# Calculating EMF of a cell using Potentiometer.

1. May 30, 2013

### harjyot

Hello,
I'm having a bit of a hard time understanding the concept of calculating the emf of a cell (e1)
By connecting it to a potentiometer.
the thing where im getting confused is why do we connect the cell as in -++-
And not -+-+.
Kindly help

2. May 31, 2013

### 256bits

Since you did not post a picture of your circuit, I found one from Wiki
http://en.wikipedia.org/wiki/Potentiometer_(measuring_instrument [Broken])

You want zero deflection on the galvanometer. To obtain zero deflection the voltage on both sides of the galvanometer would need to be equal.

Last edited by a moderator: May 6, 2017
3. May 31, 2013

### harjyot

I'm sorry but this is actually not my question. My question wasthat why do we connect the cell in the opposite way ?

4. May 31, 2013

### Simon Bridge

Guessing that the situation in 256bit's post is, indeed, the situation you are talking about... he has answered your question. vis:
You'd want to use the [-+][+-] or [+-][-+] arrangement with a galvinometer and a load if you wanted to compare the emf of two cells... which is how the method works.
The two cells try to push the current in opposite directions ... the higher emf wins.
You adjust the potentiometer until the effective emfs balance.

Note: it is not good practice for us to start answering your question until we know what you are talking about. 256bits went beyond the call trying to figure out what you were talking about and you didn't even answer his question to confirm or deny the situation.

In future, please respond to questions and try to include all the relevant information in the first post. Please do not assume that people in other countries are familiar with the exact setup in front of you.

Last edited: May 31, 2013
5. May 31, 2013

### harjyot

I'm sorry for that . I'm new to this forum and not quite well versed. My apologies. 256bits thanks for bothering so much. I appreciate it.
here's the link to the original circuit diagram :
http://db.tt/9pGzPz0W

I'm failing to understand how the two opposite emfs are balancing each other. I know it should be simple but I'd really appreciate it if someone could explain in easy terms. Only a high school student here.

6. May 31, 2013

### Simon Bridge

Do you see that the cells in the circuit are trying to push the electric current in opposite directions?

7. May 31, 2013

### harjyot

Yes. But how is the circuit getting balanced ?
from what it was looking as , to me , it seemed as if the cell we want to measure (e1)
Is connected in parallel to the potentiometer and by moving the slider we are finding that point where the potential e1 and v'(potential drop due to the wire) comes to equal.

8. May 31, 2013

### Simon Bridge

Cool - trying to find your level of knowledge here or I risk pitching the answer too high (which would be useless) or too low (which would be insulting) ... help me out here.
Do you know Kirkoff's Laws and mesh analysis?

9. May 31, 2013

### harjyot

Yep. I know kirchhoff's laws. And yes solved a few sums with mesh anaysis. Only I didn't know that the method I used had a specific name to it. Lol.

10. May 31, 2013

### Simon Bridge

OK - then I can skip ther hand-wavey "pushing current" ideas ... just do a mesh analysis on the circuit.

Using the left hand diagram ...
there are two nodes - n1 is at the R1/R2 intersection and the other is at the bottom of R2.
there are two meshes ... left and right.

Let the current through R1 into n1 by I1, and the current out of n1 through R2 be I2 ... the current into n1 from the galvanometer is I3.

Then for n1: I1+I3=I2

For the left loop, going clockwise, Vs=I1.R1+I2.R2
For the right loop, going clockwise, Vr=I2R2 (galvanometer has no resistance)

Since R1 and R2 come from a potentiometer with total resistance R, R1+R2=R

Moving the slider adjusts the ratio of R1/R2 - you adjust until I3=0 ... so R1/R2 should depend on Vr and you have calibrated the equipment.

From the mesh analysis you should be able to see how I3 varies with R1/R2.
How would this work out if you put Vr the other way around?

Repeat for the uknown EMF - call it U.

But you should be able to see it from inspection if you suppose that Vs > Vr and Vs > U ... if the volt-drop across R2 is the same as U, then the current through the galvanometer is zero.

11. May 31, 2013

### sophiecentaur

The basis idea is:
emf is measured when there is no current taken from the cell. If you connected the cells the other ways round, there would be loads of current through the cell under test. Just what you don't want!
You want the cell emf to be equal to the PD on the potentiometer wiper - so no current flows either in or out of the test cell.

12. May 31, 2013

### harjyot

That's how I thought too but simonbridge here says that current is flowing but in the opposite direction?

Simonbridge, the loop method explains it but still I' wanted a more intuitive answer to whats happening.
thanks

13. May 31, 2013

### Simon Bridge

Me? (BTW - there's a space in my name there... just sayin'... where was I... oh yes:) No I didn't!
I said that's what the cells were trying to do. I didn't say they succeeded.

I was aiming at an intuitive feel ... the current flows one way when the voltage across R2 is higher than U and the other way when U is higher than V(R2). You adjust that voltage by adjusting R1/R2. You should have that relation via the mesh anaylsis.

14. May 31, 2013

### sophiecentaur

The term "mesh analysis" can bring on the pains for someone looking for an intuitive answer. Haha. The 'no current when balanced' idea can be sufficient for most purposes and isn't scary.
There's an Ironbridge in the UK. Site of the very first (I think) iron bridge ever built. So you just could be a place not a person. Lol

15. Jun 1, 2013

### harjyot

Haha I'm sorry about the name.
lets see I will sit with my book for a while and see of what you say makes sense . I'm sure it does to you .. but to me , well that's a different case altogether.