Voltage Drop Calculator

Written byOleh Soloninupdated 25 September 2026

Load current, A

Run length, m

Conductor area, mm²

Phase-to-neutral voltage U₀, V

Russia's GOST 29322-2014 (IEC 60038) sets 230/400 V; on a three-phase line enter the phase voltage here too, 230 or 220 V.

Conductor

Supply

What the circuit feeds

Advanced options

Supply

Power factor cos φ

0.8 is the figure IEC 60364-5-52 tells you to use when nothing better is known; a heater is 1.

Result

Voltage drop, V11.6
Drop, %5.04%
Verdict
Over 5% for other loads (supplied from the public network, IEC 60364-5-52, table G.52.1) — a larger cross-section is needed
Phase voltage at the load, V218.4
Standard area within the limit, mm²4
Cross-section
The exact figure is 2.52 mm²; the next standard IEC 60228 size is 4 mm².
Heat in the conductors, W230.4
Working
u = 2 × (0.0225 × 50 ÷ 2.5 × 0.8 + 0.00008 × 50 × 0.6) × 16 = 11.6 V; 11.6 ÷ 230 = 5.04%

Why the far end of a long run is dimmer

Works out cable voltage drop with the method in annex G of IEC 60364-5-52 — service resistivity and conductor reactance included — and checks it against the 3% limit for lighting and 5% for other loads.

Part of the voltage is lost in the cable itself: the current runs through conductors with resistance, and a voltage u drops across them. Annex G of IEC 60364-5-52 (adopted in Russia as GOST R 50571.5.52-2011) gives the formula u = b × (ρ₁ × L ÷ S × cos φ + λ × L × sin φ) × I, with b = 2 for a single-phase circuit and 1 for a three-phase one, and λ = 0.08 mΩ per metre of reactance.

The standard takes the resistivity ρ₁ at service temperature, 1.25 times the 20 °C figure: 0.0225 Ω·mm²/m for copper and 0.036 for aluminium. The percentage is worked against the phase-to-neutral voltage U₀ — 230 V on a 230/400 V supply.

How to use

Enter the load current, the one-way run length and the conductor cross-section.
Set the phase-to-neutral voltage, the conductor material, the supply type and what the circuit feeds: lighting, or sockets and other loads.
If you know the load's power factor, set it under the advanced fields; otherwise it stays at 0.8, as the standard says.
Read the drop in volts and percent, the verdict against the limit, and the standard cross-section that would stay within it.

The voltage-drop limits

Table G.52.1 of IEC 60364-5-52 limits the drop between the origin of the installation and any point of use: 3% for lighting and 5% for other uses when the installation is supplied from the public low-voltage network, 6% and 8% when it has its own supply.

For runs longer than 100 m the standard lets these limits grow by 0.005% per metre beyond 100 m, up to 0.5% more. The calculator does not add that allowance and compares with the table values.

Drop on a 16 A circuit

Each row is a separate run: single-phase, U₀ = 230 V, power factor 0.8. "Size for the limit" is the smallest IEC 60228 cross-section that keeps the drop within 5%.

All rows come from the calculator on this page. The size needed for current-carrying capacity is a separate check.
Run	Drop	Percent	5% limit	Size for the limit
10 m, copper 2.5 mm²	2.32 V	1.01%	within	0.75 mm²
20 m, copper 2.5 mm²	4.64 V	2.02%	within	1.5 mm²
30 m, copper 2.5 mm²	6.96 V	3.03%	within	2.5 mm²
50 m, copper 2.5 mm²	11.6 V	5.04%	over	4 mm²
50 m, copper 4 mm²	7.28 V	3.16%	within	4 mm²
50 m, copper 6 mm²	4.88 V	2.12%	within	4 mm²
50 m, aluminium 4 mm²	11.6 V	5.04%	over	6 mm²

What this calculator shows

The drop in volts and as a percentage of U₀
The verdict against table G.52.1: 3, 5, 6 or 8%
The smallest standard cross-section that meets the limit
How much heat the conductors give off

What to keep in mind

Method: annex G of IEC 60364-5-52 — ρ₁ = 0.0225 (copper) and 0.036 (aluminium) Ω·mm²/m, λ = 0.08 mΩ/m, power factor 0.8 unless you set another.
The length is the one-way route; the return through the neutral of a single-phase circuit is the factor of 2.
A three-phase load is taken as balanced; with only one phase loaded, the standard says to treat the circuit as single-phase.

FAQs

How much voltage drop is allowed?

Table G.52.1 of IEC 60364-5-52: 3% for lighting and 5% for other uses on a supply from the public network, 6% and 8% on a private supply.

Why is there a factor of 2 in a single-phase circuit?

The current goes out on the line conductor and comes back on the neutral, and the drop builds up on both. For a balanced three-phase circuit the standard sets b = 1.

Why does copper use 0.0225 rather than 0.0172?

0.0225 is the annex G value for a conductor at service temperature, 1.25 times the 20 °C figure. Tables of resistivity at 20 °C give about 0.0172 for annealed copper, which understates the drop in a working cable.

How much worse is aluminium than copper?

Annex G gives it 0.036 against copper's 0.0225 — 1.6 times as much. A 4 mm² aluminium run of 50 m at 16 A loses the same 11.6 V as 2.5 mm² copper.

How much heat does the voltage drop cost?

The energy the line loses goes into heating the conductors, at a rate of I²R. A 50 m copper run of 2.5 mm² at 16 A dissipates 230 W along its length.

Can I size a cable on voltage drop alone?

No: the cable must also carry the current without overheating. For a 10 m run at 16 A the drop sum asks for only 0.75 mm², while Russia's PUE table 1.3.6 needs at least 1.5 mm² of copper. Take the larger of the two.

Worked example

A run to the garage

Input: 16 A, 50 m, copper 2.5 mm², 230 V

Output: A drop of 11.6 V, or 5.04% — over the 5% limit

Note: 4 mm² brings it within the limit — the exact figure is 2.52 mm². With 4 mm² the drop is 7.28 V, or 3.16%.

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