Fault through impedance at bus 2

The fault current is found from the Thevenin (driving point) impedance at bus 2, $I_f = V_f/(Z_{22}+Z_f)$, with prefault voltage 1 pu. Bus voltages during the fault then give the generator currents.

Assumptions: all values are reactances (lines 1-2 = j0.75 and 3-2 = j1.0 pu), $Z_f = j0.088$ pu, no prefault load, prefault voltage 1 pu everywhere. Base: 20 MVA, 11 kV (generator side), so 110 kV on the line side.

Step 1: Reactance network

Generator + transformer branches: bus 1 to reference $= 0.15 + 0.10 = 0.25$; bus 3 to reference $= 0.15 + 0.225 = 0.375$.

      (ref)               (ref)
        |                   |
      j0.25               j0.375
        |                   |
  bus 1 o------j0.5---------o bus 3
         \                 /
        j0.75           j1.0
           \             /
            o---- bus 2 (fault, Zf=j0.088)

Step 2: Bus admittance matrix and Z-bus (values of $-j$ / $j$ dropped, all reactive)

$$Y_{bus} = -j\begin{bmatrix} 4+2+1.333 & -1.333 & -2 \\ -1.333 & 1.333+1 & -1 \\ -2 & -1 & 2.667+2+1 \end{bmatrix} = -j\begin{bmatrix} 7.333 & -1.333 & -2 \\ -1.333 & 2.333 & -1 \\ -2 & -1 & 5.667 \end{bmatrix}$$

Inverting:

$$Z_{bus} = j\begin{bmatrix} 0.1884 & 0.1473 & 0.0925 \\ 0.1473 & 0.5788 & 0.1541 \\ 0.0925 & 0.1541 & 0.2363 \end{bmatrix}$$

So the Thevenin impedance at bus 2 is $Z_{22} = j0.5788$ pu.

Step 3: Fault current

$$\begin{aligned} I_f &= \frac{1}{Z_{22}+Z_f} = \frac{1}{j(0.5788+0.088)} = -j1.4998\ \text{pu} \\ I_{b,110} &= \frac{20\times10^6}{\sqrt3 \times 110\times10^3} = 104.97\ \text{A} \\ |I_f| &= 1.4998 \times 104.97 = 157.4\ \text{A} \end{aligned}$$

Step 4: Bus voltages during fault ($V_i = 1 - Z_{i2} I_f$)

$$\begin{aligned} V_1 &= 1 - 0.1473 \times 1.4998 = 0.7791\ \text{pu} \\ V_2 &= 1 - 0.5788 \times 1.4998 = 0.1320\ \text{pu} \\ V_3 &= 1 - 0.1541 \times 1.4998 = 0.7689\ \text{pu} \end{aligned}$$

Check: $V_2 = Z_f I_f = 0.088 \times 1.4998 = 0.132$ pu.

Step 5: Generator currents

$$\begin{aligned} I_{G1} &= \frac{1 - V_1}{0.25} = \frac{0.2209}{0.25} = 0.8834\ \text{pu} \\ I_{G2} &= \frac{1 - V_3}{0.375} = \frac{0.2311}{0.375} = 0.6163\ \text{pu} \end{aligned}$$

Check: $0.8834 + 0.6163 = 1.4997 \approx I_f$.

Generator currents flow on the 11 kV side, base current $I_{b,11} = 20\times10^6/(\sqrt3 \times 11\times10^3) = 1049.7$ A:

$$I_{G1} = 0.8834 \times 1049.7 = 927.4\ \text{A}, \qquad I_{G2} = 0.6163 \times 1049.7 = 647.0\ \text{A}$$

Answer: Fault current at bus 2 = 1.50 pu = 157.4 A (110 kV side); G1 supplies 0.883 pu = 927 A and G2 supplies 0.616 pu = 647 A (11 kV side), all lagging the prefault voltage by 90°.