Bandwidth of the power cable table. Calculation of the cable cross-section by power

Below I will give a table of wire cross-sections, but I recommend that you be patient by reading this small theoretical part to the end. This will allow you to be more conscious in choosing wires for wiring, in addition, you can independently make wire section calculation, moreover, even "in the mind."

The passage of current through the conductor is always accompanied by the release of heat (respectively, heating), which is directly proportional to the power dissipated in the wiring section. Its value is determined by the formula P=I 2 *R, where:

• I - the value of the flowing current,
• R is the resistance of the wire.

Excessive heat may cause insulation failure, resulting in a short circuit and/or fire.

The current flowing through the conductor depends on the load power (P), defined by the formula

I=P/U

(U is the voltage that for household electrical network is 220V).

The wire resistance R depends on its length, material and section. For electrical wiring in an apartment, cottage or garage, the length can be neglected, but the material and cross section must be taken into account when choosing wires for electrical wiring.

CALCULATION OF THE SECTION OF THE WIRE

The cross section of the wire S is determined by its diameter d as follows (hereinafter, I will simplify the formulas as much as possible):
S=π*d 2 /4=3.14*d 2 /4=0.8*d 2.

This may come in handy if you already have a wire, and without a marking that immediately indicates the cross section, for example, VVG 2x1.5, here 1.5 is the cross section in mm 2, and 2 is the number of cores.

The larger the cross section, the greater the current load the wire can withstand. With the same sections of copper and aluminum wires - copper can withstand more current, in addition, they are less brittle, oxidize worse, therefore they are most preferable.

Obviously, with a hidden installation, as well as wires laid in a corrugated hose, an electrical box, due to poor heat transfer, they will heat up more strongly, which means that their cross section should be chosen with a certain margin, so it's time to consider such a value as current density (let's denote it Iρ).

It is characterized by the amount of current in amperes flowing through the unit section of the conductor, which we will take as 1mm 2. Since this value is relative, it is convenient to use it to calculate the cross section using the following formulas:

1. d=√ 1.27*I/Iρ =1.1*√I/Iρ- get the value of the wire diameter,
2. S \u003d 0.8 * d 2 - previously obtained formula for calculating the section,

We substitute the first formula into the second, round everything that is possible, we get a very simple ratio:

S=I/Iρ

It remains to determine the value of the current density Iρ), since the operating current I) is determined by the power of the load, I gave the formula above.

The permissible value of the current density is determined by many factors, the consideration of which I will omit and give the final results, and with a margin:

Calculation example:

We have: the total load power in the line is 2.2 kW, the wiring is open, the wire is copper. For the calculation we use the following units of measurement: current - Ampere, power - Watt (1kW = 1000W), voltage - Volt.

All materials presented on this website are for informational purposes only and should not be used as guidelines or normative documents

Content:

Of great importance in electrical engineering is such a value as the cross section of the wire and the load. Without this parameter, it is impossible to carry out any calculations, especially those related to the laying of cable lines. The table of dependence of power on the wire section, used in the design of electrical equipment, helps to speed up the necessary calculations. Correct calculations ensure the normal operation of devices and installations, contribute to the reliable and long-term operation of wires and cables.

Rules for calculating the cross-sectional area

In practice, calculations of the cross section of any wire do not present any difficulty. It is enough just with a caliper, and then use the resulting value in the formula: S = π (D / 2) 2, in which S is the cross-sectional area, the number π is 3.14, and D is the measured diameter of the core.

Currently used mainly copper wires. Compared to aluminum, they are more convenient to install, durable, have a much smaller thickness, with the same current strength. However, as the cross-sectional area increases, the cost of copper wires begins to increase, and all advantages are gradually lost. Therefore, with a current value of more than 50 amperes, the use of cables with aluminum conductors is practiced. Square millimeters are used to measure the cross section of wires. The most common indicators used in practice are areas of 0.75; 1.5; 2.5; 4.0 mm2.

