Determine the voltage loss on the device using the formula. Online calculation of voltage losses in a cable. What does the loss depend on?

At home, we often use portable extension cords - sockets for temporary ( usually remaining permanently) turning on household appliances: electric heater, air conditioner, iron with high current consumption.
The cable for this extension cord is usually selected according to the principle of whatever comes to hand, and this does not always correspond to the required electrical parameters.

Depending on the diameter (or cross-section of the wire in mm2), the wire has a certain electrical resistance for the passage of electric current.

The larger the cross-section of the conductor, the lower its electrical resistance, the lower the voltage drop across it. Accordingly, there is less power loss in the wire due to its heating.

Let us carry out a comparative analysis of the power loss for heating in the wire depending on its transverse sections. Let's take the most common cables in everyday life with a cross-section: 0.75; 1.5; 2.5 mm.sq. for two extension cords with cable length: L = 5 m and L = 10 m.

Let's take as an example a load in the form of a standard electric heater with electrical parameters:
- supply voltageU = 220 Vol T ;
— electric heater power P = 2.2 kW = 2200 W ;
— current consumption I = P/U = 2200 W / 220 V = 10 A.

From reference literature, let's take resistance data for 1 meter of wire of different cross sections.

A table of resistances of 1 meter of wire made of copper and aluminum is given.

Let's calculate the loss of power spent on heating for the cross-section of the wire S = 0.75 mm.sq. The wire is made of copper.

Resistance of 1 meter wire (from the table) R 1 = 0.023 Ohm.
Length of cable L=5 meters.
Length of wire in cable (round trip)2 L =2 · 5 = 10 meters.
Electrical resistance of a wire in a cable R = 2 · L · R 1 = 2 · 5 · 0.023 = 0.23 Ohm.

Voltage drop in the cable when current passes I = 10 A will: U = I R = 10 A 0.23 Ohm = 2.3 V.
The power loss due to heating in the cable itself will be: P = U I = 2.3 V 10 A = 23 W.

If the cable length L = 10 m. (same cross-section S = 0.75 mm2), the power loss in the cable will be 46 W. This is approximately 2% of the power consumed by the electric heater from the network.

For cables with aluminum conductors of the same section S = 0.75 mm.sq.. the readings increase and amount to L = 5 m-34.5 W. For L = 10 m - 69 W.

All calculation data for cables with a cross section of 0.75; 1.5; 2.5 mm.sq. for cable length L = 5 and L = 10 meters are given in the table.
Where: S – wire cross-section in mm2;
R 1– resistance of 1 meter of wire in Ohm;
R - cable resistance in Ohms;
U – voltage drop in the cable in Volts;
P – power loss in the cable in watts or as a percentage.

What conclusions should be drawn from these calculations?

• — With the same cross-section, a copper cable has a greater margin of safety and less electrical power loss due to heating of the wire P.
• — As the cable length increases, losses P increase. To compensate for losses, it is necessary to increase the cross-section of the cable wires S.
• — It is advisable to choose a cable with a rubber sheath, and the cable cores should be multi-core.

For the extension cord, it is advisable to use a Euro socket and Euro plug. The pins of the Euro plug have a diameter of 5 mm. A simple electric plug has a pin diameter of 4 mm. Euro plugs are designed to carry more current than a simple socket and plug. The larger the diameter of the plug pins, the larger the contact area at the junction of the plug and socket,hence lower contact resistance. This contributes to less heating at the junction of the plug and socket.

When designing electrical networks with low currents, voltage losses in conductors are often calculated. The results obtained are then used to determine the optimal cross-section of current-carrying conductors. If a mistake is made during the selection of wires and cables, the electrical system will quickly fail or will not start at all. To carry out the necessary calculations, special formulas or online calculators are used.

Reasons for losses

Every electrician knows that cables are made up of cores. They are made of copper or aluminum and covered with an insulating layer. To protect against mechanical damage, the conductors are placed in an additional polymer sheath. Since the current-carrying conductors are densely located and compressed by a protective coating, when the line is long, they begin to work on the principle of a capacitor. To put it simply, a charge is created in the cores that has capacitive reactance.

The voltage loss diagram in the wires looks like this:

If this process is represented graphically, then the segment AD will be an indicator of losses.

Performing such calculations manually is quite difficult and an online calculator is now often used. Voltage losses calculated with its help turn out to be quite accurate, and the error is minimal.

