Cable Size Calculations: The Six-Step Method for the 18th Edition Exam

Cable sizing is the calculation that separates candidates who understand BS 7671 from those who can only look things up in it. It appears in the 18th Edition exam, in Level 3 written assessments, and in every real design job you’ll ever do — and the method never changes.

 

The good news is that it’s a fixed procedure. Six steps, in the same order, every time. Get the order right and the numbers fall into place. Get it wrong — for example, picking a cable before applying the rating factors — and you’ll produce an undersized circuit that looks convincing on paper.

 

This guide walks the full method twice: once for a single-phase circuit and once for three-phase, using nothing but the regulations book.

The Six Steps at a Glance

Every cable sizing question — single-phase or three-phase, exam or real job — follows this sequence:

 

Step	What You Do	Formula / Source
1	Calculate the design current Ib	P ÷ U (1φ) or P ÷ (√3 × UL) (3φ)
2	Select the protective device rating In	Standard ratings — Table 41.3
3	Calculate the minimum It	It ≥ In ÷ (Ca × Cg × Ci × Cf)
4	Select the cable CSA	Appendix 4 tables (4D5, 4D4A, etc.)
5	Calculate the actual voltage drop	(mV/A/m × Ib × L) ÷ 1000
6	Confirm Ib ≤ In ≤ Iz and volt drop ≤ limit	Reg. 433.1.1 and Table 4Ab

 

Key point: Steps 1 to 4 size the cable for current. Step 5 checks it for voltage drop. These are two separate requirements and a cable must satisfy both — passing one does not excuse the other.

 

The Golden Rule: Ib ≤ In ≤ Iz

Regulation 433.1.1 is the foundation of the whole method. Learn what each symbol means and half the confusion disappears:

 

Symbol	Name	What It Actually Is
Ib	Design current	The current the load will actually draw — from the equipment rating or the designer’s calculation
In	Nominal rating of the protective device	What’s printed on the front of the MCB or fuse
Iz	Current-carrying capacity in service	What the cable can carry in its installed conditions, after derating
It	Tabulated current-carrying capacity	The figure straight out of the Appendix 4 table, before derating

 

The chain reads left to right: the load must not exceed the device rating, and the device rating must not exceed what the cable can safely carry. The cable must always be the electrically strongest part of the circuit — the protective device should operate long before the conductor reaches its limiting temperature.

 

Remember: Iz and It are not the same thing. It is the table value. Iz is It multiplied by all the rating factors. Confusing the two is the single most common source of wrong answers in cable sizing questions.

 

The second half of Regulation 433.1.1 — I2 ≤ 1.45 × Iz — is automatically satisfied for BS EN 60898 circuit breakers and BS EN 61009 RCBOs, because their I2 (conventional operating current) is 1.45 In by design. It matters for BS 3036 rewireable fuses, where the 0.725 factor has to be applied.

 

The Tables and Pages You Need

Cable sizing questions send you to a small, predictable set of tables. Tab these before the exam — they’ll cover almost every calculation you’re asked to do.

 

What You Need	Where to Find It
Standard device ratings and max Zs	Table 41.3 (Chapter 41)
Ambient temperature factor Ca	Table 4B1 (Appendix 4)
Grouping factor Cg	Table 4C1–4C5 (Appendix 4)
Thermal insulation factor Ci	Appendix 4, Section 2.6 (and Table 52.2)
BS 3036 fuse factor Cs	0.725 — Appendix 4
Maximum permitted voltage drop	Table 4Ab (Appendix 4)
Twin & earth capacity and mV/A/m	Table 4D5 (Appendix 4)
Multicore armoured (SWA) capacity	Table 4D4A (Appendix 4)
Multicore armoured mV/A/m	Table 4D4B (Appendix 4)
Reference methods (A, B, C, E, F, 100–103)	Table 4A2 (Appendix 4)

 

Exam tip: The single biggest time-waster in a calculation question is hunting for the right table. Knowing that twin and earth lives in 4D5 and armoured in 4D4A/4D4B turns a two-minute search into a five-second flick. Our guide to installation reference methods and correct cable sizes covers how to read the column headings correctly.

 

Worked Example 1: Single-Phase Water Heater

The question: A single-phase 230 V AC circuit supplies a 6 kW water heater, wired in 70°C thermoplastic insulated and sheathed flat cable with protective conductor (twin and earth). The copper cable is clipped direct throughout its length of 25 m. Ambient air temperature is 35°C, and the cable passes through a thermally insulating wall 100 mm thick. Protection is by a Type B BS EN 60898 circuit breaker of suitable rating.

 

Before touching a calculator, sketch the circuit and list what you know. It doesn’t need to be a work of art — a rough line with the length, the load and the ambient temperature written on it is enough. The act of writing it down stops you misreading the question, and it takes twenty seconds.

