Overview
The NEC Short Circuit Analysis calculator computes available fault currents at equipment locations to verify that protective devices and equipment have adequate interrupting and withstand ratings per the National Electrical Code (NEC).
The calculator provides:
- Transformer fault contribution — from kVA rating, %Z, and system voltage
- Parallel source contribution (optional) — another source connected at the fault point (e.g. a generator or a second service); its fault current is added to the transformer path. It is not the utility on the transformer primary — the transformer is always fed from an infinite bus.
- Conductor impedance reduction (optional) — fault current reduction along feeder runs
- Motor contribution (optional) — motor backfeed during faults (4–6× FLA)
- Multiple fault types — 3-phase bolted, line-to-ground, line-to-line, or all
- Minimum breaker AIC rating — required interrupting capacity
Input Parameters
Transformer Parameters
| Parameter | Default | Range | Description |
|---|---|---|---|
transformer_kva | 75 kVA | ≥ 1 | Transformer rated power |
transformer_impedance_percent | 5.0% | 0.1 – 20 | Nameplate percent impedance (%Z) |
system_voltage | 208 V | 208 – 34500 | Line-to-line system voltage |
system_configuration | Wye Grounded | Wye/Delta variants | System grounding arrangement |
Optional Parameters
| Parameter | Default | Description |
|---|---|---|
include_utility | No | Add a source in parallel at the fault point |
utility_fault_current | 10,000 A | Fault current of that parallel source at the system voltage |
include_conductor_impedance | No | Account for feeder impedance |
conductor_length | 100 ft | One-way feeder run length |
conductor_size | 4/0 AWG | Conductor size (AWG/kcmil) |
conductor_material | Copper | Copper or Aluminum |
fault_type | 3-Phase Bolted | Type of fault to calculate |
x_r_ratio | 3.0 | System X/R ratio (0.5 – 20) |
motor_contribution | No | Include motor backfeed |
safety_factor | 1.0 | Additional safety margin (1.0 – 1.5) |
Output Results
| Output | Unit | Description |
|---|---|---|
| Transformer Fault Current | kA | Fault contribution from transformer alone |
| Total Fault Current | kA | Sum of all contributions (transformer path + parallel source + motors) × safety factor |
| Min Breaker AIC Rating | kAIC | Minimum required interrupting capacity |
| Fault Current at Point | kA | Reduced fault current after conductor impedance (if included) |
Formulas
Transformer Fault Current
Three-phase short circuit current from the transformer (infinite primary; N identical units in parallel):
IFL = (kVA × 1000) / (√3 × VLL) IT = 100 × IFL / (%Z / N)
Transformer Impedance
ZT = VLL² / (kVA × 1000) × %Z/100 / N RT = ZT / √(1 + (X/R)²) XT = RT × X/R
Conductor Impedance Reduction
Feeder impedance (one-way, positive sequence) is added in series with the transformer:
Zconductor = (R + jX) × Lengthft / (1000 × nparallel) Ipoint = VLL / (√3 × |(RT + RC) + j(XT + XC)|)
With a Parallel Source at the Fault
The source's fault current is converted to an impedance magnitude and combined in parallel with |Z|. Because magnitudes (not phasors) are combined, this is the same as adding the two currents — exact when both paths have the same X/R, conservative otherwise:
ZS = VLL / (√3 × IS) Zcombined = |Z| × ZS / (|Z| + ZS) I = Ipoint + IS
Motor Contribution and Safety Factor
Imotor = FLAtotal × multiplier (4–6, default 5) I3φ = (I + Imotor) × safety factor
Other Fault Types (estimates)
Line-to-ground and line-to-line currents are fixed multiples of the three-phase current, not a sequence-impedance calculation:
IL-G ≈ 1.25 × I3φ (solidly grounded wye) or 0.87 × I3φ (other systems) IL-L ≈ 0.87 × I3φ (√3/2 = 0.866)
For a delta-wye transformer terminal fault the true line-to-ground current is about 1.0 × I3φ (Z0 ≈ Z1), so the 1.25 multiplier is conservative; on an ungrounded system a single line-to-ground fault draws little current. Use an IEC 60909 / symmetrical-component study where the ground-fault duty matters.
