MWBC 2 Pole GFCI Breaker / vector analysis

[MD84]

Thanks a lot. Both of the explanations above confirm my understanding. I appreciate both the practical guidance as well as the more technical knowledge. I am always learning...

 

[MD84]

Thanks. I can see how finding those right triangles allows the use of trigonometry. It is fun to see how the answer can be derived from drawing a picture and measuring lines or then derive equations based on those shapes to solve with more accuracy.

151213-2145 EST

MD84:

From your first post.I get 17.32 . An interesting result from
sq-root of ( (10*0.5*sq_root 3)^2 + (20-10*0.5)^2 ) )
sq-root of ( ( 25*3) + 225 ) = sq-root (100) * sq-root 3 = 10*1.732 = 17.32

sine 60 deg = 0.5*sq_root 3 = 1.723/2 = 0.866
cosine 60 deg = 0.5
sine 30 deg = 0.5

I am giving you verification on your current. So from your vector diagram you should be able to figure out the angles of interest. Note: you have a right triangle with sides 10*0.5*sq_root 3 = 8.866 and 20-10*0.5 = 15 .

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[kwired]

Thanks a lot. Both of the explanations above confirm my understanding. I appreciate both the practical guidance as well as the more technical knowledge. I am always learning...

If you clamped a ammeter on those three conductors it should read zero if no current is taking any external paths, you probably realized that going into this but maybe didn't give much thought as to why on a wye system :blink: The GFCI is essentially doing just this but is sensitive enough to detect imbalance in the milli-amp ranges.