Three-phase fault current using the per-unit method

Both ratings already share the same MVA and voltage bases, so the reactances add directly.

Worked answer.

Step 1: Choose bases. Take S_base = 20 MVA. Voltage bases: 11 kV on the generator side and 66 kV on the line side (they follow the transformer ratio). Because both machines are rated at 20 MVA and their rated voltages match the bases, the given per-unit values need no conversion.

Step 2: Fault impedance. The generator and transformer are in series from the source EMF to the fault:
X_total = 0.2 + 0.1 = 0.3 pu.

Step 3: Pre-fault voltage. On no load at rated voltage, E = 1.0 pu.

Step 4: Fault current in per unit. I_f = E / X_total = 1.0 / 0.3 ≈ 3.333 pu.

Step 5: Base current on the 66 kV side.
I_base = S_base / (√3 × V_base) = 20 × 10⁶ / (√3 × 66 × 10³) ≈ 174.95 A.

Step 6: Fault current in amperes. I_f = 3.333 × 174.95 ≈ 583 A on the 66 kV side.

Step 7: Fault MVA. Fault MVA = S_base / X_total (with 1.0 pu voltage) = 20 / 0.3 ≈ 66.7 MVA.

Check. √3 × 66 kV × 583.2 A ≈ 66.7 MVA, which matches Step 7. On the 11 kV side, the same fault draws 3.333 × [20 × 10⁶ / (√3 × 11 × 10³)] ≈ 3.333 × 1049.7 ≈ 3499 A, which is exactly 6 times the 66 kV current, as expected from the 11/66 kV turns ratio.

Why per unit helps. In per unit the transformer's impedance is the same on both sides, so you can add series impedances without referring them across the turns ratio.

Common mistake. Using the 11 kV base current to convert a fault located on the 66 kV side. Always use the base current of the zone where the fault occurs.

Likely follow-up. "What is the fault current if a 0.15 pu reactance line is added before the fault?" X_total becomes 0.45 pu, giving about 2.222 pu, or roughly 389 A at 66 kV.