How to correctly calculate the cable cross-section for the load. Calculation of the load on the foundation Installed electrical appliances

Calculation of the load on the foundation is necessary for the correct selection of its geometric dimensions and the area of ​​the foundation base. Ultimately, the strength and durability of the entire building depends on the correct calculation of the foundation. The calculation is reduced to determining the load per square meter of soil and comparing it with the permissible values.

To calculate, you need to know:

• The region in which the building is being built;
• Soil type and depth of groundwater;
• The material from which the structural elements of the building will be made;
• Building layout, number of storeys, type of roof.

Based on the required data, the calculation of the foundation or its final verification is carried out after the design of the structure.

Let's try to calculate the load on the foundation for a one-story house made of solid solid brick with walls 40 cm thick. The dimensions of the house are 10x8 meters. The overlap of the basement is made of reinforced concrete slabs, the overlap of the 1st floor is made of wood on steel beams. The roof is gable, covered with metal tiles, with a slope of 25 degrees. Region - Moscow region, soil type - wet loam with a porosity coefficient of 0.5. The foundation is made of fine-grained concrete, the thickness of the foundation wall for the calculation is equal to the wall thickness.

Determination of the depth of the foundation

The depth of placement depends on the depth of freezing and the type of soil. The table shows the reference values ​​of the depth of soil freezing in different regions.

Table 1 - Reference data on the depth of soil freezing

Reference table for determining the depth of the foundation by region

The depth of the foundation in the general case should be greater than the freezing depth, but there are exceptions due to the type of soil, they are indicated in table 2.

Table 2 - Dependence of the depth of the foundation on the type of soil

The depth of the foundation is necessary for the subsequent calculation of the load on the soil and determining its dimensions.

Determine the depth of soil freezing according to table 1. For Moscow, it is 140 cm. According to table 2, we find the type of soil - loam. The depth of the laying should be at least the calculated freezing depth. Based on this, the depth of the foundation for the house is chosen 1.4 meters.

Roof load calculation

The roof load is distributed between those sides of the foundation on which the rafter system rests through the walls. For a conventional gable roof, these are usually two opposite sides of the foundation, for a four-pitched roof, all four sides. The distributed load of the roof is determined by the area of ​​the roof projection, referred to the area of ​​the loaded sides of the foundation, and multiplied by the specific gravity of the material.

Table 3 - The proportion of different types of roofing

Reference table - The proportion of different types of roofing

1. Determine the projection area of ​​the roof. The dimensions of the house are 10x8 meters, the projection area of ​​the gable roof is equal to the area of ​​the house: 10 · 8 = 80 m 2.
2. The length of the foundation is equal to the sum of its two long sides, since the gable roof rests on two long opposite sides. Therefore, the length of the loaded foundation is defined as 10 2 = 20 m.
3. The area of ​​the foundation loaded with the roof with a thickness of 0.4 m: 20 · 0.4 = 8 m 2.
4. The type of coating is metal, the slope angle is 25 degrees, which means that the calculated load according to table 3 is 30 kg / m 2.
5. The load of the roof on the foundation is 80/8 · 30 = 300 kg / m 2.

Snow load calculation

The snow load is transmitted to the foundation through the roof and walls, therefore the same sides of the foundation are loaded as in the calculation of the roof. The area of ​​snow cover is calculated equal to the area of ​​the roof. The resulting value is divided by the area of ​​the loaded sides of the foundation and multiplied by the specific snow load determined from the map.

Table - calculation of snow load on the foundation

1. The length of the slope for a roof with a slope of 25 degrees is (8/2) / cos25 ° = 4.4 m.
2. The area of ​​the roof is equal to the length of the ridge multiplied by the length of the slope (4.4 · 10) · 2 = 88 m 2.
3. The snow load for the Moscow region according to the map is 126 kg / m 2. We multiply it by the roof area and divide by the area of ​​the loaded part of the foundation 88 126/8 = 1386 kg / m 2.

Floor load calculation

Ceilings, like the roof, usually rest on two opposite sides of the foundation, therefore the calculation is carried out taking into account the area of ​​these sides. The floor area is equal to the area of ​​the building. To calculate the floor load, you need to take into account the number of floors and the basement overlap, that is, the floor of the first floor.

The area of ​​each floor is multiplied by the specific gravity of the material from Table 4 and divided by the area of ​​the loaded part of the foundation.