 

Step 1 — Calculate Ib

For a single-phase resistive load:

 

Ib = P ÷ U = 6000 ÷ 230 = 26.1 A

 

Step 2 — Select In

The device must be rated equal to or greater than Ib. From the standard ratings in Table 41.3, the Type B options either side of 26.1 A are 20 A and 32 A. A 20 A device would trip on normal load, so:

 

In = 32 A

 

Step 3 — Calculate the minimum It

Now apply the rating factors. Two apply here:

 

Factor	Condition	Value	Source
Ca	35°C ambient, 70°C thermoplastic	0.94	Table 4B1
Ci	100 mm thermal insulation	0.78	Appendix 4, Section 2.6

 

It ≥ In ÷ (Ca × Ci) = 32 ÷ (0.94 × 0.78) = 32 ÷ 0.7332 = 43.65 A

 

Important: You divide by the factors, you don’t multiply. Rating factors always make the required tabulated capacity larger, because hostile conditions mean the cable can carry less than the table suggests. If your answer comes out smaller than In, you’ve divided the wrong way round.

 

Step 4 — Select the cable size

Go to Table 4D5 (twin and earth). Read down the clipped direct column — reference method C — for the first value equal to or greater than 43.65 A. That’s 47 A. Traverse left to the conductor size:

 

6 mm² copper, It = 47 A

 

Key point: The load hasn’t changed. Ib is still 26.1 A and the heater still draws 26.1 A. What changed is the conditions — hot ambient air and thermal insulation — so we need a bigger conductor to tolerate them without overheating.

 

Step 5 — Calculate the voltage drop

First the limit. Table 4Ab gives 5% for other than lighting on a public low-voltage supply. A water heater is “other uses”:

 

Maximum permitted = 230 × 0.05 = 11.5 V

 

Now the actual drop. The mV/A/m figure for 6 mm² twin and earth is in the right-hand column of Table 4D5: 7.3 mV/A/m.

 

Vd = (mV/A/m × Ib × L) ÷ 1000 = (7.3 × 26.1 × 25) ÷ 1000 = 4.76 V

 

Note that the volt drop calculation uses Ib, the actual design current — not In and not It. This trips up a surprising number of candidates.

 

Step 6 — Confirm everything

 

Check	Values	Result
Ib ≤ In ≤ Iz	26.1 ≤ 32 ≤ 47	✓
Vd ≤ maximum	4.76 V ≤ 11.5 V	✓

 

Answer: 6 mm² twin and earth. Ib = 26.1 A, In = 32 A, It = 47 A, actual volt drop = 4.76 V.

 

Worked Example 2: Three-Phase Pump

Three-phase sizing follows exactly the same six steps. Only three things change: the Ib formula gains a √3, you use different tables, and the voltage drop limit is based on 400 V.

 

The question: A new supply to a three-phase 9 kW, 400 V AC pump is to be installed using 70°C thermoplastic multicore armoured cable (SWA), clipped direct throughout its length of 55 m. The cable passes through a 200 mm thick thermally insulating wall. Ambient temperature is 40°C. A suitable Type C BS EN 60898 circuit breaker is to be selected.

 

Step 1 — Calculate Ib

 

Ib = P ÷ (√3 × UL) = 9000 ÷ (1.732 × 400) = 9000 ÷ 692.8 = 12.99 A ≈ 13 A

 

Step 2 — Select In

From Table 41.3, the smallest Type C rating at or above 13 A:

 

In = 16 A

 

Step 3 — Calculate the minimum It

 

Factor	Condition	Value	Source
Ca	40°C ambient, 70°C thermoplastic	0.87	Table 4B1
Ci	200 mm thermal insulation	0.63	Appendix 4, Section 2.6

 

It ≥ 16 ÷ (0.87 × 0.63) = 16 ÷ 0.5481 = 29.19 A

 

Step 4 — Select the cable size

Armoured cable means Table 4D4A. Use the reference method C (clipped direct) column for three or four loaded conductors — reading the single-phase column here is a classic error. The first value at or above 29.19 A is 33 A:

 

4 mm² SWA, It = 33 A

 

Step 5 — Calculate the voltage drop

Table 4Ab, 5% of 400 V:

 

Maximum permitted = 400 × 0.05 = 20 V

 

For armoured cable, mV/A/m values live in Table 4D4B. Take the three-phase column for 4 mm²: 9.5 mV/A/m.

 

Vd = (9.5 × 13 × 55) ÷ 1000 = 6.79 V

 

Step 6 — Confirm everything

 

Check	Values	Result
Ib ≤ In ≤ Iz	13 ≤ 16 ≤ 33	✓
Vd ≤ maximum	6.79 V ≤ 20 V	✓

 

Answer: 4 mm² SWA. Ib = 13 A, In = 16 A, It = 33 A, actual volt drop = 6.79 V.

