# Misunderstandings about current and voltage

1. Jul 23, 2015

### himmellaeufer

I have a hard time understanding the concept of electric current and voltage in a circuit. Most of all I do not understand the physics behind the "sudden" voltage drop due to a resistor (although I do know Kirchhoffs and Ohms law).
I think I may have the same problem as discussed here:https://www.physicsforums.com/threads/why-does-an-element-in-a-circuit-cause-a-voltage-drop.823346/
We are probably doing the same thinking mistake. What I don't get is the following example.

Lets say we have a + and - pole (i.e. a simple battery) connected by a wire. Now of course, due to the difference in electric potential of the two poles there will be an electric current. According to
E=-∇φ
this is due to the electric field creating a force on the electrons F=qE (which doesn't yield in infinte drift velocity due to scattering - in the simple Drude modell).
If we consider a uniform E-field and a one-dimensional problem, we can also say that φ(x)=-Ex+C. According to this, there would be a linear voltage drop with the spatial coordinate x, no? So even if there was a load there must not be a jump in the electric potential because this is clearly a spatial property.
In this example the E-field should not be altered in x direction due to the flowing current (the E-field components in x-direction of the travelling electrons will cancel each other out and the other don't affect the travel in x direction). I also don't get the analogy of water falling on a wheel - here the potential energy changes due to the distance to earth's surface. The wheel does not change anything about that. The water just lost some kinetic energy, but it will accelerate again (as would the electron due to F=qE).
Is my understanding of the E-field and the electrostatic potential in a circuit wrong?
Right now I am thinking that if you have a battery and connect the two poles with a wire, the E-field somehow travels through the wire (with c) and the current flows due to φ(x)=-Ex+C. This would also mean that the electric field in the wire would get smaller with longer distances and the same voltage, doesn't it?

2. Jul 23, 2015

### vanhees71

Let's consider the DC case and only the stationary state. Then you have a constant current density $\vec{j}=\sigma \vec{E}$ along the wire (directed in tangential direction). The potential is $\phi=-E s$ along the wire, where $s$ is the arc length of the wire measured from an arbitrary point on the wire. The voltage difference is given by the battery.

3. Jul 23, 2015

### himmellaeufer

Yes, this is all clear to me and I agree with it. However I don't see how this adresses my problem.
Isn't this what I have mentioned (in my case however one dimensional, so no arc)? According to this formula the potential would decrease linearly the farther away your arbitray point from the reference point is, i.e. the longer you choose your arc.

Edit: As it is my understanding from the posts of user DaleSpam in this ongoing discussion, I think he states the same that I have said here. The electric potential energy of the electron (only differs by the factor -e from the electric potential) in a circuit only differs due to the location of the electron in the circuit (same thing with his stone analogy). Why would a resistor change anything about that fact (resulting in a location dependent voltage drop)?

Last edited: Jul 23, 2015
4. Jul 24, 2015

### vanhees71

No, the voltage difference is the same (for an ideal voltage source), no matter how long your wire is. The longer the wire, the higher its resistance and the lower your current density and the lower the electric field according to Ohm's Law. The resistance is proportional to the length of the wire, and this makes everything consistent.

5. Jul 24, 2015

### himmellaeufer

Again, this is all clear to me (I hope I'm not getting on your nerves too much).
I think I have never stated something else.
I agree, I have said the same at the end of my first post (see my very last sentence).

What I don't get is the following:
Electrostatics clearly states that that the decrease in potential in an electric field is linear. And the potential in the electric field is only a function of position (User DaleSpam clearly states the same in my referred thread at the edit of my last post). All according to $\phi(x)=-Ex+\phi_{0}$. However according to Kirchhoff's law this is not true anymore, because if you have a load/resistor the voltage already drops (the electric potential decreases to $\phi_{0}$ ) right after the resistor and this independently of where the resistor is. This doesn't make any sense to me, because as I understand it, the potential in an electric field is well defined by its spatial coordinate. Again, why would the resistor change anything about that?

Hmmm I think I cannot explain my confusion any clearer than that:(

6. Jul 24, 2015

### jbriggs444

For any given configuration of circuit elements, the field defines a potential as a pure function of spatial position. But if you change the circuit elements -- for instance, move the wires and the resistor around, you now have a different field. The potential is still a pure function of spatial position. But it is not the same function.

7. Jul 24, 2015

### himmellaeufer

Ok, so for one resistor I would get a step function (if the conductor is ideal), the step being at the resistor. If we take a look at the E-field vectors (for simplicity still one dimensional), this would mean, that from the resistor to the minus pole there would have to be an E-field pointing from the minus pole to the resistor. Such that the external E-field (from the battery) and the E-field between resistor and minus pole cancel each other out.

Why is that, I mean where would that "deelectrifing" E-field come from?

8. Jul 24, 2015

### vanhees71

It's not electrostatics but the stationary case, because you have a current. To get the details, you have to solve the appropriate boundary-value problem for the stationary Maxwell equations.

Qualitatively the solution should be as follows: If you have a wire consisting of different materials, you still have a constant current density along the wire, but the electric field becomes discountinuous according to $\vec{E}=\sigma \vec{j}$, because $\sigma$ changes, where the material changes.

9. Jul 24, 2015

### himmellaeufer

Thanks, I think I can see the flaw in my argumentation since I am probably arguing with electrostatics. So I cannot just simply say, that every x-component of the moving charges will cancel each other out, s.t. only the external E-field of the battery preserves. Does my thinking not work because moving charges create electromagnetic waves thus "complicating" the problem by having the need to solve Maxwell equations? So the E-field cannot be continuous as I have argued?

