What is 1.5V, 9V or 12V really means?

Discussion in 'General Electronics Chat' started by dtvonly, May 13, 2015.

1. WBahn Moderator

Mar 31, 2012
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Keep in mind that it is a thought experiment. Its presence and its charge (both sign and magnitude) are a given.

2. BR-549 AAC Fanatic!

Sep 22, 2013
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Voltage is nothing more than a measurement of electric field density. This magnitude of density has direction, so it is a vector. Only one source is needed for voltage.
Voltage is the gradient of the electric field density.

Something to think about. All of our experiments, study and measurements has been done with negative charges and negative fields. The so called positive fields that we use are not positive. They are depleted negatives. We have always used electrons against electrons, not electrons against protons. We have really never worked with a positive field before. We just change the direction of a negative, and call that positive. We can not separate and than accumulate protons like we can electrons.
Many think that if we could, the same formulas and circuits would work, just change the polarity signs. This would not work.

There are many more differences than polarity between + and - charge.

Negative charge is normally 300 to 500 larger in physical size than positive charge, and the size difference can be much larger.
Because of this size difference, the electron is much more reactive to a field than a proton is.
This along with small mass is why electrons do most of the moving in the universe.

Even though the proton is much smaller, the proton is almost 2000 times heavier than an electron. Along with this added weight and the small size for reaction......it will take magnitudes more voltage to move a coulomb of positive charge than it does to move a coulomb of negative charge.

Electrons are normally much colder than protons(hot). And again, because of mass and size, the electrons do most of the temperature transfer in the universe.

We have much to learn.

3. studiot AAC Fanatic!

Nov 9, 2007
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As some know, that I am currently working away in Scotland.

Well this week, when visiting Perth museum, I saw the largest salmon ever caught here ( by a lady in 1922) and it inpired this example.

The length of this salmon is 54 inches.

So taking a range of salmon lengths (all measured in inches) I have 54, 44, 34, 24, 14, 04.

So I can establish a second quantity "length difference" also measured in inches.

From my table I can derive the following length differences 10, 20, 30, 40 and 50 between salmon.

So it is with voltage.

My single charge establishes an electric field where I can observe a voltage at evry point.
I can establish voltage differences between two such points.

But what happens if I introduce a disconnected 10 volt battery?

What is now the potential between the two points in space occupied by its terminals?

What if I now connect a 1k resistor between these two points?

4. Kermit2 AAC Fanatic!

Feb 5, 2010
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all very good theoretical material and it is sure to be helpful to someone, somewhere, sooner or later. Our little hook baiter, however, has yet to check back in and give his thanks for the deep thoughts we have provided him. Hmmmmm?
I stand firmly on my original assesment of this thread...

5. WBahn Moderator

Mar 31, 2012
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You salmon lengths are all length differences -- the difference between the position of the tip of the head to the tip of the tail. You do NOT have a single quantity "length".

So too with you voltage at every point in space due to a single charge. It gives the DIFFERENCE in electrical potential energy, per unit charge, between that point and whatever reference point you HAD to choose in order to evaluate the integral of the electric field from that reference point to that point in space. Again, if you say that the voltage at some point in space is 10V, or 10 J/C, what does that mean? If you place a 1C test charge there then it will have 10 J of potential energy. What does that mean? It is the same with a gravitational field. I can assign a gravitational potential, J/kg, to every point in space. But to do so I HAVE to explicitly or implicitly define a location at which the gravitational potential energy is zero, because the gravitational potential, just like the electrical potential, is the integral of gravitational field dotted with the path vector between the reference point and the point in question.

6. studiot AAC Fanatic!

Nov 9, 2007
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All of which conveniently avoids my question.

And no, length is not length difference.

Length is a property of a single salmon.

Length-difference require two salmon.

This is a subtle point

but

Potential is the property of a single point. There can be only one potential at one point so a difference or comparison cannot be made.

Potential-difference requires two points.

Zero is of course just another single unique value, of no more significance than any other.

7. wayneh Expert

Sep 9, 2010
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Not to me, or Wikipedia. Potential is undefined without a reference.

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8. studiot AAC Fanatic!

Nov 9, 2007
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Not undefined, undetermined.

Your Wikipedia reference is to potential difference, not potential.

The appropriate wiki reference is to electric scalar potential.

http://en.wikipedia.org/wiki/Electric_potential

An electric field has both a vector potential and a scalar potential if you want to use those terms.

I do not disagree that definitions appropriate to the vector potential are not applicable to the scalar potential and vice versa.

But I am saying you need to distinguish.

I can assign a number (scalar) to every point in space and it could map an electric field.

And yes I could choose any point in that field as the base or reference.

But if I changed base I would have to do so in such a way as to preserve the difference between any and every pair of points.

