# Electromotive force and potential difference

1. Oct 14, 2011

### Ravalanche

hi everyone, i would like to ask some quick questions that i find confusing.
first, electromotive force is the energy supplied by a battery / power source, to ''push'' the charge through a circuit right? its unit is in Volts.
then, potential difference is defined as the work done to push 1 Coulomb of charge from point B back to point A. and its units is also in Volts.
So imagine a complete circuit. when a charge leaves the battery, lets say it contains 9 Joules of electrical energy / charge ( this is the E.M.F) then when it crosses a resistor. it drops from 9 Joules of Electrical energy to 6Joules. the drop is 3V, therefore potential difference is 3V.
now what do i call the remaining 6 Joules of electrical energy per charge? is it stilled called the EMF? I asked my teachers, they say its just simply called voltage. which i don't really understand what is the definition of voltage.

i've read wiki but it doesn't help , the term voltage is confusing. i'm not sure they are referring to EMF or P.d. thanks for the help in advance and i apologize for any mistakes in my statement as I am still learning.

2. Oct 14, 2011

### sankalpmittal

Hiii !
Potential difference (PD)is defined as the amount of work done per unit positive charge (we assume positive but they are electrons !) on bringing it from infinity to the particular location.
We assume infinity because it defines a system to be simplified. It is also defined as the difference of potential energy or charge difference per coulomb of charge between two points in an electric circuit.

Electromotive force (EMF)is the potential difference existing between two terminals of the battery in a closed circuit , when no current is drawn from it. It is also defined as the amount of electrodynamic force (or energy supplied by battery) required to drive 1 coulomb of charge across a circuit. In other words you can also call it - Total Potential Difference.

EMF > PD , this condition is always true in a circuit.

See , suppose you have a closed circuit. In positive terminal you have 6 units of charge and in negative terminal you have 15 units of charge. The potential difference existing is 15-6 = 9 V which is electromotive force . So EMF = 9 V.

The difference in Voltages is equal to Work done on unit charge / Total Charges

V2-V1 = W/Q
OR
V=W/Q we assume PD

where V = V2-V1

Now

We knew EMF = 9 V

When driving a charge(electrons they are we just assume charge.) suppose due to a resister we do X J of work an there is PD - 3 V.
Then remaining = 9-3 = 6V. This drop is manifested in form of energy.

Always remember that EMF is total PD.

EMF = Total drop in voltage.
or
EMF = V1+V2+V3 +.....Vn

In your case , suppose energy used up = work done = 9-6 = 3 J
So V=W/Q
3=3/Q ==> Q=1 coulomb
You haven't assumed what amount of PD the circuit had when it was closed. Well if 6 J of energy remains then you cannot call it EMF. Its total voltage.

EMF = W/Q , total energy supplied is 9 J , So EMF = 9/Q say Q=1 or EMF = 9 V

If you assume that total voltage drop along a complete circuit is 9 V.

Last edited: Oct 14, 2011
3. Oct 14, 2011

### Ravalanche

Thanks for the detailed explaination. However, still i'm confused by the term voltage. What does voltage refer to? The p.d or emf?

4. Oct 14, 2011

### Delta²

I think your confusion is because we use the term voltage for

1) Voltage between two points A and B $$V_{AB}$$ which is the potential difference between these Points. It is $$V_{AB}=V_A-V_B$$ where $$V_A$$ is the potential at point A and $$V_B$$ the potential at point B. Both potentials $$V_A,V_B$$ and voltage $$V_{AB}$$ are measured in units of volts. The potential at a point A $$V_A$$ is the amount of work needed (or provided by the electric field) to drive a unit charge from point A to infinity.

In your example where the charge voltage drops from 9 to 6 it is actually that the charge is moved between one point with potential 9Volt and another point with potential 6volt. The potential difference between these points is 9V-6V=3V.

5. Oct 14, 2011

### Ravalanche

so you are saying that if at a particular point, a charge has lets say again 6 Joules of electrical energy, then I can say that the potential at that particular point is 6volts?
or should I say the voltage at that point is 6volts?

6. Oct 14, 2011

### Delta²

Both are correct but more accurate and correct is the first that is to say that the potential at that point is 6volts.

7. Oct 14, 2011

### Ravalanche

alright, thanks alot.
cheers.

8. Oct 14, 2011

### sophiecentaur

The term Voltage (although we use it all the time) is too sloppy to use in a formal argument. People also use that horrid term "Amperage" (ugh!) when they mean Current. (Also 'footage', meaning length - but never 'pintage' for amount of beer consumed?)
Potential difference is the only really meaningful term and it is strictly defined in terms of energy transferred by a charge. Hence when there is a PD of 1, volt, every coulomb that flows, transfers 1 Joule of energy.
1V = 1Joule per Coulomb
PD is, OF COURSE always a difference in potential between two points. 'Voltage' may be used as shorthand to refer a PD relative to Ground.
emf is a term which refers to an 'ideal' potential which would be available from a source of electrical power if there were no losses of energy inside the source. Once current starts to flow - supplying energy to a circuit, there is always energy lost inside any source (internal resistance) So, as has been said already, you can sometimes measure this 'emf' if you take no current. It may not be actually possible in all instances, though, because you can't actually 'get inside' the works.

9. Oct 14, 2011

### Ravalanche

haha, yeah its true, I see the term voltage everywhere when dealing with electricity, its kinda confusing though, but thanks to you guys I finally got it.
btw, another side question, thought i would post here instead of starting a new thread.
anyway, how does current actually flow? we say that current is the flow of charge right? and we group these charges into couloumbs with consist of electrons. But I know that its not that simple. Can someone please share their knowledge on how does current actually ''move''? current flows opposite to the direction of electron flow, in electronics, holes also move in the opposite direction, does that mean current = holes?

