The smaller cross-section of the copper wire allows high currents to pass and, accordingly, is designed for increased power or load.
This feature is due to low resistance values, which makes it possible to use a copper core for household use in the presence of not only 220 V, but also 380 Volts.
Electrical cable products differ in the type of insulation, cross-sectional diameter and conductor material.
These parameters determine not only the area of use, but also the basic operating conditions.
The working element of any copper cable product is represented by a conductive core made on the basis of electrical copper.
At the same time, several insulated conductors are enclosed in one common shell. The outer coating is represented by the so-called "armor", or a special protective screen.
The indisputable advantages of copper cable products are presented:
- high thermal conductivity;
- good indicators of current conductivity;
- plasticity and flexibility;
- resistance to kinks in bending or twisting;
- ease of self-assembly;
- the duration of operation;
- resistance to corrosive changes;
- minimal risk of fire.
Copper vein
When choosing a cable product, you should pay attention to the marking. Installation in tunnels, in the open air and in the ground, is carried out with armored copper cable with durable double insulation. The ng-LS mark indicates high rates fire safety products.
It should be noted that single-core copper products are most often used in the installation of stationary wiring, and a stranded conductor is in demand if it is necessary to use increased flexibility and elasticity, as well as vibration resistance.
The cross-section of the copper wire is marked with the first number following the letter designation of the conductor type.
Wiring with copper conductors is used for indoor and outdoor installation in residential premises and office buildings, in industrial and industrial complexes, due to its high technical and quality characteristics.
Selection of wire cross-section
Copper is a reliable material with sufficient bend resistance, elevated level electrical conductivity, as well as a slight susceptibility to corrosive changes. It is for this reason, in conditions of the same level, that a smaller cross-section of the copper core is provided in comparison with aluminum cable products.
The purchase of a copper-type electrical wire is carried out with a certain cross-sectional margin, which reduces the risk of overheating as a result of an increase in the load when new volatile devices are connected.
VVGng cable 4x4 0.66 kV
It is important that the cross-section fully corresponds to the maximum load indicators, as well as the current value for which the automatic protective devices are designed.
The current value is one of the main indicators affecting the calculation of the wire cross-sectional area in copper cable products. A certain area determines the throughput of the current for a long time. This parameter is called long-term permissible load. In this case, the cross-section of the copper core is the total cut-off area of the central part that conducts current to consumers.
Determined by the basic dimensions, measured with a caliper:
- for a circle - S = πd 2/4;
- for a square - S = a 2;
- for a rectangle - S = a × b;
- for a triangle - πr 2/3.
16-wire power cable
Standard design designations: radius (r), diameter (d), width (b) and length (a) of the section, as well as π = 3.14. As a rule, the standard cross-section of the input cable is 4-6 mm 2, the wiring for connecting the socket group is 2.5 mm 2, and the cross-sectional area for connecting the main lighting system is about 1.5 mm 2.
Before copper conductors, it is necessary to take into account the specific operating conditions and the estimated maximum current load, which will flow through the electrical wiring for a long time.
Calculation of wire cross-section
To independently determine the rated current value, you need to calculate the maximum power of all connected volatile devices.
With the already known indicators of the power consumed by the devices, the current strength is calculated.
The standard calculation formula for a single-phase 220 V network:
I = P × K and / U × cos φ
- Р - total indicators of power consumed by all connected electrical devices (W);
- U - indicators of the voltage of the mains (V);
- K and - coefficient of simultaneity equal to 0.75;
- cos φ - indicator for connected household volatile appliances.
The standard calculation formula for a 380 V electrical network:
I = P / √3 × U × cos φ
After calculating the current value, you can easily determine the cross-section of the copper wire, using tabular data for this purpose.
It is necessary to choose the cable cross-sectional area taking into account the current value and indicators of the required power, applying the table and rounding the obtained values upwards with the addition of 15-20% of the margin.
