Short answer: voltage drop is the current multiplied by the loop resistance of the run — twice the one-way length for a single-phase circuit, √3 times for a three-phase circuit. In symbols: ΔU = 2 × I × L × (R cos φ + X sin φ) for single-phase, and ΔU = √3 × I × L × (R cos φ + X sin φ) for three-phase, with R and X in ohms per metre. Keep the result within 3% for lighting and 5% for other loads at the origin of the installation, as IEC 60364-5-52 recommends for a public low-voltage supply. On runs over a few tens of metres this check, not the cable’s current rating, is usually what decides the cross-section.
The calculation is simple and skipped constantly, because a cable that carries the current feels like a solved problem. It is not: a pump that starts on 12% drop will trip its contactor, an LED driver at 10% under its rated voltage will fail early, and a 120 V site has a fifth of the absolute budget that a 400 V site has for the same percentage. We write as an independent trading company — we are not a manufacturer and not an appointed distributor of any brand — so this is the method we work to when a buyer sends a load list and a route length.
The formula, and where the 2 and the √3 come from
Current goes out to the load and comes back, so the loop length is twice the route length in a two-wire circuit. In a balanced three-phase circuit the return path is the other two phases, and the geometry of three vectors 120° apart reduces the factor to √3. Same physics, different constant.
| Circuit | Formula | Loop factor | Notes |
|---|---|---|---|
| Single-phase, two-wire (L + N) | ΔU = 2 × I × L × (R cos φ + X sin φ) | 2 | L is one-way route length; the neutral carries the same current |
| Three-phase, balanced | ΔU = √3 × I × L × (R cos φ + X sin φ) | 1.732 | Neutral current is near zero, so no return factor |
| Three-phase with a loaded neutral | Add the neutral drop vectorially | — | Harmonic-rich loads: treat the neutral as a current-carrying conductor, not as a reference |
| DC two-wire | ΔU = 2 × I × L × R | 2 | No reactance term; relevant on PV strings and battery feeds |
| Percentage form | ΔU% = ΔU / Un × 100 | — | Un is the nominal system voltage, line-to-line for three-phase |
For cables up to about 16 mm² at power frequencies, X is small enough that the reactive term is often dropped and the result stays conservative. Above that, and on long runs at low power factor, X becomes material and should come from the cable data rather than be assumed away.
Resistance comes from the conductor, not from memory
R is the maximum DC conductor resistance of the cable at 20 °C, tabulated in IEC 60228 by cross-section and conductor class. Using the standard limit rather than a measured value keeps the calculation on the safe side, and it removes the temptation to use a rounded number that is 10% optimistic.
| Cross-section (mm²) | Copper, class 2 (Ω/km at 20 °C) | Typical use |
|---|---|---|
| 1.5 | 12.1 | Lighting final circuits, short drops |
| 2.5 | 7.41 | Socket circuits, small single-phase feeders |
| 4 | 4.61 | Common step up when volt drop fails at 2.5 |
| 6 | 3.08 | Sub-mains up to about 40 A on moderate runs |
| 10 | 1.83 | Long single-phase runs, small three-phase feeders |
| 16 | 1.15 | Three-phase feeders, 60–80 A range |
| 25 | 0.727 | Distribution runs to sub-boards |
| 35 | 0.524 | Main feeders where the route is long |
| 50 | 0.387 | Incoming mains on medium sites |
| 70 | 0.268 | Long mains runs |
| 95 | 0.193 | Heavy feeders; reactance now matters |
Aluminium is roughly 1.6 times the resistance for the same cross-section, which is why an aluminium run that “carries the current” can still fail on drop. Temperature matters too: resistance rises with conductor temperature, so a cable running hot has more drop than the 20 °C table suggests — one more reason the derating and volt drop checks belong together.
