What this calculator does
A cable is a resistor you are obliged to install. Every metre of it takes a slice of the supply voltage and turns it into heat, and by the time the current reaches the load the voltage there is lower than the voltage at the board. This calculator works out how much lower, in volts and as a percentage, for a single-phase, three-phase or DC circuit — from the conductor area, the one-way route length, the design current and the conductor’s operating temperature.
It also answers the question you usually ask second: what is the smallest conductor that still meets my limit? Enter your permissible drop as a percentage and the tool inverts the whole calculation and returns the minimum cross-sectional area in mm². Conductor size can be entered in mm², kcmil or cmil, or picked straight off the AWG ladder.
The formula
The drop along a run is the phasor difference between the sending and receiving end voltages. For the load angles seen in distribution work the standard approximation — the projection of the impedance drop onto the load voltage — is accurate to well under half a percent:
ΔU = k · I · (R · cos φ + X · sin φ)
ΔU— voltage drop, Vk— 2 for single-phase (2-wire) and DC, √3 for a balanced three-phase circuitI— design (load) current, AR— resistance of one conductor over the one-way route length, ΩX— reactance of one conductor over the one-way route length, Ωcos φ— load power factor;sin φ = √(1 − cos²φ)for a lagging load
The k factor is where most hand calculations go wrong. On a two-wire run the current flows out along one conductor and back along the other, so the loop resistance is twice the one-way resistance — hence 2, and hence the instruction to enter the one-way length, not the loop length. On a balanced three-phase circuit there is no return current in the neutral, and the line-to-line drop works out to √3 times the per-conductor drop.
Conductor resistance comes from resistivity and geometry, corrected to the temperature the conductor actually runs at:
R = ρ_T · L / A
ρ_T = ρ_20 · (1 + α · (T − 20 °C))
ρ_20— resistivity at 20 °C: 0.0172 Ω·mm²/m for copper, 0.0282 Ω·mm²/m for aluminiumα— temperature coefficient of resistance: 0.00393 /K for copper, 0.00403 /K for aluminiumT— conductor operating temperature, °C — not ambientL— one-way route length, m;A— area of one conductor, mm²
Those are the standard values for the annealed copper and EC-grade aluminium used in IEC 60228 conductor classes. The temperature correction is not optional bookkeeping: a copper conductor at its 70 °C PVC rating is about 20 % more resistive than the same conductor at 20 °C, and a 90 °C XLPE conductor about 28 % more. Cable tables quote mV/A/m at the insulation’s rated conductor temperature for exactly this reason.
Power lost as heat is I²R summed over the current-carrying conductors — 2·I²·R on a two-wire run, 3·I²·R on balanced three-phase.
The minimum-area output inverts the resistive term while holding the reactive term fixed, because X per unit length is set by conductor spacing rather than by conductor area and does not shrink when you upsize:
A_min = k · I · cos φ · ρ_T · L / (ΔU_perm − k · I · X · sin φ)
Reading the result
The percentage is the number that gets checked; the volts are the number that causes the problem. A 3 % drop is 6.9 V on a 230 V circuit and 1.44 V on a 48 V DC circuit — the same percentage, wildly different consequences for anything with a fixed dropout threshold.
Voltage at the load tells you whether equipment will actually work. Motors develop torque proportional to the square of terminal voltage, so a 10 % drop costs about 19 % of starting torque. Contactor coils release somewhere around 70–80 % of nominal. Power lost in the cable is the operating-cost side of the same physics: a circuit dropping 3 % at full load is throwing away roughly 3 % of the delivered power as heat, worth converting to kWh per year before you accept a marginal cable size.
When the resistive-only figure is good enough. Below roughly 16 mm² (about 6 AWG), reactance is typically under 10 % of the impedance and the resistive-only answer is fine at any power factor. Between 16 and 35 mm² it starts to matter below about 0.9 power factor. Above 35 mm², or on long runs of spaced single-core cables, the reactive term can dominate and ignoring it understates the drop badly. Leave the power factor at 1.0 and the reactance term vanishes entirely (sin φ = 0), which is the correct behaviour for DC and for purely resistive loads.
Typical limits
Permissible voltage drop is not physics — it belongs to the wiring rules in force where the installation is built, and this tool deliberately makes you type it in rather than assuming one country’s number. As a guide to the values in circulation:
- 3 % for a final circuit, 5 % total from the origin of the installation is the most widely used pairing, and appears in one form or another in IEC 60364-5-52 Annex G, AS/NZS 3000 Clause 3.6, and BS 7671 Appendix 4 (which splits it 3 % for lighting and 5 % for other uses on a public-supply installation).