Consequences of voltage reduction

In accordance with regulatory documentation, losses on the main line from the transformer to the most remote point for public facilities should not exceed 9%. As for possible losses at the point where the line enters the end user, this figure should be no more than 4%.

In case of deviation from the specified limits, the following consequences are possible:

• Volatile equipment will not be able to function normally.
• If the input voltage is low, electrical appliances may malfunction.
• The current load will not be distributed evenly between consumers.

High demands are placed on the characteristics of power lines. When designing them, it is necessary to calculate possible losses not only in the main networks, but also in the secondary ones.

Several methods can be used to calculate voltage losses. It’s worth considering everything so that every electrician can choose the most attractive one depending on the situation.

Using tables and formulas

In practice, when installing electric mains, copper or aluminum conductors are used. Knowing the resistivity of these materials, as well as the current strength and wire resistance, you can use the following voltage drop formulas:

A home master and even a specialist can use special tables. This is a fairly convenient and simple way to carry out the necessary calculations. However, in some cases it is necessary to obtain the most reliable result, taking into account the indicators of active and reactance. In such a situation, you have to use a more complex formula:

To ensure optimal load in a three-phase network, each phase must be loaded evenly. To solve this problem, electric motors should be connected to linear conductors, and lamps should be connected between the neutral line and phases.

Online services

The use of formulas, graphs and tables is a rather labor-intensive process. It is not always necessary to get the most accurate results, and in such a situation it is worth using online calculators. These services work as follows:

• Current indicators, conductor material, cross-section of current-carrying conductors and line length are entered into the program.
• You will also need to provide information on the number of phases, network voltage, power and line temperature during operation.
• After entering all the necessary data, the program will automatically perform all the necessary calculations.

At the preliminary design stage, it is worth using several services and then determining the average value. It should be recognized that there is a certain error in calculations when using online calculators.

Reduce losses

It is quite obvious that losses depend on the length of the conductor in the line. The higher this parameter is, the more the voltage drops. Several methods can be used to reduce losses:

The latter method works great in power networks that have several backup lines. It should also be remembered that the voltage may drop as the temperature of the cable increases. If additional thermal insulation measures are used during cable installation, losses can be reduced.

In the energy industry, calculating the voltage drop on the main line is one of the most important tasks. If all calculations were carried out correctly, then the consumer will not have problems with the operation of electrical equipment.

Electrical appliances require certain network parameters to operate. Wires have resistance to electric current, so when choosing a cable cross-section, it is necessary to take into account the voltage drop in the wires.

What is voltage drop

When measuring in different parts of a wire through which electric current flows, a change in potential is observed as it moves from the source to the load. The reason for this is the resistance of the wires.

How is voltage drop measured?

There are three ways to measure fall:

• Two voltmeters. Measurements are taken at the beginning and end of the cable;
• Alternately in different places. The disadvantage of the method is that during transitions the load or network parameters may change, which will affect the readings;
• One device connected in parallel to the cable. The voltage drop in the cable is small, and the connecting wires are long, which leads to errors.

Important! The voltage drop can be from 0.1V, so devices are used with an accuracy class of at least 0.2.

Resistance of metals

Electric current is the directed movement of charged particles. In metals, this is the movement of free electrons through a crystal lattice, which resists this movement.

In calculations, resistivity is denoted by the letter “p” and corresponds to the resistance of one meter of wire with a cross-section of 1 mm².

For the most common metals used to make wires, copper and aluminum, this parameter is 0.017 and 0.026 Ohm*m/mm², respectively. The resistance of a piece of wire is calculated by the formula:

R=(p*l)/S, where:

• l – length,
• S – cable section.

For example, 100 meters of copper wire with a cross section of 4mm² has a resistance of 0.425 Ohms.

If the cross section S is unknown, then, knowing the diameter of the conductor, it is calculated as:

S=(π*d²)/4, where:

• π – number “pi” (3.14),
• d – diameter.

How to calculate voltage loss

According to Ohm's law, when current flows through a resistance, a potential difference appears across it. In this section of cable, at a current of 53A, permissible with open installation, the drop will be U=I*R=53A*0.425Ohm=22.5V.

For normal operation of electrical equipment, the network voltage should not exceed ±5%. For a household network 220V is 209-231V, and for a three-phase network 380V the permissible fluctuation limits are 361-399V.

