Calcady™

Transformer Secondary Short-Circuit Current

Estimate the infinite-bus short-circuit current at a transformer secondary from kVA, secondary voltage and nameplate impedance, with secondary full-load current and the 100 / %Z multiplier shown.
Trades & Construction
Standard: Infinite-bus method; NEC 110.9 interrupting-rating context

Nameplate Parameters

Accepted range 0.1–25%. Values outside are clamped.

Equipment Rating Check

Overcurrent devices at the transformer secondary should have an interrupting rating at least equal to the available fault current at their line terminals (NEC 110.9), estimated here as 12.03 kA. This infinite-bus value is an upper bound for the transformer alone; it does not include motor contribution, and it is not an arc-flash incident-energy result (that requires an NFPA 70E / IEEE 1584 study).

Infinite Bus Estimate

Bolted Fault
Available Fault Current (Isc)
12028 A
12.03 kA
Interrupting Rating Band
10–22 kA band: common rating steps 14, 18 and 22 kA
Secondary Full-Load Current
601.4 A
M Factor
Fault Multiplier
20.0x
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Worked Example: 500 kVA, 480 V, 5.0% Impedance

Calculator default inputs, infinite-bus method

Design Scenario

A 500 kVA three-phase transformer with a 480 V secondary and 5.0% nameplate impedance feeds a main distribution panel.

Design Parameters
Rating500 kVA
Secondary voltage480 V, 3-phase
Impedance5.0 %Z
SourceInfinite bus
Mathematical Solution
1Secondary full-load current
500,000 / (480 x 1.732) = 601.4 A
2Multiplier
100 / 5.0 = 20
3Infinite-bus fault current
601.4 A x 20 = 12,028 A (12.03 kA)
4Tolerance check at 4.5 %Z

A 10% reduction in impedance raises the estimate by about 11%.

601.4 A x (100 / 4.5) = 13,365 A
Engineering Conclusion: The transformer alone can deliver about 12.0 kA at its secondary terminals, or about 13.4 kA if impedance is 10% below nameplate. Equipment at the secondary needs an interrupting rating at least equal to the available fault current at its line terminals (NEC 110.9); a 10 kA-rated device is not adequate here.

What is Transformer Infinite-Bus Fault Current?

Transformer percent impedance is the percentage of rated voltage needed to drive full-load current through a shorted secondary. Its reciprocal therefore sets how many times full-load current can flow during a bolted fault when the primary source is unlimited. This quick estimate is commonly used to screen interrupting ratings before a full short-circuit study.

Mathematical Foundation

Secondary full-load current
IFLA,3ϕ=kVA×1000VLL×3IFLA,1ϕ=kVA×1000VI_{FLA,3\phi} = \frac{kVA \times 1000}{V_{LL} \times \sqrt{3}} \qquad I_{FLA,1\phi} = \frac{kVA \times 1000}{V}
kVAkVA= Transformer nameplate rating (kVA)
VLLV_{LL}= Secondary line-to-line voltage for three-phase units (V)
VV= Secondary voltage for single-phase units (V)
Infinite-bus short-circuit current
ISC=IFLA×100%ZI_{SC} = I_{FLA} \times \frac{100}{\%Z}
%Z\%Z= Nameplate percent impedance; the calculator accepts 0.1% to 25%
ISCI_{SC}= Bolted symmetrical fault current at the secondary terminals, assuming unlimited primary source capacity (A)

Laws & Principles

  • NEC 110.9: equipment intended to interrupt current at fault levels must have an interrupting rating at least equal to the available current at its line terminals.
  • NEC 110.24: service equipment in other than dwelling units must be field-marked with the available fault current and the date it was determined.
  • Lower %Z means higher fault current. Halving impedance doubles the infinite-bus estimate.
  • Motor contribution, utility source impedance and conductor impedance all change the actual value; a short-circuit study by a qualified person accounts for them.

Step-by-Step Example Walkthrough

" A 25 kVA single-phase pole-mounted transformer with a 240 V secondary and 2.0% impedance serves a dwelling. "

  1. 1. Secondary full-load current: 25,000 / 240 = 104.17 A.
  2. 2. Multiplier: 100 / 2.0 = 50.
  3. 3. Infinite-bus fault current: 104.17 A x 50 = 5,208 A (5.21 kA).
Final Result: About 5.2 kA at the transformer terminals, in the below-10 kA band. Service drop length reduces the value at the service equipment, which the Point-to-Point calculator can estimate.

Quick Answer

Divide 100 by the nameplate percent impedance and multiply by the secondary full-load current. A 500 kVA, 480 V, 5% transformer gives 601.4 A x 20 = 12,028 A. This is an upper bound for the transformer alone; utility impedance lowers it and motor contribution raises it.

