Chapter 54 Earthing and Bonding: Exam Questions and Answers Explained

Chapter 54 is one of the densest parts of BS 7671 — five tables, an equation, and a set of regulations that all deal with the same broad subject: earthing arrangements and protective conductors. It is also guaranteed exam material. Every 18th Edition paper contains Chapter 54 questions, and most candidates lose marks not because they don’t know the content but because they open the wrong table.

 

Here’s the point that changes how you should prepare: the exam is not testing how much of a 600-page book you can memorise. It is testing whether you can find the answer. Chapter 54 rewards that skill more than almost any other chapter, because nearly every question is answerable in seconds — once you’re on the right page.

 

This guide walks through each Chapter 54 table, explains what the exam question wording tells you about which table to use, and finishes with practice questions and worked answers.

Earthing Is Not Bonding

Before touching a single table, get this distinction straight. Examiners test it directly, and candidates who blur the two lose marks on definition questions in Part 2 as well as on Chapter 54 questions.

 

	Earthing	Bonding
Purpose	Provides a low-impedance path for fault current so the protective device disconnects	Equalises potential between metallic parts so no dangerous voltage appears between them
Mechanism	Enables automatic disconnection of supply (ADS)	Prevents potential differences developing
Protects against	Shock and equipment damage from earth faults	Shock from touching two conductive parts simultaneously
Typical conductor	Earthing conductor, circuit protective conductor (CPC)	Main protective bonding, supplementary bonding

 

Key point: Earthing is not bonding, and bonding is not earthing. They are two different methods of achieving electrical safety. Earthing gets the fault cleared; bonding makes sure nothing dangerous appears between metalwork while it is being cleared.

 

If you want the full background on how the two work together across TN-S, TN-C-S and TT systems, our guide to earthing and bonding for the 18th Edition exam covers the systems in detail.

 

Naming the Conductors Correctly

Exam questions use the precise BS 7671 name for each conductor, and each name points to a different table. Get the name wrong and you’ll size the wrong thing.

 

Conductor	Runs Between	Sized By
Earthing conductor	Main earthing terminal and the means of earthing (cut-out, earth electrode)	Table 54.1 (if buried) or Reg. 543.1 / adiabatic
Main protective bonding conductor	Main earthing terminal and incoming metallic services (water, gas, oil, structural steel)	Table 54.8 (TN-C-S) or Reg. 544.1.1
Circuit protective conductor (CPC)	Exposed-conductive-parts and the main earthing terminal	Table 54.7 or the adiabatic equation
Supplementary bonding conductor	Two conductive parts within a location (e.g. tap to radiator)	Reg. 544.2

 

One point about supplementary bonding that trips people up: it does not run all the way back to the consumer unit on its own. It links local conductive parts together, and where it connects to an exposed-conductive-part — the shower, a shaver socket, a socket outlet — it uses that circuit’s CPC as the return path. That’s by design, not a shortcut.

 

Table 54.1 — Buried Earthing Conductors

Page location aside, the trigger words in the exam question are “buried earthing conductor”. If you see those, go straight to Table 54.1. Nothing else in the book answers that question.

 

The table is a simple grid: find your row (protected or not protected against corrosion), find your column (protected or not protected against mechanical damage), read where they cross.

 

Condition	Mechanically Protected	Not Mechanically Protected
Protected against corrosion	2.5 mm² Cu / 10 mm² Fe	16 mm² Cu / 16 mm² Fe
Not protected against corrosion	25 mm² Cu / 50 mm² Fe	25 mm² Cu / 50 mm² Fe

 

Exam tip: Every clue you need is in the question. “Copper”, “not protected against corrosion”, “not protected against mechanical damage” — three clues, three coordinates, one answer. Read the question twice and pick out the conditions before you look at the table.

 

There’s a second technique worth learning here. Regulations that relate to a table usually sit next to that table. A question asking what an earthing conductor must comply with where PME conditions apply is answered by Reg. 542.3, which sits just above Table 54.1 and directs you to Reg. 544.1.1. When you’re already on the right page, scan up and down before flicking elsewhere.

 

The Adiabatic Equation

If there is one calculation certain to appear, it’s this one. “Adiabatic” is a thermodynamic term meaning no heat enters or leaves the system — for a conductor, it assumes every joule of fault energy stays in the conductor and heats it. That’s the worst case, which is exactly what we design for. In reality some heat escapes to the surroundings, so a conductor that passes the adiabatic check has margin in hand.

 

S = √(I²t) / k

 

Symbol	Meaning	Units
S	Minimum cross-sectional area	mm²
I	Fault current (RMS)	A
t	Operating time of the protective device	s
k	Factor for conductor material and insulation	— (Tables 54.2 to 54.6)

 

Important: Only the I²t term sits under the square root — not the whole expression. Dividing by k comes after the root. Getting this the wrong way round is the single most common calculation error in the exam.

