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Current in Potentiometer

  1. Dec 12, 2014 #1
    potentiometer_zps5127cf2e.png

    Based on above figure, balanced point is point N. If the jockey is put at the left or right of point N, there will be current flowing through the galvanometer. On my note, when jockey is at point L, VAL < E (where E is potential difference of cell at lower part of the circuit). When the jockey is at point R, VAR > E.

    I want to ask several questions:
    1.
    when the jockey is at L, there will be current flowing through galvanometer (let's call this P) and the direction is clockwise (in my opinion). So the current flowing through AL will be the sum of P and the current produced by upper cell X, directed counterclockwise (let's call this Q). Am I right?

    2.
    when the jockey is at N, there is no current through galvanometer. The current flowing through AN will be equal to Q. When the jockey is at R, there will be current flowing through galvanometer in counterclockwise direction with value less than P and again the current passing through AR will be the sum of both current from both cells. Am I right?

    3. Why:
    a. at point L, VAL < E
    b. at point R, VAR > E
    c. at point N, VAN = E

    Thanks
     
  2. jcsd
  3. Dec 12, 2014 #2

    sophiecentaur

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    This reads very much like homework and the answers to a,b and c can be found if you know the basic theory of the potential divider. What do you know of it?
     
  4. Dec 12, 2014 #3
    No, this is not homework. This is written in the handout given by the teacher and he didn't explain anything. But if this type of question considered as homework-type questions, feel free to move it to homework thread.

    I know the basic theory of potential divider, but I don't know how to apply it here. When connected to L, VAL = (RAL / RAB ) X. Then I don't know how to proceed.

    Can you verify whether my view for no.1 and 2 is correct?

    Thanks
     
  5. Dec 13, 2014 #4

    CWatters

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    You are basically correct in 1 and 2 .

    It might be easier to understand what's going on if you know that a galvanometer is a very sensitive instrument. Perhaps assume that the current through it is negligible (zero) compared to the current flowing around the potentiometer loop. If you do that then the galvanometer part of the circuit doesn't affect the current flowing in the potentiometer loop. This "cheat" isn't always true!

    If it takes very little current any small difference between the voltage on it's terminals will make it swing one way or the other. The direction depending on the polarity of that voltage difference. So in this circuit think of the galvanometer as more like a voltage comparator.

    Note that the +ve terminals of the batteries are connected together at point A so...

    One terminal of the galvanometer (D) is at the fixed voltage -E (with respect to node A).

    The other terminal is at the voltage of the potential divider -(RAN/RAB)X (with respect to node A).
     
  6. Dec 14, 2014 #5
    I am not sure I get your hint. Maybe you want to point out why VAN = E. Because the +ve terminals of the batteries are connected together at point A, so VA = X - E?

    One terminal of the galvanometer (D) is at the fixed voltage -E and the other terminal is at the voltage of the potential divider -(RAN/RAB)X so if the current through galvanometer is zero then -E = -(RAN/RAB)X and we get the equation when the potentiometer is at balance point. But why the current through galvanometer is zero?

    My best guess is like this:
    When the jockey is at L, VAL = (P + Q) . RAL [P is current produced by E and Q is current produced by X] is less then E so current flows through galvanometer.

    As we move the jockey to the right, the resistance of wire connected to E increases thus lowering the current produced by E and at point N, VAN = I times RAN (where I is the current produced by X) equals E. There is no potential difference between E and VAN thus no current flows through galvanometer. What I am not sure about is the increasing of E as the jockey moves from L to N. The resistance increases but the total current (P + Q) decreases because E produces less current. How can we be sure that VAN will be greater than VAL?

    Now I think I get the hint of sophiecentaur. If at N the value of E = (RAN/RAB)X, then it is obvious (by using potential divider) that VAL = (RAL/RAB)X will be less then E because RAL < RAN. But this can be said for sure only if the current flowing through potentiometer AB is constant, which is not as I have mentioned in post #1

    I am very certain I miss something. I am not sure how I can explain my confusion. Maybe I just don't understand why the current flowing through the galvanometer can be zero?

    Thanks
     
  7. Dec 14, 2014 #6

    sophiecentaur

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    If the volts on either side are the same then the PD is Zero so how could any current flow?
     
  8. Dec 14, 2014 #7
    I understand when there is no potential difference, current can't flow. So my next question is how can the volts on either side be the same?