Table of cable section by core diameter

The basic principle of calculations is the sufficiency of the cross-sectional area for normal flow through it electric current. That is, the permissible current should not heat the conductor to a temperature above 60 degrees. The voltage drop must not exceed the allowable value. This principle is especially relevant for long-distance transmission lines and high strength current. Ensuring the mechanical strength and reliability of the wire is carried out due to the optimal thickness of the wire and protective insulation.

Wire cross section for current and power

Before considering the ratio of cross section and power, one should dwell on an indicator known as the maximum operating temperature. This parameter must be taken into account when choosing the cable thickness. If this indicator exceeds its permissible value, then due to strong heating, the metal of the cores and the insulation will melt and collapse. Thus, the operating current for a particular wire is limited by its maximum operating temperature. An important factor is the time during which the cable will be able to function in such conditions.

The main influence on the stable and durable operation of the wire is the power consumption and. For the speed and convenience of calculations, special tables have been developed that allow you to choose required section according to the intended operating conditions. For example, with a power of 5 kW and a current of 27.3 A, the cross-sectional area of ​​\u200b\u200bthe conductor will be 4.0 mm2. In the same way, the cross section of cables and wires is selected in the presence of other indicators.

It is also necessary to take into account the influence environment. When the air temperature is 20 degrees higher than the standard, it is recommended to choose larger section next in order. The same applies to the presence of several cables contained in one bundle or the value of the operating current approaching the maximum. Ultimately, the table of dependence of power on the wire section will allow you to choose the appropriate parameters in case of a possible increase in the load in the future, as well as in the presence of large starting currents and significant temperature drops.

Formulas for calculating the cable section

Correct selection electric cable important to ensure enough level security, cost-effective use of the cable and fully exploit the full potential of the cable. A properly dimensioned cross-section must be able to operate continuously under full load without damage, withstand short circuits in the network, provide a load with an appropriate voltage (without excessive voltage drop) and ensure the operation of protective devices during a lack of grounding. That is why scrupulous and exact calculation cable cross-sections by power, which today can be done using our online calculator quickly enough.

Calculations are made individually according to the formula for calculating the cable section separately for each power cable, for which you need to select a specific section, or for a group of cables with similar characteristics. All cable sizing methods to some extent follow the main 6 points:

• Collecting data about the cable, its installation conditions, the load it will carry, etc.
• Definition minimum size cable based on current calculation
• Determining the minimum cable size based on consideration of the voltage drop
• Minimum cable size determination based on short circuit temperature rise
• Determination of the minimum cable size based on the loop impedance with insufficient grounding
• The choice of the most large sizes based on the calculations of points 2, 3, 4 and 5

Online calculator for calculating the cable cross-section by power

To apply the online calculator for calculating the cable section, it is necessary to collect the information necessary to perform the sizing calculation. As a rule, you need to obtain the following data:

• Detailed characteristics of the load that the cable will supply
• Cable purpose: for three-phase, single-phase or direct current
• System and (or) source voltage
• Total load current in kW
• Total load power factor
• Starting power factor
• Cable length from source to load
• Cable design
• Cable laying method

Line length (m) / Cable material:		Copper Aluminum
Load power (W) or current (A):
Mains voltage (V):		Power	1 phase
Power factor (cosφ):		Current	3 phase
Permissible voltage loss (%):
Cable temperature (°C):
Cable laying method:	Open wiring Two single-core in a pipe Three single-core in a pipe Four single-core in a pipe One two-core in a pipe One three-core in a pipe Gr. laying in boxes, 1-4 cables Gr. laying in boxes, 5-6 cables Gr. laying in boxes, 7-9 cables Gr. laying in boxes, 10-11 cables Gr. laying in boxes, 12-14 cables Gr. laying in boxes, 15-18 cables
Cable cross section not less than (mm²):
Current Density (A/mm²):
Wire resistance (ohm):
Load voltage (V):
Voltage loss (V / %):

Section tables for copper and aluminum cables

Copper cable section table
Aluminum cable section table

When determining most of the calculation parameters, the cable cross-section calculation table presented on our website is useful. Since the main parameters are calculated based on the needs of the current consumer, all initial ones can be calculated quite easily. However, the brand of cable and wire, as well as an understanding of the cable design, also plays an important role.