When the power consumption and current in electrical cables change, the voltage drop in the conductors and its value near the consumer changes. These fluctuations must be taken into account when designing power supplies.

Selection based on acceptable losses

When calculating losses, it is necessary to take into account that a single-phase network uses two wires, Accordingly, the formula for calculating the voltage drop changes:

In a three-phase network the situation is more complicated. With a uniform load, for example, in an electric motor, the powers connected to the phase wires compensate each other, the current does not flow through the neutral wire, and its length is not taken into account in the calculations.

If the load is uneven, as in electric stoves, in which only one heating element can be turned on, then the calculation is carried out according to the rules of a single-phase network.

In long-distance lines, in addition to active ones, inductive and capacitive reactance is also taken into account.

The calculation can be done using tables or using an online calculator. In the previously given example, in a single-phase network and at a distance of 100 meters, the required cross-section will be at least 16 mm², and in a three-phase network - 10 mm².

Selection of cable cross-section for heating

The current flowing through the resistance releases energy P, the value of which is calculated by the formula:

In the cable from the previous example, P=40A²*0.425Ohm=680W. Despite the length, this is enough to heat the conductor.

When the wire is heated above the permissible temperature, the insulation fails, which leads to a short circuit. The amount of permissible current depends on the material of the conductor, insulation and installation conditions. To select, you must use special tables or an online calculator.

How to reduce the voltage drop in a cable

When laying electrical wiring over long distances, the cable cross-section selected for the permissible voltage drop is many times greater than the choice made for heating, which leads to an increase in the cost of power supply. But there are ways to reduce these costs:

• Increase the potential at the beginning of the supply cable. This is only possible when connected to a separate transformer, for example, in a holiday village or microdistrict. If some consumers are disconnected, the potential in the sockets of the rest will be overestimated;
• Installation near the stabilizer load. This requires costs, but guarantees constant network parameters;
• When connecting a 12-36V load through a step-down transformer or power supply, place them near the consumer.

Reference. When the voltage decreases, the current in the network, the voltage drop and the required wire cross-section increase.

Ways to reduce cable losses

In addition to disrupting the normal operation of electrical appliances, a voltage drop in the wires leads to additional energy costs. These costs can be reduced in different ways:

• Increasing the cross-section of the supply wires. This method requires significant costs for cable replacement and careful feasibility testing;
• Reducing line length. A straight line connecting two points is always shorter than a curve or broken line. Therefore, when designing power supply networks, lines should be laid as short as possible;
• Decrease in ambient temperature. When heated, the resistance of metals increases, and electricity losses in the cable increase;
• Reducing the load. This option is possible if there are a large number of consumers and power sources;
• Bringing cosφ to 1 near the load. This reduces current consumption and losses.

Important! All changes must be reflected on diagrams.

For your information. Improving ventilation in cable trays and other structures reduces temperature, resistance and line losses.

To achieve maximum effect, it is necessary to combine these methods with each other and with other energy saving methods.

Calculation of voltage drop and electricity losses in a cable is important when designing power supply systems and cable lines.

Video

Wires and cables are designed to transmit electricity to consumers. In this case, the voltage in an extended conductor drops in proportion to its resistance and the magnitude of the passing current. As a result, the voltage supplied to the consumer is slightly less than it was at the source (at the beginning of the line). Along the entire length of the wire, the potential will change due to losses in it.

Voltage losses in home lighting

The cable cross-section is selected to ensure its operability at a given maximum current. In this case, one should take into account its length, on which another important parameter depends - the voltage drop.

Power lines are selected according to the normalized value of the economic current density and the voltage drop is calculated. Its deviation from the original must not exceed the specified values.

The amount of current passing through the conductor depends on the connected load. As it increases, heating losses also increase.

The figure above shows a circuit for supplying voltage to lighting, where voltage losses are indicated at each section. The furthest load is the most important, and most of the voltage loss occurs for it.

Voltage loss

Calculation of voltage loss ∆Uon a section of chain lengthLdo according to the formula:

∆U = (P∙r 0 +Q∙x 0)∙L/ U nom, where

• P and Q – power, W and var (active and reactive);
• r 0 and x 0 – active and reactive resistance of the line, Ohm/m;
• U nom – rated voltage, V.
• U nom is indicated in the characteristics of electrical appliances.

