# Voltage drop in a Wire

1. Jun 29, 2013

### qitara

Hi

Why is there a voltage drop in a wire when there is a high current flowing ?. Is this due to the electrons colliding with the lattice which causes the kinetic energy that the electrons have gained trough the voltage to be transfer to heat energy ?

2. Jun 29, 2013

### Staff: Mentor

The collisions need energy (and heat the wire), so you need some force on them (=> electric field => voltage difference) to maintain a current.

3. Jun 29, 2013

### DrZoidberg

The current doesn't cause the voltage drop. It's the other way around.

4. Jun 30, 2013

### CWatters

Good point, it's easy to forget that.

Perhaps better to ask why a conductor has resistance.

5. Jun 30, 2013

### Staff: Mentor

Both are connected via V=RI. I don't think it is meaningful to talk about "x causes y".

I can set a power supply to 1mA, and measure the voltage afterwards. Did voltage cause a current, or vice versa? Or did I just get both at the same time?

6. Jun 30, 2013

### sophiecentaur

Imo, the notion of electrons 'bashing into atoms' on their way through the metal is a bit naive. That model would not account for the increase in resistance with temperature as the cross section of an atom would not be likely to increase just because it's moving about a bit more. Electrons interact with the whole lattice.

7. Jun 30, 2013

### sophiecentaur

I think that, as voltage is Energy per Coulomb, it is justifiable to say that the Energy source is the 'cause' of what is happening. There are always plenty of mobileelectrons around in a metal but net current will only flow when there is a Potential difference.

8. Jul 1, 2013

### cabraham

But how is a potential difference realized? By separating charges, i.e. an E field. So in order to move e-, an E field is needed. To get an E field, other charges must be moved first. So the E field which forces e- to move, was generated by moving other e-. It's a vicious circle for sure.

A battery is a good illustration. Outside the battery charges move in accordance with E force. But inside it is the opposite. How does current exist inside the battery. Positive ions move towards the positive terminal which is going against the E field. An E field cannot sustain a current unless another current is steadily replenishing E field. Likewise negative ions move against the E field towards negative terminal.

This is a classic scenario where it is impossible to say that V causes I nor I causes V. Energy is converted from chemical to electrical, or mechanical, optical, etc. Regarding resistance, the classical model is that of charges colliding with lattice structure. Electrons in the conduction band which transition to valence band are going to a lower energy state. Conservation of energy results in photonic emission in the IR region, and the wire feels warm.

The voltage drop across the resistance can be said to be a result of the collisions, i.e. current. Heaviside transmission line theory supports this model. A switxh closes, connecting a generator w/ 75 ohm internal R to a 75 ohm t-line. At the other end of the t-line is a mismatched load resistor of 100 ohm. A reflection occurs. The entire current cannot be absorbed in the 100 ohm load due to Ohm's law. The collisions with lattice ions generates a local E field opposing the original one. The next wave of e- encounter this barrier and current is reduced.

The current gives rise to the voltage drop, then the voltage drop influences the current, reducing it. The 2 are truly interactive. Either one can affect the other. Neither is the general cause nor effect.

They are inclusive.

Last edited: Jul 1, 2013
9. Jul 1, 2013

### sophiecentaur

That's a point of view but the Volts are still there, on the battery, when there's no current passing. It is true, though, that there is no energy loss in a load if no current is passing so, there is no Voltage drop. I think that's about fifteen all (?). Topical reference and back to the TV.

10. Jul 1, 2013

### cabraham

But the "volts are still there" because of a previous current. That is my point. If one asks the question "which drives other?", my response would be along the lines of your remark "about 15 all". It is impossible using what we currently know to establish 1 as the cause of the other universally. There are applications where one determines the other if we disregard past history.

An unloaded battery, or unloaded charged cap, has a V, then if an R is placed across it, and I exists. This I is given by V/R. Here one can say that I was determined by V. But we ignored past history, which can often be done since it has no direct bearing on the problem.

But an energized inductor is shorted with superconducting winding and terminals. An R is placed across the inductor as well as a shunt SC switch. When opened the current will continue in the R. A voltage develops based on I*R. I already existed, V was determined by I and R.

Of course, we ignored past history. In order to energize the inductor, a voltage per V = L*dI/dt, was employed. So if we consider negative time/past history, as well as forward looking analysis, it's not possible to say which is the driver, which is driven since it can vary based on conditions.

Just thought this was worth mentioning.