Worked example: 20 A over 30 m at 230 V
Single-phase, 20 A, 30 m one-way, copper, power factor 0.95. The cable chosen on current rating might well be 2.5 mm². Run the numbers on three sizes and the decision changes.
| Size | Calculation | ΔU | ΔU% of 230 V | Verdict at 5% |
|---|---|---|---|---|
| 1.5 mm² | 2 × 20 × 30 × 0.0121 × 0.95 | 13.8 V | 6.0% | Fails |
| 2.5 mm² | 2 × 20 × 30 × 0.00741 × 0.95 | 8.4 V | 3.7% | Fails the 3% lighting limit, passes 5% for power |
| 4 mm² | 2 × 20 × 30 × 0.00461 × 0.95 | 5.3 V | 2.3% | Passes both |
| 6 mm² | 2 × 20 × 30 × 0.00308 × 0.95 | 3.5 V | 1.5% | Passes with margin for future load |
So the honest answer for that circuit is 4 mm², not 2.5 mm² — not because of current, which 2.5 mm² handles, but because of distance. This is the pattern that surprises buyers most often: at 20 A the current rating barely enters the decision, and at 30 m the drop decides it.
The same calculation on three phases, and on 120 V
Three-phase, 63 A, 80 m, 16 mm², 400 V, power factor 0.9: ΔU = 1.732 × 63 × 80 × 0.00115 × 0.9 ≈ 9.0 V, which is 2.3% — comfortably inside the limit. The same cable at 25 mm² would drop about 1.4% and might be chosen anyway to keep margin for a future increase in load.
Now move to a 120 V market. A 16 A load on 25 m of 2.5 mm² in a 120 V single-phase system gives ΔU = 2 × 16 × 25 × 0.00741 × 0.95 ≈ 5.6 V, or 4.7% — inside 5%, outside 3%, and only just inside for a load that starts. A 120 V system has one third of the absolute volts per percent that a 400 V system has, so every long run in a 110–127 V market needs the calculation done rather than borrowed from a 230 V project. The remedy is the same everywhere: a larger cross-section, a shorter route, or a distribution point moved closer to the load.
Interactions people forget
Voltage drop and breaker operation. A long run raises the earth-loop impedance, which lengthens disconnection time on a fault. A circuit can pass the 5% running check and still fail the fault-disconnection check; both use the same cable data but ask different questions.
Starting currents. A motor drawing six times its running current on start sees six times the drop. If the contactor coil is fed from the same run, the drop can pull it below its hold-in threshold and the starter chatters — a volt drop problem presenting as a control problem.
Frequency and power factor. Reactance scales with frequency, so a 60 Hz system has slightly more reactive drop than 50 Hz. On small cross-sections that is negligible; on large feeders, include X properly. And assuming unity power factor “to be conservative” is not conservative at all on a 95 mm² feeder — use the actual figure.
What to send with the enquiry
Load current and whether it is continuous or motor start; system voltage, frequency and number of phases; one-way route length and installation method; conductor material and preferred insulation; the acceptable drop at the load, or the local code limit; and the destination country, since both the installation rule and the certificate requirement follow the market. With the route length in hand the size can be checked against drop as well as current — the check that usually changes the quote.
FAQ
Is it 3% or 5% for voltage drop?
IEC 60364-5-52 recommends 3% for lighting and 5% for other uses, measured from the origin of the installation when the supply is a public low-voltage network. Some national codes are stricter or split the allowance between the installation and the final circuit, so the destination decides. Where the code is silent, 3% for lighting and 5% for power is a defensible default.
Do I include the neutral in the calculation?
In a two-wire circuit the neutral carries the same current as the line, which is why the factor is 2. In a balanced three-phase circuit the neutral current is near zero and only the phase conductors count. With harmonic-rich loads the neutral can carry more than the phase current, and then it has to be calculated as a current-carrying conductor in its own right.
Why did the electrician use 2.5 mm² when 4 mm² was needed?
Because the cable was chosen on current rating alone. At 20 A, 2.5 mm² is entirely adequate for current and fails on distance. The fix is not a bigger breaker, it is a bigger conductor, and it costs far less before the cable is pulled than after.
Does voltage drop affect the breaker’s trip point?
Not directly — the thermal element responds to current, not voltage. It affects the fault path: a lower voltage at the far end means less current flows into a short circuit there, so disconnection is slower. That is the earth-loop impedance check, and it tightens as the run gets longer.
Can I compensate with a higher transformer tap?
Tapping up raises the voltage at the origin, which shifts the whole installation above nominal and stresses everything connected to it. It is legitimate on a transformer with an off-load tap, but it is not a substitute for sizing the cable: the drop along the run is still there, and equipment at the origin now sees an overvoltage.
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