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How to use the calculator
- Pick the system: single-phase AC, three-phase AC or DC. For three-phase, enter the line-to-line voltage (400 V, 480 V, 208 V).
- Enter the load as a current in amps, or switch to Power and enter kW with the power factor. The tool works out the current.
- Enter the one-way length, from the supply to the load, in metres or feet. Do not double it: the formula does that.
- Choose copper or aluminium and the conductor size, in mm² or in AWG and kcmil.
- Choose the wiring code and the type of circuit. The limit fills in, and you can type your own.
The result shows the drop in volts and percent, the voltage at the load, the power lost in the cable, a pass or fail against the limit and the smallest size that passes. More options holds the conductor temperature, parallel sets and reactance. The Maximum cable length tab works the other way round: how far can a given size run within the allowed drop? "Compare every size" lists all sizes at once.
Copy link gives a URL that reopens the same problem, and you can write one by hand:
| Parameter | Meaning | Example |
|---|---|---|
sys | dc, 1ph or 3ph | sys=3ph |
v | supply voltage in volts | v=400 |
i, or p and pf | load current in A, or power in kW with its power factor | p=22&pf=0.85 |
len, lu | one-way length, and its unit m or ft | len=100&lu=ft |
size | mm² (2.5), AWG (12awg, 1/0) or kcmil (250kcmil) | size=12awg |
mat | cu or al | mat=al |
code, use, lim | iec, bs, as or nec; the circuit type; your own limit in % | code=nec |
t, par, x | conductor temperature in °C, parallel sets, reactance per km or per 1000 ft | par=2 |
mode | len opens the maximum length tab | mode=len |
For example, ?sys=1ph&v=230&i=16&len=30&size=2.5&mat=cu opens the first worked example below.
Why voltage drop matters
Every conductor has resistance, so the equipment at the far end gets less voltage than the supply provides. A little is normal. Too much shows up as lights that dim when a motor starts, motors that run hot, drives that trip on undervoltage, and wasted energy.
The lost voltage is lost power, turned into heat in the cable. In the first worked example, a 2.5 mm² cable carrying 16 A for 30 m loses 126.7 W, which is 3.44% of the 3,680 W the circuit carries. For a resistive load the share of power lost in the cable equals the percentage voltage drop, so a 5% drop wastes about 5% of the energy for as long as the load runs.
Voltage drop formulas
The tool first works out the resistance of one conductor from its material, area and temperature:
R = ρ / A × (1 + α (T − 20))
- ρ is the resistivity at 20 °C: 1/58 = 0.017241 Ω·mm²/m for annealed copper (100% IACS, a conductivity of 58 MS/m) and 0.028264 Ω·mm²/m for aluminium (61% IACS).
- α is the temperature coefficient: 0.00393 per °C for copper, 0.00403 per °C for aluminium.
- A is the cross-section in mm² and T the conductor temperature. Use the cable's maximum operating temperature: 70 °C for PVC, 90 °C for XLPE, 75 °C to match the NEC tables. At 70 °C copper has 1.1965 times its 20 °C resistance.
Then it applies the formula for the system, where I is the current and L the one-way length:
| System | Voltage drop | Why |
|---|---|---|
| DC | VD = 2 × I × L × R | current flows out and back on two conductors |
| Single-phase AC | VD = 2 × I × L × R | the same two conductors |
| Three-phase AC, balanced | VD = √3 × I × L × R | the drop between lines; the neutral carries no current |
With n conductors in parallel per phase, divide by n. The percentage is VD / V × 100, using the line-to-line voltage for three-phase.
When reactance matters
AC cables also have inductive reactance X. With a power factor cos φ below 1, the effective impedance is Z = R cos φ + X sin φ, and Z replaces R. This is the effective Z of the notes to NEC Chapter 9 Table 9 and of IEC 60364-5-52 Annex G. It only matters for large conductors, where R is small. At a power factor of 0.8 with a typical X of 0.08 Ω/km, Z is 2% above R for 95 mm² copper, but 36% above R for 240 mm². Turn reactance on under More options for 50 mm² (1/0 AWG) and up, or whenever you have the value from the cable datasheet.