Minimum Interrupting Rating
The next standard rating at or above the highest of I3φ, IL-G and IL-L: 10, 14, 18, 22, 25, 35, 42, 50, 65, 100, 125, 150, 200 kA.
Conductor Table (Ω per 1000 ft, 60 Hz)
| Size | R copper | R aluminum | X conduit | X cable tray | X open air |
|---|---|---|---|---|---|
| 14 AWG | 3.07 | 5.06 | 0.058 | 0.052 | 0.048 |
| 12 AWG | 1.93 | 3.18 | 0.056 | 0.050 | 0.046 |
| 10 AWG | 1.21 | 2 | 0.054 | 0.048 | 0.044 |
| 8 AWG | 0.764 | 1.26 | 0.052 | 0.046 | 0.042 |
| 6 AWG | 0.491 | 0.808 | 0.051 | 0.045 | 0.041 |
| 4 AWG | 0.308 | 0.508 | 0.050 | 0.044 | 0.040 |
| 3 AWG | 0.245 | 0.403 | 0.049 | 0.043 | 0.039 |
| 2 AWG | 0.194 | 0.319 | 0.048 | 0.042 | 0.038 |
| 1 AWG | 0.154 | 0.253 | 0.047 | 0.041 | 0.037 |
| 1/0 | 0.122 | 0.201 | 0.046 | 0.040 | 0.036 |
| 2/0 | 0.0967 | 0.159 | 0.045 | 0.039 | 0.035 |
| 3/0 | 0.0766 | 0.126 | 0.044 | 0.038 | 0.034 |
| 4/0 | 0.0608 | 0.1 | 0.043 | 0.037 | 0.033 |
| 250 kcmil | 0.0515 | 0.0847 | 0.042 | 0.036 | 0.032 |
| 300 kcmil | 0.0429 | 0.0707 | 0.041 | 0.035 | 0.031 |
| 350 kcmil | 0.0367 | 0.0605 | 0.041 | 0.035 | 0.031 |
| 400 kcmil | 0.0321 | 0.0529 | 0.040 | 0.034 | 0.030 |
| 500 kcmil | 0.0258 | 0.0424 | 0.039 | 0.033 | 0.029 |
| 600 kcmil | 0.0214 | 0.0353 | 0.038 | 0.032 | 0.028 |
| 750 kcmil | 0.0171 | 0.0282 | 0.037 | 0.031 | 0.027 |
| 1000 kcmil | 0.0129 | 0.0212 | 0.036 | 0.030 | 0.026 |
R is the NEC Chapter 9 Table 8 DC resistance at 75 °C. The reactances are the calculator's simplified values for copper (aluminium is 0.002 Ω/kft higher); NEC Chapter 9 Table 9 gives, for example, 0.041 (PVC/aluminium conduit) and 0.051 (steel conduit) Ω/kft for 4/0 against 0.043 here.
Worked Examples
The 8 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_short_circuit; worked examples for every calculation).
| # | Case | I3φ (kA) | IL-G (kA) | X/R | Min rating (kA) | Rating (kA) | Governing | Result |
|---|---|---|---|---|---|---|---|---|
| 1 | 75 kVA, 208 V transformer terminals | 4.164 | 5.204 | 3.00 | 10 | – | minimum interrupting rating | PASS |
| 2 | 75 kVA, 208 V plus 100 ft of 4/0 Cu in conduit | 3.419 | 4.274 | 2.08 | 10 | – | minimum interrupting rating | PASS |
| 3 | 500 kVA, 480 V delta — terminal fault | 12.028 | 10.464 | 3.00 | 14 | – | minimum interrupting rating | PASS |
| 4 | Two 1000 kVA transformers in parallel at 208 V | 96.547 | 83.996 | 3.00 | 100 | – | minimum interrupting rating | PASS |
| 5 | 1000 kVA, 480 V, 2 × 500 kcmil Al in tray | 15.664 | 19.580 | 2.60 | 22 | 25 | interrupting rating | PASS |
| 6 | 750 kVA, 480 V with 400 A of running motors — 14 kA rating fails | 17.289 | 15.041 | 3.00 | 18 | 14 | interrupting rating | FAIL |
| 7 | 1500 kVA, 480 V with a 10 kA source in parallel and a 1.1 safety factor | 45.516 | 39.598 | 5.00 | 50 | 50 | interrupting rating | PASS |
| 8 | 150 kVA, 600 V, 75 m of 2 AWG in open air | 3.100 | 2.697 | 0.79 | 10 | – | minimum interrupting rating | PASS |
Example 1 — 75 kVA, 208 V transformer terminals
The simplest ohmic case — a small dry-type transformer on a solidly grounded wye — and the minimum interrupting rating it needs.