Table 4 - Specific gravity of floors

1. The floor area is equal to the area of ​​the house - 80 m 2. The house has two floors: one made of reinforced concrete and one made of wood over steel beams.
2. We multiply the area of ​​the reinforced concrete floor by the specific gravity from Table 4: 80 * 500 = 40,000 kg.
3. We multiply the area of ​​the wooden floor by the specific gravity from table 4: 80 200 = 16000 kg.
4. We sum them up and find the load per 1 m 2 of the loaded part of the foundation: (40,000 + 16,000) / 8 = 7000 kg / m 2.

Wall load calculation

The wall load is defined as the volume of the walls multiplied by the specific gravity from Table 5, the result is divided by the length of all sides of the foundation multiplied by its thickness.

Table 5 - Specific gravity of wall materials

Table - Specific weight of walls

1. The area of ​​the walls is equal to the height of the building multiplied by the perimeter of the house: 3 · (10 · 2 + 8 · 2) = 108 m 2.
2. The volume of the walls is the area multiplied by the thickness, it is equal to 108 · 0.4 = 43.2 m 3.
3. We find the weight of the walls by multiplying the volume by the specific gravity of the material from Table 5: 43.2 · 1800 = 77,760 kg.
4. The area of ​​all sides of the foundation is equal to the perimeter multiplied by the thickness: (10 2 + 8 2) 0.4 = 14.4 m 2.
5. The specific load of the walls on the foundation is 77760 / 14.4 = 5400 kg.

Preliminary calculation of the foundation load on the ground

The load of the foundation on the ground is calculated as the product of the volume of the foundation and the specific density of the material from which it is made, divided by 1 m 2 of its base area. The volume can be found as the product of the depth to the foundation thickness. The thickness of the foundation is taken at a preliminary calculation to be equal to the thickness of the walls.

Table 6 - Specific density of foundation materials

Table - specific gravity of the material for soil

1. The area of ​​the foundation is 14.4 m 2, the depth of the foundation is 1.4 m. The volume of the foundation is 14.4 · 1.4 = 20.2 m 3.
2. The mass of a foundation made of fine-grained concrete is: 20.2 · 1800 = 36360 kg.
3. Ground load: 36360 / 14.4 = 2525 kg / m 2.

Calculation of the total load on 1 m 2 of soil

The results of the previous calculations are summed up, while calculating the maximum load on the foundation, which will be greater for those sides on which the roof rests.

Conditional design soil resistance R 0 is determined according to tables SNiP 2.02.01-83 "Foundations of buildings and structures".

1. We summarize the weight of the roof, the snow load, the weight of the ceilings and walls, as well as the foundation on the ground: 300 + 1386 + 7000 + 5400 + 2525 = 16 611 kg / m 2 = 17 t / m 2.
2. We determine the conditional design soil resistance according to the tables of SNiP 2.02.01-83. For wet loams with a porosity coefficient of 0.5, R 0 is 2.5 kg / cm 2, or 25 t / m 2.

The calculation shows that the load on the ground is within the permissible range.

Why is it necessary to calculate the load on the cable?

One of the main parameters that determine the cost of a cable is its cross-section. The larger it is, the higher its price. But if you buy an inexpensive wire, the cross section of which does not correspond to the loads in the circuit, the current density increases. Because of this, resistance increases and the release of thermal energy during the passage of electricity. Electricity losses increase, and the efficiency of the system decreases. Throughout the entire period of operation, the consumer pays for significant losses of electricity.

But this is not the only disadvantage of installing a cable with an incorrectly selected section. Due to the increased heat generation, the insulation of the wires heats up excessively - this shortens the life of the wires and often causes a short circuit.

Calculation of the load on the cable allows:

• Reduce electricity bills;
• Increase the service life of the wiring;
• Reduce the risk of short circuits.

What are the losses caused by the passage of electric current?

When calculating the load on the cable, you need to consider:

1. Loss of electric current when passing through the wires

The movement of electricity from a current generator to receivers (household appliances, electrical equipment, lighting fixtures) is accompanied by the release of thermal energy. This physical process is not beneficial. The generated heat heats up the insulating shells, which leads to a reduction in their service life. They become more fragile and break down quickly. Violation of the integrity of the insulation can cause a short circuit when the wires touch each other, and in case of contact with a person - dangerous injury.

