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 a cable 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 areaat the junction of the plug and socket,

hence lower contact resistance. This contributes to less heating at the junction of the plug and socket.

Content:

Any cable is limited in its bandwidth. For this reason, conditions may arise in the electrical network when the voltage is insufficient for normal operation of the equipment. This phenomenon often occurs, and for this reason it deserves a more detailed consideration, which will be done later in our article.

Main causes of voltage drop

  • So, the cable throughput is influenced by its two main parameters:
  • cross-sectional area;

length.

But the current strength in the conductors is precisely the physical quantity with which the listed parameters are inextricably linked according to Ohm’s law for a section of the electrical circuit:

But the phenomena considered are fully consistent with direct current cables used mainly for electric transport. Otherwise, this is only part of what is included in the concept of voltage drop (ΔU) along the length of a cable operating in an industrial power supply network in which alternating voltage operates. Under these conditions, any conductor is characterized by an impedance that takes into account its inductance and capacitance, which form the reactive component of voltage and current. Therefore, in general, we get a complex problem, which, in essence, comes down to electricity losses. And ΔU is their objective manifestation (see explanatory image below):

We remind you that in electrical engineering complex numbers are used to calculate voltages and currents involving a load calculated by impedance. Inductance and capacitance cause a shift between current and voltage. Therefore, a complex number can be represented graphically. One vector is an active component, the other is a reactive one. The shift between current and voltage is characterized by the angle between the two vectors mentioned, emerging from a common point. In the image above, the above is represented by vector diagrams in red.

Options for determining ΔU

Vector method

During the design of an electrical network, the basis is the load, the operability of which must be ensured. If the cable is not selected correctly, the ΔU on it will not allow the load to operate correctly. Asynchronous motors will not reach the specified speed, transformers on the secondary windings will not provide rated voltages, etc., etc. For a single-phase network, the load is divided into active and reactive components.

A three-phase network is represented as three independent single-phase networks. They are called equivalent circuits. This method provides fairly accurate results if the load is symmetrical. If the symmetry is broken, then an analysis of the reasons that caused this process can also be performed using this method. Based on known quantities, you can construct a vector diagram and, by changing the length of the vectors according to the task at hand, determine the quantities that are necessary.

For example, the parameters that are necessary for normal operation of the load are known. The line parameters are also known. Consequently, the task comes down to determining the vector voltage U1. The steps leading to the appearance of the desired vector are shown below.

The length of the vector and its direction are determined based on Ohm's law and the direction of the voltage vector that determines the current (the current and voltage vectors coincide in direction). The voltage vector, which is obtained as a result of the addition of the active and reactive components of the load (IR + IХ), is ΔU in the line connecting the voltage source U1 to the load. From the obtained vectors it is easy to also obtain the voltage losses. To do this, vectors U1 and U2 are combined so that the direction of both is the same as that of vector U2. The difference between them is in length - this will be a voltage loss.

Knorring tables

But constructing vectors is quite tedious. Moreover, since the need for designing electrical networks has existed, faster solutions have been invented for standard situations. These include Knorring tables. The standard situation for them is the constant voltage at the input of the cable or other conductor (alternating voltage with an effective value of 220 V). This is important for both single phase and three phases. That is, in a three-phase electrical network, the load must be symmetrical.

It is also necessary to have the cross-sectional area of ​​the conductor (in square millimeters), the length of the conductor (in meters) and the load power (in kilowatts). We get the product of power and length, in the column starting with the appropriate cross-section of the core, we find this value, and in the leftmost column we look at ΔU on the cable. That's all. Two versions of tables for the voltage of a single-phase and three-phase electrical network, as well as one for a voltage of 12 V, shown below, can be used by the reader for calculations.

Consumers of electrical energy operate normally when their terminals are supplied with the voltage for which the electric motor or device is designed. When transmitting electricity through wires, part of the voltage is lost due to the resistance of the wires and as a result, at the end of the line, i.e. at the consumer, the voltage is lower than at the beginning of the line.

A decrease in consumer voltage compared to normal affects the operation of the pantograph, be it a power or lighting load. Therefore, when calculating any power transmission line, voltage deviations should not exceed permissible norms; networks selected by load current and designed for heating are, as a rule, checked by voltage loss.

