A circuit can be correctly rated, correctly protected, correctly connected and still not work. The reason is voltage drop, and it is the one limit that depends on length rather than on current alone. It is also the limit most often left out of hand calculations, because the simplified form of Ohm's law hides everything that matters.
Why the resistance formula is R = ρL/A
Most people meet voltage drop as "V = I × R" with a resistance read off a catalogue. That is fine when the cable is short and the conductor is the only resistance in the circuit. It stops being adequate as soon as you need the resistance of the cable itself, because that resistance is not a catalogue item — it depends on how long the cable is, how thick it is, what it is made of and how hot it is.
Those four dependencies are packaged in one expression:
R = ρ × L ÷ A
where ρ is the resistivity of the material, L is the length of the conductor and A is its cross-sectional area. Deriving it takes three lines, and it is the same equation that came out of the electron-drift argument in the ohm's law and power guide. The resistivity depends only on the substance and its temperature; length and area are geometry.
The consequences are worth stating explicitly, because they drive every sizing decision:
- Resistance is proportional to length. Doubling the run doubles the resistance and doubles the drop. Distance is not a detail.
- Resistance is inversely proportional to area. Doubling the cross-section halves the resistance and halves the drop. This is the lever you actually pull when you upsize a cable.
- Resistivity is fixed by the material. You cannot change copper into something better; you can only change the amount of it.
- Resistance rises with temperature. Copper increases about 0.39% per degree Celsius above 20 °C.
So the full voltage drop is Vdrop = I × ρ × L ÷ A, and the design question becomes: what area A do I need so that this stays under my limit?
Material resistivities and the copper versus aluminium trade
At 20 °C, the two figures you need are:
- Copper: ρ ≈ 1.72 × 10-8 Ω·m
- Aluminium: ρ ≈ 2.65 × 10-8 Ω·m
Aluminium is 2.65 ÷ 1.72 = 1.54 times more resistive than copper, so for the same drop you need about 54% more cross-sectional area. Aluminium conductors of the same gauge are therefore larger, more expensive in the cable, harder to terminate reliably, and have a higher contact resistance at every joint. The economic argument for aluminium is only compelling on long, high-current runs where the conductor cost dominates — and the termination risk is what keeps copper the default for most work. The voltage drop calculator takes the resistivity as an input precisely because this comparison comes up constantly.
Why the length is doubled
This is the error that most consistently produces a result that is too optimistic by exactly a factor of two.
Current does not flow one way. It leaves the supply on the live conductor, passes through the load, and returns on the neutral or protective conductor. The current path is a loop, and the loop is twice the one-way distance. Both conductors have resistance, both dissipate energy, and both contribute to the drop.
So for a 40 m one-way run, the length to use in the formula is 80 m. A common variant of the mistake is to halve the result afterwards, or to compare a 16.5 V drop against a limit and conclude it is acceptable because "the return conductor is only contributing half of that anyway" — it is contributing the other half, which is the part you omitted.
There is one important exception. In a three-phase system the return paths are shared and the geometry differs; there the factor is √3 rather than 2, which is derived below. For single-phase, and for the line-to-neutral voltage in a three-phase four-wire system, the factor is 2.
AWG and square millimetres
Two numbering systems describe conductor size, and mixing them up is a routine source of error. The metric system (mm²) increases with size. AWG runs in the opposite direction: a higher number means a thinner wire. 14 AWG is thinner than 10 AWG, which is thinner than 2 AWG. There is no simple arithmetic relationship between the two, so use a table:
14 AWG ≈ 2.08 mm² · 12 AWG ≈ 2.88 mm² · 10 AWG ≈ 5.26 mm² · 8 AWG ≈ 8.37 mm² · 6 AWG ≈ 13.3 mm² · 4 AWG ≈ 21.2 mm² · 2 AWG ≈ 33.6 mm²
The metric sizes that come up most often in cable work are 1.5, 2.5, 4, 6, 10, 16, 25, 35 and 50 mm². Note that 2.5 mm² sits between 12 and 14 AWG, and 6 mm² is slightly thinner than 8 AWG. A rough bridge worth remembering: 1.5 mm² ≈ 14 AWG, 4 mm² ≈ 8 AWG, 10 mm² ≈ 6 AWG, 35 mm² ≈ 2 AWG — close enough for a quick sanity check when a drawing gives you one scale and the order book gives you the other.
The 3% and 5% limits
Voltage drop is usually expressed as a percentage of the nominal supply voltage, and the limits in common use are:
- 3% for lighting circuits. Light output falls with voltage, lamps age faster, and electronic drivers behave poorly. This is the tighter limit and it applies where the load is sensitive.
- 5% for power circuits. Motors, heaters and appliances tolerate more, and a 5% drop is the usual ceiling for final circuits.
Some standards work differently, allowing a total of 5% for the whole installation — for example 2% in the sub-main and 3% in the final circuit. Which convention applies is a local matter, but the reasoning is the same: the point of the limit is that the load receives close to its rated voltage, and every percent of drop is a percent of voltage that the equipment no longer has to work with.
