Symmetrical fault at bus 3

A three-phase fault is balanced, so only the positive-sequence network is used; the transformer connections (Y/Δ) and grounding do not affect it. Pre-fault voltage is 1.0 pu and load current is neglected.

Positive-sequence network

• Source 1 to bus 1: $0.1 + 0.1 = 0.2$ pu.
• Source 2 to bus 2: $0.2 + 0.2 = 0.4$ pu.
• Lines: $X_{12} = 0.8$, $X_{13} = 0.4$, $X_{23} = 0.4$ pu (a delta between buses 1, 2, 3).

Delta-star conversion of the line delta

$$\begin{aligned} X_1 &= \frac{X_{12}X_{13}}{X_{12} + X_{13} + X_{23}} = \frac{0.8 \times 0.4}{1.6} = 0.20 \\ X_2 &= \frac{X_{12}X_{23}}{1.6} = \frac{0.8 \times 0.4}{1.6} = 0.20 \\ X_3 &= \frac{X_{13}X_{23}}{1.6} = \frac{0.4 \times 0.4}{1.6} = 0.10 \end{aligned}$$

 E=1-j0.2-bus1-j0.2-+
                    N--j0.1--bus3 (F)
 E=1-j0.4-bus2-j0.2-+

Thevenin reactance at bus 3

$$\begin{aligned} X_{path1} &= 0.2 + 0.20 = 0.40, \quad X_{path2} = 0.4 + 0.20 = 0.60 \\ X_{th} &= \frac{0.40 \times 0.60}{0.40 + 0.60} + 0.10 = 0.24 + 0.10 = 0.34\ \text{pu} \end{aligned}$$

(The same value is $Z_{33}$ of the bus impedance matrix.)

Fault current

$$\begin{aligned} I_f &= \frac{V_f}{jX_{th}} = \frac{1}{j0.34} = -j2.9412\ \text{pu} \\ \text{Fault MVA} &= \frac{100}{0.34} = 294.1\ \text{MVA} \end{aligned}$$

Answer: $I_f = 2.941$ pu (fault level 294.1 MVA on the 100 MVA base). The current in amperes is $I_f \times I_{base}$ once the bus voltage base is known.