How much current will actually flow into a dead short at the end of your cable run? From the
transformer nameplate and the conductor you ran, this gives the symmetrical RMS fault current,
the peak that busbars and contacts have to survive, and the fault level at
every point along the run. Fault current is not a constant — it collapses with distance, and
that is the whole reason this page exists.
Where the numbers come from
A transformer nameplate gives impedance as a percentage, which is a per-unit figure. Turn it
into ohms with the base impedance of the machine, then treat the transformer and the cable as
two impedances in series. Nothing else is needed for a first-pass answer.
Z_base = V² ÷ S (V in volts, S in volt-amperes)
Z_tx = (Z% ÷ 100) × Z_base
R_tx = Z_tx ÷ √(1 + (X/R)²) · X_tx = (X/R) × R_tx
R_c = ρ(T) × L ÷ A · X_c = x · L ÷ 1000
Z_tot = (R_tx + R_c) + j (X_tx + X_c)
Isc = V_LL ÷ (√3 × |Z_tot|) (three-phase)
Isc,LN = (V_LL ÷ √3) ÷ |Z_tx + 2 × Z_c| (return path doubles the cable term)
κ = 1.02 + 0.98 · e^(−3 R/X) · i_p = κ × √2 × Isc
A_min = Isc × √t ÷ k
The line-to-neutral case is the one people get wrong. The fault loop runs out along the phase
and back along the neutral, so the cable term counts twice — and if the neutral is
reduced, its own resistance and reactance have to be added separately rather than doubled.
That is why a reduced neutral can drop the L–N fault current far below the three-phase value,
which matters when a device has to clear a line-to-neutral fault at the far end of a run.
Worked example
1000 kVA, 400 V transformer, 6 % impedance, X/R = 5; 30 m of 50 mm² copper at 70 °C,
0.08 Ω/km reactance.
- Z_base = 400² ÷ 1,000,000 = 0.16 Ω → Z_tx = 0.06 × 0.16 =
9.60 mΩ
- Split at X/R = 5: R_tx = 1.88 mΩ, X_tx = 9.41 mΩ
- ρ(70 °C) = 0.017241 × (1 + 0.00393 × 50) = 0.02063 Ω·mm²/m → R_c = 0.02063 × 30 ÷ 50 =
12.38 mΩ; X_c = 0.08 × 30 ÷ 1000 = 2.40 mΩ
- Z_tot = 14.26 + j 11.81 mΩ → |Z| = 18.52 mΩ
- Isc = 400 ÷ (1.732 × 0.01852) = 12.47 kA — barely half the 24.1 kA the
transformer can deliver at its own terminals
- R/X = 1.21 → κ = 1.02 + 0.98 · e^(−3.62) = 1.046 → i_p =
18.4 kA
- L–N: cable term doubles → 26.64 + j 14.21 mΩ → |Z| = 30.19 mΩ →
(400 ÷ √3) ÷ 0.03019 = 7.65 kA
The lesson in one line: the same transformer that puts 24 kA into a fault at
its terminals puts 12.5 kA into one 30 metres away, and only 7.6 kA into a line-to-neutral
fault at that distance. A device rated 16 kA is fine at the end of that cable and hopeless
next to the transformer.
Limitations
- Transformer positive-sequence impedance only. No utility source
impedance, no motor contribution, no parallel transformers or generators. All of these
raise the real fault current, so treat the result as a floor, not a ceiling,
unless you are at the far end of a very long cable where the cable dominates anyway.
- The L–N figure uses the positive-sequence impedance for the return path as well. Real
zero-sequence impedance differs, especially with a reduced neutral or a delta-wye
transformer — the direction of the error depends on the earthing arrangement.
- Cable reactance is a single number you supply. Spacing, armouring and conductor shape all
move it; for long runs it can matter more than resistance.
- Conductor temperature is assumed uniform along the run. In practice the cable starts at
operating temperature and heats further during the fault, so the later part of a long fault
sees a little more resistance than this model uses.
- Nothing here selects a protective device. Breaking capacity,
discrimination, disconnection time and the applicable code's requirements are all outside
this page. k factors are offered as a convenience and must be confirmed against the code
that governs your installation.
How do I calculate fault current from a transformer nameplate?
Convert Z% to ohms with Z = (Z%/100) × V²/S, then Isc = V ÷ (√3 × Z) for a three-phase bolted fault. A 1000 kVA, 400 V unit at 6 % gives 9.6 mΩ and about 24.1 kA at its terminals. Add the cable impedance in series for a fault further away.
Why is fault current lower at the end of a long cable?
The cable impedance is in series with the transformer. 50 mm² copper is about 0.39 mΩ/m at 70 °C, so 30 m adds 12.4 mΩ of resistance — comparable to or larger than the transformer impedance itself in many LV systems. Distance, not the transformer, sets the fault level at the far end.
What is peak asymmetrical current?
The DC offset in the first loop makes the peak higher than √2 × RMS. IEC 60909 gives κ = 1.02 + 0.98·e^(−3R/X): about 1.75 at X/R = 10, about 2.4 as X/R approaches 1. Busbar bracing and contact ratings are quoted against this peak.
Is this a substitute for a short-circuit study?
No. It models one transformer and one cable. Source impedance, motor contribution, parallel sources and zero-sequence paths are all ignored. Use it for a first-pass device rating or to sanity-check a number, not to sign off a design.
Does this check the cable survives the fault?
It shows the adiabatic minimum, I·√t/k, as an informational figure only. k depends on material and insulation and is set by your code. Device selection, discrimination and disconnection times are not checked.