Secondary Fault Current: Use 0.15 pu, Not 15 pu Impedance

The calculated secondary fault current collapses to about 0.154 kA when transformer impedance is entered as 15 pu. The transformer value is 15%, which equals 0.15 pu; correcting that conversion produces approximately 12.9 kA at the 25 kV secondary bus for the stated three-phase case.

What calculation basis applies to this fault?

Start with the fault definition. The supplied utility data are a 132 kV bus and 15 kA fault current, while the transformer is rated 100 MVA, 132/25 kV, with 15% impedance. The use of sqrt(3) in the supplied calculations defines the working case as a balanced three-phase fault with line-to-line voltage and line current.

The calculation finds the fault at the transformer secondary terminals. It includes the finite utility impedance and transformer impedance but no secondary cable, bus, reactor, generator, or motor impedance. Add those elements before applying the result at a downstream location.

Use one common base throughout the per-unit calculation. A 100 MVA base is convenient because it matches the transformer rating, so the transformer percent impedance converts directly to per unit without a base-MVA correction. The voltage base changes across the transformer according to its 132/25 kV ratio; impedance expressed in per unit remains comparable across the transformer when the voltage bases follow that ratio.

Check 1: What short-circuit MVA does the utility supply?

Convert the utility fault level into three-phase short-circuit MVA:

S_sc = sqrt(3) × V_LL × I_sc

Using the stated values:

S_sc = 1.732 × 132 kV × 15 kA = approximately 3429.36 MVA

If this result is not near 3429 MVA, inspect the voltage and current units before proceeding. Entering volts with kiloamperes, or kilovolts with amperes, introduces a factor-of-1000 error. Using secondary voltage in this first calculation without first referring the source current to that side also gives the wrong utility fault level.

The utility short-circuit MVA represents the admittance of the upstream network at the primary bus. A high short-circuit MVA means low source impedance. It is not added directly to the transformer MVA rating; each component must first be represented as an impedance or as its own short-circuit MVA.

Check 2: Is the transformer impedance 0.15 pu or 15 pu?

Convert percent impedance by dividing by 100:

Z_tx,pu = 15% / 100 = 0.15 pu

Entering 15 pu treats the transformer as having 1500% impedance. That makes the calculated fault impedance 100 times too large in the transformer term and drives the result down to approximately 0.154 kA. The implausibly low current is a unit-conversion symptom, not a characteristic of the stated transformer.

The transformer short-circuit MVA from an infinite bus is:

S_sc,tx = 100 MVA / 0.15 = 666.7 MVA

At 25 kV, that transformer-only value corresponds to approximately:

I_sc,infinite = 666.7 MVA / (1.732 × 25 kV) = 15.40 kA

This is a useful boundary check. Adding finite upstream impedance must reduce the secondary fault current below 15.40 kA. A result above that boundary means an impedance was omitted, subtracted, or placed in parallel incorrectly.

Check 3: What do the measured inputs say when values are wrong?

Look at the signal chain: the utility fault level defines upstream impedance, the transformer impedance limits transferred fault power, and the secondary voltage converts the combined fault MVA into line current. Tuning a numerical result does not fix a wrong unit or base.

Signal or value	Source to read	Wrong-value symptom
132 kV primary voltage	Utility fault data and one-line diagram	Using 25 kV here without a complete impedance referral corrupts the utility short-circuit MVA.
15 kA primary fault current	Utility short-circuit data	Mixing amperes and kiloamperes produces a factor-of-1000 error.
3429.36 MVA utility fault level	Calculated from primary voltage and current	A much smaller value makes the utility appear weaker and suppresses the secondary current.
15% transformer impedance	Transformer rating data	Entering it as 15 pu yields about 0.154 kA instead of about 12.9 kA.
100 MVA transformer rating	Transformer rating data	Using a different calculation base without conversion gives the wrong transformer per-unit impedance.
25 kV secondary voltage	Transformer ratio and one-line diagram	Using 132 kV in the final current conversion understates secondary line current.

Check 4: Does the per-unit calculation close?

