Part 4ProtectionCalculations

Circuit Design and Overload Protection: Revision Notes for the 18th Edition

IET Wiring Regulations Team ·

Circuit design is where the regulations stop being theoretical and start producing numbers. Cable size, device rating, voltage drop, disconnection time — get the design right and the circuit protects itself. Get it wrong and the cable cooks under load, or the protective device fails to clear a fault.

 

In the 18th Edition exam, circuit design and overload protection sit at the intersection of Part 4 (the rules) and Part 5 (the implementation), with Appendix 4 providing the data. Between them, these topics account for a substantial share of the paper — and they’re where calculation-based questions cluster.

 

These revision notes pull together the rules, the formulae, and the worked logic you need under exam conditions.

The Goal of Circuit Design

Every final circuit you design has to satisfy four things simultaneously:

 

RequirementDriven byReference
Safe under normal loadCable current-carrying capacity (Iz)Reg. 523
Protected against overloadCoordination of Ib, In, IzReg. 433
Protected against fault currentBreaking capacity, energy let-throughReg. 434
Compliant disconnection timeEarth fault loop impedance (Zs)Reg. 411.3
Acceptable voltage at the loadVoltage drop calculationReg. 525

 

Key point: A cable size that passes the current-carrying check can still fail on voltage drop, and a protective device that’s right for overload may not be fast enough for fault disconnection. You have to check all the criteria, not just the obvious one.

 

Overload vs Fault Current — Know the Difference

The regulations treat these as two separate problems:

 

Overload CurrentFault Current
CauseExcess load on a healthy circuitInsulation failure (L-L, L-N, L-E)
MagnitudeModest — typically 1.1 to 5 × InVery high — hundreds to thousands of amps
OnsetGradualInstantaneous
Governed byRegulation 433Regulation 434
Device part usedThermal element / inverse-time curveMagnetic element / instantaneous trip

 

A 32 A Type B MCB on a B6 curve will tolerate 3 to 5 × In before instantaneous trip — that’s 96 to 160 A. Below that, it relies on the thermal element to clear an overload over seconds or minutes.

 

The Overload Coordination Rule: Ib ≤ In ≤ Iz

This is the single most important inequality in Part 4. Regulation 433.1 states:

 

Ib ≤ In ≤ Iz

 

SymbolMeaning
IbDesign current of the circuit — the current the circuit is intended to carry
InRated (nominal) current of the protective device
IzCurrent-carrying capacity of the cable, after correction factors

 

Read in plain English: the cable must be able to carry the design current, and the protective device must be small enough that it trips before the cable is overloaded.

 

Exam tip: A common wrong answer pairs a 32 A MCB with a cable whose Iz, after derating, is only 27 A. Even though the cable’s tabulated capacity is 32 A or more, the corrected Iz is what counts. Always derate first.

 

The I2 ≤ 1.45 × Iz Check

The second condition in Regulation 433.1 is:

 

I2 ≤ 1.45 × Iz

 

Where I2 is the operating current of the device — the current at which it’s guaranteed to disconnect within conventional time.

 

Device TypeI2 ValueAutomatic?
BS EN 60898 MCB1.45 × InYes — automatically satisfied if Ib ≤ In ≤ Iz
BS EN 61009 RCBO1.45 × InYes
BS 88 HRC fuse1.6 × InYes — well within 1.45 × Iz when derated
BS 3036 rewireable fuse2 × InNo — apply a 0.725 factor to Iz

 

Remember: For modern devices (MCBs, RCBOs, HRC fuses) the I2 check is automatic — you only need to verify Ib ≤ In ≤ Iz. The exam likes to test the BS 3036 edge case because that’s the one where the 0.725 correction factor must be applied.

 

Applying Correction Factors to Iz

The tabulated current-carrying capacity in Appendix 4 (typically called It) is the capacity under reference conditions: 30 °C ambient, one circuit, no grouping, no thermal insulation. Real-world installations rarely meet those conditions, so we derate:

 

Iz = It × Ca × Cg × Ci × Cc × Cs

 

FactorWhat It Corrects ForWhere Found
CaAmbient temperature ≠ 30 °CTable 4B1
CgGrouping of circuitsTable 4C1
CiThermal insulation surrounding the cableReg. 523.9
CcBuried cables / BS 3036 fusesNote in 433.1.101
CsSoil thermal resistivity (buried only)Table 4B3

 

The procedure is also covered in detail in our cable size calculation guide — work through a couple of full examples and the steps stick.

 

Important: When multiple factors apply, multiply them all together. A 2.5 mm² T&E with It of 27 A, in an ambient of 40 °C (Ca = 0.87), grouped with two other circuits (Cg = 0.7), gives Iz = 27 × 0.87 × 0.7 = 16.4 A. That’s only suitable for a 16 A device — not the 20 A you might assume from the table.

