Difference between emf and potential difference

[cupid.callin]

So you don't agree with tiny-tim that by moving a resistor we can produce current in a resistor ... because his explanation sounds very satisfactory to me!

 

[vish_al210]

I am not disagreeing with him or anyone else as the context he is proposing the theory in, is different... It is a supposition..
but if you connect one end of the ac mains to one end of an ammeter and the other end of the ammeter to one end of a resistor, leaving the other end of the resistor and the other end of the AC mains open, you would find that no current flowed through the ckt. and that no real power was consumed...
I am also perplexed by this.. but reality s___s...It never let's our million dollar plans realize...
Or else regardless of you leaving a switch open or in off condition, the power would still be consumed... and that would not be a good thing for ur bill and the energy conservationists..."Reality is generally fluid...until we fall face first on it... then we reaise that it actually is rock hard..., alas realisation is a little too late always..." by me...

 

[Studiot]

Tiny Tim you are correct, many thanks for pointing me in the right direction.

Here is a scanned extract.

 

[tiny-tim]

Thanks, Studiot.

I still get really confused about emf …

people keep saying that there's a difference between emf and pd, but I've never managed to grasp it.

 

[Studiot]

I have been doing some calculations, based upon what you said, about moving a dipole from infinity to A_B in the field of an isolated charge.

I have no time tonight as it is past the witching hour here, but am quite clear about the difference between the terms. I was trying to be too clever introducing potential fields and tripped myself up.

The difference really has nothing to do with fields. I will post again.

 

[facenian]

Here's another approach: Consider that the emf concept refer to electric circuits only therefore it is not a general concept from the physicist stand point,ie, electromagnetic theory. EMF is usued in the cuasi-stationary regime and can be defined as:
emf=\oint\vec{E}\cdot\vec{dr}.
Now if we remember
\vec{E}=-\nabla\phi-\frac{\partial\vec{A}}{\partial t}
so if we consider only electromagnetic forces the only source of emf is of magnetic origen.
Now if we consider as E in the integral the total force per unit charge in the circuit(and this would includ quemical forces in a battery for intance) we may call this an "efective electic field" if you wish you may reconcile what is ussually defined as emf in more elementary textbooks.

 

[Loro]

No one has actually said what I think is most important, so I'll add it:
I think the difference has got a lot to do with fields.

Emf is defined in general as a line integral of E*dL along a path. If we choose two endpoints we may or may not get the same emf when we compute it along two different paths.

If the emf calculated between two points is independent of path, we call that E-field "conservative". In such fields we define potential at a point - as an integral of E*dr from infinity to that point. Then we can talk about the potential difference (= voltage) between two points A and B as an integral of E*dr from A to B.

If the emf calculated between two points depends on the path (this means that it can be nonzero, around a closed loop), then that E-field is called "non-conservative". In this case we can't talk about potential differences, because potential is not well-defined (You could get different values for different paths of line integration from infinity to a point).

An example of a non-conservative E-field is an E-field caused by a changing magnetic field passing through a loop of wire. This magnetic field induces an emf around the loop. And in this example we can't talk about potential differences, because the E-field is non-conservative.

But if there are no changing magnetic fields around, all E-fields turn out to be conservative.

So all potential differences are emfs, but not all emfs are potential differences. Potential differences make sense only in conservative fields.

-----------------------------
But in this case of the battery we're dealing with conservative E-fields, so all emfs are in fact potential differences. But there's a distinction between the voltage measured between the electrodes, and this "ideal voltage" which would be there if it had not been for the internal resistance. They needed another name for this ideal voltage so they called it emf.

Perhaps it's been called that because when we talk about this "true" voltage we actually talk about the potential difference between the electrodes - two physical pieces of metal. And the "ideal" voltage is not a potential difference between any two physical points.
It's a voltage of an ideal voltage source, which in parallel with the internal resistance, can model a non-ideal voltage source.

 

[Studiot]

OK first a little bit of history and some definitions.

There are many quantities in Physics that qualify for the term "Potential".
Potential difference simply means exactly that. The difference between potential A and potential B
ie P_A - P_B.
In what follows I will use the term voltage to mean the difference in electrical potential and suppose we can identify such a difference between two points or terminals A and B, in some part of an electrical circuit or arrangement.

