Electrical cable size calculator

Size a new circuit or verify an existing cable using the technical criteria applied in electrical installation design. Results appear instantly in your browser, and every equation is explained in the Methodology section.

Voltage drop in normal operation
Voltage drop at motor start
Continuous current rating (I_z ≥ I_n)
Short-circuit thermal withstand

Load

Active power

Nominal power factor (cos φ)

Allowed voltage drop

Network and cable

Nominal voltage

System

Parallel conductors

Conductor material

Cable construction

Route length

Conductor cross-section

Allow cross-sections above 240 mm² (up to 630 mm²)By default, above 240 mm² the tool switches automatically to 2 parallel conductors per phase, then to 3 if needed.

Installation conditions

Insulation

Installation method

Ambient temperature

°C

Grouped circuits On the same tray or in the same trench, including this one

Motor starting The load is an electric motorThe voltage drop at start-up is checked as well.

Short circuit Include short-circuit thermal withstandAdiabatic heating until the fault is cleared.

Methodology and equations

The calculation follows the basic electrotechnical principles of IEC 60364-5-52 and the Romanian standard I7-2011. The cross-section is derived analytically from the voltage-drop equation, rounded up to the next standard size, then checked for continuous current rating and, optionally, short-circuit withstand.

1. Load current

I = P / (√3 · U · cos φ) (three-phase)

I = P / (U · cos φ) (single-phase)

2. Voltage drop

ΔU = K · I · L · (r₀ · cos φ + x₀ · sin φ)

K = √3 for three-phase systems and K = 2 for single-phase systems. L is the length in km, r₀ = 1000 / (γ · S) [Ω/km], with γ_Cu = 56 and γ_Al = 34 m/(Ω·mm²). The specific reactance x₀ is 0.08 Ω/km for multicore cables and single-core cables in trefoil, 0.095 Ω/km for single-core cables laid flat and touching, and 0.13 Ω/km for single-core cables laid flat with 1×D spacing.

3. Cross-section from the voltage-drop limit

S = 1000 · cos φ / ( γ · [ ΔU_max / (K · I · L) − x₀ · sin φ ] )

If the reactive term alone exceeds the allowed drop, no cross-section can meet the limit. In that case the tool switches to parallel conductors.

4. Continuous current rating

I_z = I_z0 · k₁ · k₂ ≥ I_n

k₁ = √( (θ_max − θ_a) / (θ_max − θ_ref) )

θ_max is 70 °C for PVC and 90 °C for XLPE/EPR; θ_ref is 30 °C in air and 20 °C in ground. For XLPE/EPR insulation the I_z0 values are multiplied by 1.18. k₂ is the grouping factor: 1.00 for 1 circuit, 0.80 for 2, 0.70 for 3, 0.65 for 4, 0.60 for 5 and 0.57 for 6 or more circuits.

5. Short-circuit thermal withstand

S_min = I_sc · √t / k

Factor k	PVC	XLPE / EPR
Copper	115	143
Aluminium	76	94

6. Sizing algorithm

The required cross-section is computed for each active criterion: normal voltage drop, starting voltage drop and short circuit.
The next standard size above the largest value is selected.
If I_z < I_n, the size is increased until the current rating is met.
The maximum size is 240 mm², or 630 mm² when explicitly allowed. Above it, the calculation is repeated automatically with one more conductor per phase (2, then 3) until a solution is found.

In verification mode the limits are ΔU ≤ 5 % in normal operation and ΔU ≤ 10 % at motor start.

Base current ratings I_z0 [A] (PVC insulation, in air, 30 °C)

S [mm²]	1.5	2.5	4	6	10	16	25	35	50	70	95	120	150	185	240	300	400	500	630
Copper	19.5	27	36	46	63	85	112	138	168	213	258	299	344	392	461	530	634	730	830
Aluminium	–	21	28	36	49	66	83	103	125	160	195	226	261	298	352	406	485	560	640

Note. This tool is intended for teaching and preliminary estimates. Its results do not replace design by a qualified engineer using the cable manufacturer's data and the applicable standards.

Frequently asked questions

How do you choose the cross-section of an electrical cable?

The cross-section must meet several conditions at once: the voltage drop must stay below the allowed limit, the cable's current rating (corrected for ambient temperature and grouping with other circuits) must be at least equal to the load current, and, where required, the cable must withstand the heating caused by the short-circuit current until the fault is cleared. The largest of the resulting sizes is kept and rounded up to the next standard cross-section.

How is voltage drop calculated on a three-phase cable?

Use ΔU = √3 · I · L · (r₀ · cos φ + x₀ · sin φ), with L in km and r₀, x₀ in Ω/km. For example, a 15 kW load at 400 V with a power factor of 0.85 draws 25.5 A. On a 100 m copper cable of 4 mm², the voltage drop is about 16.9 V, or 4.23 %.

What cable size do I need for 15 kW at 400 V?

It depends on the length, the conductor material and the installation method. For a 15 kW three-phase load at 400 V with a power factor of 0.85, supplied through an XLPE-insulated copper cable installed in air at 30 °C over 100 m, the result is 4 mm². The limiting criterion is the 5 % voltage drop, not heating. For other lengths or conditions, enter your data in the calculator.

Why check motor starting separately?

At start-up an induction motor draws several times its rated current (often 5 to 7 times) at a low power factor. The resulting voltage drop can prevent the motor from starting or disturb other loads, so the calculator checks it separately, with a default limit of 10 %.

What is the k factor in the short-circuit check?

The k factor depends on the conductor material and insulation and appears in S_min = I_sc · √t / k. The values used are 115 for copper with PVC, 143 for copper with XLPE, 76 for aluminium with PVC and 94 for aluminium with XLPE. For example, a 6 kA fault lasting 0.1 s requires at least 13.27 mm² of XLPE-insulated copper.