Potentiometers

Discussion in 'General Electronics Chat' started by paul_alan, Dec 1, 2011.

  1. paul_alan

    Thread Starter Member

    Nov 5, 2011
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    I'm working with a potentiometer in class. I'm using a 10V source: I have 2 1kΩ resistors in series: and a 1kΩ resistor and a potentiometer set to 2.56kΩ in series. And those two series are in parallel with each other. I don't understand why the measured value of the potentiometer is 2.56kΩ but when I measure it in the circuit, it reads 1.374kΩ. Why is there such a dramatic difference when it's read by itself and when it's connected to the circuit?
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    This is a question of resistances in parallel. The total resistance can be easily calculated. Have you not been taught this?

    Either Rtotal = 1/(1/R1 + 1/R2 + 1/R3...1/RN), or for two resistances Rtotal = R1*R2/(R1+R2).

    For two equal resistances in parallel the total resistance is thus half the individual value.

    http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Resistors_2
     
  3. JMac3108

    Active Member

    Aug 16, 2010
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    Paul,

    As Adjuster says, if you measure your resistance in circuit, there are other resistances in parallel that affect your measurement.
     
  4. paul_alan

    Thread Starter Member

    Nov 5, 2011
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    We've gone over calculating total resistance. Why does the Potentiometer read 2.56kΩ when I measure it seperately (not hooked up) and when I connect it to the circuit (with no current) it measures 1.434kΩ.
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If you measure the potentiometer connected into a circuit, you will expect to get the total value that you would calculate for such a circuit.

    I hope that you are not asking obvious questions for the sake of it.
     
  6. strantor

    AAC Fanatic!

    Oct 3, 2010
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    your potentiometer is just a resistor.
    when you measure the resistance of a resistor alone, the measurement will correspond with the color bands; when you put it in parallel with another resistor, the measurement will be different. this is no different for a potentiometer.

    also note that it does not matter where you take the measurement. you say you are measuring the potentiometer; I assume that means you have your leads on the potentiometer legs. it doesn't matter; you coul put your leads on the resistor legs (resistors in parallel) and get the same reading, because these are electrically common points.
     
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  7. JMac3108

    Active Member

    Aug 16, 2010
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    Paul,

    As everyone is teling you, the potentiometer measures less in circuit because there are other resistances in parallel with it that you are including in your measurement.

    Perhaps another way of looking at things will help you understand...

    Your ohm-meter has to source a small current to make the resistance measurement. When your potentimeter is out of the circuit, all of the current flows through the potentiometer. When the potentiometer is in the circuit, the current splits up between the potentiometer and the parallel paths through other resistances in your circuit.
     
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  8. smanus

    New Member

    Jan 3, 2012
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    What confuses me is that if a potentiometer is to be used to provide a specific voltage lower than the voltage of the power source to a circuit and that circuit is placed in parallel so that it experiences the same voltage drop, the actual voltage drop will vary depending upon the resistance of the added circuit. So of what use is the potentiometer? What am I missing?
     
  9. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    A buffer.

    <pause for humerus effect>

    Actually you are missing nothing, you seem to understand the effect of adding a load. Now if you add some sore of amplifier (or a buffer) you can lower or maybe eliminate the effects of adding that load.

    Now I'm curious, so do tell us what your professor has you do next.
     
  10. Adjuster

    Well-Known Member

    Dec 26, 2010
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    In many cases the device connected to the output of the potentiometer will be of a relatively high impedance, so the loading may be fairly insignificant. Even if the loading is quite heavy, provided that it is predictable a consistent attenuation may still result. Often a potentiometer is a user control adjusted to obtain a particular result, so that the actual setting may not be of great concern to us.

    Of course, the word "potentiometer" sounds as if the device referred to ought to be something for measuring with, like other ...meter words. Do you know why?
     
  11. smanus

    New Member

    Jan 3, 2012
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    Thanks ErnieM and Adjuster. You may be confusing me with the original poster on this thread. I've been reading through the lessons in this sight and found the section on voltage dividers a bit confusing. I think I now get the idea that the metering capability of the potentiometer lets you dial in the voltage you need AFTER you put a load on it. It's actually an elegant concept.
     
  12. stahta01

    Member

    Jun 9, 2011
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    Thank you for the info on this link; the above posts made me think I never looked for info on potentiometers on this site.

    http://www.allaboutcircuits.com/vol_1/chpt_6/1.html

    Second link; that I found on potentiometer.

    http://www.allaboutcircuits.com/vol_6/chpt_3/7.html

    I am Teaching Assistant for a class and it seems like every year I get several students who have poor understanding of potentiometers. I real do not understand the cause of the problem because most of them understand voltage dividers perfectly. Note, they are at least a year past the point where the idea of potentiometers should have been taught. I told the department Chair about it and he said he would do something about it.

    Tim S.
     
  13. Adjuster

    Well-Known Member

    Dec 26, 2010
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    No, I'm sorry, but that is not what I meant. The original sense of a potentiometer was as a potential (voltage) measuring device, typically made in the form of a resistance wire set against a graduated scale, tapped by a sliding contact.

    The ends of the wire would be connected to a battery (or power supply), and the potential per unit length would be found by means of a sensitive galvanometer (current indicator) connected to the slider in conjunction with a reference voltage by finding the length along the wire at which the detector gives zero indication.

    Any unknown potential of some fraction of the total wire voltage drop could then be found by connecting it in place of the reference voltage, and finding the new length along the wire giving zero indication. The ratio of the the unknown to the known potential is then given by the ratio of the lengths along the wire giving balance in each case.

    Here is a link to a description of a potentiometer lab practical. http://ualr.edu/dcwold/phys2122/p23man/p23man.html
     
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