What a drop actually costs is more than a slightly dimmer lamp. On an induction motor, a 5% voltage deficit at constant load means the current rises — the motor draws more current to develop the same torque, dissipating more as heat in the windings, running hotter and shorter-lived. On a resistive heater, the power falls with the square of the voltage, so a 5% drop costs very nearly 10% of the heat. The ohm's law calculator is useful for quantifying these follow-on effects.
Worked example 1: 40 m of 2.5 mm² copper carrying 30 A
A single-phase load on a 230 V supply is 40 m from the origin. The cable is 2.5 mm² copper. The design current is 30 A. Is the drop acceptable?
Step 1 — convert the area to square metres. 2.5 mm² = 2.5 × 10-6 m².
Step 2 — double the length for the return path. L = 2 × 40 = 80 m.
Step 3 — calculate the loop resistance.
R = ρ × L ÷ A = (1.72 × 10-8 × 80) ÷ (2.5 × 10-6)
R = (1.376 × 10-6) ÷ (2.5 × 10-6) = 0.550 Ω
Step 4 — calculate the drop. Vdrop = I × R = 30 × 0.550 = 16.5 V
Step 5 — express it as a percentage. 16.5 ÷ 230 = 0.0717 = 7.2%
Step 6 — compare against the limits. 7.2% exceeds even the 5% power limit, and the load would receive 230 − 16.5 = 213.5 V instead of 230 V. 2.5 mm² is not adequate for this run.
Worked example 2: choosing the next size up
Rather than testing candidate sizes one at a time, work out the area you actually need. Rearranging Vdrop = I × ρ × L ÷ A for the area:
A = (I × ρ × L) ÷ Vdrop,max
Against the 5% limit: Vdrop,max = 0.05 × 230 = 11.5 V
A = (30 × 1.72 × 10-8 × 80) ÷ 11.5 = (4.128 × 10-5) ÷ 11.5 = 3.59 × 10-6 m² = 3.59 mm²
So anything at or above 3.59 mm² passes. The next standard size is 4 mm². Check it: R = (1.72 × 10-8 × 80) ÷ (4 × 10-6) = 0.344 Ω, so Vdrop = 30 × 0.344 = 10.3 V = 4.5%. Under the limit, with a little margin. ✓
Against the 3% lighting limit: Vdrop,max = 0.03 × 230 = 6.9 V
A = (4.128 × 10-5) ÷ 6.9 = 5.98 × 10-6 m² = 5.98 mm²
So 6 mm² is the minimum for a lighting circuit on this run. Check it: R = (1.72 × 10-8 × 80) ÷ (6 × 10-6) = 0.229 Ω, so Vdrop = 30 × 0.229 = 6.88 V = 3.0%. Exactly at the limit, and within rounding. ✓
The recommendation: 6 mm² copper satisfies the 3% limit with margin and is the right choice if the circuit is lighting or mixed. 4 mm² is sufficient for a pure power load on the 5% limit. Both calculations are done in the voltage drop calculator, which reports the required area as well as the drop for a chosen size.
Note that this whole answer is set by the copper. Substituting aluminium at 2.65 × 10-8 gives a required area of 3.59 × 1.54 = 5.53 mm² for the 5% limit, and 9.22 mm² for the 3% limit — which on a standard size list means 6 mm² and 10 mm² respectively, and 10 mm² for the same duty copper handles with 4 mm².
What the drop costs in money, not volts
The energy dissipated in the cable is the same I²R as any resistive loss, and it is paid for on every electricity bill for the life of the installation.
At 30 A in the 2.5 mm² cable: I²R = 900 × 0.550 = 495 W lost continuously.
In 4 mm²: 900 × 0.344 = 310 W lost.
In 6 mm²: 900 × 0.229 = 206 W lost.
Run for 8 hours a day and 300 days a year, the difference between 2.5 mm² and 6 mm² is 289 W saved, which is 0.289 kW × 2,400 h = 694 kWh per year — around 104 per year at 0.15 per kWh. Over twenty years that pays for the larger cable several times over. The watt calculator handles the arithmetic once you have the two I²R figures.
This is the strongest argument for upsizing, and it is worth raising early, because the extra cost of a larger cable at installation is small compared with twenty years of losses. The quadratic dependence is the key: cutting the resistance in half halves the loss, but cutting the current in half cuts it to a quarter, which is why reducing the load is always the better first move.
Voltage drop is not the same as overheating
These are two independent limits and a cable must satisfy both. Conflating them is the most consequential error in cable sizing, because it can go in either direction.
Voltage drop depends on length, current and cross-section, and it limits how much of the supply voltage is lost in transit. On a short run it is almost never the binding constraint — a 2.5 mm² cable running 3 m has essentially no drop and would pass any voltage-drop check.
Current carrying capacity depends on the conductor size, the insulation temperature rating, the ambient temperature, the number of conductors sharing the enclosure and the installation method. It limits how much current the cable can carry before it overheats. It is completely independent of length.
