studiot,

When you started this thread you clearly stated that you know what EMF is and that you were posting to help others understand.

At that time, I thought that EMF was synonymous with voltage. I was presenting what I thought was a clearer, more direct, and more descriptive way to think about voltage.

You also decried the use of the term EMF

Yes, I could not and still do not understand what that term means, although I automatically translate it mentally into the word "voltage" and consider it to be a distinction without a difference.

We have to wait until post#47 to discover that you don't actually understand EMF at all.

That is too much tail chasing.

I did not think the meaning of the name "EMF' was relevant to the discussion.

If you were to say EMF is confusingly named and ask for explanation, I would agree with you and try to give a suitable account of what it is and how it came to be so named.

I already did in responses #13 and #47.

I do not know what your technical background is or where in the world you are, but your command of English is such that I guess it to be your first language.

You would be correct.

When I took my first level secondary school exams I thought I knew everthing about geometry.

You probably had a very good knowledge of Euclidian geometry.

What a shock was in store when I started co-ordinate geometry.
So by my second level exams I again thought I knew all about geometry and now, vectors and scalars.

There lots of different geometries and applications.

What a shock when I started my applied maths degree and started linear algebra, vector and scalar fields, Poisson's equation, the Navier-Stokes equations.......

You must have known that you would learn new and wonderful things from those courses. Otherwise you would not have bothered to enroll in them.

After 35 years applying maths in engineering in several parts of the world , postgraduate qualifications and membership of an august Engineering Institution -

I am still learning.

As are we all.

So tell us what you don't understand and what level you would like the answer pitched at and we will try to help, don't just decry what others have said.

See response #13. Keep it below the tensor analysis level.

Otherwise this thread contains apparantly authoritative mis-statements and misconceptions by several posters ...

Which get discussed and corrected.

and so I think, reluctantly, should be closed.

I don't. As long as it stays civil. That would be sweeping it under the rug. It will just show up elsewhere under a different venue. Besides, this discussion is interesting and informative. Ratch

 

Otherwise this thread contains apparantly authoritative mis-statements and misconceptions by several posters and could well be misleading and so I think, reluctantly, should be closed.

That's a crap argument to close a thread. Let the reader beware. This is a forum, not a textbook.

Why is it that simple discussions that help beginners are encouraged, but higher level discussions like this one that go to a more fundamental level are pushed to be closed?

 

Provided the topic is both constructive and civil then there is no reason for it to be closed.

Dave

 

Lots of folks wanting to continue this thread, so why the silence?

I have often found when trying to grasp a concept that it helps to get familiar with the calculations and then revisit the concepts. So here is a simple calculation, let us see what we can get out of it.

I have already shown by direct measurement in the earlier example that there are at least two types of components which will interact differently with a voltmeter.

So we need terminology to distinguish these. Working through the example, in the light of Kirchoffs Voltage Law, as already mentioned, will show that we actually need more than this.

Consider the circuit in the attachment. It contains two batteries and two resistors. The batteries are perfect so they have no internal resistance.

Kirchoffs law says that if we start at any point and work our way around the loop and back to our starting point, counting voltage rises as positive and voltage drops as negative.

The sum of the EMFs plus the sum of the PDs equals zero.

So here is the first difference EMFs and PDs have the opposite sign. You can see this from the positions of the + and - signs on the diagram.

Noting that the same current, I, flows throughout the circuit we can calculate the PDs across the resistors from Ohms law.

Starting from the negative (right hand) side of battery B1 and working in a clockwise direction K's law says

and here we have the second difference because the IR drop across the battery is zero since the resistance of the battery is zero

So we allocate an EMF equal to the battery voltage +V1. This is consistant with our convention of a rise in voltage as positive, as we move across the battery from the negative terminal to the positive.

Similary we get V2 as we pass across the second battery.

When we reach R1 we arrive at the +ve end first so as we pass across R1 the voltage falls so the PD is -(I times R1). The voltage falls again across R2 by -(I times R2). Then we are back at the starting point.