But what if the material stays the same (2 wires connected of the same material)? Due to Kirchhoff $\vec{j}$ must not change and $\sigma$ shouldn't change because it is the same material. Therefore $\vec{E}$ would have to be continuous again (resistor between the wires or not). Please don't hit me, I'm really confused.

10. Jul 24, 2015

### nasu

The electric field is continuous. The sharp changes are just idealizations, if they are shown as such. The potential varies continuously both along the wires with small resistance and the resistors with high resistance. The resistors are not point objects. The potential varies from one end to the other of them. This produces the electric field that drives the electrons through the resistors. So the potential drop across a resistor is not a discontinuity but a gradual drop. In ideal circuits, the potential is considered constant along the connection wires and dropping along the resistors.
Your confusion may come from schematic representation of potential. It may be useful to show the link.

11. Jul 24, 2015

### himmellaeufer

Ok, but not constant as it is defined by the "external" E-field $E=\frac{V}{x}$ where V is the voltage of the battery and x is the length of the wire?

Because this would otherwise contradict $\vec{E}=\sigma \vec{j}$. (resistor -> small $\sigma$ -> small $|\vec{E}|$.

But if I take a look at the current after the resistor, as I have explained the E-field would have to be the same as in front of the resistor (because Kirchoff and the same resistance).
So the E-field decreases gradually into the resistor, where it is constant at a lower value and then increases again until it has the same value as it had in front of the resistor? Like a function that looks like a bathtub

Last edited: Jul 24, 2015
12. Jul 24, 2015

### nasu

No, the potential decreases gradually in the resistor. The field is constant, if the resistor is homogeneous, with same resistivity and cross-section everywhere.

13. Jul 24, 2015

### himmellaeufer

Yes, I agree. I was talking about the E-field decreasing at the wire-resistor interface, not the potential in the resistor itself.
I'll try to list my thoughts so you can tell which one is false, if there is one that's not correct. I'm considering the simplest circuit like this (one dimensional, ideal wire, battery voltage V, lenght d):
plus-Pole ----(wire)-----resistor-----(wire)----- minus-Pole

• the battery is the source of the external E-field, which is present and constant in the whole circuit. This value is a constant given by $E=\frac{V}{d}$
• The E-field travels with light speed c from the plus to the minus pole and is always tangential to the wire.
• The electrostatic potential of this E-field is $\phi(x)=-Ex+\phi_{0}$.
• however I'm not implying that it is the actual E-field in the circuit, because it cannot be since that would mean that the potential drop (voltage) is only dependent on x (a resistor in the circuit wouldn't change that). Moreover a current is not something electrostatic.
• The E-field in the wire left of the resistor is the same as it is in the wire right of the resistor : $E=\sigma j$. This follows from Kirchhoff's laws ($I$ is constant) and from the fact that $\sigma$ is the same for the wire left and right of the resistor.
• Due to $E=\sigma j$ the E-field in the resistor, where $\sigma$ is smaller, gets smaller/weaker.
• This gives an E-field profile $E_{wire},E_{resistor},E_{wire}$ (like a bathtub...)
• I am probably stupid

Is this correct so far?

Edit: corrected my first point. This probably confirms the last point...

Last edited: Jul 24, 2015
14. Jul 24, 2015

### gleem

Lets start here. this is incorrect E = dV/dx

15. Jul 24, 2015

### nasu

Starting with the first item, "constant" usually means "in time". So with this meaning, yes, the field is constant in time, in a DC circuit, after the transient regime dies out.
But the field does not have to be uniform, to have the same values in all segments of the circuit. Usually it is not uniform. It changes from one resistor to another and so on.
So you cannot use a linear (E=Vx) relation between field and potential globally, for the whole circuit, as E is not a constant.

You also got Ohm's law wrong.
It is j=σE and not as you wrote it.
If you have resistors with the same cross-section, the larger resistance one will have the higher field. But it does not have to be this way, the current density does not have to be uniform along the circuit. So the field may change in a quite complex way from point to point, depending on the actual geometry and resistivity of various areas.
It is not an efficient way to do circuit analysis based on it.
I still do not understand what you are trying to understand. You did not show the reference that started your confusion.

I don't see why you care about propagation of field at this point. I thought you are thinking about a DC, steady situation.

16. Jul 25, 2015

### himmellaeufer

Yes, thank you this helps!

stationary case -> $\frac{\delta \vec{E}}{\delta t}=0$
$\frac{\delta \vec{E}}{\delta \vec{r}} \neq 0$
I think this exactly adresses my confusion and solves my problem, thanks! Sorry for not being able to word my problem better

Heh, while I should know this I blame this guy
(no I don't, take it easy)

Sure, but it is nice to know.

Anyway, thanks again!

PS: is there a way to quote LaTex-Syntax without having to rewrite it? Simply embedding it in numbersigns doesn't work somehow

17. Jul 25, 2015

### vanhees71

Argh ;-)). That's of course a typo. It must read $\vec{E}=\sigma \vec{j}$.

18. Jul 25, 2015

### nasu

Only if by σ you mean resistivity. More often σ is used for conductivity ( scalar or tensor). And the relationship is writen as
j=σE.

19. Jul 25, 2015

### Staff: Mentor

Be careful here. Electrostatics states that the decrease in potential in a uniform electric field is linear. In most circuits the field is not uniform.

20. Jul 25, 2015