Voltage Drop and Cable Resistance: Exam Practice Questions with Worked Answers

Two calculations come up in almost every 18th Edition paper and every inspection and testing assessment: cable resistance (R1+R2) and voltage drop. They look similar. Both involve a figure from a table, a cable length, and a division by 1000. That is exactly why candidates mix them up.

 

Get the wrong table or the wrong formula and every answer that follows is wrong. This guide sets the two calculations side by side, shows you which table to open for each, and then works through six exam-style questions, including the ones where you have to run the formula backwards.

 

If you want the wider picture on how these formulas fit with the adiabatic equation and rating factors, read our guide to the three essential formulas for the 18th Edition exam alongside this one.

Resistance vs Voltage Drop: Why They Are Not the Same

The two calculations answer two different questions about the same cable.

 

	R1+R2 (Resistance)	Voltage Drop
Conductors involved	Line and CPC	Line and neutral
Circuit condition	Fault between line and CPC	Healthy, fault-free, under normal load
What it tells you	How much fault current will flow, and therefore Zs	How many volts are lost before the load
Where to find the figures	Table B1 (GN3) or Table I1 (On-Site Guide)	Appendix 4 of BS 7671 (Table 4D5 for twin and earth)
Units in the table	Milliohms per metre (mΩ/m)	Millivolts per amp per metre (mV/A/m)
Compared against	Maximum Zs (Table 41.3 and the 80% rule)	Table 4Ab limits (3% lighting, 5% other)

 

Key point: R1+R2 is a fault calculation, so the CPC is in and the neutral is out. Voltage drop is a load calculation, so the neutral is in and the CPC is out. Decide which one the question is asking before you open a single table.

 

Calculating R1+R2: Formula and Tables

Resistance and R1+R2 anticipate what happens if a fault occurs between the line conductor and the CPC. The longer the cable, the higher the resistance, and the lower the fault current that will flow. That is why R1+R2 feeds directly into Zs and the disconnection time checks in Chapter 41.

 

The Formula

R1+R2 (Ω) = (mΩ/m × L) ÷ 1000

 

Symbol	Meaning
R1+R2	Combined resistance of line conductor and CPC, in ohms
mΩ/m	Combined resistance per metre from Table B1 or Table I1
L	Circuit length in metres (one way, from the board to the point of use)
÷ 1000	Converts milliohms to ohms

 

The Table

Table B1 in Guidance Note 3 (Table I1 in the On-Site Guide) gives the resistance of copper conductors at 20°C. The first entry tells you that a 1 mm² copper conductor has a resistance of 18.1 mΩ per metre. For a 1 mm² line conductor with a 1 mm² CPC, the combined figure doubles to 36.2 mΩ/m because the fault current travels out along one conductor and back along the other.

 

Line / CPC (mm²)	Combined Resistance (mΩ/m)
1.0 / 1.0	36.20
1.5 / 1.0	30.20
2.5 / 1.5	19.51
4.0 / 1.5	16.71
6.0 / 2.5	10.49
10.0 / 4.0	6.44

 

The table is headed “line conductor and protective conductor” but you can use it for any pair of conductors as long as you know their individual cross-sectional areas.

 

What Changes R1+R2

Change	Effect on R1+R2
Longer circuit	Increases
Smaller CSA	Increases
Higher conductor temperature	Increases
Shorter circuit or larger CSA	Decreases
Cooler conductors	Slight decrease

 

The temperature effect is the reason the 80% rule exists. Our guide to Maximum Zs and the 80% Rule explains how a cold R1+R2 measurement is corrected before you compare it with Table 41.3.

 

Calculating Voltage Drop: Formula and Tables

Voltage drop is what happens in a healthy circuit. Current flows out along the line and back along the neutral, and some voltage is lost across the resistance of those two conductors. A short circuit might lose 2 V, leaving 228 V at the lamp. A longer run might lose 4 V, leaving only 226 V at the point of use.

 

The Formula

Voltage drop (V) = (mV/A/m × Ib × L) ÷ 1000

 

Symbol	Meaning
mV/A/m	Millivolts per amp per metre from the Appendix 4 table for your cable type
Ib	Design current in amps
L	Circuit length in metres
÷ 1000	Converts millivolts to volts

 

The Tables

Two tables in Appendix 4 do the work. Table 4D5 covers flat profile thermoplastic cable with copper conductors, which is twin and earth. The left-hand columns list the cable sizes, and the voltage drop column gives the mV/A/m figure. Notice that the figure falls as the conductor size increases, because a bigger conductor has less resistance.

