Wire Resistance Calculator
About Wire Resistance Calculation
This wire resistance calculator determines the electrical resistance of a copper or aluminium conductor from its length and cross-sectional area or diameter. It also performs cable ohm calculation for loop resistance, resistance per metre and conductance, and can calculate voltage drop and power loss for a single-phase or balanced three-phase line when current and voltage are specified.
The calculation is based on the material resistivity at 20 °C. If required, the resistance is adjusted for the conductor temperature.
Guidelines and recommendations
Material resistivity
Reference resistivity is taken at a temperature of 20 °C. The calculator uses ρ20 = 0.0175 Ω·mm2/m for copper and ρ20 = 0.0283 Ω·mm2/m for aluminium. The higher the resistivity of the material, the higher the resistance of a conductor with the same length and cross-sectional area.
Temperature correction is calculated linearly relative to 20 °C:
ρT = ρ20 × [1 + α × (T - 20)]
Here T is the conductor temperature in °C and α is the temperature coefficient of resistance. The calculator uses α = 0.00393 1/°C for copper and α = 0.00403 1/°C for aluminium. If temperature correction is not enabled, all calculations are performed for 20 °C.
Conductor cross-section and diameter
Conductor geometry is treated as circular. If the diameter d is known in mm, the cross-sectional area is calculated as:
S = π × d2 / 4
If the cross-sectional area S is given in mm2, the equivalent diameter is calculated as:
d = √(4 × S / π)
The calculation uses π ≈ 3.14159. It is the cross-sectional area, not the diameter itself, that is used directly in the electrical resistance formula.
Resistance of one conductor and loop resistance
Resistance of one conductor is calculated from the length L of one conductor in metres and its cross-sectional area S in mm2:
R = ρT × L / S
The result is expressed in ohms. Resistance is directly proportional to length, so doubling the length doubles the resistance. Cross-sectional area has the opposite effect, so doubling the area halves the resistance.
Resistance per metre is calculated without using the total line length:
R1m = ρT / S
Loop resistance corresponds to two identical conductors of length L, for example the phase and neutral conductors in a single-phase circuit. The calculator therefore uses:
Rloop = 2 × R
Electrical conductance is the reciprocal of the resistance of one conductor and is expressed in siemens:
G = 1 / R
Voltage drop
For a single-phase line, current flows through two conductors, so the voltage drop is calculated as:
ΔU = 2 × I × R
For a balanced three-phase line, the factor √3 ≈ 1.732 is used:
ΔU = √3 × I × R
Here I is the line current in amperes and R is the resistance of one conductor over the full specified length. The relative voltage drop is calculated from the specified supply voltage:
ΔU% = ΔU / U × 100%
Nominal voltages of 230 V for single-phase systems and 400 V for three-phase systems are commonly used in Europe. These are practical reference values, while the percentage voltage drop is calculated from the actual voltage entered.
Power loss
Resistive power loss is the electrical power dissipated as heat in the conductors. For a single-phase line, two conductors are included:
Ploss = 2 × I2 × R
For a balanced three-phase line, three phase conductors are included:
Ploss = 3 × I2 × R
Because current is squared in the formula, doubling the current increases resistive power loss by a factor of four when resistance remains unchanged.
Calculation assumptions
Alternating current is calculated using a simplified resistive model. For voltage drop calculations, cos φ = 1 is assumed and conductor reactance is not included. In three-phase mode, the load is assumed to be balanced between the phases.
Temperature dependence is calculated using a linear formula with a constant temperature coefficient. Resistance of joints, terminals and other components of the electrical circuit is not added to the conductor resistance.
European standards and reference documents
HD 60364-5-52 "Low-voltage electrical installations - Part 5-52: Selection and erection of electrical equipment - Wiring systems" is used in Europe for the design of wiring systems and includes requirements and guidance related to voltage drop. The calculated ΔU% can be compared with the requirements of the applicable national implementation of this harmonised document.
IEC 60228 "Conductors of insulated cables" specifies nominal conductor cross-sectional areas and requirements for electrical resistance of cable conductors. A calculation based on resistivity gives a theoretical value for the selected material, geometry and temperature, so the actual characteristics of a specific cable should be compared with its technical documentation and the applicable standard.
FAQs
Why is the resistance of aluminium wire higher than that of copper wire?
Aluminium has a higher electrical resistivity. The calculator uses 0.0283 Ω·mm2/m for aluminium and 0.0175 Ω·mm2/m for copper at 20 °C. Therefore, with the same length and cross-sectional area, an aluminium conductor has higher resistance, voltage drop and resistive power loss.
How much does temperature affect wire resistance?
The resistance of copper and aluminium increases with temperature. For example, the copper coefficient of 0.00393 1/°C corresponds to an increase in resistivity of approximately 0.393% per 1 °C above 20 °C in the linear model used by the calculator.
Why is voltage drop calculated differently for single-phase and three-phase lines?
In a single-phase circuit, the calculation includes the current path through two conductors, so a factor of 2 is used. In a balanced three-phase system, the relationship between phase quantities results in the factor √3 ≈ 1.732, so the voltage drop formula is different even when current, length and conductor resistance are the same.
Which has a greater effect on power loss, conductor length or current?
Increasing conductor length increases resistance and power loss linearly. Current has a stronger effect because it is squared in the formula, so doubling the current increases power loss by a factor of four.
Why can a calculation based on diameter differ from one based on the nominal cable cross-section?
When diameter is entered, the calculator assumes a solid circular cross-section and calculates it as S = π × d2 / 4. The nominal cross-sectional area of a cable conductor is a standardised characteristic, and for stranded or compacted conductors it does not necessarily match the area obtained from a simple measurement of the conductor diameter.