The Adiabatic Equation Explained: How to Check if Your CPC Is Big Enough

The adiabatic equation is one of those formulas that looks intimidating but is actually straightforward once you understand what it’s doing. It answers one simple question: is the circuit protective conductor (CPC) big enough to handle the energy released during an earth fault — without melting?

If the CPC is too small, the fault current will heat it beyond its safe limit before the protective device trips. The insulation could melt, the conductor could burn through, and the protective earth path would be lost — exactly when you need it most.

The Formula

The adiabatic equation

S = √(I²t) ÷ k

Where:

S = minimum cross-sectional area of the CPC in mm²
I = prospective earth fault current in amps
t = operating time of the protective device in seconds
k = a constant depending on the conductor material and insulation type

The equation calculates how much energy (I²t) the fault will dump into the CPC before the protective device disconnects, and whether the conductor can absorb that energy without exceeding its temperature limit.

When Do You Need It?

You use the adiabatic equation when:

The CPC is not part of the same cable as the line conductor (e.g., a separate earth wire run alongside the circuit cable)
You want to verify that the CPC within a standard T&E cable is adequate for the circuit
The exam asks you to prove that a protective conductor is correctly sized

For standard T&E cable, the manufacturer has already designed the CPC to work with the cable’s rated circuit — but you still need to be able to demonstrate the calculation in an exam, and there are situations (long runs, high fault currents) where the standard CPC may not be sufficient.

Understanding the Variables

Fault Current (I)

The fault current is calculated from:

I = Uo ÷ Zs

Where Uo is the nominal voltage (230V) and Zs is the earth fault loop impedance at the furthest point. The lower the impedance, the higher the fault current, and the more energy the CPC must handle.

Disconnection Time (t)

This is how long the protective device takes to trip at the calculated fault current. For an MCB, you read this from the time-current characteristic curve:

Type B MCB trips instantaneously (magnetically) at 3–5 × In
Type C MCB trips magnetically at 5–10 × In
Type D MCB trips magnetically at 10–20 × In

If the fault current is well above the magnetic trip threshold, t is typically 0.1 seconds or less. If it’s below, the MCB operates on its thermal element, which is slower.

The k Value

The k value represents how much energy per mm² the conductor material and insulation combination can absorb before reaching its maximum permissible temperature.

k values for different conductor and insulation combinations

For the exam, the most important k value to remember is 115 — copper conductor with PVC insulation (Table 54.4). This covers the vast majority of domestic installations using standard T&E cable.

Worked Example

Let’s work through a real scenario:

Circuit: 32A ring final circuit, Type B MCB, TN-C-S supply. The cable is 2.5 mm² T&E with a 1.5 mm² CPC. Measured Zs at the furthest socket is 0.62 Ω.

Worked example: is the CPC big enough?

Step-by-Step Calculation

Step 1 — Calculate the fault current:

I = Uo ÷ Zs = 230 ÷ 0.62 = 371A

Step 2 — Determine the disconnection time:

A 32A Type B MCB trips magnetically at 5 × 32 = 160A. Our fault current of 371A is well above this, so the MCB will trip on its magnetic element. From the manufacturer’s time-current curve, t ≈ 0.1 seconds.

Step 3 — Get the k value:

Copper conductor with PVC insulation → k = 115 (Table 54.4)

Step 4 — Apply the formula:

S = √(I²t) ÷ k = √(371² × 0.1) ÷ 115 = √(13,764) ÷ 115 = 117.3 ÷ 115 = 1.02 mm²

Step 5 — Compare with actual CPC:

Minimum CPC required = 1.02 mm². Actual CPC in the cable = 1.5 mm².

1.5 mm² ≥ 1.02 mm² — the CPC is adequate. ✓

What If the CPC Is Too Small?

If the calculation shows the minimum S is larger than the actual CPC, you have three options:

Increase the cable size — use a cable with a larger CPC
Reduce the fault current — this isn’t usually practical (it would mean increasing Zs)
Reduce the disconnection time — use a more sensitive protective device (e.g., lower rated MCB or an RCD)
Run a separate CPC — install an additional earth conductor alongside the cable

Quick Reference

Variable	What it is	Where to find it
I (fault current)	Uo ÷ Zs	Calculate from measured Zs
t (disconnection time)	Time for device to trip at current I	MCB time-current curves
k (material constant)	Conductor/insulation thermal capacity	BS 7671 Tables 54.2–54.6
S (minimum CPC)	Required CPC cross-sectional area	Calculate using S = √(I²t) ÷ k

Key Regulations

Reg. 543.1.1 — CPC sizing using Table 54.7 (simplified method)
Reg. 543.1.3 — CPC sizing using the adiabatic equation (calculation method)
Table 54.2 — k values for protective conductors not in a cable
Table 54.4 — k values for protective conductors as a core in a cable (most common)
Table 54.5 — k values for aluminium conductors
Reg. 434.5.2 — Adiabatic equation for line conductors (same formula, different application)

Practice and Further Study

The adiabatic equation appears in Part 5: Selection and Erection of Equipment of BS 7671. Test your knowledge:

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