ElectroHub

Short Circuit Fault Current Calculator

Short the terminals of a transformer and it delivers its own full-load current multiplied by 100/%Z — a 5 % transformer gives twenty times, for as long as it takes something upstream to clear. That number, not the load current, is what every breaker downstream has to be able to interrupt, and it is the one figure a switchboard cannot be designed without.

Prospective short-circuit current

A transformer multiplies its own full-load current by 100 / %Z when its terminals are shorted. That figure — not the load current — decides the breaking capacity every device downstream has to carry, and it is the one number a switchboard cannot be designed without.

Cable to the board (optional)

Full-load current
Fault level
At the terminals
At the board
Peak (κ = 1.75)
Breaking capacity
  • No upstream fault level was given, so the HV network is treated as an infinite bus. That is the standard conservative assumption: a real source can only reduce the current, never increase it.
Common trap: Percentage impedance is the whole calculation. A 5 % transformer gives 20× full-load current; a 4 % transformer of the same rating gives 25×. Reading the nameplate wrong by one percentage point moves the required breaking capacity by a whole standard size.
Three-phase symmetrical fault only. Peak factor by IEC 60909, κ = 1.02 + 0.98 e−3R/X, with the supply X/R taken as 10 unless you change it. A single-phase or earth fault has a different loop impedance and is not what this returns.

The formula

I_sc = I_fl × 100 / %Z ; MVA_sc = MVA_rating × 100 / %Z ; κ = 1.02 + 0.98 e^(−3R/X)

I_fl
transformer full-load current = kVA × 1000 / (√3 × V)
%Z
nameplate percentage impedance — 4 to 6 % for LV distribution
MVA_sc
fault level; sources in series combine like parallel resistors
κ
IEC 60909 peak factor, 1.0 to 2.0, set by R/X at the fault
i_p
peak asymmetrical current = κ √2 I_sc — the make duty

Worked example

A 1000 kVA transformer, 415 V secondary, 5 % impedance, on a strong supply.

  1. I_fl = 1 000 000 / (√3 × 415) = 1391 A
  2. Multiplier = 100 / 5 = 20
  3. I_sc = 1391 × 20 = 27 824 A
  4. Fault level = 1.0 MVA × 20 = 20 MVA

27.8 kA symmetrical — which needs a 36 kA breaking capacity, the next standard size up.

Where you'll use it

Specifying breaking capacity for a switchboard, checking that an existing board is not under-rated after a transformer upgrade, and the symmetrical-fault question in every power systems paper. A cable run between the transformer and the board lowers the current usefully — but it lengthens the disconnection time too, so both ends have to be checked.

The laws behind it

Parts this applies to

Resistor (fixed)Potentiometer / Rheostat / TrimmerResistor Array / NetworkThermistor (NTC / PTC)Varistor / MOVSupercapacitorLED (light-emitting diode)MOSFET

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