Given: 208 V solidly grounded wye system fed by 75 kVA, 5 % Z, X/R 3 (infinite primary).
| Transformer let-through | IFL = 75 × 1000 / (√3 × 208) = 208.2 A; IT = 100 × IFL / (%Z) = 100 × 208.2 / 5.000 = 4,164 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 208² / 75,000 × 0.05 = 0.028843 Ω; split at X/R 3: RT = Z / √(1 + 3²) = 0.0091209 Ω, XT = 0.027363 Ω |
| Fault currents | I3φ = 4.164 kA; IL-G ≈ 1.25 × I3φ = 5.204 kA (solidly grounded wye estimate); IL-L ≈ 0.87 × I3φ = 3.622 kA; S = √3 × 208 × 4,164 = 1.500 MVA |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 5.204 kA = 10 kA → 5.204 vs limit 10.000 kA PASS |
| Result | 1/1 checks pass → PASS; I3φ 4.164 kA, highest 5.204 kA; minimum interrupting rating 10 kA; governing: minimum interrupting rating |
| Reference | Hand calculation: I_FL = 75 kVA / (√3 × 208 V) = 208.2 A; Z_T = 208² / 75 000 × 0.05 = 0.028843 Ω; I3φ = 208 / (√3 × 0.028843) = 4164 A (= 100 × 208.2 / 5); L-G estimate for a grounded wye 1.25 × 4164 = 5204 A (the calculator's stated multiplier); next standard rating ≥ 5.20 kA = 10 kA (validation case c0029-nec-75kva-208v-terminal) |
Example 2 — 75 kVA, 208 V plus 100 ft of 4/0 Cu in conduit
Feeder R + jX added in series to the transformer R + jX (X/R 3) and the resulting fault reduction at the downstream panel.
Given: 208 V solidly grounded wye system fed by 75 kVA, 5 % Z, X/R 3 (infinite primary); feeder 4/0 Cu in conduit, 30.48 m.
| Transformer let-through | IFL = 75 × 1000 / (√3 × 208) = 208.2 A; IT = 100 × IFL / (%Z) = 100 × 208.2 / 5.000 = 4,164 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 208² / 75,000 × 0.05 = 0.028843 Ω; split at X/R 3: RT = Z / √(1 + 3²) = 0.0091209 Ω, XT = 0.027363 Ω |
| Feeder impedance | 4/0 Cu, conduit: R = 0.0608 Ω/kft (NEC Ch. 9 Table 8, 75 °C), X = 0.0430 Ω/kft; L = 30.48 m = 100.00 ft → RC + jXC = (0.0608 + j0.0430) × 100.00/1000 = 0.006080 + j0.004300 Ω |
| Fault at the feeder end | Z = (RT + RC) + j(XT + XC) = 0.01520 + j0.03166 Ω, |Z| = 0.035122 Ω; I = V / (√3 |Z|) = 208 / (√3 × 0.035122) = 3,419 A (17.88 % below IT); X/R = 2.083 |
| Fault currents | I3φ = 3.419 kA; IL-G ≈ 1.25 × I3φ = 4.274 kA (solidly grounded wye estimate); IL-L ≈ 0.87 × I3φ = 2.975 kA; S = √3 × 208 × 3,419 = 1.232 MVA |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 4.274 kA = 10 kA → 4.274 vs limit 10.000 kA PASS |
| Result | 1/1 checks pass → PASS; I3φ 3.419 kA, highest 4.274 kA; minimum interrupting rating 10 kA; governing: minimum interrupting rating |
| Reference | Hand calculation (ohmic method, IEEE 141 ch. 4): Z_T = 0.028843 Ω at X/R 3 → R_T = 0.009121, X_T = 0.027363 Ω; 4/0 Cu: R = 0.0608 Ω/kft (NEC Ch. 9 Table 8, 75 °C), X = 0.043 Ω/kft (calculator conductor table, conduit) × 100 ft → 0.00608 + j0.00430 Ω; |Z| = |0.015201 + j0.031663| = 0.035122 Ω; I = 208 / (√3 × 0.035122) = 3419 A; reduction (4164 − 3419)/4164 = 17.88 %; X/R = 2.083; L-G 1.25 × 3419 = 4274 A → 10 kA (validation case c0029-nec-75kva-208v-4-0-feeder) |
Example 3 — 500 kVA, 480 V delta — terminal fault
An ungrounded delta secondary: the L-G estimate falls to 0.87 × and the three-phase current sets the rating.
