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

  1. Jul 15, 2014 #1
    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 learnt 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.
  2. jcsd
  3. Jul 16, 2014 #2


    User Avatar
    Science Advisor

    Welcome to PF.
    Is this a homework problem?
    Can you draw the circuit diagram showing how you connect the cells, potentiometer and galvanometer.
  4. Jul 18, 2014 #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:

  5. Jul 18, 2014 #4


    User Avatar
    Science Advisor

    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.
  6. Jul 19, 2014 #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.
  7. Jul 20, 2014 #6


    User Avatar

    Staff: Mentor

    So, you've worked this out now?
  8. Jul 21, 2014 #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?

    Attached Files:

    • PF.jpg
      File size:
      17.6 KB
  9. Jul 22, 2014 #8


    User Avatar
    Science Advisor

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

    Attached Files:

  10. Jul 24, 2014 #9
    Thank you. :approve:
  11. Jul 24, 2014 #10


    User Avatar
    Science Advisor
    Gold Member
    2015 Award

    It may help to think in terms of Kirchoff 2 in problems like this one.

Have something to add?
Draft saved Draft deleted
Similar Discussions: Finding the ratio of the emfs of two cells by a potentiometer