Voltage drop calculator
Voltage lost along a cable run, from its length, cross-section and material.
Fill in the fields and the result will appear here automatically.
Compare voltage losses as cable length, cross-section or material changes. This is a resistive model: both conductors are counted for a two-wire single-phase circuit; balanced three-phase uses line-to-line voltage. Current is per conductor, displayed resistance is for one conductor on the one-way run, and power loss totals the current-carrying conductors. Reactance, imbalance, heating and contact resistance are not calculated.
How it works
Formula and logic
Rc = ρL/S with adopted 20 °C model coefficients: copper 0.0175 and aluminium 0.0282 Ω·mm²/m. ΔU = 2IRc for two wires; ΔU = √3IRc for balanced three-phase at power factor 1. Total heat loss is 2I²Rc or 3I²Rc respectively. Percentage = 100ΔU/U; load voltage = U − ΔU.
Example
Copper, 16 A, 20 m, 2.5 mm², 230 V: Rc = 0.0175 × 20/2.5 = 0.14 Ω; ΔU = 4.48 V = 1.95%; load voltage 225.52 V; loss 71.68 W. Aluminium, 32 A, 50 m, 6 mm², three-phase, 400 V: Rc = 0.235 Ω; ΔU = 13.025 V; total loss = 3 × 32² × 0.235 = 721.92 W.
Fields and units
- Current — A
- One-way run — m
- Conductor cross-section — mm²
- Conductor material — list option
- Supply — list option
- Nominal voltage — V
How to use
- — Enter amperes and one-way length in metres; do not double the length yourself.
- — Enter one conductor’s cross-section in mm², not its diameter, and choose copper or aluminium.
- — Use voltage between the two wires for single-phase and between phases for three-phase, such as 400 V rather than 230 V.
- — Zero current gives zero loss. If the drop exceeds nominal voltage, those inputs fall outside this supply model’s applicable range.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- Schneider Electric: voltage drop; this model omits reactance OpenStax: AC power and a purely resistive load
- Limitation
- Resistive estimate at 20 °C and power factor 1, not code-compliant cable selection. The adopted resistivities are model coefficients, not a product-specific resistance table.
FAQ
Why is three-phase loss not current times the voltage drop?
The drop √3IRc is line-to-line. Three conductors dissipate heat, giving 3I²Rc = √3IΔU. The voltage-drop factor √3 cannot replace the heating factor 3.
What does the displayed cable resistance mean?
It is one conductor’s resistance from source to load. A two-wire loop doubles it; three-phase has no single common loop resistance.
Can voltage-drop percentage alone select a cable?
No. Ampacity, protection, installation method and applicable local rules are also needed. This tool does not check them or set a regulatory limit.
Why can a warm cable differ from this result?
The coefficients are fixed at 20 °C. Actual cable temperature, tolerances and connections differ; operating temperature and reactive loads need a fuller model.
What changes if cross-section rises from 2.5 to 5 mm²?
At fixed current, length and material, R = ρL/S halves; voltage drop and I²R losses halve too. The 16 A, 20 m single-phase copper example becomes 2.24 V and 35.84 W instead of 4.48 V and 71.68 W. This comparison does not check ampacity.