5% vs 6% Transformer Impedance: Fault Current Compared

Power Infra Lab · Technical explainer · Updated October 8, 2026

For the same transformer rating and secondary voltage, increasing impedance from 5% to 6% reduces the idealized three-phase terminal fault-current estimate by one sixth, or about 16.7%. It does not reduce current by just 1%. The impedance change is one percentage point, but fault current varies inversely with the impedance fraction.

This article works through a hypothetical 1,500 kVA, 480 V example using the transformer fault current calculator. It is an educational sensitivity comparison, not a switchgear selection, protection-setting recommendation or arc-flash assessment.

Define the calculation boundary first

Both cases assume a balanced three-phase bolted fault at the transformer secondary terminals, nominal voltage, an infinite primary source and no other sources feeding the fault. Feeder impedance and impedance tolerance are not included. The two transformers are identical for this exercise except for their entered percent impedance.

Eaton’s short-circuit calculation reference gives the transformer multiplier as 100 divided by percent impedance. It is applied to rated secondary current. That establishes the model used here; it does not make every real installation an infinite-source system.

Step 1: calculate rated secondary current

For the balanced three-phase case, rated current in amperes equals kVA × 1,000 ÷ (√3 × line-to-line volts). With the example inputs, 1,500 × 1,000 ÷ (√3 × 480) = 1,804.22 A.

This is the nameplate-based current corresponding to the transformer’s rating, not a measured operating current. A load currently drawing half the transformer rating does not make the transformer rating in this formula half as large. Keep load-demand calculations and available-fault-current calculations in different worksheet columns.

Step 2: compare the two impedance inputs

The 5% calculation is 1,804.22 ÷ 0.05 ÷ 1,000 ≈ 36.084 kA. The 6% calculation is 1,804.22 ÷ 0.06 ÷ 1,000 ≈ 30.070 kA. The difference is approximately 6.014 kA. Results shown here are rounded only after evaluating the formula.

When entering values into this site’s calculator, enter 5 or 6 in the field labelled impedance percent. Entering 0.05 would mean 0.05%, not 5%; it is also below the tool’s permitted range. The conversion from percentage to fraction belongs inside the formula, not in that input field.

Why the reduction is 16.7%, but the reverse increase is 20%

Taking the 5% case as the starting point, the remaining current fraction is 5 ÷ 6 = 0.83333. Therefore the reduction is 1 − 0.83333 = 0.16667, or 16.667%.

Reverse the comparison and the denominator changes. Moving from 6% impedance to 5% makes the current 6 ÷ 5 = 1.20 times its previous value: a 20% increase. Both statements describe the same pair of results. A comparison note should always identify which case is the baseline.

Two useful sensitivity checks

A hypothetical impedance change

If a separate scenario uses 4.5% instead of 5%, leaving all other inputs unchanged, the result becomes 1,804.22 ÷ 0.045 ÷ 1,000 ≈ 40.094 kA. That is 11.111% above the 5% result. Here 4.5% is an arbitrary sensitivity input, not a claim about a universal manufacturing tolerance. Obtain the tolerance and applicable test basis for the exact equipment.

A different secondary voltage

At 415 V instead of 480 V, a hypothetical 1,500 kVA transformer has a rated current of about 2,086.81 A. With 5% impedance, the same simplified method gives 41.736 kA. The 480 V answer therefore cannot be copied into a 415 V worksheet. Likewise, a 277 V line-to-neutral value must not replace 480 V line-to-line in this three-phase calculation.

Why this is not the final system result

Schneider Electric’s Electrical Installation Guide calculates fault current from the combined resistance and reactance of the relevant network sections. A transformer-only estimate omits the upstream network and the path from the transformer to a downstream fault. It also omits contributions from other connected sources.

Do not call 30.070 kA a guaranteed maximum for every arrangement, or infer incident energy from it. The site’s calculator does not model source strength, parallel transformers, motors, generators, inverter controls, fault type, X/R effects or protective-device clearing time. A lower terminal number alone is not proof of a safer or better system.

Make the handoff reproducible

1. Record the actual nameplate rating, secondary line-to-line voltage and impedance source.
2. Run the calculator with 1,500 kVA, 480 V and 5%, then repeat with 6% to reproduce this example.
3. Label the exported result as a nominal-voltage, infinite-source, three-phase terminal estimate.
4. Provide a qualified reviewer with the single-line arrangement and utility, cable and other-source data before equipment duty is assessed.

For the separate question of current at a chosen apparent load, use the kVA-to-amps calculator. Keeping those two tasks separate prevents a valid arithmetic result from acquiring a meaning it does not have.

Sources & further reading

1. Eaton: Short-Circuit Current Calculations ↗
2. Schneider Electric Electrical Installation Guide: Three-phase short-circuit current at any point ↗

Sources checked October 8, 2026. Examples are hypothetical unless explicitly identified as published product data.