EMF and voltage the same ? NO ? How ?

[nishantve1]

Okay so are EMF and Potential Difference the same thing ? I know they are now but I am not able to understand how ?
I came through this Derivation today if someone could explain it to me it would be great

The figure shows a battery connected to a resistor
V(A) - V(B) = V(b) = ε (emf)

what the Derivation says is if we Remove the resistor R the resistance becomes infinitely large (HOW?) . So R = ∞ therefore the current
I = V/R = 0 So the current flowing would be zero

Now , the derivation says that there is always an internal resistance (r(i)) present inside the battery . So now if we again Connect the Resistor R
so ε = I(R+r(i)) (HOW? IS OHM's law used here ?)
Further it says , now V(b) = IR = ε - I(i) (Why is V(b) not I(R+r(i)) ? )
So the ε is greater than the V(b) . Please someone explain this Derivation .

 

[Zubeen]

The first fault here is that when the resistor is removed the resistance becomes 0 and not infinite. Please read it again. As the resistance becomes 0 an infinite amount of current flows by- I=V/R ; R=0 => I= infinite.
That is why, such conditions lead to short circuit. As the large current produces large heat by- H=I^2 * RT.
Now further,
See a movie in your mind as i do tell you -
Circuit as you attached here. (Note that battery attached her and normally used in theories is a battery that provides continuous and steady potential difference.)
The battery provides potential difference because of the difference in the charges of the 2 electrodes of the battery.
In that circuit see the current flowing, as it flows the -ve charge from the -ve electrode flows into the +ve electrode and hence after a long time the when the charge difference between them will become 0 because of the flow of the -ve charge, the battery's potential difference will also become 0. But as mentioned earlier, the potential difference of the battery must be constant.
For potential difference to be constant, there must be some force inside the battery that must be bringing back the negative charge to the -ve electrode and hence doesn't let it to nullify the +ve charge on +ve electrode. The work done per unit charge by this inside force is EMF.
This all EMF work occurs in the liquid of cell, but the wires that act as elecrodes also have resistance thus till the current comes to the terminals of battery it has to do work against that resistance therefore its output potential (V(b) in your diagram ) will be lesser than EMF.
Potential difference of a cell is the potential difference across the terminals of the cell.
Further, V(b) is not = I(R + r(i)) because it is the potential difference at the liquids of cell and not at the terminals of the cell.
---------
Zubeen

 

[nishantve1]

Thanks for the answer but I am still Kinda Confused . Here I have Walter Lewin's Lecture on the same thing
http://www.youtube.com/watch?list=PLABFEA1D1AC3EF062&v=RQX8I9ZWtPQ&feature=player_detailpage#t=446s
The video will start where he starts Explaining the same thing and it goes for like 4 to 5 minutes . What do you say about the stuff he says .

 

[Doc Al]

The figure shows a battery connected to a resistor
V(A) - V(B) = V(b) = ε (emf)

Not exactly. The EMF is the chemical potential difference of the battery. The voltage across A to B will equal the EMF minus any IR drop due to the internal resistance of the battery.

what the Derivation says is if we Remove the resistor R the resistance becomes infinitely large (HOW?) . So R = ∞ therefore the current
I = V/R = 0 So the current flowing would be zero

They mean if you disconnect the resistor, leaving a gap in the circuit. No current will flow.

Now , the derivation says that there is always an internal resistance (r(i)) present inside the battery . So now if we again Connect the Resistor R
so ε = I(R+r(i)) (HOW? IS OHM's law used here ?)

Yep, just Ohm's law. The emf drives the current through the total resistance.

Further it says , now V(b) = IR = ε - I(i) (Why is V(b) not I(R+r(i)) ? )

V(b) is the voltage across R, so V(b) = IR (Ohm's law). It's also the voltage across the battery, so V(b) = ε - Ir(i). (Ohm's law again.)

So the ε is greater than the V(b) .

Sure. The effective voltage across the battery is reduced by the internal resistance.

 

[Zubeen]

I appreciate your interest, but i gave you answer that was just the same but with less detailing. In that video the professor also discussed about the inside of a chemical cell, which falls in domain of chemistry and u seem to be a student of 12th so you will soon learn about it also.
But even after it, the thing that professor was explaining is same as i explained.
If even then you have some thing going in your mind, then post it.

 

[Zubeen]

I saw the video just for 5:30 minuets.
And just keep in mind that energy is always conserved in this universe and Voltage is associated with energy.

 

[nishantve1]

Thanks all its much clearer now . Thanks for the Detailed Explanation @Doc Al

 

[truesearch]

The essential thing to note is that EMF is where Energy is supplied to the flow of charge and can relate to chemical cells, dynamoes, solar cells etc.
Potential difference relates to where energy is dissipated.
The sum of the potential differences equals the sum of the EMFs is an example of conservationof energy.
It is energy that brings out the subtle difference between EMF and PD