The main characteristics of the cable design are:

• conductor-material
• Conductor Shape
• conductor type
• Conductor Surface Coating
• Insulation type
• Number of cores

The current flowing through the cable creates heat due to losses in the conductors, losses in the dielectric due to thermal insulation and resistive losses from the current. That is why the most basic is the load calculation, which takes into account all the features of the power cable supply, including thermal ones. The parts that make up the cable (eg conductors, insulation, sheath, armour, etc.) must be able to withstand the temperature rise and heat radiating from the cable.

The capacity of the cable is maximum current, which can continuously flow through the cable without damaging the cable insulation and other components. It is this parameter that is the result when calculating the load, to determine the total cross section.

Cables with larger zones cross section conductors have lower resistance losses and can dissipate heat better than thinner cables. Therefore, a cable with a 16 mm2 section will have a large throughput current than 4 mm2 cable.

However, this difference in cross-section is a huge difference in cost, especially when it comes to copper wiring. That is why it is necessary to make a very accurate calculation of the wire cross-section in terms of power so that its supply is economically feasible.

For systems alternating current A commonly used method is to calculate voltage drops based on the power factor of the load. Typically the full load currents are used, but if the load was high at start (e.g. a motor), then the voltage drop based on the starting current (power and power factor if applicable) must also be calculated and accounted for as low voltage It is also the reason for the failure of expensive equipment, despite the modern levels of its protection.

Video reviews on the choice of cable section

Take advantage of others online calculator mi.

Calculation of the cable (wire) section - not less than milestone when designing electrical circuit apartments or houses. From the right choice and quality electrical work depends on the safety and stability of the operation of electricity consumers. At the initial stage, it is necessary to take into account such initial data as the planned power consumption, the length of the conductors and their type, the type of current, the method of wiring. For clarity, consider the methodology for determining the section, the main tables and formulas. Also, you can use the special calculation program presented at the end of the main material.

Calculation of the power section

The optimal cross-sectional area allows current to pass without possible overheating of the wires. Therefore, when designing electrical wiring, first of all, they find the optimal wire cross-section depending on the power consumption. To calculate this value, you should calculate the total power of all devices that you plan to connect. At the same time, take into account the fact that not all consumers will be connected at the same time. Analyze this frequency to choose optimal diameter conductor strands (more details in the next paragraph "Calculation by load").

Table: Approximate power consumption of household electrical appliances.

Name	Power, W
Lighting	1800-3700
TVs	120-140
Radio and audio equipment	70-100
Refrigerators	165-300
Freezers	140
Washing machines	2000-2500
Jacuzzi	2000-2500
Vacuum cleaners	650-1400
electric irons	900-1700
Electric kettles	1850-2000
Dishwasher with hot water	2200-2500
Electric coffee makers	650-1000
Electric meat grinders	1100
Juicers	200-300
Toasters	650-1050
Mixers	250-400
Electric hair dryers	400-1600
microwaves	900-1300
Above plate filters	250
Fans	1000-2000
Grill ovens	650-1350
Stationary electric stoves	8500-10500
Electric saunas	12000

For a home network with a voltage of 220 volts, the current value (in amperes, A) is determined by the following formula:

I=P/U,

where P is the electrical full load (presented in the table and also indicated in technical passport devices), W (watt);

U is the voltage of the electrical network (in this case 220), V (volts).

If the voltage in the network is 380 volts, then the calculation formula is as follows:

I \u003d P / √3 × U \u003d P / 1.73 × U,

where P is the total power consumption, W;

U - voltage in the network (380), V.

