Sequence reactances from fault currents

Assumptions: the faults are at the terminals of the unloaded alternator, so the emf $E = 1$ pu (rated voltage); resistance neglected; neutral solidly grounded ($Z_n = 0$).

Base current

$$I_{base} = \frac{50\times10^6}{\sqrt3 \times 12\times10^3} = 2405.6\ \text{A}$$

Fault current formulas

$$I_{3\phi} = \frac{E}{X_1}, \qquad I_{LL} = \frac{\sqrt3 E}{X_1 + X_2}, \qquad I_{LG} = \frac{3E}{X_1 + X_2 + X_0}$$

Positive sequence reactance

$$X_1 = \frac{1}{0.7773} = 1.2864\ \text{pu}$$

Negative sequence reactance

$$\begin{aligned} X_1 + X_2 &= \frac{\sqrt3}{1.0766} = 1.6088 \\ X_2 &= 1.6088 - 1.2864 = 0.3223\ \text{pu} \end{aligned}$$

Zero sequence reactance

$$\begin{aligned} X_1 + X_2 + X_0 &= \frac{3}{1.7168} = 1.7474 \\ X_0 &= 1.7474 - 1.6088 = 0.1387\ \text{pu} \end{aligned}$$

Answer: $X_1 = 1.286$ pu, $X_2 = 0.322$ pu, $X_0 = 0.139$ pu.

The large $X_1$ (it is the steady-state synchronous reactance here, since sustained currents were measured) and small $X_0$ explain why the L-G current is the highest and the 3-phase current the lowest.