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Phasor to complex units

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Several of the voltages associated with a certain circuit are given by V12=9∠30° V, V32=3∠132° V, V14=2∠10° V. Determine V21, V13, V34, and V24.


    2. Relevant equations



    3. The attempt at a solution
    I am just using this problem as an example as I have no idea how to go from phasor to complex. So V13=V12+V32=9∠30°+3∠132°=? I dont really know here I can only imagine Magnitude*cos(θ)=Real and Magnitude*sin(θ)=imaginary I know this but how do I ultilize these beacuse everything I have done does not work?
     
  2. jcsd
  3. Sep 19, 2013 #2

    UltrafastPED

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    Your first example is magnitude 9, phase angle 30 degrees. So the complex number is 9 x [cos(30) + j sin(30)]; this is what you have proposed. Make sure your calculator/computer is set to degrees, or else convert them to radians.

    The alternative expression (in radians) may be more useful: use the Euler identity and write it as
    9 exp(j*30/360*2pi).


    Here is an example: http://people.clarkson.edu/~jsvoboda/eta/phasors/AddPhasors10.html
     
  4. Sep 19, 2013 #3
    Ok thanks man
     
  5. Sep 19, 2013 #4
    hey for V13 I get 5.865+j6.798 V
     
  6. Sep 19, 2013 #5
    back in phasor I get 8.978∠49.21°
     
  7. Sep 19, 2013 #6
    V21=V32+V13=3.936+j9.096
     
  8. Sep 19, 2013 #7
    V34=V13+V14=7.834+j7.145 and V24=V32+V34=5.906+j9.443
     
  9. Sep 19, 2013 #8

    gneill

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    Staff: Mentor

    As a reality check, note that you are given a value for V12. Shouldn't V21 = -V12?
     
  10. Sep 19, 2013 #9
    oh heh so it should be V21=-9∠30° V
     
  11. Sep 19, 2013 #10
    Also are the rest of my answers correct?
     
  12. Sep 19, 2013 #11

    gneill

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    Right. So if the method you employed previously doesn't result in that value, something may be amiss with your method. Better check into that.

    Note that it's common practice to roll the "negative" into the angle in order to leave the magnitude positive. Add + or - 180° to the angle.
     
  13. Sep 19, 2013 #12

    gneill

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    To keep things straight I'd suggest creating a little circuit fragment containing the "known" points and their given potential differences. Something like:

    attachment.php?attachmentid=61989&stc=1&d=1379638976.gif

    The voltage supply polarities reflect the subscript convention, namely Vab would be the potential at a with respect to b, so that by convention a is where your meter's "+" lead would go and b would be where the "-" lead would go, if you were to make the measurement Vab. The actual values assigned to the supplies may be positive or negative depending upon what you're given.

    Then, to find any potential difference between points, perform the "KVL walk" between them and sum the potential changes.
     

    Attached Files:

  14. Sep 19, 2013 #13
    so for V13=V12+V32
     
  15. Sep 19, 2013 #14

    gneill

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    Staff: Mentor

    By convention V13 would be the potential at 1 with respect to 3. So start at 3 and "walk" to 1. Pay attention to the polarities of the sources along the way.
     
  16. Sep 19, 2013 #15
    Ahhh woops V13=V12-V32
     
  17. Sep 19, 2013 #16
    V34=sqrt((V14)^2+(V32)^2+(V12)^2) right?
     
  18. Sep 19, 2013 #17
    or should it be V34=sqrt(-(V14)^2+(V32)^2+(V12)^2)
     
  19. Sep 19, 2013 #18
    hmmmmmmm V31=V32-V12 so that V34=V31-V41
     
  20. Sep 19, 2013 #19

    gneill

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    Staff: Mentor

    What are you trying to accomplish? The individual voltages are in complex form and you can sum them by summing their real and imaginary components separately.

    If you want the magnitude of the result, then do the square root of the sum of the squares of its components.
     
  21. Sep 19, 2013 #20
    so I solved V13=9.801+j2.271
     
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