Copper wire cross-section by power: table
Tabular data are the most convenient to use and the most accurate, therefore, experts recommend determining the cross-section of a copper cable product in accordance with the power indicators in the table.
| For voltage 220 V | For voltage 380 V | Copper section | ||
| Power | Current | Power | Current | |
| 4.1 kW | 19 A | 10.5 kW | 16 A | 1.5 mm |
| 5.9 kW | 27 A | 16.5 kW | 25 A | 2.5 mm |
| 8.3 kW | 38 A | 19.8 kW | 30 A | 4.0 mm |
| 10.1 kW | 46 A | 26.4 kW | 40 A | 6.0 mm |
| 15.4 kW | 70 A | 33.0 kW | 50 A | 10,0 mm |
| 18.7 kW | 80 A | 49.5 kW | 75 A | 16.0 mm |
| 25.3 kW | 115 A | 59.4 kW | 90 A | 25.0 mm |
| 29.7 kW | 135 A | 75.9 kW | 115 A | 35.0 mm |
| 38.5 kW | 175 A | 95.7 kW | 145 A | 50.0 mm |
| 47.2 kW | 215 A | 118.8 kW | 180 A | 70.0 mm |
| 57.2 kW | 265 A | 145.2 kW | 220 A | 95.0 mm |
| 66.0 kW | 300 A | 171.6 kW | 260 A | 120 mm |
How to determine the cross section for a stranded wire?
Stranded type copper wires are conductors, the cross-section of which is represented by several cores, which in some brands of cable products are intertwined with each other. any stranded wire is calculated using the standard formula S = π × d² / 4.
In this case, the total cross-sectional area of a copper cable product will be the sum of the cross-sectional area of all its veins.
Assessment of the load capacity of a stranded wire can be performed without measuring the diameter of each individual conductor.
In this case, you need to measure the total diameter of the cable multicore product, and then use the standard multiplying factor of 0.91 in the formula.
The diameter of the copper conductors can be measured with a caliper or a micrometer.
Maximum flexibility and high level elasticity is noted in copper conductors, the cores of which are woven into a dense thread.
As a result of the use of special terminals, the connection of stranded conductors acquires high reliability and lower current resistance, but the use of such cable products in high-frequency electrical circuits is limited.
Calculations performed manually by yourself are not always accurate for the correct determination of the long-term permissible loads on the electrical network.
The table of loads for the cable cross-section belongs to the category of refined calculations, and allows you to correctly determine the choice of external or external wiring.
Conductor cross-section
The core of a cable product is a conductive copper or aluminum core of the wire, protected by an insulating material.
The nominal cross-sectional values of the conductor are in the conductive part of the cable product, and are indicated in the marking on the insulation.
An independent calculation of the actual cross-section of the core is relevant in several situations:
- checking the cable product for compliance with the actual cross-sectional parameters declared by the manufacturer;
- assessment of the quality and technical characteristics of an unmarked cable product.
S = πD 2/4
- π - number "Pi" equal to 3.14;
- D - the results of measurements of the diameter of the core in mm;
- S - the required indicators of the cross-section of the cable core in mm 2.
In multi-wire cable products, the cross-section of one core is measured, after which the result is multiplied by the number of all elements. The calculation of segment cables is more complex.
The calculation of the cross-section of a single-wire conductor is carried out most often using a caliper, and a multi-wire cable product - with a micrometer.
Wire based on different materials
Electrical cable products can be represented by wires with aluminum or copper conductors. The second option is more preferable due to lower resistance and durability. However, it is the aluminum cable that is more affordable.
Power cable aluminum 4-core section 38 sq. Mm
A cable product consists of several basic elements:
- veins - parts responsible for conducting electric current;
- insulation - a protective dielectric cable surface.
Monolithic veins are represented by one wire, and composite ones - several twisted into a bundle, which has a positive effect on their flexibility. The connection of the main elements of electrical wiring is most often carried out with special clamps - terminals.
Copper type
The indisputable advantages of a cable product with a conductor copper type are:
- insignificant indicators of electrical resistance;
- high level of flexibility;
- mechanical stability;
- suitability for soldering and tinning;
- ease of welding and twisting.
The oxidized surface on the contacts has insignificant indicators of transition resistance, and during installation and crimping there is no need to lubricate the surfaces, which makes it easier to work with the material. Most popular brands:
- PV- single-core wire with a cross section of 0.5-95 mm 2;
- PPV- two- or three-core wire with a cross section of 0.75-4.0 mm 2;
- NS- single-core with a cross section of 0.75-120 mm 2.
The main disadvantage of wiring with copper conductor is the high cost of the raw material, and, accordingly, all cable products containing copper.
Aluminum type
The main advantages of a cable product with an aluminum core are presented:
- lower weight of the mounted electrical wiring;
- wide selection and affordable cost.
