Voltage Drop Calculator

The voltage a cable loses at your load current — in volts and as a percentage — with the voltage left at the load and the power wasted as heat, for mm² or AWG, copper or aluminium, DC, single-phase or three-phase.

Last updated 22 Sep 2026

Voltage drop in a cable

Size + length + current → ΔV and %

Size given asEnter the size in one system; the other box shows the equivalent.

Conductor size (mm²)Standard IEC sizes are 1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120 mm² and up. The nominal area of one conductor, not the cable’s outside diameter.

Conductor size (AWG)Solid-conductor area from the ASTM B258 formula. Stranded conductors of the same gauge are within about 1% of it.

Conductor

Conductor temperature (°C)Resistance rises with temperature: 70 °C is a PVC-insulated conductor at full load, 20 °C the cold figure. The difference is about 20%.

Cable length, one way (m)Supply to load. The return conductor is counted by the formula.

Load current (A)

System

Supply voltage (V)For three-phase, the line-to-line voltage (400 V or 415 V).

The cable drawn as two resistors, one for the conductor out and one for the return, each with the voltage it drops; the load gets what is left. For three-phase the drop shown is between lines (√3 × I × R), so the two resistors together stand for that. The conductors turn amber past 3% and red past 5%, the usual recommended limits. The dots show the size of each RMS current; a real 50 Hz or 60 Hz current reverses a hundred or a hundred and twenty times a second, so the direction drawn is only a convention. The supply is drawn with the battery symbol, the only source symbol in this set; on AC the + marking is just a reference direction.

2.87%Example

2.5 mm² copper, 25 m one way, 16 A single-phase at 230 V, conductor at 70 °C

Voltage drop in a cable

ΔV = k × I × ρ_T × L ÷ A; ρ_T = ρ₂₀ (1 + α (T − 20)); k = 2 (DC, single-phase) or √3 (balanced three-phase); drop % = ΔV ÷ V × 100

L: one-way length in metres — the formula counts the return conductor
A: conductor cross-section in mm² (or the AWG equivalent)
ρ₂₀, α: copper 0.017241 Ω·mm²/m and 0.00393 /°C; aluminium 0.028264 and 0.00403
P_loss: I²R in every conductor: 2 I²R, or 3 I²R three-phase

Worked example

2.5 mm² copper, 25 m one way, 16 A single-phase at 230 V, conductor at 70 °C
ρ at 70 °C = 0.017241 × (1 + 0.00393 × 50) = 0.020629 Ω·mm²/m
R of one conductor = 0.020629 × 25 ÷ 2.5 = 206.3 mΩ
ΔV = 2 × 16 × 0.2063 = 6.60 V = 2.87% of 230 V
The load sees 223.4 V, and 105.6 W is lost as heat in the cable
This cable reaches 3% at 16.7 A and 5% at 27.9 A; the wire size calculator picks 2.5 mm² for exactly this circuit

The same circuit, other choices

All computed by this page’s own formula at 16 A over 25 m at 230 V. Doubling the length doubles the drop; doubling the area halves it.
Change	Drop	Voltage at the load
As entered: 2.5 mm² copper at 70 °C	2.87% (6.60 V)	223.4 V
Same cable cold (20 °C)	2.40% (5.52 V)	224.5 V
Aluminium instead of copper	4.72% (10.87 V)	219.1 V
4 mm² copper at 70 °C	1.79% (4.13 V)	225.9 V
Twice the length (50 m)	5.74% (13.20 V)	216.8 V
12 AWG (3.309 mm²) copper at 70 °C	2.17% (4.99 V)	225.0 V

How much voltage a cable loses

Every conductor has resistance, so current flowing along it drops voltage that never reaches the load. The drop is the current times the resistance of the whole loop — out along one conductor and back along the other — and the resistance of a conductor is its resistivity times its length divided by its cross-section. Warm conductors are worse than cold ones: copper’s resistance rises about 0.39% per degree, so a cable at its 70 °C working temperature has roughly 20% more resistance than the handbook’s 20 °C figure.

The example. 2.5 mm² copper, 25 m each way, 16 A at 230 V: each conductor is 206.3 mΩ, the drop is 6.60 V or 2.87%, the load sees 223.4 V, and 106 W disappears into warming the cable. That is 925 units a year if the circuit ran flat out continuously — the reason voltage drop is an energy question as well as a performance one.

Three-phase is different. In a balanced three-phase circuit the return currents in the three lines cancel, so the line-to-line drop is √3 × I × R rather than 2 × I × R. Enter the line-to-line voltage and the line current; the percentage is against the line-to-line figure.

How much is too much? IEC 60364-5-52’s informative Annex G suggests 3% for lighting and 5% for other uses from a public low-voltage network; the NEC’s informational notes suggest 3% on a branch circuit and 5% overall. Both are recommendations, not rules, and your local code, your distribution company and the equipment’s own tolerance decide. Motors are the loads that suffer most: torque falls with the square of the voltage, and a motor that starts sluggishly on a long run will run hot for the rest of its life.

Drop is not the only limit. A cable also has to carry its current without overheating — its ampacity — and that depends on insulation, installation method, grouping and ambient temperature, all from your code’s tables. This page says nothing about it. To go the other way and pick a size from an allowed drop, use the wire size calculator; for the physical size of a gauge, the AWG wire size calculator; for the protective device, the MCB size calculator.

Frequently asked questions

How do I calculate voltage drop?

Multiply the current by the resistance of the loop. For DC and single-phase: ΔV = 2 × L × I × ρ ÷ A, with L the one-way length in metres, A the area in mm² and ρ about 0.0172 Ω·mm²/m for cold copper (0.0206 at 70 °C). For balanced three-phase, replace the 2 with √3.

What is an acceptable voltage drop?

3% to a final circuit and 5% overall are the usual recommendations (IEC 60364-5-52 Annex G, NEC informational notes). They are advisory; equipment tolerances and your local code decide.

Why does aluminium drop more than copper?

Its resistivity is about 64% higher, so the same size drops about 64% more — 4.72% against 2.87% in the example. For the same drop it needs roughly 1.6 times the area.

Does voltage drop waste energy?

Yes. The power lost is I² × R in each conductor — heat in the wall, paid for on your meter. Halving the drop by doubling the area halves that loss.

Should I use the hot or the cold resistance?

The temperature the conductor actually reaches. Use 70 °C (or 90 °C for XLPE) for a circuit at full load, which is the conservative choice, and 20 °C only for lightly loaded runs.

Related calculators

References

IEC 60364-5-52:2009, Annex G (informative) voltage-drop recommendations — 3% for lighting and 5% for other uses from a public low-voltage network — as tabulated in Schneider Electric’s Electrical Installation Guide.
IEC 60228:2004 (conductor cross-sections), IEC 60028 (annealed copper, 0.017241 Ω·mm²/m at 20 °C, α = 0.00393 /K) and IEC 60889 (hard-drawn aluminium, 0.028264 Ω·mm²/m, α = 0.00403 /K).
NFPA 70, National Electrical Code: 210.19(A)(1) and 210.20(A) (noncontinuous load plus 125% of the continuous load) and the informational notes to 210.19(A) and 215.2(A) (3% branch, 5% total voltage drop, advisory).
IS 732:2019, Code of practice for electrical wiring installations (Bureau of Indian Standards).