how do you?

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rattus

Senior Member
Smart,

I should say that the phase angles of the phase voltages were not defined, but that is not important since the key is the angle between the load currents. 120 degrees is convenient because it allows the use of your formula.

The graphical solution is more of a learning tool than than it is a tool to be used in the field. It is helpful even to those who understand phasors. No question though that a simple formula is quicker and more accurate than a manually done graphic solution. If we do the graphics with a CAD program, then the accuracy should be the same, but that not a practical tool either.

True, you did not compute In, but others did with a misapplied formula. That is the reason for my comment.

PS: I do understand all the angles involved. I just got one of them backward. Didn't feel right at the time, but I did it anyway.
 
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bob

Senior Member
Location
Alabama
rattus said:
Someone please post a diagram showing the path of this neutral current.

There is no such path with only L-L loads.

You must have loads returned to the neutral to have any neutral current.
Rattus
I hate it when your are right. I seemed to have overlooked that insignificant point.
 

rattus

Senior Member
The Magic Number for the Trigonometrically Impaired:

The Magic Number for the Trigonometrically Impaired:

Just for fun now:

Consider any perfectly balanced 3-phase sytem. What is the line current for 1A L-L loads? We already know of course, but let's apply the magic formula,

Iline = sqrt(1^2 + 1^2 + 1*1) = sqrt(3) = 1.732

Of course this works with any current.
 

rattus

Senior Member
JohnConnolly said:
This is why I bend conduit, hang boxes and pull wire.

30 plus 60 is 90. If I am close, that's all I want to know

Hey John, someone has to do the real work. More power to you!
 
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