ripplenwinder

 Profile Answers by ripplenwinder Questions by ripplenwinder

• Dec 10th, 2013

 

Good guestion Ill get back to you on that one.

ripplenwinder

 Profile Answers by ripplenwinder Questions by ripplenwinder

• Dec 10th, 2013

 

We can define the 3-phase voltages as:

1) v1 = Vp x sin(u);
2) v2 = Vp x sin(u +120);
3) v3 = Vp x sin(u -120)
4) vn^2 = v1^2 + v2^2 +v3^2 = K
5) sin(u +/- v) = sin(u)cos(v) +/- cos(u)sin(v);
6) sin^2(u) + cos^2(u) = 1
7) vn^2 = sin^2(u) + sin^2(u)cos^2(120) + cos^2(u)sin^2(120) + sin^2(u)cos^2(-120) + cos^2(u)sin^2(-120)
= sin^2(u) + sin^2(u)(0.25) + cos^2(u)(0.75) + sin^2(u)(0.25) + cos^2(u)(0.75)
= 1.5( sin^2(u) + cos^2(u)) = 1.5

I got the anwser from someone else. I still have to check his work! He looks smart doesnt he. No but thank you for asking this question I am intrested in better understanding how this works.