Calculation of power from the wire section. How to calculate the required wire cross-section according to the load power

In theory and practice, the choice of the area of ​​the transverse wire cross-sections for current(thickness) given Special attention. In this article, analyzing the reference data, we will get acquainted with the concept of "sectional area".

Calculation of the cross section of wires.

In science, the concept of "thickness" of the wire is not used. Literary sources use terminology - diameter and cross-sectional area. Applicable to practice, the thickness of the wire is characterized sectional area.

Pretty easy to calculate in practice. wire section. The cross-sectional area is calculated using the formula, having previously measured its diameter (can be measured using a caliper):

S = π(D/2)2 ,

• S - wire cross-sectional area, mm
• D is the diameter of the conductive core of the wire. You can measure it with a caliper.

A more convenient form of the formula for the cross-sectional area of ​​\u200b\u200bthe wire:

S=0.8D.

A small correction - is a rounded factor. The exact calculation formula:

In electrical wiring and electrical installation, in 90% of cases it is used copper wire. Copper wire has a number of advantages over aluminum wire. It is more convenient to install, with the same current strength, it has a smaller thickness, and is more durable. But the larger the diameter cross-sectional area), the higher the price of copper wire. Therefore, despite all the advantages, if the current strength exceeds 50 amperes, aluminum wire is most often used. In a particular case, a wire having an aluminum core of 10 mm or more is used.

Measured in square millimeters wire area. Most often in practice (in household electrics), there are such cross-sectional areas: 0.75; 1.5; 2.5; 4 mm.

There is another system for measuring cross-sectional area (wire thickness) - the AWG system, which is used mainly in the USA. Below is section table wires according to the AWG system, as well as the conversion from AWG to mm.

It is recommended to read the article about the choice of wire cross-section for direct current. The article presents theoretical data and reasoning about the voltage drop, about the resistance of wires for different sections. Theoretical data will orient which wire section for current is most optimal for different allowable voltage drops. Also on real example object, in the article on the voltage drop on three-phase cable lines of great length, formulas are given, as well as recommendations on how to reduce losses. The losses on the wire are directly proportional to the current and the length of the wire. And they are inversely proportional to the resistance.

There are three main principles that selection of wire section.

1. For the passage of electric current, the cross-sectional area of ​​the wire (wire thickness) must be sufficient. The concept enough means that when the maximum possible, in this case, electricity, the heating of the wire will be permissible (no more than 600C).

2. Sufficient cross-section of the wire so that the voltage drop does not exceed the allowable value. This mainly applies to long cable lines (tens, hundreds of meters) and large currents.

3. The cross section of the wire, as well as its protective insulation, must ensure mechanical strength and reliability.

For power, for example, chandeliers, they mainly use light bulbs with a total power consumption of 100 W (a current of just over 0.5 A).

When choosing the thickness of the wire, it is necessary to focus on the maximum operating temperature. If the temperature is exceeded, the wire and the insulation on it will melt and, accordingly, this will lead to the destruction of the wire itself. The maximum operating current for a wire with a certain cross section is limited only by its maximum operating temperature. And the time that the wire can work in such conditions.

The following is a table of wire cross-sections, with the help of which, depending on the current strength, you can select the cross-sectional area of ​​​​copper wires. The initial data is the cross-sectional area of ​​the conductor.

Maximum current for different thicknesses of copper wires. Table 1.

Conductor cross section, mm 2	Current, A, for wires laid
	open	in one pipe
		one two core	one three core

The ratings of the wires that are used in electrics are highlighted. "One two-wire" - a wire that has two wires. One Phase, the second - Zero - this is considered a single-phase power supply to the load. "One three-wire" - used for three-phase power supply of the load.

The table helps to determine at what currents, as well as under what conditions it is operated wire of this section.

For example, if “Max 16A” is written on the outlet, then a wire with a cross section of 1.5 mm can be laid to one outlet. It is necessary to protect the socket with a switch for a current of not more than 16A, even better 13A, or 10 A. This topic is covered by the article “About replacing and choosing a circuit breaker”.