Voltage drop limits by country
Each code measures the drop from the origin of the installation, the meter or point of supply, to the equipment. Feeders and final circuits share one budget.
| Code | Circuit | Limit |
|---|---|---|
| NEC (US), informational notes to 210.19 and 215.2 | Branch circuit, or feeder alone | 3% |
| NEC (US) | Feeder plus branch circuit, to the farthest outlet | 5% |
| BS 7671 (UK), Appendix 4 | Public supply: lighting, other uses | 3%, 5% |
| BS 7671 (UK), Appendix 4 | Private supply: lighting, other uses | 6%, 8% |
| IEC 60364-5-52, Annex G, Table G.52.1 | Public supply (type A): lighting, other uses | 3%, 5% |
| IEC 60364-5-52, Annex G, Table G.52.1 | Private supply (type B): lighting, other uses | 6%, 8% |
| AS/NZS 3000:2018, clause 3.6.2 | Point of supply to any point | 5%, or 7% from a substation on the premises |
The NEC figures are informational notes, which are advice rather than enforceable rules (NEC 90.5(C)). One NEC voltage drop rule is mandatory: 647.4(D) holds the sensitive electronic equipment systems of Article 647 to 1.5% on the branch circuit and 2.5% for feeder plus branch circuit. BS 7671 deems its voltage requirement met when the drop from the origin to the socket-outlet or equipment terminals stays within the Appendix 4 values, which match IEC 60364-5-52. In Australia and New Zealand the 5% is a requirement of the Wiring Rules.
Because the budget is shared, split it. A common split is 2% on the feeder and 3% on the final circuit, but any split that keeps the total within the limit works.
AWG, kcmil and mm²
American Wire Gauge sizes come from a formula: the diameter is d = 0.127 mm × 92^((36 − n)/39), with 1/0 as n = 0, 2/0 as −1, 3/0 as −2 and 4/0 as −3. A smaller number is a thicker wire, and the area doubles every three gauge numbers. Above 4/0, sizes are in kcmil (thousands of circular mils), and 1 kcmil is 0.5067 mm². Metric cables are sold by nominal area, and the two series do not line up:
| AWG | Diameter | Area | Nearest metric size |
|---|---|---|---|
| 14 | 1.628 mm | 2.08 mm² | 2.5 mm² |
| 12 | 2.053 mm | 3.31 mm² | 4 mm² |
| 10 | 2.588 mm | 5.26 mm² | 6 mm² |
| 8 | 3.264 mm | 8.37 mm² | 10 mm² |
| 6 | 4.115 mm | 13.3 mm² | 16 mm² |
| 4 | 5.189 mm | 21.2 mm² | 25 mm² |
| 2 | 6.544 mm | 33.6 mm² | 35 mm² |
| 1/0 | 8.251 mm | 53.5 mm² | 50 mm² |
| 4/0 | 11.684 mm | 107.2 mm² | 120 mm² |
The mm² and AWG buttons beside the size list switch between the two series and pick the nearest size this way.
How these resistances compare with published tables
The calculator computes resistance from the constants above instead of copying a table, and shows the value it used. For copper at 75 °C, its figures are about 2% below the stranded-conductor values in NEC Chapter 9 Table 8 for every size from 18 AWG to 300 kcmil, and within 0.1% of the table's solid 14 AWG value. The gap is the stranding: the strands spiral, so each one is slightly longer than the cable. Aluminium shows the same 2% gap.
Metric cables are made to IEC 60228, which caps each size's resistance instead of fixing its exact area. The cap is 3% to 12% above the value for the exact nominal area: for 2.5 mm² copper it is 7.41 Ω/km at 20 °C, against 6.90 Ω/km. The simplified method in IEC 60364-5-52 Annex G uses 0.0225 Ω·mm²/m for copper, 9% above this tool's 70 °C value. A check made with a code table can therefore show a slightly larger drop. When a result is within 10% of the limit, the tool says so; confirm it with the table your inspector uses.
Worked examples
1. A 230 V socket circuit on 2.5 mm²
A 16 A load sits at the end of a 30 m single-phase run of 2.5 mm² copper with PVC insulation (70 °C).
- R at 20 °C = 0.017241 / 2.5 = 6.897 Ω/km. At 70 °C, × 1.1965 = 8.252 Ω/km.
- VD = 2 × 16 × 30 m × 8.252 Ω/km / 1000 = 7.92 V.
- 7.92 / 230 × 100 = 3.44%, leaving 222.08 V at the load.
That passes the 5% limit for power circuits in BS 7671 and IEC 60364. On a lighting circuit, with a 3% limit, it would fail, and 4 mm² is the smallest size that passes.
2. A 120 V branch circuit on 12 AWG
Two 12 AWG copper conductors carry 16 A to a load 100 ft away. In NEC mode the tool works at 75 °C.
- d = 0.127 × 92^(24/39) = 2.0525 mm, so A = 3.3088 mm² and R = 1.932 Ω per 1000 ft at 75 °C.
- VD = 2 × 16 × 100 ft × 1.932 / 1000 = 6.18 V, which is 5.15% of 120 V.
That is over the 3% NEC recommendation. 10 AWG drops 3.89 V (3.24%), still over, and 8 AWG drops 2.44 V (2.04%), so 8 AWG is the smallest size that passes. A well-known EC&M worked example by Mike Holt gets 6.4 V for the same circuit, because it takes 2 Ω per 1000 ft from NEC Chapter 9 Table 9, which includes stranding.
