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Potentiometer circuit

  1. Jul 31, 2017 #1
    1. problem statement, all variables and given/known data

    1) What is the relationship between Ix
    and Rx ?
    2) Prove that on varying resistance Rx , Ix reduces to zero .

    ?temp_hash=58f9f9179e62bd430e2c624d6094b137.png

    2. Relevant equations


    3. The attempt at a solution

    Assume resistance of Galvanometer is ##r## .
    Total resistance in middle branch is ##R## , Variable resistance is ##R_x##

    Applying KVL in right loop clockwise ,

    ##-E_x - I_xr + (I-I_x)R_x =0##

    Applying KVL in the left loop clockwise ,

    ##E_0 - (I-I_x)R = 0##

    Solving the above two equations ,

    $$I_x = E_0 \frac{R_x}{rR} - \frac{E_x}{r}$$

    Now when ##R_x## decreases , the first term decreases and consequently ##I_x## decreases .Finally at a particular value of ##R_x## , ##I_x## becomes zero .The current in the Galvanometer is zero and the circuit is said to be balanced .

    Is this correct ?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Jul 31, 2017 #2

    cnh1995

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    That's not correct.
    Current through R is not I-Ix throughout.
    What happens at point d?
     
  4. Jul 31, 2017 #3
    Sorry .

    The total resistance R in the middle branch can be broken up in two parts such that ##R_x + R' = R## .

    Applying KVL in right loop clockwise ,

    ##-E_x - I_xr + (I-I_x)R_x =0##

    Applying KVL in the left loop clockwise ,

    ##E_0 - (I-I_x)R_x - I(R-R_x) = 0##

    Is it OK now ?
     
  5. Jul 31, 2017 #4

    cnh1995

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    Yes.
     
  6. Jul 31, 2017 #5
    $$I_x = \frac{E_0R_x -E_x R}{RR_x -R_x^2+Rr}$$

    Denominator will always be positive since ##R >R_x## .

    As ##R_x## decreases ,first term in numerator decreases .As a consequence ##I_x## decreases . Finally for a particular value of ##R_x## , ##I_x## becomes 0 .
     
  7. Jul 31, 2017 #6

    cnh1995

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    Looks good to me.
     
  8. Jul 31, 2017 #7
    Thank you .

    Just to keep things simple , I think there was no need to include resistance ##r## .The resistance of Galvanometer could have been neglected .

    The final expression will be $$I_x = \frac{E_0R_x -E_x R}{RR_x -R_x^2}$$

    Is there anything you would like to add to the analysis I have done ?
     
  9. Jul 31, 2017 #8
    No . I think resistance ##r## has to be included . If resistances of both battery ##E_x## and Galvanometer are neglected then potential difference across points 'ad' would be equal to emf ##E_x## at all times . Including ##r## would mean that only when ##I_x = 0## the potential difference across 'ad' would equal emf ##E_x## .
     
  10. Jul 31, 2017 #9

    cnh1995

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    As far as the potentiometer experiment is concerned, you need not use r as the null point does not depend on r.
    But then, Rx=R becomes an invalid condition (two ideal unequal voltage sources in parallel).
    So, I believe you should include r to make the circuit practical and free from contradictions such as infinite current and unequal voltage sources in parallel.
     
  11. Jul 31, 2017 #10
    Thank you
     
  12. Aug 2, 2017 #11
    From this equation I can also infer that just when the product of current from battery ##E_0## times resistance ##R_x## becomes equal to EMF ##E_x## , current ##I_x## becomes zero and null point is achieved .

    In other words , ##I_x =0## , as and when ##E_x=IR_x## condition is achieved .

    Can we put it this way ?
     
  13. Aug 2, 2017 #12

    cnh1995

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    Yes.
     
  14. Aug 2, 2017 #13
    I hope you are not reasoning it backwards in the sense , when ##I_x =0## , then ##E_x=IR_x## .

    It's the other way round .
    It may look like I am over thinking this , but actually I need to explain potentiometer to someone . So I am just trying to simplify the analysis as much as I can .Since you are an expert in circuit , I just want to make sure my thinking is correct :smile:
     
  15. Aug 2, 2017 #14

    cnh1995

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  16. Aug 3, 2017 #15
    Thanks .

    One last clarification .

    In the figure given in OP , as we move the jockey along potentiometer wire , current ##I## in the primary circuit changes .

    But in the textbook I have to refer states that , there is also a rheostat in the primary circuit maintaining a constant current flow in the potentiometer wire all the time (while jockey moves on the wire)

    Is this the way ?
     
    Last edited: Aug 3, 2017
  17. Aug 3, 2017 #16
    @gneill , do you believe a constant current flowing in the potentiometer wires is a necessity ? Is a rheostat really required in the primary circuit ?
     
  18. Aug 3, 2017 #17

    cnh1995

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    I don't see how a rheostat can maintain a constant current in the circuit. Constant current is not a requirement here. I believe it is used to adjust the potential gradient along the potentiometer wire.
     
  19. Aug 3, 2017 #18
    I think the rheostat comes in picture when we are required to take more than one reading .First time any suitable value of rheostat would work .While taking second reading , it's value could change resulting in different value of primary current .Potential gradient would change . If we are comparing EMF's , balance lenths would also change , but the ratio of EMF's would very nearly be same as the previous result . Then we could find the mean value of the results .
     
  20. Aug 3, 2017 #19

    cnh1995

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    Yes, that's how it is performed in laboratory.
    But the rheostat's function is still the same.
     
  21. Aug 3, 2017 #20
    But potential gradient is adjusted by altering current in primary circuit .
     
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