9. WBahn Moderator

Mar 31, 2012
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I am not avoiding your questions, all of which have very simple answers having nothing to do with salmon:

You are the one that is avoiding a very simple question that I have asked repeatedly.

If you say that the voltage at a particular point is 10V, what does that mean?

How was the voltage determined?

What does it mean for a 1C charge sitting at that point to have 10J of potential energy?

Sure. So what?

How are you going to do this in a universe that only contains a single charge?

Let's assume that you have relaxed this requirement.

By introducing the battery you change the electric fields because you have change the charge distribution in this universe. The new distribution of the fields is such that if you integrate the electric field dotted with the path vector along any path between the negative terminal to the positive terminal you will get a difference of 10V. If you want to know what the voltage at the negative terminal is, then you have to specify the reference, just as you did when you came up with the voltage at that same point before introducing the battery.

Then you get a current flowing in the resistor of 10 mA and you also establish an electric field along the resistor that changes the electric field distribution in that region which, in turn, changes the voltage field. As before, you can calculate the "voltage" at any given point given the knowledge of the field distribution AND only after specifying, implicitly or explicitly, a reference point for that voltage field having an arbitrarily defined value.

10. WBahn Moderator

Mar 31, 2012
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Which you proceed to contradict when you say:

[/QUOTE]

If potential is the property of a single point, then it can't matter what we choose to call the value at some other point. If it does, then it is not a property of a single point any longer.

[/QUOTE]
Zero is of course just another single unique value, of no more significance than any other.[/QUOTE]

Agreed. So what? That doesn't change the fact that to compute the potential at a single point, you have no choice but to use a reference, implicitly or explicitly. Where that reference is located and what voltage you choose to assign to it are arbitrary.

Again, it comes down to the very simple question that you refuse to answer.

If you way that the voltage at a particular point is 10 J/C, what does that mean?

You've just agreed that you can use whatever reference you want, so what if you choose a reference such that the voltage at that point is now 1000 J/C. What does that mean?

You keep insisting that the voltage at that point doesn't need a reference at all (since you keep insisting that it isn't a potential difference of any kind). So what is the voltage at that point if no reference is used? And how do you then change the base or reference of something that, according to you, HAS no reference?

11. BR-549 AAC Fanatic!

Sep 22, 2013
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All definitions have a context. I look at it this way.

The source is the reference.

Distance provides the potential.

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12. WBahn Moderator

Mar 31, 2012
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The source, unfortunately, doesn't provide a suitable reference because it is a singularity at its origin. In the absence of an explicitly defined reference, the only reasonable one to use -- and the one that is used by almost all physicists -- is to use a reference of zero energy at an infinite distance from the source.

13. BR-549 AAC Fanatic!

Sep 22, 2013
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I know, but that's an administrative problem, not a physical one.
We don't need to know the value of the source to use it.
And besides, zero is only as good as your measurement is.
You can try to make a zero.....but theoretically, that zero could only exist at one point.
It's just the way you look at it.
I believe the field gradient is there, and therefore a potential difference is there, whether you can measure the work done or not.

14. studiot AAC Fanatic!

Nov 9, 2007
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But I did answer it, several times in several ways, all consistent with each other.

It means that I have assigned the value 10V to that point.

I also pointed out that assignation affects the potential at every other point in space and if I assign a different value say -10 volts or 0 volts or 20 volts then every other point in space will have to be reassigned accordingly to preserve the potential difference between the two points.

This is a sort of Galilean relativity of potential. Many physical properties are like this. There is no such thing as an 'absolute' value. You can add an arbitrary constant.

Returning to my battery, that you rather peremptorily dismiss.

Let me place my zero of voltage at point A in space, and let me connect one terminal of my voltmeter to it.
Due to the charge in the system I measure a small potential Vb and Vc at points B and C in space.
Let me now introduce an unconnected 10V battery so one terminal is at point B and the other at point C.

What voltage will my meter now read if I touch one terminal only of the battery with the free lead of my voltmeter?

15. WBahn Moderator

Mar 31, 2012
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Which means that you have now picked a reference to which all other points are referred. That you assigned that reference point to be 10V instead of 0V merely makes the bookkeeping a bit harder, but as you said, zero is not more special than any other number.

All of which makes my case and destroys yours.

You mean that you measure a small potential difference between Point B and Point A and you measure a small potential difference between Point C and Point A.

But don't forget that the introduction of the meter changes the fields that you are attempting to measure and unless your source that is creating the background field is dominant, the meter might measure virtually no voltage difference between its probe tips because it might dominate the local fields. Just like if you would if you simply put a piece of wire between points B and C.

