Transformer fault current from %Z: a 75 kVA worked example

For a 75 kVA, 208 V three-phase transformer with 5.0% nameplate impedance, the infinite-source estimate at the secondary terminals is about 4.16 kA symmetrical RMS. First calculate rated secondary current, 75,000 VA / (√3 × 208 V) = 208.18 A. Then convert 5.0% to 0.050 per unit and divide: 208.18 A / 0.050 = 4,163.5 A. This is a screening calculation at a declared point, not a complete short-circuit study or a protective-device selection.

Scope and terms before the arithmetic

This example models a fictional 75 kVA, three-phase transformer with a 480 V primary, a 208Y/120 V secondary, and 5.0% nameplate impedance. The calculation point is the secondary terminals. It assumes an infinite upstream source and zero impedance between the transformer and F1, so the transformer's impedance is the only current-limiting element in the model.

Rated secondary current is the line current delivered at rated three-phase apparent power and rated line-to-line voltage. Percent impedance, written %Z, is the transformer short-circuit impedance expressed against its own base values. Eaton's calculation guide describes it through the factory test concept: a fraction of rated primary voltage is applied while the secondary is shorted until rated secondary current flows. A 5% transformer therefore reaches rated current at roughly 5% of rated applied voltage under that test condition.

That relationship explains the reciprocal. If 0.05 per unit voltage produces 1.0 per unit current, then 1.0 per unit voltage produces 1 / 0.05 = 20 per unit current in the simplified model.

The current calculated here should be labeled estimated three-phase symmetrical RMS short-circuit current at F1. It is not the instantaneous peak, an arc current, an incident-energy result, or the current at an unspecified downstream panel. Those quantities require additional system data and methods.

Inputs and assumptions

The 120 V in 208Y/120 V is the nominal line-to-neutral voltage. It is not used in this three-phase line-current equation. Substituting 120 V would mix a phase voltage with a three-phase apparent-power formula and overstate the rated line current by approximately √3.

Step 1: calculate full-load secondary current

For balanced three-phase apparent power,

S = √3 × VLL × I

I_FL = S / (√3 × VLL)
     = 75,000 VA / (1.7320508 × 208 V)
     = 208.18 A.

The unit check is VA / V = A; √3 is dimensionless. Schneider Electric's current FAQ uses the same 75 kVA, 208Y/120 V example and rounds the full-load result to 208 A.

Step 2: convert percent impedance to per unit

Percent is not a whole-number multiplier:

Z_pu = 5.0% / 100
     = 0.050.

Dividing by 5 instead of 0.05 would produce 41.6 A, a result below the transformer's normal full-load current. That fails an immediate plausibility check: a bolted terminal fault under the model cannot be only one-fifth of rated current.

Step 3: estimate the terminal fault current

Eaton's point-to-point procedure expresses the transformer multiplier as 100 / %Z, which is the same as 1 / Z_pu. Apply it to the rated current:

Multiplier = 100 / 5.0 = 20

I_sc = I_FL × Multiplier
     = 208.18 A × 20
     = 4,163.5 A
     ≈ 4.16 kA symmetrical RMS.

The same calculation can be written directly:

I_sc = 208.18 A / 0.050 = 4,163.5 A.

Rounding the intermediate full-load current to Schneider's 208 A gives 4,160 A. The 3.5 A difference is only rounding; both results support a reported value of 4.16 kA given the input precision.

Cross-check with short-circuit apparent power

An independent base-power check should return the same result. Dividing the transformer rating by per-unit impedance gives the terminal short-circuit apparent power:

S_sc = 75 kVA / 0.050
     = 1,500 kVA
     = 1.50 MVA.

Convert that back to current at 208 V:

I_sc = 1,500,000 VA / (√3 × 208 V)
     = 4,163.5 A.

The current-base method and the power-base method agree. This checks the arithmetic and voltage basis; it does not validate the input impedance or the infinite-source assumption.

Reproducible Schematex source

sld "75 kVA transformer fault-current boundary"

SOURCE = utility [
  voltage: "480 V",
  label: "Infinite source assumption"
]
T1 = transformer_dy [
  label: "T1",
  voltage: "480 V / 208Y-120 V",
  rating: "75 kVA",
  impedance: "5.0% Z"
]
F1 = bus [
  voltage: "208 V",
  label: "F1 | Isc 4.16 kA RMS sym."
]
CB1 = breaker [
  rating: "Interrupting rating: verify",
  label: "CB-1"
]
LOAD1 = load [label: "Downstream system"]

SOURCE -> T1
T1 -> F1
F1 -> CB1
CB1 -> LOAD1

Schematex 1.0.13 strictly parsed and rendered this SLD on September 2, 2026 with no diagnostics. The current Schematex SLD syntax reference documents the utility, transformer_dy, bus, breaker, and load nodes, arbitrary transformer nameplate attributes, and directed source-to-load connections.

The SVG makes the calculation boundary explicit, but Schematex does not run the short-circuit equation or select the breaker. Recompute the values independently and treat the verify rating as an unresolved engineering input, not a recommendation.

Impedance tolerance sensitivity

Eaton notes that listed transformers of this size can have impedance tolerance and recommends using the actual nameplate value or appropriate worst-case data. To show the direction—not to assign a product's tolerance—vary the stated 5.0% input by ±10%:

Lower impedance produces higher available current. Therefore a single rounded %Z copied from a generic table may not be conservative for interrupting-duty review. Obtain the transformer's actual nameplate or certified data and apply the governing study method.

Checks and invariants

1. Voltage basis: use 208 V line-to-line in the three-phase formula, not 120 V line-to-neutral.
2. Percent conversion: 5.0% = 0.050 per unit.
3. Multiplier: 100 / 5.0 = 20; the estimated fault current is 20 times rated current.
4. Current result: 208.18 A × 20 = 4,163.5 A.
5. Power cross-check: 75 kVA / 0.050 = 1.50 MVA, and 1.50 MVA / (√3 × 208 V) = 4.1635 kA.
6. Sensitivity direction: reducing %Z increases I_sc; increasing %Z reduces it.
7. Point identity: the 4.16 kA label belongs to F1 at the transformer terminals, not automatically to every downstream device.
8. Current category: report symmetrical RMS separately from asymmetrical peak, arc current, and incident energy.

Failure modes and review boundary

The infinite-source shortcut can overstate the transformer's source contribution because real utility and primary-feeder impedances limit current. Yet ignoring downstream motor contribution can understate current at a common bus. Secondary conductor and bus impedance normally reduce current with distance, while source voltage, transformer impedance tolerance, parallel transformers, generator operation, fault type, grounding, and system configuration change the result. X/R ratio affects asymmetry and peak duty. None of those effects is captured in I_FL / Z_pu alone.

OSHA 29 CFR 1910.303(b)(4) requires equipment intended to interrupt fault-level current to have sufficient interrupting rating for the nominal voltage and current available at its line terminals. Paragraph (b)(5) separately addresses total circuit impedance, component short-circuit current ratings, and coordination. The worked estimate supplies neither a compliance finding nor every input those reviews require.

Take the actual transformer's kVA, secondary line-to-line voltage, and nameplate %Z, then repeat the two independent calculations. Put the result on the one-line at the exact calculation point, date the source data, and list what is still excluded. You can paste the source into the Schematex playground to draft the topology; complete the engineering study before using the drawing to review equipment ratings.