# Formula for calculating the current in the neutral of an unbalanced 3 phase system

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

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.

• 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). Commented Mar 25, 2022 at 11:46

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

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

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.

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