Short-circuit capacity and impedance — quick reference

Initial symmetrical short-circuit current at the fault point (IEC 60909-0): the rms current the network drives into a bolted three-phase fault, with the voltage source replaced by the c·Un/√3 equivalent.

Why it matters: It is the number every device rating is compared with — Icw, breaking capacity and the cable thermal check all read against Ik. Under-estimate Ik and the panel you quote will be destroyed on the first fault; over-estimate it and the client pays for switchgear he does not need.

Single-phase-to-earth short-circuit current (IEC 60909-0), calculated from the zero-sequence impedance Z0 in addition to Z1 and Z2.

Why it matters: It sizes the earth-fault protection and the rated withstand of the neutral / earthing path. Depending on the transformer vector group and the earthing arrangement Ik1 can be smaller or larger than the three-phase Ik, so it must never be assumed equal to it.

Peak short-circuit current ip = κ·√2·I″k (IEC 60909-0): the highest instantaneous value of the fault current, reached about half a cycle after inception.

Why it matters: It is the number that generates the electrodynamic force on busbars, cable cleats and device terminals. Compare it with the rated peak withstand Ipk — quoting only Ik leaves the mechanical strength of the switchboard unchecked.

Peak factor κ = ip / (√2·I″k): the ratio of the first peak to the rms value, calculated in IEC 60909-0 from the R/X ratio of the equivalent impedance.

Why it matters: It is what converts a short-circuit current into a mechanical force. An inductive network (LV cables, transformers) pushes κ towards 2.0 and raises the peak force, which is why ip cannot be derived from Ik alone.

Ratio of the equivalent reactance to the equivalent resistance seen from the fault point (IEC 60909-0).

Why it matters: It fixes two things at once: the peak factor κ (hence ip) and the decay rate of the DC component. A high X/R (inductive LV feeds, generator sources) gives both a higher peak and a longer DC transient, which the breaker must interrupt.

Short-circuit apparent power of the upstream network at the connection point, Ssc = √3·Un·I″k, or the equivalent source impedance behind it.

Why it matters: It is the single input that says how "stiff" the grid is. Without it no fault-current figure can be computed, so it must be requested from the utility (or taken from a stated assumption) before any switchgear can be quoted.

Voltage factor c for short-circuit calculation (IEC 60909-0 Table 1): the ratio between the voltage assumed at the fault location and the nominal voltage — c > 1 for maximum currents, c < 1 for minimum currents (for LV: 1.05 / 0.95; for HV: 1.10 / 1.00).

Why it matters: It is the explicit safety margin of the calculation. The maximum-current factor drives the device ratings you quote; the minimum-current factor drives protection sensitivity. Using the wrong one shifts Ik by several percent in a way that is invisible in the report unless the factor is printed.

Aperiodic (DC) component of the short-circuit current, i_dc = √2·I″k·e^(−ωt·R/X) (IEC 60909-0): the decaying offset that rides on top of the symmetrical component.

Why it matters: It is the extra current the breaker has contact to interrupt at the moment of contact parting, and it is why the required breaking capacity is higher than I″k for fast (generator-close) faults. In a quotation it appears as the rated breaking capacity at that % DC component.

Protection clearing time: the total time from fault inception to arc extinction — relay detection + intentional delay + breaker operating time.

Why it matters: In an arc-flash study the incident energy is almost proportional to the clearing time, and the PPE category is selected from it. Halving the clearing time usually halves the energy and can drop the required PPE by one step — which is a direct operating-cost item.

Short-circuit impedance voltage of a transformer: the primary voltage, in percent of rated voltage, that drives rated current through the short-circuited secondary winding (IEC 60076-1).

Why it matters: It fixes the LV fault level (approximately I″k ≈ In / uk) and at the same time the voltage drop under load. This is a real quotation trade-off: a larger uk lowers the fault level (cheaper switchgear downstream) but increases voltage drop and losses. Always quote uk together with the transformer rating.

Transformer vector group: the winding connections of the two sides plus the phase displacement between them, for example Dyn11 = delta HV, star LV with neutral brought out, 11 × 30° displacement (IEC 60076-1 connection symbols).

Why it matters: It decides the phase shift, whether a neutral is available for earth-fault protection, and where zero-sequence current can flow — the same kVA transformer with YNyn0 instead of Dyn11 gives a different single-phase fault current and needs a different protection scheme.