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Cables

Cable Voltage Drop Calculator

Calculate the voltage lost along a cable run for single-phase, three-phase or DC circuits. Conductor resistance is corrected to the operating temperature, reactance and load power factor are included, the result is checked against your % limit, and the smallest IEC or AWG size that meets the limit is found.

For estimation and education. Not a substitute for a licensed engineer, the applicable codes (IEC 60364, NEC/NFPA 70, BS 7671, local regulations) or manufacturer data. Read the disclaimer.

Circuit

Conductor

Material
Size system
°C

70 °C PVC, 90 °C XLPE at full load

Ω/km

From IEC 60228 or maker data; overrides resistivity

Circuit

A
m
Length unit
V
mΩ/m

Default 0.08 mΩ/m

%

Common design limits: 3 % lighting, 5 % other

How to use this calculator

  1. Pick the circuit type and the conductor material (copper or aluminium).
  2. Choose the size system (mmآ² or AWG/kcmil), the conductor size, and how many conductors run in parallel per phase.
  3. Enter the conductor operating temperature: 70 آ°C for PVC-insulated or 90 آ°C for XLPE-insulated cables at full load is a common conservative choice.
  4. Enter the load current, the one-way route length (m or ft), the nominal voltage (line-to-line for three-phase) and the load power factor.
  5. Set your design limit (for example 3 % or 5 %) and read the drop, the pass/fail status and the smallest size that meets the limit.

How it works: the formula

R_T = دپ20 أ— (1 + خ± أ— (T − 20)) / A [خ©/m] Single-phase: خ”V = 2 أ— I أ— L أ— (R cos د† + X sin د†) Three-phase: خ”V = âˆڑ3 أ— I أ— L أ— (R cos د† + X sin د†) DC: خ”V = 2 أ— I أ— L أ— R خ”V% = خ”V / V_nominal أ— 100 AWG diameter: d = 0.127 أ— 92^((36 − n) / 39) mm, A = د€ dآ² / 4

The factor 2 counts both the outgoing and the return conductor of a single-phase or DC circuit. In a balanced three-phase circuit no current returns in the neutral. The âˆڑ3 factor turns the per-phase drop into the line-to-line drop, which is then compared with the line-to-line voltage. The R cos د† + X sin د† term is the component of the cable's impedance drop that lines up with the supply voltage. It is the standard approximation used in IEC and NEC-based practice and is accurate for drops of a few percent.

Variables and units

SymbolMeaningUnit
دپ20Resistivity at 20 آ°C: Cu 0.017241, Al 0.028264خ©آ·mmآ²/m
خ±Temperature coefficient: Cu 0.00393, Al 0.004031/K
TConductor operating temperatureآ°C
AConductor cross-sectional areammآ²
R, XResistance and reactance per metre of one conductorخ©/m
ILoad currentA
LOne-way route lengthm
cos د†Load power factor—

Worked example

Three-phase, 400 V, copper 16 mmآ² at 70 آ°C, 32 A, 50 m, PF 0.85, X = 0.08 mخ©/m:

R_70 = 0.017241 أ— (1 + 0.00393 أ— 50) / 16 = 0.0012893 خ©/m sin د† = âˆڑ(1 − 0.85آ²) = 0.5268 خ”V = 1.73205 أ— 32 أ— 50 أ— (0.0012893 أ— 0.85 + 0.00008 أ— 0.5268) = 2771.3 أ— 0.0011381 = 3.154 V خ”V% = 3.154 / 400 أ— 100 = 0.79 %

Single-phase check: 2.5 mmآ² copper at 20 آ°C, 16 A, 30 m, 230 V, PF 1: خ”V = 2 أ— 16 أ— 30 أ— 0.017241 / 2.5 = 6.62 V = 2.88 %. Against a 2 % limit that fails, and the solver picks 4 mmآ² (1.80 %).

Assumptions

  • A single load at the far end, so current is constant along the run.
  • Reactance is the same for every size in the solver (default 0.08 mخ©/m).
  • Resistance comes from resistivity and nominal area unless you enter a tabulated value at 20 آ°C.

Limitations

  • Voltage drop is only one of the sizing checks. Current-carrying capacity (IEC 60364-5-52 or NEC 310.16 with correction factors), short-circuit withstand and protective-device disconnection times must also be met.
  • Resistance from nominal area is slightly lower than the maximum values in IEC 60228 for stranded conductors. For compliance-grade work, enter the tabulated resistance in the override field.
  • For large conductors (about 150 mmآ² and above) AC resistance is higher than DC because of skin and proximity effects. Use the cable maker's AC resistance and reactance.
  • Motor starting causes much larger momentary drops than the running value calculated here.
  • Permitted drop limits vary by code and application. Check the rules that apply to your installation.

References

Standards and texts that cover this topic. Use the edition adopted in your jurisdiction.

  • IEC 60228, Conductors of insulated cables — nominal cross-sections and maximum conductor resistance.
  • IEC 60364-5-52, Low-voltage electrical installations — Selection and erection of wiring systems (current-carrying capacity; its informative annex covers voltage drop).
  • BS 7671, Requirements for Electrical Installations (IET Wiring Regulations) — voltage drop limits in Appendix 4 for UK installations.
  • NFPA 70, NEC Chapter 9 Table 8 (conductor properties) and Table 9 (AC resistance and reactance); informational notes in Articles 210 and 215 suggesting 3 % / 5 % voltage drop.
  • IEC 60028, International standard of resistance for copper (annealed copper دپ = 1/58 خ©آ·mmآ²/m at 20 آ°C).

Frequently asked questions

Should I enter the one-way length or the total length?

One-way route length, from the supply point to the load. The formula already counts the return path: the factor 2 for single-phase and DC, and âˆڑ3 for balanced three-phase.

Why does temperature matter?

Copper's resistance rises about 0.39 % per آ°C. A conductor at 70 آ°C has roughly 20 % more resistance than at 20 آ°C, so its voltage drop is about 20 % higher. Using the operating temperature avoids underestimating the drop.

Is 3 % or 5 % the legal limit?

It depends on where you are. The NEC mentions 3 % for a branch circuit and 5 % total in informational notes, which are recommendations, not requirements. BS 7671 Appendix 4 gives 3 % for lighting and 5 % for other uses from a public supply. Other countries set their own values.

Why does the solver ignore ampacity?

Ampacity depends on installation method, grouping, ambient temperature and insulation type, which a voltage-drop tool cannot know. Choose the larger of the voltage-drop size and the ampacity size from your code's tables.

How accurate is the AWG area?

The area comes from the exact AWG diameter formula, d = 0.127 أ— 92^((36 − n)/39) mm. That gives 5.26 mmآ² for 10 AWG and 107.2 mmآ² for 4/0. Stranded conductors share the nominal area, but their tabulated resistance in NEC Chapter 9 Table 8 is slightly higher.

Related calculators

How these formulas were checked: methodology and testing.