 

Entering It on a Calculator

Work in this order: square I, multiply by t, take the square root, then divide by k. On most basic calculators that’s straight-line entry. Some models sequence differently, so check with the actual calculator you’ll take into the exam — it must be a standalone basic calculator, not a phone. If you’re at all unsure, break the sum into chunks and write out your workings; there’s no prize for doing it in one keystroke sequence.

 

Worked Example

What is the minimum CSA for a copper protective conductor if 750 A of fault current flows, the protective device operates in 0.4 s, and k = 115?

 

Step	Working	Result
1	I² = 750 × 750	562,500
2	× t = 562,500 × 0.4	225,000
3	√225,000	474.34
4	÷ k = 474.34 ÷ 115	4.12 mm²

 

The calculated minimum is 4.12 mm². You must now select a standard cable size equal to or greater than that figure — so 6 mm², not 4 mm².

 

Common trap: 4.12 is tantalisingly close to 4 mm², and plenty of candidates round down. Don’t. The regulations say equal to or greater than, and 4 mm² is less by 0.12 mm². In the exam, work to exactly what the book says.

 

For a fuller treatment including k value selection and where the equation sits within Reg. 543.1.3, see our guide to the adiabatic equation explained.

 

mm² or mm Squared?

A genuine point of confusion. For conductor CSA we say square millimetres — how many 1 mm squares would cover the cut end of the conductor. A 10 mm² conductor is one that ten 1 mm squares would cover: a circle just under 4 mm in diameter. “10 mm squared” would mean a 10 mm × 10 mm area — one hundred 1 mm squares, a circle nearly 12 mm across. You won’t be fitting that into an earth bar.

 

Table 54.7 — Tabulated CPC Sizes

There are two legitimate ways to size a CPC: calculate it with the adiabatic equation, or read it from Table 54.7. The exam tells you which it wants.

 

Exam tip: If the question asks for a tabulated value and gives you only the line conductor CSA, it wants Table 54.7. If it gives you a fault current, a disconnection time and a k value, it wants the adiabatic equation. The data in the question tells you the method.

 

In Table 54.7, S is the CSA of the line conductor. There are only three rows:

 

Line Conductor CSA (S)	Minimum CPC CSA	Example
S ≤ 16 mm²	S (same as line)	2.5 mm² line → 2.5 mm² CPC
16 mm² < S ≤ 35 mm²	16 mm²	25 mm² line → 16 mm² CPC
S > 35 mm²	S / 2	50 mm² line → 25 mm² CPC

 

The middle row has no choices in it — anything over 16 mm² and up to 35 mm² gets a 16 mm² CPC, full stop. That fixed value is worth remembering because it’s the row candidates most often misread.

 

Table 54.8 — Main Protective Bonding

Table 54.8 applies where the supply is TN-C-S (PME). In a PME supply the protective earth and neutral are combined in the incoming service cable — the PEN conductor — so the main bonding is sized in relation to the PEN conductor’s CSA.

 

PEN Conductor CSA	Minimum Main Bonding CSA (Cu)
≤ 35 mm²	10 mm²
> 35 mm² and ≤ 50 mm²	16 mm²
> 50 mm² and ≤ 95 mm²	25 mm²
> 95 mm² and ≤ 150 mm²	35 mm²
> 150 mm²	50 mm²

 

Worked example: a PEN conductor of 25 mm² — that’s less than 35 mm², so the top row applies, and the minimum main protective bonding conductor is 10 mm². Straightforward once you’re on the right table.

 

Remember: 10 mm² is the everyday domestic answer, which is why most electricians can recite it. Don’t let familiarity make you skip the table when the question gives a larger PEN — the exam does ask about 50 mm² and 95 mm² supplies. Note too that Reg. 544.1.1 requires the main bonding to be at least half the earthing conductor CSA, subject to a 6 mm² minimum, for non-PME supplies.

 

Supplementary Bonding Sizes

This is where Chapter 54 stops being a table lookup. Section 544.2 has no table — the sizes are written into the text of the regulations, which is exactly why candidates struggle to find them under time pressure.

 

In practice you’re choosing between two sizes, 2.5 mm² and 4 mm², and the deciding factors are what you’re connecting and whether the conductor has mechanical protection.

 

First, the two terms you must have clear:

 

Term	Definition	Example
Exposed-conductive-part	Conductive part of the installation, touchable, not normally live but may become live under fault	Metal casing of a fluorescent luminaire
Extraneous-conductive-part	Not part of the installation, but liable to introduce a potential (usually earth potential)	Central heating radiator, metal water pipe

 

Now the sizing logic:

 

Connecting	Mechanically Protected	Not Mechanically Protected	Regulation
Exposed to exposed	Not less than the smaller CPC	Minimum 4 mm²	Reg. 544.2.1
Exposed to extraneous	Half the CPC CSA of the exposed part	Minimum 4 mm²	Reg. 544.2.2
Extraneous to extraneous	2.5 mm² Cu	4 mm² Cu	Reg. 544.2.3

 

Worked example: a copper supplementary bonding conductor connects two extraneous-conductive-parts and has no mechanical protection. Find the extraneous-to-extraneous row, find the no-mechanical-protection column — 4 mm².