Given: 480 V delta system fed by 500 kVA, 5 % Z, X/R 3 (infinite primary).
| Transformer let-through | IFL = 500 × 1000 / (√3 × 480) = 601.4 A; IT = 100 × IFL / (%Z) = 100 × 601.4 / 5.000 = 12,028 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 480² / 500,000 × 0.05 = 0.023040 Ω; split at X/R 3: RT = Z / √(1 + 3²) = 0.0072859 Ω, XT = 0.021858 Ω |
| Fault currents | I3φ = 12.028 kA; IL-G ≈ 0.87 × I3φ = 10.464 kA (delta estimate); IL-L ≈ 0.87 × I3φ = 10.464 kA; S = √3 × 480 × 12,028 = 10.000 MVA |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 12.028 kA = 14 kA → 12.028 vs limit 14.000 kA PASS |
| Result | 1/1 checks pass → PASS; I3φ 12.028 kA, highest 12.028 kA; minimum interrupting rating 14 kA; governing: minimum interrupting rating |
| Reference | Hand calculation: I_FL = 500 kVA / (√3 × 480 V) = 601.4 A; I3φ = I_FL × 100 / 5 = 12 028 A; I_LL = √3/2 × I3φ = 10 417 A; next standard rating ≥ 12.03 kA = 14 kA (85.9 %) (validation case c0029-nec-500kva-480v-terminal) |
Example 4 — Two 1000 kVA transformers in parallel at 208 V
Paralleled transformers halve the source impedance and double the fault current — here to the 100 kA class.
Given: 208 V delta system fed by 2 × 1,000 kVA, 5.75 % Z, X/R 3 (infinite primary).
| Transformer let-through | IFL = 1,000 × 1000 / (√3 × 208) = 2,775.7 A; IT = 100 × IFL / (%Z ÷ 2) = 100 × 2,775.7 / 2.875 = 96,547 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 ÷ 2 = 208² / 1,000,000 × 0.0575 ÷ 2 = 0.0012438 Ω; split at X/R 3: RT = Z / √(1 + 3²) = 0.00039334 Ω, XT = 0.0011800 Ω |
| Fault currents | I3φ = 96.547 kA; IL-G ≈ 0.87 × I3φ = 83.996 kA (delta estimate); IL-L ≈ 0.87 × I3φ = 83.996 kA; S = √3 × 208 × 96,547 = 34.783 MVA |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 96.547 kA = 100 kA → 96.547 vs limit 100.000 kA PASS |
| Result | 1/1 checks pass → PASS; I3φ 96.547 kA, highest 96.547 kA; minimum interrupting rating 100 kA; governing: minimum interrupting rating |
| Reference | Hand calculation: I_FL = 1000 kVA / (√3 × 208 V) = 2775.7 A per transformer; two in parallel halve %Z: I3φ = 2 × 2775.7 × 100 / 5.75 = 96 547 A; next standard rating 100 kA (validation case c0029-nec-2x1000kva-208v-parallel) |
Example 5 — 1000 kVA, 480 V, 2 × 500 kcmil Al in tray
Aluminium conductors in cable tray, two sets per phase, X/R 6; on a grounded wye the L-G estimate governs the rating.