It can be seen from the data in the table that a single-core wire means that no more wires pass near (at a distance of less than 5 wire diameters). When two wires are nearby, as a rule, in one common insulation - a two-wire wire. There is a heavier thermal regime here, so less maximum current. The more collected in a wire or bundle of wires, the lower the maximum current should be for each individual conductor, due to the possibility of overheating.

However, this table is not very convenient from a practical point of view. Often, the initial parameter is the power of the consumer of electricity, and not the electric current. Therefore, you need to choose a wire.

We determine the current, having the power value. To do this, we divide the power P (W) by the voltage (V) - we get the current (A):

I=P/U.

To determine the power, having a current indicator, it is necessary to multiply the current (A) by the voltage (V):

P=IU

These formulas are used in cases of active load (consumers in residential premises, light bulbs, irons). For a reactive load, a coefficient from 0.7 to 0.9 is mainly used (for the operation of powerful transformers, electric motors, usually in industry).

The following table proposes the initial parameters - current consumption and power, and the determined values ​​​​- wire cross-section and tripping current of the protective circuit breaker.

Based on power consumption and current - selection area cross section wires and automatic switch.

Knowing the power and current, in the table below you can select wire size.

Table 2.

Max. power, kW	Max. load current, BUT	cross section wires, mm 2	machine current, BUT

Critical cases in the table are highlighted in red, in these cases it is better to play it safe without saving on the wire by choosing a thicker wire than indicated in the table. And the current of the machine, on the contrary, is smaller.

From the table, you can easily select wire cross section for current, or wire cross-section by power. Select a circuit breaker for the given load.

In this table, all data are given for the following case.

• Single phase, voltage 220 V
• Temperature environment+300C
• Laying in the air or a box (located in a closed space)
• Three-core wire, in common insulation (wire)
• Uses the most common TN-S system with a separate ground wire
• In very rare cases, the consumer reaches the maximum power. In such cases, the maximum current can act permanently without negative consequences.

Recommended to choose larger section (next in a row), in cases where the ambient temperature will be 200C higher, or there will be several wires in the bundle. This is especially important in cases where the value of the operating current is close to the maximum.

In dubious and controversial points, such as:

large starting currents; possible future increase in load; fire hazardous premises; large temperature differences (for example, the wire is in the sun), it is necessary to increase the thickness of the wires. Or, for reliable information, refer to formulas and reference books. But basically, tabular reference data is applicable for practice.

Also, the thickness of the wire can be found out by the empirical (experimentally obtained) rule:

The rule for choosing the cross-sectional area of ​​\u200b\u200bthe wire for maximum current.

Needed cross-sectional area for copper wire, based on the maximum current, can be selected using the rule:

The required wire cross-sectional area is equal to the maximum current divided by 10.

Calculations according to this rule without a margin, so the result must be rounded to big side to the nearest size. For example, you need wire section mm, and the current is 32 amperes. It is necessary to take the nearest, of course, in the big direction - 4 mm. It can be seen that this rule is well within the tabular data.

It should be noted that this rule works well for currents up to 40 Amps. If the currents are greater (outside the living quarters, such currents are at the input) - you need to choose a wire with an even greater margin, and divide it not by 10, but by 8 (up to 80 A).

The same rule is for finding the maximum current through a copper wire, if its area is known:

Maximum current equal to area section, multiply by 10.

About aluminum wire.

Unlike copper, aluminum is less conductive to electricity. For aluminum ( wire of the same size, as copper), at currents up to 32 A, the maximum current will be less than for copper by 20%. At currents up to 80 A, aluminum passes current worse by 30%.

Rule of thumb for aluminum:

The maximum current of the aluminum wire is sectional area, multiply by 6.

With the knowledge gained in this article, you can choose a wire according to the ratios “price / thickness”, “thickness / operating temperature”, as well as “thickness / maximum current and power”.