The permissible load for a copper cable is 10 A / mm², and for aluminum - 8 A / mm². For the calculation, the obtained current value is necessary ( I) divided by 10 or 8 (depending on the chosen conductor). The resulting value will be approximate size required section.

Load calculation

On the initial stage it is recommended to make an adjustment for the load. This was mentioned above, but still we repeat that in everyday life situations rarely arise when all energy consumers turn on at the same time. Most often, some devices work, while others do not. Therefore, for clarification, the resulting cross-sectional value should be multiplied by the demand factor ( Kc). If you are sure that you will operate all the devices at once, then you do not need to use the specified coefficient.

Table: Coefficients of demand for various consumers (Kc).

Influence of conductor length

The length of the conductor is important in the construction of industrial scale networks, when the cable must be pulled over considerable distances. During the passage of current through the wires, power losses (dU) occur, which are calculated using the following formula:

where I is the current strength;

p - resistivity (for copper - 0.0175, for aluminum - 0.0281);

L is the cable length;

S is the calculated cross-sectional area of ​​the conductor.

According to specifications, maximum value voltage drop along the length of the wire should not exceed 5%. If the drop is significant, then another cable should be selected. This can be done using tables, which already reflect the dependence of the power and current on the cross section.

Table: Selection of wire at a voltage of 220 V.

Wire core cross 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

Calculation example

When planning the wiring diagram in the apartment, you first need to determine the places where the sockets and lighting. It is necessary to determine which devices will be involved and where. Next, you can compose general scheme connection and calculate the cable length. Based on the data obtained, the size of the cable section is calculated according to the formulas given above.

Suppose we need to determine the size of the cable to connect washing machine. We take the power from the table - 2000 W and determine the current strength:

I=2000 W / 220 V=9.09 A (round up to 9 A). To increase the margin of safety, you can add a few amperes and select the appropriate section depending on the type of conductor and the laying method. Under the considered example, a three-core cable with a copper core cross section of 1.5 mm² is suitable.

Cross-section of the copper core of the conductor, mm²	Permissible continuous load current, A	Maximum power of a single-phase load for a voltage of 220 V, kW	Rated current of the circuit breaker, A	Limiting current of the circuit breaker, A	Possible consumers
1,5	19	4,1	10	16	lighting and signaling groups
2,5	27	5,9	16	25	socket groups and electric floors
4	38	8,3	25	32	water heaters and air conditioners
6	46	10,1	32	40	electric stoves and ovens
10	70	15,4	50	63	introductory supply lines

cable 2.1 calculation program

After reviewing the calculation methodology and special tables, for convenience, you can use this program. It will save you from independent calculations and select the optimal cable section according to the specified parameters.

There are two types of calculation in the cable 2.1 program:

1. Calculation of the cross section for a given power or current.
2. Calculation of the maximum current and power over the cross section.

Let's consider each of them.

In the first case, you need to enter:

• Power value (in the considered example 2 kW).
• Select the type of current, type of conductor, laying method and number of cores.
• By pressing the "Calculate" button, the program will give the required cross section, current strength, recommended circuit breaker and residual current device (RCD).

Calculation of the cross section for a given power or current

In the second case, according to a certain section of the conductor, the program selects the maximum allowable:

• Power.
• The strength of the current.
• Recommended circuit breaker current.
• Recommended RCD.

Calculation of the maximum current and power by section

As you can see, the interface of the calculator is quite simple, and the end results are useful and informative.

Installation is not required. Open the archive and run the "cable.exe" file.

Video on this topic

It is impossible to pass more than a certain amount of current through the cable. When designing and installing electrical wiring in an apartment or house, select the correct cross-section of the conductor. This will allow you to avoid overheating of the wires, short circuits and unplanned repairs in the future.

