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Determine the position of the slider on the pot

  1. Dec 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A circuit shows a 10kΩ potentiometer with a 5kΩ load. Determine the position of the pot across point XX (taken across the 5kΩ load) when the voltage across the load is 3v


    2. Relevant equations



    3. The attempt at a solution

    I DONT HAVE A CLUE.....
    Really stumped on this one an could do with some explaining this to me in as simpler terms as possible. sorry to sound thick...
    I understand we split the 10kΩ resister into 2 so we have 2 5kΩ resisters in series and that we are looking for a ration but that is as far as i seem to understand.
    Ive gone through previous posts but seem to get confused with x10k and (1-x)10k.
    Please help as i have been on this for weeks and getting peed off with myself for not understanding it.
     
  2. jcsd
  3. Dec 19, 2012 #2

    gneill

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    You haven't stated what voltage is supplying the potentiometer (and there's no circuit diagram attached).

    The 10kΩ "pot" is divided into two pieces, but not always 5k + 5k; the potentiometer allows a continuous selection of splits, so long as the total of the two parts is always 10kΩ.

    That's why you make one of them be x10kΩ and the other (1-x)10kΩ, letting x range from zero to 1. The two always sum to 10kΩ and x "selects" the sizes of the pieces.
     
  4. Dec 20, 2012 #3
    Ok gneil sorry bout that, I have a supply voltage and at the current moment i cant upload the picture.

    Where can i read up about the x10kΩ and (1-x)10kΩ because i havent been able to find examples of it used anywhere.
     
  5. Dec 20, 2012 #4

    gneill

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    The x and (1-x) are just a convenient mathematical way to specify that the potentiometer resistance is being divided into two parts. It's like taking a plank of length L and cutting into two pieces; If one of the pieces is of length x*L, then what's left must be of length (1-x)*L. The same thing is being done only with the resistance of the potentiometer.

    Potentiometers are just voltage dividers, which you should have encountered plenty of examples of. The difference is that potentiometers allow one to manually select the ratio of the resistances comprising the divider. You might find some interesting background material if you do a web search on "potentiometer theory".

    Here's a diagram to help you along:

    attachment.php?attachmentid=54119&stc=1&d=1356016842.gif
     

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  6. Dec 21, 2012 #5
    Thanks Gneil.

    I havent really covered many pots, i'm doing my HNC via open learning and the one thing i'm finding is that sometimes its hard to visualise things without it physically in front of you.

    So, because i have a pot with a wire connected to the wiper to some other point in the circuit, a resistor that goes to ground, what we really have is a Rheostat?

    Could i not say take the 10k divide it by 9volt to give me 1111.1Ω per volt, times this by 6 to give me 6666.6 for the volt drop which would then give me the 3v i require. I would there fore have 3333.3Ωon the remained of the pot. This would give me a ratio of 3/6 which would equal 1/2 which could be expressed on a percentage scale of 0-100% as 50% or on a scale of 0-1 as 0.5.
     
  7. Dec 21, 2012 #6

    NascentOxygen

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    You can think of it as a rheostat but with 3 terminals.
    You can do that sort of thing ONLY if all of the current through the top resistance of the pot also goes through the lower part of the pot. But in your case, it doesn't. Some of the current diverts through the load, RL. You need to derive an equation for the output voltage in terms of Ra, Rb, and RL. That sounds like 3 unknowns, but it's only 2 really, because you know a relationship between Ra and Rb.
     
  8. Dec 24, 2012 #7
    how bad is this, xmas eve and im frying mt brain tyring to get my head around this question.

    Ok, i under stand the equation and the working practice behind it all but the part im struggling with is the ratio part. how do i go about working out the correct ratio of r1 and r2?
    Whatever figure i get for r2 i have to involve the r3 (5kΩ) load (resistors in parrallel) and combine them to make 1.

    Any help making me understand the ratio would be much appreciated.
     
  9. Dec 24, 2012 #8

    gneill

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    Take a look at the right-hand circuit in the image I posted. Can you write an expression for Vout in terms of E, Ra, Rb, and RL?
     