Given: 480 V solidly grounded wye system fed by 1,000 kVA, 5.75 % Z, X/R 6 (infinite primary); feeder 2 × 500 Al in cable tray, 60 m; equipment interrupting rating 25 kA.
| Transformer let-through | IFL = 1,000 × 1000 / (√3 × 480) = 1,202.8 A; IT = 100 × IFL / (%Z) = 100 × 1,202.8 / 5.750 = 20,918 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 480² / 1,000,000 × 0.0575 = 0.013248 Ω; split at X/R 6: RT = Z / √(1 + 6²) = 0.0021780 Ω, XT = 0.013068 Ω |
| Feeder impedance | 500 Al, cable tray: R = 0.0424 Ω/kft (NEC Ch. 9 Table 8, 75 °C), X = 0.0350 Ω/kft; L = 60 m = 196.85 ft → RC + jXC = (0.0424 + j0.0350) × 196.85/1000 ÷ 2 sets = 0.004173 + j0.003445 Ω |
| Fault at the feeder end | Z = (RT + RC) + j(XT + XC) = 0.006351 + j0.01651 Ω, |Z| = 0.017692 Ω; I = V / (√3 |Z|) = 480 / (√3 × 0.017692) = 15,664 A (25.12 % below IT); X/R = 2.600 |
| Fault currents | I3φ = 15.664 kA; IL-G ≈ 1.25 × I3φ = 19.580 kA (solidly grounded wye estimate); IL-L ≈ 0.87 × I3φ = 13.628 kA; S = √3 × 480 × 15,664 = 13.023 MVA |
| Interrupting rating | NEC 110.9: highest fault current 19.580 kA ≤ equipment rating → 19.580 vs limit 25.000 kA PASS |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 19.580 kA = 22 kA → 19.580 vs limit 22.000 kA PASS |
| Result | 2/2 checks pass → PASS; I3φ 15.664 kA, highest 19.580 kA; minimum interrupting rating 22 kA; governing: interrupting rating |
| Reference | Hand calculation (ohmic method): Z_T = 480²/1 000 000 × 0.0575 = 0.013248 Ω at X/R 6 → 0.002178 + j0.013068 Ω; 500 kcmil Al: R = 0.0424 Ω/kft (NEC Ch. 9 Table 8), X = 0.035 Ω/kft (calculator table, cable tray) × 196.85 ft ÷ 2 sets = 0.004173 + j0.003445 Ω; |Z| = 0.017692 Ω; I3φ = 480 / (√3 × 0.017692) = 15 664 A (25.12 % below 20 918 A); L-G 1.25 × 15 664 = 19 580 A governs → 22 kA minimum; 19.58 / 25 kA = 78.3 % (validation case c0029-nec-1000kva-aluminium-tray-parallel) |
Example 6 — 750 kVA, 480 V with 400 A of running motors — 14 kA rating fails
Motor back-feed (FLA × 4) is added to the transformer let-through and pushes the fault past a 14 kA rating.
Given: 480 V delta system fed by 750 kVA, 5.75 % Z, X/R 3 (infinite primary); running motors 400 A FLA × 4; equipment interrupting rating 14 kA.
| Transformer let-through | IFL = 750 × 1000 / (√3 × 480) = 902.1 A; IT = 100 × IFL / (%Z) = 100 × 902.1 / 5.750 = 15,689 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 480² / 750,000 × 0.0575 = 0.017664 Ω; split at X/R 3: RT = Z / √(1 + 3²) = 0.0055858 Ω, XT = 0.016758 Ω |
| Motor contribution | IM = 400 A × 4 = 1,600 A → I = 17,289 A |
| Fault currents | I3φ = 17.289 kA; IL-G ≈ 0.87 × I3φ = 15.041 kA (delta estimate); IL-L ≈ 0.87 × I3φ = 15.041 kA; S = √3 × 480 × 17,289 = 14.374 MVA |
| Interrupting rating | NEC 110.9: highest fault current 17.289 kA ≤ equipment rating → 17.289 vs limit 14.000 kA FAIL |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 17.289 kA = 18 kA → 17.289 vs limit 18.000 kA PASS |
| Result | 1/2 checks pass → FAIL; I3φ 17.289 kA, highest 17.289 kA; minimum interrupting rating 18 kA; governing: interrupting rating |
| Reference | Hand calculation: I_FL = 750 kVA / (√3 × 480) = 902.1 A; transformer I = 902.1 × 100/5.75 = 15 689 A; motors 400 A × 4 = 1600 A; I3φ = 17 289 A > 14 kA (123.5 %) — fails NEC 110.9; next standard rating 18 kA (validation case c0029-nec-750kva-motor-contribution-fail) |
Example 7 — 1500 kVA, 480 V with a 10 kA source in parallel and a 1.1 safety factor
A second source at the fault point (e.g. a generator bus tie) adds its current, then the result is scaled by the safety factor.
