L-G fault in phase B

For an L-G fault on phase B, the sequence currents of phase B are equal, $I_{b0} = I_{b1} = I_{b2} = I_b/3$, so the three sequence networks are connected in series. In terms of phase-a reference components they are joined through phase shifts: $I_{a1} = aI_{a0}$, $I_{a2} = a^2I_{a0}$.

Generator sequence equations (phase a reference)

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

Boundary conditions (bolted fault on B)

$$I_a = 0, \qquad I_c = 0, \qquad V_b = 0$$

Sequence currents

$$\begin{aligned} I_{a0} &= \tfrac13(0 + I_b + 0) = \tfrac13I_b \\ I_{a1} &= \tfrac13(0 + aI_b + 0) = \tfrac13aI_b = aI_{a0} \\ I_{a2} &= \tfrac13(0 + a^2I_b + 0) = \tfrac13a^2I_b = a^2I_{a0} \end{aligned}$$

Phase-b components are $I_{b0} = I_{a0}$, $I_{b1} = a^2I_{a1} = I_{a0}$, $I_{b2} = aI_{a2} = I_{a0}$. So

$$I_{b0} = I_{b1} = I_{b2} = \frac{I_b}{3}$$

The same current flows in all three networks when they are written for phase b, which means a series connection.

Voltage condition

$$V_b = V_{a0} + a^2V_{a1} + aV_{a2} = V_{b0} + V_{b1} + V_{b2} = 0$$

Substituting:

$$-Z_0I_{a0} + a^2(E_a - Z_1aI_{a0}) - aZ_2a^2I_{a0} = 0 \;\Rightarrow\; I_{a0} = \frac{a^2E_a}{Z_0 + Z_1 + Z_2} = \frac{E_b}{Z_0 + Z_1 + Z_2}$$

since $E_b = a^2E_a$. Hence

$$I_f = I_b = 3I_{a0} = \frac{3E_b}{Z_1 + Z_2 + Z_0}$$

Interconnection of sequence networks

Drawn with phase b as reference (EMF $E_b$ in the positive network):

      Ib1 -->
  +--(Eb)--Z1--F1  N2--Z2--F2  N0--Z0--F0--+
  |             |__|        |__|           |
  N1 ---------------------------------------+
     Ib0 = Ib1 = Ib2 = Ib/3

With phase a as reference the same series loop is used, but the currents entering the networks are linked by phase shifts: $I_{a1} = aI_{a0}$ and $I_{a2} = a^2I_{a0}$ (ideal phase-shifting transformers of $+120^\circ$ and $+240^\circ$ in the positive and negative networks).

Justification: the series connection satisfies both conditions: equal phase-b sequence currents and zero total phase-b voltage. The result is the same as an L-G fault on phase a, only shifted by $-120^\circ$, as expected for a symmetrical machine.