Voltage and Potential: Understanding the Difference

[Chenkel]

I've heard in this video

that the voltage is the electrical potential difference, for two points (A, and B) you measure their voltage and you subtract to find the difference. If you measure the voltage at A isn't that a 'electrcal potential difference' between two points? So is the voltage at A the electrical difference between the voltage at that point and 0 volts. Kind of like the way a pressure gauge measures relative to atmospheric pressure? Except in this case it would be measuring pressure relative to a vacuum?

Is this a valid way to think about it:

V = (Vb - 0) - (Va - 0)

Let me know what you guys think, thank you!

 

[DaveE]

Yes. I didn't watch the video, but you've got the basic (and very important) concept. Voltage is ALWAYS a potential difference between two locations. This can be confusing because it is most convenient to keep one of those locations fixed. This fixed location is often called "ground" in electronics. Then, people get so used to always doing this that they just assume that you know that voltage is a measurement between two points, and that you already know, by the context, what that second reference point is. Then they (we) start talking about the voltage at a (single) point. This is just sloppy and wrong; but really, really common.

If you are ever confused about this, just remember what a voltmeter looks like; it has two leads. Always.

 

[DaveE]

Also, as it relates to your equation...

Electricity and Magnetism are conservative forces (fields). This means that in any electric field the potential difference (voltage) between two points is not affected by the path you take in traveling from one point to the other. This means that if you have different points A, B, C, and D, the voltage between A and D $V_{AD} = V_{AB} + V_{BD} = V_{AC} + V_{CB} + V_{BD}$ etc.

You do have to be careful about the signs though, $V_{AB} + V_{BD} \neq V_{AB} + V_{DB}$ . You'll want to think of measuring voltage sequentially (in a fixed direction) along a path. Any directed path will do.

 

[Chenkel]

Yes. I didn't watch the video, but you've got the basic (and very important) concept. Voltage is ALWAYS a potential difference between two locations. This can be confusing because it is most convenient to keep one of those locations fixed. This fixed location is often called "ground" in electronics. Then, people get so used to always doing this that they just assume that you know that voltage is a measurement between two points, and that you already know, by the context, what that second reference point is. Then they (we) start talking about the voltage at a (single) point. This is just sloppy and wrong; but really, really common.

If you are ever confused about this, just remember what a voltmeter looks like; it has two leads. Always.
View attachment 302379

So when voltage is measured at a single point, the second point is omitted and considered to be ground.

I've heard voltage is joules per coulumb,
If you have 12 volts and that implies 12 joules per coulumb, if we up the voltage to 24 that means 24 joules per coulumb, so one coulumb is a set amount of electrons (charges), but the energy stored by them when considering points A and B has increased, is this analogous to a bunch of tennis balls being lifted higher in the air, the amount of tennis balls remains constant but the potential energy increases whenever we increase the height; joules per coulumb have been increased and joules per tennis ball has increased.

Another thing I noticed in the video is

W = -q*dV=-dPE

W is the work done across A and B, where dV is the voltage in between the points A and B, and dPE is the change in potential energy, so if there is positive voltage difference there will be a negative change in potential energy, if there is a negative voltage there will be a positive change in potential energy. I can imagine electrons moving that create a net decrease in potential energy just like a ball falling loses potential energy and gains kinetic energy, but a negative voltage sounds like a ball falling in reverse to increase PE, so I am wondering what the motivation of matter is to increase PE.

Let me know what you think, thanks!

 

[berkeman]

@Chenkel please use the "Reply" feature to quote other posts. Just click the Reply link in the lower right of a post to quote the whole thing, or click-drag what you want to quote and select Reply from the pop-up menu. Thank you.

DaveE said:
Yes. I didn't watch the video, but you've got the basic

 

[Chenkel]

@Chenkel please use the "Reply" feature to quote other posts.

I just tried quoting your post and I notice that the dragging feature is not there. I'll try to be more careful and only quote the relevant part of the post in the future by back spacing some of the quote instead of quoting the whole post, I think that's what you wanted, if there's anything else I can do let me know.

 

[berkeman]

Looks perfect!

 

[Baluncore]

I've heard voltage is joules per coulumb, ...

That will lead you astray for everything other than a perfect resistor.
You must forget it, or you will be unable to advance beyond ohm's law.

 

[Chenkel]

W = -q*dV=-dPE

W is the work done across A and B, where dV is the voltage in between the points A and B, and dPE is the change in potential energy, so if there is positive voltage difference there will be a negative change in potential energy, if there is a negative voltage there will be a positive change in potential energy. I can imagine electrons moving that create a net decrease in potential energy just like a ball falling loses potential energy and gains kinetic energy, but a negative voltage sounds like a ball falling in reverse to increase PE, so I am wondering what the motivation of matter is to increase PE.

Let me know what you think, thanks!

I think I made a mistake here, correction: when dV is positive, dPE is positive, when dV is negative, dPE is negative. So positive increase in PE is mystifying to me, how does positive voltage indicate an increase in potential energy if work is being done? What does it mean when voltage between two points is negative?