Rating Bands Shown by the Calculator

Calculated fault current Common equipment rating steps in band
Below 10 kA10 kA
10 to 22 kA14, 18, 22 kA
22 to 42 kA25, 35, 42 kA
42 to 65 kA50, 65 kA
65 to 85 kA100 kA
85 kA and above100 kA, 200 kA or current-limiting equipment

Select a device rating at or above the calculated value. Series-rated combinations must be listed and marked as such; check manufacturer documentation.

Application to Service Equipment & Distribution

This calculator determines the maximum available symmetrical short-circuit current delivered at utility service entrance transformer secondary terminals under worst-case infinite-bus conditions. It displays the secondary full-load current (FLA), fault multiplier (100 / %Z), and the corresponding standard equipment interrupting rating band (AIC).

For downstream distribution panels, use the Point-to-Point Fault Current calculator to evaluate how feeder conductor impedance reduces available fault levels along run lengths. To evaluate industrial control equipment compliance with NEC 409.110, use the SCCR Panel Sizing calculator.

Frequently Asked Questions

What does infinite bus mean in a transformer fault calculation?

It assumes the utility can supply unlimited current to the transformer primary, so the transformer impedance is the only limit on fault current. Real utility sources have some impedance, so the actual fault current is lower than this value, which is why the result is treated as an upper bound for the transformer alone.

How is the interrupting rating band chosen?

The calculator places the result in a band (below 10 kA, 10 to 22 kA, 22 to 42 kA, 42 to 65 kA, 65 to 85 kA, 85 kA and above) and lists common equipment rating steps in that band. NEC 110.9 requires interrupting equipment to have a rating at least equal to the available current at its line terminals, so select a rating at or above the calculated value, not just within the band.

Why might I use a lower impedance than the nameplate?

Actual impedance can differ from the nameplate value within manufacturing tolerance. Many short-circuit guides reduce %Z by up to 10% for a conservative estimate. For a 500 kVA, 480 V transformer, 5.0% gives 12,028 A while 4.5% gives 13,365 A.

Is this an arc flash calculation?

No. The bolted fault current is one input to an arc flash study, but incident energy and arc flash boundaries require an NFPA 70E or IEEE 1584 analysis that includes arcing current, clearing time and working distance.

How is utility service entrance transformer secondary fault current determined?

Service entrance fault current is determined by evaluating the transformer secondary bolted short-circuit capacity using the infinite-bus method. This calculation establishes the maximum available symmetrical fault current at the secondary terminals based on transformer kVA, secondary voltage, and nameplate impedance (%Z). This value provides the foundational baseline for sizing main service equipment interrupting ratings (NEC 110.9) and complying with service entrance marking requirements (NEC 110.24).

Related Calculations

Calculation Provenance & Validation Record

Method

Infinite-bus method: secondary full-load current multiplied by 100 / %Z gives the bolted symmetrical fault current at the transformer secondary terminals.

Formula
I_{SC} = rac{kVA imes 1000}{V imes k} imes rac{100}{%Z}
Assumptions
  • Infinite-bus primary: utility source impedance is taken as zero, which gives an upper-bound value for the transformer alone.
  • Bolted fault at the secondary terminals; no conductor impedance downstream.
  • Motor contribution is not included and can add to the available fault current.
  • Nameplate %Z is used as entered; manufacturing tolerance is not applied automatically.
References
  • NFPA 70, National Electrical Code (NEC 2023) 110.9 - interrupting rating
  • NEC 2023 110.24 - available fault current marking at service equipment
  • IEEE Std 551 (Violet Book) - short-circuit current calculation methods
Last substantive review:
Automated test status: 5 golden test vectors passing (TSC-01, TSC-02, TSC-03, TSC-04, TSC-05)
Method, assumptions & governing standards
Standard Sourced

Calculation Methodology

Infinite-bus method: secondary full-load current multiplied by 100 / %Z gives the bolted symmetrical fault current at the transformer secondary terminals.

Governing Standard 2023 NEC / IEEE Std 141

Standard:NFPA 70 (NEC) / IEEE 141

Statutory building, electrical, and mechanical codes vary by jurisdiction. Confirm local municipality amendments before installation.

Key Assumptions & Constraints

  • Infinite-bus primary: utility source impedance is taken as zero, which gives an upper-bound value for the transformer alone.
  • Bolted fault at the secondary terminals; no conductor impedance downstream.
  • Motor contribution is not included and can add to the available fault current.
  • Nameplate %Z is used as entered; manufacturing tolerance is not applied automatically.
Regression Tests: 5 golden vectors
Last Verified:
Primary References: 3 documented
Field Trade Notice: For trade planning and engineering estimates. Final installations must conform to project blueprints, authority having jurisdiction (AHJ) code approvals, and site-specific inspections.