Voltage loss Δ U called the voltage difference at the beginning and end of the line (section of the line). ΔU is usually determined in relative units - in relation to the rated voltage. Analytically, the voltage loss is determined by the formula:

where P is active power, kW, Q is reactive power, kvar, ro is active line resistance, Ohm/km, xo is inductive resistance of the line, Ohm/km, l is line length, km, Unom is rated voltage, kV.

The values ​​of active and inductive resistance (Ohm/km) for overhead lines made with A-16 A-120 grade wire are given in the reference tables. The active resistance of 1 km of aluminum (grade A) and steel-aluminum (grade AC) conductors can also be determined by the formula:

where F is the cross-section of the aluminum wire or the cross-section of the aluminum part of the AC wire, mm 2 (the conductivity of the steel part of the AC wire is not taken into account).

According to the PUE (“Rules for the Construction of Electrical Installations”), for power networks the voltage deviation from normal should be no more than ± 5%, for electrical lighting networks of industrial enterprises and public buildings - from +5 to - 2.5%, for residential electrical lighting networks buildings and outdoor lighting ±5%. When calculating networks, they are based on the permissible voltage loss.

Taking into account the experience of designing and operating electrical networks, the following permissible values ​​of voltage loss are accepted: for low voltage - from the busbars of the transformer room to the most remote consumer - 6%, and this loss is distributed approximately as follows: from the station or step-down transformer substation and to the entrance to the room depending on the load density - from 3.5 to 5%, from the input to the most remote consumer - from 1 to 2.5%, for high voltage networks under normal operating conditions in cable networks - 6%, in air networks - 8%, during network emergency mode in cable networks - 10% and in air networks - 12%.

It is believed that three-phase three-wire lines with a voltage of 6-10 kV operate with a uniform load, that is, that each of the phases of such a line is loaded evenly. In low voltage networks, due to the lighting load, it can be difficult to achieve uniform distribution between the phases, so a 4-wire three-phase current system of 380/220 V is most often used. With this system, electric motors are connected to linear wires, and lighting is distributed between linear and zero wires. In this way, the load on all three phases is equalized.

When calculating, you can use both given powers and current values ​​that correspond to these powers. In lines that extend for several kilometers, which, in particular, applies to lines with a voltage of 6-10 kV, it is necessary to take into account the influence of the inductive reactance of the wire on the voltage loss in the line.

For calculations, the inductive reactance of copper and aluminum wires can be taken equal to 0.32-0.44 Ohm/km, and a smaller value should be taken for small distances between wires (500-600 mm) and wire sections above 95 mm2, and a larger value for distances 1000 mm and above and sections 10-25 mm2.

The voltage loss in each wire of a three-phase line, taking into account the inductive resistance of the wires, is calculated using the formula

where the first term on the right side represents the active, and the second - the reactive component of the voltage loss.

The procedure for calculating a power transmission line for voltage loss with wires made of non-ferrous metals, taking into account the inductive reactance of the wires, is as follows:

1. We set the average value of inductive reactance for an aluminum or steel-aluminum wire to 0.35 Ohm/km.

2. We calculate the active and reactive loads P, Q.

3. Calculate the reactive (inductive) voltage loss

4. The permissible active voltage loss is defined as the difference between the specified line voltage loss and the reactive one:

5. Determine the wire cross-section s, mm2

Where γ is the reciprocal of resistivity (γ = 1/ro - conductivity).

6. We select the nearest standard value s and find for it from the reference table the active and inductive reactance per 1 km of line (ro, xo).

7. We calculate the adjusted value using the formula.

The resulting value should not be greater than the permissible voltage loss. If it turns out to be more than permissible, then you will have to take a wire of a larger (next) cross-section and perform the calculation again.

For direct current lines, there is no inductive reactance and the general formulas given above are simplified.

Calculation of networks constant current based on voltage losses.

Let the power P, W, be transmitted along a line of length l, mm, this power corresponds to the current

where U is the rated voltage, V.

Line wire resistance at both ends

where p is the resistivity of the wire, s is the cross-section of the wire, mm2.

Line voltage loss

The last expression makes it possible to make a test calculation of the voltage loss in an existing line when its load is known, or to select a wire cross-section for a given load

Power lines transport current from the switchgear to the final consumer along current-carrying conductors of varying lengths. The voltage will not be the same at the entry and exit points due to losses resulting from the large length of the conductor.

