L-L fault on 40 MVA generator

Data: 40 MVA, 11 kV, $Z_1 = Z_2 = j0.3$ pu, $Z_0 = j0.4$ pu. Generator unloaded, $E_a = 1\angle0^\circ$ pu. Fault between phases b and c.

Fault conditions

$I_a = 0,\ I_b = -I_c,\ V_b = V_c$, which give $I_{a0} = 0$ and $I_{a2} = -I_{a1}$: the positive and negative sequence networks are connected in parallel (opposing), and $Z_0$ does not appear.

Sequence currents

$$I_{a1} = -I_{a2} = \frac{E_a}{Z_1 + Z_2} = \frac{1}{j0.6} = -j1.667\ \text{pu}$$

Fault current

$$\begin{aligned} I_b &= (a^2 - a)I_{a1} = -j\sqrt3\, I_{a1} = -j\sqrt3(-j1.667) = -2.887\ \text{pu} \\ I_c &= -I_b = 2.887\ \text{pu} \end{aligned}$$

Base current:

$$I_{base} = \frac{40\times10^6}{\sqrt3 \times 11\times10^3} = 2099.5\ \text{A}$$ $$|I_f| = 2.887 \times 2099.5 = 6061\ \text{A}$$

Answer: L-L fault current = 2.887 pu ≈ 6.06 kA (equal and opposite in phases b and c; $I_a = 0$).

For comparison, a 3-phase fault would give $1/0.3 = 3.333$ pu, so $I_{LL} = \frac{\sqrt3}{2} I_{3\phi}$ when $Z_1 = Z_2$.