# Current and Resistance

1. Sep 23, 2015

### plain stupid

I've read all the posts about this, but I still can't get this one thing: Why does current decrease if a resistor is put into the circuit?

1) there's a circuit without an 'official' resistor, only a wire that has small resistance.
2) there's a circuit with an 'official' resistor, and a wire that has small resistance.

Both have a source.

Now, in case 1), electrons simply bump into some atoms while accelerated by Electric field, then pick up speed, again. In case 2), E-field is still the same, there aren't more atoms for electrons to bump into before or after the resistor, only in it, and I'm not talking about electrons immediately leaving/entering the resistor, but the ones that are, say, 2 meters from a resistor (since they're pretty slow).

The analogy the book gives is a pipe and incompressible flow; I can't accept this, as electrons are pretty small, don't practically repel each other in order to flow, and don't bump into each other (as to make any significant advances/pushes). The only explanation I've heard is that "the mean free path" is smaller, but I don't again see how that affects electrons that aren't near the resistor.

I can only imagine them being slowed down exactly at the resistor's entrance, and then the rate will be lower, as they leave the resistor. That however implies making a full circle, because otherwise the electrons way after the resistor ought to be faster and denser (just like in case 1)).

[I know about drift velocity, Voltage I believe is simply some newly made measure that's basically work, and is implemented by E-field, and I know that electrons do work while passing through a resistor, thus reduce their Potential Energy and thus Voltage.]

Just one more example so you can tell me where's a flaw in my assumptions:

Let's say we have a circuit like 1) with 10 electrons. They pass through some point A at the rate 10e/2s = 5 electrons per sec. Now let's make a circuit like 2) but put 5 electrons on both sides of the resistor. Now the group that's after the resistor must travel at same rate of 5e per sec, nothing is obstructing their flow (except the wire's resistance that I made obvious).

2. Sep 23, 2015

### BvU

Hello PS, welcome to PF !

Nope. They just wiggle around. You push one in from the source, then another has to pop out of the circuit and back into the source on the other side. No accelerating or slowing down. Pipe and incompressibe flow is a fair analogy (up to a point, but here it works ok). Or marbles in a rainpipe.

The difference between your circuit 1 and 2 is only that in case 2 you have to push harder (= need more voltage) to get the same current. Conversely, if you push with the same voltage, you get less current.

In ordinary matter (a copper wire, for instance) you have about 9 g/cm3 x 29 electrons/atom x NAvogadro atoms/Mole / 63.5 gram/Mole $\approx$ 2.5 x 1024 electrons/cm3. Extra electrons, even if only one in a million, generate enormous voltages. So they behave decently and carefully distribute themselves over e.g. the wire: a nice smooth voltage drop going from source cathode (-) to source anode (+)

3. Sep 23, 2015

### plain stupid

Thanks.

As far as I know, there's a Force, namely Electric Force (from the battery's Electric field) acting on electrons making them accelerate and bump into several atoms' nuclei, as the wire has some resistance. When they do hit the atoms (=nuclei) they ~stop, then they pick up speed again, averaging to some ~constant velocity. If this isn't so, I don't even know how they pick up speed in the first place.

I don't like pipe analogies as water is incompressible while electrons aren't, which is, I think, a crucial assumption to begin my question with.
How would electrons "know", it's the battery that, for each received electron gives one out — it's not the electrons' worry (?).

4. Sep 23, 2015

### Drakkith

Staff Emeritus
The electrons do in fact repel each other. That's how they 'know' what's going on in the circuit. Electrons may be small, but there are a lot of them, and since they are electrically charged, they don't have to 'collide' with each other directly like two baseballs colliding in order to interact. Just being nearby is sufficient.

The reality of the situation is much, much more complicated. The electrons in a conductor at room temperature with no applied voltage move about randomly in many different directions by virtue of having thermal energy (ignoring quantum effects). They are constantly interacting with both the other electrons in the conductor and with the ionic lattice, colliding, scattering, having tea... etc. When you apply a voltage, this random motion is slightly altered so that there is a net motion in one direction in this random sea of moving electrons. This is the current you see in a circuit.

5. Sep 23, 2015

### plain stupid

Thank you very much. So, let me get this straight (please correct me): when there's a resistor in the circuit, the current starts flowing, but it flows slower in the resistor segment, this mass of electrons slows down, they then repel electrons back and so forth. That explains the part before the resistor; what about the wire after the resistor — do they slow down because of lack of some push from the electrons in the back? Etc.

Why I think it's not plausible is because if there was a heap of electrons creating at the beginning of the resistor, the small E-field originating from this newly-created heap would push back some electrons before them, and push electrons already in the resistor further, therefore making them go faster. (Potentially negating the slowing-down effect of those in the resistor.)
... So I'm back at the beginning.

6. Sep 23, 2015

### Drakkith

Staff Emeritus
Pretty much. If the electrons after the resistor tried to flow out of the circuit faster (faster meaning more charges per second, not velocity or speed) than they could flow through the resistor, then you'd end up with an imbalance of charges, which would reduce the electric field until the current flows were equal.

