What is the voltage drop formula for a cable run?

For a single-phase circuit: V_drop = 2 × L × I × ρ / A. For a balanced three-phase circuit: V_drop = √3 × L × I × ρ / A. Here L is the one-way run length in metres, I is the load current in amps, ρ is the conductor resistivity (1.724×10⁻⁸ Ω·m for copper at 20°C, 2.65×10⁻⁸ Ω·m for aluminium), and A is the conductor cross-sectional area in m². The factor of 2 on a single-phase circuit accounts for both the live and neutral conductors carrying the full current; the √3 factor on a three-phase circuit comes from the phase-to-neutral geometry of a balanced load.

What percentage voltage drop is acceptable?

NEC 210.19 (USA) recommends no more than 3% drop on a branch circuit and a combined feeder-plus-branch total of no more than 5%. UK wiring regulations (BS 7671 / IET Wiring Regulations) use similar thresholds: 3% for lighting circuits and 5% for power circuits measured from the origin of the installation. Both standards treat these as design targets rather than hard limits, but exceeding 5% wastes significant energy and can cause equipment to malfunction or overheat.

Why does cable length matter so much?

Voltage drop scales linearly with run length. A cable carrying 20 A that loses 1% over 10 m will lose 5% over 50 m with no other changes. Long runs to outbuildings, garden sockets, solar inverters, EV chargers or industrial machinery almost always require a conductor larger than the minimum ampacity table dictates. Use the "Find minimum wire size" mode to account for run length from the start of a project, rather than retrofitting a heavier cable after installation.

How do I convert between AWG and mm²?

AWG diameter follows the formula d = 0.127 × 92^((36−AWG)/39) mm. Cross-sectional area is then A = π/4 × d². As a practical reference: AWG 14 ≈ 2.08 mm², AWG 12 ≈ 3.31 mm², AWG 10 ≈ 5.26 mm², AWG 8 ≈ 8.37 mm², AWG 6 ≈ 13.3 mm². The calculator accepts both systems and shows the mm² equivalent alongside every AWG selection.

Does conductor temperature affect the result?

Yes. This calculator uses resistivity values at 20°C (the standard reference temperature). As a conductor heats up under load, copper resistivity rises by roughly 0.4% per °C. A cable running near its rated ampacity at 70°C will have a resistivity about 20% higher than the 20°C value, increasing voltage drop by the same proportion. For critical installations apply a temperature correction factor: ρ_T = ρ_20 × (1 + 0.00393 × (T − 20)) for copper. The calculator gives the conservative 20°C baseline — multiply the result by the correction factor if your cable will run hot.

What is the difference between single-phase and three-phase voltage drop?

In a single-phase circuit both the live and neutral conductors carry the same current in opposite directions, so the total resistive length is 2 × the one-way run, giving a factor of 2 in the formula. In a balanced three-phase circuit the currents in the three phases are 120° apart; the vector sum means the effective factor is √3 ≈ 1.732 instead of 2. This makes three-phase distribution inherently more efficient for long runs — the same cable carrying the same total power loses about 13% less voltage than a single-phase equivalent.

What is the Voltage Drop Calculator?

Free voltage drop calculator. Enter wire size, current, run length and system voltage to get drop in volts and percent, with step-by-step working. Or flip to find-wire mode to solve for the minimum conductor cross-section that keeps drop within 3% (or any target). Supports single-phase and three-phase, copper and aluminium, metric mm² and AWG sizes. Everything runs in your browser — nothing is uploaded. It runs free in your browser on Gera Tools, with nothing uploaded.

Voltage Drop Calculator

Name: Voltage Drop Calculator
Creator: Gera Tools
License: https://creativecommons.org/licenses/by/4.0/

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Voltage drop is the reduction in electrical potential that occurs as current flows through the resistance of a conductor. Even a perfectly good cable has a small but non-zero resistance, and that resistance turns some of the supply voltage into heat before it reaches the load. For short runs carrying light currents the loss is negligible; for long runs, high-current loads, or thin conductors the drop can become large enough to trip protection devices, cause lights to flicker, or make motors run hot and slow.

This calculator solves both directions of the problem. In Calculate voltage drop mode you pick a conductor and the tool tells you exactly how many volts — and what percentage of the system voltage — are lost in the run. In Find minimum wire size mode you set the maximum drop you will accept (3% is the standard design target) and the tool works backwards to give you the minimum theoretical cross-section, plus the nearest standard metric IEC size or AWG gauge that meets it. Both modes show the full step-by-step working so every number is traceable.

How it works

The core formula is derived directly from Ohm’s law. The resistance of one conductor is:

R_one = ρ × L / A

where ρ is the conductor resistivity in Ω·m, L is the one-way run length in metres, and A is the cross-sectional area in m². For a single-phase circuit the current must travel out on the live conductor and return on the neutral, so the total resistive path is twice as long:

V_drop = 2 × L × I × ρ / A

For a balanced three-phase circuit the geometry of the rotating phasors reduces the effective factor from 2 to √3:

V_drop = √3 × L × I × ρ / A

The percentage drop is simply V_drop / V_system × 100. Rearranging the formula to solve for area gives the minimum cross-section for a target drop:

A_min = factor × ρ × L × I / V_drop_max

The calculator keeps everything in SI base units internally (metres, amps, ohms, m²) and only converts to mm² or feet for display.

Worked example

Scenario: A 230 V single-phase supply feeds a workshop 50 m from the consumer unit. The load draws 20 A and you are considering 4 mm² copper cable.

Step 1 — Calculate drop:

ρ(copper) = 1.724 × 10⁻⁸ Ω·m
A = 4 mm² = 4 × 10⁻⁶ m²
R_one = 1.724×10⁻⁸ × 50 / 4×10⁻⁶ = 0.2155 Ω
R_total = 2 × 0.2155 = 0.431 Ω (round trip)
V_drop = 20 × 0.431 = 8.62 V
Drop% = 8.62 / 230 × 100 = 3.75% — exceeds the 3% design target

Step 2 — Find the right size (3% limit = 6.9 V):

A_min = 2 × 1.724×10⁻⁸ × 50 × 20 / 6.9 = 4.99 × 10⁻⁶ m² = 4.99 mm²
Nearest standard size: 6 mm² copper

At 6 mm², V_drop = 20 × (2 × 1.724×10⁻⁸ × 50 / 6×10⁻⁶) = 5.75 V = 2.5% — within limits.

Cable (copper, 50 m, 20 A, 230 V)	Drop (V)	Drop (%)	Pass 3%?
2.5 mm²	13.8 V	6.0%	No
4 mm²	8.6 V	3.75%	No
6 mm²	5.8 V	2.5%	Yes
10 mm²	3.4 V	1.5%	Yes

Formula note

The resistivity values used are at 20°C (standard reference temperature). A conductor at operating temperature will have slightly higher resistance — roughly +20% for copper at 70°C — which increases the actual drop proportionally. For final design always apply a temperature correction factor: ρ_T = ρ_20 × (1 + 0.00393 × (T − 20)) for copper. The calculator provides the baseline; all figures run locally in your browser and nothing is uploaded or stored.