Calculation of cable cross-section and power of automatic wiring machines
Calculation of cable cross-section and power of automatic wiring machines
0.8-0.9 для бытовых нагрузок, 0.7-0.85 для двигателей
Обычно 3% для силовых сетей, до 5% для освещения
Enter parameters for calculation
An online electrical calculator will help you calculate the required cable cross-section and the rating of the circuit breaker for electrical wiring. Load power, current, voltage, cable length, conductor material (copper or aluminum), network type (single-phase or three-phase) and permissible voltage drop are taken into account.
The electrical calculator uses exact formulas and standards of the PUE (Electrical Installation Rules) to calculate the cable cross-section and select the machine. Suitable for calculating electrical wiring in apartments, houses, offices and industrial premises. Formulas: Cable Section = I × √(L × ΔU) / (k × U) or Section = I / J, where I - current, L - length, ΔU - voltage drop, k - material coefficient, U - voltage, J - current density. Machine denomination = I × 1.25 (with a margin of 25%).
Let's look at practical examples of calculating the cable cross-section and power of the machines for various types of loads:
Group of sockets in the apartment, power 2.5 kW
Входные данные:
Power: 2.5 kW
Voltage: 220 V
Cable length: 15 m
Material: Copper
Network type: Single phase
Voltage drop: 3%Расчёт:
Current: 2500 / (220 × 0.9) = 12.6 A
Current Density (Copper): 4 A/mm²
Current cross-section: 12.6 / 4 = 3.15 mm²
Standard cross-section: 2.5 mm² is not enough, choose 4 mm²
Machine rating: 12.6 × 1.25 = 15.75 A → 16 A (type C)
Voltage drop: 0.8%Результат:
Cable cross-section: 4 mm² (copper), Automatic: 16 A (type C)
Тип:
Sockets in the apartment
For sockets, a copper cable with a cross-section of at least 2.5 mm² is recommended, but for a length of more than 10 m it is better to use 4 mm²
Lighting group, power 0.5 kW
Входные данные:
Power: 0.5 kW
Voltage: 220 V
Cable length: 25 m
Material: Copper
Network type: Single phase
Voltage drop: 3%Расчёт:
Current: 500 / (220 × 0.95) = 2.4 A
Current Density (Copper): 4 A/mm²
Current cross-section: 2.4 / 4 = 0.6 mm²
Standard cross-section: 1.5 mm²
Machine rating: 2.4 × 1.25 = 3 A → 6 A (type C)
Voltage drop: 0.5%Результат:
Cable cross-section: 1.5 mm² (copper), Automatic: 6 A (type C)
Тип:
Lighting
For lighting, a 1.5 mm² copper cable and a 6-10 A circuit breaker are sufficient.
Electric stove with a power of 7.5 kW
Входные данные:
Power: 7.5 kW
Voltage: 220 V
Cable length: 8m
Material: Copper
Network type: Single phase
Voltage drop: 3%Расчёт:
Current: 7500 / (220 × 0.9) = 37.9 A
Current Density (Copper): 4 A/mm²
Current cross-section: 37.9 / 4 = 9.5 mm²
Standard cross-section: 10 mm²
Machine rating: 37.9 × 1.25 = 47.4 A → 50 A (type C)
Voltage drop: 1.2%Результат:
Cable cross-section: 10 mm² (copper), Automatic: 50 A (type C)
Тип:
Electric stove
Powerful consumers require a large cross-section cable and a corresponding circuit breaker.
Three-phase asynchronous motor 5.5 kW
Входные данные:
Power: 5.5 kW
Voltage: 380V
Cable length: 30 m
Material: Copper
Network type: Three-phase
Voltage drop: 3%Расчёт:
Current: 5500 / (√3 × 380 × 0.85) = 9.8 A
Current Density (Copper): 4 A/mm²
Current cross-section: 9.8 / 4 = 2.45 mm²
Standard cross-section: 2.5 mm²
Machine rating: 9.8 × 1.25 = 12.25 A → 16 A (type D)
Voltage drop: 2.1%Результат:
Cable cross-section: 2.5 mm² (copper), Automatic: 16 A (type D)
Тип:
Three-phase motor
For motors, type D circuit breaker is used for protection against inrush currents
Input cable to a private house, power 10 kW
Входные данные:
Power: 10 kW
Voltage: 220 V
Cable length: 50 m
Material: Aluminum
Network type: Single phase
Voltage drop: 3%Расчёт:
Current: 10000 / (220 × 0.9) = 50.5 A
Current density (aluminum): 3 A/mm²
Current cross-section: 50.5 / 3 = 16.8 mm²
Standard cross-section: 16 mm²
Machine rating: 50.5 × 1.25 = 63.1 A → 63 A (type C)
Voltage drop: 2.8%Результат:
Cable cross-section: 16 mm² (aluminum), Automatic: 63 A (type C)
Тип:
Input cable
Aluminum cable requires a larger cross-section, but is cheaper than copper
Street lighting, power 1 kW, long line
Входные данные:
Power: 1 kW
Voltage: 220 V
Cable length: 100 m
Material: Copper
Network type: Single phase
Voltage drop: 3%Расчёт:
Current: 1000 / (220 × 0.95) = 4.8 A
Current Density (Copper): 4 A/mm²
Current cross-section: 4.8 / 4 = 1.2 mm²
Voltage drop cross-section: 4 mm² required
Standard cross-section: 4 mm²
Machine rating: 4.8 × 1.25 = 6 A → 10 A (type C)
Voltage drop: 2.9%Результат:
Cable cross-section: 4 mm² (copper), Automatic: 10 A (type C)
Тип:
Street lighting
With long lines, the deciding factor is the voltage drop rather than the current density
The electrical calculation includes several stages to obtain an accurate result, taking into account all load parameters and installation conditions.
