Transformers

Transformer Fault & Motor-Starting Study

Available fault current at a transformer secondary, from nameplate %Z and the utility's contribution — with symmetrical and asymmetrical duty from the combined X/R ratio, and the voltage dip a motor start causes on that same bus.

linked calculations: 4linked calculations
inputs, entered once: 8inputs, entered once
value derived, not retyped: 1value derived, not retyped

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The system this study analyses

Drawn from the study’s own definition and annotated with the example values below. Every quantity on it is one the study entered or computed.

Network configuration (example)

Single-line arrangement analysed by this study, annotated with the values it used and computed.

›Diagram values as a table

Single-line diagram: Utility source (source), then Transformer (transformer), then Secondary bus (bus), then Motor (motor).
Element	Quantity	Value
Utility source	Primary	13.8 kV
Utility source	Fault power	250 MVA
Utility source	X/R	15
Transformer	Rating	1500 kVA
Transformer	Impedance	5.75 %
Transformer	X/R	6
Transformer	Secondary FLA	1804.2 A
Secondary bus	Voltage	480 V
Secondary bus	Symmetrical Icc	28413 A
Secondary bus	Peak asymmetrical	64737 A
Secondary bus	Fault power	23.622 MVA
Motor	Full-load current	250 A
Motor	Locked-rotor current	1500 A
Secondary bus to Motor	Voltage dip at start	5.01 %
Secondary bus to Motor	Bus voltage during start	455.9 V

The fault level at a transformer secondary is the number half a dozen other decisions hang off. It is also the number people most often carry between spreadsheets by hand, which is exactly where a percent impedance gets transcribed wrong and quietly changes every figure downstream.

This study takes the transformer's rating, impedance and X/R, and the utility's available fault power at the primary, and computes the whole set once: full-load current, symmetrical fault current, the fault power available at the secondary, the peak and half-cycle asymmetrical currents, and the voltage dip a motor starting on that bus would cause.

The fault power is not entered twice. Step 02 computes it, and the motor-starting step takes that value — so changing the transformer impedance moves the starting dip, which is the physical relationship a spreadsheet usually loses.

Every shared value is entered once

The system voltage is not typed into four calculators. It is defined once for the study and carried into every step that needs it — so a change reaches all of them, and no two steps can describe different systems.

The study’s shared inputs and the steps each one reaches
Input	Example value	Used by
Transformer ratingTransformer	1500 kVA	Step 01, Step 02, Step 03
Secondary voltageTransformer	480 V	Step 01, Step 02, Step 03, Step 04
Transformer impedanceTransformer	5.75 %	Step 02, Step 03
Transformer X/R ratioTransformer	6	Step 03
Primary voltageSource	13.8 kV	Step 01
Available fault power at the primarySource	250 MVA	Step 02, Step 03
Source X/R ratioSource	15	Step 03
Motor full-load currentMotor	250 A	Step 04
Available fault power at the secondaryDerived	23.622047244094485 MVA	Computed by Step 02, used by Step 04

The 4 steps

Worked through below with an example transformer fault and motor-starting study. Every number shown was produced by the same calculators the library already ships — nothing on this page is illustrative.

