# Is my understanding of voltage correct?

I read that a voltmeter measures the voltage difference between two
points, but this doesn't make much sense to me. Let's take the following
picture:

Would it make sense if the voltmeter was receiving current from both
sides, so that it could measure the kinetic energy difference of the
electrons it was receiving from each side, working out the difference in
voltage?

Would it make more sense to say that since components in parallel
receive the same voltage, we can assume that whatever the voltmeter is
getting, the bulb is also getting (in this case I don't see how the
voltmeter is actually measuring anything between two points but is
instead measuring what it's receiving and we assume the component is
getting the same voltage)?

Can you also assume that since voltage is divided between all components
in a series circuit, (and that the component eats all the voltage it is provided
with?), so you can assume that whatever the voltmeter is getting, the
measured component is using 100% of that voltage showing on the voltmeter?

Since kinetic energy (which I was taught was voltage) is converted into
different forms such as heat, would you say the voltage is actually used/eaten
by a component/resistor, but since I = V/R if voltage is reduced by a component
then the current in the circuit is reduced too (which would explain why
resistance lowers current)?

jbriggs444
Homework Helper
Voltage is not about the kinetic energy of electrons. The electrons in a current flow have an average "drift velocity" on the order of millimeters per second. You would be better served to conceptualize an electrical current as an abstract flow (current in amperes) driven by an abstract pressure (the electric field potential in volts) and ignore electrons entirely.

Current is divided among the various components in a parallel circuit. The entire voltage applies across each individual component.

Andrew Mason
Homework Helper
Welcome to PF, John.

I am not sure where to begin. Let's get a few concepts clear:

Voltage is a difference in potential energy per unit charge measured between two points. It is a measure of the difference in electrical potential energy that a unit charge has at the first compared to the potential energy that the same unit charge has at the second by virtue of the electric field between those points. If the charges are free to move between those two points, a positive potential difference between two points is equal to the change in kinetic energy that a unit of positive charge would have after moving from the first to the second; a negative potential difference between the two points is the change in kinetic energy that a unit of negative charge would have after moving from the first to the second.

In a conductor, things are not so simple. Electrons are free to move initially but they keep running into atoms and lose the kinetic energy gained from the electric field. As a result, they "drift" through the conductor. But the idea is still the same: the electric field in the conductor provides a force on the electrons that causes them to gain energy. Some of that energy may be given up when the electron hits an atom and bounces around but the electric field keeps giving the electron energy.

AM

Karmo
I had this question too, please correct me if I'm wrong, I imagined an analogy with magnets e.g if we have two magnets in a table with low friction at a given distance, the potential force that attracts/ repeals them could be same as voltage, taking account of velocity and acceleration till they both touches or stay on repose (repeal), where table's friction represents resistance, and if i think on current flow i would have to add more magnets and count the number of flowing magnets that are moving between d.

It's this analogy useful?
Thanks this is my first entrance

K

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Is my understanding of how the voltmeter works correct at least?

CWatters
Homework Helper
Gold Member
Would it make more sense to say that since components in parallel receive the same voltage, we can assume that whatever the voltmeter is getting, the bulb is also getting
Correct.

(in this case I don't see how the voltmeter is actually measuring anything between two points but is instead measuring what it's receiving and we assume the component is getting the same voltage)?
That is normally correct but not always...

The resistance (impedance) of an Volt meter is normally large so very little current flows through the meter. This means that a volt meter shouldn't affect the operation of the circuit itself and it will normally display the voltage across the component.

However all things are relative. Not all voltmeter have a very high impedance and in some cases the impedance of the volt meter can change the operation of the circuit. For example if you were to charge up a small capacitor by connecting it to a 9V battery (briefly), and then try measuring the voltage you might find that the voltage on the capacitor disappears quickly because the capacitor is discharged by the small current flowing though the meter.

sophiecentaur
Gold Member
2020 Award
Would it make sense if the voltmeter was receiving current from both
sides, so that it could measure the kinetic energy difference of the
electrons it was receiving from each side, working out the difference in
voltage?
The Kinetic Energy of the electrons is totally dominated by their thermal motion. They have a range of speeds from zero to hundreds of m per second. The fractional increase in KE when a voltage of, perhaps 0.01V is applied across a piece of Copper wire will not be detectable. It just doesn't work to try to describe the way 'electricity works' in terms of the KE of electrons. Current (sometimes a very very small amount) will flow through a Voltmeter. The amount flowing will be determined by the Potential Difference across it and its Resistance. I = V/R
A voltmeter will not receive current from both sides -the current will pass from the more positive terminal to the less positive terminal.

