Cable Fault Withstand Size Calculator
Enter fault current, clearing time and conductor type to get the minimum conductor area and next standard size.
Step-by-step
- Enter valid values to begin.
Adiabatic method, valid for clearing times up to about 5 s. The k values are typical IEC 60364 / BS 7671 figures typed from memory: verify against IEC 60364-4-43 Table A.1 and BS 7671 Table 43.1 and the cable datasheet before relying on them. Use the protective device let-through I²t where known, not the prospective current. AWG matches are by circular mils at 1973.5 per mm².
Cable Fault Withstand: Will the Conductor Survive the Clearing Time?
Core Engineering Principles
A short circuit dumps energy into the conductor far faster than the insulation can shed heat, so for faults shorter than about five seconds we treat the heating as adiabatic: all of the I²t goes into raising the copper or aluminium temperature. Area therefore scales with current but only with the square root of time. The constant k bundles the heat capacity of the metal and the permitted rise from starting to final temperature. PVC tolerates roughly 160 °C and XLPE 250 °C, hence 143 for XLPE copper against 115 for PVC.
The number that sizes the cable is the energy the protective device lets through, not the prospective current on the drawing. A current-limiting breaker or fuse can clear a 50 kA fault in milliseconds and let through a fraction of the full-value I²t, so read it from the manufacturer’s let-through curve. Doubling the clearing time raises the required area by √2, about 41%, so a slow upstream breaker costs copper.
1 mm² = 1973.5 circular mils
NEC & Standard References
IEC 60364-4-43 requires devices to break a short circuit before conductors reach a damaging temperature and gives the adiabatic equation and k values. IEC 60364-5-54 applies the same method to protective conductors. IEC 60949 is the method for calculating thermally permissible short-circuit currents, including non-adiabatic effects. BS 7671 Section 543 and the Chapter 43 tables give the UK k factors. In North America, NEC 110.10 and 250.4 address the same concern. Verify k against your adopted edition and the cable datasheet.1. I²t = 10,000² × 0.5 = 50,000,000 A²s.
2. S = √50,000,000 / 143 = 7,071 / 143 = 49.45 mm², so the next standard size is 50 mm² (about 97,586 circular mils, nearest 1/0 AWG).
3. A 50 mm² cable withstands 143 × 50 / √0.5 = 10.1 kA for 0.5 s, or 0.51 s at exactly 10 kA, so it just passes.
4. A 35 mm² cable manages only 0.25 s at 10 kA and fails.
5. If the clearing time doubles to 1.0 s, S = 69.9 mm², so 70 mm².
- Prospective is not let-through. Using full prospective current with a limiting device oversizes the cable.
- Check the slowest clearing case. A far-end fault may be too low for the instantaneous element, so a slower element sets the time.
- Long-delay and selectivity settings. A selectivity delay of 0.4 s or more changes the answer completely.
- Bunching and joints. Hot or bunched cables start warmer, and lugs and joints have their own short-circuit rating.
- Long clearing times. Beyond about five seconds the adiabatic equation turns conservative; use IEC 60949.