It should be noted that the electrical conductivity of aluminum is one and a half times lower than that of a copper cable, and a material amorphous in its characteristics during long-term operation is able to "flow" from the crimps.
Insulated aluminum cable
Over time, the aluminum surface oxidizes, and as a result of this natural process there is a noticeable loss of current conductivity. Most popular brands:
- APV
- APPV- two- or three-core wire with a cross section of 2.5-6.0 mm 2;
- APR- single-core wire with a cross section of 2.5-120 mm 2;
- PRN- single-core wire with a cross section of 2.5-120 mm 2.
Some difficulties arise during the installation of aluminum cable products, which is explained by the need to use gas welding and soldering using fluxes and solders.
The design and installation of any electrical circuit presupposes the correct choice of cable cross-section, taking into account the maximum power consumption or load.Measured in mm 2 or "squares", the conductor cross-section has different highest throughput for a long time, and also differs in the heating period:
- with an aluminum conductor - 4.0 A;
- with copper core - 10 A.
For example, a volatile consumer using 4 kW or 4000 W in a single-phase 220 V network needs 4000/220 = 18.18 A + 15%, which is provided by a wire with a copper core of 2.0 mm 2.
When using an aluminum conductor, the core of the cable product to be installed must have a thickness of at least 4.5-5.0 mm 2.
The current load values are most often determined in accordance with the power of volatile consumers declared in the product passport, as well as according to the formula: I = P / 220.
Table of dependence of wire cross-section on load
The most popular and common wire cross-sectional indicators currently used in practice are the cable core areas of 0.75, 1.5, 2.5 and 4.0 mm 2. When choosing a section depending on the load parameters, it is advisable to use standard tabular data.
| Cross section | Open wiring | Closed wiring | ||||||||||
| Aluminum conductor | Copper vein | Aluminum conductor | Copper vein | |||||||||
| Current | Power | Current | Power | Current | Power | Current | Power | |||||
| 380 | 220 | 380 | 220 | 380 | 220 | 380 | 220 | |||||
| 0.5mm 2 | – | – | – | 11A | – | 2.4V | – | – | – | – | – | – |
| 0.75 mm 2 | – | – | – | 15A | – | 3.3V | – | – | – | – | – | – |
| 1.0 mm 2 | – | – | – | 17A | 6.4V | 3.7V | – | – | – | 14A | 5.3V | 3.0V |
| 1.5 mm 2 | – | – | – | 23A | 8.7V | 5.0V | – | – | – | 15A | 5.7V | 3.3V |
| 2.0 mm 2 | 21A | 7.9V | 4.6V | 26A | 9.8V | 5.7V | 14A | 5.3V | 3.0V | 19A | 7.2V | 4.1V |
| 2.5 mm 2 | 24A | 9.1V | 5.2V | 30A | 11B | 6.6V | 16A | 6.0V | 3.5V | 21A | 7.9V | 4.6V |
| 4.0 mm 2 | 32A | 12V | 7.0V | 41A | 15V | 9.0V | 21A | 7.9V | 4.6V | 27A | 10V | 5.9V |
| 6.0 mm 2 | 39B | 14B | 8.5V | 50A | 19V | 11B | 26A | 9.8V | 5.7V | 34A | 12V | 7.4V |
| 10.0 mm 2 | 60V | 22B | 13B | 80A | 30V | 17B | 38A | 14B | 8.3V | 50A | 19V | 11B |
| 16.0 mm 2 | 75V | 28V | 16V | 100A | 38V | 22B | 55A | 20V | 12V | 80A | 30V | 17B |
| 25.0 mm 2 | 105V | 39B | 23B | 140A | 53V | 30V | 65A | 24V | 14B | 100A | 38V | 22B |
| 35.0 mm 2 | 130V | 49B | 28V | 170A | 64V | 37B | 75A | 28V | 16V | 135A | 51B | 29B |
A prerequisite the right choice the cross-section of the core in power cable products is taking into account the value of the maximum current consumed in the load.
Only high-quality wires are capable of withstanding a sufficient load, therefore, when choosing, it is necessary to attach importance to the marking, which contains information about GOST and TU, the manufacturer and the type of cable product.
Along the entire length of the cable product, directly on the insulating layer, the manufacturer must indicate the brand of the wire and its cross-section. In the absence of information even about one of the listed parameters, it is recommended to refuse to purchase a cable product.
