The formula
S = √(I² t) / k ; table rule: S ≤ 16 → S, 16–35 → 16, above 35 → S/2
- I
- prospective earth fault current in amperes
- t
- disconnection time of the protective device, in seconds
- k
- 143 for PVC-insulated copper, 95 for aluminium
- S
- conductor cross-section in mm²
Worked example
16 mm² copper phase conductor, 6 kA prospective earth fault, cleared in 0.4 s.
- Table rule: 16 mm² is at the boundary, so the protective conductor is also 16 mm²
- Adiabatic: S = 6000 × √0.4 / 143
- S = 6000 × 0.632 / 143 = 26.54 mm²
- The adiabatic value is larger, so it governs
35 mm² — the next standard size above 26.54 mm².
Where you'll use it
Sizing the protective conductor in any installation, and showing why a large fault current can demand more than the table suggests. This sizes the conductor, not the earth electrode or the pit — those are a separate design under IS 3043.