3. A 12 V DC run at 20 A
A 12 V battery feeds a 20 A load 6 m away on 4 mm² cable, worked at 70 °C. VD = 2 × 20 × 6 m × 5.157 Ω/km / 1000 = 1.24 V, a 10.31% drop that leaves 10.76 V. 6 mm² still drops 6.88%, and 10 mm² brings it down to 0.495 V, or 4.13%. Low voltage circuits need much bigger conductors for the same current, because each volt lost is a larger share of 12 V.
4. Maximum length of a lighting circuit
How far can a 10 A lighting circuit on 2.5 mm² copper run before it reaches the 3% BS 7671 lighting limit at 230 V? The allowed drop is 3% × 230 = 6.9 V, so L = 6.9 / (2 × 10 × 8.252 / 1000) = 41.8 m, or 137.2 ft.
Voltage drop is not ampacity
This tool checks one thing: how much voltage the cable loses. It does not check whether the cable can carry the current without overheating (its ampacity, or current-carrying capacity), derating for grouping and ambient temperature, short-circuit withstand or the installation method. Those come from separate tables: NEC 310.16 in the US, the current-carrying capacity tables in BS 7671 Appendix 4, and AS/NZS 3008.1.1 in Australia and New Zealand, whose 2025 edition, published on 19 December 2025, replaced the 2017 one.
On a long run, voltage drop usually sets the size. On a short run, heat does: 30 A over 3 m of 1 mm² copper drops only 1.61%, yet whether 1 mm² can carry 30 A without overheating is a question this tool cannot answer. So the smallest size shown here is the smallest by voltage drop alone. Use the larger of that size and the one your code's ampacity table requires, and have the work done by a licensed electrician.
Common mistakes
- Doubling the length twice. Enter the one-way distance. The 2 in the DC and single-phase formulas already counts the return conductor.
- Phase voltage on a three-phase circuit. The √3 formula gives the drop between lines, so compare it with 400 V, not 230 V.
- Resistance at 20 °C. A loaded cable is warm. Using the 20 °C value instead of the 70 °C one understates the drop by 16.4%.
- Breaker rating instead of load. Use the current the load really draws, or the design current. AS/NZS 3000 lets you use half the protective device rating for circuits whose load is spread along the run, such as socket-outlet and lighting circuits.
- Forgetting the feeder. The limit is for the whole path. A final circuit at 4% on top of a 2% submain is a 6% installation.
Embed the Voltage Drop Calculator on your website
Teachers, bloggers and course sites can add this tool to any web page for free. Paste the code below into an HTML or embed block (WordPress, Google Sites, Canvas, Moodle, Blogger, Wix). It loads the same tool you see above, without our menus or ads. Please keep the credit line under the frame. Sizes, a live preview and step-by-step guides for each platform are on the calculator widget page.
Frequently asked questions
What is the maximum allowed voltage drop?
It depends on the code. The NEC recommends 3% for a branch circuit and 5% for feeder plus branch circuit. BS 7671 and IEC 60364 use 3% for lighting and 5% for other uses on a public supply, and 6% and 8% on a private supply. AS/NZS 3000 requires 5% from the point of supply.
How do I calculate voltage drop on a three-phase circuit?
Multiply √3 (1.732) by the current, the one-way length and the resistance per metre: VD = √3 × I × L × R. The result is the drop between lines, so divide it by the line-to-line voltage for the percentage. For example, a 22 kW motor at 400 V and power factor 0.85 draws 37.36 A, and 80 m of 10 mm² copper at 70 °C drops 10.68 V, which is 2.67%.
Is the cable length one way or there and back?
One way, from the supply to the load. The formulas already include the return path: the factor 2 for DC and single-phase circuits, and √3 for three-phase.
Why is my result a little lower than the NEC or BS 7671 table?
This calculator works out resistance from the resistivity of annealed copper or aluminium and the exact nominal area. Tables describe real stranded cables: NEC Chapter 9 Table 8 values are about 2% higher, and IEC 60228 lets a metric conductor have 3% to 12% more resistance than its nominal area gives. Near the limit, confirm with your code's table.
Does this calculator size cables for current carrying capacity?
No. It checks voltage drop only. Ampacity, derating and protection come from NEC 310.16, the BS 7671 Appendix 4 tables or AS/NZS 3008.1.1, and a cable must pass both checks.
How do I convert AWG to mm²?
Work out the diameter with d = 0.127 mm × 92^((36 − n)/39), then the area is π/4 × d². 12 AWG is 3.31 mm², 10 AWG is 5.26 mm² and 4/0 is 107.2 mm². For kcmil sizes multiply by 0.5067, so 250 kcmil is 126.7 mm².