How to correctly select cable cross-sections for electrical circuits

The cross-section is selected based on the design current of the line: the current is calculated using the load power (for a 230 V network – I = P / (U × cosφ)), then the next standard cross-section is selected from the tables in HD 60364-5-52 where the permissible current is higher than the design current – applying corrections for the installation method, the number of conductors in a bundle, and the ambient temperature. Typical "copper cross-section – MCB" pairs are: 1.5 mm² – 10–16 A, 2.5 mm² – 16–20 A, 4 mm² – 25 A, 6 mm² – 32 A, 10 mm² – 50 A. You can calculate your specific case using the online cross-section and voltage drop calculators provided directly in this article.

Selecting by current is only the first step. Long lines must additionally be checked for voltage drop (not exceeding 5% at the consumer, or 3% for lighting), and the cross-section must be checked for thermal withstand during a short circuit. For long extension leads and temporary construction networks, voltage drop is most often the reason for choosing a cross-section one step larger than the current alone would require.

Why cross-section matters: conductor heating

The conductor cross-section determines how much the wire heats up under load, which in turn affects the fire safety of the line and energy losses. When current flows through a conductor, heat is generated; this process is described by Joule's Law:

Where: Q is the amount of heat released (J); I is the current flowing through the conductor (A); R is the electrical resistance of the conductor (Ω); t is the duration of the current flow (s). The resistance of the conductor, in turn, depends on its geometric parameters and the material of the cores:

Where: ρ is the electrical resistivity of the material (for copper at 20°C ρ ≈ 0.0172 Ω·mm²/m, for aluminium ρ ≈ 0.028 Ω·mm²/m); L is the length of the conductor (m); S is the cross-sectional area of the core (mm²). Together, these two formulas provide a simple relationship: the smaller the cross-section S, the higher the resistance R, and the heat increases quadratically with the increase in current. If the conductor temperature exceeds the maximum permissible values for the insulation material (typically 70°C for PVC and 90°C for XLPE), thermal degradation of the dielectric will occur, leading to a short circuit and fire.

Step 1. Calculating current from power

The process begins not with tables, but with the design current of the line. The input data is the rated power of the connected equipment.

Calculation for a single-phase network (230 V):

The formula for determining the current for a single-phase load is:

Example: A 4.5 kW (4500 W) heater on a 230 V network with cosφ = 1. The current is:

Calculation for a three-phase network (400 V):

The formula for a balanced three-phase load is:

Example: An industrial distribution board with a total load of 22 kW (22,000 W) at cosφ = 0.85 on a 400 V network. The current is:

Step 2. Current carrying capacity tables

Once the design current is calculated, refer to the current carrying capacity standards (such as HD 60364-5-52). These standards define the continuous current ratings for various conductor materials, insulation types, and installation methods (e.g., in air, in ground, in conduits, installed singly or in groups).

According to the standard tables for copper conductors with PVC or rubber insulation, each cross-section corresponds to a specific allowable current. For the single-phase example with a current of 19.57 A: a 1.5 mm² conductor only allows 17.5 A when installed in a conduit or 21 A when installed in free air. To account for thermal rise and potential grouping of cables, a 2.5 mm² cross-section is selected (allowing up to 27 A in free air).

For the three-phase example with a current of 37.36 A: according to the standard tables for copper cables, a 4 mm² three-core cable installed in air can handle up to 32 A (installation in a cable trunking is typically treated as installation in air). This is insufficient. The next standard cross-section, 6 mm², provides an allowable current of 41 A, which covers the calculated requirement of 37.36 A. In accordance with HD 60364, the use of aluminium conductors in internal residential distribution networks is generally restricted to larger cross-sections (typically 16 mm² or greater), which is why copper is used for smaller cross-sections.

Copper conductor cross-section (mm²)	Allowable current (A) in a single conduit (single-phase, 2 cores)	Allowable current (A) installed openly in air	Maximum load power (kW) at 230V (openly, cosφ=1)	Maximum load power (kW) at 400V (openly, cosφ=0.85)
1.5	19	23	5.3	13.5
2.5	27	30	6.9	17.7
4.0	38	41	9.4	24.1
6.0	46	50	11.5	29.4
10.0	70	80	18.4	47.1
16.0	85	100	23.0	58.9

Step 3. Correction factors for installation conditions

Using standard current-carrying capacity values from tables without applying correction factors is a serious error. The continuous allowable current decreases if cables are installed in groups rather than individually, or if the ambient temperature differs from the reference temperature (+25°C for air).

When multiple cables are installed together in the same tray or trunking, heat dissipation conditions are reduced. According to HD 60364-5-52, when more than four loaded conductors are placed in a conduit, trunking, or bundle, the current-carrying capacity must be adjusted using reduction factors (e.g., 0.68 for 5–6 conductors, 0.63 for 7–9, and 0.6 for 10–12). If the base current for a 4 mm² cross-section in open air is 41 A, then in a bundle of 5 cables, it would be:

Similarly, if the ambient air temperature at the site reaches +40°C during the summer, a temperature correction factor must be applied. Standard current-carrying capacity tables are typically based on a conductor temperature of +65°C at a reference ambient temperature of +25°C; for an ambient temperature of +40°C, the correction factor is 0.79. This coefficient is determined by the relationship between the rated conductor temperature and the design ambient temperature, rather than the insulation material alone. Failure to apply these factors leads to accelerated polymer ageing and the degradation of the protective sheath.