Since kinetic energy (which I was taught was voltage)
Now, where were you taught that? It is a seriously wrong notion.

davenn
Andrew Mason
Homework Helper
I had this question too, please correct me if I'm wrong, I imagined an analogy with magnets e.g if we have two magnets in a table with low friction at a given distance, the potential force that attracts/ repeals them could be same as voltage, taking account of velocity and acceleration till they both touches or stay on repose (repeal), where table's friction represents resistance, and if i think on current flow i would have to add more magnets and count the number of flowing magnets that are moving between d.

It's this analogy useful?
Welcome to PF Karmo.

Not only is it not useful but it is not correct. I am afraid you are confusing many concepts here.

AM

Karmo
I learnt it in A level physics

I was taught that one end of the battery is slightly positive, and one end is slightly negative - causing electrons to flow from - to +. I was taught that when you add a supply of electrons (like a battery) and complete the circuit, the electrons gain kinetic energy and start to move through the circuit. The electron's kinetic energy is used by an appliance and converted into other forms such as heat, light, etc.

I was taught that a flow of electrons is current, and the more you increase the voltage (by increasing the supply of electrons on the negative end) - the more you increase the electron's kinetic energy. Is there a better explanation for voltage that doesn't use abstract concepts?

jbriggs444
Homework Helper
The electrons do not have kinetic energy by virtue of being at the - terminal. They have potential energy. Hence the term "potential difference". The electrons that you supply are not the (only) electrons that are moving. The electrons that are moving are the ones already present in the wires and other circuit elements. The presence of the excess electrons at the negative terminal (which are typically being continually replaced by a power source) causes an electrostatic force which causes the electrons already present in the circuit to begin drifting.

The electrons in the wires rearrange themselves so that the potential difference is just right to maintain a constant current all the way around the circuit and back to the positive terminal. Low potential difference across areas with low resistance and high potential difference across areas with high resistance. If the current flowing into an area in the circuit did not match the current flowing out, an excess or deficit of electrons would build up there and current would naturally flow so as to erase the excess or deficit. In this respect, an electrical circuit behaves very much like a closed plumbing loop or hydraulic system with a pump acting as the battery and the hoses acting like wires.

Andrew Mason explained it well in post #3. The electrons all along the wire gain kinetic energy as they flow from higher potential upstream to lower potential downstream. They give that energy off as they bounce off the atoms in the conductor.

davenn
sophiecentaur
Gold Member
2020 Award
The electrons all along the wire gain kinetic energy as they flow from higher potential upstream to lower potential downstream. They give that energy off as they bounce off the atoms in the conductor.
I wonder whether that simplistic explanation really helps in this discussion. I thought that the present ideas involve interaction of electrons with the whole lattice. A model of snooker balls and marbles has big limitations. I doubt that the total amount of KE in the mass of (classically described) electrons is anything like enough to transfer the kWs of power in an ordinary heating element. The KE is just not enough at speeds of mm/s

Drakkith
Staff Emeritus
Unfortunately, understand the details of voltage requires that you dig into some very complicated topics that haven't even been introduced to you yet. The way voltage is explained in most basic electronics, basic physics, and other similar courses is very simplified. This is okay because unless you want to talk about what's happening on the molecular level, or with extremely small or large voltages/currents, then it works just fine.

sophiecentaur
Gold Member
2020 Award
If you want to get a good grasp of EE then you need to split things into layers - like an onion. The outer structure of this onion is more than adequate for most of EE - which is quite complicated enough, in any case. When you come up against a particular bit that you find problematical - say the Potential Divider Formula, the very last thing you need to do is to delve into what may or may not be happening to the electrons in there. Treat it all classically (with the appropriate Maths, of course) and you can become an ace circuit designer. I feel that many people, given the option, reach out for an electron model in an attempt to understand circuit theory as soon as it gets a bit hard. They often do that to avoid (as they think) the nasty bits of conventional 'Electricity'. You can't avoid it, I'm afraid, despite what some elementary teacher tried to tell you in School - that's a total illusion.
It is possible to use a musical instrument to play music to the very highest level without having a clue about the physics of how it actually works. Conversely, there are guys who can make the most sublime quality Guitars but who can barely play more than three chords. You really don't need to (can't) know it all.

