# I Why is voltage constant in a wire?

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1. Oct 9, 2016

### MagicPandaLol

Hey! I have a question and a drew up a diagram to help out. I drew up a circuit that includes just a voltage source and a resistor. I know the voltage is constant in each of the circled portions of the circuit. As in, as you move through the wiring, the voltage doesn't change from one point to the next. Really, the voltage only changes when you move through the load. This much I know, haha.

The source of my confusion is I don't know why this happens. If voltage is supposed to be the driving force that moves the current, how can the electrons move from one point to the next if the voltage isn't changing? After all, there isn't a difference in energy states from one point to the next within the wiring, so what's causing the motion?

This has been what's bothering me. Any and all help is appreciated!

2. Oct 9, 2016

### Pete_L

Connect one lead of a voltmeter to the negative terminal of your voltage source. Now connect the other lead of the voltmeter to first one circled section of wire, and then to the other circled section. Do you get the same voltage reading for each of the two different connections to the circuit?

3. Oct 9, 2016

### Jonathan Scott

Consider the water level in a slow flowing river. It's not exactly level, but near enough. The same applies to the voltage in a good conductor.

4. Oct 9, 2016

### rbelli1

To simplify circuit design we assume that the voltage across a wire is zero. In reality it is a small nonzero value. Unless you are talking about a superconductor.

If you were building your circuit in a classroom setting you would often be using 22guage wire. This is about 0.053 ohms per meter. If you were to have a small lamp that draws 0.1 Amps through say 0.1 meter wire you would get 530 micro-volts. Even if you were using a single cell at 1.5V that is an insignificant amount and we can safely call it zero in most cases.

BoB

5. Oct 9, 2016

### Staff: Mentor

When we say that the voltage is the same at both ends of the wire, that's a very good approximation, not absolutely true. The wire has some very very small resistance (for example, ten centimeters of 14 gauge has a resistance of about .0001 ohms), so by $V=IR$ if there is any current flowing there will be some minuscule voltage drop along the wires.

In many more practical problems we can just neglect the resistance of the wire altogether. It's a good exercise to calculate the voltage at both ends of the wires in your diagram, assuming a 10 volt power supply, a 1000 ohm resistor, and the wires having a resistance of .0001 ohms as in my example above.

Even when we cannot neglect the resistance, we often care only about the current flow through the circuit, and then we can treat the wire as having as having zero resistance while adding the resistance of the wires to the resistor itself - we get the same answer and the calculation is easier.

Last edited: Oct 9, 2016
6. Oct 9, 2016

### MagicPandaLol

Been away from this for awhile, but, if I recall correctly, you'd get the same voltage for each reading with just different signs (one negative and the other positive). Am I close?

I see, so it's just an approximation. There is a change in voltage as the electrons move through the wire, and that change is going to be a function of the wire length among other things.

There is another thing that is still bothering me though. I know that current is constant throughout a circuit, but how can it remain constant if the voltage is dropping? These electrons move with an amount of potential energy that changes as you move throughout the circuit, and yet, they still move at the same speed throughout. How is this possible?

7. Oct 9, 2016

### Pete_L

The OP says, "Been away from this for awhile, but, if I recall correctly, you'd get the same voltage for each reading with just different signs (one negative and the other positive). Am I close?"

No, connecting a 1st lead of the voltmeter to the negative terminal of the voltage source, then the 2nd voltmeter lead to the upper circled wire gives a reading equal to the voltage of the voltage source. Alternately connecting the 2nd lead to the lower circled wire would give you a 0 Volts reading. This is because usually wire resistance is insignificant relative to resistance of the resistor, and essentially all of the work done by the voltage source consists of forcing current through the resistor.

The OP next says, "These electrons move with an amount of potential energy that changes as you move throughout the circuit, and yet, they still move at the same speed throughout. How is this possible?"

The electrons have a constant kinetic energy as they are moving at the same rate anywhere in the circuit. What does change is voltage drop, with the resistor hogging all of it.

-Interesting questions which are not as easy to answer in detail as one might suspect.

Pete

Last edited: Oct 9, 2016
8. Oct 9, 2016

### davenn

Pete_L, PLEASE learn how to quote .... I had a whole response written to correct your comments till I realised half of your comments were actually unquoted comments from some one else's posts, then I had to delete my response

Dave

9. Oct 9, 2016

### davenn

that is dreadfully vague ... if talking about voltage, then it's incorrect, if about current in a series DC circuit then it's correct

Last edited: Oct 9, 2016
10. Oct 9, 2016

### davenn

this is incorrect ... see my response in post #9

11. Oct 9, 2016

### Staff: Mentor

The current can remain constant if the voltage difference between any two points in the circuit is given by $\Delta{V}=IR$ where $R$ is the resistance between those two points (this is just Ohm's law; the $\Delta{V}$ instead of $V$ is to make it clear that it's the voltage difference, not the absolute level, that matters).