Today there is a wide range of cable products, with a cross-section of conductors from 0.35 mm.kv. and higher.
If you choose the wrong cable cross-section for household wiring, then the result can have two results:
- An overly thick vein will "hit" your budget, because its running meter will cost more.
- With the wrong conductor diameter (less than necessary), the conductors will begin to heat up and melt the insulation, which will soon lead to a short circuit.
As you understand, both results are disappointing, therefore, in front of the apartment, it is necessary to correctly calculate the cable cross-section, depending on the power, current and line length. We will now take a closer look at each of the techniques.
Calculation by the power of electrical appliances
For each cable there is a certain amount of current (power) that it is able to withstand during the operation of electrical appliances. If the current (power) consumed by all devices exceeds the permissible value for a conductive core, then an accident cannot be avoided soon.
To independently calculate the power of electrical appliances in the house, you need to write down the characteristics of each device separately on a sheet of paper (stove, TV, lamps, vacuum cleaner, etc.). After that, all the values are added up, and the resulting number is used to select the cable with the conductors with the optimal cross-sectional area.
The calculation formula is:
Ptotal = (P1 + P2 + P3 +… + Pn) * 0.8,
Where: P1..Pn - power of each device, kW
We draw your attention to the fact that the resulting number must be multiplied by the correction factor - 0.8. This coefficient means that only 80% of all electrical appliances will work at the same time. This calculation is more logical, because, for example, you will definitely not use a vacuum cleaner or hairdryer for a long time without interruption.
Tables for selecting the cable cross-section by power:
These are given and simplified tables, you can find more precise values in p. 1.3.10-1.3.11.
As you can see, for each specific type of cable, the table values have their own data. All you need is to find the closest power value and see the corresponding cross-section of the conductors.
So that you clearly understand how to correctly calculate the power cable, we will give a simple example:
We calculated that the total power of all electrical appliances in the apartment is 13 kW. This value must be multiplied by a factor of 0.8, resulting in 10.4 kW actual load. Further in the table, we look for a suitable value in the column. We are satisfied with the figure "10.1" for a single-phase network (voltage 220V) and "10.5" for a three-phase network.
This means that you need to choose such a cross-section of the cable cores that will power all the calculating devices - in an apartment, a room or any other room. That is, such a calculation must be carried out for each outlet group powered from one cable, or for each device if it is powered directly from the panel. In the example above, we have given the calculation of the cross-sectional area of the conductors of the lead-in cable for the entire house or apartment.
In total, we stop the choice of cross-section on a 6 mm conductor with a single-phase network or 1.5 mm with a three-phase network. As you can see, everything is quite simple and even a novice electrician can cope with such a task on his own!
Calculation for current load
Calculation of the cable cross-section for current is more accurate, so it is best to use it. The essence is similar, but only in this case it is necessary to determine the current load on the wiring. To begin with, we calculate the current strength for each of the devices using the formulas.
If the house has a single-phase network, you must use the following formula for the calculation:For a three-phase network, the formula will look like:
Where, P is the power of the electrical appliance, kW
cos phi power factor
You can read more about formulas related to calculating cardinality in the article:.
We draw your attention to the fact that the values of the tabular values will depend on the conditions for laying the conductor. At the permissible current loads and power will be significantly higher than at.
Again, any calculation of the cross-section is carried out for a specific device or a group of them.
Table for selecting the cable cross-section for current and power:
Length calculation
Well, the last way to calculate the cable cross-section is by length. The essence of the following calculations is that each conductor has its own resistance, which contributes with an increase in the length of the line (the greater the distance, the greater the loss). In the event that the value of losses exceeds the 5% mark, it is necessary to choose a conductor with larger cores.
The following technique is used for calculations:
- It is necessary to calculate the total power of electrical appliances and amperage (we have provided the corresponding formulas above).
- The calculation of the resistance of the wiring is performed. The formula is as follows: conductor resistivity (p) * length (in meters). The resulting value must be divided by the selected cable cross-section.
R = (p * L) / S, where p is a tabular value
We draw your attention to the fact that the length of the current passage must be doubled, since the current initially flows through one core, and then returns back through the other.
- Voltage losses are calculated: the current is multiplied by the calculated resistance.
