Finding the ratio of the emfs of two cells by a potentiometer

In summary: The cells have anode (E1) and cathode (E2) and the polarity of the cells is important for the circuit to work. The jockey is what connects the cells and the galvanometer measures the voltage across the jockey. The balance of the emfs is important because it affects the voltage across the jockey. In the first case, E1 is connected to the jockey and E2 is disconnected. In the second case, E1 is connected to the jockey and E2 is connected to the galvanometer. The ratio of E1 and E2 is 4:3 because the voltages across the jockey are 4 times greater when E1 is connected to the jockey than when E2 is connected
  • #1
betaleonis
8
0
I don't have a clear perception of how the following question is answered:

"Two primary cells of emfs E1 and E2 respectively are joined in series with (i) the same polarity, and (ii) opposite polarities. The combination is connected through a galvanometer and a jockey to a potentiometer. The balancing lengths in the two cases are found to be 350 cm and 50 cm respectively. Find the ratio of E1 and E2."

With a little bit of calculation I get the answer, 4:3. With the knowledge of the basic principle of potentiometer and from what I have learned from my physics textbook, the sum of the two emfs is proportional to the length given in the first case, and when E2 is subtracted from E1, the result is proportional to the length given in the second case. I don't quite understand why the emfs are added when they are joined with the same polarity and subtracted when they are joined the other way round.
 
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  • #2
Welcome to PF.
Is this a homework problem?
Can you draw the circuit diagram showing how you connect the cells, potentiometer and galvanometer.
 
  • #3
Thanks for the reply.

Yes, it can be viewed as a homework problem (though answering the question isn't exactly part of my homework).

I have roughly drawn the following circuit diagrams in reference to the question:

https://www.physicsforums.com/attachment.php?attachmentid=71449&stc=1&d=1405709293
 
  • #4
To attach a file, while posting by “Quick Reply” or “Edit” a recent post, click “Go Advanced”,
then scroll down to “Additional Options”, “Attach Files”, click “Manage Attachments”.
Click “Browse”, select your file, then click “Upload”. Wait till done, then “Close this window”.
“Preview Post” and you will see your file attached.
 
  • #5
I did that at first; but ended up opening the file in another tab, copying its URL, inserting it in that post, and deleting the attachment. :redface: My apologies.
 
  • #6
So, you've worked this out now?
 
  • #7
I know the answer to the question and how to answer it, but I don't have an intuitive explanation as to why I answered it like the way I did. I still don't know why E2 is added to and subtracted from E1 in the first case and second case respectively. Also, is there any other way to come up with a solution?
 

Attachments

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  • #8
Does the attached file help?
Think of the V axis as the potentiometer wire.
 

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  • Sum&Diff.jpg
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  • #9
Thank you. :approve:
 
  • #10
It may help to think in terms of Kirchoff 2 in problems like this one.
 

1. What is a potentiometer and how does it work?

A potentiometer is an electrical instrument used to measure the electromotive force (EMF) of a cell or battery. It works by comparing the unknown EMF of a cell to a known EMF of a standard cell, and then adjusting the resistance of a wire until the two EMFs are equal. This resistance value can then be used to calculate the ratio of the two EMFs.

2. What is the purpose of finding the ratio of the EMFs of two cells?

The purpose of finding the ratio of the EMFs of two cells is to determine the relative strength and efficiency of the cells. This information can be useful in determining which cell is better suited for a specific application or in troubleshooting any issues with the cells.

3. How accurate is the potentiometer method for finding the ratio of EMFs?

The potentiometer method is considered to be a very accurate way of finding the ratio of EMFs. The accuracy can be improved by using a high-quality potentiometer and ensuring that all connections are secure. However, it is important to note that the accuracy may be affected by external factors such as temperature and humidity.

4. Are there any limitations to using a potentiometer for finding the ratio of EMFs?

Yes, there are a few limitations to using a potentiometer for finding the ratio of EMFs. The main limitation is that it can only be used for comparing the EMFs of two cells at a time. Additionally, the potentiometer method may not be suitable for very high or very low EMF values, as it relies on adjusting the resistance of a wire to match the two EMFs.

5. Can the potentiometer method be used for all types of cells and batteries?

The potentiometer method can be used for most types of cells and batteries, including alkaline, lithium-ion, and lead-acid cells. However, it may not be suitable for specialized cells or batteries with very high or low EMF values. It is always best to consult the manufacturer's instructions or specifications before using the potentiometer method for a specific type of cell or battery.

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