# 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
20,064
5,665
This looks like a homework problem and will get more attention over there.

S850 likes this.

Jan 3, 2011
600
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4. ### MrAl Distinguished Member

Jun 17, 2014
3,615
758
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
3,922
601
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.

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.