Why is voltage constant in a wire?

[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!

 

[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?

 

[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.

 

[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 22gauge 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

 

[Nugatory]

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.

 

[MagicPandaLol]

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?

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?

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 resistence 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.

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?

 

[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

 

[davenn]

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.

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

Dave

 

[davenn]

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.

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

 

[davenn]

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.

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

 

[Nugatory]

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?

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).

 

[Pete_L]

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 matter).

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

 

[Nugatory]

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

That's right - it's what it's what @rtbell1 said in #4 and what I was trying to say in #5.

...but they are essential to a functioning circuit.

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.

 

[Fervent Freyja]

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? View attachment 107191

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

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

Dave

Is someone getting frustrated?

 

[sophiecentaur]

If voltage is supposed to be the driving force

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.

 

[MagicPandaLol]

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

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

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.

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

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.

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!

 

[sophiecentaur]

So, the kinetic energy remains the same throughout the circuit since the current remains the same.

What "kinetic Energy" would you be referring to?

 

[MagicPandaLol]

What "kinetic Energy" would you be referring to?

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

 

[davenn]

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

no, I didn't say that

 

[davenn]

What causes a potential difference though? As in, what causes the electrons at one point to have more potential energy than at another point?

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

 

[MagicPandaLol]

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

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

 

[davenn]

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

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

 

[sophiecentaur]

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

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.

 

[sophiecentaur]

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

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).

 

[MagicPandaLol]

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)

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?

 

[davenn]

As in, how does the separation of unlike charges in a battery lead to a voltage difference in a circuit?

did you do a google search on how a battery works, as I suggested ?

 

[MagicPandaLol]

did you do a google search on how a battery works, as I suggested ?

I did, and I was confused when I typed that. I think I might get it now though after thinking about it a bit longer. We were able to convert charge separation into voltage just through measurement. We took different materials that would undergo a redox reaction, and just measured their voltage with a voltmeter. Haha, am I close?

 

[Mark Harder]

Explanatory models for electronic circuits assume that all conductors connecting the parts of a circuit are perfect, with no resistance, capacitance or inductance. To predict the circuit's real behavior, one adds virtual resistors, capacitors and inductors to account for the non-idealities. Estimating the resistances, inductances and capacitances of these virtual components and applying the appropriate mathematical models (eg. Ohm's law), one can predict the behavior of the circuit. To get a notion of the convenience of using fictitious components, take the case of a battery: An electrochemical voltage source is modeled as having an internal resistance, pictured as a resistor in series with an ideal voltage/current source. In turn, 'internal resistance' is an idealization of more than one factor that opposes the flow of electrons into and out of the battery. Perhaps the most significant of these is the finite speed at which the chemical system in the battery can donate and receive electrons. As a battery ages, the concentrations of reducing and oxidizing agents in its component cells decrease and the reaction between them slows down. ( Same goes for temperature: As we know, batteries don't 'work' when they're too cold.) When you place a voltmeter across a battery's terminals, the current drawn from the battery must pass through the internal resistance of the meter. As long as the battery's internal resistance is negligible compared to the meter resistance, the latter will read the voltage defined by the standard electrochemical potentials of the battery reactions. So even a 'dead' battery will show a voltage when measured by a good voltmeter. Connected to a real circuit, however, and this condition is no longer met - the internal resistance of the 'dead' battery exceeds the resistance of the circuit and very little current will flow. This is why a 'battery tester' meter contains a low-value resistance in parallel with the meter - to simulate a realistic load on the battery.

 

[sophiecentaur]

It's all to do with the different layers of complexity needed to get predictions about a particular situation 'right'. Resistances in circuits is a quantity that can vary from much less than 1Ω to Hundreds of MΩ. It is quite excusable (and very practical) to make approximations in design calculations where such a wide range of a variable is found.
This can be confusing and annoying to people who are unfamiliar with 'electricity' but it's really only the same in mechanical problems where we consider massless strings and pulleys in many situations; the results of calculations can be well within the experimental error of final measurements.

