I understand that Kirchhoff's current law says that the current, $I$, is constant throughout a resistor, i.e. there is no build up of charge in a resistor. All charge going in to the resistor is the same as all charge coming out. In other words, Coulombs/sec going in = Coulombs/sec coming out.

However, suppose we have an ideal wire, i.e. a wire with no resistance (or a physical wire with very little resistance; but lets use an ideal one), if there is a battery providing a voltage, the current through the wire is infinite. Once we get to the resistor the current is a finite amount $I = dq/dt = V/R,$ in other words, the current has decreased. So, from what I understand there should be a little charge pileup at the entrance to the resistor, and in fact, this is what gives the resistor voltage to drive the current through. Is this true? If not, please explain.

  • $\begingroup$ Why would there be an infinite current through the wire before the resistor? $\endgroup$ – ACuriousMind Nov 2 '14 at 20:51
  • $\begingroup$ @ACuriousMind I am assuming an ideal wire meaning a wire with no resistance. If resistance is zero, and we have a battery providing a voltage, the current should be infinite. $\endgroup$ – user35687 Nov 2 '14 at 20:54
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    $\begingroup$ Only if the resistance of the entire circuit were zero, which is not the case. $\endgroup$ – ACuriousMind Nov 2 '14 at 20:58
  • $\begingroup$ Well, then what is the current through the wire itself. It can't be equal to that of the resistor because the drift velocity of the resistor is smaller. Why is this different from the case of resistance of an entire circuit being zero? $\endgroup$ – user35687 Nov 2 '14 at 21:12
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    $\begingroup$ @user35687, you're not thinking about this clearly. One doesn't add a resistor at a point - one inserts a resistor into the circuit in series with the connecting wires. But, by KCL (Kirchhoff's current law), the current through series connected circuit elements is identical. This, it is impossible (in steady state) for the current through the wire to be different from the current through the resistor. And, the current through the resistor is given by Ohm's law. If ideal wires are used to connect the resistor to the battery, the battery voltage and resistor voltage are identical. $\endgroup$ – Alfred Centauri Nov 5 '14 at 22:54

The voltage that drives the current is different in different parts of the circuit. The sum of the voltages driving current through each part of the circuit, wire and resistor must be equal to the voltage applied with a power source or battery.

In the case of the circuit with the resistor and wire then nearly all the voltage will be pushing electrons through the resistor and only a tiny bit will be pushing the electrons through the wire. If the resistance of the wire is zero then there would be no voltage pushing electrons through the wire - they would just flow on their own (as electrons flow in a superconductor).

Charge does sometimes pile up, but we call that capacitance, as I am sure you know. I mention it here because there can be 'stray' unintended capacitance (and also inductance) which people need to worry about when they use really high frequency AC circuits in the high RF or microwave region.

So it is possible that charge would build up in front of a resistor, but this would be due to stray capacitance mostly but in answer to your question....

Final point, as you know the voltage drop is over the resistor so the potential is different at each end of the resistor. It will be that the surface of the wire on different sides of the resistor will have a slightly different charge density and thus slightly different potential, thus there will be a slight difference in charge density each side of the resistor. I think that this pile up of charge would be very tiny though, but you are correct about this.

To estimate this final effect one could make a guess at the capacitance of the wire and then use $Q=CV$ to determine the difference in charge between the two wires. I would make a guess that we are talking about C of less than $10^{-12}~F$ - if say $10^{-15}~F$ and $10 V$ the pile up would be about $10^{-14}~Coulombs$ or about $100,000$ electrons.

** Final edit **

the capacitance of a wire depends on its length, radius and distance to nearest 'earth' the value chosen above may be on the small side, but I expect the capacitance would be less than $10^{-12}~F$. See here to calculate wire capacitance.

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    $\begingroup$ On a molecular level, don't you need charge build up to have a voltage? "In the case of the circuit with the resistor and wire then nearly all the voltage will be pushing electrons through the resistor" what's causing the voltage? I understand the battery is, but how does the effect of the battery get to the resistor. Is it not by small charge build up? $\endgroup$ – user35687 Nov 2 '14 at 21:20
  • $\begingroup$ @user35687 YES you are correct - I just edited as you put in comment - This is very interesting insight into the physics of what is happening, but I expect a very small effect. $\endgroup$ – tom Nov 2 '14 at 21:21
  • $\begingroup$ Thank you. Just to clarify is it that difference in charge density that directly causes the potential difference over the resistor? $\endgroup$ – user35687 Nov 2 '14 at 21:23
  • $\begingroup$ @user35687 - I would say that the battery or power source causes a slight imbalance of charge density to change the potential and that this then leads to the potential over the resistor and current being driven through the resistor and wires. $\endgroup$ – tom Nov 2 '14 at 21:28

Crudely, electrons repel each other and even out the charge. While the influence of the electrons travels at a good fraction of the speed of light the electrons themselves do not move much. From this link

" The electricity that is conducted through copper wires in your home consists of moving electrons. The protons and neutrons of the copper atoms do not move. The actual progression of the individual electrons in a given direction through the wire is quite slow. The electrons have to work their way through the billions of atoms in the wire and this takes considerable time. In the case of a 12 gauge copper wire carrying 10 amperes of current (typical of home wiring), the individual electrons only move about 0.02 cm per sec or 1.2 inches per minute (in science this is called the drift velocity of the electrons.). "

  • $\begingroup$ The wire in the question is an ideal wire, not a "12 gauge copper wire" $\endgroup$ – user35687 Nov 2 '14 at 21:14
  • $\begingroup$ Even if we assume it is a superconductor, the actual charge carriers move at a vastly slower rate than the electric field. Unless we are into talking about a hollow vacuum tube stuffed full of electrons. In which case they are still going to move incredibly slowly compared to the field $\endgroup$ – user56903 Nov 2 '14 at 21:40
  • $\begingroup$ Here is another effect: en.wikipedia.org/wiki/Space_charge $\endgroup$ – user56903 Nov 2 '14 at 21:41

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