What the Voltage Drop Calculator does
This calculator works out the voltage lost along a cable run in volts and as a percentage, for single-phase or three-phase circuits in copper or aluminium. It corrects the conductor's resistance for its operating temperature - which changes the answer by around 20% and is the step most online calculators skip - shows the resistance and resistivity it used, and gives the longest run that stays within your drop limit.
It also lists the same run in the next eight larger conductor sizes, because the useful question is rarely "what is the drop?" but "what does it take to fix it?". Everything runs in your browser.
How to use it
- Choose single or three phase. The multiplier is 2 for a single-phase circuit, because the current goes out and comes back, and root-three for a balanced three-phase one.
- Enter the supply voltage, the current and the one-way run length. Do not double the length for the return path - the formula already does.
- Enter the conductor size in mm², or pick an AWG size from the list and the exact area is used.
- Open the advanced section and set the operating temperature. 70 °C suits a fully loaded PVC cable, 90 °C an XLPE one, and 20 °C matches a cold resistance measurement.
- For cables above about 25 mm², enter the reactance in ohms per kilometre - around 0.08 for a multi-core cable - and the load's power factor, because at a poor power factor the reactive term is a real part of the drop.
- Read the percentage against your limit, then use the size table to see what the next size up would buy you.
Reading the results
The percentage is what design limits are written in. 3% for lighting and 5% for other uses, measured from the origin of the installation, is the long-standing guidance in IEC-derived practice; NEC 210.19 and 215.2 recommend 3% on a branch circuit and 5% overall in informational notes rather than as requirements.
Millivolts per amp per metre is the figure published in cable tables, so it is the number to compare against a manufacturer's data sheet.
The maximum length is the run at which this conductor, at this current, reaches your limit. It is the fastest way to see whether the problem is the cable or the route.
The power lost is a continuous cost. A 2% drop on a 100 A three-phase circuit is around 700 W of heating in the cable, every hour it runs.
Worked example: a 120 m submain in 95 mm² copper carrying 120 A
At 70 °C the copper resistivity is 1.724 x 10⁻⁸ x (1 + 0.00393 x 50) = 2.0628 x 10⁻⁸ ohm.m. Over 120 m of 95 mm² that is 2.0628 x 10⁻⁸ x 120 / 95 x 10⁻⁶ = 0.02606 ohm per conductor.
With 0.08 ohm/km of reactance, the run has 0.0096 ohm of reactance. At a power factor of 0.9, sinφ is 0.436, so the effective impedance in the drop formula is 0.02606 x 0.9 + 0.0096 x 0.436 = 0.02764 ohm.
Three-phase drop is root-three x 120 x 0.02764 = 5.74 V, which is 1.44% of 400 V. Comfortably inside a 5% limit, with the run able to stretch to about 416 m before it runs out.
Had the resistance been taken at 20 °C instead of 70 °C, it would have been 0.02177 ohm and the drop 4.86 V - about 15% optimistic. On a marginal circuit that is the difference between passing and failing.
The continuous loss is 3 x 120² x 0.02606 = 1,126 W. Going up to 120 mm² would cut that to about 892 W and the drop to 1.20%, which on a heavily used feeder can pay for the larger cable in a few years.
Formulas and scoring rules
- Single phase
Vd = 2 x I x (R cos(phi) + X sin(phi))The 2 accounts for the line and the neutral, both of which carry the current.- Three phase (balanced)
Vd = root-three x I x (R cos(phi) + X sin(phi))Not 3 and not 2: the line-to-line drop of a balanced system is root-three times the per-conductor drop.- Conductor resistance
R = rho(T) x L / AL is the one-way length; the multiplier above supplies the return path.- Temperature correction
rho(T) = rho20 x (1 + alpha x (T - 20))Copper: rho20 = 1.724e-8 ohm.m, alpha = 0.00393 per degree. Aluminium: 2.826e-8 and 0.00403.- Percentage drop
Vd% = Vd / V_nominal x 100Against the nominal supply voltage, not the measured one.- Maximum length
L_max = L x limit% / actual%Exact, because the drop is directly proportional to length at a fixed current.- Power lost
single phase 2 I^2 R; three phase 3 I^2 RResistive loss only; reactance stores and returns energy rather than dissipating it.
Why the multiplier is root-three and not 2
On a single-phase circuit the same current flows out along the line and back along the neutral, so it meets the conductor resistance twice and the drop is 2IR. On a balanced three-phase circuit there is no return current in the neutral at all; each line carries current that returns through the other two.
The drop that matters is between lines, and working it through with the 120-degree phase relationships gives root-three x IR - about 1.73 rather than 2. That is why the same cable, the same length and the same current produce a smaller drop on a three-phase circuit, before you even account for the current being lower for the same power.
