208 three phase

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domnic

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Electrical Contractor
If I have a 208 three phase service and have a 10 amp load on A and B on a MWB circuit the neutral will pull 10 amps . what is the trigonometric function for this circuit?
 
The neutral current, assuming that the phase angles are 120°, is equal to the square root of (a² + b² + c² - (a*b) - (b*c) -(a*c)).

I think the trig function is the law of cosines, but not sure.
 
What is mwb?
so
a = 10 ang 0
b = 10 ang 120
c = 0
n = 10 ang x
??
you need an expression to calc c magnitude?
 
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don_resqcapt19 said:
The neutral current, assuming that the phase angles are 120°, is equal to the square root of (a² + b² + c² - (a*b) - (b*c) -(a*c)).

I think the trig function is the law of cosines, but not sure.
Click to expand...
Yes, it is an application of the Law of Cosines. Your formula is for full boat. For only two loads it reduces to the square root of (a² + b² - (a*b)).
 
don_resqcapt19 said:
The neutral current, assuming that the phase angles are 120°, is equal to the square root of (a² + b² + c² - (a*b) - (b*c) -(a*c)).

I think the trig function is the law of cosines, but not sure.
Click to expand...

The law of cosines is a modified Pythagorean theorem for cases when the angle opposite side c is not 90 degrees.

Instead of the famous a^2 + b^2 = c^2, you have:
c^2 = a^2 + b^2 - 2*a*b*cos(C)

where C is the angle opposite side c.

In our applications, C = 120 degrees, the angle between phases, and its cosine is 1/2. So when C=120 deg, you get:
c = sqrt(a^2 + b^2 - a*b)

This looks familiar for the neutral current (c) corresponding to an MWBC on phases A and B.
 
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