Допустимые токи для медного кабеля при открытой прокладке:
19 А
27 А
38 А
50 А
70 А
100 А
140 А
175 А
Различные типы автоматов и их применение:
Для слабоиндуктивных нагрузок
Для бытовых нагрузок
Для высоких пусковых токов
Расчёт сечения кабеля по току нагрузки и падению напряжения с выбором стандартного сечения
Расчёт номинала автоматического выключателя с запасом 25% и выбором типа (B, C, D)
Учёт падения напряжения при выборе сечения кабеля, проверка соответствия нормам ПУЭ
Выбор материала проводника (медь или алюминий) с автоматическим учётом плотности тока
Высокая точность расчётов на основе нормативов ПУЭ и стандартов электромонтажа
Using an electrical calculator provides many advantages when designing electrical wiring: accurate calculation of cable cross-section, correct choice of machine, safety, saving on materials.
Точный расчёт сечения кабеля и номинала автомата с учётом всех параметров и нормативов
Обеспечение безопасности электропроводки путём правильного выбора сечения и автомата
Оптимизация затрат на материалы за счёт точного расчёта без избыточного запаса
Удобный и быстрый расчёт без необходимости сложных вычислений и таблиц
To get an accurate result, follow our recommendations when calculating electrical wiring.
Carefully measure the cable length with a margin of 10%
Choose copper for new wiring, aluminum only for budget options
Always choose a machine with a margin of 25% of the rated current
Monitor the voltage drop, no more than 3-5%
For sockets, calculate the load current: I = P / (U × cos φ). Select the current cross-section: for copper, the current density is 4 A/mm², for aluminum 3 A/mm². Example: power 2.5 kW, current 12.6 A → cross-section 4 mm² (copper) or 6 mm² (aluminium).
Machine rating = Design current × 1.25 (25% margin). For sockets, 16 A or 25 A (type C) circuit breakers are usually used. Example: current 12.6 A → automatic 16 A (12.6 × 1.25 = 15.75 A).
For lighting, a copper cable with a cross section of 1.5 mm² and a 6-10 A circuit breaker are usually sufficient. For line lengths of more than 30 m, it is better to use 2.5 mm² to compensate for the voltage drop.
Copper: current density 4 A/mm², conducts current better, more expensive, but more reliable. Aluminum: current density 3 A/mm², cheaper, but requires a larger cross-section and is less reliable in connections. Copper is recommended for new wiring.
For a 7.5 kW stove: current = 7500 / (220 × 0.9) = 37.9 A. Copper cross-section: 37.9 / 4 = 9.5 mm² → choose 10 mm². Automatic: 37.9 × 1.25 = 47.4 A → 50 A (type C).
Type B: for slightly inductive loads (lamps, heaters). Type C: for household loads (sockets, lighting) - the most common. Type D: for high starting currents (motors, compressors).
The voltage drop should not exceed 3-5%. For long lines (>30 m), choose a larger cable cross-section. Formula: ΔU = (2 × ρ × L × I) / (S × U) × 100%, where ρ is resistivity (0.0175 for copper), L is length, I is current, S is cross-section, U is voltage.
For a three-phase network, the current is calculated: I = P / (√3 × U × cos φ), where √3 ≈ 1.73. Then the cross section is selected according to the current density: for copper 4 A/mm², for aluminum 3 A/mm². Example: 5.5 kW, 380 V → current 9.8 A → cross-section 2.5 mm² (copper).
Yes, with hidden installation the permissible current density is lower due to worse heat dissipation. For copper in a hidden gasket: 3 A/mm² instead of 4 A/mm². For aluminum: 2 A/mm² instead of 3 A/mm².