11. Jul 1, 2013

### sophiecentaur

Chicken and egg, in principle. But most of us buy a battery (a Voltage source) and attach a load. In that instance, we start off with a load of energy in our Duracell, ready to drive current through a load when we want it. Who knows but that, one day we will have a super-cooled magnetic store of energy in our cars and your scenario will just be commonplace and totally unremarkable. Super Caps are turning people on these days after a hundred years of tiddlers only being available.

12. Jul 1, 2013

### cabraham

Chicken and egg, in principle, is exactly my point. I agree that we buy a battery, or plug into a wall outlet, and that source is constant voltage. Our thinking has adopted a voltage source plus series resistance model, i.e. "Thevenin" view of the world.

You are correct that a "Norton" source may be in the future. Nuclear cells, (search using "nucell") provide constant current source, and AC at that instead of conventional DC cells. But being nuclear, we may never see them in commercial use due to safety issues.

I think we agree to the letter. It is indeed chickens and eggs.

13. Jul 1, 2013

### technician

'Naive'......'chicken and egg'.......'that's a point of view'.....
This is supposed to be a PHYSICS forum !!!!!! With PHYSICS explanations to suit ALL levels....even the 'naive'....
What is the answer to the original post?

14. Jul 1, 2013

### sophiecentaur

What's your suggestion then? Was there something wrong with the second post?
Just because the sums give the right answer it doesn't mean that there's only one approach to a problem.

15. Jul 1, 2013

### technician

Second post is fine...no mention of chickens or eggs or naivety

16. Jul 1, 2013

### sophiecentaur

And have you any comments to add, involving the Physics content or an opinion about 'cause and effect'? Perhaps you are implying it's not worth discussing? That is surely a matter of opinion.

17. Jul 1, 2013

### cabraham

Not really. Physics observes that light displays wavelike nature, and also particle-like nature. So which is it? Wave or particle? Here is the answer physics produced - light is a wave-icle? Is that naive? Did they duck the issue? Tell me, why must the right answer be 1 of 2, why not no. 3?

Post #2 says that charges move because of force due to E field, thus E forces J. This view cannot withstand scrutiny because of the nature of E fields. E fields cannot sustain a current at all, they can only provide a temporary transfer of energy while giving up their energy. A battery connected to a lamp is our example.

The current I is the same inside and outside the battery, since the currents are in series. Inside the battery ions are moving AGAINST the E field/force. These internal ions cannot be acting under the influence of E. E is created by the internal battery current I. Outside the battery, charges move with the E field. In moving charges E field gives up energy. If not replenished, E would lose all its energy and I would cease.

What replenishes the E field? It is I, the internal current, generated by the chemical reaction reduction/oxidation (redox) inside battery. Redox produces I, producing E, which then tranmits energy to free charges outside battery moving charges through lamp. In doing so E keeps losing energy, but gets replenished by redox creating I which restores E field/energy.

The line integral of E over a path is the voltage along that path. So V & I are inter-related and neither is universally the "cause" of the other. Like the "wave-icle" description of light, that is all we have. Those with limited knowledge always place V at the top of the pecking order then treat I as the result of V across R. This model is incomplete and does not account for action inside battery, nor does it account for how E field energy gets replenished.

Post no. 2 is grossly lacking.

18. Jul 1, 2013

### Crazymechanic

Well I could agree that on many occasions here on PF someone derails the thread but this is not the case , as the latter posts where also good and accurate , and a good addition to the OP question , so I think we are quite fine with this thread if I might add.

19. Jul 1, 2013

### Staff: Mentor

$$\sigma E= j$$
Does E cause j, or j cause E? I don't think those are meaningful questions.

Consider an example from classical mechanics:
$$p=mv$$
Would you say that momentum "causes" velocity, or that velocity "causes" momentum?

Indeed, you have to take electrochemical potentials into account in a battery.

There is nothing to "replenish". Electric fields are not losing their energy just because they are there. The electric field strength can (but does not have to) reduce if charges move somewhere, as the charge distribution changes.
To be honest, I don't think the following posts made it better.

20. Jul 1, 2013

### cabraham

E fields lose energy when a charge gains energy via the E force. Conservation is immutable. When external electrons are driven towards the positive battery terminal, the electrons combine with positive charge and weaken the E field and energy. Likewise for the negative terminal receiving positive charge. If not replenished E field diminishes as does current.

"Electrochemical potentials" are due to what? Please explain how EC potentials affect this example. Thanks.

Claude