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Mesh Analysis by Inspection

  1. Feb 24, 2016 #1
    1.
    I have a question regarding the treatment of voltage sources in Mesh Analysis by inspection. I hope this is the correct place to post.

    In my course textbook (Fundamentals of Electric Circuits 4e p. 101) it says that the voltage is the algebraic sum taken clockwise of all independent voltage sources in the related mesh. Yet in an example problem they do not stay consistent with this.

    I found the same example problem in some lecture slides online:
    64c5d045795fc82d6e97b0fcbd2cefb3.png

    So I'm left wondering:
    - Does the example have an error and they actually meant to do clock wise drop is positive? My textbook makes no mention of treating a drop as positive for Mesh analysis by inspection.
    - If it isn't an error, and they did CW with rise positive, is that any different than just taking it CCW with drop being positive?
     
  2. jcsd
  3. Feb 24, 2016 #2

    gneill

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    Staff: Mentor

    In loop ##i_1## going clockwise you "walk" from - to + though the 4 V source. So that's a rise of 4 V going in that direction and the voltage vector reflects this with a "4" entry.

    In loop ##i_2## again going clockwise, there's a rise of 10 V followed by a drop of 4 V for a net of 6 V, again reflected in the voltage vector.

    So, no mistake there.

    P.S. Please be sure to retain the formatting template headings in future. The template is required for threads in the homework areas.
     
  4. Feb 24, 2016 #3
    Apologies for not using the template. It just didn't seem to fit my question.
    So instead of viewing it as CW with rise being positive can I take it as CCW with drop being positive?
     
  5. Feb 25, 2016 #4

    gneill

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    Staff: Mentor

    [Sorry for the delay in responding ... we had a power failure here that lasted several hours]

    You need to respect the potential change that occurs in the direction of your "KVL walk", just as you must respect the rise/drop across resistors that occurs in the direction of the walk. The latter depends upon the assumed current direction. So if you're "walking with the current", resistors show a potential drop. If you're "walking against the current" then resistors show a potential rise.

    Consider that by placing the source potential changes in a separate vector on the other side of the equals sign, this implicitly changes the sign of the entries already. So if you are summing potential drops as positive values on the LHS, then a potential rise through a source would have a negative entry on the LHS. Moved to the RHS it becomes positive.

    Why don't you create a very simple one-loop test case to play with, say two sources with different orientation and a couple of resistors. Write the KVL loop equation manually and verify the results, then apply the "mesh by inspection" rules to do write the equation. You should end up with the same solution. Then you can try out your CW/CCW rise/drop theories to see what happens.
     
  6. Mar 11, 2016 #5
    I'm very sorry for replying so late to you. I really appreciate you taking your time to provide me with such a detailed response. I got ill after I posted this question, and that caused me to have very limited time for extra activities like the internet.

    Thanks again for taking the time to solve this. I completely understand it now. Once I took the time to understand what the matrix of inspection meant it hit me what I needed to do.
     
  7. Mar 11, 2016 #6

    gneill

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    Staff: Mentor

    No worries, glad I could help. Hope you're feeling better!
     
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