Fault Current Calculator

Calculate simplified single- or three-phase symmetrical RMS fault current from transformer nameplate data or known equivalent system impedance.

Calculator is for informational purposes only. Terms and Conditions

Transformer secondary — three-phase

\[ I_{\mathrm{sc}}=\frac{S}{\sqrt{3}\,V_{LL}\,z_{\mathrm{pu}}},\qquad z_{\mathrm{pu}}=\frac{Z_{\%}}{100} \]

Choose the calculation setup

Select the available data and electrical phase arrangement.

Calculation Method

Use transformer mode for a terminal estimate from nameplate data, or impedance mode when the equivalent impedance at the fault point is known.

System Phase

Three-phase calculations use line-to-line voltage and per-phase equivalent impedance.

Enter the known values

Only fields required by the active method are enabled.

Transformer Rating

Use the transformer nameplate apparent-power rating.

Secondary Voltage (line-to-line)

For three-phase systems, enter line-to-line voltage. For single-phase systems, enter the voltage across the fault path.

Transformer Impedance

Enter the transformer nameplate percent impedance at rated kVA.

Advanced Options

Answer Units

Significant Figures

Solution

Live result, fault level checks, warnings, and calculation steps.

Symmetrical RMS Fault Current

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Enter the required values to calculate.

Quick checks

Fault level—

Show solution steps Review conversions, equation, substitution, assumptions, and result

Enter valid values to see the complete solution.

Source, Standards, References, and Assumptions

Calculation basis, authoritative references, limitations, and verification requirements.

Simplified symmetrical RMS method

Symmetrical RMS Bolted fault Screening estimate

Uses standard transformer percent-impedance and equivalent-impedance relationships for a simplified fault-current estimate; it is not a complete IEC 60909 short-circuit study.

References

Transformer mode assumes an infinite upstream source and uses the entered nameplate impedance.
Known-impedance mode requires the equivalent impedance seen at the fault point.
DC offset, peak asymmetry, arc resistance, and dynamic motor or generator contribution are not separately modeled.
Final equipment ratings and protection settings must be checked against applicable codes, standards, utility data, manufacturer data, site conditions, and qualified engineering judgment.

Calculator guide

How the Fault Current Calculator Works

The Fault Current Calculator above estimates symmetrical RMS bolted-fault current for single-phase or three-phase systems. Use transformer mode when you know transformer rating, secondary voltage, and nameplate percent impedance. Use known-impedance mode when you already know the equivalent impedance seen from the fault location.

The primary result is available symmetrical RMS fault current in amperes or kiloamperes. Transformer mode also checks full-load current, the impedance current multiplier, equivalent impedance, and fault level. The result applies to the electrical point represented by the inputs; it is not automatically the fault current at every downstream panel or piece of equipment.

Best for: Transformer-terminal and known-impedance short-circuit screening
Primary output: Symmetrical RMS fault current in A or kA, plus fault level
Key dependency: Equivalent impedance between the source and the selected fault location

How to Use the Calculator Correctly

Choose the method that matches the data you actually have, identify the exact fault location, and confirm the phase and voltage convention before calculating.

Choose transformer or known-impedance mode

Use Transformer kVA and %Z for a simplified terminal estimate from transformer nameplate data. Use Known system impedance when the equivalent impedance at the selected fault point is already known.
Select three-phase or single-phase

Three-phase mode uses line-to-line voltage and a per-phase equivalent impedance. Single-phase mode uses the voltage across the complete fault path represented by the entered loop impedance.
Enter values in the units shown

Transformer mode accepts apparent-power rating, secondary voltage, and percent impedance. Known-impedance mode accepts voltage and equivalent impedance. Confirm kVA versus MVA, V versus kV, and ohms versus milliohms before calculating.
Review the supporting checks

Use fault level and equivalent impedance as scale checks. In transformer mode, verify that the fault-current multiplier equals \(1/z_{pu}=100/Z_{\%}\).
Carry the result only to the location it represents

A transformer-terminal result does not include downstream conductor, busway, or connection impedance. For downstream equipment, use the equivalent impedance through that location or a suitable point-to-point or full short-circuit study.

Fault Current Inputs and Outputs

The calculator changes the required fields based on method and phase selection. Each input must describe the same electrical source boundary and fault location.