So a 2.5 mm² cable over 3 m passes the voltage-drop check comfortably but still has a current limit — roughly 20 A or so for PVC-insulated copper in conduit, depending on ambient temperature and grouping. Running 30 A through it would be a thermal problem, not an electrical one, and the fix is a larger conductor or a lower current, not a shorter cable. Conversely, a 1.5 mm² cable over 200 m will fail on voltage drop long before it fails on heating.
Two rules of thumb for copper in PVC conduit, to be treated as starting points rather than as authority — actual values depend on installation method, insulation rating and local code: 2.5 mm² carries roughly 20–21 A, 4 mm² around 28–32 A, 6 mm² around 34–40 A, 10 mm² around 50–57 A and 16 mm² around 64–76 A. The power factor guide explains why a cable sized on real power in watts can still be undersized, because a low power factor raises the current above what the wattage suggests.
Temperature correction
The resistivity figures quoted are all at 20 °C. Copper's resistance rises about 0.39% per degree Celsius, so:
RT = R20 × (1 + 0.00393 × (T − 20))
Taking the example cable at a working conductor temperature of 70 °C: the multiplier is 1 + 0.00393 × 50 = 1.1965. Applied to 2.5 mm²: 0.550 × 1.1965 = 0.659 Ω, giving Vdrop = 30 × 0.659 = 19.8 V = 8.6% rather than 7.2%.
This is why a marginal design that passes on cold numbers can fail once it is loaded. The correction is small — about 20% in this case — but on a run already close to the limit it is the difference between passing and failing. Design calculations are usually done on the cold figures, with the thermal check done separately against the current rating, which is where the correction effectively reappears.
Three-phase systems and the √3 factor
In a three-phase system the relationship between line and phase voltage is not 1:1, and this catches people out regularly.
Vline = √3 × Vphase = 1.732 × Vphase
So a 400/230 V system is exactly that: 230 V phase to neutral, and 400 V line to line. Check it: 230 × 1.732 = 398.4 V, which rounds to 400 V. Conversely, 400 ÷ 1.732 = 231 V, which is the 230 V neutral voltage. Both directions of the conversion appear in practice, and the √3 factor is the same number used for the three-phase power equation.
For voltage drop in a balanced three-phase system, the drop in the line-to-line voltage is:
Vdrop = √3 × I × R × cos φ
The √3 comes from the phase relationship between the currents in the three conductors, and the cos φ because the current is not in step with the voltage in a balanced three-phase load. On a resistive or unity-power-factor load the cos φ term is 1.
Worked example: a balanced three-phase 400 V load drawing 30 A at power factor 0.8, fed by 50 m of 10 mm² copper. R = (1.72 × 10-8 × 50) ÷ (10 × 10-6) = 0.086 Ω. Vdrop = 1.732 × 30 × 0.086 × 0.8 = 3.58 V, which is 0.89% of 400 V — comfortably inside any limit. That load is 30 × 1.732 × 400 × 0.8 = 16.6 kW, and delivering the same power from a single-phase 400 V supply at the same power factor would need 16,600 ÷ (400 × 0.8) = 52 A through one conductor. Three-phase distribution carries that power in 30 A of conductor instead, which is why it is standard above about 10 kW.
Common mistakes
Using the one-way length. The most consequential single error in voltage-drop work, and it silently halves every result. Draw the loop and count both conductors.
Applying the wrong limit. 3% for lighting, 5% for power, and a check on whether a local standard caps the whole installation instead. Using 5% for a lighting circuit is a real design error, not a rounding preference.
Reading AWG in the wrong direction. Assuming a larger AWG number means more capacity. It is the reverse: 10 AWG carries more current than 14 AWG, and always has.
Treating the voltage-drop check as a safety check. It is a performance check. A cable can pass it comfortably and still be grossly overloaded thermally, which is the subject of the next section.
Frequently asked questions
Why is the cable length doubled when calculating voltage drop?
Because current flows out along the live conductor and back along the neutral or protective conductor, so the current travels the whole loop. A 40 m one-way run gives an 80 m current path, and the resistance — and therefore the loss — doubles. Forgetting the factor of two halves the calculated drop and is the most common error in cable sizing.
What is the maximum acceptable voltage drop?
Common practice is 3% for lighting circuits and 5% for power circuits, measured from the origin of the supply to the point of utilisation. Some standards instead limit the total to 5% across the installation, allowing 2% in the main supply and 3% in the final circuit. A motor that receives 6% less than rated voltage runs hotter and draws more current.
Is a wire that passes the voltage drop check safe from overheating?
Not necessarily, because the two checks are different. Voltage drop limits the fraction of voltage lost, and it scales with length, so a short run rarely fails it. Current carrying capacity limits heating, and it depends on the conductor size, insulation rating, ambient temperature and how many circuits share the enclosure. A cable must satisfy both limits independently.
How do I convert AWG to square millimetres?
Use a conversion table rather than a formula, because the two scales are not related by any simple arithmetic. Going down the AWG numbers makes the wire thicker, which is the opposite of most other numbering. Useful anchors are 14 AWG at 2.08 mm², 10 AWG at 5.26 mm², 6 AWG at 13.3 mm² and 2 AWG at 33.6 mm².