Categories: Electricity

Calculation of current strength and selection of power cable for mains voltage 220 and 380 V

To ensure safety in the operation of household appliances, it is necessary to correctly calculate the cross section of the power cable and wiring. Because an incorrectly selected cable core cross section creates a risk of ignition wiring due to a short circuit. This threatens to cause a fire in the building. This also applies to the choice of cable for connecting electric motors.

current calculation

The amount of current is calculated by power and required at the design stage (planning) housing - apartments, houses.

  • The choice of power cable depends on the value of this value (wire), which can be used to connect electricity appliances to the grid.
  • Knowing the mains voltage and the full load of electrical appliances, you can calculate the current strength by the formula, which will need to be passed through the conductor (wire, cable). According to its size, choose the cross-sectional area of ​​the veins.

If known consumers in the apartment or house, it is necessary to perform simple calculations, to properly mount the power supply circuit.

Similar calculations are performed for production purposes: determining the required cross-sectional area of ​​the cable cores when connecting industrial equipment (various industrial electric motors and mechanisms).

Single-phase mains voltage 220 V

Current strength I (in amperes, A) calculated by the formula:

I = P / U,

where P is the electric full load (must be specified in the technical data sheet of the device), W (watts);

U - mains voltage, V (volt).

The table below shows the load values ​​of typical household appliances and the current consumed by them (for voltage 220 V).

Electrical appliance Power consumption, W Amperage, A
Washing machine 2000 — 2500 9,0 — 11,4
jacuzzi 2000 — 2500 9,0 — 11,4
electric floor heating 800 — 1400 3,6 — 6,4
Stationary electric stove 4500 — 8500 20,5 — 38,6
Microwave oven 900 — 1300 4,1 — 5,9
Dishwashers 2000 — 2500 9,0 — 11,4
Freezers, refrigerators 140 — 300 0,6 — 1,4
Electric meat grinder 1100 — 1200 5,0 — 5,5
electric kettle 1850 — 2000 8,4 — 9,0
electric coffee maker 6з0 - 1200 3,0 — 5,5
Squeezer 240 — 360 1,1 — 1,6
toaster 640 — 1100 2,9 — 5,0
mixer 250 — 400 1,1 — 1,8
hair dryer 400 — 1600 1,8 — 7,3
iron 900 — 1700 4,1 — 7,7
Vacuum cleaner 680 — 1400 3,1 — 6,4
fan 250 — 400 1,0 — 1,8
TV 125 — 180 0,6 — 0,8
radio equipment 70 — 100 0,3 — 0,5
lighting devices 20 — 100 0,1 — 0,4

The figure shows ??the scheme of the device of power supply of the apartment at single-phase connection to a network with voltage 220 V.

As can be seen from the figure, various consumers of electricity are connected through appropriate machines to electricity meters and then to the general machine, which must be designed to load the devices, what the apartment will be equipped with. Wire, supplying power must also meet the load of energy consumers.

Below is a table for concealed wiring with a single-phase connection diagram of the apartment for the selection of wires at voltage 220 V

Intersection of wire wires, mm2 The diameter of the conductor core, mm copper cores aluminum cores
Current, A Power, W Current, A Power, kw
0,50 0,80 6 1300
0,75 0,98 10 2200
1,00 1,13 14 3100
1,50 1,38 15 3300 10 2200
2,00 1,60 19 4200 14 3100
2,50 1,78 21 4600 16 3500
4,00 2,26 27 5900 21 4600
6,00 2,76 34 7500 26 5700
10,00 3,57 50 11000 38 8400
16,00 4,51 80 17600 55 12100
25,00 5,64 100 22000 65 14300

As can be seen from the table, the cross section of the veins depends not only on the load but also on the material, of which the wire is made.

Three-phase mains voltage 380 V

In three-phase power supply current I (in amperes, A) calculated by the formula:

I = P / 1,73 U,

where P -power consumption, W;

U - mains voltage, V,

since the voltage in a three-phase power supply circuit 380 V, the formula will take the form:

I = P / 657, 4.

In the case of supply to the house three-phase power supply voltage 380 The wiring diagram will look like this.

The section lived in the power cable at different loads in a three-phase voltage circuit 380 In for hidden wiring is presented ??in the table.

Intersection of wire wires, mm2 The diameter of the conductor core, mm copper cores aluminum cores
Current, A Power, W Current, A Power, kw
0,50 0,80 6 2250
0,75 0,98 10 3800
1,00 1,13 14 5300
1,50 1,38 15 5700 10 3800
2,00 1,60 19 7200 14 5300
2,50 1,78 21 7900 16 6000
4,00 2,26 27 10000 21 7900
6,00 2,76 34 12000 26 9800
10,00 3,57 50 19000 38 14000
16,00 4,51 80 30000 55 20000
25,00 5,64 100 38000 65 24000

To calculate the current in the load supply circuits, characterized by high reactive full power, which is typical of the use of electricity in industry:

  • electric motors;
  • chokes of lighting devices;
  • welding transformers ,;
  • induction furnaces.

This phenomenon must be taken into account in the calculations. In powerful devices and equipment the share of reactive loading is higher and therefore for such devices in calculations the power factor is accepted equal 0,8.

In practice, it is considered, that when calculating electrical loads for domestic purposes, the power reserve is taken 5%. In the case of calculation of electrical networks for industrial production, the power reserve is accepted 20%.

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