Calculation Method: Transformer kVA and %Z estimates transformer-limited terminal fault current. Known system impedance calculates current from the equivalent impedance seen at the selected fault point.
System Phase: Three-phase mode uses line-to-line voltage and a \(\sqrt{3}\) factor. Single-phase mode uses the voltage across the modeled fault loop.
Transformer Rating: Transformer apparent-power nameplate rating in kVA or MVA. This field is used only in transformer mode.
System or Secondary Voltage: For three-phase calculations, use RMS line-to-line voltage. For single-phase calculations, use the voltage across the fault path represented by the model.
Transformer Impedance: Enter the nameplate percent impedance as a percentage. For example, enter 5.75 for \(5.75\%\), not 0.0575.
Equivalent System Impedance: In three-phase mode, use the positive-sequence per-phase equivalent impedance seen from the fault point. In single-phase mode, use the complete source-and-return loop impedance.
Symmetrical RMS Fault Current: The primary calculated bolted-fault current magnitude in A or kA. It is not instantaneous peak current, arcing current, or arc-flash incident energy.
Result Checks: The calculator reports fault level and equivalent impedance. Transformer mode additionally reports full-load current and the impedance current multiplier.

Fault Current Formulas

Transformer mode converts nameplate percent impedance to per unit and scales rated secondary current. Known-impedance mode applies the AC magnitude form of Ohm’s law at the selected fault point.

Three-phase transformer method

\[ I_{FL}=\frac{S}{\sqrt{3}V_{LL}},\qquad z_{pu}=\frac{Z_{\%}}{100},\qquad I_{sc}=\frac{I_{FL}}{z_{pu}} =\frac{S}{\sqrt{3}V_{LL}z_{pu}} \]

Calculate rated secondary current from transformer VA and line-to-line voltage, then divide by transformer impedance in per unit.

This simplified transformer-terminal estimate assumes an effectively infinite upstream source and does not include downstream conductor impedance.

Single-phase transformer method

\[ I_{FL}=\frac{S}{V},\qquad I_{sc}=\frac{S}{Vz_{pu}} \]

Use the single-phase transformer rating and the voltage across the modeled secondary fault path.

Known equivalent impedance

\[ I_{sc,3\phi}=\frac{V_{LL}}{\sqrt{3}\lvert Z_{eq}\rvert}, \qquad I_{sc,1\phi}=\frac{V}{\lvert Z_{loop}\rvert} \]

The three-phase form uses the positive-sequence per-phase equivalent impedance. The single-phase form uses the complete loop impedance.

Fault level

\[ S_{sc,3\phi}=\sqrt{3}V_{LL}I_{sc},\qquad S_{sc,1\phi}=VI_{sc} \]

Fault level expresses the same short-circuit condition as apparent power and is commonly reported in MVA.

\(I_{sc}\): Symmetrical RMS short-circuit current in amperes.
\(I_{FL}\): Transformer full-load secondary current in amperes.
\(S\): Transformer apparent-power rating in volt-amperes.
\(V_{LL}\): Three-phase RMS line-to-line voltage in volts.
\(z_{pu}\): Transformer impedance in per unit, equal to nameplate percent impedance divided by 100.
\(\lvert Z_{eq}\rvert\): Magnitude of the relevant equivalent system impedance in ohms.

Worked Example: 750 kVA Transformer

Estimate the three-phase symmetrical RMS fault current at the secondary terminals of a 750 kVA transformer with a 480 V line-to-line secondary and 5.75% nameplate impedance.

Given values

Transformer rating: \(S=750\,\mathrm{kVA}=750{,}000\,\mathrm{VA}\)
Secondary voltage: \(V_{LL}=480\,\mathrm{V}\)
Nameplate impedance: \(Z_{\%}=5.75\%\)
Find: Three-phase symmetrical RMS terminal fault current

Convert percent impedance

\[ z_{pu}=\frac{5.75}{100}=0.0575 \]

Calculate full-load current

\[ I_{FL}=\frac{750{,}000}{\sqrt{3}(480)} \approx902.11\,\mathrm{A} \]

Calculate fault current

\[ I_{sc}=\frac{902.11}{0.0575} \approx15{,}688.87\,\mathrm{A} =15.6889\,\mathrm{kA} \]

Result

Symmetrical RMS fault current \(\approx15.69\ \mathrm{kA}\)

The calculator also reports approximately 902.1 A full-load current, 13.04 MVA fault level, 17.66 mΩ equivalent impedance, and a 17.39× impedance current multiplier.

Where the Calculated Fault Current Applies

Available fault current is location-specific. The value at transformer secondary terminals is normally higher than the value farther downstream when additional conductor, busway, and connection impedance is introduced.

Transformer secondary terminals

Transformer mode estimates current at the transformer secondary with the upstream source treated as effectively infinite. This is useful as a screening boundary but does not include downstream feeder impedance.

