L-L fault current in kA at bus 3

Base: 50 MVA, 11 kV (generators), 220 kV (lines); all data are on this base. Pre-fault voltage at bus 3: $V_f = 0.95$ pu, no load.

Single line diagram

 G1      T1               Line-1              T2      G2
 (~)-(1)-)(-(2)======================(3)-)(-(4)-(~)
 Yg      D  Yg     Line-2 (parallel)     Yg   D     Yg
                                         |          |
                                       j0.03     j0.024

(i) Sequence networks

Lines in parallel: $X_1 = X_2 = 0.12/2 = 0.06$, $X_0 = 0.42/2 = 0.21$ pu.

Positive sequence (emfs $E_1$, $E_2$):

 E1-j0.2-(1)-j0.1-(2)-j0.06-(3)-j0.075-(4)-j0.15-E2
                             |
                             F

Negative sequence: same reactances (0.2, 0.1, 0.06, 0.075, 0.15), no emfs, all ends to reference.

Zero sequence:

• G1 is behind the Δ of T1, so it is isolated; T1 connects bus 2 to reference through $j0.1$.
• T2 star side (bus 3) grounded through $j0.03$: bus 3 to reference through $j(0.075 + 3\times0.03) = j0.165$; G2 is behind the Δ, so isolated.

 ref-j0.1-(2)-j0.21-(3)-j0.165-ref
                     |
                     F
 G1 zero seq: j0.05 isolated;  G2: j0.03+3(j0.024) isolated

$Z_0 = j0.31 \| j0.165 = j0.1077$ pu (not needed for L-L fault).

(ii) L-L fault at bus 3

Thevenin impedances:

$$\begin{aligned} Z_1 &= j(0.2+0.1+0.06)\ \|\ j(0.075+0.15) = j0.36\ \|\ j0.225 = j0.1385\ \text{pu} \\ Z_2 &= Z_1 = j0.1385\ \text{pu} \end{aligned}$$

Sequence currents:

$$I_{a1} = -I_{a2} = \frac{V_f}{Z_1+Z_2} = \frac{0.95}{j0.2769} = -j3.431\ \text{pu}$$

Fault current:

$$|I_f| = |I_b| = |I_c| = \sqrt3 \times 3.431 = 5.942\ \text{pu}$$

Base current at 220 kV:

$$I_{base} = \frac{50\times10^6}{\sqrt3 \times 220\times10^3} = 131.2\ \text{A}$$ $$I_f = 5.942 \times 131.2 = 779.7\ \text{A} = 0.780\ \text{kA}$$

Answer: L-L fault current at bus 3 ≈ 5.94 pu ≈ 0.78 kA.