- NEC does not mandate a limit in its enforceable text at all: Art. 210.19(A) and 215.2(A) carry informational notes recommending 3 % on a branch circuit and 5 % combined feeder-plus-branch. Notes are not requirements, but specifications routinely make them contractual.
- Motor and large-load circuits are often held tighter — 2 % running, with a separate starting-current check that may allow 10–15 % transiently.
- Extra-low-voltage DC (48 V telecom, 12/24 V vehicle and solar) is usually specified in volts, not percent, because the load’s dropout voltage is the real constraint.
Typical reactance values for the X field: around 0.08 Ω/km for small multicore cables, 0.07–0.09 Ω/km for larger multicore, and 0.09–0.20 Ω/km for single-core cables depending on spacing and formation. Always prefer the manufacturer’s figure.
Worked example
A 230 V single-phase final circuit runs 30 m from the board to the load and carries a design current of 20 A. The cable is 4 mm² copper with PVC insulation, so the conductor is taken at its 70 °C rating. The load is resistive, so cos φ = 1. The permissible drop is 3 %.
Resistivity at the operating temperature:
ρ_70 = 0.0172 × (1 + 0.00393 × 50) = 0.0172 × 1.1965 = 0.0205798 Ω·mm²/m
Resistance of one conductor over the one-way length:
R = 0.0205798 × 30 / 4 = 0.617394 / 4 = 0.1543485 Ω
At unity power factor sin φ = 0, so the reactance term drops out and the two-wire multiplier gives:
ΔU = 2 × 20 × 0.1543485 = 6.17 V
That is 6.17 V, or 6.17394 / 230 × 100 = 2.68 % — inside the 3 % limit with about 0.32 percentage points of margin. The load sees 223.83 V, and the cable dissipates 2 × 20² × 0.1543485 = 123.5 W along its length while the circuit is at full load.
The permissible drop of 3 % is 6.9 V, so the smallest conductor that still fits the limit is:
A_min = 2 × 20 × 0.0205798 × 30 / 6.9 = 24.69576 / 6.9 = 3.58 mm²
There is no such thing as a 3.58 mm² cable, which is the point: 2.5 mm² would fail and 4 mm² is the next size up that passes. Note also that had the temperature correction been skipped and 20 °C resistivity used, the drop would have come out at 5.16 V (2.24 %) — comfortably inside the limit, and comfortably wrong.
FAQ
Do I enter the one-way length or the total conductor length? One-way — the route length from source to load. The multiplier (2 or √3) already accounts for the return path. Entering a doubled length is the single most common way to get an answer that is twice as large as it should be.
How do I handle cables in parallel? Enter the total area of the parallel set as the conductor area and the total circuit current. Two 95 mm² cables per phase behave, for voltage drop purposes, exactly like one 190 mm² conductor — provided the parallel cables are the same length, size and material, which the wiring rules require anyway.
Which temperature should I use — ambient or conductor? Conductor. Entering 30 °C ambient when the conductor runs at 70 °C understates the resistance by about 15 %. If you do not know the operating temperature, using the insulation’s rated temperature (70 °C for PVC, 90 °C for XLPE/EPR) is the conservative choice and matches how cable tables are compiled.
What is the relationship between AWG and mm²?
American Wire Gauge is a geometric ladder: the diameter is defined as d = 0.127 mm × 92^((36 − n)/39), so every 6 gauge numbers roughly halves the diameter and every 3 roughly halves the area. Larger numbers mean smaller wire, and above 4/0 the scale is abandoned in favour of thousands of circular mils (kcmil), where 1 kcmil = 0.5067 mm². The area field accepts kcmil and cmil directly for exactly that reason.
Why does a lower power factor sometimes reduce the drop in this tool?
Because cos φ scales the resistive term down while sin φ scales the reactive term up. On a small cable, where R ≫ X, dropping the power factor from 1.0 to 0.85 reduces the calculated drop. On a large cable, where X is comparable to R, the reactive term wins and the drop increases. That crossover is real, and it is why the reactance field exists.
This tool provides indicative figures for design and checking. Cable selection must also satisfy current-carrying capacity, short-circuit withstand, earth-fault loop impedance and protective-device coordination under the wiring rules applicable to your installation.