# Circuit Questions -- Electrons flowing in wires and a resistor

1. Jul 22, 2015

### r0306

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have selected what I thought were the right choices in the original question. Resistors reduce current so electron flow should have been greater on the right. Also, the electrons flow opposite to current so potential is higher on the left. Potential energy is lost over resistors as heat so electrons have a higher potential energy when they start off but kinetic energy should not increase since energy was converted to heat instead.

2. Jul 22, 2015

### Dr. Courtney

Equations?

3. Jul 22, 2015

### SammyS

Staff Emeritus

Not all of those answers are correct.

4. Jul 22, 2015

### r0306

Yes. That was what I put but my answers were incorrect.

5. Jul 22, 2015

### SammyS

Staff Emeritus
Sorry. I misread the Original Post.

If more electrons flow in the right hand wire than in the left hand wire, in what location do the excess electrons accumulate?

6. Jul 22, 2015

### r0306

Wouldn't they accumulate in the right hand wire then?

7. Jul 22, 2015

### r0306

I still don't get what this entails. Does this mean that current will be less in the resistor?

8. Jul 22, 2015

### Nathanael

It's not that resistors reduce current in a circuit, but if you have two similar circuits and you add a resistor to one of them, then the current will be less than in the other circuit.

If electrons began accumulating somewhere, what do you think might happen?

9. Jul 22, 2015

### r0306

This would cause a greater potential difference between that point and the end of the circuit.

10. Jul 22, 2015

### Nathanael

I was thinking more on a small scale... if you have charges accumulating wouldn't you expect them to repel each other? This would cause the charges to spread back out if they began to accumulate.

11. Jul 22, 2015

### r0306

I'm having trouble putting this in perspective. If the charges spread back out, wouldn't they just push backwards towards the x direction?

12. Jul 22, 2015

### Nathanael

Which direction they push depends on which points you are considering. The main idea is that the charges resist accumulation. If they begin to accumulate, they will repel until the charge density is uniform again. I'm sure there are some technical points which are beyond me, but I'm just giving you a hand-wavey explanation of why the current tends to be uniform throughout a circuit.

If you meant that the charges will repel and then began accumulating in the opposite direction, well then the electric force would began opposing that accumulation. So which ever way it accumulates, the electric force tries to set it even again. Couple this with the fact that a non-uniform current implies a non-uniform charge density and the logical result is that current tends to be uniform throughout a circuit.

13. Jul 22, 2015

### r0306

I see what you mean now! Since current through a circuit is uniform, the number of electrons per second should not change either. Thank you!