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Homework Help: Potential Dividers - Help : (

  1. Feb 18, 2010 #1
    I have a simple series circuit with one potentiometer, and I'm supposed to find the V(output) for various V(input). However, I think that V(input) is constant isn't it, because it's the voltage of the battery...right? I know I can change the resistance and make V (output) change, but V(input) should be unaffected right? I'm stuck : (
     
  2. jcsd
  3. Feb 18, 2010 #2

    rock.freak667

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    For a potential divider with two resistances R1 and R2

    the voltage across R1 is

    V1=R1/(R1+R2)V
     
  4. Feb 18, 2010 #3
    i know but, isn't v(input) constant?
     
  5. Feb 18, 2010 #4

    rock.freak667

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    Yes, but if you change V(input), V(output) changes appropriately for given resistances.
     
  6. Feb 18, 2010 #5
    how do i change V(input) if it comes from a constant voltage source such as a battery

    sorry i don't mean to bother you, i really appreciate your help
    but i'm still a bit confused
     
  7. Feb 18, 2010 #6

    rock.freak667

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    You'd need to change the battery itself.
     
  8. Feb 18, 2010 #7

    collinsmark

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    You could replace the battery with a variable voltage source such as a programmable bench-top power supply. :rofl:

    No, seriously though. Circuits similar to what you describe do have applications. Examples of such circuits are; or are likely used within:

    (a) Battery testers. These things can be used to test whether or not a battery is fully charged. If you have one you can decide which batteries to keep and which to recycle.

    (b) Battery chargers. This one is pretty obvious. If the voltage of the battery's terminals reaches a certain point, the charging algorithm changes or stops altogether.

    (c) Low-voltage drop-out (LDO) regulators. These are in nearly all portable electronic devices these days. Basically the regulator shuts off power to other circuits if the battery voltage (or even if output voltage) gets too low, to prevent unwanted behavior (such as memory getting corrupt or something).

    I suppose the general idea here is that battery voltage does change a little depending on how fully charged the battery is. And in practical applications, it's useful to account for this. So often times, it's generally not a good idea to assume that batteries always have constant terminal voltages.
     
    Last edited: Feb 18, 2010
  9. Feb 18, 2010 #8

    Ouabache

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    Is that all you were given (is this the whole question)?
    If they gave a diagram of your circuit, could you please upload it?
    If not, to help us understand your thought process, could you please draw and upload your own
    diagram (labelling Vin, Vout, gnd, resistances on either side of the potentiometer wiper, etc..)?
     
  10. Feb 19, 2010 #9
  11. Feb 19, 2010 #10
    If you're turning the knob on the potentiometer, you're changing the resistance. You can never change two things at the same time during your experiment (ie. you wouldn't turn the knob on the potentiometer AND vary the input voltage, you'd do one or the other). If your lab manual says to do so you may want to check with your prof, because it could be a mistake.

    Are you sure you're supposed to be finding Vout vs Vin, not Vout vs R?
     
  12. Mar 2, 2010 #11

    Ouabache

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    According to what you have described and what your circuit shows, you are right, you are only changing the resistances of your voltage divider (potentiometer), not the supply voltage (Vin). The only way you can find Vout for various Vin, is to be able to vary the power supplied. And if you are varying the power supplied, I would do that for one setting on the pot at a time.
     
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