L-G, L-L and L-L-G faults at bus q

Assumptions: $E_1 = E_2 = 1\angle0^\circ$ pu, no pre-fault load, all values on a common base, $Z_f = 0$. The values printed as "X1" are the zero-sequence reactances ($X_0$).

Sequence networks seen from bus q

Positive (negative similar, no emf):
 E1-j0.16-(p)-T1 j0.1-line j0.11-T2 j0.1-(q)-j0.2-E2
                                          |
                                          F
Zero:
 T1 delta on p side, T2 delta on q side
 -> line + transformers form a path that does not
    reach q.   q --j0.15-- ref  (G2 solidly grounded)

$$\begin{aligned} Z_1 &= j(0.16+0.1+0.11+0.1)\ \|\ j0.2 = j0.47 \| j0.2 = j0.1403\ \text{pu} \\ Z_2 &= j(0.17+0.1+0.11+0.1)\ \|\ j0.22 = j0.48 \| j0.22 = j0.1509\ \text{pu} \\ Z_0 &= j0.15\ \text{pu (only G2; the Δ of T2 blocks the rest)} \end{aligned}$$

(i) L-G fault (phase a)

$$\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{1}{j(0.1403+0.1509+0.15)} = \frac{1}{j0.4412} = -j2.267\ \text{pu} \\ I_a &= 3I_{a1} = -j6.800\ \text{pu}, \quad I_b = I_c = 0 \end{aligned}$$

(ii) L-L fault (phases b, c)

$$\begin{aligned} I_{a1} = -I_{a2} &= \frac{1}{j(0.1403+0.1509)} = -j3.435\ \text{pu}, \quad I_{a0} = 0 \\ I_b &= -j\sqrt3\,I_{a1} = -5.949\ \text{pu} = 5.949\angle180^\circ \\ I_c &= 5.949\angle0^\circ\ \text{pu}, \quad I_a = 0 \end{aligned}$$

(iii) L-L-G fault (phases b, c to ground)

$$\begin{aligned} Z_2\|Z_0 &= \frac{0.1509\times0.15}{0.3009} = j0.0752 \\ I_{a1} &= \frac{1}{j(0.1403+0.0752)} = -j4.640\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j2.313\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j2.327\ \text{pu} \\ I_b &= I_{a0}+a^2I_{a1}+aI_{a2} = 6.960\angle{149.91^\circ}\ \text{pu} \\ I_c &= I_{a0}+aI_{a1}+a^2I_{a2} = 6.960\angle{30.09^\circ}\ \text{pu} \\ I_n &= 3I_{a0} = j6.980\ \text{pu}, \quad I_a = 0 \end{aligned}$$

Summary

The L-G current is high because only G2's small $X_0$ (0.15) is in the zero-sequence path.