Step 01
Transformer full-load current
Transformer Sizing & Full-Load Current Calculator
Computes the primary and secondary full-load currents from the transformer's rating and its two voltages, and reports the nearest standard rating.
Inputs and their sources
- SystemThree-phaseFixed by the study
- Transformer rating1500 kVAYou enter
- Primary voltage13.8 kVYou enter
- Secondary voltage480 VYou enter
Example results
- Secondary full-load current1804.2A
- Primary full-load current62.76A
- Nearest standard size1500kVA
›Why this step is here
The study starts here because the secondary voltage established at the transformer is the voltage every fault and dip figure below is referred to. The full-load current is the continuous rating; the numbers in the steps that follow are the fault and inrush conditions the same transformer sees, which are a different question entirely.
›Worked solution
1. Primary full-load current
  $I_{FL,P} = \dfrac{S}{\sqrt{3}\,V_P}$
  $I_{FL,P} = \dfrac{1.500e+6}{\sqrt{3} \times 1.380e+4}$
  $I_{FL,P} = 62.76\ \text{A}$
2. Secondary full-load current
  $I_{FL,S} = \dfrac{S}{\sqrt{3}\,V_S}$
  $I_{FL,S} = \dfrac{1.500e+6}{\sqrt{3} \times 480}$
  $I_{FL,S} = 1804\ \text{A}$
Calculator assumptions
- Self-cooled (ONAN / dry-type AA) nameplate rating; forced-air or other cooling stages would raise the FLA basis.
- Balanced loading on both sides.
- Standard-size list reflects common IEEE/ANSI distribution transformer ratings — verify against the manufacturer's actual offered sizes.
References
- IEEE Std C57.12.00 — General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers
- NEC Table 450.3 — Overcurrent Protection for Transformers
Step 02
Available fault current and power
Symmetrical Fault Current Calculator
Safety-sensitive
Combines the source and transformer impedances on a per-unit basis to give the symmetrical fault current at the secondary, and the fault power available there.
Inputs and their sources
- Source impedanceInclude finite source impedanceFixed by the study
- Available fault MVA at source250 MVAYou enter
- Transformer rating1500 kVAYou enter
- Transformer impedance5.75 %You enter
- Secondary voltage480 VYou enter
Example results
- Symmetrical fault current28413A
- Available fault power23.622MVA
- Total impedance to fault0.0635p.u.
- Source impedance0.006p.u.
- Transformer impedance0.0575p.u.
- Fault current (infinite bus, transformer only)31378A
›Why this step is here
This step produces two different things. The fault current is what the fault study is about; the available fault power in MVA is what step 04 needs, because a bus that can deliver more fault power sags less when a motor starts. The infinite-bus figure is reported alongside so the source's contribution to the total is visible rather than buried.
›Worked solution
1. Source impedance on the transformer's kVA base
  $Z_{source,pu} = \dfrac{S_{base}}{S_{sc,source}}$
  $Z_{source,pu} = \dfrac{1500\ \text{kVA}}{2.5000e+5\ \text{kVA}}$
  $Z_{source,pu} = 0.006\ \text{p.u.}$
2. Transformer impedance (nameplate)
  $Z_{xfmr,pu} = \dfrac{\%Z}{100}$
  $Z_{xfmr,pu} = \dfrac{5.75}{100}$
  $Z_{xfmr,pu} = 0.0575\ \text{p.u.}$
3. Total impedance to the fault point
  $Z_{total,pu} = Z_{source,pu} + Z_{xfmr,pu}$
  $Z_{total,pu} = 0.006 + 0.0575$
  $Z_{total,pu} = 0.0635\ \text{p.u.}$
4. Available fault power at the secondary
  $S_{sc} = \dfrac{S_{base}}{Z_{total,pu}}$
  $S_{sc} = \dfrac{1500\ \text{kVA}}{0.0635}$
  $S_{sc} = 23622\ \text{kVA}$
5. Symmetrical fault current
  $I_{sc} = \dfrac{S_{sc}}{\sqrt{3}\,V_{LL}}$
  $I_{sc} = \dfrac{2.3622e+7\ \text{VA}}{\sqrt{3} \times 480\ \text{V}}$
  $I_{sc} = 28413\ \text{A}$
Calculator assumptions
- Single transformer, radial system, bolted three-phase fault at the transformer secondary terminals.
- No motor contribution to fault current is included.
- Cable/busway impedance between the transformer and the fault point is neglected.
- Symmetrical RMS fault current only — asymmetrical peak current (X/R-adjusted) is not computed.
References
- IEEE Std 141 (Red Book), Ch. 2 — Short-Circuit Current Calculations
- IEEE Std 242 (Buff Book), Ch. 2
Step 03
Peak and asymmetrical duty
Transformer Impedance & Short-Circuit Current Calculator
Safety-sensitive
Combines the same two impedances through their X/R ratios to give the symmetrical RMS current, the peak asymmetrical current and the half-cycle asymmetrical RMS current.
Inputs and their sources
- Source impedanceInclude finite source impedanceFixed by the study
- Transformer rating1500 kVAYou enter
- Transformer impedance5.75 %You enter
- Transformer X/R ratio6You enter
- Secondary voltage480 VYou enter
- Available fault MVA at source250 MVAYou enter