 

[Chenkel]

That will lead you astray for everything other than a perfect resistor.
You must forget it, or you will be unable to advance beyond ohm's law.

Are you saying the only place it rings true is a circuit with a perfect resistor?

Suppose V is 12 volts and R is 1 ohm then A is 12 amps, does the
"joules per coulumb" only work in this circuit? What does it tell us? Isn't the power requirement of this circuit 144 watts?

 

[Baluncore]

Are you saying the only place it rings true is a circuit with a perfect resistor?

That is correct. It applies to heat generation.
Forget it as soon as possible, or you will continue to waste your time and ours.

 

[DaveE]

I've heard voltage is joules per coulumb,

In general yes. If you move a charge Q (i.e. coulombs) through and electric field, let's say from point A to point B, the electric potential V (volts) will result in an energy change E. That energy is $E=QV$. It takes work to move a charge against a voltage gradient (or you get work done for the other polarity). So you can think of voltage as "potential energy per coulomb moved". That's one reason voltage requires two points in it's definition. This is basic physics, which underlies all of EE, of course. When you start dealing with real world components it can get a bit more complicated because energy is sometimes lost as heat or stored in magnetic fields. But for basic electrostatics, then, yes, what you heard was correct.

 

[DaveE]

What does it mean when voltage between two points is negative?

This is just a question of working out the polarities. If the voltage from A to B is V>0, and the charge on your particle is Q>0 (like a proton), and it starts at point A it will experience a force pushing it from A towards B. In that process it will gain kinetic energy* and lose potential energy. That energy change will be E=QV. If you change one of the polarities then you will have to do work to move it and it will gain potential energy.

*OK, for the real physicists out there, yes we know there is also energy stored in a magnetic field, but that isn't B level stuff, is it?

 

[Chenkel]

That is correct. It applies to heat generation.
Forget it as soon as possible, or you will continue to waste your time and ours.

I'm sorry that you feel I've wasted your time, If I can avoid it in the future I will by being more careful with my questions, no sense in making it a pain for someone to teach.. I try not to throw anything with a mathematical use away, so when you tell me to forget about it I feel that only applies when it has no use at all. Hopefully if something is stated so much in mathematical literature such as energy being joules per coulomb the authors stating such things will say the specific places it applies.

Thanks for the help you've given me so far, I appreciate it! :)

 

[Borek]

Volts being joules over coulombs is just the SI definition of the unit. Practical applications/realizations of this exact definition can be tricky, but it doesn't make it invalid nor wrong.

 

[Chenkel]

Volts being joules over coulombs is just the SI definition of the unit. Practical applications/realizations of this exact definition can be tricky, but it doesn't make it invalid nor wrong.

I'm concerned that I'll accidentally use it when I shouldn't, and it will give me an incorrect result. I'm keeping my fingers crossed that my mathematical intuition will be good enough to know where to apply it and when not to.

As long as C is the charge pushed through the circuit, and the voltage across the circuit is V, can we always use V=J/C to figure out how much energy the circuit uses for a given amount of charge, or is there a hidden pitfall that only seasoned electrical engineers will be able to avoid?

 

[Baluncore]

As long as C is the charge pushed through the circuit, and the voltage across the circuit is V, can we always use V=J/C to figure out how much energy the circuit uses for a give amount of charge, or is there a hidden pitfall that only seasoned electrical engineers will be able to avoid?

C represents capacitance.
The symbol used for a quantity of charge is Q. The unit is coulombs, C.
One pitfall is that the voltage must remain constant during the charge transfer.

 

[Borek]

I wonder if in practice it is not much easier to work not in terms of charge, but in terms of current and time.

 

[Baluncore]

I wonder if in practice it is not much easier to work not in terms of charge, but in terms of current and time.

Yes.
It is easiest to use; power = voltage * current.
Then multiply by time to get energy.

 

[Chenkel]

C represents capacitance.
The symbol used for a quantity of charge is Q. The unit is coulombs, C.
One pitfall is that the voltage must remain constant during the charge transfer.

In the case of a changing voltage can you use calculus to integrate amps with respect to time to get total coulombs, and then use V=J/Q to get total joules for a given time period, and then divide those joules by that time to get watts?

 

[Baluncore]

In the case of a changing voltage can you use calculus to integrate amps with respect to time to get total coulombs, and then use V=J/Q to get total joules for a given time period, and then divide those joules by that time to get watts?

There is always a longer way of doing it.
You could integrate the product of voltage and current, over a period of time.

 

[sophiecentaur]

how does positive voltage indicate an increase in potential energy if work is being done?

I think you may be seeing some sort of paradox when there is none.

A gravitational equivalent. You increase the Gravitational Potential of a ball that you roll up hill. That ball will be able to do more work when rolling up hill.

Work is 'done'/ put into the system. We call that positive. More (positive) work can be done, according to the extra work is put in. Would it help to remember that both Force and Displacement have different signs going up and going down. Work (Energy) is scalar but Energy Conservation only applies when there are no losses or additions / subtractions from the system (which is what happens when there's an external push or force against the ball).