U losses = I load * R wires
LOSS = (U loss / U nom) * 100%
- The amount of losses is determined: voltage losses are divided by the voltage in the network and multiplied by 100%.
- The final number is analyzed. If the value is less than 5%, we leave the selected core section. Otherwise, we select a "thicker" conductor.
Suppose we calculated that the resistance was 0.5 Ohm, and the current was 16 Amperes, then:
U loss = 16 * 0.5 = 8 Volts
LOSS = (8/220) * 100% = 0.03636 * 100% = 3.6%
Resistivity table:
The correct selection of electrical cable is important to ensure an adequate level of safety, to use the cable cost-effectively and to fully utilize all the cable's capabilities. A well-calculated cross-section must be able to constantly work 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 performance of protective devices during a lack of grounding. That is why a scrupulous and accurate calculation of the cable cross-section in terms of power is performed, which today can be done quickly using our online calculator.
Calculations are made individually according to the formula for calculating the cable cross-section separately for each power cable for which you need to select a certain cross-section, or for a group of cables with similar characteristics. All methods for determining the size of the cable to one degree or another follow the main 6 points:
- Collecting data about the cable, its installation conditions, the load it will carry, etc.
- Determination of the minimum cable size based on the calculation of the amperage
- Determination of minimum cable size based on consideration of voltage drop
- Determination of the minimum cable size based on the rise in short-circuit temperature
- Determining the minimum cable size based on the loop impedance with insufficient grounding
- Choosing the most large sizes based on calculations of points 2, 3, 4 and 5
Online calculator for calculating cable cross-section by power
To use the online calculator for calculating the cable cross-section, it is necessary to collect the information necessary to perform the size calculation. As a rule, you need to obtain the following data:
- Detailed characteristic of the load that the cable will supply
- Purpose of the cable: for three-phase, single-phase or direct current
- System and / or source voltage
- Total load current in kW
- Full load power factor
- Starting power factor
- Cable length from source to load
- Cable construction
- Cable routing method
Copper and aluminum cable section tables
Copper cable section table
Aluminum cable section table
When determining most of the calculation parameters, the cable section calculation table presented on our website will come in handy. Since the main parameters are calculated based on the needs of the current consumer, all the initial ones can be quite easily calculated. However, the brand of the cable and wire, as well as the understanding of the cable design, also play 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 conductor losses, dielectric losses due to thermal insulation and resistive current losses. That is why the most basic is the calculation of the load, which takes into account all the features of the power cable supply, including thermal ones. The parts that make up the cable (for example, conductors, insulation, sheathing, armor, etc.) must be able to withstand the temperature rise and heat radiated from the cable.
Cable carrying capacity is the maximum current that 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 conductor cross-section areas have lower resistance losses and can dissipate heat better than thinner cables. Therefore, a cable with a 16 mm2 cross-section will have a higher current carrying capacity than a 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 AC systems, the voltage drop calculation method is usually used based on the load power factor. Generally, full load currents are used, but if the load was high at start-up (e.g. a motor), then the voltage drop based on the starting current (power and power factor, if applicable) should also be calculated and accounted for, since the 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 cross-section
Take advantage of others online calculator mi.
Standard apartment wiring is designed for a maximum current consumption with a continuous load of 25 amperes (the circuit breaker that is installed at the input of the wires to the apartment is also chosen for this current strength) is made 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 PUE the cross-section of the copper conductor for apartment 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 devices 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 you can see from the formula, the cross-section of the wire is easy on 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 diameter of the conductor can be determined with a vernier caliper with an accuracy of 0.1 mm or a micrometer with an accuracy of 0.01 mm. If there are no instruments at hand, then an ordinary ruler will help out.
Section selection
copper wire electrical wiring by current
The magnitude of the electric current is indicated by the letter " A»And is measured in amperes. When choosing, a simple rule applies, the larger the wire cross-section, the better; therefore, the result is rounded up.
| Table for the selection of 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 given in the table is based on personal experience and guarantee the reliable operation of electrical wiring under the most unfavorable conditions of its laying and operation. When choosing a wire cross-section according to the value of the current, it does not matter whether it is alternating current or direct. The magnitude and frequency of the voltage in the wiring also do not matter, it can be the on-board network of a DC car for 12 V or 24 V, an aircraft for 115 V with a frequency of 400 Hz, an electrical wiring of 220 V or 380 V with a frequency of 50 Hz, a high-voltage power line for 10,000 V.