Step 4. Checking Voltage Drop

The cross-section selected based on current-carrying capacity and power must also be verified for voltage drop at the end of the line. This is particularly critical for temporary construction networks, where the cable length from the distribution unit to the final tool can reach 100–150 metres. The voltage drop must not exceed 5% of the nominal network voltage (in accordance with HD 60364-5-52); furthermore, the voltage deviation at the consumer must not exceed ±10% according to standard voltage quality requirements.

The calculation for linear voltage drop (ΔU) in volts for a single-phase circuit is performed using the following formula:

Example: A current of 16 A (I) flows through a copper cable with a cross-section of 2.5 mm² (S) and a length of 80 metres (L). The voltage drop will be:

As a percentage of a 230 V network: (17.61 / 230) × 100% = 7.66%, which significantly exceeds the 5% limit. Operating power tools at such a voltage (approx. 212.4 V instead of 230 V) will lead to a drop in torque, excessive heating of the stator windings, and potential equipment failure. A 4 mm² cross-section does not solve the problem: ΔU = 44.032 / 4 = 11.01 V, which is 4.8%—right on the limit. A 6 mm² cross-section meets the requirement: ΔU = 44.032 / 6 = 7.33 V (3.2%). For lighting circuits, the limit is stricter—3% (per HD 60364-5-52).

You can calculate the voltage drop for your specific line using the calculator below.

Step 5. Checking Short-Circuit Current Rating

When designing cable routes and selecting conductor cross-sections, calculating based on the continuous current-carrying capacity (steady-state heating) is only the first, fundamental step. Engineering practice, based on the requirements of the HD 60364 series, requires that the selected cross-section must also be verified for thermal stability under short-circuit conditions. A short circuit is characterised by an instantaneous, step-like surge in current reaching thousands of amperes. At such levels, heat is generated within the copper or aluminium conductor so rapidly that heat dissipation to the surrounding environment can be entirely neglected. This process occurs adiabatically (without heat exchange).

The purpose of this verification is to ensure that the conductor does not melt and the insulation does not ignite during the milliseconds or seconds required by the circuit breaker's magnetic trip unit to interrupt the circuit. The minimum allowable cross-section for thermal stability is determined using the Joule integral equation according to the following mathematical algorithm:

Step 1: Determine the three-phase short-circuit current. The short-circuit current ($I_{sc}$) is calculated based on the total loop impedance ($Z_{loop}$) from the transformer to the most remote point of the distribution board. For this example, let us assume the calculated or measured short-circuit current ($I_{sc}$) is 1500 A (1.5 kA).

Step 2: Determine the protection operating time. The total time ($t$) during which the short-circuit current acts upon the cable consists of the circuit breaker's own operating time and the time required for the electric arc to extinguish within the device's arc chute. For modern miniature circuit breakers (MCBs), this time is typically around 0.02 seconds (at high currents) or is determined by the time-current characteristic. For calculation reliability, we will assume $t = 0.1$ seconds.

Step 3: Select the thermal coefficient. The coefficient $K$ is a constant depending on the conductor material and the type of insulation. It accounts for the initial and maximum permissible temperatures. For a copper conductor with PVC insulation (which can withstand short-term heating up to 160 °C), the coefficient $K = 115\text{ A}\cdot\sqrt{\text{s}}/\text{mm}^2$. For copper with XLPE insulation, which can withstand up to 250 °C, $K = 143\text{ A}\cdot\sqrt{\text{s}}/\text{mm}^2$.

Step 4: Calculate the minimum cross-section. The adiabatic heating formula is:

Substituting the obtained values:

Step 5: Analysis of results. The calculation shows that to prevent the conductor from being damaged by a short-circuit current of 1500 A for 0.1 seconds, its cross-section must be at least 4.12 mm². Rounding up to the nearest standard value, we select a 6 mm² cable. A 2.5 mm² cable, chosen based solely on a 20 A operating current, will not withstand this stress: the conductor will overheat, and the PVC insulation may ignite before the circuit breaker trips the line.

Cross-section and Circuit Breaker

A circuit breaker protects the cable, not the load: the breaker rating is selected so that the continuous current rating of the line is not exceeded. Typical combinations for copper conductors, in accordance with HD 60364-5-52, are: 1.5 mm² – 10–16 A breaker, 2.5 mm² – 16–20 A, 4 mm² – 25 A, 6 mm² – 32 A, 10 mm² – 50 A. You must not select a breaker with a rating that is too high relative to the cross-section: in the event of an overload, it will fail to trip the line, causing the cable to overheat beyond its permissible limits.

You can select the copper conductor cross-section and the circuit breaker rating for your specific application using the calculator below.

Sources

• Note: The following references relate to technical requirements for selection of cross-sections, wiring, and electrical equipment for residential and public buildings in accordance with the HD 60364 / IEC 60364 series.
• Note: Requirements for power cables with plastic insulation for rated voltages of 0.6/1 kV are governed by HD 603 S1 and IEC 60502-1.
• Note: Requirements for current-carrying conductors used in cables and wires are governed by the HD 308 S2 and relevant IEC standards.