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*sigh* volatage divider ?

  1. Jul 1, 2008 #1
    *sigh* volatage divider....?

    I've been going over basic circuitry and got to the voltage divider. I don't understand the proof here:

    http://en.wikipedia.org/wiki/Voltage_divider#General_case

    The first line makes sense. From Vi to Ground there are two resistors in series, so you add them, and V = IR and I is the same all around the whole circuit.

    Then the 2nd line I don't get. Why is it R2 and not R1? I thought current would go From Vi through R1 and then to Vo, not Ground to Vo.
     
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  3. Jul 1, 2008 #2

    Defennder

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    Re: *sigh* volatage divider....?

    The potential difference across R2 (or Z2) is given by [itex]V_{out} - 0[/itex], where 0 is the ground voltage. For a resistor, V = IR so this means [itex]V_{out} - 0 = IR_2[/tex]
     
  4. Jul 1, 2008 #3
    Re: *sigh* volatage divider....?

    Let me see if I have this right.

    Current goes from Vin through R1, which gives a voltage drop. Then since there is a branch, i.e. two things are in parallel, V across both is equal, but current probably isn't. So then you exploit the fact that at V at the branching point = Vout, and then you do this:

    I think I got it. Thanks a lot. :D
     
  5. Jul 1, 2008 #4

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    Re: *sigh* volatage divider....?

    Actually there isn't any "branch" in between R1 and R2. In fact I would say that if there is a branch there, the voltage divider principle does not work. It works only if the circuit elements are connected in series, which they would not be if there was a branch in between.
     
  6. Aug 19, 2008 #5
    Re: *sigh* volatage divider....?

    Sorry to revive this. I'm back in "electronics" mode and still stuck on this damn problem again...

    By branch I mean it branches off between R1 and R2 to Vout.

    I guess my next question is how do they know Iin = Iout? Wouldn't it be Vin = I_in(R1 + R2) and Vout = Iout*R2? Why do they use the same value for I?

    EDIT: Or are you assuming it's an open circuit, so that no current actually goes to the right into Vout? Then how would that work with a Load?
     
  7. Aug 19, 2008 #6

    Defennder

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    Re: *sigh* volatage divider....?

    The diagram is a little misleading. Vout does not denote a wire connected in between R1 and R2. It's just a line which indicates that the potential between the two resistors is V2. The same goes for Vin.

    Iin = Iout is the assumption they make in order to derive the voltage divider principle. As said above, if that assumption is not true, ie. if there is a wire branching off from in between 2 resistors, then the potential divider principle does not work.

    EDIT: Yes you're right here. I don't know what you mean by "work with a load". It depends on whether you designate R2 as the load or something else.
     
  8. Aug 19, 2008 #7
    Re: *sigh* volatage divider....?

    Okay okay, that much makes sense. So how do you apply a voltage divider if you're not allowed to have anything between R1 and R2? I thought that was the whole point, where you kind of siphon off some of the current.
     
  9. Aug 19, 2008 #8

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    Re: *sigh* volatage divider....?

    The voltage divider principle is just a principle of circuits for easy evaluation of voltages across a circuit element, not a technique for getting a desired voltage out of something. To get a desired voltage across some complicated linear network, you first reduce it to it's Thevenin equivalent, then add resistors in the appropriate manner until you get the desired voltage.
     
  10. Aug 19, 2008 #9
    Re: *sigh* volatage divider....?

    I see... but if I had Rload at Vout >> R1 and R2, wouldn't that be approximately correct still?

    So if I used a voltmeter I could still read Vout from in between R1 and R2, right?

    I guess that's what confused me above all. I had thought this was something practical, not just a thought experiment type of thing.
     
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