Downstream panel or equipment

Use the total equivalent impedance through the equipment location. Longer or smaller conductors, busway, and other series impedance generally reduce available current compared with the transformer terminals.

What must be represented for different fault-current locations
Fault location	Minimum calculation model	Common missing data
Transformer secondary	Transformer rating, secondary voltage, and percent impedance	Finite utility/source strength and other connected sources
Downstream panel	Source plus transformer plus feeder/bus equivalent impedance	Conductor length, size, material, parallel paths, busway, connections
Complex facility bus	Complete network short-circuit model	Generators, motors, inverter sources, X/R ratios, multiple transformers, sequence networks

Eaton’s short-circuit calculation guidance describes point-to-point calculations by representing utility, transformer, cable, and busway components as impedances and calculating available current at different points in the distribution system.

How to Interpret the Result

The result is the symmetrical RMS AC component of a simplified bolted-fault calculation. Use it as an electrical-duty value at the modeled point, not as a normal load current or an arc-flash result.

What 15.69 kA means

A result of 15.69 kA means the simplified model predicts approximately 15,690 A RMS of symmetrical current at that fault location before protective clearing is considered.

Impedance controls the current

For fixed voltage, \(I\propto1/Z\). Increasing impedance by 10% reduces current to \(1/1.10\approx90.9\%\) of the original value; decreasing impedance by 10% increases current to \(1/0.90\approx111.1\%\).

Transformer sanity check

In transformer mode, \(I_{sc}/I_{FL}=100/Z_{\%}\). A 5.75% transformer should therefore have an idealized terminal multiplier of about 17.39.

Fault Current, AIC, Interrupting Rating, and SCCR

Available fault current is a system condition. Breaker or fuse interrupting rating and equipment SCCR are equipment ratings. They answer different questions and should not be used interchangeably.

How available fault current differs from common equipment ratings
Term	What it describes	How to use it
Available fault current	Current the connected system can deliver to a fault at a specific location	Calculate or obtain it for the equipment location
Interrupting rating / AIC	Fault-current magnitude an interrupting device is rated to interrupt at a stated voltage	Verify the device rating is suitable for the available current and circuit voltage
SCCR	Short-circuit current rating of an equipment assembly or component under its applicable rating conditions	Verify the installed equipment rating is suitable for the available fault current and its application requirements
Breaker ampere rating	Normal-current rating or setting	Do not confuse it with interrupting capability

For U.S. workplace installations, OSHA 29 CFR 1910.303(b)(4) requires equipment intended to interrupt current at fault levels to have an interrupting rating sufficient for the nominal circuit voltage and the current available at its line terminals. The next paragraph also addresses circuit impedance, component short-circuit ratings, and coordination of protective devices.

Units and Input Checks

Fault-current arithmetic is simple enough that unit and definition errors are often more dangerous than algebra errors. Confirm the source value and unit before entering each field.

Percent versus per unit

\(5.75\%=0.0575\ \mathrm{pu}\). Enter 5.75 in the calculator’s percent field. Entering 0.0575 would represent only 0.0575%.

Ohms versus milliohms

\(1\ \Omega=1000\ \mathrm{m}\Omega\). A value of 25 mΩ is 0.025 Ω; treating it as 25 Ω would reduce the calculated current by a factor of 1000.

kVA versus MVA

\(1\ \mathrm{MVA}=1000\ \mathrm{kVA}\). A 0.75 MVA transformer is the same apparent-power rating as 750 kVA.

Line-to-line voltage

Three-phase mode uses \(V_{LL}\). For a 480Y/277 V system, enter 480 V for the balanced three-phase equation, not 277 V.

Use transformer kVA, voltage, and percent impedance from the same nameplate basis and tap condition.
Identify the exact electrical point where the available current is needed.
In known-impedance mode, verify that the equivalent includes every series path required by the selected fault type.
Determine whether motors, generators, parallel sources, or inverter-based resources contribute to the modeled fault.

Common Mistakes and Troubleshooting

If the number looks implausible, first check the meaning and scale of the inputs. A correctly solved equation can still represent the wrong system.

Using line-to-neutral voltage in three-phase mode

This understates the line-to-line voltage basis of the balanced three-phase equation. Use the system’s line-to-line voltage.

Applying transformer-terminal current downstream

Downstream series impedance normally reduces available current. Recalculate using the equivalent impedance through the actual equipment location.

Confusing RMS current with peak or arcing current

The calculator reports symmetrical RMS bolted-fault current. Peak asymmetrical duty, arcing current, and incident energy are different quantities.