I'm assuming one lead of your meter is still at Point A? If so, then you will be in the same boat you were before. For simplicity, let's put the positive terminal of the battery at C. What you will NOT get is a measurement between C and A that is 10V greater than you get between Points B and A, even though you, hopefully, agree that the potential at Point C is 10V greater than the potential at Point B. By reconfiguring the meter you change the fields and, hence, the voltage at Point A has changed (compared to any point that was not affected by the change in meter configuration). But use two meters and put the negative probe tip of each at Point A and the positive tips on the two battery terminals and now you will read Point C and being 10V higher than Point B. You've still changed the voltage at Point A (relative to unaffected points), but now you are at least using the same voltage at Point A in both measurements.

16. studiot AAC Fanatic!

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C'mon,
You have already acknowledged to someone else that this is a thought experiment so we can have perfect voltmeters, and components.

You know as well as I do that with the conditions stated the introduction of an unconnected (as already stated but ignored) battery will make no difference to the readings on a perfect voltmeter.
Yet, as you rightly point out, if we make a suitable connection we can measure 10 volts difference between the terminals.

This is because the other point you dispute is that the statement that any and every point could be used as a reference is equivalent to stating there is no origin or zero point, just as in ( again already stated but ignored) Galilean relativity. This becomes even more poignant if we move electrodynamics to Einstinian relativity, were even the pds are not preserved.

So my comments are in perfect keeping with current Physics

17. WBahn Moderator

Mar 31, 2012
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Okay, I'm willing to stipulate a perfect voltmeter.

So. What's your point. You introduced the battery and, with it, a change in the charge distribution of your universe such that electric fields changed everywhere such that there is a 10V difference between points C and B.

Saying that the reference point is arbitrary is NOT the same thing as saying that there is no reference point. Look at highway mileage markers on the vast majority of U.S. roads. When you are at mileage marker 163, what does that mean? It means you are 163 miles from the arbitrary reference point that was chosen for that metric.

Why won't you answer the question about what it means to say that the voltage at a particular point in space is 10V?

Why won't you answer the question about how you determined that the voltage at that particular point in space is 10V to begin with?

You say that if you choose to define the voltage at point A as being 0V, or whatever, that you have to add an appropriate constant to the voltage at every other point in space, to which I agree because, in doing so, you are defined the reference to which all other points in space are measured. How, after defining the voltage at Point Z to be 2.52741V, do you determine the voltage at Point R? Simple, you find the voltage DIFFERENCE between Point Z and Point R. Why? Because voltage is always a difference in potential between two points. Whenever you say that the voltage AT a particular point is some value, you are ALWAYS (again, with the caveat that we are talking about conservative electric fields) saying that the voltage DIFFERENCE between that point and the explicit or implicit reference point is that voltage.

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18. studiot AAC Fanatic!

Nov 9, 2007
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All this says is that a voltage difference is a voltage difference and is measured in volts.

I said that in my first post in this thread and have been saying that ever since.

But enough of this nonsense

The Electric scalar potential, phi, is defined by the equation

$E = - \nabla \phi$

So yes of course the E field is in volts/m

So what are the units of phi?

19. WBahn Moderator

Mar 31, 2012
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phi are in units of volts. So? Are you trying to say that because you have chosen to express the relationship between the electric field and the voltage field in differential form that this somehow means that the voltage at a point in the voltage field is not relative to a reference value somewhere?

Again (and again and again), using YOUR example. You have a point charge at the origin of your universe. Let's give that charge substance and say that it is 10 coulombs at the origin. Starting with that and using only Maxwell's equations, what is the voltage at a point on the x-axis 10 meters away from the origin and what is the voltage at a point on the y-axis 20 meters away from the origin. Show that you can get those values without relying on the fact that the voltage at a point is the voltage difference between that point and a reference point.

20. studiot AAC Fanatic!

Nov 9, 2007
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As a teacher of the subject you made the above comment.

Because I heartily agree with this, and although not myself a teacher, observe generation upon generation of new starters struggling with the concepts of voltage, current, charge etc I made the following post

I see nothing wrong with the brief comments and the fact I identified the principal use of the volt unit (potential difference) as the likely candidate as the subject of th OP's question.

However you chose to flatly deny Physics recognises a second quantity, also measured in volts and we have spent the next 30 something posts squabbling about this.

You finally agree that phi is also measured in volts.

If you now integrate the differential equation that I used to define phi, you end up with an arbitrary function (since it is a PDE).

It is convenient to eliminate this by choosing a reference as you say and working in terms of phi difference or potential difference so that is what we do practically.
It is even more convenient that since curl(E) = 0 the field is conservative and we can do this in a way that sets the three arbitrary constants that arise in splitting the PDE into three ODEs to zero.

But the OP asked for the underling physical significance of phi (as they all do at some stage).

And before anyone asks, the fact about the curl and the conservative field is enough to make consideration of phi desirable at a deeper level.