If you do not know the current consumption of an electrical appliance, but you know the supply voltage and power, 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 consists in the fact that with an increase in 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 carried out according to other laws.
Determination of the load capacity of 220 V wiring
made of aluminum wire
In houses built long ago, electrical wiring is usually made of aluminum wires. If the connections in the junction boxes are correct, the service life of aluminum wiring can be up to one hundred years. After all, aluminum practically does not oxidize, and the service life of the 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 to withstand additional power by the cross-section or diameter of the wires of the wires. This is easy to do according to the table below.
If your wiring in the apartment is made of aluminum wires and there is a need to connect the newly installed outlet to junction box copper wires, then such a connection is performed in accordance with the recommendations of the article Connecting aluminum wires.
Calculation of the cross-section of the wire of electrical wiring
by the power of the connected electrical appliances
To select the cross-section of the conductors of the cable wire when laying electrical wiring in an apartment or house, you need to analyze the fleet of available electrical appliances from the point of view of their simultaneous use. The table provides a list of popular household electrical appliances with an indication of the consumed current depending on power. You can find out the power consumption of your models yourself from the labels on the products themselves or in the passports, often the parameters are indicated on the packaging.
If the strength of the current consumed by the electrical appliance is unknown, 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 electrical appliance | Power consumption, kW (kBA) | Power consumption, 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 the operating mode |
| Microwave | 1,5 – 2,2 | 7 – 10 | Periodically |
| Electric meat grinder | 1,5 – 2,2 | 7 – 10 | Depends on the 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 the operating mode |
| Coffee maker | 0,5 – 1,5 | 2 – 8 | Constantly |
| Electric oven | 1,0 – 2,0 | 5 – 9 | Depends on the operating mode |
| Dishwasher | 1,0 – 2,0 | 5 – 9 | |
| Washer | 1,2 – 2,0 | 6 – 9 | Maximum from the moment of switching on to heating the water |
| Drying machine | 2,0 – 3,0 | 9 – 13 | Constantly |
| Iron | 1,2 – 2,0 | 6 – 9 | Periodically |
| Vacuum cleaner | 0,8 – 2,0 | 4 – 9 | Depends on the operating mode |
| Heater | 0,5 – 3,0 | 2 – 13 | Depends on the operating mode |
| Hair dryer | 0,5 – 1,5 | 2 – 8 | Depends on the operating mode |
| Air conditioning | 1,0 – 3,0 | 5 – 13 | Depends on the operating mode |
| Stationary computer | 0,3 – 0,8 | 1 – 3 | Depends on the operating mode |
| Power tool (drill, jigsaw, etc.) | 0,5 – 2,5 | 2 – 13 | Depends on the operating mode |
The current is also consumed by a refrigerator, lighting devices, a radiotelephone, chargers, and a TV set on duty. But in total, this power is no more than 100 W and it can be ignored in the calculations.
If you turn on all the electrical appliances in the house at the same time, then you will need to choose a wire cross-section capable of passing a current of 160 A. You need a finger-thick wire! 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 will accordingly be 7 A + 8 A + 3 A + 4 A = 22 A. Taking into account the switched on lighting, refrigerator and in addition, for example, a TV, the current consumption can reach 25 A.
for 220 V network
You can choose the cross-section of the wire not only by the current strength but also by the amount of power consumed. To do this, you need to make a list of all electrical appliances planned for connection to this section of the 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 (kBA) | 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 it is necessary to determine the wire cross-section for each of them from the tables, and then add the results obtained.
The choice of the cross-section of the copper wire by power
for 12 V vehicle on-board network
If, when connecting additional equipment to the vehicle's on-board network, only its power consumption is known, then you can determine the cross-section of the additional wiring using the table below.
| Table for selecting the cross-section and diameter of copper wire by power for vehicle electrical system 12 V |
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 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 |
Selection of wire cross-section for connecting electrical appliances
to a three-phase 380 V network
When electrical appliances, for example, an electric motor, are connected to a three-phase network, the current consumed no longer flows through two wires, but through three and, therefore, the amount of current flowing in each individual wire is slightly less. This allows you to use a smaller wire to connect 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 220 V.