Standard apartment wiring is calculated for a maximum current consumption at a continuous load of 25 amperes (a circuit breaker is also selected for this current strength, which is installed at the input of wires to the apartment) is carried out with a copper wire with a cross section of 4.0 mm 2, which corresponds to a wire diameter of 2.26 mm and load power up to 6 kW.

According to the requirements of clause 7.1.35 of the PUE the cross section of the copper core for residential wiring must be at least 2.5 mm 2, which corresponds to a conductor diameter of 1.8 mm and a load current of 16 A. Electrical appliances with a total power of up to 3.5 kW can be connected to such wiring.

What is wire cross section and how to determine it

To see the cross section of the wire, it is enough to cut it across and look at the cut from the end. The cut area is the cross section of the wire. The larger it is, the more current the wire can transmit.

As can be seen from the formula, the cross section of the wire is light in its diameter. It is enough to multiply the diameter of the wire core by itself and by 0.785. For the cross section of a stranded wire, you need to calculate the cross section of one core and multiply by their number.

The conductor diameter can be determined with a vernier caliper to the nearest 0.1 mm or a micrometer to the nearest 0.01 mm. If there are no instruments at hand, then in this case an ordinary ruler will help out.

Section selection
copper wire electrical wiring by current strength

The magnitude of the electric current is indicated by the letter " BUT” and is measured in Amperes. When choosing, a simple rule applies, the larger the cross section of the wire, the better, so the result is rounded up.

Table for selecting the cross section and diameter of the copper wire depending on the current strength
Maximum current, A	1,0	2,0	3,0	4,0	5,0	6,0	10,0	16,0	20,0	25,0	32,0	40,0	50,0	63,0
Standard section, mm 2	0,35	0,35	0,50	0,75	1,0	1,2	2,0	2,5	3,0	4,0	5,0	6,0	8,0	10,0
Diameter, mm	0,67	0,67	0,80	0,98	1,1	1,2	1,6	1,8	2,0	2,3	2,5	2,7	3,2	3,6

The data I have provided in the table is based on personal experience and guarantee reliable operation of electrical wiring under the most unfavorable conditions of its laying and operation. When choosing a wire cross-section according to the magnitude of the current, it does not matter whether it is alternating current or direct current. The magnitude and frequency of the voltage in the wiring also do not matter, it can be an on-board network of a 12 V or 24 V DC vehicle, aircraft 115V 400Hz, 220V or 380V 50Hz wiring, 10,000V high voltage power line.

If the current consumption of an electrical appliance is not known, but the supply voltage and power are known, then you can calculate the current using the online calculator below.

It should be noted that at frequencies of more than 100 Hz in the wires, when an electric current flows, the skin effect begins to appear, which means that with increasing frequency, the current begins to “press” against the outer surface of the wire and the actual cross section of the wire decreases. Therefore, the choice of wire cross-section for high-frequency circuits is performed according to other laws.

Determination of the load capacity of electrical wiring 220 V
made of aluminum wire

In older homes, electrical wiring is usually made of aluminum wires. If the connections in the junction boxes are made correctly, the service life aluminum wiring may be a hundred years. After all, aluminum practically does not oxidize, and the service life of the electrical wiring will be determined only by the service life of the plastic insulation and the reliability of the contacts at the points of connection.

In the case of connecting additional energy-intensive electrical appliances in an apartment with aluminum wiring, it is necessary to determine the ability of it to withstand additional power by the cross section or diameter of the wire cores. The table below makes this easy.

If your wiring in the apartment is made of aluminum wires and it became necessary to reconnect installed socket in junction box copper wires, then such a connection is made in accordance with the recommendations of the article Connection of aluminum wires.

Calculation of the cross section of the electrical wiring
by power of connected electrical appliances

To select the cross section of the cable wires when laying electrical wiring in an apartment or house, it is necessary to analyze the fleet of existing electrical appliances in terms of their simultaneous use. The table provides a list of popular household electrical appliances with an indication of the current consumption depending on the power. You can find out the power consumption of your models yourself from the labels on the products themselves or passports, often the parameters are indicated on the packaging.

