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Electric Circuits: Nodal/Mesh Analysis

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    http://edugen.wileyplus.com/edugen/courses/crs4797/dorf1571/dorf1571c04/image_n/n04f090.gif

    Determine values of the node voltages, v1, v2 v3, v4, and v5 in the circuit shown.

    3. The attempt at a solution

    Doing nodal analysis I came up with the following four equations but I'm missing one for v5.

    (v1 - v5)/2Ω + (v1 - v3)/5Ω + (v2 - v4)/4 = 0

    v2 - v1 = -8v

    v3/8Ω + (v3 - v2)/10Ω + (v3 - v1)/5 = 0

    v4 - v3 = 16

    Is there an easy way to find v5 or do I have to attempt mesh current analysis to get v5?
     
  2. jcsd
  3. Feb 10, 2013 #2

    The Electrician

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    I don't see any circuit.
     
  4. Feb 10, 2013 #3
    Sorry, must be a login issue. Please see attached.
     

    Attached Files:

  5. Feb 10, 2013 #4

    gneill

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    Your first equation seems to be missing a contribution by the current from v2 to v3.

    The potential v5 is set by the controlled voltage source. How might you determine ##i_x##?
     
  6. Feb 10, 2013 #5
    Yes sorry, I missed that when transcribing it.

    (v1 - v5)/2Ω + (v1 - v3)/5Ω + (v2 - v3)/10Ω + (v2 - v4)/4Ω = 0

    I'm not seeing how I get the ix without doing a complete mesh current analysis of the circuit. Is that what I have to do?
     
  7. Feb 10, 2013 #6

    gneill

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    Consider that your node equations are a sum of terms, each of which represents a current between nodes. KCL applied at the right location would seem to be a promising strategy :smile:
     
  8. Feb 11, 2013 #7
    So would ix = (v3 - v2)10Ω + (v3 - v1)/5Ω ?
     
  9. Feb 11, 2013 #8

    gneill

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    It would be the negative of that --- consider the directions of the currents that you calculate.
     
  10. Feb 11, 2013 #9

    The Electrician

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    These 3 currents don't add up to zero unless there is no current in the 16 volt source.
     
  11. Feb 11, 2013 #10
    v3/8Ω + (v3 - v2)/10Ω + (v3 - v1)/5Ω + 16v = 0 ?

    Also am I missing something by not incorporating the 2A source in my equations?
     
  12. Feb 11, 2013 #11

    gneill

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    Yup. You need to include it in the equation as it pushes current into that supernode. The 16v at the end of the equation doesn't make sense --- 16v is not a current.
     
  13. Feb 11, 2013 #12
    So the actual third equation or the v3 v4 supernode would be as follows.

    v3/8Ω + (v3 - v2)/10Ω + (v3 - v1)/5Ω + v4/8 + (v4 - v2)/4 - 2A = 0

    The 16v in the supernode is take care of by the v4 - v3 = 16v equation.

    Is that correct?
     
  14. Feb 11, 2013 #13

    The Electrician

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    When I said "These 3 currents don't add up to zero unless there is no current in the 16 volt source." I didn't mean to just add the quantity "16V" to your equation!

    But one end of the 16V is connected to the V3 node and you have to include all the currents into or out of a node. The 16V is connected to things that might provide a current into the left end of the 16V, and that current must also be present at the right end.

    Edit: You posted while I was composing my post. It looks like you have got it. Post your final numerical results.
     
    Last edited: Feb 11, 2013
  15. Feb 11, 2013 #14
    @The Electrician - Does the new equation that I wrote for the supernode satisfy the currents going in and out of that node or am I still missing something?
     
  16. Feb 11, 2013 #15

    The Electrician

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    I think so. I don't see any other paths to the V3 node.

    Post your numerical results so I can see how they compare to mine.

    What I would typically do in a complicated case like this is to solve in two different ways (typically nodal and mesh) and compare results. I set up for a nodal solution only so far, so if I get the same numbers you do, I'll likely think it's ok.
     
  17. Feb 11, 2013 #16
    See attached for matrix. This is what I got but I think my polarities might be wrong.

    v1 = -8.91v
    v2 = -16.91v
    v3 = -14.63v
    v4 = 1.37v
    v5 = -16.23v
     

    Attached Files:

  18. Feb 11, 2013 #17

    The Electrician

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    I'm not getting those numbers. I'll work on this further in a few hours.
     
  19. Feb 12, 2013 #18

    The Electrician

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    Where did the term in red come from? Leave that term out.

    You didn't show your equation for node 5, so I'll just derive it.

    The voltage at node 5 is4*((v2 - v3)/10Ω + (v1 - v3)/5Ω) so your fiifth equation is just a constraint equation:

    4*((v2 - v3)/10Ω + (v1 - v3)/5Ω)=v5

    The numerical result if you make these changes is:

    v1 =11.321v
    v2 = 3.3208v
    v3 = 2.1132v
    v4 = 18.113v
    v5 = 7.8491v

    I verified this with Spice. Does this mean I lacked confidence in my result?

    This problem was more complicated than the usual ones we see here.
     
    Last edited: Feb 12, 2013
  20. Feb 13, 2013 #19
    Hehe thank you so much! You were right on the money. When I had my professor look at my work, he noticed the v4/8ohm error but not the v5 error. Thank you for catching that.

    Ya, wasn't that an awesome problem? This is a hw problem in the forth week of intro to circuits. I guess this university believes in baptism by fire.
     
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