Simple Voltage Divider Algebra problem

Discussion in 'Homework Help' started by ECE101, Apr 14, 2012.

  1. ECE101

    Thread Starter New Member

    Nov 25, 2011
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    1: Vout = Vin.R2 / R2 + R1

    2: (R2 + R1). Vout = Vin. R2

    3: (-R2. Vin + R2) + R1. V out = 0

    4: R2(-Vn +1) + R1.Vout = 0

    5: R2 = (R1.Vout) / (Vin + 1)


    How does one go from line 2, to line 3?

    I end up with:

    2: (R2 + R1). Vout = Vin. R2

    3: (-R2. Vin) + R2. Vout + R1. Vout = 0

    Could someone please explain?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Loks like there is and error between line 2 and line 3 above.


    Your result is correct.

    hgmjr
     
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  3. ECE101

    Thread Starter New Member

    Nov 25, 2011
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    2: (R2 + R1). Vout = Vin. R2

    3: (-R2. Vin) + R2. Vout + R1. Vout = 0

    So in terms of R2:

    3: (-R2. Vin) + R2. Vout + R1. Vout = 0

    4: R2(Vout - Vin) + R1. Vout = 0

    5: R2 = -R1. Vout / (Vout - Vin)

    Is this correct?

    If so how is the following derived here?
    [​IMG]
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    All you should need to do is divide the numerator and the denominator by the value Vout. Remember that whatever you do the denominator, you must do to the numerator to perserve the equality.

    hgmjr
     
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  5. ECE101

    Thread Starter New Member

    Nov 25, 2011
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  6. ECE101

    Thread Starter New Member

    Nov 25, 2011
    29
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    Thanks again for the reply hgmjr

    I'm still having a problem getting the wiki answer though!

    1: Vout = Vin.R2 / R2 + R1

    2: (R2 + R1). Vout = Vin. R2

    3: (R2 + R1). Vout/Vin = R2

    4: R2(Vout/Vin - 1) + R1(Vout/Vin) = 0

    5: R2 = -R1(Vout/Vin)/(Vout/Vin - 1) ?
     
  7. hgmjr

    Moderator

    Jan 28, 2005
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    I can see why.

    The transition from 3 to 4 is a case of taking too big a bite. It should have been split up into two seperate steps.

    hgmjr


    hgmjr
     
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  8. ECE101

    Thread Starter New Member

    Nov 25, 2011
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    0
    Got it! Cheers:

    Vout/Vin = R2/R1+R2
    (R1 + R2) (Vout/Vin) = R2
    R1 (Vout/Vin) + R2 (Vout/Vin) - R2 = 0
    R1 (Vout/Vin) = R2 - R2 (Vout/Vin)
    R1 = R2(Vin/Vout) - R2
    R1 = R2((Vin/Vout) - 1)
    R2 = R1/((Vin/Vout) - 1)
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    A couple of times now you have written:

    The second line doesn't follow from your first line, except when whoever is doing the work opts to replace the first line with what you meant to write and not what you actually wrote.

    What you wrote is:

    Vout/Vin = (R2/R1)+R2

    Since division has precedence over addition. It's an easy mistake to make, which makes it all the more critical that you develop the habit of not making it very often.
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,792
    Look at line 3 carefully. Do the units work out? No. The first term has units of volt-ohms while the second just has units of ohms. You can't add the two. You KNOW the equation is incorrect. There is no point even looking at what comes after it, because it is guaranteed to be WRONG (unless another error just happens to fix the first).

    Having said that, let's look at what comes next. In equation three, you have (-Vn + 1). Most people, since they don't track units, won't be bothered by this at all because, being sloppy, they are used to seeing crap like this all the time. But, again, the units don't work. Vn (should be Vin, BTW) has units of volts and 1 is dimensionless and, if you were in the habit of tracking units, just seeing this would SCREAM out to you ERROR! ERROR! ERROR!
     
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