According to the PUE, the permissible voltage deviations from the norm are as follows:

• power circuits – no higher than ±5%;
• lighting schemes for residential premises and outside buildings – up to ±5%;
• lighting of enterprises and public buildings – from +5% to -2.5%.

The total voltage loss from transformer substations to the most remote load in public and residential buildings should not exceed 9%. Of these, 5% relates to the section up to the main input and 4% from the input to the consumer. In accordance with GOST 29322-2014, the voltage rating in three-phase networks is 400 V. In this case, a deviation from it of ±10% is allowed under normal operating conditions.

It is necessary to ensure a uniform load in three-phase lines at 0.4 kV. It is important here that each phase is loaded evenly. To do this, electric motors are connected to linear wires, and lighting is connected between phases and neutral, thus equalizing the loads across phases.

Current or power values ​​are used as initial data. For long lines, inductive reactance is taken into account when ∆U in the line is calculated.

Resistance x 0 wires are taken in the range from 0.32 to 0.44 Ohm/km.

Calculation of losses in conductors is carried out using the previously given formula, where it is convenient to divide the right side into active and reactive components:

∆U = P∙r 0 ∙L / U nom + Q∙x 0 ∙L/ U nom,

Load connection

The load is connected in different ways. The most common are the following:

• connecting the load at the end of the line (Fig. a below);
• uniform distribution of loads along the length of the line (Fig. b);
• line L1, to which another line L2 is connected with uniformly distributed loads (Fig. c).

Diagram showing how to connect loads from the electrical panel

Calculation of power lines for voltage loss

1. Selecting the average value of reactance for conductors made of aluminum or steel-aluminum, for example, 0.35 Ohm/km.
2. Calculation of loads P, Q.
3. Calculation of reactive loss:

∆U p = Q∙x 0 ∙L/U nom.

Determination of the permissible active loss from the difference between the voltage loss, which is specified, and the calculated reactive one:

∆U a = ∆U – ∆U p .

The wire cross-section is found from the relation:

s = P∙L∙r 0 /(∆U a ∙U nom).

Selecting the nearest cross-section value from the standard series and determining the active and reactive resistance per 1 km of line from the table.

The figure shows a number of cross-sections of cable cores of different sizes.

Cable cores of different sections

Based on the obtained values, the adjusted voltage drop value is calculated using the formula given earlier. If it exceeds the permissible value, you should take a larger wire from the same row and make a new calculation.

Example 1. Cable calculation under active loads.

To calculate the cable, first of all, you should determine the total load of all consumers. P = 3.8 kW can be taken as the initial value. The current strength is determined by the well-known formula:

If all loads are active, cosφ=1.

By substituting the values ​​into the formula, you can find the current, which will be equal to: I = 3.8∙1000/220 = 17.3 A.

According to the tables, the cross-section in the cable is found, for copper conductors it is 1.5 mm 2.

Now you can find the resistance of a cable 20 m long: R=2∙r 0 ∙L/s=2∙0.0175 (Ohm∙mm 2)∙20 (m)/1.5 (mm 2)=0.464 Ohm.

The formula for calculating resistance for a two-core cable takes into account the length of both wires.

Having determined the cable resistance value, you can easily find the voltage loss: ∆U=I∙R/U∙100% =17.3 A∙0.464 Ohm/220 V∙100%=3.65%.

If the rated voltage at the input is 220 V, then the permissible deviations to the load are 5%, and the result obtained does not exceed it. If the tolerance had been exceeded, it would have been necessary to take a larger wire from the standard range, with a cross-section of 2.5 mm 2.

Example 2. Calculation of the voltage drop when power is supplied to the electric motor.

The electric motor consumes current under the following parameters:

• I nom = 100 A;
• cos φ = 0.8 in normal mode;
• I starting = 500 A;
• cos φ = 0.35 at start-up;
• The voltage drop across an electrical panel distributing 1000 A current is 10 V.

In Fig. and below is a diagram of the power supply of the electric motor.

Power supply circuits for the electric motor (a) and lighting (b)

To avoid calculations, tables that are sufficiently accurate for practical use are used with already calculated ∆U between phases in a cable 1 km long at a current value of 1 A. The table below takes into account the cross-sectional values ​​of the cores, conductor materials, and type of circuit.

