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Potentiometer question.

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A circuit shows a 10kΩ Pot with a 5k Ω load. Determine the position of the slider on the pot when the voltage across the 5kΩ load is 3 volts. The input voltage is 9 volts.


    2. Relevant equations



    3. The attempt at a solution Input voltage = 9 volts. Output voltage = 3 volts. Therefore the ratio of the resistance must be 3/9. Which equates to 1/3. So 10Ω/3 = 3.3Ω is the position of the Pot.
    Am i on the right track? Any help would be appreciated.
     
  2. jcsd
  3. Nov 9, 2013 #2

    UltrafastPED

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    If by "output voltage" they mean the voltage at the node between the pot and the load ...

    Then the 5k load is taking up 1/3 of the voltage, leaving 2/3 to the pot. So what is the resistance of the pot?
     
  4. Nov 9, 2013 #3
    6.6k ohms?
     
  5. Nov 9, 2013 #4

    UltrafastPED

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    Have you studied voltage dividers?

    Or you can just apply the Kirchoff Voltage loop rule.
     
  6. Nov 9, 2013 #5

    gneill

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    There are several different ways the components could be connected. Is it one of these?

    attachment.php?attachmentid=63809&stc=1&d=13840246056.gif
     

    Attached Files:

  7. Nov 9, 2013 #6

    UltrafastPED

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    Really creative! I just assumed (d) - though I have seen most of the variations when assisting in the circuits lab.
     
  8. Nov 9, 2013 #7
    It is circuit C. The voltage across the 10k pot is 9v. The voltage is 3 volts across the 5k resistor/load. Therefore positioning the slider at a certain point will result in the voltage output to be 3 volts. not sure how to determine this. voltage output = r1/r1+r2 which according to my calculations equals 6 kilohms. I am currently trying to get my head around kirchoffs law etc. I apologise for my lack of understanding. This is an open university course and i am struggling!
     
  9. Nov 9, 2013 #8

    UltrafastPED

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    So for circuit (c) the slider is positioned 2/3 of the way down - the 5k load is in parallel with the portion of the pot which is past the slider; that is, they both see 3 volts.

    So the upper portion of the pot drops 6 volts. See your result from message #3.
     
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