L-L fault: I1 and I2 equal and opposite

In a line-to-line fault, $I_{a2} = -I_{a1}$: the positive and negative sequence currents have equal magnitude and are 180° out of phase, while $I_{a0} = 0$.

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

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

Now $a - a^2 = j\sqrt3$ and $a^2 - a = -j\sqrt3$, so

$$I_{a1} = \frac{j\sqrt3}{3}I_b = \frac{I_b}{\sqrt3}\angle90^\circ, \qquad I_{a2} = \frac{-j\sqrt3}{3}I_b = \frac{I_b}{\sqrt3}\angle{-90^\circ}$$

Both have magnitude $|I_b|/\sqrt3$ and differ in angle by 180°: they are equal and opposite.

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: $I_f = I_b = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} = \dfrac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f}$.

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.

The positive network drives $I_{a1}$ out of F1 and the same current enters F2 of the negative network, so $I_{a2} = -I_{a1}$.