Assuming this does happen (I don't know if it does) then it only matters when current initially starts to flow. Very quickly the circuit establishes an equilibrium situation where the current into and out of any point in the circuit is equal.

7. Sep 23, 2015

### plain stupid

That makes sense. Thank you very much. The thing is, I've actually finished the electric engineering intro and electronics course but this has bugged me since like the 8th grade...
Yes indeed. Thank you for everything. :)

8. Sep 23, 2015

### willem2

There has to be a heap of electrons at the beginning of the resistor (or more accurately at the side connected to the negative side of the battery).
There also has to be a shortage of electrons on the side of the resistor connected to the positive side of the battery.

The size of these excess charges regulates itself. If there isn't enough charge on both sides of the resistor, the electric field in the resistor would be too small, less current would go through the resistor than go through the wire, but this would make the charges on both side of the resistor increase.
If the charges would be too big, there would be more current through the resistor than through the wires and this would make the charges decrease.
An equilibrium sets in very quickly, because the amount of charge needed to produce the required electric field in the resistor is very small.

9. Sep 23, 2015

### Staff: Mentor

I am not sure if the question is more about the microscopic origins of Ohm's law or about how Kirchoff's current law works in a circuit.

10. Sep 24, 2015

### plain stupid

I don't see how classification is (ever) relevant?

11. Sep 24, 2015

### Staff: Mentor

Clarifying a question is usually relevant. I don't know if you are asking for a mechanistic explanation why $J=\sigma E$ or if you don't understand why $\Sigma I =0$. They are substantially different topics.

12. Sep 25, 2015

### plain stupid

Alright. I was asking about current being the same throughout the whole circuit: that explained using a microscopic approach. However I didn't actually realize it's the application of Kirchoff's law, but a consequence of Ohm's law.

Why it confused me is because I can't believe current is incompressible, therefore I can't understand why the current in the whole circuit (before and after the resistor) would be lower than without a resistor (assuming some minimal wire resistance in both cases). Also, I think the electrons don't repel other electrons before and after them, but that there's Voltage that acts on each electron independently, thus the incompressibility thing (unlike water). Apparently, that's not the case, and the system has some sort of self-regulation.

13. Sep 26, 2015

### Staff: Mentor

OK, so to me that seems like you are more interested in why $\Sigma I =0$ (Kirchoff's Current Law, or KCL).

Here is what I think that you are envisioning. Let's suppose that we have a battery with a wire connected to the + terminal and another wire connected to the - terminal. You understand that when those wires are directly connected to each other there is a large current going through the wires. You also understand that when there is a resistor connected between the wires there is a small current through the resistor. What you don't understand (if I am getting it) is why when the resistor is connected there is a small current through the wire. Your gut feeling is that there should still be a large current in each wire, and a small current in the resistor. This would mean that $\Sigma I \ne 0$ at the boundary between the wire and the resistor, but since you view current as compressible that should be OK.

So, if I am understanding you correctly, let's just do a little calculation to see what would happen in such a circuit. Let's suppose that we have a $10 V$ battery, and a $10 k\Omega$ resistor, so the current is $1 mA$. Let's further say, that the wires themselves have a $1 \Omega$ resistance so that in the absence of the resistor the current is $1 A$. So then the compressible current idea is that the wires still have their $1 A$ current and the resistor has its $1 mA$ current. Because of charge conservation, this means that there is a buildup of $999 mC/s$ of charge on the + end of the resistor and a buildup of $-999 mC/s$ of charge on the - end of the resistor.

If we assume that the resistor is $1 cm$ in length, then we can use Coulomb's law to calculate the resulting forces. I won't walk through the details, after $1 s$ I get a force of $9E13N$, or approximately 9 billion metric tons. And that force increases another 9 billion metric tons each second. Such an enormous force would certainly crush the resistor after a millisecond or less. Also, the voltage across the resistor would be about $900 GV$ in the opposite direction of the battery.

So, essentially that is why we cannot have the situation where $\Sigma I \ne 0$. If KCL were violated the forces involved would be truly staggering. And those forces would work to oppose the violation of KCL.

14. Sep 26, 2015

### plain stupid

Yes indeed. Thank you for "the numbers", they truly show the strength of the "electron repulsion is real" argument.

15. Sep 26, 2015

### sophiecentaur

In Electrical Current flow in wires, we are dealing with something more analogous to an unstretchable bicycle chain or a hydraulic ram system than a set of marbles falling through a loose arrangement of nails on a ramp.
However, for a current flowing through a rarified gas, the idea of electrons speeding up and then being brought to a halt by collisions with heavy ions does apply. This is because the spaces between electrons are massive in comparison with a solid. You can, indeed, get bunching and rarefactions of an electron beam in a gas discharge tube, which can produce bands of light (striations). There is a temptation to apply this idea to what goes on in solids - because of the way that things are sometimes explained at School.