 

Important: You will not have a neat summary table in the exam. The information is in the wording of Regs 544.2.1 to 544.2.3. Practise reading those three regulations and extracting the number, because that’s the actual exam skill being tested here.

 

Supplementary bonding also has a “when is it required at all?” dimension, which matters for Part 7 questions on bathrooms. Our article on supplementary bonding and touch voltage covers the conditions under which it can be omitted.

 

Reading the Question to Find the Table

Everything above collapses into one habit: read the question for the words that identify the table. Here’s the mapping.

 

Question Contains	Go To	Because
”buried earthing conductor”	Table 54.1	Only table covering buried earthing conductors
Fault current + time + k value	Adiabatic equation (Reg. 543.1.3)	You’ve been given calculation data
”tabulated” + line conductor CSA only	Table 54.7	Tabulated CPC sizing from line conductor
PEN conductor CSA + main bonding	Table 54.8	PME main bonding sizing
Supplementary bonding + exposed/extraneous	Reg. 544.2	Text, not a table
”k value” for a material/insulation combination	Tables 54.2–54.6	k factor tables

 

Key point: The question is written to point you at one specific place in the book. Train yourself to spot the pointer before you start turning pages. That habit alone is worth several marks.

 

If you want to build this skill more broadly, using the contents page and index to find answers fast applies the same approach across the whole book.

 

Ten Practice Questions

Work through these before looking at the answers. Write your answers down — the act of committing to one matters.

 

1. A buried copper earthing conductor is protected against corrosion but not protected against mechanical damage. What is the minimum CSA?
2. Use the adiabatic equation with I = 380 A, t = 5 s, k = 143. What minimum CSA is required, and which standard size would you select?
3. What is the tabulated minimum CSA of a protective conductor where the associated line conductor is 100 mm²?
4. Which regulation must an earthing conductor comply with where PME conditions apply?
5. A copper supplementary bonding conductor connects an exposed-conductive-part to an extraneous-conductive-part and is not mechanically protected. What is the minimum CSA?
6. What size copper protective conductor is required if a fault current of 680 A may flow, disconnection time is 0.4 s, and k = 115?
7. Same data as Q6, but the disconnection time is 5 s. What minimum CSA is now required?
8. What is the minimum CSA of a main protective bonding conductor where the PEN conductor of the supply is 50 mm²?
9. What is the tabulated minimum CPC size for a 25 mm² line conductor?
10. Which of these is the adiabatic equation: (a) S = I²t / k, (b) S = √(I²t) / k, (c) S = √(I²t / k), (d) S = k√(I²t)?

 

Answers and Where to Find Them

 

Q	Answer	Where to Find It
1	16 mm²	Table 54.1 — protected against corrosion row, not mechanically protected column
2	√(380² × 5) ÷ 143 = √722,000 ÷ 143 = 849.7 ÷ 143 = 5.94 mm² → select 6 mm²	Reg. 543.1.3
3	50 mm² (S > 35 mm², so S/2)	Table 54.7
4	Reg. 544.1.1	Reg. 542.3, immediately above Table 54.1
5	4 mm²	Reg. 544.2.2 — no mechanical protection
6	√(680² × 0.4) ÷ 115 = √184,960 ÷ 115 = 430.07 ÷ 115 = 3.74 mm² → select 4 mm²	Reg. 543.1.3
7	√(680² × 5) ÷ 115 = √2,312,000 ÷ 115 = 1520.5 ÷ 115 = 13.22 mm² → select 16 mm²	Reg. 543.1.3
8	16 mm² (50 mm² PEN falls in the >35 and ≤50 band)	Table 54.8
9	16 mm² (S is above 16 and up to 35 mm²)	Table 54.7
10	(b) S = √(I²t) / k	Reg. 543.1.3

 

Compare questions 6 and 7 carefully. The only thing that changed was the disconnection time — 0.4 s to 5 s — and the required conductor jumped from 4 mm² to 16 mm². That’s a fourfold increase in CSA from a longer disconnection time, and it’s a favourite way for examiners to test whether you actually ran the calculation or just recognised the numbers.

 

Exam tip: With roughly 2 minutes per question, a Chapter 54 table lookup should take you 30 seconds and leave time in the bank. The adiabatic questions are the ones that eat the clock, so practise the calculator sequence until it’s automatic.

 

Practice and Further Study

Chapter 54 sits inside Part 5, the second-highest-weighted part of the exam. The tables reward repetition — the more times you’ve located Table 54.7 under pressure, the faster you’ll be on the day.

 

Test yourself on the parts where Chapter 54 material appears:

• Part 5 — Selection and Erection of Equipment quiz
• Part 4 — Protection for Safety quiz
• Part 6 — Inspection and Testing quiz

 

Our app includes 690+ practice questions across all 8 parts, each with an explanation citing the specific regulation or table, plus full timed mock tests that match the real exam’s weighted question distribution — so you get repeated practice at the Chapter 54 lookups under genuine time pressure.

 

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