The main points about the cross-sectional area of ​​\u200b\u200bthe wires are highlighted, but if something is not clear, or there is something to add, write and ask in the comments. Subscribe to SamElectric's blog to receive new articles.

To the maximum current, depending on the cross-sectional area of ​​\u200b\u200bthe wire, the Germans have a slightly different attitude. The recommendation for choosing a circuit breaker (protective) is located in the right column.

Table of dependence of the electric current of the circuit breaker (fuse) on the section. Table 3

This table is taken from the "strategic" industrial equipment, perhaps therefore it may give the impression that the Germans are reinsured.

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 site are for informational purposes only and cannot be used as guidelines and normative documents.

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 section 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 electrical wiring also do not matter, it can be an on-board network of a DC vehicle for 12 V or 24 V, 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 the current can be calculated using the following online calculator.

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 long-standing houses, electrical wiring is usually made of aluminum wires. If the connections in the junction boxes are made correctly, the service life of aluminum wiring can be up to a hundred years. After all, aluminum practically does not oxidize, and the life of the electrical wiring will be determined only by the 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 your apartment is made of aluminum wires and you need 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 the refrigerator, lighting, 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 = 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. Therefore, a stranded wire is used to connect non-stationary electrical appliances to the mains, such as an electric hair dryer, an electric razor, an 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.

When repairs are planned in a house or apartment, the replacement of wiring is one of the most important jobs. It is from the correct choice of the wire section that not only the durability of the electrical wiring depends, but also its functionality. The correct calculation of the cable cross-section by power can be carried out by a qualified electrician who can not only select the appropriate cable, but also carry out installation. If the wires are chosen incorrectly, they will heat up, and at high loads they can lead to negative consequences.

As you know, when a wire overheats, its conductivity decreases, which results in even greater overheating. When the wire overheats, its insulation can be damaged and cause a fire. So that after installation new wiring do not worry about your housing, you should initially perform the correct calculation of the cable power and pay special attention to this issue, as well as attention.

Why carry out cable calculations for load current?

Wires and cables through which electric current flows are the most important part of electrical wiring. The calculation of the wire cross section must be made in order to make sure that the selected wire meets all the requirements for reliability and safe operation wiring.

An incorrectly selected cable cross-section will lead to overheating of the wire and, as a result, after a short time you will have to call the wizard to troubleshoot the electrical wiring. Calling a specialist today costs a lot, so in order to save money, you need to do everything right from the beginning, in which case you can not only save money, but also save your home.

It is important to remember that the electrical and fire safety of the room and those who live or live in it depend on the correct choice of cable section.

Safe operation lies in the fact that if you choose a section that does not correspond to its current loads, this will lead to excessive overheating of the wire, melting of the insulation, short circuit and fire.

Therefore, the issue of choosing a wire cross-section must be taken very seriously.

What affects the calculation of the cross section of a wire or cable

There are many factors affecting, which are fully described in paragraph 1.3 of the EMP. This item provides for the calculation of the cross section for all types of conductors.

In this article, dear readers of the site "Electrician in the House", the calculation of the wire cross-section by power consumption for copper conductors in PVC and rubber insulation will be considered. Today, such wires are mainly used in houses and apartments for wiring.

The main factor for cable section calculation the load used in the network or current is considered. Knowing the power of electrical equipment, we will obtain the rated current as a result of a simple calculation using the formulas below. Based on this, it turns out that the cross section of the wires is directly related to the calculated power of the electrical installation.

The choice of conductor material is also important when calculating the cable cross-section. Perhaps every person knows from physics lessons at school that copper has a much higher conductivity than the same wire made of aluminum. If we compare copper and aluminum wires of the same cross section, then the former will have higher rates.

Also important when calculating the cable cross-section is the number of cores in the wire. A large number of the vein heats up much higher than a single-core wire.