So the full equation is

V1 + V2 -IR1 -IR2 = 0 or V1 + V2 = IR1 +IR2

So we see that even at an elementary level EMFs and PDs need to be treated differently.
A further complication needs to be accounted for in passing through an EMF as the IR drop is zero, but the voltage across it definitely is not.

I have done all this without reference to charge, fancy mathematics or the Rain God. To proceed further you will have to allow me a little calculus. To help prepare I leave you with a question.

Ohms law states that

Voltage = Current x Resistance = Current x resistivity x length / area

What happens (to the voltage) in the limit when we shrink area and length to zero?
i.e. how does Ohms law work at a point - What does it mean?

 

I've just re-read this entire thread. Twice.

Ratch, I am going to make an assumption. (If you've seen the movie Under Siege 2: Dark Territory then you know what "assumption" is.) I am going to assume the snagging point in your understanding is the use of the word "force" in "electromotive force."

The word "force" in this particular context does not mean "that which accelerates a mass." Consider metallurgy, if you will. In modern technical dialog, the term "temper" means quite specifically "to heat hardened steel in order to relieve stress and precipitate very small carbides." Not much more than a century ago, however, it meant "to heat treat a steel in any undefined manner." The study of electricity is far younger than the passions of the metallurgist. Some of our terminology is still somewhat quaint.

Joules per Coulomb. Ohm-Amperes. Po-TAAAAAAY-to. Po-TAAAAAH-to.

What is the definition of "definition?" If the term "EMF" bothers you, continue to say "applied voltage" instead. The rest of the world might catch up to you in a century or two. Be patient.

Simple and straight forward is my way.

You may not realize it, Sir, but you are quite in error here. Quite in error indeed. Simple and straightforward would have been something along the lines of: "Greetings! This is my fifth post here. Would someone be so kind as to explain why the word 'force' is used in 'EMF?' I don't see how it could apply."

Want to try for definitions of the word "mole" next?

It is the ugly brown thing on my thigh. The one my hound killed in the back yard a few moments ago and brought in to drop in my lap. For years I had suspected it was simply a normal American squirrel, and had trusted it implicitly. Imagine my surprise when my hound provided irrefutable proof of the squirrel's Soviet allegiances.

 

thingmaker3,

Ratch, I am going to make an assumption. (If you've seen the movie Under Siege 2: Dark Territory then you know what "assumption" is.) I am going to assume the snagging point in your understanding is the use of the word "force" in "electromotive force."

Never saw the movie, but you are correct about me not understanding the use of the word "force" in EMF. Using the force when you mean voltage is confusing, especially when physicists define the strength and direction of an electrostatic field by the physical force exerted on a unit charge.

The word "force" in this particular context does not mean "that which accelerates a mass."

Yes, that is what I am concerned about. Using force to mean voltage in one instance, and then using physical force to define an electric field in another.

Joules per Coulomb. Ohm-Amperes. Po-TAAAAAAY-to. Po-TAAAAAH-to.

Not quite. It is possible to have joules/coulomb (notice the lower case) when ohm-amperes have no meaning. because no current is present. For instance, a perfect capacitor without leakage does not exist. But if it did, one could energize it to a respectable voltage and observe that it had energy (joules) and had a imbalance of charge carriers (coulombs). But its leakage resistance is infinite and its amperage is zero. So ohm-amps is undefined.

What is the definition of "definition?"

Confusing in this case.

If the term "EMF" bothers you, continue to say "applied voltage" instead. The rest of the world might catch up to you in a century or two. Be patient.

I certainly hope it does not take that long.

You may not realize it, Sir, but you are quite in error here. Quite in error indeed. Simple and straightforward would have been something along the lines of: "Greetings! This is my fifth post here. Would someone be so kind as to explain why the word 'force' is used in 'EMF?' I don't see how it could apply."