If the strength of the current consumed by the appliance is not known, then it can be measured using an ammeter.

Table of power consumption and current strength of household electrical appliances
at supply voltage 220 V

Typically, the power consumption of electrical appliances is indicated on the case in watts (W or VA) or kilowatts (kW or kVA). 1 kW=1000 W.

Table of power consumption and current strength of household electrical appliances
household appliance	Power consumption, kW (kVA)	Consumed current, A	Current consumption mode
Incandescent light bulb	0,06 – 0,25	0,3 – 1,2	Constantly
Electric kettle	1,0 – 2,0	5 – 9	Up to 5 minutes
electric stove	1,0 – 6,0	5 – 60	Depends on operating mode
Microwave	1,5 – 2,2	7 – 10	Periodically
Electric meat grinder	1,5 – 2,2	7 – 10	Depends on operating mode
Toaster	0,5 – 1,5	2 – 7	Constantly
Grill	1,2 – 2,0	7 – 9	Constantly
coffee grinder	0,5 – 1,5	2 – 8	Depends on operating mode
Coffee maker	0,5 – 1,5	2 – 8	Constantly
Electric oven	1,0 – 2,0	5 – 9	Depends on operating mode
Dishwasher	1,0 – 2,0	5 – 9
Washing machine	1,2 – 2,0	6 – 9	Maximum from the moment of inclusion before heating of water
Dryer	2,0 – 3,0	9 – 13	Constantly
Iron	1,2 – 2,0	6 – 9	Periodically
A vacuum cleaner	0,8 – 2,0	4 – 9	Depends on operating mode
Heater	0,5 – 3,0	2 – 13	Depends on operating mode
Hair dryer	0,5 – 1,5	2 – 8	Depends on operating mode
Air conditioner	1,0 – 3,0	5 – 13	Depends on operating mode
Desktop computer	0,3 – 0,8	1 – 3	Depends on operating mode
Power tools (drill, jigsaw, etc.)	0,5 – 2,5	2 – 13	Depends on operating mode

The current is also consumed by a refrigerator, lighting devices, a radiotelephone, charging device, TV in standby condition. But in total, this power is no more than 100 W and can be ignored in calculations.

If you turn on all the electrical appliances in the house at the same time, then you will need to select a wire section that can pass a current of 160 A. You will need a wire as thick as a finger! But such a case is unlikely. It is hard to imagine that someone is able to grind meat, iron, vacuum and dry hair at the same time.

Calculation example. You got up in the morning, turned on the electric kettle, microwave, toaster and coffee maker. The current consumption, respectively, will be 7 A + 8 A + 3 A + 4 A \u003d 22 A. Taking into account the included lighting, refrigerator and in addition, for example, a TV, the current consumption can reach 25 A.

for 220 V network

You can choose the wire section not only by the current strength, but also by the amount of power consumption. To do this, you need to compile a list of all electrical appliances planned for connection to this section of electrical wiring, determine how much power each of them consumes separately. Then add up the data and use the table below.

for 220 V network
Appliance power, kW (kVA)	0,1	0,3	0,5	0,7	0,9	1,0	1,2	1,5	1,8	2,0	2,5	3,0	3,5	4,0	4,5	5,0	6,0
Standard section, mm 2	0,35	0,35	0,35	0,5	0,75	0,75	1,0	1,2	1,5	1,5	2,0	2,5	2,5	3,0	4,0	4,0	5,0
Diameter, mm	0,67	0,67	0,67	0,5	0,98	0,98	1,13	1,24	1,38	1,38	1,6	1,78	1,78	1,95	2,26	2,26	2,52

If there are several electrical appliances and for some the current consumption is known, and for others the power, then you need to determine the wire cross-section for each of them from the tables, and then add the results.

Selection of the copper wire cross-section by power
for 12 V vehicle electrical system

If, when connected to the vehicle's on-board network additional equipment only its power consumption is known, then you can determine the cross section of additional wiring using the table below.

