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Finding nodal voltages using KCL.

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data

    jayn7s.jpg

    Where XX is 60V

    I'm trying to find the node voltages at point A and B, however each time I've tried this and then checked using circuit wizard it has come up incorrect. Yes my circuit in circuit wizards is correct.

    2. Relevant equations



    3. The attempt at a solution

    V1: I1=I2+I3+Iground

    60-V1/25 = V1-V2/40 + V1-40/30 + V1-0/10

    then i multiplied by 600 to remove the denominators

    1440-24V1=60V1 +15V1-15V2+2-V1-800
    2240=119V1-15V2 (Eqn 1)

    V2: I2+I3=2.5+Iground2

    V1-V2/40 +V1-40/30=2.5 + V2-0/20

    then i multiplied by 120 to remove the denominators

    3V1 - 3V2 + 4V1 - 160 = 300 + 6V2
    7V1-9V2=460

    I then put them into a matrix and tried using cramers rule to solve for v1 and v2

    119 -15 l V1 l l 2240 l
    7 -9 l V2 l l 460 l

    119x-9 - -15 x 7
    Doriginal= -966

    2240 -15
    460 -9

    2240 x -9 - -15 x 460
    D1= -13260

    119 2240
    7 460

    119 x 460 - 2240 x 7
    =39060

    V1= -13260/-966 = 13.7V (Circuit wizard shows it as being 2.37V)
    V2= 39060/-966 = -40.43V (Circuit wizard shows it as being -9.71V)

    please help!
     
  2. jcsd
  3. Aug 18, 2010 #2
    I would put everything on the left side of an equation, and equate to zero, include parentheses and fix a sign error and an omission:

    (V1-V2)/40 + (V1-V2+40)/30 + (V1-0)/10 + (V1-60)/25 = 0

    Multiply through by 600, collect terms and get Eqn 1:

    119V1 - 35V2 = 640

    For the second equation:

    (V2-V1-40)/30 + (V2-V1)/40 + (V2-0)/20 + 2.5 = 0

    Multiply through by 120 and get Eqn 2:

    -7V1 + 13 V2 = -140

    Solving, I get:

    V1 = 570/217 = 2.6267
    V2 = -290/31 = -9.3548
     
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