You are correct, I don't realize how I have been in error. You are confusing how I would like to challenge something with the time and effort expended to convince others that my point is valid. Studiot first mentioned EMF in post #6 of this thread. My first reply to studiot, asking about EMF was in post #13. Between those events other members posted and were engaged. I had no control over that. So I was as prompt as was possible in asking about EMF. Ratch

 

studiot,

Consider the circuit in the attachment. It contains two batteries and two resistors. The batteries are perfect so they have no internal resistance.

Unfortunately, I don't see anything in the attachment except a blank rectangular box labeled "Attached Thumbnails". I am going to assume a series circuit with both batteries summing the voltage.

Kirchoffs law says that if we start at any point and work our way around the loop and back to our starting point, counting voltage rises as positive and voltage drops as negative.

The sum of the EMFs plus the sum of the PDs equals zero.

You appear to be concerned about the way the voltages occur in a circuit, i.e., whether they are caused by energy gains from elements like batteries, or energy losses due to passive elements like resistors. My understanding of K's law is that the sum of the voltages around a loop circuit where charge is flowing is zero. This has to be because the charge flow is the same at all points in a loop circuit. If the voltage was not zero, then a constant energy gain/loss would be present with no accountability as to where it is coming/going. I would simply sum up the voltages around the loop with no regard as to whether they were batteries or resistors.

So we see that even at an elementary level EMFs and PDs need to be treated differently.

Why?

I have done all this without reference to charge, fancy mathematics or the Rain God. To proceed further you will have to allow me a little calculus. To help prepare I leave you with a question.

Well, I can apply K's law without regard to "EMFs" and PDs.

Ohms law states that

Voltage = Current x Resistance = Current x resistivity x length / area

Not true. The formula above is correct, but that is not Ohm's law. I will shortly start another thread explaining what I mean and providing proof of what I say.

What happens (to the voltage) in the limit when we shrink area and length to zero?

Voltage across what? If no charge is moving, then K's law does not apply. But voltage can still be present with no current existing and resistance at infinity, as I have shown in the previous post #38.

i.e. how does Ohms law work at a point - What does it mean?

You really mean the resistance/impedance formula. It does not apply if no charge is moving.

In conclusion, I don't see the difference between what is meant by EMF and voltage. Whether a voltage difference is caused by a perfect battery or a resister is a distinction without a difference as far as I can understand it.

Ratch

 

This is what i think about EMF and potential difference.

Isn't it true that for an electric cell the EMF is the potential difference between the half cells.

Another way of looking at KVL is:
Total EMF = Total PDs (?)

AFAIK EMF is a misnomer.
Potential difference is just what its name implies. If taken across a resistance a small change in resistance will change the magnitude of PD(*potential difference), but PD is not a term unique to voltage across resistances(or is it?).

Edit: I am not claiming anything here. As stated, this is what I have learned intuitively. I'd like to be corrected if I am found wrong.

 

Not quite. It is possible to have joules/coulomb (notice the lower case) when ohm-amperes have no meaning. because no current is present. For instance, a perfect capacitor without leakage does not exist. But if it did, one could energize it to a respectable voltage and observe that it had energy (joules) and had a imbalance of charge carriers (coulombs). But its leakage resistance is infinite and its amperage is zero. So ohm-amps is undefined.

If I put an infinite amount of charge (coulombs) in a field with infinite energy (joules), then I would have infinity over infinity. And don't tell me that that situation cannot exist so it is an invalid argument. A capacitor without leakage doesn't exist either.
All that ohm-amps means is that you would get 1V of potential if you applied 1Amp to a 1 Ohm resistance. It also shows implicitly ohm's law. Now explain \(V = \frac{kg\cdot m^2}{A\cdot s^3}\)

 

Can anyone else not see the the attached diagram in post 64?

It is a simple 15K jpg file.

 

a perfect capacitor without leakage does not exist.

Correct. This is why V=I*R

But if it did,

Hypothesis contrary to fact.

ou are confusing how I would like to challenge something with the time and effort expended to convince others that my point is valid.

No. Please re-read my statement. I am teasing you for taking the long way around what should be a simple statement or question, and then claiming to be "direct."