The transformation of electrical energy into heat occurs due to resistance, which increases as the density of the passing current increases. This value is calculated by the formula:

Ј = I / S a / mm2

• I is the current strength;

When installing internal wiring, the current density should not exceed 6 A / mm2. For other works, the calculation of the current cable cross-section is made on the basis of the tables contained in the Rules for the Construction and Technical Operation of Electrical Installations (PUE and PTEEP).

If the calculated density value is higher than the recommended one, you need to buy a cable with a large wire cross-section. Despite the increase in the cost of wiring, this decision is justified from an economic point of view. Choosing a cable for wiring with the optimal cross-sectional size will increase its safe operation several times and reduce electricity losses when passing through the wires.

2. Losses due to electrical resistance of materials

The resistance of materials arising in the process of transferring electric current leads not only to the release of thermal energy and heating of wires. There is also a loss of voltage, which negatively affects the operation of electrical equipment, household appliances and lighting fixtures.

When installing electrical wiring, it is also necessary to calculate the value of the line resistance (Rl). It is calculated by the formula:

• ρ is the specific resistance of the material from which the wire is made;
• l - line length;
• S is the cross-section of the wire.

The voltage drop is defined as ΔUl = IRl, and its value should be no more than 5% of the original, and for lighting loads - no more than 3%. If it is larger, it is necessary to choose a cable with a larger cross-section or made of a different material with a lower resistivity. In most cases, both from a technical and economic point of view, it is advisable to increase the cable cross-sectional area.

Choice of cable material

Our catalog of cable products in Brest includes a large selection of cables made of various materials:

• Copper

Copper has a very low resistivity (lower only for gold), so the conductivity of copper wires is much higher than that of aluminum. It does not oxidize, which significantly increases the effective service life. The metal is very flexible and the cable can be folded and rolled many times. Due to the high plasticity, it is possible to manufacture thinner conductors (copper conductors are made from 0.3 mm2, the minimum size of an aluminum conductor is 2.5 mm2).

A lower resistivity reduces the release of thermal energy when current passes, therefore, when laying internal wiring in residential premises, only copper wires are allowed.

• Aluminum

The resistivity of aluminum is higher than that of gold, copper and silver, but lower than that of other metals and alloys.

The main advantage of an aluminum cable over copper is that its price is several times lower. It is also much lighter, which makes it easier to install power grids. When installing long power grids, these characteristics are of decisive importance.

Aluminum does not corrode, but when it comes into contact with air, a film forms on its surface. It protects the metal from atmospheric moisture, but practically does not conduct current. This feature makes it difficult to connect cables.

The main types of section calculation

The calculation of the loads on the wire must be performed for all significant characteristics:

By power

The total power of all devices that consume electricity in a house, apartment, in a production workshop is determined. The power consumption of household appliances and electrical equipment is indicated by the manufacturer.

It is also necessary to take into account the electricity consumed by the lighting fixtures. All electrical appliances at home rarely work at the same time, but the calculation of the cable cross-section for power is performed with a margin, which makes the wiring more reliable and safe. For industrial facilities, a more complex calculation is performed using demand and simultaneity factors.

By voltage

The calculation of the cable cross-section for voltage is based on the type of electrical network. It can be single-phase (in apartments in multi-storey buildings and most individual cottages) and three-phase (in enterprises). The voltage in a single-phase network is 220 V, in a three-phase network - 380 V.

If the total power of electrical appliances in the apartment is 15 kW, then for single-phase wiring this indicator will be 15 kW, and for three-phase it will be 3 times less - 5 kW. But when installing three-phase wiring, a cable with a smaller cross-section is used, but containing not 3, but 5 cores.

By load

Calculation of the cable cross-section for the load also requires the calculation of the total power of the electrical equipment. It is desirable to increase this value by 20-30%. Wiring is carried out for a long time, and the number of household appliances in the apartment or equipment in the workshop may increase.

Then you should determine which equipment can be turned on at the same time. This figure can vary significantly from home to home. Some have a large number of household appliances or electrical equipment that are used several times a month or a year. Others have only necessary but frequently used electrical appliances in their homes.

Depending on the value of the coefficient of simultaneity, the power can differ insignificantly or several times from the load.