Attention, when choosing the cross-section of the wire for connecting the electric motor in terms of power, it should be taken into account that the maximum mechanical power that the motor can create on the shaft, and not the consumed electrical power, is indicated on the nameplate of the electric motor. The electrical power consumed by the electric motor, taking into account the efficiency and cos φ, is approximately two times greater than that created on the shaft, which must be taken into account when choosing the wire cross-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 an electric motor with a power of 2.0 kW to a three-phase 380 V network, you will need a copper three-core cable with a cross-section of each core of 0.5 mm 2.
It is much easier to choose the cross-section of the wire for connecting a three-phase motor, based on the value of its current consumption, which is always indicated on the nameplate. For example, in the nameplate shown in the photo, 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) only 0.7 A. Taking the current indicated on the nameplate, according to the table to select the wire cross-section for apartment wiring, select a wire with a cross section of 0.35 mm 2 when connecting the motor windings according to the "triangle" or 0.15 mm circuit 2 with star connection.
About choosing a brand of cable for home wiring
At first glance, it seems cheaper to make apartment wiring from aluminum wires, but operating costs due to the low reliability of contacts over time will many times exceed the cost of copper wiring. I recommend making the wiring exclusively from copper wires! Aluminum wires are indispensable when laying overhead wiring, as they are lightweight and cheap and, if properly connected, 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 of cross-section and installation, single-core is better. So for home wiring, you only need to use a single core wire. A stranded conductor 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 shaver, an electric iron and all the rest.
After deciding 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.
Cable PUNP since 1990, in accordance with the decision of Glavgosenergonadzor "On the prohibition of the use of wires such as APVN, PPBN, PEN, PUNP, etc., produced according to TU 16-505. 610-74 instead of wires АПВ, АППВ, PV and ППВ in accordance with GOST 6323-79 * "for use is prohibited.
VVG and VVGng cable - copper wires in double PVC insulation, flat shape. Designed to operate at ambient temperatures from -50 ° C to + 50 ° C, for wiring inside buildings, 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 with a cross-section of conductors from 1.5 to 35.0 mm 2 are produced. If the letter A (AVVG) is in front of VVG in the cable designation, then the wires in the wire are aluminum.
NYM cable (its Russian analogue is VVG cable), with copper conductors, round, with non-combustible insulation, complies with the German standard VDE 0250. Specifications and the area of application, almost the same with the VVG cable. Two-, three- and four-core with a cross-section of conductors from 1.5 to 4.0 mm 2 are produced.
As you can see, the choice for laying electrical wiring is not great and is determined depending on which shape of the cable is more suitable for installation, round or flat. A round cable is more conveniently laid through walls, especially if it is entered 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 wiring wires
There are desperate situations when you urgently need to lay the wiring, and the wires of the required cross-section are not available. In this case, if there is a wire of a smaller than necessary cross-section, then you can make the wiring from two or more wires by connecting them in parallel. The main thing is that the sum of the cross-sections of each of them is not 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 the calculations, 10 mm 2 is needed. Connect them all in parallel and the wiring will handle up to 50 amps. Yes, you yourself have seen many times the parallel connection of more thin conductors to transmit high 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 parallel-connected thin copper wires. In a car, the battery is also connected to the on-board network using the same flexible stranded wire, since when the engine is started, the starter consumes up to 100 A. ...
The method of increasing the cross-section of an electric wire by connecting several wires of different diameters in parallel can be used only as a last resort. When laying household wiring, it is permissible to connect in parallel only wires of the same cross section, taken from the same coil.
Online calculators for calculating the cross-section and diameter of the wire
With the help of the online calculator presented below, you can solve the opposite problem - to determine the diameter of the conductor by the cross-section.
How to calculate the cross section of a stranded wire
A 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.
Let's look at an example. There is a stranded flexible wire, in which 15 cores with a diameter of 0.5 mm. The cross-section of one core is 0.5 mm × 0.5 mm × 0.785 = 0.19625 mm 2, after rounding we get 0.2 mm 2. Since we have 15 wires in the wire, these numbers must be multiplied to determine the cable cross-section. 0.2 mm 2 × 15 = 3 mm 2. It remains to determine from the table that such a stranded wire will 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 the total diameter of all stranded wires. But since the wires are round, there are air gaps between them. To exclude the area of gaps, you need to multiply the result of the wire cross-section obtained by the formula by a factor of 0.91. When measuring the diameter, make sure that the stranded wire does not flatten.
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.