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Mesh analysis with dependent voltage source

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the:
    -Complex power delivered by source
    -Power factor of total load
    -Is the power factor leading, lagging, or in unity?

    upload_2016-10-24_22-39-39.png

    2. Relevant equations
    KVL/KCL

    3. The attempt at a solution
    I want to solve this via mesh analysis to get currents.
    Mesh1: -10V + V1 + 0.75V1 = 0
    V1 = 40/7V
    4I1 = 40/7
    I1 = 10/7 A

    Mesh2: -0.75V1 = I2(1-j1.5) = 0
    I2(1-j1.5) = 0.75 * 40/7
    I2 = 30/(7(1-j1.5))

    Did I do this right?
     
  2. jcsd
  3. Oct 25, 2016 #2

    gneill

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    Staff: Mentor

    The controlled source depicted is a current source, not a voltage source. So either employ a supermesh or introduce another variable to represent the potential across it (and you'll need another equation so that you have as many equations as variables in your equation set).
     
  4. Oct 25, 2016 #3
    -10V + 4(I1) + I2(1 - j1.5) = 0
    I2 = 3/4V1 + I1
    V1 = 4(I1)

    Are these correct?
     
  5. Oct 25, 2016 #4

    gneill

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    Staff: Mentor

    Yes, those look good.
     
  6. Oct 25, 2016 #5
    i1 = 0.8 + 0.6i
    i2 = 3.2 + 2.4i
    correct?
     
  7. Oct 25, 2016 #6

    gneill

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    Staff: Mentor

    That looks good, too.
     
  8. Oct 25, 2016 #7
    How would I find complex power supplied by the source?
     
  9. Oct 25, 2016 #8

    gneill

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    Staff: Mentor

    Use the current supplied by the source and the source voltage.
     
  10. Oct 26, 2016 #9
    Is current supplied by source I1 - 3/4V1?
    How would I find the power factor?
     
  11. Oct 26, 2016 #10

    gneill

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    Staff: Mentor

    Presumably the source referred to is the 10 V voltage source. It supplies just I1.

    For the power factor you need to look at the power triangle. The complex power will give you the two "legs" of the triangle.
     
  12. Oct 27, 2016 #11
    So complex power = 10 * I* = 10 * (0.8 - 0.6i)?
    S = 8-6i
    Power factor = P/S = 8/10 = 0.8 leading? (Q<0)
    Does this triangle look right?
    upload_2016-10-26_23-5-16.png
     
  13. Oct 27, 2016 #12

    cnh1995

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    Homework Helper

    Yes. Also, write 'j' instead of 'i' in the imaginary part of the complex form (e.g
    S=8-j6). That's a convention followed in order to avoid confusion with the symbol 'I' for current.
     
  14. Oct 27, 2016 #13
    Can (should) I convert to VA form? (sqrt(P^2+Q^2))
     
  15. Oct 27, 2016 #14

    cnh1995

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    Homework Helper

    Not necessary if apparent power is not asked. You can keep it in the complex form.
     
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