Installed power (kW) for open cables

Core section, mm2	Copper cables		Aluminum cables
	Voltage 220 V	Voltage 380 V	Voltage 220 V	Voltage 380 V
0,5	2,4	-	-	-
0,75	3,3	-	-	-
1	3,7	6,4	-	-
1,5	5	8,7	-	-
2	5,7	9,8	4,6	7,9
2,5	6,6	11	5,2	9,1
4	9	15	7	12
5	11	19	8,5	14
10	17	30	13	22
16	22	38	16	28
25	30	53	23	39
35	37	64	28	49

Installed power (kW) for cables laid in a groove or pipe

Core section, mm2	Copper cables		Aluminum cables
	Voltage 220 V	Voltage 380 V	Voltage 220 V	Voltage 380 V
1	3	5,3	-	-
1,5	3,3	5,7	-	-
2	4,1	7,2	3	5,3
2,5	4,6	7,9	3,5	6
4	5,9	10	4,6	7,9
5	7,4	12	5,7	9,8
10	11	19	8,3	14
16	17	30	12	20
25	22	38	14	24
35	29	51	16	-

By current

To calculate the rated current, the value of the total load power is used. Knowing it, the maximum permissible current load is calculated by the formula:

• I - nominal current;
• P - total power;
• U is the voltage;
• cosφ - power factor.

Based on the obtained value, we find the optimal size of the cable cross-section in the tables.

Permissible current loads for a cable with copper conductors laid hidden

Core section, mm	Copper conductors, wires and cables
	Voltage 220 V	Voltage 380 V
1,5	19	16
2,5	27	25
4	38	30
6	46	40
10	70	50
16	85	75
25	115	90
35	135	115
50	175	145
70	215	180
95	260	220
120	300	260

Important nuances for the correct calculation of the cable load

When designing any electrical circuits, a power calculation is performed. On its basis, the selection of the main elements is made and the permissible load is calculated. If the calculation for the DC circuit is not difficult (in accordance with Ohm's law, it is necessary to multiply the current by the voltage - P = U * I), then with the calculation of the AC power, it is not so simple. For an explanation, you will need to turn to the basics of electrical engineering, without going into details, we will give a summary of the main theses.

Full power and its components

In AC circuits, the power is calculated taking into account the laws of sinusoidal changes in voltage and current. In this regard, the concept of total power (S) has been introduced, which includes two components: reactive (Q) and active (P). A graphical description of these quantities can be done through the power triangle (see Fig. 1).

The active component (P) means the power of the payload (irreversible conversion of electricity into heat, light, etc.). This value is measured in watts (W), at the household level it is customary to calculate in kilowatts (kW), in the industrial sphere - in megawatts (MW).

The reactive component (Q) describes the capacitive and inductive electrical load in the AC circuit, the unit of this value is Var.

Fig. 1. Triangle of powers (A) and voltages (V)

In accordance with the graphical representation, the ratios in the triangle of powers can be described using elementary trigonometric identities, which makes it possible to use the following formulas:

• S = √P 2 + Q 2, - for full power;
• and Q = U * I * cos⁡ φ, and P = U * I * sin φ - for reactive and active components.

These calculations are applicable for a single-phase network (for example, a household 220 V), to calculate the power of a three-phase network (380 V), you need to add a multiplier to the formulas - √3 (with a symmetrical load) or add up the powers of all phases (if the load is unbalanced).

For a better understanding of the process of influence of the components of the total power, let's consider the "pure" manifestation of the load in active, inductive and capacitive form.

Active load

Take a hypothetical circuit that uses a "pure" resistor and an appropriate AC voltage source. A graphical description of the operation of such a circuit is shown in Figure 2, which displays the main parameters for a certain time range (t).

Figure 2. Power of an ideal active load

We can see that the voltage and current are synchronized in both phase and frequency, while the power is at twice the frequency. Note that the direction of this value is positive, and it is constantly increasing.

Capacitive load

As you can see in Figure 3, the graph of capacitive load characteristics is slightly different from the active one.

Figure 3. Graph of an ideal capacitive load

The oscillation frequency of the capacitive power is twice the frequency of the sinusoid of the voltage change. With regard to the total value of this parameter, during one harmonic period, it is equal to zero. In this case, an increase in energy (∆W) is also not observed. This result indicates that it moves in both directions of the chain. That is, when the voltage increases, there is an accumulation of charge in the container. With the onset of a negative half-cycle, the accumulated charge is discharged into the circuit circuit.

In the process of accumulating energy in the load capacity and the subsequent discharge, no useful work is performed.

Inductive load

The graph below shows the nature of a “clean” inductive load. As you can see, only the direction of power has changed, as for the increase, it is equal to zero.