Voltage drop along cable length occurs due to the passage of high current, causing an increase in the resistance of the conductor.

On lines of considerable length, losses will be higher than when current passes through short conductors of the same cross-section. To ensure that the required voltage is supplied to the final object, it is necessary to calculate the installation of lines taking into account losses in the current-carrying cable, starting from the length of the conductor.

Result of undervoltage

According to regulatory documents, losses on the line from the transformer to the most remote energy-loaded area for residential and public facilities should be no more than nine percent.

Losses of 5% are allowed to the main input, and 4% - from the input to the final consumer. For three-phase, three- or four-wire systems, the rating should be 400 V ± 10% under normal operating conditions.

Deviation of a parameter from the normalized value can have the following consequences:

  1. Incorrect operation of volatile installations, equipment, lighting devices.
  2. Failure of electrical appliances to operate when the input voltage is reduced, equipment failure.
  3. Reduced acceleration of the torque of electric motors at starting current, loss of energy taken into account, shutdown of motors when overheated.
  4. Uneven distribution of current load between consumers at the beginning of the line and at the remote end of a long wire.
  5. Lighting devices operate at half heat, resulting in underutilization of current power in the network and loss of electricity.

In operating mode the most acceptable indicator voltage loss in the cable is considered 5%. This is the optimal calculated value that can be accepted as acceptable for power grids, since in the energy industry currents of enormous power are transported over long distances.

Increased demands are placed on the characteristics of power lines. It is important to pay special attention to voltage losses not only on main networks, but also on secondary lines.

Causes of voltage drop

Every electromechanic knows that a cable consists of conductors - in practice, conductors with copper or aluminum cores wrapped in insulating material are used. The wire is placed in a sealed polymer shell - a dielectric housing.

Since the metal conductors are located too tightly in the cable and are additionally pressed by layers of insulation, when the power line is long, the metal cores begin to work on the principle of a capacitor, creating a charge with capacitive resistance.

The voltage drop occurs according to the following scheme:

  1. The conductor carrying the current overheats and creates capacitance as part of the reactance.
  2. Under the influence of transformations occurring on the windings of transformers, reactors, and other circuit elements, the power of electricity becomes inductive.
  3. As a result, the resistive resistance of the metal cores is converted into active resistance of each phase of the electrical circuit.
  4. The cable is connected to a current load with a total (complex) resistance along each current-carrying core.
  5. When operating a cable in a three-phase circuit, the three current lines in the three phases will be symmetrical, and the neutral core passes a current close to zero.
  6. The complex resistance of conductors leads to voltage loss in the cable when current passes with vector deviation due to the reactive component.

Graphically, the voltage drop diagram can be represented as follows: a straight horizontal line emerges from one point - the current vector. From the same point, the input voltage vector U1 and the output voltage vector U2 come out at an angle to the current at a smaller angle. Then the voltage drop along the line is equal to the geometric difference between the vectors U1 and U2.

Figure 1. Graphical representation of voltage drop

In the figure shown, the right triangle ABC represents the voltage drop and loss along a long cable line. Segment AB is the hypotenuse of a right triangle and at the same time the drop, legs AC and BC show the voltage drop taking into account active and reactance, and segment AD demonstrates the amount of losses.

It is quite difficult to make such calculations manually. The graph serves to visually represent the processes occurring in a long-distance electrical circuit when a current of a given load passes.

Calculation using formula

In practice, when installing trunk-type power lines and distributing cables to the end consumer with further distribution on site, copper or aluminum cable is used.

The resistivity for conductors is constant, for copper p = 0.0175 Ohm*mm2/m, for aluminum conductors p = 0.028 Ohm*mm2/m.

Knowing the resistance and current strength, it is easy to calculate the voltage using the formula U = RI and the formula R = p*l/S, where the following values ​​are used:

  • Wire resistivity - p.
  • The length of the current-carrying cable is l.
  • Conductor cross-sectional area - S.
  • Load current in amperes - I.
  • Conductor resistance - R.
  • The voltage in the electrical circuit is U.

Using simple formulas on a simple example: it is planned to install several sockets in a detached extension of a private house. A copper conductor with a cross section of 1.5 square meters was selected for installation. mm, although for an aluminum cable the essence of the calculations does not change.

