Morning All, my first post here! I am looking at 3 phase calculations and how to find the current in the neutral for an unbalanced 3 phase system. I am aware that it can be drawn out to find the current but I've come across this formula Calculation of the neutral current

Can anyone point me at where this formula derives from? I am sure it's to do with vector addition but can't find anything that goes from first principles. Thanks for your time, I'm sure its obvious but I can't seem to figure it out.

  • 1
    \$\begingroup\$ Not a valid question. If you have a formula then you should know where it came from. Asking others to figure that out is asking for opinions (off-topic). \$\endgroup\$
    – Andy aka
    Commented Mar 25, 2022 at 11:46

1 Answer 1


Yes, it has to do with vector addition. I'm gonna take this from a math-based approach and define our three line currents as \$L_{1}, L_{2}, L_{3}\$. If I orient my three vectors such that \$L_{1}\$ is oriented on the vertical axis in our 2D plane, then \$L_{2}, L_{3}\$ are both oriented with a minor angle of 30 degrees from the horizontal axis as they each have a 120 degree phase difference enter image description here

Now, if I break up each vector into their horizontal/vertical components I get $$ L_{1y} = L_{1}\\ L_{1x} = 0\\ L_{2y} = -L_{2}sin(30^\circ) = -\frac{1}{2} L_{2}\\ L_{2x} = -L_{2}cos(30^\circ) = -\frac{\sqrt{3}}{2} L_{2}\\ L_{3y} = -L_{3}sin(30^\circ) = -\frac{1}{2} L_{3}\\ L_{3x} = L_{3}cos(30^\circ) = \frac{\sqrt{3}}{2} L_{3}\\ $$

Now, we can form a right triangle with the sum of the x components and y components. The hypotenuse is the neutral current enter image description here

Basically, going through the algebra, we end up at the equation you posted.

$$ L_{N}^2 = (\frac{\sqrt{3}}{2} L_{3} - \frac{\sqrt{3}}{2} L_{2})^2 + (L_{1} - \frac{1}{2} L_{2} -\frac{1}{2} L_{3})^2 $$

Factoring out \$ \frac{\sqrt{3}}{2} \$ and \$ -\frac{1}{2} \$

$$ = \frac{3}{4} ( L_{3} - L_{2} )^2 + (L_{1} - \frac{1}{2} (L_{2} + L_{3}))^2\\ $$ Expanding the squares $$ = \frac{3}{4}[L_{2}^2 + L_{3}^2 - 2L_{2}L_{3}] + [L_{1}^2 + \frac{1}{4} (L_{2}^2 + L_{3}^2+2L_{2}L_{3}) - L_{1}(L_{2}+L_{3})]\\ = \frac{3}{4}L_{2}^2 + \frac{3}{4}L_{3}^2 - \frac{3}{2}L_{2}L_{3}+L_{1}^2 + \frac{1}{4}L_{2}^2 + \frac{1}{4}L_{3}^2 + \frac{1}{2}L_{2}L_{3} - L_{1}L{2} - L_{1}L_{3} $$

And finally, combining like terms and we get the square of your result

$$ L_{N}^2 = L_{1}^2 + L_{2}^2 + L_{3}^2 - L_{1}L{2} - L_{1}L_{3} - L_{2}L_{3} $$

Square rooting both sides and you have your equation.

  • \$\begingroup\$ That's fantastic thank you so much for your assistance really appreciate it. \$\endgroup\$
    – Terry70
    Commented Mar 28, 2022 at 8:56

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