Table for selecting the cross section and diameter of copper wire by power for on-board vehicle network 12 V
Appliance power, watt (BA)	10	30	50	80	100	200	300	400	500	600	700	800	900	1000	1100	1200
Standard section, mm 2	0,35	0,5	0,75	1,2	1,5	3,0	4,0	6,0	8,0	8,0	10	10	10	16	16	16
Diameter, mm	0,67	0,5	0,8	1,24	1,38	1,95	2,26	2,76	3,19	3,19	3,57	3,57	3,57	4,51	4,51	4,51

The choice of wire cross-section for connecting electrical appliances
to a three-phase network 380 V

When operating electrical appliances, such as a motor, connected to three-phase network, the consumed current no longer flows through two wires, but through three and, therefore, the magnitude of the current flowing in each separate wire somewhat less. This allows you to use a smaller wire for connecting electrical appliances to a three-phase network.

To connect electrical appliances to a three-phase network with a voltage of 380 V, for example, an electric motor, the wire cross-section for each phase is taken 1.75 times less than for connecting to a single-phase network of 220 V.

Attention, when choosing the wire section for connecting the electric motor in terms of power, it should be taken into account that the maximum mechanical power, which the engine can create on the shaft, and not consumed electric power. The electric power consumed by the electric motor, taking into account the efficiency and cos φ, is approximately two times greater than that generated on the shaft, which must be taken into account when choosing the wire section based on the motor power indicated on the plate.

For example, you need to connect an electric motor that consumes power from a network of 2.0 kW. The total current consumption by an electric motor of such power in three phases is 5.2 A. According to the table, it turns out that a wire with a cross section of 1.0 mm 2 is needed, taking into account the above 1.0 / 1.75 = 0.5 mm 2. Therefore, to connect a 2.0 kW electric motor to a 380 V three-phase network, you will need a three-core copper cable with a cross section of each core of 0.5 mm 2.

It is much easier to choose the wire size for connection three-phase motor, based on the magnitude of the current of its consumption, which is always indicated on the nameplate. For example, in the nameplate shown in the photograph, the current consumption of a motor with a power of 0.25 kW for each phase at a supply voltage of 220 V (the motor windings are connected according to the "triangle" scheme) is 1.2 A, and at a voltage of 380 V (the motor windings are connected according to “star” circuit) is only 0.7 A. Taking the current strength indicated on the nameplate, according to the table for selecting the wire cross-section for apartment wiring, we select a wire with a cross section of 0.35 mm 2 when connecting the motor windings according to the “triangle” scheme or 0.15 mm 2 when connected according to the "star" scheme.

About choosing a brand of cable for home wiring

At first glance, it seems cheaper to make residential electrical wiring from aluminum wires, but operating costs due to the low reliability of contacts over time will many times exceed the costs of electrical wiring from copper. I recommend doing wiring exclusively from copper wires! Aluminum wires are indispensable when laying overhead wiring, as they are light and cheap and correct connection serve reliably for a long time.

And which wire is better to use when installing electrical wiring, single-core or stranded? From the point of view of the ability to conduct current per unit section and installation, single-core is better. So for home wiring, you need to use only single-core wire. Stranded allows multiple bends, and the thinner the conductors in it, the more flexible and durable it is. That's why stranded wire used to connect non-stationary electrical appliances to the mains, such as an electric hair dryer, electric razor, electric iron and all the others.

After making a decision on the cross section of the wire, the question arises about the brand of cable for electrical wiring. Here the choice is not great and is represented by only a few brands of cables: PUNP, VVGng and NYM.

PUNP cable since 1990, in accordance with the decision of the Glavgosenergonadzor “On the prohibition of the use of wires of the type APVN, PPBN, PEN, PUNP, etc., manufactured according to TU 16-505. 610-74 instead of APV, APPV, PV and PPV wires in accordance with GOST 6323-79 * "is prohibited for use.

