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Load on Nuetral

Load on Nuetral

Load on Nuetral

(OP)
120/208 wye system

Phase
A-13 A
B-7 A
C-5.5 A
N-?

What calculations are made to find the load on the nuetral conductor in this unbalanced system.

RE: Load on Nuetral

In order to do this properly you need the phase angles of each current as well as the absolute value of the current.
Then you find the zero sequence current this is also the current flowing through the neutral of your system.

I0=1/3(Ia+Ib+Ic)  (they are all vectors)   

RE: Load on Nuetral

consider them as three vectors , 120 degree apart..and add them up. Analytically or graphically....

For more read text books..

RE: Load on Nuetral

(OP)
So I0 is the zero sequence current?  My phase angles are 120 degrees.  How is that factored into the equation?

That is roughly 8.5 A from that formula.  

RE: Load on Nuetral

It's like adding three vectors together.  The vectors are separated by 120 degrees.  If you add these three vectors together vectorially, the resultant is your neutral current.  

RE: Load on Nuetral

Neutral current is 3*I0. No need for the 1/3 factor. The 8.5 answer came from numerical rather than vector addition, and was wrong even before the 1/3 factor was applied.

RE: Load on Nuetral

(OP)
How do I add 13, 7, 5.5 vectors 120 degrees apart.
Does someone have a  solution to this that I can compare my math to.  I dont think I'm adding these vectors correctly.


 

RE: Load on Nuetral

Take the square root of  (a^2 + b^2 + b^2) - (a*b + b*c + a*c)

a^2 = a squared

That comes to 6.87 amps in this case.

Don

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