Fault at load end of overhead line

Base: 15 MVA (transformer rating), 11 kV on the generator side and 33 kV on the line side. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit reactances (15 MVA)

 G1 j0.150--+
               +--T1 j0.06--line j0.275--F (load end)
 G2 j0.225--+

Thevenin reactance

$$\begin{aligned} X_G &= \frac{0.150 \times 0.225}{0.150 + 0.225} = 0.090 \\ X_{th} &= 0.090 + 0.06 + 0.2755 = 0.4255\ \text{pu} \end{aligned}$$

Short-circuit MVA and fault current

$$\begin{aligned} \text{SC MVA} &= \frac{15}{0.4255} = 35.25\ \text{MVA} \\ I_f &= \frac{1}{0.4255} = 2.3503\ \text{pu} \\ I_{base,33} &= \frac{15 \times 10^6}{\sqrt{3} \times 33 \times 10^3} = 262.43\ \text{A} \Rightarrow I_f = 616.8\ \text{A at 33 kV} \end{aligned}$$

Current supplied by each generator

On the 11 kV side, $I_{base,11} = \dfrac{15 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 787.30$ A, so the fault current there is $2.3503 \times 787.30 = 1850.4$ A. It divides inversely as the generator reactances:

$$\begin{aligned} I_{G1} &= 1850.4 \times \frac{0.225}{0.150 + 0.225} = 1110.2\ \text{A} \\ I_{G2} &= 1850.4 \times \frac{0.150}{0.150 + 0.225} = 740.1\ \text{A} \end{aligned}$$

Answer: SC MVA $\approx 35.25$ MVA; fault current $\approx 617$ A at 33 kV; G1 supplies $\approx 1110$ A and G2 $\approx 740$ A (at 11 kV).