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schematic

simulate this circuit – Schematic created using CircuitLab

From what I know, current is the flow of charges, and charges move because of the potential difference, i.e. voltage. The thing is, charges in a circuit are affected by voltage drops, and total voltage drop is equal to the initial voltage. So, once the charge passes through the last resistor, it's voltage is 0! There is no potential difference anymore, so why is it moving towards the positive terminal?

I have 2 guesses, one is that those charges are still negatively charged so there is still some potential difference. Voltage from the battery is just some sort of additional energy (is this correct?) My other guess is that they are being pushed away by incoming charges.

You wanted me to sketch it so here it is, but I think it works for most circuits...

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    \$\begingroup\$ It's really hard to understand the question as it is worded using some quasi-technical language. But apparently yes, battery is an energy source. \$\endgroup\$ – Eugene Sh. Nov 18 '16 at 22:14
  • \$\begingroup\$ Please really use the circuit editor and draw the circuit. The question editor has a schematic editor button for a purpose! (@EugeneSh.: M. Wother asked this in a by-sentence in another question, so I asked her/him to ask this separately, but I also already explained that a good schematic is absolutely necessary when talking about something) \$\endgroup\$ – Marcus Müller Nov 18 '16 at 22:20
  • \$\begingroup\$ show your circuit and assumptions. The assumptions may be wrong. \$\endgroup\$ – Sunnyskyguy EE75 Nov 18 '16 at 22:37
  • \$\begingroup\$ In the presense of no external voltage, charges will try and move away from each other - look up gold-leak electroscope. This isn't a real question so I'm voting to close. \$\endgroup\$ – Andy aka Nov 18 '16 at 22:38
  • \$\begingroup\$ I don't understand everyone's issue with this question. The circuit can be a battery and a resistor. Between the resistor and the negative battery terminal in normal circuit analysis, it's assumed that the voltage is 0. He/she is asking how can charge move between across the wire between the resistor and the battery if there's 0 volts to push the charge. Seems straight forward enough. The issue is the idealization of the wire having 0 ohms. If they represent it as 0.001 ohms as it is closer to, then all of the confusion should go away. \$\endgroup\$ – horta Nov 18 '16 at 22:44
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Look at this problem from a different perspective because what you have here is similar to a Zeno's arrow paradox. You have effectively stopped time and examined a static position without considering the dynamic situation.

You don't have a stationary single electron, you always have billions of electrons in motion. (either as a drift current or as random Fermi motion - see https://en.wikipedia.org/wiki/Drift_velocity ) Even your single electron is always in motion.

Let's assume our electron has the same potential as the 0V terminal of the battery and has no inclination to move (totally ignoring the fact the electron would have momentum). If there is a drift current (the actual movement of electrons) then very soon a second electron will find itself in the same position near the first, then a third, then a fourth etc.

This accumulation of electrons (increased electron density) constitutes a more negative potential than the 0V so creating an electric field which would sweep the electrons towards the 0V electrode.

Let's not ignore momentum - The electron has been subjected to an electric field (from +V to 0V when you complete the circuit) - it has been drifting in the direction of the 0V terminal at a speed of about 2.3 x 10^-5 m/s. It will continue to do so because there is nothing to stop it.

Electron drift is not the same thing as electric current. See https://en.wikipedia.org/wiki/Speed_of_electricity

Does this make sense in the real world? - Kirchoff's law states that charge cannot accumulate at a node - current going in must equal current going out. In other words - what goes in must come out.

One final point - The charge on an electron is a Universal constant (-1.6 (0217662) x 10^-19 Coulombs), it always has the same value. Charge cannot be created or destroyed, it is always conserved.

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A potential difference sets up an electric force which causes conduction electrons to move. A battery is a piece of complicated chemistry, with anions and cations, and I don't want to delve into the details there. Plenty of better sources for that, than me.

But the fundamental problem with your imagination is illustrated where you write, "it's voltage is 0!" Up until that point, you were talking about potential differences (appropriate.) Suddenly, right there you skip tracks and then conflate (confuse) the idea of a voltage at a single point with the idea of potential differences. These aren't even close to the same thing.

