Voltage Drop Calculator

V_drop = 2 × L × I × ρ / A, where ρ is 0.0172414 Ω·mm²/m for copper at 20 °C

L

The one-way length of the cable run, in metres or feet — the distance from the supply to the load, not the total length of copper. This is the single input on this page that is easy to get wrong without noticing: a 30 metre run needs 60 metres of conductor, because the current travels out and back, and the formula's leading two already accounts for that. Type 60 instead of 30 and every answer on the page doubles, which looks like a slightly long run rather than a mistake

I

The current the load draws, in amps. It is the same current that flows in both conductors, which is why it multiplies the whole loop resistance rather than half of it. Voltage drop is directly proportional to current, so doubling the load doubles the loss — and because the loss also costs power, a heavily loaded long run wastes energy as well as dropping volts

A

The conductor cross-sectional area, in square millimetres, which is the number printed on the cable you buy. It sits in the denominator, so it is the lever you have: to halve the voltage drop, double the cross-section. The table below shows what that does across the range from 1.5 mm² to 25 mm² on the same load, and it is the reason a long run needs a heavier cable than a short one carrying the same current

ρ

The resistivity of the conductor, 0.0172414 ohm square millimetres per metre for annealed copper at 20 °C, which is the value behind the IACS 100% conductivity standard. Aluminium is about 1.64 times higher, so an aluminium conductor of the same size drops about 64% more volts. Copper does not change resistance with temperature here: a warm conductor has a higher resistance than the table value, so this calculator returns a best case

2

The two conductors in the loop, one out and one back. It is not a safety margin and not a correction factor, and leaving it out underestimates the drop by exactly half. Single-phase and DC circuits both need it; the arithmetic for a balanced three-phase circuit is different again and this page does not cover it

Use it before running cable any distance at all, and especially on low-voltage DC, where the supply is small enough that the loss matters as a fraction of it. A 12 V circuit that drops 1 V has lost 8% of its supply; the same cable carrying the same current on 230 V drops the same 1 V, which is 0.4% and nobody cares. That is why the calculators for garden lighting, boat wiring and solar panels ask for the run length, and it is why the answer here is a percentage and not only a voltage. The percentage is what you compare against the recommendation.

1. The defaults: 230 V, 20 A, a 30 m run on 2.5 mm²

   1. Supply 230 V, load current 20 A, run 30 m one way, conductor 2.5 mm²
   2. Loop resistance is 2 × 30 × 0.0172414 / 2.5 = 0.4138 ohms
   3. Voltage drop is 20 × 0.4138 = 8.28 V, which is 8.28 / 230 = 3.6% of the supply
   4. The load sees 230 − 8.28 = 221.72 V, and the two conductors together dissipate 8.28 × 20 = 165.52 W

   Read the two middle numbers together, because they are the page's whole point. 3.6% is above the 3% branch-circuit recommendation and below the 5% overall one, so the badge comes out amber, and this is the most useful thing a voltage drop calculation does: it tells you not that the drop exists but that it is at the edge. The four rows also close on each other — 8.28 plus 221.72 is 230, and 8.28 times 20 is 165.52 — so if you change one input you can re-derive the rest and check the panel against itself.

2. The same cable at half the load and two thirds the length

   1. Supply 230 V, load current 10 A, run 20 m one way, conductor 2.5 mm²
   2. Halving the current and cutting the length from 30 m to 20 m multiplies the drop by 0.5 × 0.667
   3. 8.28 V × 0.333 = 2.76 V, which is 1.2% of 230 V
   4. The load sees 227.24 V and the cable wastes 27.59 W instead of 165.52 W

   The waste falls much faster than the voltage does, and that is worth noticing: the drop went down by a factor of three while the power lost went down by a factor of six, because power lost is the drop times the current and both of those fell. This is the arithmetic behind the advice that long, thin, heavily loaded runs are the expensive ones. It also shows that the percentage is the number to watch, not the volts — 2.76 V sounds like more than 1.2% does.

3. Low voltage DC: 12 V, 1 A, 5 m on 1.5 mm²

   1. Supply 12 V, load current 1 A, run 5 m one way, conductor 1.5 mm²
   2. Loop resistance is 2 × 5 × 0.0172414 / 1.5 = 0.1149 ohms
   3. Voltage drop is 1 × 0.1149 = 0.11 V, and the load sees 11.89 V
   4. As a percentage of 12 V that is 0.96%, comfortably inside the recommendation

   Short and lightly loaded, so the cable is barely doing anything — but the percentage is already close to 1% on a 12 V rail, where the identical cable on 230 V would be at 0.05%. This is the comparison that explains why low-voltage installations need surprisingly fat cable: the drop in volts is set by the current and the cable, and it does not know or care what the supply voltage is. Shrinking the supply voltage shrinks the denominator, and the percentage is what fails.