Now for the history part. The term PD used to stand for Potential Drop, not Potential Difference, when I first learned it.
So we actually had three terms, EMF, and two confusingly with the same initials, PD. Both potential drop and EMF referred to an identifiable voltage as above.

So why were there two types of voltage identified?

Well imagine we isolate the section of circuit, a bit like a free body diagram in mechanics, and ask the question what happens?

1) If the voltage remains on the isolated section, without the rest of the circuit in place, the voltage is an EMF.

2) If the voltage disappears from the isolated section, ie is only present by virtue of the rest of the circuit then it is a potential drop (PD)

So, for instance,

1)Take a 9 volt battery away from a circuit and there is still 9 volts between the battery terminals.

2)Take a resistor with 9 volts across its ends, in circuit, away from a circuit and all voltage disappears.

 

[cupid.callin]

So does it mean that EMF is self sustaining but PD is caused due to something with emf?

 

[Studiot]

In essence, yes.
But by self sustaining, that just means that whatever causes an EMF voltage is not part of the circuit.

 

[cupid.callin]

OK!
Now i understand EMF a lot better that before ...

Thanks a lot sir,
you've been a big help

 

[Studiot]

There are many implications to this simple viewpoint, generally consistent with the previous posts by others.
In particular the energy view by Dadface is worth discussing further.

 

[facenian]

In essence, yes.
But by self sustaining, that just means that whatever causes an EMF voltage is not part of the circuit.

I have to agree with that, emf are ussually caused by forces allien to the(eletromagnetic) field and are related to things like "external forces" and "source of energy".
I said ussually because it is not always so, for intance if you suddenly cancel the magnetic flux concatenated by a circuit there appears an emf induced according to Faraday's law.
So all this is really confussing not eassy to grasp

 

[Studiot]

There's always one isn't there?

Magnetically sourced EMFs are more difficult to handle, because you have to remove a whole loop (or loops) from the circuit, not just a two port network, to complete the task I described.

A further corollary of 1 and 2 is as follows

1) If you place an EMF in another (different) circuit it will still be the same voltage. So if you connect a 9 volt battery to a different circuit, there is still a 9 volts between its terminals. This is because the voltage is determined by the EMF alone.

2) If you place the resistor with 9 volts across it into a different circuit you will find the voltage across it is, in general, not now 9 volts. This is because the votlage is determined by the rest of the circuit, as well as the resistor.

 

[vish_al210]

In essence, yes.
But by self sustaining, that just means that whatever causes an EMF voltage is not part of the circuit.

I am a little confused with your view point... Please do clarify...
If a battery is a source of EMF, then well it is a part of the circuit.
Every operational ckt has a source of emf attached to it. If there is a transformer then too the primary is a part of the circuit, though electrically isolated. (By electricaly isolated is that faults from the ckt are not observed as is by the primary, but only the electro magnetic effects are felt by the primary, the isolation acts as a surge absorber.)
So how can it be said that the emf source is not a part of the circuit.P.S.: Does a floating capacitor or resistor (by floating I am referring to one end of the element connected to the ckt an the other end left open) consume power, is there a current flow (event 10% of what it is when you connect the element comletely) when connected to an AC ckt?
I don't think so. Please do correct me if I am wrong.

And EMF is not self sustaining, so as to say it dies out or is consumed over time depending on the power stored and rate at which it is consumed.

 

[Studiot]

then well it is a part of the circuit

Not if you remove it!

Of course you no longer have that circuit.

 

[Kaldanis]

This is a simple way that it was explained to me, is it correct?

The EMF is the total value of the power sorce, for example, when a 12V battery is connected to a circuit you would say the EMF is 12V.

The terminal potential difference is the voltage left once you remove the voltage lost to the internal resistance of the battery. So if the 12V battery loses 1V to internal resistance, then the t.p.d. will be 11V?

 

[Whakataku]

To simplify and sum it up, the emf of a cell is the total potential difference the cell can produce around the circuit, including any potential wasted in driving the current through the cell itself.

V = I*R

ε = I *(R+r)
I = V/R
ε = V/R *(R+r)

Where V is the potential difference, ε is the electromotive force, R is the resistance of an external load resistor, and r is the internal resistance of the electrochemical cell.

Say if the resistor of load resistance is close to 0 but not entirely 0, say 0.0000000000000000000000000000000011 Ω,
then ε ~ V *(r)

Correct me if I'm wrong. >: )