- Source X/R ratio15You enter
Example results
- Symmetrical RMS fault current28425A
- Peak asymmetrical current64737A
- RMS asymmetrical current (1/2 cycle)37551A
- Combined X/R ratio6.36
- Total impedance0.06347p.u.
›Why this step is here
This step and step 02 describe the same fault from the same impedance data, and their symmetrical currents land within about a twentieth of a percent of each other rather than matching exactly. That is the two methods, not an error: step 02 adds the per-unit impedances as scalars, while this step combines them as vectors through their X/R ratios. What the vector method adds is the part a scalar sum cannot produce — the DC offset that rides on the first cycles of a fault, which is governed by the combined X/R and which sets the peak and half-cycle currents that momentary and closing-and-latching ratings are read against.
›Worked solution
1. Resistance/reactance components from %Z and X/R
  $R = \dfrac{Z_{pu}}{\sqrt{1+(X/R)^2}},\quad X = R \times (X/R)$
  $\text{applied to the transformer, and the source if included}$
  $R_{xfmr} = 0.009453,\ X_{xfmr} = 0.05672,\ R_{source} = 0.0003991,\ X_{source} = 0.005987$
2. Combined series impedance
  $R_{total}=R_{source}+R_{xfmr},\ X_{total}=X_{source}+X_{xfmr}$
  $R_{total} = 0.0003991+0.009453,\ X_{total} = 0.005987+0.05672$
  $Z_{total} = 0.06347\ \text{p.u.},\quad X/R = 6.365$
3. Symmetrical fault current
  $I_{sym} = \dfrac{S_{xfmr}/Z_{total}}{\sqrt{3}\,V_{LL}}$
  $I_{sym} = \dfrac{1500/0.06347\ \text{kVA} \times 1000}{\sqrt{3} \times V_{LL}}$
  $I_{sym} = 2.842e+4\ \text{A} \quad (S_{sc} = 2.363e+4\ \text{kVA})$
4. Peak asymmetrical current
  $I_{peak} = I_{sym} \times \sqrt{2}\left(1+e^{-\pi/(X/R)}\right)$
  $I_{peak} = 2.842e+4 \times 2.277$
  $I_{peak} = 6.474e+4\ \text{A}$
5. RMS asymmetrical current (first half-cycle)
  $I_{rms,asym} = I_{sym} \times \sqrt{1+2e^{-2\pi/(X/R)}}$
  $I_{rms,asym} = 2.842e+4 \times 1.321$
  $I_{rms,asym} = 3.755e+4\ \text{A}$
Calculator assumptions
- Single transformer, radial system, bolted three-phase fault at the transformer secondary terminals.
- No motor contribution to fault current is included.
- Peak and RMS asymmetrical factors use standard closed-form approximations (IEEE Std 551), not the full ANSI C37.010 multiplying-factor curves.
References
- IEEE Std 551 (Violet Book), Ch. 4 — Short-Circuit Currents
- ANSI/IEEE C37.010 — Application Guide for AC High-Voltage Circuit Breakers
Step 04
Motor-starting voltage dip
Motor Starting Current & Voltage Dip Estimator
Estimates the voltage dip on the secondary bus while a motor draws locked-rotor current, using the fault power derived in step 02 as the bus stiffness.
Inputs and their sources
- Motor full-load current250 AYou enter
- Locked-rotor current multiple6You enter
- Bus voltage480 VYou enter
- Available fault MVA at the bus23.62 MVADerivedStep 02 · Available fault current and power
- Maximum acceptable dip10 %You enter
Example results
- Voltage dip during starting5.01%
- Locked-rotor current1500A
- Starting apparent power1247.1kVA
- Bus voltage during start455.9V
›Why this step is here
Fault level and starting dip are the same property of a bus seen from two directions: how much current the source can push into it. The fault power computed in step 02 is not re-entered here — it arrives from that step, so changing the transformer impedance moves the dip without anyone retyping a number. This assumes the motor starts on the transformer secondary itself; a motor further down a feeder sees a weaker bus than this.
›Worked solution
1. Locked-rotor (starting) current
  $I_{LR} = I_{FL} \times \dfrac{I_{LR}}{I_{FL}}$
  $I_{LR} = 250 \times 6$
  $I_{LR} = 1500\ \text{A}$
2. Starting apparent power
  $S_{start} = \sqrt{3}\,V\,I_{LR}$
  $S_{start} = \sqrt{3} \times 480 \times 1500$
  $S_{start} = 1247\ \text{kVA}$
3. Voltage dip (MVA method)
  $VD\% = \dfrac{S_{start}}{S_{start}+S_{sc}} \times 100$
  $VD\% = \dfrac{1247}{1247+2.362e+4} \times 100$
  $VD\% = 5.015\%$
4. Bus voltage during the start
  $V' = V(1 - VD\%/100)$
  $V' = 480(1 - 5.015/100)$
  $V' = 455.9\ \text{V}$
Calculator assumptions
- MVA method approximation: starting current treated as predominantly reactive, source and motor impedance angles assumed similar enough to combine as scalar MVA.
- Single motor start, no other simultaneous starting or switching events on the bus.
- Cable impedance between the source and the motor is neglected — the dip shown is at the source bus, not necessarily at the motor terminals.
References
- IEEE Std 141 (Red Book), Ch. 5 — Motor-starting voltage dip, the MVA method
- NEMA MG 1, Motors and Generators — locked-rotor current code letters

