Adding phasors

  • Thread starter James889
  • Start date
  • #1
192
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?
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
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].
 
  • #3
192
1
Okay, thanks.

I have just one last question.

How would you carry out a division such as: (10[tex]\angle[/tex]-90)/(250+j250)
 
  • #4
Defennder
Homework Helper
2,591
5
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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