Re: 3 hots one not
Gmreynol,
In 310.15(B)(4)(b)it refers to the common wire as a neutral,
actually it is refering to the Neutral Conductor of a wye system being a Common Conductor when connected to two phases since it is no longer a Neutral.
This would also be true if connected to one phase.
To expand on Don's post (only I will use 5 amps) it is as follows
To figure the neutral load on a wye system the formula is as follows;
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\/ (AI^2+BI^2+CI^2) - (AxB)+(BxC)+(CxA)
In otherwords it is the square root of
A phase squared + B phase squared + c phase squared - A phase x B phase + B phase x C phase + C phase x A phase
Give all three phases of value of 5 amps and figure the Neutral load as shown above.
5x5 + 5x5 + 5x5 = 75 - 5x5 + 5x5 + 5x5 = 75
75 - 75 = 0 the square root of Zero is Zero, so the Grounded Conductor is indeed a Neutral Conductor.
Now using just two phases A & B figure it again.
5x5 + 5x5 = 50 - 5x5 = 25
50 - 25 = 25 the square root of 25 is 5, so in this example you can see the Grounded Conductor carries the same current as the two phase conductors.
Charlie b could probably show it better.
Roger