Last edited: Oct 9, 2016
12. Oct 9, 2016

### Pete_L

Yes that is absolutely true. My initial thought was, okay, if the entire voltage of the voltage source drops across the resistor, then what is the motive force driving the electrons towards and away from the resistor? Following E=IR, there must be a very small voltage drop across each "feed" and "return" wire to the resistor. Indeed very small voltage drops are there, but they are essential to a functioning circuit.

I thought that this was kind of interesting and hope it is somewhat interesting to others, and it helps and does not confuse the OP.

Pete

13. Oct 9, 2016

### Staff: Mentor

That's right - it's what it's what @rtbell1 said in #4 and what I was trying to say in #5.
For any real circuit built out of real wires that have non-zero resistance, yes. The zero resistance ideal wire is a very useful idealization that when applied thoughtfully works for an enormous number of problems and gives the same answer as the more painful calculation that recognizes that the wires are not really ideal. That's why we teach it, and that's why everyone studying circuits should have it down cold before they move on to the harder problems. But once you've got that baseline, you do have to learn to recognize and properly handle the situations in which the zero resistance wire idealization won't work - the electrical code has an entire chapter devoted to the resistance of wires and why it matters.

14. Oct 10, 2016

### Fervent Freyja

An analogy might be helpful here: https://en.wikipedia.org/wiki/Hydraulic_analogy

Is someone getting frustrated?

15. Oct 10, 2016

### sophiecentaur

It isn't, except in arm waving and sometimes misleading 'explanations'. Voltage Is Not A Force; the Voltage (Potential Difference) across any circuit element (including the connecting wires) tells you the Energy Needed to move charges through it. If they will go through without dissipating energy then the PD is Zero. We approximate the PD along a length of wire to zero but, once the wire is long enough or thin enough (or made of something other than Copper or Silver) there may be a significant amount of energy dissipated in it and we have to include the Voltage Drop.

16. Oct 10, 2016

### MagicPandaLol

Are you saying that the current in a series DC circuit is not exactly constant throughout the circuit but only approximately so?

It is really interesting. I had forgotten that the zero-voltage drop across a wire in a circuit diagram was only an approximation, haha.

So, the kinetic energy remains the same throughout the circuit since the current remains the same. The potential energy is what changes due to heat and work. When neither of those two things are happening between two points, the potential difference is zero. What causes a potential difference though? As in, what causes the electrons at one point to have more potential energy than at another point?

Thank you for all the responses!

17. Oct 10, 2016

### sophiecentaur

What "kinetic Energy" would you be referring to?

18. Oct 10, 2016

### MagicPandaLol

The kinetic energy of the motion of the electrons in the circuit

19. Oct 10, 2016

### davenn

no, I didn't say that

20. Oct 10, 2016

### davenn

a voltage difference (potential difference) between 2 points in a circuit

21. Oct 10, 2016

### MagicPandaLol

Aren't a voltage difference and a potential difference the same thing?

22. Oct 10, 2016

### davenn

yes, that's why I put one of them in brackets .. indicating they were the same thing

Maybe what you are really asking is ... "where does the initial potential difference come from" ?

the answer to that is charge separation .... where electrons are separated from the protons ( the rest of the atom)

in a battery, it is done chemically
in a generator at a power station, it is done using (usually) a stationary coil of wire and a moving magnetic field

you can google both of those methods and come back with any queries

there are some other methods .... you could look up "how does a solar panel work"?

Dave

23. Oct 10, 2016

### sophiecentaur

A Voltage is the informal term for a Potential Difference. It is sometimes the Potential above Earth.
A Voltage Difference would be the difference in the voltages produced by, say, two different batteries. It's not an 'official' term.

24. Oct 10, 2016

### sophiecentaur

Although quite 'popular', that is not a valid model for how energy is transferred along a wire. The average speed of the electrons in a wire (the drift velocity) is a matter of around 1mm per second. The mass of electrons in a wire will be a tiny fraction of the total mass of the wire. That would be 1/1800n where n is the atomic number of the metal (29 for copper). Do the sum for KE with that and you'll see that the electrons that arrive at the end of a wire with virtually no KE at all.
As with a bicycle chain that transfers power, the chain links are going slowly and their KE is negligible but they transfer all the cyclist's power to the wheels.
Edit: The drift velocity of the electrons does change on the way round the circuit. The same number go past all points around the loop per second but (as with water through pipes) for a given number per second, the speed will be slower for a thicker wire (fat pipe and faster for a narrow wire (narrow pipe).

Last edited: Oct 10, 2016
25. Oct 10, 2016

### MagicPandaLol

I see, so the buildup of negative charge at one end and positive charge at another is what leads to the voltage difference. How exactly does this translate to a voltage difference? As in, how does the separation of unlike charges in a battery lead to a voltage difference in a circuit?