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Posted By: u2slow Unbalanced delta calculations - 11/02/04 06:44 AM
I'm in 3rd year school right now, and some of the 3-phase math is getting me down... unbalanced delta loads specifically - trying to figure out the line currents.

Our instructor's method works but the values he chooses and when he uses them seems arbitrary. Put the wrong two phasors head to tail, and the resultant in meaningless.

Anyone have some tips or tricks?
Posted By: crash Re: Unbalanced delta calculations - 11/07/04 02:14 AM
There are no "Tricks" in 3 phase, phasor calculations. Just math.

Can You post a specific problem with (or with-out) your instructors solution? And also what your instructor choose in an "arbitrary" manor?
Posted By: u2slow Re: Unbalanced delta calculations - 11/07/04 04:01 AM
I figured it out [Linked Image] (Thanks for responding!)

My instructor was busy disecting each phasor into X and Y values with sin/cos functions and subtracting them. It just didn't make sense until he showed what he was doing with a phasor diagram and finally had a 3-blade propellor drawn on the board. *Now* that made a whole lot more sense! I worked through a few more problems with the diagrams only... and solved them all. Now the original purely-arithmetic method is starting to make sense. Go figure [Linked Image]

What I found arbitrary was how he assigned the A, B, & C phase subscripts to the current phasors. I now know it doesn't matter - you get same answers no matter how you do it. [Linked Image]

Here is one of the problems. I'll see if I can transcribe the solution and post it up.


The phase currents in a delta-connected, unbalanced load are 20A, 32A, and 40A. Calculate the three line currents, if all loads are resistors.


52.92A, 62.35A, 45.43A
Posted By: crash Re: Unbalanced delta calculations - 11/08/04 06:13 AM
I got the same answers as you on that problem. It took a little while for me to remember how to do it though.
Posted By: Peter Re: Unbalanced delta calculations - 11/10/04 03:50 AM
I am confused here. What is the equation? I thought it would be amps times the square root of three.
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