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Using node-voltage analytical technique

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the voltage at each of the extraordinary nodes (V1 – V3), and the current through each of the six resistors.

    http://i.imgur.com/VbZQH.png


    2. Relevant equations

    KCL: ƩIn = 0
    V=ir
    I=[itex]A^{-1}[/itex]B
    [itex]\Delta[/itex] = a11C11 + a12C12 + a13C13

    [itex]A^{-1}[/itex] = [itex]\frac{adjA}{\Delta}[/itex]
    adjA = [itex][C_{jk}]^{T}[/itex]


    3. The attempt at a solution



    For I1+I2+I3=0:

    [itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V1-V2}{0.5}[/itex] + [itex]\frac{V1-2}{0.1}[/itex] = 0

    For I4+I5+I6=0:

    [itex]\frac{V2-V1}{0.5}[/itex] + [itex]\frac{V2-V3}{0.5}[/itex] + [itex]\frac{V2}{0.1}[/itex] = 0

    For I7+I8+I9=0:

    -[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V3-V2}{0.5}[/itex] + [itex]\frac{V3-3}{0.1}[/itex] = 0


    ([itex]\frac{1}{0.2}[/itex] + [itex]\frac{1}{0.5}[/itex] + [itex]\frac{1}{0.1}[/itex])V1 - ([itex]\frac{1}{0.2}[/itex] + [itex]\frac{1}{0.5}[/itex])V2 + 0V3 = 40

    ([itex]\frac{1}{0.5}[/itex]+[itex]\frac{1}{0.5}[/itex]+[itex]\frac{1}{0.1}[/itex])V2 - ([itex]\frac{1}{0.5}[/itex])V1 + (-[itex]\frac{1}{0.5}[/itex])V3 = 0

    (-[itex]\frac{1}{0.2}[/itex])V1 + (-[itex]\frac{1}{0.5}[/itex])V2 + [itex]\frac{1}{0.5}[/itex] + [itex]\frac{1}{0.1}[/itex])V3 = 10

    And then I simplified that to:

    17V1 - 7V2 + 0V3 = 40
    -2V1 + 14V2 -2V3 = 0
    -5V1 - 2V2 + 12V3 = 10


    My book says I should use Cramer's rule or matrix inversion to solve this. I chose to use matrix inversion because I thought solving a simultaneous equation would take a while.


    I ended up with I1 = 2.627, I2 = 2/3, I3 = 2.039

    That's obviously wrong since the sum should equal zero. I'm not sure what I did wrong, I just followed the steps in my book.
     
  2. jcsd
  3. Sep 25, 2012 #2

    Simon Bridge

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    Cannot tell without your working.
     
  4. Sep 25, 2012 #3

    The Electrician

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    You must include the effect of V3 in this equation, as shown:

    For I1+I2+I3=0:

    [itex]\frac{V1-4-V3}{0.2}[/itex] + [itex]\frac{V1-V2}{0.5}[/itex] + [itex]\frac{V1-2}{0.1}[/itex] = 0

    And this one:

    For I7+I8+I9=0:

    -[itex]\frac{V1-4-V3}{0.2}[/itex] + [itex]\frac{V3-V2}{0.5}[/itex] + [itex]\frac{V3-3}{0.1}[/itex] = 0
     
  5. Sep 25, 2012 #4
    Okay, I redid my work and I'm still getting the wrong answer. In my previous post I mistakenly wrote my answers as current values instead of voltages sinceI'm calculating the voltages at the extraordinary nodes.

    17V1 - 2V2 - 2V3 = 40
    -2V1 + 14V2 -2V3 = 0
    -2V1 - 2V2 + 17V3 = 50

    After doing the matrix inversion, I ended up with V1 = 2.85556V, V2= 0.891089V and V3 = 3.3819V.

    I built the circuit on my computer and I got v1 = 2.8693V, v2 = 0.625V , v3 = 1.5057V
    EDIT:

    I messed up on my calculations above. I'll re-do them again.
     
    Last edited: Sep 25, 2012
  6. Sep 25, 2012 #5

    The Electrician

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    You haven't done your algebra right. You should have:

    17V1 - 2V2 - 5V3 = 40
    -2V1 + 14V2 -2V3 = 0
    -5V1 - 2V2 + 17V3 = 10
     
  7. Sep 25, 2012 #6
    Yeah, I realized that after I posted. I got the correct values now, thanks for the help.

    What confused me was the top part of the circuit, specifically the 4V source between the two extraordinary nodes. When I look at circuit diagrams I usually make the mistake of assuming the corners are all nodes. So I treated the corner right above V2 as a node. I make a lot of careless mistakes.
     
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