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Potentiometer Problem

  1. Nov 29, 2013 #1
    Hi, I have a question which goes like this:
    There is a potentiometer consisting of a 2.0V driver cell, slide wire and galvanometer with a 2.0kΩ series resistor used to find the emf of a Daniel cell. The emf of the Daniel cell is found to be 1.08 V.

    Now, approximately what current flows through the galvanometer if the protective resistor is in use and the sliding contact is moved to
    (i) one end
    (ii) the other end
    of the slide wire?
    (The resistances of the galvanometer and of the cells may be neglected)

    The answers my teacher gave me are
    (i) I=V/R
    =1.08/2000
    = 0.54 A
    (ii) I=(E-V)/R
    =(2.0-1.08)/2000
    =0.46x10^3 A

    My problem is that I don't understand why in (i), current=(voltage of Daniel cell)/(resistance of the protective resistor) and in (ii) the similar doubt. Can anyone explain to me? Thanks in advance! :smile:
     
  2. jcsd
  3. Nov 29, 2013 #2
    In the first instance the galvanometer is connected back to the Daniel cell. So its emf determines the current through the galvanometer. In the second instance the two emfs oppose each other so their difference determines the driving voltage for the current through the galvanometer. Best to make a drawing with the sliding contact first on the same side as the Daniel cell and then another drawing with the sliding contact at the other end of the wire.
     
  4. Nov 29, 2013 #3
    ok but the thing is, isn't it that the galvanometer and the 2.0kΩ resistor are on different branches? then why does the calculation involve the 2.0kΩ resistance?
     
  5. Nov 29, 2013 #4

    NascentOxygen

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    Staff: Mentor

    Please attach a neat schematic of the circuit you are discussing.
     
  6. Dec 2, 2013 #5
    Galvanometer protection

    This is for the 2nd case. The two cells oppose each other. The difference in their emfs will drive the current.
     

    Attached Files:

  7. Dec 2, 2013 #6

    NascentOxygen

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    Staff: Mentor

    It was cheah10 who I was asking to post his schematic.
     
  8. Dec 2, 2013 #7
    I got that yes, but I don't think the student knows what the circuit looks like. I was trying to make the knowledge step smaller. It seems the students needs more help than in the past.
     
  9. Dec 4, 2013 #8

    NascentOxygen

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    Staff: Mentor

    By being encouraged to come up with the circuit it is quite likely the OP would have discovered he could answer his own question after all. If spoonfed, s/he is deprived of that valuable learning opportunity.

    Quite possibly they were hoping to arrive at the answer without even drawing a schematic.
     
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