jbriggs444
Homework Helper
I doubt that the total amount of KE in the mass of (classically described) electrons is anything like enough to transfer the kWs of power in an ordinary heating element. The KE is just not enough at speeds of mm/s
The total amount of KE picked up and then deposited by a hypothetical electron going through the filament of a 3 volt (2 cell) flashlight bulb would indeed be 3 electron volts.

sophiecentaur
Gold Member
2020 Award
That's a nearly convincing argument and it has legs. Except that it still assumes that only KE is involved. The eV is certainly transferred in the resistor but is it necessarily only to KE? For instance, in a motor, the measured resistance, when running is due to EM interaction with the rotor. I'm in the car ( not driving!!) and can't do the sums but you could calculate the total KE of all electrons in a wire. You could then calculate the number of collisions per second to deduce the power dissipated this way. Added problem is the thermal activity too. It has made me think again. Thanks.

nasu
Gold Member
Indeed a good question. I did not think about it before either. :)
Considering relaxation times (in metals) of $10^{-15}$ seconds and free electron concentrations of the order $10^{22}$ cm^3, you can easily obtain power densities of the order of W per cm^3 with drift velocities of the order of 1mm/s.
The drift KE is of the order $10^{-30}kg \times (10^{-3}m/s)^2 =10^{-36}J$
Assuming that it is transferred to the lattice after each relaxation time, one electron will transfer $10^{-21}$ W
In one cm^3 this will amount to a few watts transferred to the lattice.

The drift KE of one electron is indeed very small, negligible compared with the thermal KE. But there are so many of them and they transfer it so many times per second...

sophiecentaur
Gold Member
2020 Award
An analogy with a bicycle chain is even more naive but I could use it to explain my unease at just using the KE idea. A bicycle chain can be made as light as you like (in my thought experiment) and its KE would be correspondingly as small as you like. So no power could be transmitted via the tiny proportion of KE. Also, the same amount of Power could be transmitted and the same losses could occur within the chain due to friction. The power loss in the chain is due to the friction forces times the speeds of the rubbing surfaces and, as there is no relative motion is an ideal chain, there would be no KE involved with the loss either.
The power transmitted through a wire to an electrical device is no way to do with Kinetic Energy - as stated earlier, it's Potential Energy that does the biz. So why would you reckon that the power dissipated by a resistive length of wire would be due to KE, if the electrical and mechanical models are to be analogous?
Your "eV' argument seems compelling but I suspect that 'someone' could point out a fallacy there. Not me (yet) though.

jbriggs444
Homework Helper
An analogy with a bicycle chain is even more naive but I could use it to explain my unease at just using the KE idea. A bicycle chain can be made as light as you like (in my thought experiment) and its KE would be correspondingly as small as you like. So no power could be transmitted via the tiny proportion of KE.
I certainly agree that one would not want to want to claim that power is "transmitted" by KE in a bicycle chain. Or that power is "transmitted" by KE in an electrical circuit. But one might be able to envision a bicycle chain fitted with tiny paddles and immersed in fluid. One could further consider that the tension gradient in the chain is allowing the paddles to be alternately accelerated by the tension (thus gaining KE) and decelerated by a fluid (thus losing KE) that that this KE is thereby, in some sense, mediating the transfer of power from chain to fluid.

Such an imagining is rather strained and of no particular help in analyzing the situation. But not out and out wrong.

The power transmitted through a wire to an electrical device is no way to do with Kinetic Energy - as stated earlier, it's Potential Energy that does the biz. So why would you reckon that the power dissipated by a resistive length of wire would be due to KE, if the electrical and mechanical models are to be analogous?
Your "eV' argument seems compelling but I suspect that 'someone' could point out a fallacy there. Not me (yet) though.
Agreed. KE is not how power is transmitted in a wire.

sophiecentaur
Gold Member
2020 Award
I certainly agree that one would not want to want to claim that power is "transmitted" by KE in a bicycle chain. Or that power is "transmitted" by KE in an electrical circuit. But one might be able to envision a bicycle chain fitted with tiny paddles and immersed in fluid. One could further consider that the tension gradient in the chain is allowing the paddles to be alternately accelerated by the tension (thus gaining KE) and decelerated by a fluid (thus losing KE) that that this KE is thereby, in some sense, mediating the transfer of power from chain to fluid.