 

Conductor CSA (mm²)	Voltage Drop (mV/A/m)
1.0	44
1.5	29
2.5	18
4.0	11
6.0	7.3
10.0	4.4
16.0	2.8

 

Every cable type has its own table in Appendix 4, so check the question tells you what cable is in use. All six questions below use twin and earth.

 

Table 4Ab is what you compare the result against. For an installation supplied directly from a public low voltage network:

 

Circuit Type	Limit	At 230 V
Lighting	3%	6.9 V
All other circuits	5%	11.5 V

 

Remember: Regulation 525 points you to Appendix 4 for these limits. A calculated drop above the Table 4Ab figure is unacceptable, and the fix is a larger conductor, a shorter run, or a lower design current.

 

What Changes Voltage Drop

Change	Effect on Voltage Drop
Smaller CSA	Increases (worse)
Larger Ib	Increases (worse)
Longer circuit	Increases (worse)
Larger CSA, smaller Ib, or shorter run	Decreases (better)

 

Question 1: R1+R2 for a Lighting Circuit

A lighting circuit is installed using 1.5/1.0 mm² twin and earth cable. If the circuit is 25 m in length, what is R1+R2?

 

This is a resistance question, so the CPC is involved and you need Table B1 or Table I1, not Appendix 4.

 

Step	Working
Find the table figure	1.5 mm² line with 1.0 mm² CPC = 30.2 mΩ/m
Apply the formula	(30.2 × 25) ÷ 1000
Multiply	755 ÷ 1000
Answer	0.755 Ω

 

Exam tip: If it helps, sketch the circuit as a line out and a CPC back. It takes five seconds and it stops you reaching for the wrong table.

 

Question 2: Working Backwards to Find Circuit Length

A lighting circuit is wired in 1.5/1.0 mm² twin and earth cable. R1+R2 has been measured at 1.359 Ω. Calculate the circuit length.

 

Same cable, same table, but this time the length is the unknown. Rearrange the formula so that L becomes the subject:

 

L (m) = (R1+R2 × 1000) ÷ mΩ/m

 

Step	Working
Find the table figure	1.5 / 1.0 mm² = 30.2 mΩ/m
Convert R1+R2 to milliohms	1.359 Ω × 1000 = 1359 mΩ
Divide by the per-metre figure	1359 ÷ 30.2
Answer	45 m

 

Transposing formulas is a skill on its own, and the exam expects it. If moving terms across the equals sign still feels shaky, our guide to the transposition of electrical formulae walks through the method step by step.

 

Question 3: Voltage Drop in a Heater Circuit

A single-phase heater circuit has a design current (Ib) of 8 A and is installed using 2.5/1.5 mm² twin and earth cable with a length of 15 m. Calculate the voltage drop at full load.

 

Now we are in a fault-free circuit under load, so the neutral matters and the CPC does not. The 1.5 mm² CPC is a distraction. Open Table 4D5 and find 2.5 mm².

 

Step	Working
Find the table figure	2.5 mm² twin and earth = 18 mV/A/m
Apply the formula	(18 × 8 × 15) ÷ 1000
Multiply	2160 ÷ 1000
Answer	2.16 V

 

The question did not ask you to compare against a limit, but 2.16 V is comfortably inside the 11.5 V allowed for a non-lighting circuit.

 

Question 4: Lighting Voltage Drop Against the 3% Limit

A 230 V single-phase lighting circuit is supplied from a public low voltage supply and has a design current of 4 A. It is wired in 1.0/1.0 mm² twin and earth cable and has a length of 45 m. Calculate the voltage drop and compare it with Table 4Ab.

 

This is the classic two-part exam question. Calculate first, then judge.

 

Step	Working
Find the table figure	1.0 mm² twin and earth = 44 mV/A/m
Apply the formula	(44 × 4 × 45) ÷ 1000
Multiply	7920 ÷ 1000
Calculated drop	7.92 V
Table 4Ab limit for lighting	3% of 230 V = 6.9 V
Verdict	7.92 V exceeds 6.9 V. Not acceptable.

 

Important: You will not be asked to calculate distributed lighting loads along the circuit in the 18th Edition or inspection and testing exams. Treat the design current as flowing to the far end of the cable.

