Voltage Drop Calculator
Metric cable sizes. Enter the circuit, get the drop in volts and percent, checked against the IEC limit.
Line to neutral
IB, not the breaker rating
≈ 12 AWG
One way, not the loop
0.8 if you don’t know it
Advanced
0.08 is the default when no cable data is available
Derated value from IEC 60364-5-52 Annex B or the cable datasheet
- Drop
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- Voltage at load
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- Permitted
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- Power lost
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Calculated with the Annex G method: resistivity in service taken as 1.25 × the value at 20 °C. Ampacity, protection and fault withstand are separate checks.
How the calculation works
Every conductor has resistance, and current through resistance costs you voltage. The load sees less than the supply puts out. Push it too far and motors run hot, contactors chatter, and lamps dim at the end of the run.
This calculator uses the method in IEC 60364-5-52:
u = b × ( ρ1 × L / S × cosφ + λ × L × sinφ ) × IB
Two terms, added together. The first is resistance, which dominates on small cables. The second is reactance, which only starts to matter on large cross-sections and low power factors. Below about 25 mm² you could drop the reactive term and barely notice.
What each input does
b is 1 for three-phase and 2 for single-phase or DC. Two-wire circuits have a return path in the drop; balanced three-phase does not.
ρ1 is resistivity at operating temperature, not at 20 °C. The standard sets it at 1.25 times the cold value, which lands on 0.0225 Ω·mm²/m for copper and 0.036 for aluminum. A warm cable has more resistance than the datasheet figure, so calculating cold gives you an answer that is optimistic and wrong.
λ is reactance per meter, taken as 0.08 mΩ/m when the cable data is not available. Real values vary with spacing and formation. If you have the manufacturer figure, enter it under Advanced.
L is the one-way run. The formula already accounts for the return conductor through b.
IB is design current, the load the circuit actually carries. Not the breaker rating. Sizing on breaker rating inflates the drop and pushes you into cable you don’t need.
Why three-phase divides by 231, not 400
The percentage is referenced to line-to-neutral voltage. On a 400 V system that is 231 V, so the same volts dropped shows up as a bigger percentage than most people expect. Calculators that reference 400 V report roughly 58% of the real figure. It’s the most common error in this calculation and it always errs toward undersized cable.
The limits
For an installation fed from the public low voltage network, the drop from the origin to any load point should stay under 3% for lighting and 5% for everything else. For a private low voltage supply, a transformer you own, the figures relax to 6% and 8%.
Runs longer than 100 m get an extra 0.005% per meter beyond the first 100, capped at half a percent. A 300 m feeder on a 5% limit becomes 5.5%. The calculator applies this automatically and shows you what it added.
A worked example
16 mm² copper, 50 m, 60 A, 400 V three-phase, power factor 0.8.
Resistive term: 0.0225 × 50 ÷ 16 = 0.0703 Ω, times 0.8 gives 0.0563.
Reactive term: 0.00008 × 50 = 0.004 Ω, times 0.6 gives 0.0024.
Sum is 0.0587 Ω, times 60 A, times b = 1, giving 3.52 V.
Against 231 V that is 1.52%. Comfortably inside the 5% limit.
What this does not check
Voltage drop is one of four checks. The cable still has to carry the current after derating for ambient temperature and grouping. The protective device has to clear a fault at the far end. And the conductor has to survive the fault current for as long as the device takes to open. A cable can pass voltage drop and fail all three.
Motor starting is a separate case. Inrush can be six or seven times running current for a few seconds, and the standard allows a larger drop during that window as long as the equipment tolerates it. Size for the running load, then check the starting dip if the motor is large or the run is long.
FAQ
Do I enter the one-way length or the total length of both conductors?
One way. Measure from the distribution board to the load. The formula already doubles it for single-phase and DC circuits through the b coefficient, so entering the loop length would double the drop a second time.
Should I use the breaker rating or the actual load current?
The actual load current. Design current is what the circuit carries in normal service. A 32 A breaker protecting a 22 A load is sized for the cable, not for the load, and using 32 A here gives you a drop figure that never happens.
What power factor should I use if I do not know it?
Use 0.8. That is the default the standard assumes when no details are available, and it is a reasonable stand-in for a mixed load. Resistive loads such as heaters and incandescent lighting are close to 1. Lightly loaded motors can fall below 0.7.
Is the limit 3% or 5%?
3% for lighting circuits, 5% for everything else, measured from the origin of the installation to the load. Those figures apply when you are fed from the public network. If the installation runs off its own transformer, the limits become 6% and 8%.
Does copper always beat aluminum?
For a given cross-section, yes. Aluminum has about 60% more resistivity, so 95 mm² of aluminum drops roughly what 60 mm² of copper drops. Aluminum wins on cost and weight for large feeders, which is why you see it on service entrances and long runs. Size it up and check the terminations.
Can I use this instead of an ampacity check?
No. They answer different questions. Ampacity tells you whether the cable survives the current. Voltage drop tells you whether the load gets usable voltage. Long runs usually fail voltage drop first, short runs usually fail ampacity first, and both have to pass.
Why does the calculation use 1.25 times the resistivity?
Because a working cable is warm. Resistivity is normally quoted at 20 °C, but a conductor carrying its design current sits well above that, and warm copper resists more. The 1.25 multiplier puts you at operating temperature. Calculate cold and every answer comes out low, which is the wrong direction to be wrong in.
Can I use this for a job in the United States?
The physics is the same but the conventions are not. North American practice sizes in AWG and kcmil, uses resistance and reactance values published at 75 °C, and references the percentage to the nominal system voltage rather than line-to-neutral. Use the NEC voltage drop calculator for that work. Running the same cable through both methods gives two different percentages, and both are correct for their own limits.
Does the 0.08 mΩ/m reactance figure matter?
Below about 25 mm² you could delete the reactive term and barely see it. Above 95 mm², especially at low power factor, it becomes a real part of the total. The 0.08 default is the value to use when you have no cable data. If the manufacturer publishes a figure for your cable and formation, enter it under Advanced.
Does the allowance for long runs apply to every circuit?
It applies to the main wiring system of the installation, not to every final circuit as a matter of course. If your 150 m figure is a long feeder to a remote board, the allowance is appropriate. If it is a final circuit inside a building where the mains are short, leave the checkbox off. The calculator shows it as a separate line so you can see exactly what was added.
What if only one phase of a three-phase circuit is loaded?
Treat it as single-phase. A three-phase circuit carrying load on a single phase has current returning through the neutral, so the drop behaves like a two-wire circuit and the coefficient becomes 2 rather than 1. Select single-phase in the calculator and use the line-to-neutral voltage.
How do I check the drop during motor starting?
Run the calculation again with the starting current, typically six or seven times full load, and a starting power factor around 0.3. The result will look alarming, and that is expected — a larger drop is acceptable for the few seconds of acceleration as long as the contactor holds in and the motor develops enough torque. Size the cable for the running load, then use the starting figure as a sanity check on long runs and large machines.