Given: 480 V delta system fed by 1,500 kVA, 5.75 % Z, X/R 5 (infinite primary); a 10 kA source in parallel at the fault; safety factor 1.1; equipment interrupting rating 50 kA.
| Transformer let-through | IFL = 1,500 × 1000 / (√3 × 480) = 1,804.2 A; IT = 100 × IFL / (%Z) = 100 × 1,804.2 / 5.750 = 31,378 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 480² / 1,500,000 × 0.0575 = 0.0088320 Ω; split at X/R 5: RT = Z / √(1 + 5²) = 0.0017321 Ω, XT = 0.0086605 Ω |
| Parallel source | ZS = V / (√3 × IS) = 480 / (√3 × 10,000) = 0.027713 Ω; |Z| ∥ ZS = 0.0088320 × 0.027713 / (0.0088320 + 0.027713) = 0.0066975 Ω → I = 480 / (√3 × 0.0066975) = 41,378 A (the two currents add) |
| Safety factor | I = 41,378 A × 1.1 = 45,516 A |
| Fault currents | I3φ = 45.516 kA; IL-G ≈ 0.87 × I3φ = 39.598 kA (delta estimate); IL-L ≈ 0.87 × I3φ = 39.598 kA; S = √3 × 480 × 45,516 = 37.841 MVA |
| Interrupting rating | NEC 110.9: highest fault current 45.516 kA ≤ equipment rating → 45.516 vs limit 50.000 kA PASS |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 45.516 kA = 50 kA → 45.516 vs limit 50.000 kA PASS |
| Result | 2/2 checks pass → PASS; I3φ 45.516 kA, highest 45.516 kA; minimum interrupting rating 50 kA; governing: interrupting rating |
| Reference | Hand calculation: transformer 1804.2 × 100/5.75 = 31 378 A, |Z_T| = 480²/1 500 000 × 0.0575 = 0.008832 Ω; parallel source Z_S = 480 / (√3 × 10 000) = 0.027713 Ω; |Z| = 0.008832 × 0.027713 / (0.008832 + 0.027713) = 0.006698 Ω (magnitudes in parallel, so the currents add): I = 31 378 + 10 000 = 41 378 A × 1.1 = 45 516 A; S = √3 × 480 × 45 516 = 37.84 MVA; next standard rating 50 kA (91.0 %) (validation case c0029-nec-1500kva-parallel-source-safety-factor) |
Example 8 — 150 kVA, 600 V, 75 m of 2 AWG in open air
A long, small feeder at low X/R (1.5): the feeder resistance dominates and removes a third of the fault current.
Given: 600 V delta system fed by 150 kVA, 3 % Z, X/R 1.5 (infinite primary); feeder 2 Cu in open air, 75 m.