 

[MagicPandaLol]

Really appreciate all the help, everyone. Don't have any questions at the moment, but I'll come back if I do. Thank you for all the clear, concise, and detailed answers to my questions! :)

 

[StandardsGuy]

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.

Really? EMF (electromotive force) is measured in the units Volts. Consider a battery only. It has an EMF which is analogous to pressure in a water pipe. Considering the OP's education level your post may be confusing.

 

[Integrand]

Really? EMF (electromotive force) is measured in the units Volts. Consider a battery only. It has an EMF which is analogous to pressure in a water pipe. Considering the OP's education level your post may be confusing.

EMF is one of those tricky historical terms we have to cope with despite their potential to mislead. I believe the point being made is that despite the name, electromotive force is not to be equated in any way with mechanical force.

 

[sophiecentaur]

EMF is one of those tricky historical terms we have to cope with despite their potential to mislead. I believe the point being made is that despite the name, electromotive force is not to be equated in any way with mechanical force.

Electromotive Force is a term from way back before anyone had a clue about what Electricity is. It is an exception which has to be used with care. "emf" is a term that refers to the Potential available from a voltage source when no current passes. It is unfortunate that it has passed into modern usage because it does nothing but cause confusion - in threads like this one. Voltage is not a force. Energy and Force are two different things. As for 'Pressure' - just how could that be allowed into any serious discussion? As has been regularly stated on PF, analogies are very risky and are seldom received by the uninitiated in the way that's (albeit helpfully) intended.
I would say that the OP's apparent level means that the casual use of a Water Analogy could be harmful without a very strong caveat - which has not been included.
It has already been necessary to put the idea that the Kinetic Energy of Electrons is involved in the transfer of Electrical Energy, which is another problem when we try to use Mechanics to explain Electricity.

 

[litup]

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? View attachment 107191

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

I'm surprised nobody has mentioned the 4 point probe concept. If you measure a low resistance with a DVM you just get a minimum reading and for my meters they are all say 0.1, 0.2 ohms, something like that because they just can't measure resistance less than 1/10th ohm very well. A 4 point probe is different, it has 4 tips all in a row usually and the outer two tips inject a small current into the wire or whatever, and the two inner tips separated by maybe 1 mm measure the voltage developed across those tips. It can measure very tiny voltages and the instrument converts those into ohms readings, as low as one micro ohm and believe me, it will measure very well the resistance of any wire and those wires are definitely not zero ohms so there is a voltage drop across those wires and the higher the current in the wire the larger the voltage drop and that voltage drop leads directly to heating the wire. That is why an extension cord gets hot if you say, run a toaster taking 1500 watts with a small household extension cord, the toaster might work but some portion of the energy used to heat the elements of the toaster are also heating the wires used to power it which means the toaster is now not getting 1500 watts because there is a voltage drop across the extension cord which lowers the energy available to the toaster and the energy not used in the toaster is now being used directly to heat the wires of the extension cord and the smaller the diameter of the wires used for that job the more the wires heat up and the less is available for the toaster to the point where the wires will melt and maybe cause a fire which is why you use high current extension cords for big jobs.

 

[sophiecentaur]

@litup: Your example is a good one to demonstrate that we have a choice whether or not to include the very low resistance of connecting wires in our circuit calculations. When the resistance of the cable is near the same magnitude as the resistance of the components in a circuit, then it is vital to take it into account and its effect can be easy to find (as in your 'four point probe'). It would be another matter to try to use a four point probe on the copper strips between components of a CMOS circuit. It is often very reasonable to assume connecting wires to have zero resistance and we all start off that way in School. But with Meters that give only 3 sig figs and when component values are deliberately chosen by the teacher for the results to 'come out right', we don't even consider it - except when we find we need to jiggle the connectors a bit, to eliminate dirty contact resistance.