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Mesh Analysis

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data
    PK88BSE.png

    Use Mesh analysis to find voltage of Node A and B

    2. Relevant equations
    V=IR
    KVL

    3. The attempt at a solution
    Are my mesh equations correct?

    Mesh1:
    -j30V + (-j50 + 10)I1 - 10I2 = 0

    Mesh2:
    (10-j20)I2 - 10I1 - j50V - I3(-j20) = 0

    Mesh3:
    (-j20+30+j10)I3 - (-j20)I2 = 0
     
    Last edited: Oct 11, 2016
  2. jcsd
  3. Oct 11, 2016 #2

    gneill

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    Staff: Mentor

    Yes, your mesh current equations look fine.
     
  4. Oct 12, 2016 #3
    How would I get the voltage of Node A and B?
     
  5. Oct 12, 2016 #4

    gneill

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    Staff: Mentor

    Ohm's law. You've solved for the currents so now you can find the potentials across any components you wish. Choose a path from the reference node to the location where you wish to know the potential and sum up the PD's along the way.
     
  6. Oct 12, 2016 #5
    So if I wanted the voltage at Node A, I would multiple (I1-I2) with 10ohm resistor?
     
  7. Oct 12, 2016 #6

    gneill

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    Staff: Mentor

    That would work, yes. Any path from the reference node to a would work, and that happens to be a pretty convenient choice :smile:
     
  8. Oct 12, 2016 #7
    I ended up getting:
    14.329 < -71.74 degrees for Va
    Is that right?


    Would Vb be: (I2-I3)*(-j20) -> 4.48 + 36.39j?
     
    Last edited: Oct 13, 2016
  9. Oct 13, 2016 #8

    gneill

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    Staff: Mentor

    They look good to me.

    Note that for Vb you could also have just added j50 to Va; there is a fixed source of that amount tying them together.
     
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