 

Exam tip: In a three-phase question, always check which mV/A/m column you’re in. Tables give separate figures for two loaded conductors (single-phase) and three or four loaded conductors (three-phase) — and they are not the same number. Picking the wrong one gives a plausible but wrong answer, which is exactly what a good distractor looks like.

 

When the Voltage Drop Fails

The interesting questions are the ones where step 5 fails. A long circuit with a small load is the classic setup: the current-carrying capacity check passes comfortably, but the volt drop over the distance blows the limit.

 

Take a single-phase armoured circuit needing an It of just over 11 A. The table points you at 1.5 mm² — perfectly adequate for current. But at 80 m the calculated volt drop comes out at 18.17 V, against a 230 V limit of 11.5 V. That circuit is non-compliant.

 

The Fix

Increase the conductor CSA and recalculate. More copper means less resistance, which means fewer volts lost:

 

CSA	mV/A/m	Volt Drop at 80 m	Verdict
1.5 mm²	Higher	18.17 V	Fails (> 11.5 V)
2.5 mm²	Lower	11.275 V	Passes (< 11.5 V)

 

Remember: You only need to repeat step 5. The bigger cable automatically satisfies the current-carrying capacity requirement, because a larger CSA always has a higher It. Going up in size can never break the Ib ≤ In ≤ Iz check.

 

The relationship to internalise: voltage drop is inversely proportional to conductor cross-sectional area. Double the CSA and you roughly halve the volt drop. Double the length and you double the volt drop. Once you feel that relationship, you can estimate the answer before you calculate it — a useful sanity check under exam pressure.

 

If you want more practice with rearranging these expressions, our guide to transposition of electrical formulae covers the algebra that underpins every calculation on this page.

 

Common Mistakes That Cost Marks

 

Mistake	Why It’s Wrong
Multiplying by the rating factors instead of dividing	Rating factors reduce capacity, so the required tabulated value must go up, not down
Using In or It in the volt drop formula	Voltage drop depends on the actual current flowing — always Ib
Forgetting to divide the mV result by 1000	The tables give millivolts per amp per metre. Skip the ÷1000 and your answer is a thousand times too big
Reading the wrong column	Single-phase vs three-phase, and clipped direct vs in conduit, give completely different figures
Rounding the device down	In must be ≥ Ib. Choosing 20 A for a 26.1 A load means nuisance tripping and a wrong answer
Stopping at step 4	A cable that passes on current can still fail on volt drop. Always finish the check
Applying Ci when the cable only touches insulation	Full 0.5 derating applies for cables totally surrounded for more than 0.5 m; shorter runs and one-side contact use different figures
Missing the grouping factor Cg	If the question mentions other circuits in the same trunking or on the same tray, Cg applies too

 

Exam tip: Multiple-choice distractors in calculation questions are built from these exact errors. If your answer matches one of the options, that’s no guarantee it’s right — the examiner has usually calculated the wrong-method answers deliberately.

 

Exam Technique for Calculation Questions

With 2 minutes per question, you can’t afford to derive the method from first principles each time. Three habits make the difference:

 

1. Write the six steps down first. Before you calculate anything, list steps 1–6 on your scrap paper. It stops you skipping the rating factors when the pressure is on.
2. Extract the data before you start. Power, voltage, length, ambient temperature, installation method, insulation thickness, device type. If a number is given in the question, it’s given for a reason — an unused figure usually means a missed step.
3. Sanity check the magnitude. A 6 kW single-phase load draws roughly 26 A. If your Ib comes out at 2.6 A or 260 A, you’ve slipped a decimal point.

 

Not every exam question asks for the full six steps. Many test a single stage — “what is the minimum It?” or “which rating factor applies?” — so make sure you can execute each step in isolation as well as in sequence. For a broader look at how calculation questions are worded and marked, see our guide to 18th Edition exam questions and question types.

 

Bottom line: The method never changes. Ib, then In, then It, then the cable, then the volt drop, then the check. Practise it until you can run the sequence without thinking, and cable sizing goes from being the hardest topic on the paper to a source of reliable marks.

 

Practice and Further Study

Cable sizing draws on Part 4 (protection against overcurrent), Part 5 (selection and erection), and Appendix 4. Test yourself across all three:

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

 

Our app includes 690+ practice questions covering all 8 parts with detailed explanations that reference specific regulation numbers and table numbers, plus full timed mock tests using the same weighted question distribution as the real exam — so you can practise calculation questions under the same 2-minute-per-question pressure you’ll face on the day.

 

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