Since the current passes back and forth through the wires, you need to take into account that the distance of the cable length will have to be doubled. If we assume that the sockets will be installed forty meters from the house, and the maximum power of the devices is 4 kW with a current of 16 A, then using the formula it is easy to calculate the voltage losses:

U = 0.0175*40*2/1.5*16

If we compare the obtained value with the nominal value for a single-phase line 220 V 50 Hz, it turns out that the voltage loss was: 220-14.93 = 205.07 V.

Such losses of 14.93 V are practically 6.8% of the input (nominal) voltage in the network. A value that is unacceptable for the power group of sockets and lighting fixtures, the losses will be noticeable: the sockets will pass current at less than full power, and the lighting fixtures will operate with less heat.

The power for heating the conductor will be P = UI = 14.93*16 = 238.9 W. This is the percentage of losses in theory without taking into account the voltage drop at the connection points of the wires and the contacts of the socket group.

Carrying out complex calculations

For a more detailed and reliable calculation of voltage losses on the line, it is necessary to take into account the reactive and active resistance, which together forms a complex resistance, and power.

To carry out calculations cable voltage drop use the formula:

∆U = (P*r0+Q*x0)*L/ U nom

This formula contains the following values:

  • P, Q - active, reactive power.
  • r0, x0 - active, reactance.
  • U nom - rated voltage.

To ensure optimal load on three-phase transmission lines, it is necessary to load them evenly. To do this, it is advisable to connect power electric motors to linear wires, and power to lighting devices - between the phases and the neutral line.

There are three load connection options:

  • from the electrical panel to the end of the line;
  • from the electrical panel with uniform distribution along the cable length;
  • from the electrical panel to two combined lines with uniform load distribution.

An example of calculating voltage losses: the total power consumption of all volatile installations in a house or apartment is 3.5 kW - the average value for a small number of powerful electrical appliances. If all loads are active (all devices are connected to the network), cosφ = 1 (the angle between the current vector and the voltage vector). Using the formula I = P/(Ucosφ), the current strength is I = 3.5*1000/220 = 15.9 A.

Further calculations: if you use a copper cable with a cross section of 1.5 square meters. mm, resistivity 0.0175 Ohm*mm2, and the length of the two-core cable for wiring is 30 meters.

According to the formula, the voltage loss is:

∆U = I*R/U*100%, where the current is 15.9 A, the resistance is 2 (two wires)*0.0175*30/1.5 = 0.7 Ohm. Then ∆U = 15.9*0.7/220*100% = 5.06%.

The obtained value slightly exceeds the drop of five percent recommended by regulatory documents. In principle, you can leave the diagram for such a connection, but if the main values ​​of the formula are affected by an unaccounted factor, the losses will exceed the permissible value.

What does this mean for the end consumer? Payment for used electricity supplied to the distribution panel at full capacity when actually consuming lower voltage electricity.

Using ready-made tables

How can a home craftsman or specialist simplify the calculation system when determining voltage losses along the cable length? You can use special tables given in highly specialized literature for power line engineers.

The tables are calculated based on two main parameters - cable length of 1000 m and current value of 1 A.

As an example, a table is presented with ready-made calculations for single-phase and three-phase electrical power and lighting circuits made of copper and aluminum with different cross-sections from 1.5 to 70 square meters. mm when power is supplied to the electric motor.

Table 1. Determination of voltage loss along the cable length Sectional area, mm2 Single phase line
Three phase line Nutrition Three phase line Nutrition
Lighting Mode Lighting Mode
Start Copper Aluminum Cosine of phase angle = 0.35 Cosine of phase angle = 1 Aluminum Cosine of phase angle = 0.35 Cosine of phase angle = 1
1,5 24,0 10,6 30,0 20,0 9,4 25,0
2,5 14,4 6,4 18,0 12,0 5,7 15,0
4,0 9,1 4,1 11,2 8,0 3,6 9,5
6,0 10,0 6,1 2,9 7,5 5,3 2,5 6,2
10,0 16,0 3,7 1,7 4,5 3,2 1,5 3,6
16,0 25,0 2,36 1,15 2,8 2,05 1,0 2,4
25,0 35,0 1,5 0,75 1,8 1,3 0,65 1,5
35,0 50,0 1,15 0,6 1,29 1,0 0,52 1,1
50,0 70,0 0,86 0,47 0,95 0,75 0,41 0,77

Tables are convenient to use for calculations when designing power lines. Calculation example: the motor operates with a rated current of 100 A, but at start-up a current of 500 A is required. During normal operation, cos ȹ is 0.8, and at start-up the value is 0.35. The electrical panel distributes a current of 1000 A. Voltage losses are calculated using the formula ∆U% = 100∆U/U nominal.