Table for determining voltage loss in a cable

Section in mm 2		Single phase circuit			Balanced three-phase circuit
		Motor power		Lighting	Motor power		Lighting
		An ordinary slave. mode	Launch		An ordinary slave. mode	Launch
Cu	Al	cos = 0.8	cos = 0.35	cos = 1	cos = 0.8	cos = 0.35	cos = 1
1.5		24	10,6	30	20	9,4	25
2,5		14,4	6,4	18	12	5,7	15
4		9,1	4,1	11,2	8	3,6	9,5
6	10	6,1	2,9	7,5	5,3	2,5	6,2
10	16	3,7	1,7	4,5	3,2	1,5	3,6
16	25	2,36	1,15	2,8	2,05	1	2,4
25	35	1,5	0,75	1,8	1,3	0,65	1,5
35	50	1,15	0,6	1,29	1	0,52	1,1
50	70	0,86	0,47	0,95	0,75	0,41	0,77
70	120	0,64	0,37	0,64	0,56	0,32	0,55
95	150	0,48	0,30	0,47	0,42	0,26	0,4
120	185	0,39	0,26	0,37	0,34	0,23	0,31
150	240	0,33	0,24	0,30	0,29	0,21	0,27
185	300	0,29	0,22	0,24	0,25	0,19	0,2
240	400	0,24	0,2	0,19	0,21	0,17	0,16
300	500	0,21	0,19	0,15	0,18	0,16	0,13

The voltage drop during normal operation of the electric motor will be:

∆U% = 100∆U/U nom.

For a cross section of 35 mm, 2 ∆U for a current of 1 A will be 1 V/km. Then, with a current of 100 A and a cable length of 0.05 km, the losses will be equal to ∆U = 1 V/A km∙100 A∙ 0.05 km = 5 V. When adding to them the voltage drop on the panel of 10 V, the total losses ∆ U total = 10 V + 5 V = 15 V. As a result, the percentage losses will be:

∆U% = 100∙15/400 = 3.75%.

This value is significantly less than the permitted losses (8%) and is considered acceptable.

When the electric motor starts, its current increases to 500 A. This is 400 V more than its rated current. The load on the distribution board will increase by the same amount. It will be 1400 A. The voltage drop across it will increase proportionally:

∆U = 10∙1400/1000 = 14 V.

According to the table, the voltage drop in the cable will be: ∆U = 0.52∙500∙0.05 = 13 V. In total, the motor starting losses will be ∆U total = 13+14 = 27 V. Then you should determine how much this will be as a percentage relation: ∆U = 27/400∙100 =6.75%. The result is within acceptable limits, since it does not exceed the limit of 8%.

Protection for the electric motor should be selected in such a way that the response voltage is greater than at start-up.

Example 3. Calculation of ∆U in lighting circuits.

Three single-phase lighting circuits are connected in parallel to a three-phase four-wire supply line consisting of 70 mm 2 conductors, 50 m long, carrying a current of 150 A. Lighting is only part of the line load (Fig. b above).

Each lighting circuit is made of copper wire 20 m long, with a cross-section of 2.5 mm 2 and carries a current of 20 A. All three loads are connected to the same phase. In this case, the power line is load balanced.

It is required to determine the voltage drop in each of the lighting circuits.

The voltage drop in a three-phase line is determined by the effective load specified in the example conditions: ∆U phase line = 0.55∙150∙0.05 = 4.125 V. This is the loss between phases. To solve the problem, you need to find the losses between phase and neutral: ∆U line fn = 4.125/√3 = 2.4 V.

The voltage drop for one single-phase circuit is ∆U total = 18∙20∙0.02 = 7.2 V. If you add up the losses in the supply line and circuit, then in total they will be ∆U total = 2.4+7.2 = 9.6 V. As a percentage, it will be 9.6/230∙100 = 4.2%. The result is satisfactory since it is less than the permissible value of 6%.

Voltage check. Video

How to check the voltage drop on different types of cables can be found in the video below.

When connecting electrical appliances, it is important to correctly calculate and select the supply cables and wires so that the voltage losses in them do not exceed the permissible values. Losses in the supply network are also added to them, which should be summed up.

In order to ensure the supply of voltage from the distribution device to the end consumer, power lines are used. They can be overhead or cable and have a considerable length.

Like all conductors, they have a resistance that depends on the length and the longer they are, the greater the voltage loss.

