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Potentetiometer - position of slider

  1. Jul 29, 2013 #1
    1. The problem statement, all variables and given/known data
    https://www.physicsforums.com/attachment.php?attachmentid=60570&stc=1&d=1375128608
    Thew circuit shows a 10KΩ potentiometer with a 5KΩ load. determine the position of the slider on the 'pot' when tthe voltage across points 'XX' is 3V


    2. Relevant equations



    3. The attempt at a solution
    Vi=IR
    I=V/R = 12/10x10^3 = 0.0012A = 1.2mA
    Vo=IR
    I=Vo/Io = 3/5x10^3 = 0.0006A = 0.6mA
    Vi/Ri+Ro = 9/10x10^3 + 5x10^3 = 0.6mA
    VD Across Ri =Vi =RiVi/Ri+Ro = RiVi/R
     

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  2. jcsd
  3. Jul 29, 2013 #2

    rude man

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    ?
    Let x = fraction of distance from ground to wiper.
    Call the potentiometer resistance (10K) = R1 and the external resistor (5K) = R2.
    Call 9V = Vi and wiper voltage = Vo = 3V.
    Then redraw your circuit so that Vo/Vi consists of a voltage divider between two equivalent resistors.
    What is the equivalent resistor going from Vi to Vo?
    What is the equivalent resistor going from Vo to ground?
    Solve for x.
     
  4. Aug 3, 2013 #3

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  5. Aug 3, 2013 #4

    mfb

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  6. Aug 4, 2013 #5

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  7. Aug 4, 2013 #6

    mfb

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    That is right.
    I would align the two resistors to be in the same column, but that is just a graphical thing and does not influence the electric properties.
     
  8. Aug 4, 2013 #7
    so now to answer the question

    I need to find the new parallel resistence which is (R_1 R_2)/(R_1+R_2 )
    =33x10^3 ohms

    Then I think I have to put 3 V in the equation?
     
  9. Aug 4, 2013 #8

    mfb

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    The parallel resistance should depend on the position of the slider, and don't forget the third resistor in your analysis.
    What do you mean with "put 3 V in the equation"? Just do it.
     
  10. Aug 4, 2013 #9

    rude man

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    Your diagram is correct. But the parallel resistance is obviously a function of p.

    yes, put 3V at the output and solve the whole thing for p.
     
  11. Aug 4, 2013 #10
    I think somebody needs to get the glove puppets out :)
     
  12. Aug 6, 2013 #11
    Vl=RoVi/R1+R2

    3=Ro•9/(10•10^3)+(5•10^3)

    = 5k ohms
     
  13. Aug 6, 2013 #12

    mfb

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    I don't understand where that formula comes from. And where did you define all those parameters?
     
  14. Aug 6, 2013 #13
    Vl = 3 volts across the load
    Ro = what I'm trying to find
    Vi = voltage on the circuit
    R1 = 10k ohms
    R2 = 5k ohms
     
  15. Aug 6, 2013 #14
    That formula came from my notes
     
  16. Aug 6, 2013 #15
    voltage division.

    is it correct then?
     
  17. Aug 6, 2013 #16

    mfb

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    There is no 10kOhm-resistor in your network.
    It should come together with a circuit where you can apply it. You cannot just use a formula for some circuit here.
     
  18. Aug 6, 2013 #17
    10k ohm is the max position of the slider @ 9V
     
  19. Aug 6, 2013 #18

    mfb

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    Sure, but 10kOhm is not a single resistor in your network.
    The maximal position of your slider is not the solution to this problem.
     
  20. Aug 6, 2013 #19
    If the slider was at 5k ohm would that not give 3 volts across the load?
     
  21. Aug 6, 2013 #20

    mfb

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    It would, but I think it is an accident that the formula gives the right result.
    Try to do the same calculation for 0V, 9V and 4V, and see what you get (and show what you plugged in where please). Are the results reasonable?
     
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