Cable VVG and VVGng - copper wires in double PVC insulation, flat shape. Designed for operation at ambient temperature from -50°C to +50°C, for wiring inside buildings, on outdoors, in the ground when laying in tubes. Service life up to 30 years. The letters "ng" in the brand designation indicate the incombustibility of the wire insulation. Two-, three- and four-core are produced with a cross-section of cores from 1.5 to 35.0 mm 2. If in the designation of the cable before VVG there is the letter A (AVVG), then the conductors in the wire are aluminum.

The NYM cable (its Russian analogue is the VVG cable), with copper conductors, round shape, with non-combustible insulation, complies with the German standard VDE 0250. Specifications and scope, almost identical with the VVG cable. Two-, three- and four-core are produced with a cross-section of cores from 1.5 to 4.0 mm 2.

As you can see, the choice for wiring is not great and is determined depending on which shape of the cable is more suitable for installation, round or flat. A round-shaped cable is more convenient to lay through walls, especially if input is made from the street into the room. You will need to drill a hole slightly larger than the diameter of the cable, and with a larger wall thickness this becomes relevant. For internal wiring, it is more convenient to use a VVG flat cable.

Parallel connection of electrical wiring

There are hopeless situations when you urgently need to lay the wiring, but the wires of the required section are not available. In this case, if there is a wire of a smaller section than necessary, then the wiring can be made from two or more wires by connecting them in parallel. The main thing is that the sum of the sections of each of them should not be less than the calculated one.

For example, there are three wires with a cross section of 2, 3 and 5 mm 2, but according to calculations, 10 mm 2 is needed. Connect them all in parallel, and the wiring will withstand current up to 50 amperes. Yes, you yourself have seen a parallel connection many times more thin conductors for the transmission of large currents. For example, a current of up to 150 A is used for welding, and in order for the welder to control the electrode, a flexible wire is needed. It is made from hundreds of thin copper wires connected in parallel. In a car, the battery is also connected to the on-board network using the same flexible stranded wire, since during the engine start, the starter consumes up to 100 A from the battery. And when installing and removing the battery, it is necessary to take the wires to the side, that is, the wire must be flexible enough .

A method for increasing the cross section of an electric wire by parallel connection several wires of different diameters can only be used in last resort. When laying home electrical wiring, it is permissible to connect in parallel only wires of the same cross section, taken from one bay.

Online calculators for calculating the cross section and diameter of the wire

Using the online calculator below, you can solve the inverse problem - determine the diameter of the conductor from the cross section.

How to calculate the cross section of a stranded wire

Stranded wire, or as it is also called stranded or flexible, is a single-core wire twisted together. To calculate the cross section of a stranded wire, you must first calculate the cross section of one wire, and then multiply the result by their number.

Consider an example. There is a stranded flexible wire, in which there are 15 cores with a diameter of 0.5 mm. The cross section of one core is 0.5 mm × 0.5 mm × 0.785 \u003d 0.19625 mm 2, after rounding we get 0.2 mm 2. Since we have 15 wires in the wire, to determine the cross section of the cable, we need to multiply these numbers. 0.2 mm 2 ×15=3 mm 2 . It remains to determine from the table that such a stranded wire can withstand a current of 20 A.

It is possible to estimate the load capacity of a stranded wire without measuring the diameter of an individual conductor by measuring overall diameter all twisted wires. But since the wires are round, there are air gaps between them. To exclude the area of ​​​​the gaps, the result of the wire section obtained by the formula should be multiplied by a factor of 0.91. When measuring the diameter, make sure that the stranded wire is not flattened.

Let's look at an example. As a result of measurements, the stranded wire has a diameter of 2.0 mm. Let's calculate its cross section: 2.0 mm × 2.0 mm × 0.785 × 0.91 = 2.9 mm 2. According to the table (see below), we determine that this stranded wire will withstand a current of up to 20 A.