mivey
Senior Member
To start with, it appears you reversed R1 and R3.
120/208 3-ph, 4W panelboard amps on A phase?
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mivey
Senior Member
Just a note error in the drawing. I get the same results as you otherwise.
What is the net effect of these three delta-connected loads on line current? Express as both polar and rectangular
Drawing shows 3Ø load as three 1Ø delta-connected line-to-line loads. Those loads are expressed as 10A each... when they are only 10/sqrt(3)=5.77A each. Net effect is 10A of line current.
mivey
Senior Member
That was the intent, and following that intent, I got the same results as Ingenieur.
The first thing I noticed was that the drawing had an error. The load 1 resistor should be labeled "R3" and the load 2 resistor should be labeled "R1", going by the values listed on the right side of the page. I mentioned it because I was not sure if that was going to result in Ingenieur having the incorrect results and started to not check the calcs. I decided to run the numbers anyway and his results are correct.
Man, I can't believe it's going right over both your heads. :slaphead:
How can you get a 36.3A load on Line A when you only have three 10A loads connected to it? Doesn't matter what angle/phase the currents are, the total line current cannot exceed 30A.
Thank you for taking the time, not many would do that
andcI know my writing is terrible lol
and from your previous posts I have a high degree of confidence in your review
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The second calc considered that
but the problem stated line currents so the sum of the delta currents
I corrected myself in the edited post
You mistook the problem then. 3Ø loads are stated in terms of line current, not phase current. That is what I meant by this.
I did it both ways and posted both
1 line at 10 summed after delta
2 delta legs at 10 hence line >36
got class in 10 minutes
I'll leave it at this
I for loads 1 and 3 on phase A (A-B) must be in phase
and load 1 phase B load must be the same as load 1 phase opposite sign
load 2 phase A (the neut load) must differ by 30 deg
regardless I enjoyed this and thanks to all for indulging me
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mivey
Senior Member
Ingenieur's result is correct for what he drew. If you have a different circuit in mind, then the answer would be different.
This is where you are wrong. I(s) for load 3 (the 3Ø load) must be shifted 30°. Line current of a resistive 3Ø load is in phase with the line-to-neutral voltage. Perhaps you'll figure that out someday.
mivey
Senior Member
If you want 10 amp line currents then that is a different circuit. How far back to the post actually describing the circuit you are talking about?
I presented the problem...
And this is what he replied with originally...
OK
I believe you
where did I screw up lol
excuse the scribble
Don't need to draw the picture and I already did the hand part (via computer) back in post #45...
...using your angle convention. It agrees with my calculator, correcting for difference in angle convention.
mivey
Senior Member
If you both meant 10 amp line currents for each load then Rab=20.785, Ran=12, Rab=Rbc=Rca=36. Unless I fat-fingered it then you get:
Ia = 29.093 at 20.104d
Ib = 19.319 at 165d
Ic = 10 at -90d
In = 10 at 30d
add: which agrees with your #45. Y'all must have been talking about two different things.
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