  10. Dec 27, 2012 #9
    My expression for vout would surely be using the correct formula to give me the combined resistance of Ra & Rl but since i dont know the value of Ra how do i go about working it out?
    My equation for the Vout would be E*Ra/Rb+Ra=Vout but since we already know this is 3v there must be a way of transposing this? Are my terms of thinking correct or is it back to the drawing board..
     
  11. Dec 27, 2012 #10

    gneill

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    You're on the right track. You begin by writing equations describing and relating the things you know with the things that you don't.

    I think you already see that E*Ra/Rb+Ra=Vout isn't going to work, since Ra has RL in parallel with it and will so RL will conduct some current too, perturbing the output by some amount (more current will flow through Rb, so more voltage will be dropped across it).

    The general voltage divider consists of two resistors. Here the "bottom" resistor is made up of the parallel pair of Ra and RL. What's an expression for the net resistance that this pair represents (maybe call it "Rp" for "R parallel)?

    Now write the voltage divider formula using this expression for the bottom resistor value. The top resistor remains the same, Rb.
     
    Last edited: Dec 27, 2012
  12. Dec 27, 2012 #11
    Ok since we have realistically made our Parrallel resisters as one (Rp) our circuit has now become an open circuit or output voltage under no load condition with just 2 resitors in series, Vout=E*Rp/(Rb+Rp) but isnt this the same equation as my last post?
     
  13. Dec 27, 2012 #12

    gneill

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    It's not the same... it has a different resistance, Rp, for which you can write an expression and substitute for it. What's the expression for Rp?
     
  14. Dec 27, 2012 #13
    Well the expression for Rp is Ra+Rl which would give me an equation of Vout=E*Ra+Rl/(Rb+Ra+Rl) I think...
     
  15. Dec 27, 2012 #14

    gneill

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    Nope. That's not the expression for two resistors in parallel. That's for two resistors in series... What's the expression for two resistors in parallel?
     
  16. Dec 27, 2012 #15
    2 resistors in parallel is Rp=Ra*Rl/Ra+Rl.
     
  17. Dec 27, 2012 #16

    gneill

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    Okay, make that substitution for Rp into the voltage divider expression.

    When that's done, think about what you know about Ra and Rb and the resistance value of the potentiometer, Rpot.
     
  18. Dec 27, 2012 #17
    Think i'm starting to wander off the track a bit here but still, here goes. Vout=E*Ra*Rl/Ra+Rl/Rb+Ra/Ra*Rl/Ra+Rl

    I know that Ra and Rb must equal a total of 10k ohms and the higher the Rb figure the higher the reistance and vice versa for Ra.
     
  19. Dec 27, 2012 #18

    gneill

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    can you put some parentheses into that expression so we can tell the order of operations?
     
  20. Dec 27, 2012 #19
    Gneil, cheers for bearing with me. are you at your wits end yet?
    Ok, think i'm right here Vout=E*(Ra*(Rl/Ra+Rl))/(Rb+Ra/Ra)*(Rl/Ra+Rl)
    The only thing i'm still struggling to see here is how this will give me the values of Ra and Rb
     
    Last edited: Dec 27, 2012
  21. Dec 27, 2012 #20

    gneill

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    I'm told that I've been witless for years now :smile:
    Okay, as I suspected something's gone amiss with your voltage divider expansion/simplification. Note that terms like (Rb + Ra/Rb) can't exist due to unit issues; Rb has units of Ohms and Ra/Rb has no units. You can't add a unitless number to one with units. You'll have to try the simplification again I'm afraid. There should be pairwise products of resistors in the expression. If you are having trouble, you can always post your simplification steps for review.

    Once you have that straightened out, you'll write what you know about Ra, Rb, and their total as an equation. Then you'll have two equations in two unknowns.

    Alternatively, make the immediate substitutions Ra = x*R; Rb = (1-x)*R, where R is the resistance of the potentiometer. Solve for x, the portion of the potentiometer that represents resistor Ra.
     
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