Calculate the voltage drop along a cable run for single-phase or three-phase circuits. Conductor resistance is evaluated at 75 °C and the calculation includes the cable reactance term, so the result stays valid at low power factor and on larger conductors where a resistance-only estimate under-reads.
| Symbol | Meaning | Unit |
|---|---|---|
| ΔU | Voltage drop along the cable run | V |
| k | Phase multiplier — √3 three phase, 2 single phase | — |
| n | Loss multiplier — 3 three phase, 2 single phase | — |
| I | Load current | A |
| R | Conductor resistance of the run at 75 °C | Ω |
| X | Conductor reactance of the run | Ω |
| x′ | Reactance per metre, from the formation table | Ω/m |
| ρ20 | Resistivity at 20 °C — 1.72e-8 Cu, 2.82e-8 Al | Ω·m |
| α | Temperature coefficient — 0.00393 Cu, 0.00403 Al | 1/K |
| L | One-way route length | m |
| A | Conductor cross-sectional area | mm² |
| cos φ | Load power factor | — |
| ΔUmax | Permitted drop in volts — 5 % of U | V |
A 150 m submain feeds a distribution board drawing 150 A at 0.85 power factor from a 415 V three-phase supply. The cable is 70 mm² copper, laid in trefoil at low voltage, so x′ = 0.08 mΩ/m.
| Input | Unit | Accepted range | Default |
|---|---|---|---|
| Phase system | — | Single or three phase | Three phase |
| Supply voltage U | V | > 0 | 415 |
| Load current I | A | > 0 | 150 |
| Power factor cos φ | — | > 0 and ≤ 1 | 0.85 |
| Route length L | m (ft) | > 0 | 150 |
| Conductor CSA | mm² (kcmil) | > 0 | 70 |
| Material | — | Copper or aluminium | Copper |
| Formation | — | 0.08 to 0.30 mΩ/m | LV trefoil |
| Formation | Voltage class | x′ (mΩ/m) | Confidence |
|---|---|---|---|
| Trefoil / touching | LV ≤ 1 kV | 0.08 | ±15 % vs datasheet |
| Flat formation | LV ≤ 1 kV | 0.10 | ±15 % vs datasheet |
| Touching trefoil | MV 1–36 kV | 0.10 | Use manufacturer data |
| Spacing 1 × diameter | MV 1–36 kV | 0.13 | Use manufacturer data |
| Spacing 2 × diameter | MV 1–36 kV | 0.17 | ±30–50 % possible |
| Spacing 3 × diameter | MV 1–36 kV | 0.20 | Broad estimate only |
| XLPE | HV > 33 kV | 0.30 | IEC 60287 software required |
| Material | ρ20 (Ω·m) | α (1/K) | ρ75 (Ω·mm²/m) | Relative drop |
|---|---|---|---|---|
| Copper | 1.72e-8 | 0.00393 | 0.020918 | 1.00 |
| Aluminium | 2.82e-8 | 0.00403 | 0.034451 | 1.65 |
| Circuit type | Recommended limit | Reference |
|---|---|---|
| Lighting circuits | 3 % | IEC 60364-5-52 Annex G |
| Other uses (power, motors) | 5 % | IEC 60364-5-52 Annex G |
| Branch circuit | 3 % | NEC 210.19(A) note |
| Feeder plus branch | 5 % | NEC 215.2(A) note |
| Motor starting transient | 10–15 % | Typical specification practice |
IEC 60364-5-52 Annex G recommends 3 % for lighting circuits and 5 % for other uses, measured from the origin of the installation to the load. The NEC gives a comparable 3 % branch-circuit and 5 % combined feeder-plus-branch recommendation in informational notes to 210.19(A) and 215.2(A). Both are recommendations rather than hard limits, but most specifications adopt 5 % as the contractual ceiling. This calculator compares the computed percentage against 5 %.
Conductor resistivity rises with temperature, so voltage drop computed at 20 °C is optimistic. IEC and NEC practice is to evaluate voltage drop at the conductor operating temperature. This calculator fixes the conductor at 75 °C, which raises copper resistivity from 1.72e-8 to about 2.09e-8 Ω·m — roughly 21 % more resistance and therefore roughly 21 % more resistive voltage drop than a 20 °C calculation would predict.
Yes, and it is the term most often omitted. Reactive drop is proportional to sin φ, so at a power factor of 0.85 it carries about 53 % of the current in quadrature. For small conductors the resistive term dominates, but above roughly 70 to 95 mm² the reactance term becomes significant, and at low power factor or long runs it can decide whether a cable passes. This calculator includes the X sin φ term and reports the reactance used.
Use a multiplier of 2 instead of √3, because current flows out along the line conductor and back along the neutral, so the cable length is traversed twice. The formula becomes ΔU = 2 · I · (R cos φ + X sin φ). Select Single Phase in the calculator and it applies the factor of 2 automatically, and compares the drop against the single-phase supply voltage you enter.
Rearranging the voltage drop formula for area gives A = k · I · L · ρ · cos φ / (ΔUmax − k · I · X · L · sin φ). The calculator reports this as the minimum CSA. Note that this is the minimum needed for voltage drop only — the conductor must also satisfy current-carrying capacity after derating and the adiabatic short-circuit withstand check, and either of those can demand a larger cable.
Results are for estimation and preliminary design. Verify against project specifications, cable manufacturer data and a qualified engineer before construction.
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