No zero sequence current in delta lines

In a delta connected system there is no neutral or ground return path, so the three line currents always add to zero; hence their zero sequence component $I_{a0} = \frac13(I_a+I_b+I_c)$ is zero.

Line currents of a delta

            a
           /\
   Ia -->  /  \
     Iab  /    \  Ica
         /      \
 Ib --> b--------c <-- Ic
           Ibc

Applying KCL at the three corners:

$$\begin{aligned} I_a &= I_{ab} - I_{ca} \\ I_b &= I_{bc} - I_{ab} \\ I_c &= I_{ca} - I_{bc} \end{aligned}$$

Zero sequence line current

$$\begin{aligned} I_{a0} &= \frac13 (I_a + I_b + I_c) \\ &= \frac13\left[(I_{ab}-I_{ca}) + (I_{bc}-I_{ab}) + (I_{ca}-I_{bc})\right] \\ &= 0 \end{aligned}$$

This holds for any values of the delta currents, balanced or unbalanced. So line currents into a delta never contain zero sequence components.

Phase (delta) currents in sequence form

Let the delta currents have components $I_{ab0}, I_{ab1}, I_{ab2}$. Then

$$I_a = I_{ab} - I_{ca}$$

For the zero sequence part, $I_{ab0} = I_{bc0} = I_{ca0}$, so it cancels in every line current. For positive and negative sequence:

$$I_{a1} = (1 - a)\,I_{ab1} = \sqrt3\,\angle -30^\circ\, I_{ab1}, \qquad I_{a2} = (1 - a^2)\,I_{ab2} = \sqrt3\,\angle 30^\circ\, I_{ab2}$$

Consequences

• A zero sequence current can circulate inside the delta (for example, from induced zero sequence voltages or third harmonics) but it cannot leave the delta into the lines.
• In the zero sequence network, a delta connected load or winding is open from the line side. For a transformer, the delta winding is shown as a short to the reference on its own side (current circulates) but open to the line.
• Similarly, the line-to-line voltages of any system have no zero sequence component because $V_{ab}+V_{bc}+V_{ca} = 0$.