 

Voltage Drop — Regulation 525

A cable that’s thermally adequate can still produce unacceptable voltage drop on a long run. Regulation 525 sets the limits:

 

CircuitMax Voltage Drop230 V Equivalent
Lighting3%6.9 V
All other circuits5%11.5 V

 

The calculation uses the mV/A/m values from Appendix 4:

 

Vd = (mV/A/m × Ib × L) ÷ 1000

 

For a 25 m run of 2.5 mm² T&E (mV/A/m = 18) carrying 20 A:

 

Vd = (18 × 20 × 25) ÷ 1000 = 9.0 V

 

That’s 3.9% on 230 V — fine for a power circuit, fail for a lighting circuit.

 

Exam tip: Voltage drop is the calculation most candidates forget to check after they’ve sized the cable. Look out for long-run scenarios — submains, garden offices, outbuildings, EV chargers — where the volt-drop check, not the thermal check, drives the cable size up. See our EV charger cable calculation guide for a fully worked example.

 

Fault Protection and Disconnection Times

Overcurrent protection handles overload; fault protection ensures the device disconnects fast enough to prevent dangerous touch voltages persisting under earth fault conditions.

 

Maximum Disconnection Times (Reg. 411.3.2)

SystemFinal Circuits ≤ 32 ADistribution / > 32 A
TN0.4 s5 s
TT0.2 s1 s

 

To meet these times automatically using overcurrent protection alone, Zs × Ia ≤ U0 must hold — i.e. the earth fault loop impedance must be low enough that fault current exceeds the device’s instantaneous trip threshold (Ia) at U0 = 230 V.

 

Table 41.3 gives the maximum permitted Zs values for each device type and rating. The full open-book technique for using this table — including the 80% rule for live measurements — is covered in Maximum Zs and the 80% Rule.

 

The CPC and the Adiabatic Equation

A circuit’s protective conductor (CPC) must be large enough to carry fault current for the disconnection time without exceeding its insulation’s permitted temperature rise. Two options:

 

MethodReference
Table 54.7 — minimum CPC based on line conductor sizeQuick, conservative
Adiabatic equation — S = √(I²t) / kReg. 543.1.3 — minimum CPC for the actual fault scenario

 

Line ConductorMinimum CPC (Table 54.7)
Up to 16 mm²Same size as line
Over 16 up to 35 mm²16 mm²
Over 35 mm²Half the line conductor

 

For tight cases — typically twin-and-earth where the CPC is smaller than the line conductor — use the adiabatic to confirm adequacy. The adiabatic equation explained post walks through the maths step by step.

 

When Overload Protection Can Be Omitted

Regulation 433.3 permits omission of overload protection in specific situations:

 

ScenarioExample
Cable not capable of overloadFixed-load resistive heater on its own dedicated circuit
Omission would not cause dangerSafety circuits where disconnection is worse than overload
Cable already protected upstreamTap-off from a main where the upstream device gives overload protection

 

Important: Even when overload protection is omitted, fault protection is still required. The two are governed by different regulations and addressed by different parts of the device’s characteristic.

 

Worked Example: A 32 A Radial Circuit

A 32 A radial socket circuit, 28 m long, in 2.5 mm² T&E clipped direct, ambient 30 °C, no grouping, no insulation.

 

StepCheckResult
1Ib = design current32 A (assumed full load)
2In = device rating32 A Type B MCB
3It = Appendix 4 capacity (Method C, 2.5 mm²)27 A
4Iz after correction (Ca = Cg = Ci = 1)27 A
5Check Ib ≤ In ≤ Iz: 32 ≤ 32 ≤ 27FAIL

 

2.5 mm² is not adequate at the full load — Iz is 27 A but In is 32 A. The circuit needs 4 mm² (It = 37 A, Iz = 37 A, passes). For a typical domestic radial, designers usually accept that Ib will be well below 32 A in practice, but the strict Regulation 433 check uses In, not Ib.

 

The ring final circuit is the classic workaround — two paths in parallel raise Iz, allowing 2.5 mm² to be protected by 32 A. The catch, of course, is that breaking the ring at one point removes a current path. See overload in a ring final circuit for the full analysis.

 

Common Exam Traps

TrapWhy It Catches People Out
Skipping the I2 check for BS 3036 fusesRewireable fuses have I2 = 2 × In — the 0.725 factor on Iz is mandatory
Comparing In to It instead of IzThe corrected capacity is what matters — apply Ca, Cg, Ci first
Forgetting voltage drop on long runsThe thermal check passes but the 3%/5% limit fails
Mixing up disconnection timesTT systems need 0.2 s for final circuits, not 0.4 s
Using Table 54.7 without checking adiabaticTable 54.7 is conservative — sometimes a smaller CPC is permitted, sometimes Table 54.7 itself isn’t enough
Reading 1.45 × Iz as “the cable’s safe maximum”It isn’t — Iz is the safe maximum. 1.45 × Iz is the operating threshold for the device, not a thermal limit for the cable

 

Practice and Further Study

Circuit design questions appear right across Parts 4, 5, and 6 of the exam, and the calculations build on each other. The best preparation is repetition — work the same sums in different forms until the procedure is automatic.

 

 

Our app includes 580+ practice questions with worked explanations referencing specific BS 7671 regulations, plus full timed mock exams that mirror the real 60-question paper. Calculation questions are flagged so you can drill them in isolation when revising.

 

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