PW-A5 · Three-phase AC

Unbalance, first look: where the nonzero sum goes

Under balance (A1) the three currents sum to zero, so the neutral is empty. When one phase drifts, that sum leaves zero and the leftover flows down the Y neutral. Drag the balance apart yourself to learn where the current escapes when balance breaks.

Drag I_a to break the balance

Three current phasors are shown. Drag the tip of I_a to change its size or direction. The residual sum (gold) grows from zero. That gold vector is exactly the current flowing down the Y neutral.

Drag the tip of I_a to create unbalance.

Neutral current |I_N| = |I_a + I_b + I_c|

|I_N| ≈ 0.56   (|I_a| ≈ 1.35)

At the balanced spot |I_N| = 0. The farther off, the larger the neutral current.

Balanced means an empty neutral

As in A1, three currents of equal size 120 degrees apart sum to zero. So the three currents meeting at the Y neutral point erase each other and no current flows in the neutral. This is why a balanced system can size the neutral thin or omit it.

Off balance, the leftover leaks out

When one phase changes size or angle, the three currents no longer cancel cleanly. The leftover that left zero has only one place to go: the neutral. I_N = I_a + I_b + I_c. The gold vector you dragged is exactly this sum, and the more the load leans to one side, the larger the neutral current.

This leftover has a name · zero sequence

A third of the leftover is the zero-sequence component I_0 = (I_a + I_b + I_c)/3. Then the neutral current is I_N = 3 I_0. Balance is just the special case I_0 = 0. Delta has no neutral, so the same leftover has nowhere to exit and flows as a circulating current inside the closed loop. The systematic tool for unbalance is the sequence component, completed as positive, negative and zero sequence in the later D2.

ObserveI_N = I_a + I_b + I_c

The neutral current is the sum of the three currents.

Choose평형 → I_N = ?

When balanced the sum is 0 (A1).

Fill inI₀ = (I_a + I_b + I_c) / ?

The zero sequence is a third of the leftover.

On your ownI_N = ?

Delta has no neutral, so it becomes a circulating current.

Back to the first screen

Putting I_a at the balanced spot made the gold residual vanish, and dragging it out made the residual grow. That gold vector is the neutral current I_N = I_a + I_b + I_c = 3 I_0. The "sum is zero" of A1, where section A began, turns out to be one special point of the general situation of unbalance. Balance is not a rule but a balance we arranged.

In an unbalanced three-phase set the three currents (or voltages) do not sum to zero. The leftover flows as the Y neutral current I_N = I_a + I_b + I_c = 3 I₀ (the zero sequence). Balance (A1) is the special case I₀ = 0. Delta has no neutral, so the same leftover becomes a closed-loop circulating current.

The next step

Section A (three-phase AC) ends here. From the next unit (PW-B1) we look at the shape of power itself: the power triangle, where active, reactive and apparent power bind into a right triangle. When a load pulls its current out of step with the voltage, that misalignment becomes reactive power, and you see at a glance which side of the triangle the power factor cosφ is. The zero sequence seen here returns in the later symmetrical components (D2).