Negative impact of reactive load

In the above examples, options were considered where there is a "clean" reactive load. The active resistance factor was not taken into account. In such conditions, the reactive effect is zero, which means that you can ignore it. As you can imagine, this is impossible in real life. Even if hypothetically, such a load would exist, one cannot exclude the resistance of the copper or aluminum conductors of the cable, which is necessary to connect it to the power source.

The reactive component can manifest itself in the form of heating of active circuit components, for example, a motor, a transformer, connecting wires, a supply cable, etc. A certain amount of energy is spent on this, which leads to a decrease in the main characteristics.

Reactive power acts on the circuit as follows:

• does not do any useful work;
• causes serious losses and abnormal loads on electrical appliances;
• can cause a serious accident.

That is why, making the appropriate calculations for the electrical circuit, it is impossible to exclude the factor of influence of inductive and capacitive load and, if necessary, provide for the use of technical systems for its compensation.

Power consumption calculation

In everyday life, you often have to deal with the calculation of power consumption, for example, to check the permissible load on the wiring before connecting a resource-intensive electrical consumer (air conditioner, boiler, electric stove, etc.). Also, in such a calculation there is a need when choosing protective circuit breakers for the switchboard through which the apartment is connected to the power supply.

In such cases, it is not necessary to calculate the power by current and voltage, it is enough to sum up the consumed energy of all devices that can be turned on at the same time. Without getting involved with calculations, you can find out this value for each device in three ways:

When calculating, it should be borne in mind that the starting power of some electrical appliances may differ significantly from the rated one. For household devices, this parameter is almost never indicated in the technical documentation, therefore, it is necessary to refer to the corresponding table, which contains the average values ​​of the starting power parameters for various devices (it is advisable to choose the maximum value).

In order to properly lay the electrical wiring, ensure the uninterrupted operation of the entire electrical system and eliminate the risk of fire, it is necessary to calculate the loads on the cable before purchasing a cable to determine the required cross-section.

There are several types of loads, and for the highest quality installation of the electrical system, it is necessary to calculate the loads on the cable for all indicators. The cable cross-section is determined by load, power, current and voltage.

Calculation of the section by power

In order to produce, it is necessary to add up all the indicators of the electrical equipment operating in the apartment. Calculation of electrical loads on the cable is carried out only after this operation.

Calculation of cable cross-section by voltage

The calculation of electrical loads on the wire necessarily includes. There are several types of electrical networks - single-phase for 220 volts, as well as three-phase for 380 volts. In apartments and living quarters, as a rule, a single-phase network is used, therefore, in the calculation process, this moment must be taken into account - the voltage must be indicated in the tables for calculating the cross-section.

Calculation of the cable cross-section for the load

Table 1. Installed power (kW) for cables laid openly

Core section, mm 2	Copper cables		Aluminum cables
	220 V	380 V	220 V	380 V
0,5	2,4
0,75	3,3
1	3,7	6,4
1,5	5	8,7
2	5,7	9,8	4,6	7,9
2,5	6,6	11	5,2	9,1
4	9	15	7	12
5	11	19	8,5	14
10	17	30	13	22
16	22	38	16	28
25	30	53	23	39
35	37	64	28	49

Table 2. Installed power (kW) for cables laid in a gate or pipe

Core section, mm 2	Copper cables		Aluminum cables
	220 V	380 V	220 V	380 V
0,5
0,75
1	3	5,3
1,5	3,3	5,7
2	4,1	7,2	3	5,3
2,5	4,6	7,9	3,5	6
4	5,9	10	4,6	7,9
5	7,4	12	5,7	9,8
10	11	19	8,3	14
16	17	30	12	20
25	22	38	14	24
35	29	51	16

Each electrical appliance installed in the house has a certain power - this indicator is indicated on the nameplates of the devices or in the technical passport of the equipment. To exercise, you need to calculate the total power. When calculating the cable cross-section for the load, it is necessary to rewrite all electrical equipment, and also need to think about what equipment may be added in the future. Since the installation is carried out for a long time, it is necessary to take care of this issue so that a sharp increase in load does not lead to an emergency.

For example, you have a total voltage of 15,000 watts. Since in the overwhelming majority of residential premises the voltage is 220 V, we will calculate the power supply system taking into account a single-phase load.

Next, you need to think about how much equipment can work at the same time. As a result, you will get a significant figure: 15,000 (W) x 0.7 (70% simultaneity coefficient) = 10,500 W (or 10.5 kW) - the cable must be designed for this load.

You also need to determine what material the cable cores will be made of, since different metals have different conductive properties. In residential premises, copper cable is mainly used, since its conductive properties are much higher than those of aluminum.

