Harmonic Current

Here is the explanation:

For Fundamental components:

In a three-phase, four-wire system, single-phase line-to-neutral load currents flow in each phase conductor and return in the common neutral conductor. The three phase currents are separated by 120°; and for balanced three-phase loads, they are equal. When they return in the neutral, they cancel each other out, adding up to zero at all points. Therefore, for balanced three-phase loads, neutral current is zero.

Mathematically,

Ia = I sin (wt)

Ib = I sin (wt - 120)

Ic = I sin (wt + 120)

In = Ia + Ib + Ic = I{ sin (wt) + sin (wt – 120) + sin (wt + 120)} = 0

For 2^nd and all even Harmonic components:

For 2nd harmonic currents separated by 120°, cancellation in the neutral would also be complete with zero neutral current.

Mathematically,

Ia = I₂ sin 2(wt) = I₂ sin (2wt)

Ib = I₂ sin 2(wt - 240) = I₂ sin (2wt + 120)

Ic = I₂ sin 2(wt + 240) = I₂ sin (2wt - 120)

In = Ia + Ib + Ic = I₂ {sin (2wt) + sin (2wt + 120) + sin (2wt - 120)} = 0

This is true in the same way for all even harmonics.

For 3^rd and all multiples of 3^rd Harmonic components:

For 3^rd harmonic currents, the return currents from each of the three phases are in phase in the neutral and so the total 3^rd harmonic neutral current is the arithmetic sum of the three individual 3^rd harmonic phase currents.

Mathematically,

Ia = I₃ sin 3(wt) = I₃ sin (3wt)

Ib = I₃ sin 3(wt - 120) = I₃ sin (3wt - 360) = I₃ sin (3wt)

Ic = I₃ sin 3(wt + 120) = I₃ sin (3wt + 360) = I₃ sin (3wt)

In = Ia + Ib + Ic = I₃{ sin (3wt) + sin (3wt) + sin (3wt)} = 3I₃ sin (3wt)

This is true in the same way for all multiples of 3^rd harmonics.

-MS