4. Predicting a supply that cannot do the job

   1. Supply 12 V, load current 20 A, run 100 m one way, conductor 1.5 mm²
   2. Loop resistance is 2 × 100 × 0.0172414 / 1.5 = 2.299 ohms
   3. Voltage drop is 20 × 2.299 = 45.98 V, which is 383% of the 12 V supply
   4. The load row reads −33.98 V, a negative number

   The negative voltage is the answer, not a bug: it means a 1.5 mm² cable 100 metres long cannot deliver 20 amps from a 12 volt supply at all, so the model has been pushed past the point where it describes anything real. The page deliberately leaves the number unclamped rather than showing zero, because zero would look like a legitimate result — a load that gets no volts — when what is actually true is that this installation is impossible and needs a shorter run or a much heavier conductor. Every answer here is a warning, including the 919 W the cable would be asked to dissipate.

Resistivity is held at the 20 °C value for annealed copper, so a conductor that warms under load has a higher resistance and a larger drop than shown: the figures here are a best case. Aluminium, which is about 1.64 times more resistive, is not modelled. Balanced three-phase circuits use a different expression and are not covered. The load is treated as a fixed current, whereas a real constant-power load draws more current as the voltage falls, which makes the drop slightly worse than predicted.

Is the cable length the one-way distance or the total?

The one-way distance from the supply to the load, and this is the mistake the page is most likely to catch you making. The current leaves the supply along one conductor and returns along the other, so there are two lengths of copper in the circuit and the formula's leading 2 accounts for that. A 30 metre run therefore means 60 metres of cable. Type 60 into the length box and every answer doubles, which reads as a slightly long run rather than as an error, so the formula would still look plausible while being wrong by a factor of two.

Why does the load voltage come out negative?

Because the cable cannot carry that much current that far from that supply, so the model has been pushed past its range. In the worked example of 12 V, 20 A, 100 metres and 1.5 mm², the loop resistance is about 2.3 ohms and the drop is 46 volts — nearly four times the supply. A negative load voltage is not a reading you can act on; it is the arithmetic saying that this combination is impossible and that you need a shorter run, a heavier conductor, or a higher supply voltage. It is left visible rather than clamped to zero so the impossibility shows.

What counts as an acceptable voltage drop?

The National Electrical Code suggests no more than 3% on a branch circuit and no more than 5% overall, feeding the branch from the feeder — these are recommendations in the informational notes, not mandatory rules, and they are US practice. Other standards and other countries use different figures; IEC 60364 and the Chinese GB 50054 approach it through permissible voltage variation at the equipment instead. The 3% and 5% bands on this page are those recommendations, applied to the percentage, and they are a reasonable default rather than a universal law.

How much does a bigger cable reduce the drop?

In direct proportion to the cross-section, because the area sits in the denominator. Doubling the cross-section halves the voltage drop; going from 1.5 mm² to 25 mm² divides it by nearly seventeen. The table on this page shows both ends on the same 230 V, 20 A, 30 metre run: 13.79 V for the small cable and 0.83 V for the large one. What it does not do is change linearly with cost, which is why the practical answer is usually the smallest cable that keeps the percentage inside the recommendation rather than the largest one available.

Does the temperature of the conductor matter?

Yes, and this page does not model it. Copper's resistance rises with temperature at roughly 0.4% per kelvin, so a conductor running warm under load has a higher resistance than the 20 °C value used here and produces a larger voltage drop than the calculation shows. Continuous current ratings are set with that heating in mind, and a fully loaded cable can be tens of degrees above ambient. Treat the figure on this page as a best case at 20 °C, and add margin on any run that sits near the recommendation.

Is this the same as a wire size calculation?

It is half of it. The wire size calculator runs this calculation in reverse: instead of telling you the drop a chosen conductor produces, it tells you the smallest conductor that keeps the drop inside a limit you set. Both need the same four inputs and both use the same resistivity, so a figure from one should reproduce in the other. Sizing a cable also has to satisfy current-carrying capacity and the requirements of the installation, which are separate questions this page does not answer.

What about three-phase circuits?

They need a different expression and this page is single-phase and DC only. In a balanced three-phase circuit the return path is shared between the phases, so the loop factor is the square root of three rather than two, and the line-to-line voltage replaces the supply voltage in the percentage. Using the two-conductor arithmetic on a three-phase feeder overstates the drop. If you are working on a three-phase installation, take the figure from a calculator that asks for the phase configuration.

• NFPA 70 National Electrical Code, Article 210.19(A) Informational Note No. 4 and Article 215.2(A) — the branch-circuit 3% and total 5% voltage drop recommendations this page's bands are built on — National Fire Protection Association
• International Annealed Copper Standard — the IACS definition of 100% conductivity, which is where the 0.017241 Ω·mm²/m resistivity used here comes from — Wikipedia
• Voltage drop — the two-way conductor length convention, the derivation from R = ρL/A, and the difference between the single-phase and three-phase expressions — Wikipedia