Adiabatic equation calculator: minimum CPC size

The adiabatic equation, S = √(I²t) ÷ k, gives the smallest cross-sectional area of protective conductor that can carry a fault current I for the time t the device takes to clear it without damage to its insulation. k depends on the conductor and its insulation: 115 for a thermoplastic conductor inside a cable, 143 for thermosetting.

S = √(I²t) ÷ k, Regulation 543.1.3. Give it the fault current and how long the device takes to clear it, pick the insulation, and it gives the minimum copper and the standard size to use.

Regulation 543.1.3 of BS 7671, with k from the formula in Table 54.1 rather than from the tables. Checked against BS 7671:2018+A4:2026, updated 3 September 2026. Free, and nothing to sign in to.

The prospective fault current at the point the conductor protects. From the fault current tool, or the instrument.

Disconnection time, t
Conductor and insulation

The protective conductor of twin and earth, or a line conductor. Starts at 70 °C, may reach 160 °C.

Line conductor, for the Table 54.7 comparison (mm²)

Minimum protective conductor

4 mm²

2.75 mm² is the least that carries 1000 A for 0.1 s without damaging 70 °c thermoplastic, in a cable insulation; the next standard size is what you fit.

k for this conductor 70 °C to 160 °C, copper
115
Energy let through, I²t
100,000 A²s
Minimum area before rounding
2.75 mm²
Table 54.7 rule for a 6 mm² line Regulation 543.1.4, the alternative to calculating.
6 mm²

How does the adiabatic equation work, and where does k come from?

When a fault flows, the protective conductor heats until the device clears it. Regulation 543.1.3 sizes it so the heat does not damage the insulation: S = √(I² × t) ÷ k, where I is the fault current, t the seconds the device takes to clear it and k a constant for the conductor's material, insulation and the temperature it starts at. The equation holds for disconnection times up to 5 s.

Where k comes from

Not a table. BS 7671's Table 54.1, from IEC 60364-5-54, gives the formula: k = √(Qc(β+20)/ρ₂₀) × √ln((β+θf)/(β+θi)). The first root is copper alone: its heat capacity, its resistivity, its temperature coefficient. It comes to 226. The second is the insulation: the temperature the conductor starts at, θi, and the most it may reach for an instant, θf, 160 °C for thermoplastic and 250 °C for thermosetting. A conductor inside a cable starts at the cable's running temperature; one run on its own starts at ambient. Run those four cases through and you get 115, 143, 143 and 176, the published values, from first principles.

Or use the rule

Regulation 543.1.4 allows the protective conductor to be sized off the line conductor instead: the same size up to 16 mm², 16 mm² from 16 to 35, half the line above that. Twin and earth's protective conductor is smaller than that rule, which is why it is justified by the calculation, and why the tool shows both.

Worked example

A 6/2.5 mm² twin and earth shower circuit; the fault current at the shower is 1,000 A and the B40 clears it on its magnetic trip, taken as 0.1 s.

  1. Thermoplastic conductor inside a cable: 70 °C to 160 °C, k = 226 × √ln(394.5/304.5) = 115.
  2. I²t = 1,000² × 0.1 = 100,000 A²s.
  3. S = √100,000 ÷ 115 = 2.75 mm².
  4. The cable's 2.5 mm² cpc is a little under. The rule of Table 54.7 would ask for 6 mm²; the calculation says 4 mm² would do, so check the true clearing time or the device's I²t before condemning it. A B40 on 1,000 A clears in far less than 0.1 s.

Questions the trade asks

S = √(I² × t) ÷ k, from Regulation 543.1.3. S is the minimum cross-sectional area of the conductor in mm², I the fault current in amps, t the seconds the protective device takes to clear it, and k a constant for the conductor material and insulation. It says how much copper is needed to carry the fault without the insulation being damaged by the heat, and it holds for disconnection times up to 5 s.

Not from a table, but from the formula in Table 54.1 of BS 7671, which comes from IEC 60364-5-54: the heat capacity, resistivity and temperature coefficient of copper, and the temperature the conductor starts at and the most its insulation may reach. Run it for thermoplastic inside a cable (70 °C to 160 °C) and it gives 115; for thermosetting it gives 143; for a separate conductor starting at room temperature, 143 and 176. Those are the published figures, reached from first principles.

The time the device actually takes at that fault current, read off its time–current curve. For a circuit-breaker on its magnetic trip, 0.1 s is the usual conservative assumption and the true figure is often shorter. Better still is the device's let-through energy, I²t, from its manufacturer, which replaces I² × t directly.

Either. Regulation 543.1.4 lets you size the protective conductor off the line conductor instead: the same size up to 16 mm², 16 mm² from 16 to 35 mm², and half the line size above that. The protective conductor of twin and earth is smaller than that rule allows, which is why it is justified by the calculation.

No. Main protective bonding is sized by Regulation 544.1.1, not by the adiabatic equation: at least half the cross-sectional area of the earthing conductor and never less than 6 mm², with Table 54.8 making it 10 mm² on a typical PME supply.

The figures go straight onto the certificate

In Pascal the maximum Zs is filled in for every circuit as you pick the device, the schedule of test results carries all 32 columns, and the observation library suggests the wording, code and regulation as you type. EICs, EICRs and minor works, from £25 a month.

See the certificate software

A calculator applies a method to the numbers you give it. It does not know the installation in front of you, and it is no substitute for BS 7671, the guidance or the judgement of the person signing the certificate. Check anything you rely on.