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Mesh and nodal analysis don't quite match, but why?

  1. Sep 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Find I by mesh and nodal analysis

    Q2(a) Base drawing & values.jpg

    2. Relevant equations


    3. The attempt at a solution

    ##V_{10}## is defined at the junction of ##Z_1##, ##Z_2## and ##Z_4##.
    ##V_{20}## is defined at the junction of ##Z_2##, ##Z_3## and ##Z_5##.
    ##0V## is defined at the junction of ##V_1##, ##Z_4##, ##Z_5## and ##V_2##.

    I have completed the mesh analysis without too much difficulty (and those answers are correct according to other threads with this question). I thought I was on the right track with the nodal analysis but the answers are close enough to be wrong, but not far enough away to look like a major error somewhere.

    I have used Wolfram to do the heavy lifting, and have attached the PDF of the input and the result for ##V_{10}##. When I calculate the current through ##Z_4## (again using Wolfram entering values in complex form), I get the result of ##-8.7792 + j16.5853## when converted to polar is ##18.765## Amps compared to ##19.59## for the mesh analysis, (although the angle is off too).

    I've spend hours and hours trying to figure this out, but can't quite get the answers to agree. I've also been through the process of trying to identify compound rounding errors but I don't think it's that. I would appreciate some help or pointers :cry:
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2016 #2

    gneill

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    Staff: Mentor

    To me it looks like WolframAlpha has mucked up the signs of the two complex terms in your node equation. For example,
    $$\frac{v10}{-5j} = \frac{v10}{-5j}\cdot\frac{j}{j} = \frac{v10}{5}j$$
    They have made the term negative by writing it as
    $$- \frac{v10}{5}j$$
     
  4. Sep 7, 2016 #3
    Thank you for the tip! I've played with brackets to no effect. What does appear to work is to express the input as the entire complex number, i.e. ##0-j5## instead of just ##-j5##.

    However, although I now have the correct magnitude, the angle is 180 degrees out, which suggests to me that the signs of the real and imaginary components are reversed? Would that be a correct supposition?
     
  5. Sep 7, 2016 #4

    gneill

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    Staff: Mentor

    That could be it. You'll be the best judge though. Look at the complex values for your "good" results via mesh analysis and compare with the recent values returned by Wolfram. I can tell you that my own results for v10 and v20 have all components positive.
     
  6. Sep 7, 2016 #5

    The Electrician

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    Gold Member

    Looking at your Wolfram Alpha solution, you have in the very first line (v10/-5i) which Wolfram turns into a fraction with i in the numerator.

    You say you've played with brackets, but you don't show us what you did.

    I typed (v10/(-5i)) into Wolfram Alpha and got an expression with i in the denominator as it should be. Did you try that?
     
  7. Sep 8, 2016 #6
    It looks like I used the complete term and brackets to obtain the correct input. I'm a little unsure on equivalent terms, so the second example below looked like the one which is correct to me, but is the first entry also correct? Actually looking at it, they both give the same answer, so they must be...

    bracket_example.jpg

    bracket_example_fullterm.jpg
     
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