Calculate the total power of all sockets in the group, then the current: I = P / (U × cos φ). Machine denomination = I × 1.25. Example: 5 sockets of 2 kW = 10 kW, current 45.5 A → automatic 50 A or 63 A (type C).
Yes, but each group must have a separate circuit breaker in the distribution board. The input machine is calculated for the total load of all groups with a simultaneity coefficient of 0.7-0.8.
The input cable is calculated for the total power of all consumers with a simultaneity coefficient of 0.7-0.8. Example: house 10 kW, coefficient 0.8 = 8 kW, current 36.4 A → cable 10 mm² (copper) or 16 mm² (aluminum).
Power factor shows the ratio of active power to apparent power. For household loads cos φ = 0.8-0.9 (lamps, heaters), for engines cos φ = 0.7-0.85. At low cos φ, more current is required for the same power.
For a motor: current = Power / (√3 × U × cos φ × η), where η is efficiency (usually 0.85-0.9). Then the cross section: for copper 4 A/mm², for aluminum 3 A/mm². The machine is selected as type D with a margin of 1.5-2 from the rated current for protection against inrush currents.
According to the PUE, the voltage drop in the section from the source to the consumer should not exceed 5% for lighting networks and 3% for power networks. In practice, they try not to exceed 3% for all networks.
Cable resistance: R = (ρ × L) / S, where ρ is the resistivity (0.0175 Ohm mm²/m for copper, 0.0283 for aluminum), L is the length in meters, S is the cross-section in mm². Voltage drop: ΔU = 2 × R × I for a single-phase network, ΔU = √3 × R × I for a three-phase network.
According to PUE 7.1.34, in residential buildings with a cross-section of less than 16 mm², only copper cable should be used. Aluminum cable can only be used with a cross-section of 16 mm² and above, but copper is recommended for new wiring.
Calculate the total power of all consumers taking into account the simultaneity factor (usually 0.7-0.8). Then the current and cable cross-section. Example: consumers 8+5+3 kW, coefficient 0.8 = 12.8 kW, current 58 A → cable 16 mm² (copper).
The machine is selected with a margin of 25% of the calculated current: Rating = I × 1.25. This provides overload protection and prevents false alarms. For motors, the margin may be greater (1.5-2 times) due to starting currents.
Use PUE tables with permissible currents for various sections and laying methods. For copper open installation: 1.5 mm² - 19 A, 2.5 mm² - 27 A, 4 mm² - 38 A, 6 mm² - 50 A. For hidden installation, the currents are 20-30% lower.
Increase the cable cross-section. The voltage drop is inversely proportional to the cross-section: when the cross-section increases by 2 times, the drop decreases by 2 times. It is also possible to reduce the line length or use a higher voltage (380 V instead of 220 V).
For low-voltage lighting, the current is much higher: I = P / U. At a voltage of 12 V for a power of 100 W, current = 100 / 12 = 8.3 A. A larger cable cross-section is required: 2.5-4 mm² to compensate for the high current and voltage drop.
Yes, you can use one cable to the junction box, then separate it into groups. The cross-section of the main cable is calculated for the total load of all groups with a simultaneity factor of 0.7-0.8.
At temperatures above 25°C the permissible current density decreases. At 35°C: for copper 3.5 A/mm², at 40°C: 3 A/mm². At temperatures below 25°C, the current density can be increased, but not by more than 10%.
The grounding conductor must have a cross-section of at least: for phase up to 16 mm² - equal to phase, for phase 16-35 mm² - 16 mm², for phase over 35 mm² - half of phase, but not less than 16 mm².
For the motor, an automatic type D with a rating of 1.5-2.5 of the rated current of the motor is selected for protection against inrush currents. Example: motor 5.5 kW, current 11 A → automatic 16-25 A (type D). A thermal relay with a setting of 1.05-1.2 of the rated current is also used.
Electric heating has high power. Example: 3 kW heater, current 13.6 A → 4 mm² cable (copper), 16 A automatic (type C). For several heaters, calculate the total power with the simultaneity factor.
No, using a cable with a smaller cross-section is dangerous: overheating, fire, loss of power due to voltage drop. Always choose a standard section equal to or larger than the design one. You can take a larger section for reserve.
For a bath, consider high temperature and humidity. Use a double-insulated cable (VVGng-LS) and select a cross-section with a margin of 20-30%. Automatic device with RCD for protection against electric shock. Example: 6 kW, current 27 A → cable 6 mm² (copper), automatic 32 A + RCD.
Standard copper cable cross-sections: 0.75, 1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185, 240, 300, 400, 500 mm². The nearest larger standard section to the calculated one is selected.