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Circuit Analysis : Mesh

  1. Feb 23, 2014 #1

    dwn

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    1. The problem statement, all variables and given/known data

    I have attached an image of the problem:

    2. Relevant equations

    KCL Equation


    3. The attempt at a solution

    0 = 2i11 + 3(i1-i2)
    212 = 3(i2-i1) + 5(i2-i3)
    -122 = 3i3 + 5(i3-i2)

    5i1 - 3i2 = 0
    -3i1 +8i2 + 5i3 = 212
    -5i2 + 8i3 = -122

    Is this the correct approach?

    Also, does anybody have a good reference to learn this stuff? I'm taking this course online and I'm drowning. Thanks!
     

    Attached Files:

  2. jcsd
  3. Feb 23, 2014 #2

    gneill

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    Staff: Mentor

    The approach looks okay. Check the sign on the i3 term for your second equation. Otherwise looks okay to me.

    You should label the loops and current directions on the circuit diagram and associate your equations with the labels so it's clear which loop each equation belongs to. Granted it's pretty clear in this relatively simple case with only three loops, but it's good to get in the habit of making your work easy to interpret for those who are just seeing it for the first time.

    You can probably find many worked examples on the net by searching for "mesh analysis examples". For that matter there are plenty here on the Physics Forums. Take a look at the "Related Discussions" section at the bottom of your thread. Also investigate the forum search facilities.
     
  4. Feb 23, 2014 #3

    dwn

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    Thank you for your help and I will make sure to clearly present everything in the future.

    I'm not sure I quite understand why the second equation is incorrect..? If the less simplified equation above is correct, which as I understand it, you subtract the current adjacent to the resistor of interest (5 ohm) from the current loop you are analyzing, shouldn't it be a negative 5i3 ::: 212 = 3(i2-i1) + 5(i2-i3)
     
  5. Feb 23, 2014 #4

    gneill

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    Staff: Mentor

    Yes, it should be -5i3. That's the point I was making. Look at the relevant section from your first post...
    The 5i3 in the indicated line should be negative, right?
     
  6. Feb 23, 2014 #5

    dwn

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    Ooops...I thought you had changed it and that's why you used the red...my bad! I think I'm going cross eyed!
    Thanks again for the help gneill.
     
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