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Potentiometer position

  1. Jan 12, 2012 #1
    A circuit has 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.

    Think I'm there but not sure of my working out is excessive or I've taken the long route


    R1 in series with 2 resistors in parallel (R2 and 5k ohm)

    R1=10x, R2=10(1-x)

    R2 + 5k ohm = RC

    1/RC = 1/(10(1-x)) + 1/5.
    1/RC = (5+10(1-x))/(50(1-x))
    RC = 50(1-x) / (5+10(1-x))
    = 10(1-x) / (1+2(1-x)

    Voltage drop will be 9-3=6v
    I = V/R
    I = 6/R1 = 3/RC
    R1 = 2RC

    10x= 2x10(1-x)/ (1+2(1-x))
    x = 2(1-x)/(1+2(1-x))
    x(1+2(1-x))=2(1-x)
    x+2x-2x^2 =2- 2x
    2x^2-5x+2 = 0

    Quadratic gives
    x= 2 or x= .5

    x=0.5 (halfway point)
     
    Last edited: Jan 12, 2012
  2. jcsd
  3. Jan 12, 2012 #2
  4. Jan 12, 2012 #3

    berkeman

    User Avatar

    Staff: Mentor

    Your solution looks good. You can do a final check of your work by checking what voltage divider you get with your final configuration (5k on top, and 5//5k on the bottom)...
     
  5. Jan 12, 2012 #4
    I split the potentiometer into 2 parts , 10-R at the top and R at the bottom giving R in parallel with the 5k
    Resistance of parallel combination is then 5R/(5+R)
    The voltage across this combination = 3V and the voltage across the toppart (10-R) =
    6V
    This means that (10-R) = 2 x 5R/(5+R)
    This gives R^2 + 5R - 50 =0 with R=5 being the +ve solution....half way
    I think this is more or less the procedure you used.
     
  6. Jan 12, 2012 #5
    Thanks for both your help guys; technician: I like your method, thanks for the input:)
     
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