It should be borne in mind that the cable must necessarily have three cores, since grounding is required in the premises for the power supply system. In addition, it is necessary to determine which type of installation you will use - open or hidden (under plaster or in pipes), since the calculation of the cable cross-section also depends on this. After you have decided on the load, core material and type of installation, you can see the required cable cross-section in the table.

Calculation of the cable cross-section by current

First, you need to calculate the electrical loads on the cable and find out the power. Suppose that the power turned out to be 4.75 kW, we decided to use a copper cable (wire) and lay it in a cable duct. is produced according to the formula I = W / U, where W is the power, and U is the voltage, which is 220 V. In accordance with this formula, 4750/220 = 21.6 A. Next, we look at table 3, we get 2, 5 mm.

Table 3. Permissible current loads for a cable with copper conductors laid hidden

Core section, mm	Copper conductors, wires and cables
	Voltage 220 V	Voltage 380 V
1,5	19	16
2,5	27	25
4	38	30
6	46	40
10	70	50
16	85	75
25	115	90
35	135	115
50	175	145
70	215	180
95	260	220
120	300	260

To protect yourself when working with household electrical appliances, you must first of all correctly calculate the cross-section of the cable and wiring. Because if the wrong cable is selected, it can lead to a short circuit, which can cause a fire in the building, the consequences can be catastrophic.

This rule also applies to the choice of cable for electric motors.

Calculation of power by current and voltage

This calculation takes place on the basis of power, it must be done even before the design of your home (house, apartment) begins.

• From this value depends on the cable supplying devices that are connected to the mains.
• Using the formula, you can calculate the current strength, for this you need to take the exact mains voltage and the load of the powered devices. Its value gives us an understanding of the cross-sectional area of ​​the veins.

If you know all electrical appliances that in the future must be powered from the network, then you can easily make calculations for the power supply scheme. The same calculations can be performed for production purposes.

Single-phase network with a voltage of 220 volts

Formula for current I (A - amperes):

I = P / U

Where P is the electric full load (its designation must be indicated in the technical passport of this device), W is a watt;

U - mains voltage, V (volts).

The table shows the standard loads of electrical appliances and the current consumed by them (220 V).

Electrical appliance	Power consumption, W	Current strength, 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 cooker	4500 – 8500	20,5 – 38,6
microwave	900 – 1300	4,1 – 5,9
Dishwasher	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	6z0 - 1200	3,0 – 5,5
Juicer	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
A vacuum cleaner	680 - 1400	3,1 – 6,4
Fan	250 - 400	1,0 – 1,8
Television	125 - 180	0,6 – 0,8
Radio equipment	70 - 100	0,3 – 0,5
Lighting devices	20 - 100	0,1 – 0,4

In the figure you can see a diagram of the device for the power supply of the house with a single-phase connection to a 220 volt network.

As shown in the figure, all consumers must be connected to the appropriate machines and a meter, then to a common machine that will withstand the total load of the house. The cable that will bring the current must withstand the load of all connected household appliances.

The table below shows the hidden wiring for a single-phase connection of the dwelling for the selection of a cable at a voltage of 220 volts.

Wire core section, mm 2	Conductor core diameter, mm	Copper conductors		Aluminum conductors
		Current, A	Power, W	Current, A	power, kWt
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 shown in the table, the cross-section of the cores also depends on the material from which it is made.

Three-phase network with a voltage of 380 V

In a three-phase power supply, the current strength is calculated using the following formula:

I = P / 1.73 U

P - power consumption in watts;

U is the mains voltage in volts.

In the technical phase diagram of the 380 V power supply, the formula is as follows:

I = P / 657.4

If a three-phase 380 V network is connected to the house, then the connection diagram will look like this.

The table below shows a diagram of the cross-section of the conductors in the supply cable at different loads at a three-phase voltage of 380 V for hidden wiring.

Wire core section, mm 2	Conductor core diameter, mm	Copper conductors		Aluminum conductors
		Current, A	Power, W	Current, A	power, kWt
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

For further calculation of power supply in load circuits characterized by high reactive apparent power, which is typical for the use of power supply in industry:

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

This phenomenon must be taken into account in further calculations. In more powerful electrical appliances, the load is much higher, therefore, in the calculations, the power factor is taken as 0.8.

When calculating the load on household appliances, the power reserve should be taken 5%. For the power grid, this percentage becomes 20%.