On a 100 MVA base, calculate the utility source impedance:

Z_source,pu = S_base / S_sc,utility

Z_source,pu = 100 / 3429.36 = 0.02915996 pu

The transformer is already on the 100 MVA base:

Z_tx,pu = 0.15 pu

For a radial source and transformer, the impedances are in series:

Z_total,pu = 0.02915996 + 0.15 = 0.17915996 pu

Convert the combined impedance back to fault MVA:

S_sc,secondary = 100 / 0.17915996 = approximately 558.16 MVA

Then calculate the secondary three-phase line current:

I_sc = 558.16 MVA / (1.732 × 25 kV) = approximately 12.89 kA

Rounding total impedance to 0.179 pu, fault MVA to 559 MVA, or using 1.73 instead of 1.732 produces a result near 12.92 kA. Those differences are rounding effects; they do not change the engineering conclusion.

Check 5: Does the MVA method give the same answer?

The MVA method provides an independent arithmetic check. Express each series component as the short-circuit MVA it would pass from an infinite bus:

MVA_utility = 3429.36 MVA

MVA_transformer = (100 MVA × 100) / 15 = 666.67 MVA

Combine series MVA values as admittances:

MVA_total = (MVA_utility × MVA_transformer) / (MVA_utility + MVA_transformer)

MVA_total = (3429.36 × 666.67) / (3429.36 + 666.67) = approximately 558.16 MVA

Converting that result at 25 kV again gives approximately 12.89 kA. Parallel fault contributions add as MVA; series elements combine using the reciprocal relationship above. Motors connected to the faulted bus are parallel contributors, while cables between the transformer and fault are series impedances.

Check 6: When must resistance and X/R be included?

The scalar per-unit and MVA calculations add impedance magnitudes. That shortcut treats the source and transformer as though their impedance angles, and therefore their X/R ratios, are similar. When the angles differ materially, add resistance and reactance separately as complex quantities:

R_total = R_source + R_transformer + R_cable

X_total = X_source + X_transformer + X_cable

|Z_total| = sqrt(R_total² + X_total²)

I_sc = V_LL / (sqrt(3) × |Z_total|)

On the 25 kV side, the utility reactance approximation from its fault kVA is:

X_utility = 1000 × (secondary kV)² / utility short-circuit kVA

For the transformer, the corresponding expressions are:

X_transformer = 10 × %X × (secondary kV)² / transformer kVA

R_transformer = 10 × %R × (secondary kV)² / transformer kVA

Read %R, %X, or X/R from the utility study and transformer data rather than inventing a split from the 15% magnitude. Cable resistance and reactance become increasingly relevant when calculating a fault beyond the transformer terminals. The source impedance is smaller than the transformer impedance in this case—0.02916 pu versus 0.15 pu—so the transformer dominates the symmetrical current magnitude, but X/R still affects asymmetrical peak and equipment-duty calculations.

How should the result be calculated and verified?

1. Mark the fault location on the one-line diagram and list every source and series impedance between each source and that point.
2. Confirm that the utility value is a three-phase fault current at the 132 kV primary bus. Use the matching fault type for the required study.
3. Calculate utility fault MVA as 1.732 × 132 × 15 = approximately 3429.36 MVA.
4. Select the 100 MVA base and convert the utility impedance to 0.02915996 pu.
5. Convert transformer impedance from 15% to 0.15 pu. Do not enter 15 pu.
6. Add the radial series impedances to obtain 0.17915996 pu.
7. Calculate secondary fault power as 100 / 0.17915996 = approximately 558.16 MVA.
8. Convert at 25 kV to obtain approximately 12.89 kA.
9. Repeat the calculation with the series-MVA method. The result should again be approximately 558.16 MVA and 12.89 kA.
10. Add secondary conductors, bus, connected motors, and other sources before using the result for a fault away from the transformer terminals or for equipment-duty decisions.

Verify three boundaries: the finite-source result must be below the transformer infinite-bus value of approximately 15.40 kA; the per-unit and MVA methods must agree apart from rounding; and adding a series impedance must decrease, not increase, the calculated current. For interrupting, withstand, protection, or arc-flash work, use the required fault type and obtain the missing X/R and component impedance data.

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