| Transformer let-through | IFL = 150 × 1000 / (√3 × 600) = 144.3 A; IT = 100 × IFL / (%Z) = 100 × 144.3 / 3.000 = 4,811 A |
|---|---|
| Transformer impedance | ZT = V² / (kVA × 1000) × %Z/100 = 600² / 150,000 × 0.03 = 0.072000 Ω; split at X/R 1.5: RT = Z / √(1 + 1.5²) = 0.039938 Ω, XT = 0.059908 Ω |
| Feeder impedance | 2 Cu, open air: R = 0.194 Ω/kft (NEC Ch. 9 Table 8, 75 °C), X = 0.0380 Ω/kft; L = 75 m = 246.06 ft → RC + jXC = (0.194 + j0.0380) × 246.06/1000 = 0.04774 + j0.009350 Ω |
| Fault at the feeder end | Z = (RT + RC) + j(XT + XC) = 0.08767 + j0.06926 Ω, |Z| = 0.11173 Ω; I = V / (√3 |Z|) = 600 / (√3 × 0.11173) = 3,100 A (35.56 % below IT); X/R = 0.790 |
| Fault currents | I3φ = 3.100 kA; IL-G ≈ 0.87 × I3φ = 2.697 kA (delta estimate); IL-L ≈ 0.87 × I3φ = 2.697 kA; S = √3 × 600 × 3,100 = 3.222 MVA |
| Minimum rating | Next standard rating (10, 14, 18, 22, 25, 35, 42, 50, 65, 100 … kA) at or above 3.100 kA = 10 kA → 3.100 vs limit 10.000 kA PASS |
| Result | 1/1 checks pass → PASS; I3φ 3.100 kA, highest 3.100 kA; minimum interrupting rating 10 kA; governing: minimum interrupting rating |
| Reference | Hand calculation: I_FL = 150 kVA / (√3 × 600) = 144.3 A; I_T = 144.3 × 100/3 = 4811 A; Z_T = 600²/150 000 × 0.03 = 0.072 Ω at X/R 1.5 → 0.039938 + j0.059908 Ω; 2 AWG Cu: R = 0.194 Ω/kft (NEC Ch. 9 Table 8), X = 0.038 Ω/kft (calculator table, open air) × 246.06 ft = 0.047736 + j0.009350 Ω; |Z| = |0.087674 + j0.069258| = 0.11173 Ω; I = 600 / (√3 × 0.11173) = 3100 A (35.56 % reduction); X/R = 0.790; next standard rating 10 kA (validation case c0029-nec-150kva-600v-2awg-open-air) |
Standards & References
- NEC 110.9 — Interrupting rating: equipment must have adequate short-circuit current rating
- NEC 110.10 — Circuit impedance, short-circuit current ratings, and other characteristics
- NEC 408.6 — Short-circuit current rating for switchboards, switchgear, and panelboards
- IEEE Std 141 (Red Book) — Recommended Practice for Electric Power Distribution for Industrial Plants
- IEEE Std 551 (Violet Book) — Recommended Practice for Calculating Short-Circuit Currents
Tips & Limitations
- This calculator uses the infinite bus assumption for transformer-only calculations, which gives a conservative (high) fault current estimate.
- For accurate results, include conductor impedance for feeder runs greater than 50 ft.
- The X/R ratio significantly affects the asymmetric (first-cycle) fault current. Typical values: 2–4 for distribution, 6–15 for large industrial systems.
- Motor contribution is typically 4–6× FLA and is significant when total motor load is a large fraction of the transformer rating.
- Always select breakers with AIC ratings equal to or greater than the calculated available fault current at that point.
- For parallel transformers, total fault current increases proportionally — ensure downstream equipment is rated accordingly.
Developer Guide — Architecture
This calculator follows the standard calcs component pattern:
components/c0029_nec_short_circuit/ ├── __init__.py # Docstring ├── apps.py # Django AppConfig ├── calculations.py # Pure Python — all math, no Django ├── component.py # Schema, Calculator class, builders, registration ├── equations.py # LaTeX equation templates └── templates/.../sidebar_form.html # Vue 3 sidebar form
Data flow: Sidebar form → NECShortCircuitCalculator.calculate()
→ calculate_short_circuit() (pure Python) → results dict
→ equation_builder() / summary_cards_builder() → UI rendering
Developer Guide — Code Paths
| Purpose | File | Key Function / Class |
|---|---|---|
| Fault current calc | calculations.py | calculate_short_circuit(data) |
| Validation | component.py | NECShortCircuitCalculator.validate_parameters() |
| Registration | component.py | register_component(registry) |
Developer Guide — Testing
# Run all tests for this component pytest calcs/components/c0029_nec_short_circuit/ -v
Test coverage includes:
- Transformer-only fault current calculation
- Utility contribution addition
- Conductor impedance reduction
- Motor contribution
- Multiple fault types
- Parallel transformer configurations