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Nodal Analysis Clarification - What are some conventions?

  1. Oct 22, 2015 #1
    I've been assigned the following problem for homework in my Electrical Engineering class:
    xhx0tf3.png
    *Note, the duplicate v2 is supposed to be v3.

    But I keep getting
    v1 = 4.10V
    v2 = 4.81V
    v3 = -1.19V
    and
    i0 = 52.9mA

    Every simulator I've put this into says it's wrong, yet I've attempted solving it about 12 times (consuming almost all my remaining notebook pages) and still get the same answers.

    I can't post a copy of my written work since I currently don't have access to a scanner. However, I drew in arbitrary current directions and made sure to respect them for all calculations using KCL.

    I have no idea what I'm doing wrong at this point. My professor just seems to get agitated when I ask what the convention is and his general response is "just do the current thing" or something vague like that. I don't think anyone in my class actually understands how to do this (except the class-repeaters).

    I've been told by my friend to "always assume currents flow into a node. Ignore what's drawn. they always flow into nodes", but I've tried that at least 4 times. I've never gotten the answers my simulator provides me, regardless of how I solve it.

    Is there any "convention" to solving these? Can anyone tell me what error I'm making so I don't make the same error on my exam?
     
  2. jcsd
  3. Oct 22, 2015 #2

    SteamKing

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    Science Advisor
    Homework Helper

    We can't tell you what you are doing wrong, since you didn't post your work.

    It would help if you included your nodal equations or KCL equations. You shouldn't need a scanner for that.
     
  4. Oct 22, 2015 #3
    IqZxOZu.jpg

    There we go. the lighting in my room is terrible and I wasn't sure if a picture would turn out correctly. This one used directions I arbitrarily assigned and calculated from.
     
  5. Oct 22, 2015 #4

    NascentOxygen

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    At node 2 you are considering a current in from the left, a current in from the right, and a current exiting downwards.

    The current into node 2 from the right can be seen to be the sum of the two other currents entering node 3, viz., (V1-V3)/100 + (0-V3)/50
     
  6. Oct 22, 2015 #5

    gneill

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    Staff: Mentor

    I can see at least a couple of sign issues in your work. But the main thing is that you're making your life difficult and prone to errors by mixing currents flowing in and flowing out of a node in the same KCL equation. This can be done, but it takes a lot of concentration to keep all the signs straight.

    A better approach is to pick one direction, either flowing in or flowing out of the node, and write each term accordingly. That way every term is added, never subtracted. The single exception is for current sources where they dictate the current direction; Then you need to be careful of the sign of the term so that it matches the "in" or "out" paradigm that you've chosen.

    In the sixth line of your work you transported the "-12" to the RHS but didn't change the sign. In the next line where you write the v2 node equation you've mixed up the current directions so the sum is unlikely to be zero. And in fact I can confirm that the result is not a correct node equation.
     
  7. Oct 22, 2015 #6

    When you say to pick one direction, I should respect those defined directions at other nodes they affect (ex. if I say current flows into v1 from v2, I have to say it flows out of v2 in my KCL equation for v2 [subtracting it instead of adding it]), or can I sum all the currents at each node?
     
  8. Oct 22, 2015 #7

    gneill

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    Staff: Mentor

    You can treat every node independently with regards to choosing a direction paradigm. Multiplying an equation through by a constant doesn't change the equation, and -1 is a perfectly good constant :smile:

    So you can write "in" equations for every node, or "out" equations for every node, or even a mixture. But be consistent for individual nodes.
     
  9. Oct 22, 2015 #8
    When respecting that direction, I assume I also have to reflect it in my ohm's law formulas? (ex. if I say all currents are flowing into v2, I would have to consider v2 as the "lower potential" and v1, v3 as "higher potential" [v1 - v2/20] and so on?)
     
  10. Oct 22, 2015 #9

    gneill

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    Staff: Mentor

    Right.

    Personally (my own habit) is to always choose "out" as my direction of choice. So if I'm writing an equation for node "v1", every term that's not a fixed current will have the form ##\frac{v1 - <stuff>}{<stuff>}##.
     
  11. Oct 22, 2015 #10
    Ah. That helps. When it was explained to me, I was told I had to keep the current directions constant, so that's why I have the mix of ins and outs, but knowing I can use just addition or just subtraction is very helpful.
     
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