Of great importance when choosing a section is the method of laying wires. As you know, the earth is considered a good heat conductor, unlike air. Based on this, it turns out that the cable laid under the surface of the earth can withstand a greater electrical load, unlike those that are in the air.

When calculating the cross section, do not forget that when the wires are bundled and stacked in special trays, they can heat up against each other. Therefore, it is quite important to take this moment into account when making calculations, and, if necessary, make appropriate adjustments. If there are more than four cables in a box or tray, then when calculating the wire section, it is important to enter a correction factor.

As a rule, on right choice The cross section of the wire also affects the air temperature at which it will be operated. In most cases, the calculation is made from an average ambient temperature of + 25 degrees Celsius. If temperature regime does not meet your requirements, then in table 1.3.3 of the EMP there are correction factors that must be taken into account.

The voltage drop also affects the calculation of the cable cross-section. If a voltage drop of more than 5% is expected in an extended cable line, then these indicators must be taken into account in the calculations.

Calculation of the wire cross-section by power consumption

Each cable has its own rated power, which it is able to withstand when an electrical appliance is connected.

In the event that the power of the appliances in the house exceeds the load capacity of the wire, then in this case an emergency cannot be avoided and sooner or later the wiring problem will make itself felt.

To conduct an independent calculation of the power consumption of appliances, it is necessary to enter on a piece of paper the power of all available electrical appliances that can be connected at the same time (electric kettle, TV, vacuum cleaner, hob, computer, etc.).

After the power of each device is known, all values ​​\u200b\u200bmust be summed up to understand the total consumption.

Where K o - coefficient of simultaneity.

Consider an example wire section calculation for normal two-room apartment. Scroll necessary appliances and their approximate power is indicated in the table.

Based on the value obtained, you can continue the calculations with the choice of wire cross-section.

If the house has powerful electrical appliances, the load of which is 1.5 kW or more, it is advisable to use a separate line to connect them. When self-calculating, it is important not to forget to take into account the power of the lighting equipment that is connected to the network.

When properly produced, then about 3 kW will go out to each room, but you should not be afraid of these numbers, since all devices will not be used at the same time, and, therefore, a certain margin has such a value.

When calculating the total power consumed in the apartment, it turned out result 15.39 kW, now this indicator should be multiplied by 0.8, which will result in 12.31 kW actual load. Based on the obtained power indicator, it is possible to calculate the current strength using a simple formula.

Calculation of the cable cross-section by current

The main indicator by which the wire is calculated is its duration. Simply put, this is the amount of current that it is able to pass for a long time.

Knowing the current load, you can get more accurate calculations cable section. Besides, everything section selection tables in GOSTs and regulatory documents are built on current values.

The meaning of the calculation has a similar similarity with the power one, but only in this case it is necessary to calculate the current load. To calculate the cable cross-section for current, the following steps must be carried out:

• - select the power of all devices;
• - calculate the current that passes through the conductor;
• - according to the table, select the most suitable cable section.

To find the value of the rated current, it is necessary to calculate the power of all connected electrical appliances in the house. What we friends have already done in the previous section.

After the power is known, the calculation of the cross section of a wire or cable is reduced to determining the current strength based on this power. You can find the current strength by the formula:

1) The formula for calculating the current strength for single-phase network 220 V:

• - P - total power of all electrical appliances, W;
• - U - mains voltage, V;
• - for household electrical appliances cos (φ) = 1.

2) The formula for calculating the current strength in three-phase network 380 V:

Knowing the magnitude of the current, the cross section of the wire is found according to the table. If it turns out that the calculated and tabular values ​​​​of the currents do not match, then in this case the nearest larger value is chosen. For example, the calculated value of the current is 23 A, according to the table, we select the nearest greater than 27 A - with a cross section of 2.5 mm2 (for a copper stranded wire laid through the air).

I present to your attention the tables of permissible current loads of cables with copper and aluminum conductors with PVC insulation.

All data is taken not from the head, but from normative document GOST 31996-2012 "POWER CABLES WITH PLASTIC INSULATION".

