Need help in this conversion please

Discussion in 'Homework Help' started by S850, Jul 1, 2015.

  1. S850

    Thread Starter New Member

    Jul 1, 2015
    1
    0
    I have the following information
    three phase fault : I = 6060.08 <-175.6
    Single phase fault: I = 4618.63 <170.7

    How do I convert this to get the rectangular form, (R + jX) ??
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,804
    This looks like a homework problem and will get more attention over there.
     
    S850 likes this.
  3. Veracohr

    Well-Known Member

    Jan 3, 2011
    552
    76
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hi,

    You could use the relations:
    A=sqrt(R^2+I^2)
    Ph=atan2(I,R)

    to find R and I, which are the real and imaginary parts.
    A is the amplitude of the sine, Ph is the phase.
    So for:
    100*sin(wt+0.2)

    the amplitude A=100 and the phase Ph=0.2 in radians.
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,397
    497
    The best way to do it is to draw the right triangle.
    You are given the hypotenuse of the right triangle, and the angle from the origin.

    triangle.jpg

    You have right triangle.
    Hypotenuse is 4618.63
    The angle inside the triangle is 180°-170.7°=9.3°
    You need to find the R part and X part.

    cos(9.3°)=R part/hypotenuse
    cos(9.3°)=R part/4618.63
    R part= 4618.63*cos(9.3°)
    This gives you the magnitude of R part.
    Since R part lies on the left of the origin, it will have negative sign.

    You should be able to figure out the X part on your own.
     
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