10. Oct 14, 2011

### sophiecentaur

I always try to think of Current in the abstract / bulk sense and not to consider the actual mechanism at a lower level. This is because current can actually take different forms. At the end of the day, in electric circuits, it has to involve electrons moving into the end of wires at one end and leaving the other end. But one couldn't rule out a circuit in which part of the current path consisted only of positive ions flowing through some exotic fluid - or even positrons flowing through a vacuum or a tube of 'ionised' anti hydrogen. Who know the future of electrical circuitry?

In metals, there are many electrons (some people call them Valence electrons and there is at least one per atom) that are very mobile within the material. They are in constant random thermal motion, at normal temperatures, with a very high RMS velocity. When a current 'flows' there will be a net movement of all these electrons, at a 'drift speed' of only a mm or so per second. So it is not a helpful model to think of a 'stream of electrons' moving through the metal, any more than it is useful to think of a few tens of thousand immigrants flowing into a country when the airports, roads and railways are carrying possibly millions of travellers in and out every year. The 'net' migration figure is only a small part of what's really going on and wouldn't help you plan a transport system. So I don't like the 'water' analogy and nor do I really think much of the 'bicycle chain' analogy, in which a stream of well identified electrons are responsible for the 'flow' of charge. When the total net flow of electrons is added up, you can say that a certain charge has passed and we use the Coulomb as our unit. I reckon that's enough.

Conventional Current is a useful, intermediate, level at which to discuss a huge range of situations. You only need to 'change gear' and discuss mechanism (such as the "hole" conduction, you mentioned) as and when it becomes relevant. After all, it doesn't help an Engineer who's designing a turbine to get down to the level of individual gas molecules at the same time as calculating the best blade profile. Bulk properties are very often best dealt with at t heir own level.

If not, then why not look at the fundamental particles of which the actual electrons are made?
Ohm's law with Quarks could be a bit of a headache.

11. Oct 14, 2011

### Ravalanche

i see, i used to think of electrons as sort of like a chain with beads but closely packed,so when one ''vibrates'', the others will obtain its energy and thus transfering it from one another. still, thanks alot for the info, that helped alot.

12. Oct 14, 2011

### sophiecentaur

There are all sorts of models. Some of them take you further than others. I think that the 'school' model is potentially counter productive. So many 'Science' teachers know so little about this but feel duty bound to promote this model.

13. Oct 14, 2011

### antistrophy

What model would you suggest for relating voltage and current to things already familiar to a high school student?

14. Oct 14, 2011

### Ravalanche

meanwhile, can a single coulomb of charge be extremely high in electrical energy ( potential given a single point ) while the velocity of it is low ( current? since current is charge per second, its sort of like speed?). is this possible? high voltage but low current. Its hard to imagine though. because if it possess high electrical energy, wouldnt that also mean it has high kinetic energy so that it moves fast too?

15. Oct 15, 2011

### sophiecentaur

Rate of charge transfer is not the same as speed. The units are not the same and the analogy is not good.
This will rankle with some readers but I recommend using the maths as your model as soon as you can. Semi-mechanical analogies are no more valid. They are worse, in fact.

16. Oct 15, 2011

### Ravalanche

analogies gives u a better idea, but it does not fully describe the real situation right?

17. Oct 15, 2011

### Staff: Mentor

You need not be confused between the terms "electromotive force", "potential difference", and "voltage" -- these terms are synonymous. All are measured in volts. The term "potential difference" is most descriptive, though "voltage" is good shorthand and is understood by everyone to mean "voltage difference". Where just "voltage" is spoken of, there exists an accepted reference point conventionally considered zero volts (or ground).
That's an easy one: voltage is that which is measured in volts.

You will wear yourself out trying to imagine a distinction between the terms EMF, potential difference, and voltage. There is no difference.

18. Oct 15, 2011

### cabraham

Actually, emf has a distinct meaning vs. "voltage drop" or "pd". Consider a 12V battery powering an 11 ohm lamp, with a 1 ohm cable. The total resistance is 12 ohms, & the current is 1 amp.

Tee 12V across the battery terminals is an "emf" because charges transiting inside the battery are gaining energy.

But, the 1 volt across the cable is not an emf, but rather a drop. Voltage drop & emf are not the same. In the cable, charges are losing energy.

They have the same units, i.e. volts. A drop is energy per charge lost. An emf is energy per charge gained.

But both the voltage drop & the emf can be called a "potential difference". PD is a broad term, whereas "drop" & "emf" are specific.

I hope this helps.

Claude

19. Oct 15, 2011

### Staff: Mentor

Try putting your tongue across it. Yup, that's a potential difference.
I didn't mention voltage drop, though it's really nothing more than "voltage" with the substitution of "drop" for the negative sign. Nothing new there. It's still a potential difference.

20. Oct 15, 2011

### cabraham

I already stated that an emf & a voltage drop are both a potential difference, pd. As far as the tongue goes, it is a pd, but NOT an emf. The emf is the energy gained by charges inside the battery. You're not listening to what I'm saying.

Again for the record, a PD is a very broad term. The emf produced by the battery is a PD. The voltage drop in the cable is a PD as well. PD is the broad term.

Voltage drop is a PD incurred by charges losing energy, such as in the cable resistance. EMF is a PD incurred by charges gaining energy as in the redux action in the battery.

Voltage drops & emf are both a type of PD. But emf is not the same as a drop. That is what I wish to clarify.

One more point. If the battery is being recharged, then the 12V across its terminals are not an emf. The recharger is outputting the emf, & the battery PD is a drop. Just thought it deserves mention. Comments/questions welcome.

Claude