Reactance, and when it stops being negligible
Every cable has inductance as well as resistance. Below about 16 to 25 mm² the resistance is so much larger that the reactance changes the answer by a fraction of a per cent and can be ignored. Above that, the resistance falls with cross-section while the reactance stays roughly constant at 0.07 to 0.09 ohm per kilometre, so the two become comparable - and on very large conductors the reactance dominates.
The power factor decides how much it costs you. At unity power factor the reactive term disappears entirely. At 0.8 lagging, sin(phi) is 0.6 and the reactance contributes most of what its magnitude suggests. That is why a poor power factor makes voltage drop worse in two ways at once: more current, and more of the reactance showing up in the answer.
What too much drop actually does
Incandescent lamps dim and LED drivers usually cope until they suddenly do not. The serious victim is motors: torque falls with the square of the voltage, so a 10% drop costs 19% of the starting torque, and a motor that cannot accelerate sits drawing locked-rotor current until the overload trips. Contactor coils drop out below about 80% of nominal, which is how a long run of cable turns into an intermittent, maddening fault.
There is also a slow cost. The voltage lost in the cable is dissipated as heat in the cable, and you pay for it every hour the circuit runs.
Limitations: what the result does not prove
- It calculates a balanced three-phase or a simple single-phase circuit. Unbalanced loads produce neutral current and a different answer.
- The reactance figure is yours to supply. A cable's real reactance depends on the conductor spacing and arrangement, which for single-core cables in trefoil or flat formation differ noticeably.
- It assumes the whole current flows to the far end. A circuit with loads distributed along its length has a lower drop than this, and needs a load-moment calculation.
- Harmonic currents are ignored. They raise the effective resistance through skin and proximity effects and add drop that this arithmetic does not see.
- It is not a compliance statement. A qualified engineer must verify the design against the applicable standard and local regulations.
Privacy: where your data goes
Everything you paste, type or drop is processed in this browser tab. It is not uploaded, logged, stored or sent to analytics. Session recording and tag-manager scripts are switched off on this page.
Standards and sources
- IEC 60364-5-52 - Selection and erection of wiring systems - checked 19 Sep 2026
- IEC 60228 - Conductors of insulated cables - checked 19 Sep 2026
- NFPA 70 - National Electrical Code - checked 19 Sep 2026
Frequently asked questions
What is an acceptable voltage drop?
In IEC-derived practice the usual design guidance is 3% for lighting and 5% for other uses, measured from the origin of the installation to the point of use. The NEC gives 3% on a branch circuit and 5% overall in informational notes, which are recommendations rather than requirements. Either way, it is the total from the origin that matters, not any one leg.
Do I use the one-way length or the total length of cable?
The one-way length - the distance from the board to the load. The formula supplies the return path: a factor of 2 for a single-phase circuit, or root-three for a balanced three-phase one. Doubling the length yourself as well is a common error that doubles the answer.
Why is the drop bigger than my cable table says?
Almost always temperature. Published mV/A/m figures are quoted at the conductor's maximum operating temperature, but plenty of calculators use the 20 °C resistivity, which is about 20% lower for a 70 °C cable. This page shows the resistivity it used so you can see which one you are comparing against.
Does voltage drop depend on the power factor?
Yes, in two ways. A worse power factor means more current for the same kW, and it also brings the cable's reactance into the drop through the sin(phi) term. On a large cable feeding a poor power factor load, the reactive part can be a third of the total drop.
Will two smaller cables in parallel fix a voltage drop problem?
Yes - two identical conductors in parallel halve the resistance and halve the drop, the same as doubling the cross-section. The practical catches are that both must be the same length, size and material and be terminated identically so they actually share the current, and some rules restrict paralleling below a minimum size.
Is 5% drop on a motor circuit acceptable?
For running, usually. For starting, check separately: the starting current can be six times the full-load current, so a 5% running drop becomes a 30% starting drop, and torque falls with the square of voltage. That is how a motor at the end of a long run fails to start on a hot day while working perfectly on a cool one.
Does the calculator handle a load spread along the run?
No - it assumes the whole current travels the full length, which is the conservative case. For a busbar or a lighting run with loads at intervals, the real drop is lower and the proper method is to sum the load moments, that is each load's current multiplied by its distance from the source.
Aluminium or copper for a long run?
Aluminium has about 1.64 times the resistivity of copper, so for the same drop it needs roughly 1.6 times the cross-section - but it is far cheaper and lighter per unit of conductance, which is why long submains and distribution cables are so often aluminium. The compromises are termination technique, the larger containment needed, and the need for compatible lugs.
Last reviewed by the A2Z.Tools team against the sources listed above.
