Is this calc right?

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mbrooke

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Ok, I am a tad confused here. Shouldn't a lower power factor on one leg actually cause the neutral current to drop- rather than rise? Either my vector math is off or this spread sheet is off- but I can not figure out which one.
 

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Ok, I am a tad confused here. Shouldn't a lower power factor on one leg actually cause the neutral current to drop- rather than rise?

The angle is incorrect, but the magnitude is right...123.4A. The neutral current is larger than the balanced 100A, in this case.
 
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Kirchhoff is rolling in his grave.:D

Re-think what unbalanced current means and what it means when we say the neutral carries the unbalanced current.

nope
Ia, Ib, Ic are entering the node
In leaving
opposite signs (by convention and power flow)

Ia + Ib + Ic -In =0
Ia + Ib + Ic = In
not In = -(I a + Ib + Ic)
 
nope
Ia, Ib, Ic are entering the node
In leaving
opposite signs (by convention and power flow)

Ia + Ib + Ic -In =0
Ia + Ib + Ic = In
not In = -(I a + Ib + Ic)
The empirical value(s) get the sign change where appropriate. Not the vectorial representatives. K's Law says all current entering and leaving a node sum to zero. Making one or more variables a negative is making the result something other than the sum. Entering and leaving is merely stated as a required event for the occurrence of the anomaly.
 
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