A voltage at exactly one node is entirely arbitrary. I could look at your circuit and argue correctly that the node you identified as 0 V is really at 1,000,000 V. The number is completely arbitrary. You can make it up. So can I. And we are both right, so far as the idea goes. Such values depend only on your choice of reference. And you and I are allowed to make different choices, there. (Of course, once you've chosen a reference point and assigned it any arbitrary value, you can't do that again. You only get to chose a reference point and assign it a value, just once.)

The only question here is whether or not there is a non-zero electric field intensity (which is a vector), \$\vec{\varepsilon}=\left(\frac{\textrm{d}V}{\textrm{d}x}, \frac{\textrm{d}V}{\textrm{d}y}, \frac{\textrm{d}V}{\textrm{d}z}\right)\$, which provides a motive force.

The battery itself has a field intensity direction inside it, as well. So the point you called "0" has a different potential to one side and to the other side, so there is still a field intensity direction at that point, as well. Everywhere in the circuit, in fact, has an electric field gradient. So the electrons continue to be propelled by the force acting on them throughout the circuit.

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  • \$\begingroup\$ So what would you answer if instead of "it's voltage is zero" I wrote "the potential difference between the charge and the terminal is 0"? Are the other answers (which say that it is actually not 0) correct? \$\endgroup\$ – M. Wother Nov 19 '16 at 0:10
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    \$\begingroup\$ @M.Wother I would say that your statement is a matter of your choice of reference, which is entirely irrelevant to the question. The electric field gradient is still there, despite your choice of reference for assigning absolute values to points. Nature doesn't care about what you choose as a reference for making such statements. Your choice of perspective isn't "reality." \$\endgroup\$ – jonk Nov 19 '16 at 0:12
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So your point makes perfect sense, in an ideal world. In a schematic, a single node has the same voltage everywhere.

In the real world, components aren't ideal, and that extends even to the wires that make up the nodes in the physical circuit. They have a resistance, but it is usually negligibly small. So there is, in fact, a voltage that still causing charges to move in the wires after the charges have passed through all of the components.

In most cases you won't find a significant difference in the voltage at the two ends of a wire unless there is a lot of current moving through that wire, or the wire is sufficiently long. For instance, 16 gauge copper wire has a resistance of approximately 4 \$ \Omega \$ of resistance per 1000 feet (source: google 'resistance of copper wire'). With that, it would take about 250 amps to notice a 1 volt drop across a single foot of wire!

Let me know if anything is unclear, or if I didn't actually answer your question. It felt a little like I was rambling in there.

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  • \$\begingroup\$ I don't feel like it answered my question... see the scheme I wrote in the question. So once a charge passes through the resistor, it has 0 volts. So why does it keep moving? I don't see how the fact that wire has resistance helps here.. doesn't the resistance resist the movement of charge even more? \$\endgroup\$ – M. Wother Nov 18 '16 at 23:02
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    \$\begingroup\$ The point is that it doesn't have 0 volts, the voltage of that charge is (with most instruments) immeasurably small, but that charge does have a potential difference when compared to the negative of the source, so the charge still needs to move. Does that make more sense? \$\endgroup\$ – ambitiose_sed_ineptum Nov 18 '16 at 23:05
  • \$\begingroup\$ @M, after it passes through the resistor, it doesn't have 0 V. It has maybe 0.001 V , and that is enough to move it through the wire, which has, as an example, 1 milliohm of resistance. \$\endgroup\$ – The Photon Nov 18 '16 at 23:05
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    \$\begingroup\$ @M, or look at it another way, if the wire is perfect (0 ohms resistance), it doesn't take any voltage to force a current to move through it. \$\endgroup\$ – The Photon Nov 18 '16 at 23:06
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    \$\begingroup\$ @M.Wother, or, let's try yet another approach: When the charges move into what is called the 0V node, there are actually some very small voltage fluctuations that take place as the charges try to move away from each other. The charges very quickly spread out "evenly" along the negative wire and the negative node of the battery, which results in a 0V difference and no further current flow. \$\endgroup\$ – bitsmack Nov 18 '16 at 23:29

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