Cable Size Calculation
About Cable Size Calculation
This cable size by power and current calculator determines the required cross-sectional area of a copper or aluminum conductor based on load current or active power. The calculation takes into account a single-phase or three-phase system, line length, power factor cosφ, installation method, conductor resistance and reactance, and the allowable voltage drop.
The final cable size is selected from a standard range so that two conditions are satisfied simultaneously: the allowable current-carrying capacity of the conductor must not be lower than the calculated load current, and the voltage drop must not exceed the limit specified by the user.
Guidelines and recommendations
Determining the load current
Calculation by current. If the load current is entered directly in amperes, this value is used for further cable sizing. The corresponding active power is calculated from the system voltage and power factor.
For a single-phase system:
P = U · I · cosφ
For a three-phase system:
P = √3 · U · I · cosφ
Calculation by power. If active power P is entered in watts, the current I in amperes is calculated first. For a single-phase system, the voltage between phase and neutral is used. For a three-phase system, the line-to-line voltage is used.
For a single-phase system:
I = P / (U · cosφ)
For a three-phase system:
I = P / (√3 · U · cosφ)
For the same total power, the current in a three-phase system is usually lower than in a single-phase system. For example, with P = 3500 W, cosφ = 0.95 and a voltage of 400 V, the calculated three-phase current is approximately 5.32 A.
Power factor. The cosφ value represents the ratio of active power to apparent power. For predominantly resistive loads, the value is usually close to 1. For motors and mixed loads, a typical approximate range is 0.8-0.95. The reactive component is calculated as:
sinφ = √(1 - cos2φ)
Current-carrying capacity and installation methods
Allowable current-carrying capacity. For each standard cable size, the calculator compares the calculated current with the continuous current-carrying capacity of the conductor. The calculation uses Tables B.52.2 and B.52.4 of IEC 60364-5-52 "Low-voltage electrical installations - Part 5-52: Selection and erection of electrical equipment - Wiring systems" and the corresponding harmonized European document HD 60364-5-52.
Table B.52.2 is used for two loaded conductors, corresponding to a single-phase circuit. Table B.52.4 is used for three loaded conductors in a three-phase circuit. Therefore, the same current in single-phase and three-phase systems can correspond to different allowable current-carrying capacities for the same conductor size.
Conditions of the tabulated values. The calculation uses values for conductors with PVC insulation, a maximum conductor temperature of 70 °C and a reference ambient temperature of 30 °C. Under these reference conditions, the temperature correction factor for the tabulated current-carrying capacity is 1.00.
Five IEC 60364-5-52 reference installation methods are used:
- A1 - single-core insulated conductors in conduit in a thermally insulated wall.
- A2 - multicore cable in conduit in a thermally insulated wall.
- B1 - single-core insulated conductors in conduit on a wall.
- B2 - multicore cable in conduit on a wall.
- C - single-core or multicore cable directly on a wall.
The differences between these methods are mainly related to heat dissipation conditions. The poorer the cooling of the conductor, the lower the allowable current for the same cross-sectional area.
For example, for a 1.5 mm2 copper conductor with two loaded conductors, the allowable current ranges from 14 A for method A2 to 19.5 A for method C. With three loaded conductors, the range for the same cross-sectional area is 13 A to 17.5 A.
Standard conductor sizes. For copper, the current-carrying capacity check uses conductor sizes of 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185, 240 and 300 mm2. For aluminum, the range extends from 2.5 to 300 mm2. The nominal conductor sizes correspond to the range used in EN 60228 "Conductors of insulated cables".
Correction for grouped circuits. The tabulated currents are used as base values for a single circuit. No additional derating factor for several loaded circuits installed together is applied, so a factor of 1.00 is assumed for this effect.
Conductor resistance and temperature
Conductor resistance. Voltage drop is calculated using the resistance of the conductor over the line length. The base resistance values correspond to a temperature of 20 °C, after which the resistance is corrected for the specified conductor temperature.