For example, you have a three-phase load with a power of P = 15 kV. It is necessary to select a copper cable (air laying). How to calculate the cross section? First you need to calculate the current load based on this power, for this we use the formula for a three-phase network: I = P / √3 380 = 22.8 ≈ 23 A.

According to the table of current loads, we select a section of 2.5 mm2 (for it, the permissible current is 27A). But since you have a four-core cable (or five, there is not much difference here), according to the instructions of GOST 31996-2012, the selected current value must be multiplied by a factor of 0.93. I = 0.93 * 27 = 25 A. What is acceptable for our load (rated current).

Although, in view of the fact that many manufacturers produce cables with an underestimated section, in this case I would advise you to take a cable with a margin, with a section much higher - 4 mm2.

Which wire is better to use copper or aluminum?

Today, for the installation of both open wiring and hidden, of course, copper wires are very popular. Copper is more efficient than aluminum

1) it is stronger, softer and does not break in places of inflection compared to aluminum;

2) less susceptible to corrosion and oxidation. When connecting aluminum in a junction box, the twist points oxidize over time, this leads to loss of contact;

3) the conductivity of copper is higher than aluminum, with the same cross section, the copper wire is able to withstand a greater current load than aluminum.

As for the conductor material, in this article only copper wire is subject to consideration, since in most cases it is used as electrical wiring in houses and apartments. Among the advantages of this material, durability, ease of installation and the ability to use a smaller cross section compared to aluminum, with the same current, should be highlighted. If the wire cross section is large enough, then its cost exceeds all advantages and the best option will use aluminum cable, not copper.

For example, if the load is more than 50 A, then in order to save money, it is advisable to use cables with an aluminum core. Usually these are areas at the input of electricity to the house, where the distance exceeds several tens of meters.

An example of calculating the cable cross-section for an apartment

Having calculated the load and decided on the material (copper), consider an example wire section calculation for certain groups of consumers, on the example of a two-room apartment.

As you know, the entire load is divided into two groups: power and lighting.

In our case, the main power load will be the socket group installed in the kitchen, in the living rooms and in the bathroom. Since the most powerful appliances are installed there (electric kettle, microwave oven, refrigerator, boiler, washing machine etc.).

1. Water cable

Input cable cross section(section from the shield on the site to switchboard apartments) is selected based on the total capacity of the entire apartment, which we received in the table.

First, we find the rated current in this section with respect to a given load:

The current is 56 amperes. According to the table, we find the cross section corresponding to the given current load. We choose the nearest higher value - 63 A, which corresponds to a cross section of 10 mm2.

2. Room number 1

Here, the main load on the outlet group will be such equipment as a TV, computer, iron, vacuum cleaner. Load on the wiring section from housing shield to the junction box in this room 2990 W (round up to 3000 W). We find the rated current by the formula:

According to the table, we find the cross section, which corresponds to 1.5 mm2 and the permissible current is 21 Amperes. Of course, you can take this cable, but it is recommended to lay the socket group with a cable with a cross section of AT LEAST 2.5 mm2. This is also related to the rating of the circuit breaker that will protect this cable. It is unlikely that you will power this section from a 10 A machine? And most likely set the machine to 16 A. Therefore, it is better to take it with a margin.

Friends, as I said, we supply the socket group with a cable with a cross section of 2.5 mm2, so for wiring directly from the box to the sockets, we select it.

3. Room number 2

Here, appliances such as a computer, a vacuum cleaner, an iron, and possibly a hair dryer will be connected to the sockets.

The load in this case is 4050 watts. We find the current using the formula:

For a given current load, a wire with a cross section of 1.5 mm2 is suitable for us, but here, similarly to the previous case, we take it with a margin and take 2.5 mm2. We carry out the connection of sockets by him.

4. Kitchen

In the kitchen, the socket group powers the electric kettle, refrigerator, microwave oven, electric oven, electric stove and other appliances. Perhaps a vacuum cleaner will be connected here.