Such an imagining is rather strained and of no particular help in analyzing the situation. But not out and out wrong.
Agreed. KE is not how power is transmitted in a wire.
That makes sense and shows that there is a way of looking at that problem which could be said to involve Linear KE to Thermal KE transfer.
I think I should look in my mirror every morning and repeat, to myself "IT'S ONLY A MODEL" because my last few posts have been on 'reduced intellectual' setting. One simple explanation of the heating effect can be based on KE and electrons dashing themselves on the rocks in a river but that model and the better quantum based model are mutually exclusive. I should not have let myself be unsettled by the fact that the eV idea can be seen to work pretty well (the numbers add up ok); of course it does. A model based on interaction with the lattice as a whole - involving phonons and the rest is just a better one.
This has been a useful thread for me.

nasu
Gold Member
An analogy with a bicycle chain is even more naive but I could use it to explain my unease at just using the KE idea. A bicycle chain can be made as light as you like (in my thought experiment) and its KE would be correspondingly as small as you like. So no power could be transmitted via the tiny proportion of KE. Also, the same amount of Power could be transmitted and the same losses could occur within the chain due to friction. The power loss in the chain is due to the friction forces times the speeds of the rubbing surfaces and, as there is no relative motion is an ideal chain, there would be no KE involved with the loss either.
The power transmitted through a wire to an electrical device is no way to do with Kinetic Energy - as stated earlier, it's Potential Energy that does the biz. So why would you reckon that the power dissipated by a resistive length of wire would be due to KE, if the electrical and mechanical models are to be analogous?
Your "eV' argument seems compelling but I suspect that 'someone' could point out a fallacy there. Not me (yet) though.
Of course it is potential energy of the field. But you need the drift of the electrons to transfer that energy from the field to the lattice. If you replace the conducting wire with an insulating one, the field is still there (maybe altered in its distribution) and the PE is there. But there is no transfer, is it? The wire does not heat up.

Anyway I think that trying to determine which one is (PE versus KE) is not a very meaningful question, in terms of transfer.
What is definitely a wrong idea is that the electrons leave the battery with a lot of KE which is dissipated then along the wire.
The battery's voltage it is indeed PE basically. It does not have to be used or transferred to be there (potentially). So it is not quite right to say that voltage is related or defined (to) by KE.

After all, the math shows that the rate of energy transfer due to drifting electrons is of the order of magnitude expected. What else is there? It does not contradict the fact that the drift KE is negligible compared with the thermal energy, at any time and for any free electron in a metal.

Andrew Mason
Homework Helper
The battery's voltage it is indeed PE basically. It does not have to be used or transferred to be there (potentially). So it is not quite right to say that voltage is related or defined (to) by KE.
It would be if you replace the conductor with a vacuum tube and apply the battery terminals ( - to the heated cathode and + to the anode). The KE of the electrons at the anode is directly proportional to the potential difference of the battery.

AM

nasu
Gold Member
It would be if you replace the conductor with a vacuum tube and apply the battery terminals ( - to the heated cathode and + to the anode). The KE of the electrons at the anode is directly proportional to the potential difference of the battery.

AM
Yes, of course it is. But they acquire this KE by moving through the potential difference. It's not that the battery "gives" them that KE at one of the electrodes.
I thought that this idea was the original matter of discussion, if the electrons leave the battery with some initial amount of KE rather than potential energy.
In the tube case they still start with potential energy which is converted into KE as they move between the electrodes.

Drakkith
Staff Emeritus
I've got a question I've just thought of. Is there an electric field within the circuit? In other words, could you draw electric field/flux lines within the conductors and components that run from one terminal, through the circuit, and terminate in the other terminal?

nasu
Gold Member
I've got a question I've just thought of. Is there an electric field within the circuit? In other words, could you draw electric field/flux lines within the conductors and components that run from one terminal, through the circuit, and terminate in the other terminal?
It looks like you can. For example here:
https://www.tu-braunschweig.de/Medien-DB/ifdn-physik/ajp000782.pdf
Se the diagrams on the last few pages.

I had another paper about the field and surface charge in a circuit but I cannot find it now.