Missing source contributions

A detailed study can differ because of finite utility strength, motors, generators, inverter controls, parallel transformers, conductor impedance, voltage factors, and X/R ratio.

Extremely high result

If the tool flags a result above 200 kA, recheck mΩ versus Ω, percent format, missing series impedance, and whether the modeled source boundary is realistic. The warning is a QA prompt, not a universal system limit.

Unusual transformer impedance

The calculator flags values below 1% or above 15% as a screening warning. Verify the actual nameplate instead of treating those thresholds as design limits.

Assumptions and Study Limits

This calculator is a simplified short-circuit screening tool. It is not a full IEC 60909 study, protection-coordination study, SCCR evaluation, or arc-flash analysis.

Bolted symmetrical fault

The result represents the symmetrical RMS AC component with negligible fault resistance. DC offset and peak asymmetry are not separately calculated.

Infinite-source transformer mode

Transformer mode neglects finite upstream source impedance and downstream conductor impedance. It is intentionally a transformer-terminal screening estimate.

Known impedance must be complete

The known-impedance result is only as accurate as the entered equivalent network. Missing source or path impedances can materially change the answer.

Fault type matters

The balanced three-phase impedance relationship does not replace sequence-network calculations for unbalanced faults. Single-line-to-ground and other fault types can require zero-, positive-, and negative-sequence network data.

Dynamic source contribution is omitted

Motor, generator, and inverter-based source contributions are not dynamically modeled by the simplified calculator.

No protection or arc-flash solution

The calculator does not determine clearing time, protective-device coordination, current-limiting let-through, arcing current, working distance, or incident energy.

Sources and Calculation Basis

The calculator’s transformer relationship and location-specific short-circuit guidance were checked against current manufacturer, standards-body, and regulatory sources.

Schneider Electric — Maximum Available Fault Current at Transformer Secondary — supports the transformer method of dividing secondary full-load current by transformer per-unit impedance for single- and three-phase transformer terminal estimates.
Eaton Bussmann — Short-Circuit Current Calculations — supports point-to-point short-circuit calculations that represent utility, transformer, cable, and busway components as impedances and calculate available current at different system locations.
IEC 60909-0:2026 — Short-Circuit Currents in Three-Phase AC Systems — identifies the current IEC procedure scope for short-circuit-current calculations in low- and high-voltage three-phase AC systems at 50 Hz or 60 Hz.
OSHA 29 CFR 1910.303 — General Electrical Requirements — supports the requirement that equipment intended to interrupt fault current have sufficient interrupting rating for the nominal voltage and available current at its line terminals.

The worked example was independently checked using the transformer current multiplier, reverse impedance calculation, and fault-level calculation. These checks reproduce the calculator result to rounding precision.

Fault Current Calculator FAQ

These questions address the distinctions most likely to change a fault-current calculation or its practical use.

What is available fault current?

Available fault current is the current the connected source and electrical system can deliver to a fault at a specific location. It changes with source strength, transformer impedance, conductors, busway, connections, motors, generators, inverters, and fault type.

How do I calculate transformer fault current from percent impedance?

Calculate transformer full-load secondary current, convert percent impedance to per unit, then divide: \(I_{sc}=I_{FL}/z_{pu}\). For example, a transformer with 5.75% impedance has an idealized current multiplier of \(100/5.75\approx17.39\).

Why is fault current lower at a downstream panel?

Feeders, busway, and connections add series impedance between the source and the panel. Because fault current varies approximately inversely with equivalent impedance, adding path impedance generally reduces the available current.

Is available fault current the same as breaker AIC?

No. Available fault current is a system condition at a location. Breaker interrupting rating or AIC is an equipment rating that describes the fault current the breaker is rated to interrupt under its stated conditions.

Is fault current the same as SCCR?

No. Fault current is what the system can deliver at the equipment location. SCCR is a short-circuit current rating assigned to equipment or an assembly under applicable rating conditions. The equipment rating must be evaluated against the available current and the requirements of the installation.

Can measured loop impedance be used?

It can be used in single-phase known-impedance mode when it represents the complete energized fault loop and is appropriate for the calculation objective. A balanced three-phase fault instead requires the correct positive-sequence per-phase equivalent impedance.

Does this calculator determine arc-flash incident energy?

No. The calculator provides simplified bolted symmetrical RMS fault current. Arc-flash analysis requires additional modeling such as arcing current, equipment configuration, working distance, and protective-device clearing time under the applicable analysis method.