The total power of kitchen consumers is 6850 W, the current is:

For such a load, according to the table, we choose the nearest greater cable section - 4 mm2, with a permissible current of 36 A.

Friends above, I stipulated that it is advisable to connect powerful consumers with a separate independent line (own). The electric stove is just that, for her cable section calculation performed separately. When installing electrical wiring for such consumers, an independent line is laid from the shield to the connection point. But our article is about how to correctly calculate the cross section, and in the photo I did not do this on purpose for better assimilation of the material.

5. Bath

The main consumers of electricity in this room are st. machine, water heater, hair dryer, vacuum cleaner. The power of these devices is 6350 watts.

We find the current using the formula:

According to the table, we select the nearest higher current value - 36 A, which corresponds to a cable cross section of 4 mm2. Here, again, friends in a good way, it is expedient to supply powerful consumers with a separate line.

6. Hallway

In this room, portable equipment is usually used, for example, a hair dryer, a vacuum cleaner, etc. Particularly powerful consumers are not expected here, therefore, but the socket group is also accepted by a wire with a cross section of 2.5 mm2.

7. Lighting

According to the calculations in the table, we know that the power of the entire lighting in the apartment is 500 watts. The rated current for such a load is 2.3 A.

In this case, the entire lighting load can be powered by a wire with a cross section of 1.5 mm2.

It must be understood that the power in different sections of the wiring will be different, respectively, and the cross section of the supply wires is also different. Its greatest value will be on the introductory section of the apartment, since the entire load passes through it. The cross section of the input power wire is chosen 6 - 10 mm2.

At present, it is preferable to use cables for the installation of electrical wiring: VVGng, VVG, NYM. The indicator “ng” indicates that the insulation does not undergo combustion - “non-combustible”. You can use these brands of wires both indoors and outdoors. The operating temperature range for these wires varies from "+/-" 50 degrees Celsius. The warranty period of operation is 30 years, but the period of use may be longer.

If you know how to correctly calculate the current cross-section of the conductor, then you can install electrical wiring in the house without any problems. If all requirements are met, the guarantee of the safety and security of your home will be as high as possible. By choosing the right cross-section of the conductor, you will protect your home from short circuits and fires.

In order to properly lay the 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 the 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 section is determined by the load, power, current and voltage.

Calculation of the power section

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

Calculation of the cable cross-section by voltage

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

Calculation of the cable section according to the load

Table 1. Installed capacity(kW) for open cables

Cross-section of conductors, mm 2	Cables with copper conductors		Cables with aluminum conductors
	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

Cross-section of conductors, mm 2	Cables with copper conductors		Cables with aluminum conductors
	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 appliances or in technical passport equipment. To implement, you need to calculate the total power. When calculating the cable cross-section according to the load, it is necessary to rewrite all electrical equipment, and you also need to think about what equipment can be added in the future. Because the installation is long term, 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 get the sum of the total voltage of 15,000 watts. Since the voltage in the vast majority of residential premises is 220 V, we will calculate the power supply system taking into account a single-phase load.

Next, you need to consider how many equipment can work simultaneously. As a result, you will get a significant figure: 15,000 (W) x 0.7 (simultaneity factor 70%) = 10,500 W (or 10.5 kW) - the cable must be rated 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 areas, copper cable is mainly used, since its conductive properties far exceed those of aluminum.

It should be borne in mind that the cable must necessarily have three cores, since grounding is required for the power supply system in the premises. In addition, it is necessary to determine what type of installation you will use - open or hidden (under plaster or in pipes), since the calculation of the cable section also depends on this. After you have decided on the load, the material of the core and the type of installation, you can see the desired cable 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. Let's say that the power turned out to be 4.75 kW, we decided to use a copper cable (wire) and lay it in a cable channel. is produced according to the formula I \u003d W / U, where W is power, and U is voltage, which is 220 V. In accordance with this formula, 4750/220 \u003d 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

Cross-section of conductors, 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