Sizing a
cable run
Every metre of cable steals a little voltage from the machine at the far end. You will rearrange formulae, work in standard form and use percentages to pick a conductor that keeps a lathe running and the wiring regulations satisfied.
The job
Read the spec like an engineerA new lathe is being installed at the far end of Ashfield's machine shop. It draws 20 A from a 230 V single-phase supply, and the cable has to run from the distribution board across the workshop.
- Volt drop is found from Vd = (mV/A/m × I × L) ÷ 1000
- Tabulated volt drop: 2.5 mm² = 18, 4 mm² = 11, 6 mm² = 7.3 mV/A/m
- Volt drop must not exceed 5% of 230 V — that is 11.5 V
- Cable price per metre: 2.5 mm² £2.10, 4 mm² £3.15, 6 mm² £4.40
- Budget for cable: £120
Your question: the run measures 35 m. Which conductor size is the smallest one that stays inside both the 5% limit and the budget?
Explore the drawing
Drag the machine, throw a switchPush the lathe further from the board and watch the volt drop bar climb towards the 5% line. Thicker cable pulls it back but costs more per metre. Find the smallest conductor that clears both rules, then prove it algebraically below.
Worked solution
Open one step at a timeTake the 35 m run in 4 mm² cable and check it properly.
The tabulated figure is in millivolts per amp per metre. It tells you how many thousandths of a volt you lose for each amp, for each metre of run.
The ÷ 1000 is doing one job only: converting millivolts into volts. Miss it and your answer is a thousand times too big.
Substituting into the formula earns the method mark on its own. Write the substitution line out even when the arithmetic is easy.
3.35% is comfortably inside the 5% ceiling. The lathe sees 230 − 7.70 = 222.3 V. Passes.
Suppose you wanted to use the cheaper 2.5 mm² cable. How far could you go before breaking the 5% rule?
The run is 35 m, so 2.5 mm² is ruled out — it fails by about three metres.
Multiply both sides by 1000 first. Trying to divide by the bracket while the ÷ 1000 is still floating about is where most of the errors happen.
Volt drop is really just Ohm's law. Resistance of a conductor is R = ρL ÷ A, with copper's resistivity ρ = 1.72 × 10−8 Ω m.
Why is 6.02 V smaller than the tabulated 7.70 V? Because the table assumes the cable is running hot, at 70 °C, and copper resists more when it is warm. Two correct answers, two different assumptions — being able to say why they differ is what separates the top grades.
4 mm² is the answer. 2.5 mm² is cheap but fails the volt drop rule; 6 mm² passes electrically but breaks the budget. The middle option is the only one that clears both.
Your turn
Answers checked instantlyA 2.5 mm² cable (18 mV/A/m) carries 20 A over a 28 m run. Work out the volt drop in volts, to 2 decimal places.
Express that volt drop as a percentage of the 230 V supply, to 1 decimal place.
A 6 mm² cable (7.3 mV/A/m) carries 32 A. Work out the longest run that keeps the volt drop at or below 11.5 V, to 1 decimal place.
Exam-style question
Full mark schemeA 4 mm² copper cable supplies a machine 45 m away and carries 24 A. The resistivity of copper is 1.72 × 10−8 Ω m and resistance is given by R = ρL ÷ A. Remember the current flows out along one conductor and back along the other.
Work out the volt drop, to 2 decimal places.
Mark scheme — 5 marks
- Area converted: 4 mm² = 4 × 10−6 m² M1
- Conductor length = 2 × 45 = 90 m M1
- R = (1.72×10−8 × 90) ÷ (4×10−6) = 1.548×10−6 ÷ 4×10−6 = 0.387 Ω A1
- Use of V = IR M1
- Vd = 24 × 0.387 = 9.29 V A1
Accepted answer: 9.29 V. Two marks are lost more often than any others here: forgetting to double the length, and converting mm² to m² by dividing by 1000 instead of 1 000 000. Area scales by the square of the length factor.
Produced for EAL engineering learners.
Maps to GCSE Mathematics (9–1) — rearranging formulae, standard form, and percentages.