Cable size calculation: current capacity and voltage drop (with examples)

Written by Engin Demirel · Electrical technician, master instructor · 12 years of experience

Short answer: A cable size is chosen to meet two conditions: it must carry the circuit current without overheating (current-carrying capacity), and the voltage drop along the run must stay below the limit. Single-phase voltage drop is %e = (2 × L × P × 100) ÷ (k × S × U²), three-phase %e = (L × P × 100) ÷ (k × S × U²), with L in metres, P in watts, S in mm², k = 56 for copper and U = 230 V single-phase or 400 V three-phase. IEC 60364-5-52 recommends about 3% for lighting and 5% for other uses from the supply origin; some countries are stricter (Turkey: 1.5% lighting, 3% power). Whichever condition needs the larger cable wins.

1. Current-carrying capacity

The cable must carry more than the circuit breaker rating. Common pairings in home wiring:

1.5 mm² → 10 A (lighting)
2.5 mm² → 16 A (sockets)
4 mm² → 20–25 A
6 mm² → 32 A
10 mm² → 40–50 A

Installation in conduit, grouping and hot surroundings reduce these values (correction factors); the exact figure comes from the cable manufacturer's tables and the installation method.

2. Voltage drop

On long runs, voltage drops even when the capacity is sufficient: lights dim, motors struggle and chargers reduce power or report faults.

Single-phase: %e = (2 × L × P × 100) ÷ (k × S × U²), U = 230 V
Three-phase: %e = (L × P × 100) ÷ (k × S × U²), U = 400 V
k = 56 for copper, 35 for aluminium.

Example 1: 7.4 kW single-phase charger, 25 m

P = 7,400 W, L = 25 m, S = 6 mm²: %e = (2 × 25 × 7,400 × 100) ÷ (56 × 6 × 230²) = 37,000,000 ÷ 17,774,400 ≈ 2.1%

Within a 3% limit. At 40 m the same circuit gives 3.3%; 10 mm² brings it down to 2%. For charging circuits, aim below 3% and lower where possible, because charging runs at full load for hours.

Example 2: 11 kW three-phase charger, 30 m

P = 11,000 W, L = 30 m, S = 4 mm²: %e = (30 × 11,000 × 100) ÷ (56 × 4 × 400²) = 33,000,000 ÷ 35,840,000 ≈ 0.9%

Three-phase carries the same power at lower current, so the drop is small; here current capacity (16 A → 2.5–4 mm²) decides the size.

Example 3: Garden lighting, 230 V, 60 m, 300 W

S = 1.5 mm²: %e = (2 × 60 × 300 × 100) ÷ (56 × 1.5 × 52,900) = 3,600,000 ÷ 4,443,600 ≈ 0.8% — fine. In 12–24 V low-voltage systems U is very small, so voltage drop grows fast; the same power needs much thicker cable or several drivers.

Points to watch

Length is the actual cable length from the consumer unit to the last device, including vertical runs.
Larger cables must fit the terminals; a big cable will not fit a small terminal.
Drop on the main supply cable (meter to consumer unit) adds to the total.
The calculation is a guide; the installation design and a qualified electrician make the final choice.

In Bursa, ENGN provides cable sizing and installation for EV charging circuits, electrical installations and outdoor lighting.

Frequently asked questions

How do I calculate cable size?

Check current-carrying capacity for the circuit current, then voltage drop for the run length; choose whichever needs the larger cable.

What is the voltage drop formula?

Single-phase: %e = (2 × L × P × 100) ÷ (k × S × U²); three-phase: %e = (L × P × 100) ÷ (k × S × U²). k = 56 for copper; U = 230 V or 400 V.

What voltage drop is allowed?

IEC 60364-5-52 recommends about 3% for lighting and 5% for other uses; national rules may be stricter (Turkey: 1.5% lighting, 3% power).

What cable size for a 7.4 kW EV charger?

At least 6 mm² for 32 A. At 25 m the drop is about 2.1%; from 40 m, 10 mm² is preferred.

What cable size for an 11 kW three-phase charger?

Usually 5 × 2.5–4 mm² for 3 × 16 A; at 30 m with 4 mm² the drop is about 0.9%.

Is aluminium cable different from copper?

Yes; aluminium conducts less (k = 35 vs 56 for copper), so it needs a larger cross-section for the same current and drop.