And the longer the line, the greater the voltage loss will be. Those. The voltage at the input and at the end of the line will be different.

In order for the equipment to operate without failures, these losses are normalized. Their total value should not exceed 9%.

The maximum voltage drop at the input is five percent, and to the most remote consumer no more than four percent. In a three-phase network with a three or four wire network, this figure should not exceed 10%.

If these indicators are not met, end users will not be able to provide the nominal parameters. When the voltage decreases, the following symptoms occur:

• Lighting devices that use incandescent lamps begin to work (glow) at half incandescence;
• When the electric motors are turned on, the starting force on the shaft decreases. As a result, the motor does not rotate, and as a result, the windings overheat and fail;
• Some electrical appliances do not turn on. There is not enough voltage, and other devices may fail after switching on;
• Installations that are sensitive to input voltage are unstable, and light sources that do not have an incandescent filament may also not turn on.

Electricity is transmitted via overhead or cable networks. Overhead ones are made of aluminum, while cable ones can be aluminum or copper.

In addition to active resistance, cables contain capacitive reactance. Therefore, the power loss depends on the cable length.

Reasons leading to a decrease in voltage

Voltage losses in power lines occur for the following reasons:

• A current passes through the wire, which heats it, as a result, the active and capacitive resistance increases;
• A three-phase cable with a symmetrical load has the same voltage values ​​on the cores, and the neutral wire current will tend to zero. This is true if the load is constant and purely active, which is impossible in real conditions;
• In networks, in addition to the active load, there is a reactive load in the form of transformer windings, reactors, etc. and as a result, inductive power appears in them;
• As a result, the resistance will consist of active, capacitive and inductive. It affects voltage losses in the network.

Current losses depend on the cable length. The longer it is, the greater the resistance, which means that the losses are greater. It follows that power losses in a cable depend on the length or length of the line.

Loss Value Calculation

To ensure the operability of the equipment, it is necessary to make a calculation. It is carried out at the time of design. The current level of development of computer technology allows calculations to be made using an online calculator, which allows you to quickly calculate cable power losses.

To calculate, just enter the required data. Set the current parameters - direct or alternating. The power line material is aluminum or copper. Indicate by what parameters the power loss is calculated - by cross-section or diameter of the wire, load current or resistance.

Additionally, indicate the network voltage and cable temperature (depending on operating conditions and installation method). These values ​​are inserted into the calculation table and calculated using an electronic calculator.

You can make a calculation based on mathematical formulas. In order to correctly understand and evaluate the processes occurring during the transmission of electrical energy, a vector form of representing characteristics is used.

And to minimize calculations, a three-phase network is represented as three single-phase networks. Network resistance is represented as a series connection of active and reactive resistance to the load resistance.

In this case, the formula for calculating power loss in a cable is significantly simplified. To obtain the necessary parameters, use the formula.

This formula shows the power loss of a cable as a function of the current and resistance distributed along the length of the cable.

However, this formula is valid if you know the current strength and resistance. Resistance can be calculated using the formula. For copper it will be equal to p=0.0175 Ohm*mm2/m, and for aluminum p=0.028 Ohm*mm2/m.

Knowing the value of resistivity, calculate the resistance, which will be determined by the formula

R=р*I/S, where р is resistivity, I is line length, S is cross-sectional area of ​​the wire.

In order to calculate voltage losses along the cable length, you need to substitute the obtained values ​​into the formula and perform calculations. These calculations can be made when installing electrical networks or security systems and video surveillance.

If power loss calculations are not made, this may lead to a decrease in the supply voltage to consumers. As a result, the cable will overheat, it may become very hot, and as a result, the insulation will be damaged.

Which may cause electric shock or short circuit to people. A decrease in line voltage can lead to failure of electronic equipment.

Therefore, when designing electrical wiring, it is important to calculate the voltage loss in the supply wires and the laid cable.

Loss reduction methods

Power losses can be reduced by the following methods:

• Increase the cross-section of conductors. As a result, resistance will decrease and losses will decrease;
• Reduced power consumption. This setting cannot always be changed;
• Changing the cable length.

Reducing power and changing line length is practically impossible. Therefore, if you increase the cross-section of the wire without calculation, then on a long line this will lead to unjustified costs.

This means that it is very important to make a calculation that will allow you to correctly calculate the power losses in the cable and select the optimal cross-section of the cores.