RT = R20 · L · km · [1 + α · (T - 20)]
Here R20 is the resistance of one metre of conductor at 20 °C, L is the line length in metres, T is the specified temperature in °C, α is the temperature coefficient of resistance, and km is the material factor.
For copper, α = 0.00393 1/°C and km = 1.00 are used. For aluminum, α = 0.00403 1/°C and km = 1.58 are used. Therefore, as temperature increases, conductor resistance and voltage drop also increase.
Use of the temperature value. The entered temperature is used to correct the conductor resistance in the voltage drop calculation. It does not change the tabulated current-carrying capacity, which remains based on the IEC 60364-5-52 reference conditions: PVC insulation, conductor temperature of 70 °C and ambient temperature of 30 °C.
Voltage drop
Total line impedance. In addition to resistance R, the calculation includes reactance X. Depending on conductor size, the calculation uses specific reactance values of approximately 0.071 to 0.133 Ω/km.
For a single-phase system, the voltage drop is calculated as:
ΔU = 2 · I · (R · cosφ + X · sinφ)
For a three-phase system:
ΔU = √3 · I · (R · cosφ + X · sinφ)
The factor 2 in the single-phase formula accounts for the current path through two conductors. In a three-phase system, the factor √3 is used according to the relationship between line and phase quantities.
Voltage drop as a percentage. The calculated ΔU in volts is converted to a percentage of the nominal voltage:
ΔU% = ΔU / U · 100
The voltage available at the load is calculated as:
Uload = U - ΔU
For common low-voltage circuits, an allowable voltage drop of approximately 3-5% is often used as a practical guideline. The specific limit should be selected according to the purpose of the circuit and the requirements of the applicable national implementation of HD 60364.
Selecting the final cable size
Current-carrying capacity check. The calculator checks the standard conductor sizes in sequence and finds the smallest size for which the tabulated allowable current is not lower than the calculated load current.
Voltage drop check. Separately, the calculator determines the minimum conductor size for which the calculated voltage drop does not exceed the percentage specified by the user.
Final selection. If the two checks produce different results, the larger conductor size is selected. For example, if 10 mm2 is sufficient according to current-carrying capacity but 16 mm2 is required to meet the voltage drop limit, the final result is 16 mm2.
This method takes into account both conductor heating and electrical losses in the line. For a short line, the final cable size is often determined by current-carrying capacity, while for longer lines the voltage drop can become the limiting condition.
FAQs
Why can a three-phase system require a smaller cable size for the same power?
For the same total active power, the current in a three-phase system is calculated using the √3 factor and is usually significantly lower. Lower current reduces the requirements for both current-carrying capacity and voltage drop. However, the final conductor size cannot be smaller than the minimum size included in the applicable IEC table.
Why can the calculator recommend 1.5 mm2 copper even at a low current?
For copper conductors, the IEC 60364-5-52 tables used by the calculator start at 1.5 mm2. For example, with installation method C and three loaded conductors, this size has an allowable current of 17.5 A. Therefore, the calculation does not reduce the copper conductor below 1.5 mm2, even if the load current itself would allow a smaller cross-sectional area.
Why does the installation method significantly affect the required cable size?
The allowable current depends on how effectively the cable can dissipate heat to its surroundings. A cable installed directly on a wall usually cools better than conductors installed inside conduit or a thermally insulated construction. Therefore, the same load current can result in different cable sizes for methods A1, A2, B1, B2 and C.
Why can a long line require a larger cable size even with a low current?
Resistance increases in proportion to conductor length, so voltage drop also increases as the line becomes longer. A cable can fully satisfy the current-carrying capacity requirement but still fail the specified voltage drop limit. In that case, the calculator selects the next suitable conductor size.
How does temperature affect the cable size calculation?
As temperature rises, the electrical resistance of copper and aluminum increases, which increases the calculated voltage drop. Temperature coefficients of 0.00393 1/°C for copper and 0.00403 1/°C for aluminum are used for this correction. The tabulated current-carrying capacity remains based on the IEC reference conditions for PVC insulation: conductor temperature of 70 °C and ambient temperature of 30 °C.