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Voltage across a potentiometer?

  1. Oct 10, 2011 #1
    The circuit of (attached file) shows a 10 kΩ potentiometer with a 5 kΩ load. Determine the position of the slider on the ‘pot’ when the voltage across points ‘XX' is 3 V.

    I can't figure out how you would find the current first without knowing total resistance?
    Any help would be greatly appreciated thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2011 #2

    berkeman

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    Write an equation for the output voltage in terms of the wiper position. Use some variable like "x" to represent the fraction of the pot resistance that is in parallel with the output resistor. Make x go from 0 to 1, to represent that fraction (as the wiper goes from the bottom to the top of the pot. "x" multiples the total resistance of the pot....
     
  4. Oct 11, 2011 #3
    Do i not need to find the currents? I have 0.6 mA going through the 5kΩ load resistor. I'm not sure about that equation? Please help?
     
  5. Oct 11, 2011 #4

    MATLABdude

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    It may help to consider the potentiometer as two separate resistors, x*10k and (1-x)*10k. If you redraw the circuit, you now have three little resistors that may be a little easier to analyze.

    You should then be able to put the bottom resistors in parallel, and form a voltage divider (so no currents required!) The method I used means that you'll have to eventually solve a quadratic equation, but only one of the roots makes sense.
     
  6. Oct 11, 2011 #5
    If we call the resistor in series R1 and the resistor in parallel R2 and the 5kΩ load resistor R3. Can i have R1+R2=10kΩ? And because R2 has a voltage of 3V and therefore R1 has a voltage of 6V can i say R2/R1=3/6 and R2=0.5R1? I'm still a little unsure?
     
  7. Oct 11, 2011 #6

    berkeman

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    You cannot ignore the parallel R3 in your calculation. You have a voltage divider that has 6V across the top resistance R1 and 3V across the bottom parallel resistance R2//R3. Write the voltage divider equation, and use the formulatio suggested by MATLABDUDE to define R1 and R2. Then solve away...
     
  8. Oct 11, 2011 #7
    so if i use the voltage divider rule:
    V,out=V,in*R2/(R1+R2)
    Could i say:

    R1=10000x Ω
    R2=10000(1-x) Ω
    R3=5000Ω

    and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??
     
  9. Oct 11, 2011 #8

    berkeman

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    You are mixing your terms for R2. Keep R3 separate from R2 in your voltage divider equation (your first equation needs to have R3//R2 in it...)
     
  10. Oct 11, 2011 #9
    so if i use the voltage divider rule:
    V,out=V,in*R2/(R1+R2)
    Could i say:

    R1=10000x Ω
    R2=10000(1-x) Ω
    R3=5000Ω

    and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??
     
  11. Oct 11, 2011 #10

    berkeman

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    This looks to be a duplicate post? You didn't include R3 in your first equation yet...
     
  12. Oct 11, 2011 #11
    Is it like this:

    V,out=V,in*R3/(R1+R2)?
     
  13. Oct 11, 2011 #12
    Am i anywhere near?
     
  14. Oct 11, 2011 #13

    berkeman

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    No, sorry. Re-draw the circuit as suggested by MATLABDUDE. Show explicit resistors for R1 and R3, and divide up R2 into 2 separate resistors (divided at the wiper contact). Label the resistors as suggested, and write the voltage divider equation.

    Please show us your new figure with 4 labeled resistors...
     
  15. Oct 11, 2011 #14
    Ive got it, phew.

    Yes i used V,out = (V,in * (R2*R3/R2+R3))/(R1+(R2*R3/R2+R3))

    After solving an quadratic equation i found x=0.5. Phew!

    Thanks
     
  16. Oct 11, 2011 #15

    berkeman

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    Good job! :smile:
     
  17. Oct 28, 2012 #16
    i know this is an old thread but i was using it as a guide and im struggling to get down to a quadratic equation. i understand up to v,out = v,in*(R2*R3/R2+R3))/(R1 + (R2*R3/R2+R3)) but when simplifying this i get nowhere.
     
  18. Oct 30, 2012 #17

    MATLABdude

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    If you're just "following along", start from the circuit diagram and derive (note that what was in posts 6 and 7 aren't correct). When I was first learning, it took me a while to get it, but always, ALWAYS draw the circuit diagram and start from there. If you post an intermediate step or two, we can help you along.

    In this specific instance, R1 and R2 can be expressed in terms of something involving x (the wiper position of the 10k pot) and R3 is known.
     
  19. Oct 31, 2012 #18
    i have since tried this again i got x=0.54 i see earlier someone had got 0.5. not sure where i went wrong. am i right i saying i can simplify the 9(R2R3/R2+R3)/R1 +(R2R3/R2+R3) to 8(R2R3/R2+R3)/R1+1. i went on to multiply out and got to 3R1R2+15R1-37R2+15=0 before substituting R1=10x and R2=10(1-x). Sorry if this is all a bit confusing, could you tell me if this seems right or possibly where i've went wrong.


    Thanks
     
  20. Feb 14, 2014 #19
    Hi everyone I know this is an old thread but I have the same question on my HNC I got to the formulae Vp=Vs(R2R3)/(R2+R3)/R1+(R2R3)/(R2+R3) using R1 is 10x and R2 10(1-x) could someone show me the steps to find x I wouldn't normally ask for a direct answer but I'm going out of my mind and A4 sheets of paper trying to get the answer thanks
     
  21. Feb 14, 2014 #20

    gneill

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    Note: You need to use more parentheses in order to make the mathematical order of operations clear. Otherwise, strict left to right interpretation of your equation would make it incorrect. I've added a pair (in red) to your equation above.

    Note also that we cannot solve the problem for you here. That's against the rules, so you'll have to show an attempt that we can comment on and help you to sort out.

    Really, once you've got the equation and the definitions of the potentiometer R's in terms of x the rest is just algebra. A bit tedious, perhaps, but just algebra. As a suggestion I would plug in some of the hard numbers at this point to leave you with an equation in x alone. Since the resistances are all given in kΩ, just drop the units for the algebra. Same goes for the voltages. So the potentiometer resistance is 10 and the load resistance is 5. The input is 9 and the output is 3.

    Can you show the result after plugging in the numbers?
     
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