 

Question 5: Shower Circuit from Kilowatts

A 230 V single-phase 9 kW shower is installed using 10/4 mm² twin and earth cable with a circuit length of 20 m. What is the actual voltage drop?

 

The extra step here is that Ib is not given. You have to work it out from the power rating first, using I = P ÷ V.

 

Step	Working
Calculate Ib	9000 W ÷ 230 V = 39.13 A
Find the table figure	10 mm² twin and earth = 4.4 mV/A/m
Apply the formula	(4.4 × 39.13 × 20) ÷ 1000
Multiply	3443.4 ÷ 1000
Answer	3.44 V

 

Building a small data table before you touch the calculator is a habit worth forming. Write down what the formula needs, fill in what the question gives you, then calculate whatever is missing.

 

Question 6: The Multi-Step Question

A 240 V single-phase radial circuit in 2.5/1.5 mm² twin and earth cable supplies a 3 kW heater. The circuit R1+R2 has been measured at 0.38 Ω. What is the actual voltage drop for this circuit?

 

This one deliberately combines both calculations. Neither the length nor the design current is given, but the question contains enough to find both. Break it into steps.

 

Step	What to Do	Working
1	List what the voltage drop formula needs	mV/A/m, Ib, L
2	List what the question gives you	240 V, 3 kW, 2.5/1.5 mm², R1+R2 = 0.38 Ω
3	Find the length from R1+R2	Table B1: 2.5/1.5 mm² = 19.51 mΩ/m. L = (0.38 × 1000) ÷ 19.51 = 19.48 m
4	Find Ib from the power	3000 W ÷ 240 V = 12.5 A
5	Calculate the voltage drop	Table 4D5: 2.5 mm² = 18 mV/A/m. (18 × 12.5 × 19.48) ÷ 1000 = 4.38 V

 

Notice that Step 3 uses the resistance table and Step 5 uses the voltage drop table. Both tables in one question, each doing its own job.

 

Key point: The exam will always give you enough information to reach the answer. If a value you need is missing, it is because another value in the question lets you calculate it.

 

If the R1+R2 measurement itself is the part you are less sure about, our explanation of the continuity of protective conductors test covers how that 0.38 Ω figure is obtained on site.

 

Formula Summary and Exam Tips

Here are the formulas used in this guide, including the rearranged versions.

 

Purpose	Formula	Table
R1+R2 from length	R1+R2 = (mΩ/m × L) ÷ 1000	Table B1 (GN3) / Table I1 (OSG)
Length from R1+R2	L = (R1+R2 × 1000) ÷ mΩ/m	Table B1 (GN3) / Table I1 (OSG)
Voltage drop	Vd = (mV/A/m × Ib × L) ÷ 1000	Appendix 4, Table 4D5 for twin and earth
Design current from power	Ib = P ÷ V	Given in the question
Voltage drop limits	3% lighting, 5% other	Appendix 4, Table 4Ab

 

A few habits that separate a pass from a fail on these questions:

 

1. Identify the question type first. CPC mentioned or fault implied? Resistance. Load, design current, or “at full load”? Voltage drop.
2. Ignore the distractor conductor. The CPC size is irrelevant to voltage drop. The neutral is irrelevant to R1+R2.
3. Tab both tables. Table 4D5 and Table 4Ab in Appendix 4, plus Table B1 or I1 in your guidance book, should all be flagged before exam day.
4. Write a data table. Three symbols, three values, then calculate. It makes multi-step questions like Question 6 routine.
5. Check the units. Milliohms and millivolts both need the ÷ 1000. Forgetting it gives an answer a thousand times too big, and there is usually a distractor option waiting for exactly that mistake.

 

Exam tip: Practise a little every day rather than cramming in the week before. Write the formulas out by hand, invent your own numbers, and work them forwards and backwards until the method is automatic.

 

Practice and Further Study

Voltage drop sits in Part 5 (Regulation 525 and Appendix 4) and R1+R2 sits in Part 6 (continuity testing under Chapter 64), so both parts are worth revisiting:

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

 

Our app includes 690+ practice questions covering all 8 parts of BS 7671, with calculation questions on voltage drop, R1+R2, and Zs worked through step by step in the explanations. Topic-specific quizzes let you drill Part 5 and Part 6 until the tables feel familiar, and timed mock exams use the same weighted question distribution as the real 2382-26 paper so you can rehearse the two-minutes-per-question pace.

 

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