assume that the neutral was not earthed.
2006-12-11 19:00:25 · 5 answers · asked by sucheen02 1 in Science & Mathematics ➔ Engineering
It would depend on the type of loads? If there were only 3phase motors, no current would flow in the neutral. I think that with any single phase loads the neutral would be carrying the same as the highest single phase load at any given moment?! Easiest to measure with a clamp meter!
2006-12-11 19:51:17 · answer #1 · answered by jayktee96 7 · 0⤊ 0⤋
the voltage's are 120 degrees out of phase.
the currents lead or lag the voltages depending on the type of load so the values you have given would usually (but not always) have an angle as well as amplitude.
assuming the angles are negligible (ie the loads are not inductive or capacitive)
draw the currents at angles 0 degrees (L1 - R), 120 degrees (L2-Y) and -120 degrees (L3-B) with a length scaled to fit paper.
they should all originate from a centre point and look like a y shape.
next "move" current I(L2) onto the end of I(L1).
next move current I(L3) onto end of I(l2)
from the centre point draw a line onto the end of I(L3)
this will be the neutral current.
only in a 4 wire system star connected load.
neutral is earthed at the supply side. (dont know what you ment from the last line.
other way is to convert all to cartesian/rectangular form and add together. see "complex numbers" to calculate.
2006-12-12 08:14:15 · answer #2 · answered by Mark G 2 · 0⤊ 1⤋
because current flows back to earth through the neutral (you cannot have a neutral with out an earth as this is done at the source) and if the 3 phases are unbalenced the currents do not cancel each other out on ther return path as they do when all loads are ballenced. to calculate it you need to work out the phaser diagram but im affraid there is not enough room here to explain that or you could just use a clamp meter if you want the current reading
the new colours are
red = brown
yellow= black
blue = gray
neutral = blue
2006-12-12 03:22:50 · answer #3 · answered by jk 2 · 0⤊ 0⤋
The load will be (assuming no faults and ideal components etc) the algebraic sum of the currents, according to Kirchoff's Current Law. Take into account the relative phase differences (assuming an AC current). So therefore break the currents into components (I prefer rectangular for summation, though a good engineering calculator won't mind polar), add the real components and imaginary components, and the result is the load.
2006-12-12 03:01:03 · answer #4 · answered by cedsinsane 1 · 0⤊ 1⤋
Don't know the answer,but have to correct you on the phasing,its now L1,L2,and L3.
2006-12-11 19:27:15 · answer #5 · answered by Ken J 4 · 0⤊ 0⤋