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Circuits Question Node Voltage Method

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm having problems finding the power.

    I should my work to get the node voltages. Apparently they are correct. I showed how I got to those answers as well just in case they were wrong but I don't think they are.

    http://img547.imageshack.us/img547/8812/capturebu.png [Broken]

    2. Relevant equations



    3. The attempt at a solution

    I see from the reference node, which is at a potential of zero, that node 1 is 8 volts higher so...

    [itex]V_{1} = 8[/itex]

    I then write the equations for the rest of the nodes

    [itex]\frac{V_{2} - 8}{3} + \frac{V_{2} - 3 - V_{4}}{4} + \frac{V_{2} - V_{3}}{2} = 0[/itex]
    [itex]\frac{V_{3} - V_{2}}{2} + \frac{V_{3}}{1} + \frac{v_{3} - 6 - V{4}}{3} = 0[/itex]
    [itex]\frac{V_{4} + 3 - V(2)}{4} + \frac{V_{4} + 6 - V{3}}{3} + \frac{V_{4} - V{5}}{2} = 0[/itex]
    [itex]\frac{V_{5} - V{4}}{2} + 2 - 3 = 0[/itex]

    I than simplify these equations

    [itex](\frac{1}{3} + \frac{1}{4} + \frac{1}{2})V_{2} - \frac{1}{2}V_{3} - \frac{1}{4}V_{4} = -( -\frac{8}{3} - \frac{3}{4})[/itex]
    [itex]-\frac{1}{2}V_{2} + (\frac{1}{2} + 1 + \frac{1}{3})V_{3} - \frac{1}{3}V_{4} = -(-\frac{6}{3})[/itex]
    [itex]-\frac{1}{4}V_{2} - \frac{1}{3}V_{3} + (\frac{1}{4} + \frac{1}{3} + \frac{1}{2})V_{4} - \frac{1}{2}V_{5} = -(\frac{3}{4} + \frac{6}{3})[/itex]
    [itex]0V_{2} + 0V_{3} - \frac{1}{2}V_{4} + \frac{1}{2}V_{5} = -(2-3)[/itex]

    I than simplify these equations further

    [itex]\frac{13}{12}V_{2} - \frac{1}{2}V_{3} - \frac{1}{4}V_{4} + 0V_{5} = \frac{41}{12}[/itex]
    [itex]-\frac{1}{2}V_{2} + \frac{11}{6}V_{3} - \frac{1}{3}V_{4} + 0V_{5} = 2[/itex]
    [itex]-\frac{1}{4}V_{2} - \frac{1}{3}V_{3} + \frac{13}{12}V_{4} - \frac{1}{2}V_{5} = -\frac{11}{4}[/itex]
    [itex]0V_{2} + 0V_{3} - \frac{1}{2}V_{4} + \frac{1}{2}V_{5} = 1[/itex]

    I than define matrix A to be

    [itex]\frac{13}{12} -\frac{1}{2} -\frac{1}{4} 0[/itex]
    [itex]-\frac{1}{2} \frac{11}{6} -\frac{1}{3} 0[/itex]
    [itex]-\frac{1}{4} -\frac{1}{3} \frac{13}{12} -\frac{1}{2}[/itex]
    [itex]0 0 -\frac{1}{2} \frac{1}{2}[/itex]

    I than define matrix B to be

    [itex]\frac{41}{12}[/itex]
    [itex]2[/itex]
    [itex]-\frac{11}{4}[/itex]
    [itex]1[/itex]

    I than do

    [itex]C = A^{-1}B[/itex]
    C = 4.22
    2.26
    .1
    2.1

    So

    [itex]V_{1} = 8[/itex]
    [itex]V_{2} = 4.22[/itex]
    [itex]V_{3} = 2.26[/itex]
    [itex]V_{4} = .1[/itex]
    [itex]V_{5} = 2.1[/itex]

    Then

    [itex]I_{x} = \frac{V{2} - V_{1}}{3} = \frac{4.22 - 8}{3} = -1.26[/itex]
    [itex]I_{y} = \frac{V_{3}}{1} = 2.26[/itex]
    [itex]I_{z} = \frac{V_{4} - V_{5}}{2} = \frac{.1-2.1}{2} = -1[/itex]

    The answer key says

    [itex]I_{x} = -1.26[/itex]
    [itex]I_{y} = 2.26[/itex]
    [itex]I_{z} = -1[/itex]


    So I guess I'm close within round off error

    However for

    [itex]P_{6V} = IV = \frac{V_{3}-V_{4}}{3}6 = \frac{2.26 - .1 - 6}{3}6 = -7.68[/itex]

    The answer key says

    [itex]P_{6V} = (-1.28)(6) = -7.68[/itex]

    The answer key used the mesh current method. I don't like the mesh current method and try to use the node voltage method always. Either way I should still get the same answer which I don't. I'm not exactly sure why I don't. Thanks for any help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 26, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Recheck the coefficient for V4 in the above equation. Everything else looks okay to this point.
     
    Last edited by a moderator: May 6, 2017
  4. Dec 26, 2012 #3
    Never mind I figured it out
     
  5. Dec 26, 2012 #4
    Oh. For the final answer for the power. I forgot to subtract 6

    ((2.236-6-.016)/3)6 = -7.56

    let me check the coeffecent for V4
     
  6. Dec 26, 2012 #5
    ok i got it thanks
     
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