Wire Cross-Section Calculator
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Cable Cross-Section Calculator
Select the optimal cable cross-section taking into account voltage drop, current carrying capacity and power losses.
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Complete the form on the left and click "Calculate" to see the results, voltage drops and power losses.
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Other tools you may find usefulCable cross-section calculator - select the wire diameter, check the load capacity and voltage drop
Calculate the required cable cross-section for single-phase and three-phase installations. Enter the power, track length, voltage, material and permissible voltage drop, and the calculator will suggest the minimum conductor cross-section along with the load capacity and approximate voltage drop. The tool supports the selection of cables for lighting, sockets, drives and DC systems.
Run the cross-section calculator See formulas and examples
What the wire cross-section calculator calculates
- Minimum cross-section Sbased on power or current, length L, permissible voltage drop ΔU% and material.
- Operating currentfor single-phase and three-phase systems, with optional cos φ and efficiency.
- Voltage dropin the cable and the percentage result compared to the rated voltage.
- Selection of the closest catalog cross-sectionand approximate long-term load capacity.
- Estimation of powerand heat losses in the power supply conductors.
Key relationships used in calculations
I – load current, P – active power, U – voltage, cosφ – power factor, η – system efficiency
ρ – resistivity, L – length of the current path, S – cross-section of the conductor, R – circuit resistance
Direct formulas for the cross-section from the condition of voltage drop
ΔU in volts, L as the length of the current loop. For a 1f supply, L is often taken as the path back and forth.
Material parameters and temperature influence
| Material | Resistivity ρ [Ω·mm²/m] at 20°C | Temperature coefficient α [1/°C] | Practical notes |
|---|---|---|---|
| Copper Cu | 0.0172 | 0.00393 | High conductivity, smaller cross-sections for the same current |
| Aluminum Al | 0.0282 | 0.00403 | Lighter and cheaper, requires larger cross-sections |
| Silver Ag | 0.0168 | 0.0038 | Best conductor, rarely used economically |
Voltage drop limits and selection criteria
Allowable voltage drop
- Lighting: typically 3 percent
- Sockets and general loads: 3 to 5 percent
- Drives and heavy starting: 5 percent or less, short-delay
Long-term load capacity
After calculating S, verify the allowable current for the arrangement method, the number of wires in the bundle and the ambient temperature. Round the section up if necessary.
Step-by-step calculation examples
Single-phase 230 V - LED lighting
P = 1200 W, U = 230 V, cosφ = 0.95, η = 0.98, L = 35 m loops, ΔU% = 3 percent, Cu.
1. I = P / (U · cosφ · η) ≈ 1200 / (230 · 0.95 · 0.98) ≈ 5.58 A.
2. ΔU = 3 percent · 230 V = 6.9 V.
3. S = ρ · L · I / ΔU = 0.0172 · 35 · 5.58 / 6.9 ≈ 0.49 mm².
4. Practical selection: the nearest commercial cross-section of 1.5 mm² provides a margin for drop and load capacity.
Three-phase 400 V – 7.5 kW motor
P = 7500 W, U = 400 V, cosφ = 0.85, η = 0.9, L = 60 m, ΔU% = 5 percent, Cu.
1. I = P / (√3 · U · cosφ · η) ≈ 7500 / (1.732 · 400 · 0.85 · 0.9) ≈ 14.1 A.
2. ΔU = 5 percent · 400 V = 20 V.
3. S = √3 · ρ · L · I / ΔU = 1.732 · 0.0172 · 60 · 14.1 / 20 ≈ 1.26 mm².
4. In practice, select 2.5 mm² or 4 mm² depending on the installation method and thermal reserve.
DC 24 V line - control power supply
P = 240 W, U = 24 V, I = 10 A, L = 40 m loop, ΔU% = 3 percent, Cu.
ΔU = 0.03 · 24 = 0.72 V. S = ρ · L · I / ΔU = 0.0172 · 40 · 10 / 0.72 ≈ 9.56 mm². Selection: 10 mm².
Substitute for aluminum instead of copper
Same current and length as above, Al: SAl= SCu· ρAl/ρCu≈ 10 · 0.0282/0.0172 ≈ 16.4 mm².
Auxiliary table - approximate load capacity of Cu cables
| Cross-section S [mm²] | Ilong-term[A] - single conductor in air | Comments |
|---|---|---|
| 1.5 | 15 to 19 | lighting, short sections |
| 2.5 | 20 to 27 | sockets, general circuits |
| 4 | 26 to 36 | small drives, longer routes |
| 6 | 34 to 46 | power supply to local switchboards |
| 10 | 46 to 65 | higher loads and low drops |
The actual load capacity depends on the installation method, the number of loaded conductors, the ambient temperature and the insulation class. The table is indicative.
Practical tips for selecting the cross-section
- Calculate the current from the power and the cosφ and η coefficients. Always add reserves for starts and overloads.
- Determine the permissible voltage drop and determine S from the ΔU condition. If in doubt, increase the cross-section by one degree.
- Verify the long-term load capacity for the selected arrangement method. Apply correction factors for temperature and beams.
- For long runs or low DC voltages, voltage drop usually determines the cross-section more than thermals.
- In three-phase systems, remember about the currents in the N conductor with high harmonic content.
Frequently asked questions - FAQ
Should the length L be counted as one path or a loop
For 1f and DC systems, assume the length of the current loop, i.e. the path back and forth. For 3f, the impedance of the phase-phase loop is important according to the formula from √3.
What if the result is an unusual cross-section, for example 3.1 mm²
Select the nearest larger commercial cross-section, for example 4 mm². Check the drop and load capacity again.
How to take into account the operating temperature of the cable
You can calculate the resistivity: ρ(T) = ρ₀ · [1 + α · (T − 20°C)]. Higher temperature means greater resistance and greater voltage drop.
Is aluminum safe for indoor installations
Yes, if you keep the right sections and accessories. Requires careful crimping technique and dedicated Al or Al-Cu connectors.
How to select a cable for starting the engine
Consider a larger cross-section or a shorter section so that the voltage drop during starting does not exceed the permissible value for the drive.
Checklist for quick selection
Steps 1 to 3
- Determine I from the powers and coefficients.
- Determine ΔU% and convert to volts.
- Calculate S from the decline condition. For 1f use S = ρ · L · I / ΔU. For 3f use S = √3 · ρ · L · I / ΔU.
Steps 4 to 6
- Round S to the nearest standard section.
- Verify thermal load capacity for installation method and environment.
- Add margin for future expansion or aging insulation.
Summary:The correct selection of the cable cross-section is based on three equally important criteria: voltage drop, long-term load capacity and environmental conditions. Using the presented formulas and resistivity values, you will quickly calculate the cross-section for copper and aluminum and verify it against the permissible limits and practical requirements of the installation.