    Thanks
     
  9. Dec 14, 2014 #8

    sophiecentaur

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    And "I understand when there is no potential difference, current can't flow. So my next question is how can the volts on either side be the same?"

    Which of those statements that you made is correct? They are either not consistent or you have not thought it through.
    If the first quote is correct then you must realise that a point can be chosen on the wire that has any potential between 0 and X volts. All you have to do is choose a point for the jockey that has the same potential as Z.. . . . . . .
     
  10. Dec 14, 2014 #9

    CWatters

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    The position of the jockey can be adjusted until they are the same.

    You haven't given values for X or E but lets work an example with X = 2E and calculate how far along the potentiometer the jockey would need to be to get equal voltage on both sides of the galvanometer...

    Earlier I wrote...

    So for both sides to be at the same voltage...

    -E = -(RAN/RAB)X

    If X = 2E then..

    -E = -(RAN/RAB) 2E

    rearrange to give

    RAN/RAB = 1/2

    So if X = 2E then the jockey would have to be half way along the potentiometer to give VAN = E

    Does that make sense?
     
  11. Dec 16, 2014 #10
    The first quote is correct but I don't realise that a point can be chosen on the wire that has any potential between 0 and X volts.

    Yes, that makes sense.

    Last question: what happenes if the jockey is located at point B? By potential divider equation, E = X which is impossible since E < X

    Thanks
     
    Last edited: Dec 16, 2014
  12. Dec 16, 2014 #11

    sophiecentaur

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    How can you claim to know the theory of the potential divider if you don't see how any potential can be obtained on a continuous length of resistance wire? That is the whole basis of the metre wire potentiometer. Did you look it up anywhere else but here?6
     
  13. Dec 16, 2014 #12
    Yes, I read books and also web but I didn't find the answer I needed (or maybe I missed it or maybe I didn't understand it due to lack of knowledge). What I know as the basic theory of potential divider turns out only as a small fraction of what you claim as theory of potential divider. I admit it is a pretty sad fact.

    What I know about potential divider is two series resistors (R1 and R2) connected to power supply V. The voltage across R1 can be found using: R1 / (R1 + R2) V. Increasing the value of R1 will decrease the current thus decrease the voltage across R1.

    I just realise from your post that the same principal can be applied to potentiometer. But I have pointed out the thing I confused on my earlier post. Let me write it again:
    I learned potentiometer circuit by dividing it into two parts, lower part and upper part. Upper part consists of cell and potentiometer and lower part consists of part of potentiometer, cell, and galvanometer. The current produced by upper part will be constant while it is not for the lower part because moving the jockey to the right will increase the resistance of lower circuit thus reducing the current.

    Let me use the above figure. If the circuit is only the upper part, it is easy for me to understand (by potential divider theory) that moving the jockey from L to N will increase the voltage across it (VAL < VAN). This is due to constant current produced by the cell. By ohm's law, increasing the resistance means the increase of potential difference.

    Then, I wrote:
    There is mistake at the first bold part. "What I am not sure about is the increasing of E as the jockey moves from L to N" should be "What I am not sure about is the increasing of voltage across potentiometer as the jockey moves from L to N"

    Can you please help me addressing the question?

    Thanks
     
    Last edited: Dec 16, 2014
  14. Dec 17, 2014 #13

    CWatters

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    E is the voltage of the lower cell and X is the voltage of the upper cell. Neither change when you move the jockey so they are never equal.

    When the jockey is at point B the voltage on the top end of the galvanometer VAL = -X (with respect to node A). The calculation is as follows..

    VAL = - (RAL / RAB ) X
    but at position B, RAL = RAB so
    Val = -X

    The voltage on the other terminal is always -E (with respect to node A).

    So at position B one side of the galvanometer is -X and the other -E (both w.r.t node A).
     
  15. Dec 17, 2014 #14

    CWatters

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    That's not correct.

    When you move the jockey to the right the resistance of the "top part" (between A and the Jockey) increases, while the resistance of the "bottom part" (between Jockey and B) decreases. So the total resistance between point A and B remains constant. Therefore the current through the potentiometer remains constant.

    Note: I am assuming that negligible current flows through the galvanometer. That's not always the case but it's a reasonable assumption to make.
     
  16. Dec 18, 2014 #15
    Ok. Thank you very much for all of you
     
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