L-L fault: I0 absent and I1 = −I2

For a line-to-line fault (phases b and c) on a synchronous generator, the line currents sum to zero, so $I_{a0} = 0$; and transforming the boundary conditions gives $I_{a2} = -I_{a1}$, i.e. the positive and negative sequence currents are equal in magnitude (opposite in phase).

Generator and fault

An unloaded generator produces balanced EMFs; its sequence equations are

$$V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2}, \qquad V_{a0} = -Z_0I_{a0}$$

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

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

Since $a - a^2 = j\sqrt3$:

$$I_{a1} = \frac{j\sqrt3}{3}I_b = \frac{j}{\sqrt3}I_b, \qquad I_{a2} = -\frac{j}{\sqrt3}I_b$$

So $|I_{a1}| = |I_{a2}| = |I_b|/\sqrt3$: the positive and negative sequence currents have equal magnitude; they are 180° apart. The zero sequence current is absent because there is no ground connection and no neutral current flows ($I_n = 3I_{a0} = 0$).

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

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

Fault current

$$I_f = I_b = -I_c = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} = \frac{-j\sqrt3\,E_a}{Z_1 + Z_2 + Z_f}$$

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.

Because the two networks form one series loop, the current entering the negative sequence network equals the current leaving the positive network, which is another way of seeing $I_{a1} = -I_{a2}$ in magnitude.

Fault-point voltages (bolted fault)

$V_{a1} = V_{a2} = E_a - Z_1I_{a1}$, $V_{a0} = 0$, so $V_a = 2V_{a1}$ and $V_b = V_c = -V_{a1}$.