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Position of slider on potentiometer

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    The circuit (attached) shows a 10kOhm potentiometer with a 5 KOhm load. Determine the position of the slider on the 'pot' when the voltage across points 'XX' is 3V.

    2. Relevant equations

    Vout = Vin* (R2/(R1+R2))


    3. The attempt at a solution

    Please see attachment. Any help on where to go next would be greatly appreciated, Im completely stuck at the moment. Thanks.

    R2 = (R1 x Vout) / (Vin + 1)
    R2 = (10000 X 3) / (9 +1)
    R2 = 3000 Ohms = 3kOhms
     

    Attached Files:

    Last edited: Nov 26, 2011
  2. jcsd
  3. Nov 26, 2011 #2

    wukunlin

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    Gold Member

    I can't see point XX but I assume you mean across the 5K load.

    now, the first problem is your equation: Vout = Vin* (R2/(R1+R2))

    that is a valid approximation only if your load resistance is much larger than R2, therefore you will want to derive Vout taking load resistance into account.

    The other thing you want to recognize is that R1 + R2 = 10K.

    Now you be able to solve for R2. If you have any other questions feel free to ask
     
  4. Nov 26, 2011 #3
    Thanks for your reply.

    How come R1 + R2 = 10kOhms? sorry for probably a basic question.

    If R1 + R2 = 10Kohms, assuming R2 consists of the two resistors in parallel,

    R1 + R2 = 10
    R1 + (5+x) = 10
    10 + (5+x) = 10
    5 + x = 10 - 10
    = 0
    5x = 0
    x = 0 x 5
    x = 0

    therefore R2 = 5kOhms (5+0)

    Vout = Vin* (R2/(R1+R2)
    Vout = 9 x (5/10+5)
    Vout = 9 x 0.3333
    Vout = 3

    How can I use this to calculate the pot slider position?

    Thanks for your help.
     
  5. Nov 26, 2011 #4

    wukunlin

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    Gold Member

    okay, from what I understand from your question. You have a 10K pot right?
    a POTentiometer is basically a resistor, in the case of your question it is represented by R1 and R2. R1 + R2 = 10K, the value of R1 and R2 is manually adjustable. Since the resistance of resisters are proportional to their lengths, the shorter they are the lower their resistance. so if the slider right in the middle of the pot, you have 50% of the total resistance on each side, then R1 = R2 = 5K. If you move the slider upwards for a quarter of its total length then R1 will be 2.5K, and R2 = 7.5K. Does that make sense?

    in you question you basically connect a 5K resistor in parallel to R2. (which means R2 is JUST part of the 10K pot, not including the load)

    hope this makes more sense :)
     
  6. Nov 26, 2011 #5
    ok thanks think I understand it all a bit better now. So I can look at the 10KOhm pot as a variable rather than a fixed value as with resistor?
     
  7. Nov 26, 2011 #6

    wukunlin

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    Gold Member

    yea, think of R1 and R2 as variable resistors as long as they sum up to 10K
     
  8. Nov 26, 2011 #7
    brilliant thanks for your help there.
     
  9. Jan 12, 2012 #8
    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)

    This ok? Or have a gone the long way around to workout?
     
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