, if the remaining "wire" from the last bulb to the battery is a realistic wire, it also has resistance, therefore some voltage drop appears on it and correspondingly some energy dissipated in it. But if it is an ideal (theoretical) wire, or a superconducting wire the yes, all the energy difference of the electrons (energy at negative pole - the positive pole) is dissipated in the light bulbs. $\endgroup$ – Pooya Nov 13 '14 at 23:08
  • $\begingroup$ Hmh, in other words, I could think of the wire just as some kind of extended pole of the battery? Because, if the electron has not completed its cycle, I believe it should fall on to the positive pole? Why SHOULD bulb take all the energy? Is it because it actually IS on the ground level, so to speak, if it is the last bulb in the line? $\endgroup$ – Žarko Tomičić Nov 14 '14 at 17:24

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