No zero sequence in line-to-line fault

In a line-to-line fault there is no path to ground and the two fault currents are equal and opposite, so $I_a + I_b + I_c = 0$ and the zero sequence current $I_{a0} = 0$; the fault current depends only on $Z_1$ and $Z_2$.

Model: at the fault point on the line, the system is replaced by its Thevenin sequence networks: prefault voltage $V_f$ and impedances $Z_1$, $Z_2$, $Z_0$ seen from the fault point.

Boundary conditions (fault between phases b and c through $Z_f$)

  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib

$$I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b$$

Sequence currents

$$\begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned}$$

Hence

$$I_{a0} = 0, \qquad I_{a2} = -I_{a1}$$

So the zero sequence current is absent: with no connection to ground, current that leaves through phase b returns through phase c, and the three line currents sum to zero.

Voltage condition

$$\begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned}$$

so

$$V_{a1} - V_{a2} = Z_fI_{a1}$$

With $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$:

$$I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f}$$

Fault current

$$\begin{aligned} I_f = I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = (a^2 - a)I_{a1} \\ &= -j\sqrt3\,I_{a1} = \frac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f} \end{aligned}$$

$Z_0$ does not appear in the expression, confirming that the zero sequence network plays no part.

Interconnection of sequence networks

        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)

The positive and negative sequence networks are connected in parallel (in opposition) through $Z_f$; the zero sequence network is left open.

Consequences

• The L-L fault current does not depend on the grounding of the neutral or on line zero sequence impedance.
• With $Z_1 = Z_2$ and $Z_f = 0$: $I_f = \frac{\sqrt3}{2}\cdot\frac{V_f}{Z_1} = 0.866 \times I_{3\phi}$, i.e. smaller than the three-phase fault current.
• Also $V_{a0} = 0$ at the fault, since the zero sequence network carries no current.