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Adding phasors

  1. May 11, 2009 #1
    Hi,

    I have two voltages given as v1(t) = 20cos([tex]\omega[/tex]t - 45)
    and v2(t) = 10sin([tex]\omega[/tex]t + 60)

    My task is to add them on the single form Vcos([tex]\omega[/tex]t + [tex]\theta[/tex])

    The first part is relativley easy:
    The phasors are v1 = 20[tex]\angle[/tex]-45)
    and v2 = 10[tex]\angle[/tex]-30)

    so i have 20[tex]\angle[/tex]-45) + 10[tex]\angle[/tex]-30)
    20cos(-45) = 14.14
    20sin(-45) = -j14.14
    10cos(-30) = 8.66
    10sin(-30) = -j5

    Add those up and you have 22.8 -j19.14

    But how do i get this in polar form?
     
  2. jcsd
  3. May 11, 2009 #2

    Defennder

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    Homework Helper

    It's relatively simple if you use your calculator directly. Otherwise, you could use this:

    [tex]a+jb <=> \sqrt{a^2+b^2}e^{j\theta}[/tex] where [itex]\theta[/itex] is the angle in radians from positive x-axis to the line which connects the origin to that point (a,b) in the complex plane. In this case you can tell from the complex no. directly that the number lies in the 4th quadrant.

    Or in this case you can use [tex]\theta = \tan^{-1} \frac{b}{a}[/tex].
     
  4. May 11, 2009 #3
    Okay, thanks.

    I have just one last question.

    How would you carry out a division such as: (10[tex]\angle[/tex]-90)/(250+j250)
     
  5. May 12, 2009 #4

    Defennder

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    Homework Helper

    Convert both to polar form (one is already in that form). Then you can do it easily. In general, the a+jb form is good for adding and subtracting while polar form facilitates multiplication and division.
     
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