Why these steps, in this order

Why the source impedance matters

A transformer's percent impedance alone gives the infinite-bus fault current — the answer you get by assuming the utility can supply unlimited current. Adding the source's available fault power puts a real impedance upstream of the transformer, which lowers the result. Both figures are reported, so the size of the source's contribution is visible rather than assumed away.

Why there are two fault calculations

They describe the same fault from the same impedance data, and their symmetrical currents land within about a twentieth of a percent of each other — one adds the per-unit impedances as scalars, the other combines them as vectors through their X/R ratios, so the two are close rather than identical. The vector method is the only one that can produce the DC offset riding on the first cycles of a fault, and therefore the peak and half-cycle asymmetrical currents that momentary and closing-and-latching ratings are read against.

Why fault level and motor starting are the same question

Both are about how stiff the bus is: how much current the source can push into it. A high available fault power means a stiff bus, which means a smaller dip when a motor draws locked-rotor current. That is why this study derives the starting dip from the fault power it just computed rather than asking for a bus stiffness twice.

What the numbers are referred to

Every fault and dip figure here is at the transformer secondary. A motor starting further down a feeder sees a weaker bus than this study models, and a fault further downstream sees a lower current. Where the bus is not the secondary itself, the conductor impedance between them has to be accounted for separately.

What this study reports, and what it leaves to you

The study runs each calculation and reports the numbers with the assumptions, model boundaries and references the calculators state for themselves. It draws no conclusion about your installation. Comparing a derated ampacity against a load current, or a voltage drop against a design limit, is engineering judgement — the study puts the numbers in front of you and stops there.

Results are preliminary and educational. They are not verified for any specific installation and must be reviewed by a licensed Professional Engineer before they inform construction, equipment selection or protection settings.

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Frequently asked questions

What is a transformer fault study?

It is the calculation of how much current would flow in a short circuit at a transformer's secondary, given the transformer's rating and percent impedance and the fault power the utility can supply at the primary — together with the peak and asymmetrical values that follow from the combined X/R ratio.

What do I have to enter?

Eight values, each entered once: transformer rating, primary voltage, secondary voltage, transformer percent impedance, transformer X/R ratio, the utility's available fault power at the primary, the source X/R ratio, and the full-load current of the motor being started. The available fault power at the secondary is derived rather than entered.

Why is the motor-starting dip in a fault study?

Because it is the same property of the bus. The fault power available at the secondary is a measure of how stiff that bus is, and bus stiffness is what determines how far the voltage falls while a motor draws locked-rotor current. Computing them together means the two can never disagree about the same transformer.

Does the study choose equipment or settings for me?

No. It reports calculated currents, powers and a percentage dip with the assumptions, model boundaries and references each calculator states for itself. It reaches no conclusion about equipment ratings, protection settings or whether any value is appropriate for your installation. Every result is preliminary and educational and needs a licensed Professional Engineer's review before it informs construction, equipment selection or protection settings.

Can I change one input and run it again?

Yes. Changing a shared input marks every step it reaches, and running the study again recomputes only those steps — in dependency order, so the motor-starting step never reads a fault power that is about to change in the same run.

What do I get at the end?

A professional PDF report with every step's worked solution, the inputs each one used and where each value came from, the calculators' own assumptions and limitations, the collected references, and the standing engineering disclaimer.

Is this study clear?

Tell us if a step, an input or a worked solution needs explaining better — or which study you'd like next. Reports go straight to our team.