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In the case of a 3 phase wye unbalanced configuration such as the following: enter image description here

we can compute the amount of current going through the neutral wire using geometry, in this case it is \$\sqrt{(\cos(150^{\circ})\cdot 5 + \cos({30^{\circ}}) \cdot 20 )^{2} + (\sin({150^{\circ}}) \cdot 5 + \sin({30^{\circ}}) \cdot 20 -15)^{2}} \approx 13.23A\$

Therfore the amount of current going through the neutral wire is approximately 13.23A. So far so good, now, what would happen if I showed you this configuration instead:
enter image description here

In this case the red resistance is connected in between L1 and L2, thus increasing the voltage to \$\sqrt{3} \cdot V_{phase}\$, how much current is now flowing through the neutral wire?

How would I go about calculating the amount of current, how many degrees is it out of phase now, can I still use basic geometry as I did so in the other example?

Thanks

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    \$\begingroup\$ Ignore the 35A load. In other words, look at only the two loads connected between L2 & L3. \$\endgroup\$ – Dwayne Reid Jan 4 at 18:27
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    \$\begingroup\$ PS: I intensely dislike the word "optional" on the Neutral conductor. The Neutral is required if you don't want to over-voltage the lower current loads. \$\endgroup\$ – Dwayne Reid Jan 4 at 18:29
  • \$\begingroup\$ It only says optional in the event of all loads being balanced, but yeah, you're right, better not risk it just in case. \$\endgroup\$ – rr1303 Jan 4 at 18:46
  • \$\begingroup\$ Just use your formulas and fill in 0 for what used to be the red current. But don't let the answer confuse you since the result is the same! (with different phase though.) Call it coincidence, or what you like... :-) With different values it wouldn't have been the same though. \$\endgroup\$ – MartinF Jan 4 at 21:15
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I find it easy to use complex numbers for something like this.

In your first example:

- Red:   20*cos(120)  +j20*sin(120) =  -10+j17,32
- Green:                            =   15+j0
- Blue:   5*cos(-120) +j5*sin(-120) = -2.5-j4.33

That gives a result of:

(-10+15-2.5)+j(17.32+0-4.33) = 2.5+j12.99

Total current through neutral wire:

sqrt(2.5^2+12.99^2) = sqrt(175) = 13.23

Angle:

arctan(12.99/2.5) = 79.1 degrees

In your second example, the red current does not influence the neutral current, so the red current is essentially 0:

- Red:                              =    0+j0
- Green:                            =   15+j0
- Blue:   5*cos(-120) +j5*sin(-120) = -2.5-j4.33

That gives a result of:

(0+15-2.5)+j(0+0-4.33) = 12.5-j4.33

Total current through neutral wire:

sqrt(12.5^2+(-4.33^2)) = sqrt(175) = 13.23

Angle:

arctan(-4.33/12.5) = -19.1 degrees

Hope this helps.


  • I think in America they use the z=a+bi notation instead of z=a+jb? Not sure.
  • I should really figure out how to post real formulas, sorry
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    \$\begingroup\$ Thanks for the answer! For posting formulas you can use latex, there is a simple editor that you can use at codecogs.com/latex/eqneditor.php to help you out if don't have much practise. \$\endgroup\$ – rr1303 Jan 4 at 22:17
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    \$\begingroup\$ BTW: we use z=a+jb in North American engineering practice as well -- it's the mathematicians who use a+bi instead. (The mathematicians were first with i, us EEs use j instead to avoid confusion with I for current....) \$\endgroup\$ – ThreePhaseEel Jan 5 at 3:38

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