 

Can anyone else not see the the attached diagram in post 64?

It is a simple 15K jpg file.

I can see it.

Dave

 

Can anyone else not see the the attached diagram in post 64?

It is a simple 15K jpg file.

Golly, at this point it's difficult to even find post 64. This thread has become quite a pile of internal references, has it not?

But yes, the attached diagram is indeed there.

 

Golly, at this point it's difficult to even find post 64.

Funny you should say that because I had a problem finding post #64 in the mass of words!

Dave

 

caveman,

If I put an infinite amount of charge (coulombs) in a field with infinite energy (joules), then I would have infinity over infinity.

And one could evaluate the ratio of the two infinities by L'Hospital's theorem. However, I was speaking about charges that do not move, in which case I*R has no meaning.

A capacitor without leakage doesn't exist either

No, but it is represented by a perfect capacitor in parallel with a high resistance.

All that ohm-amps means is that you would get 1V of potential if you applied 1Amp to a 1 Ohm resistance.

Yes, provided that the charge carriers move.

It also shows implicitly ohm's law.

You mean the resistance formula. See http://forum.allaboutcircuits.com/showthread.php?t=11674

Now explain

The SI unit equivalent of voltage. Ratch

 

And one could evaluate the ratio of the two infinities by L'Hospital's theorem. However, I was speaking about charges that do not move, in which case I*R has no meaning.

It doesn't matter whether charges move. You can define it by what would happen if the resistance was there. This is a standard physics method similar to the concept of using a test charge in a field to define the field strength. The field strength exists whether or not the test charge is there. In fact I've even seen pressure described by using the concept of a test surface.

You mean the resistance formula. See http://forum.allaboutcircuits.com/showthread.php?t=11674

I was already aware of this technicality, but I'm not going to argue semantics. You know what I meant, that's all I care about.

 

thingmaker3,

Correct. This is why V=I*R

Provided the charge carriers move. I never said otherwise.

Hypothesis contrary to fact.

Which is done all the time in electrical calculations.

No. Please re-read my statement. I am teasing you for taking the long way around what should be a simple statement or question, and then claiming to be "direct."

And I am claiming that you cannot call me indirect because I take the time and effort to explain things, or that my posts are separated by many posts from others. Ratch

 

caveman,

It doesn't matter whether charges move. You can define it by what would happen if the resistance was there.

Sure, that is what measuring with a voltmeter does. It gives I*R a meaning by allowing the charge to move through the voltmeter and measuring it.

This is a standard physics method similar to the concept of using a test charge in a field to define the field strength.

Right, but until I*R is in place with a moving charge, voltage it is difficult to measure directly.

I was already aware of this technicality, but I'm not going to argue semantics. You know what I meant, that's all I care about.

Yes, I did know what you meant. I was just wishing that you would refer to it by its correct name. Ratch

 

Right, but until I*R is in place with a moving charge, voltage it is difficult to measure directly.

Many things are difficult to measure, but that doesn't mean that they don't exist. My only point was that ohm-amps is equivalent to joules/coulomb. If one is defined, so is the other, contrary to your previous statement. You may only be able to define it in a very specific mathematical situation, but that doesn't mean that it is invalid.

I was just wishing that you would refer to it by its correct name.

But then I wouldn't be as clear to everyone else. The needs of the many...

 

Hi Recca,

This is what i think about EMF and potential difference.

Isn't it true that for an electric cell the EMF is the potential difference between the half cells.

Another way of looking at KVL is:
Total EMF = Total PDs (?)

AFAIK EMF is a misnomer.
Potential difference is just what its name implies. If taken across a resistance a small change in resistance will change the magnitude of PD(*potential difference), but PD is not a term unique to voltage across resistances(or is it?).

Edit: I am not claiming anything here. As stated, this is what I have learned intuitively. I'd like to be corrected if I am found wrong.

Everything you have said is true.

Just one comment, however. It doesn't need a small change in resistance. Any change will alter the PD if the rest of the circuit remains unaltered.