The engine is designed for high power, so it is rational to use a wire with a cross-section of 35 square meters for connection. mm, for a three-phase circuit in normal engine operation, the voltage loss is 1 volt over a wire length of 1 km. If the wire length is shorter (for example, 50 meters), the current is 100 A, then the voltage loss will reach:

∆U = 1 V*0.05 km*100A = 5 V

The losses at the switchboard when starting the engine are 10 V. The total drop is 5 + 10 = 15 V, which as a percentage of the nominal value is 100 * 15 * / 400 = 3.75%. The resulting number does not exceed the permissible value, so the installation of such a power line is quite realistic.

At the time of starting the engine, the current should be 500 A, and during operating mode - 100 A, the difference is 400 A, by which the current in the distribution board increases. 1000 + 400 = 1400 A. Table 1 indicates that when starting the engine, the losses along a cable length of 1 km are equal to 0.52 V, then

∆U at startup = 0.52*0.05*500 = 13 V

∆U shield = 10*1400/100 = 14 V

∆U total = 13+14 = 27 V, as a percentage ∆U = 27/400*100 = 6.75% - permissible value, does not exceed the maximum value of 8%. Taking into account all the parameters, the installation of the power line is acceptable.

Using the service calculator

Calculations, tables, graphs, diagrams - precise tools for calculating the voltage drop along the cable length. You can simplify your work if you perform the calculations using an online calculator. The advantages are obvious, but it is worth checking the data on several resources and starting from the average value obtained.

How it works:

  1. The online calculator is designed to quickly perform calculations based on initial data.
  2. You need to enter the following quantities into the calculator - current (alternating, direct), conductor (copper, aluminum), line length, cable cross-section.
  3. Be sure to enter parameters for the number of phases, power, network voltage, power factor, line operating temperature.
  4. After entering the initial data, the program determines the voltage drop along the cable line with maximum accuracy.
  5. An unreliable result can be obtained if the initial values ​​are entered incorrectly.

You can use such a system to carry out preliminary calculations, since service calculators on various resources do not always show the same result: the result depends on the competent implementation of the program, taking into account many factors.

However, you can carry out calculations on three calculators, take the average value and build on it at the preliminary design stage.

How to cut losses

Obviously, the longer the cable on the line, the greater the resistance of the conductor when current passes and, accordingly, the higher the voltage loss.

There are several ways to reduce the percentage of losses that can be used either independently or in combination:

  1. Use a cable with a larger cross-section, carry out calculations in relation to a different conductor. An increase in the cross-sectional area of ​​current-carrying conductors can be obtained by connecting two wires in parallel. The total cross-sectional area will increase, the load will be distributed evenly, and the voltage loss will be lower.
  2. Reduce the working length of the conductor. The method is effective, but it cannot always be used. The cable length can be reduced if there is a spare conductor length. At high-tech enterprises, it is quite realistic to consider the option of re-laying the cable if the costs of the labor-intensive process are much lower than the costs of installing a new line with a large cross-section of cores.
  3. Reduce the current power transmitted through long cables. To do this, you can disconnect several consumers from the line and connect them via a bypass circuit. This method is applicable on well-branched networks with backup highways. The lower the power transmitted through the cable, the less the conductor heats up, the resistance and voltage loss are reduced.

Attention! When the cable is operated at elevated temperatures, the conductor heats up and the voltage drop increases. Losses can be reduced by using additional thermal insulation or laying the cable along another route, where the temperature is significantly lower.

Calculation of voltage losses is one of the main tasks of the energy industry. If for the end consumer the voltage drop on the line and power losses are almost unnoticeable, then for large enterprises and organizations involved in supplying electricity to facilities, they are impressive. The voltage drop can be reduced if all calculations are performed correctly.

When designing electrical networks and systems with low currents, calculations of voltage losses in cables and wires are often required. These calculations are necessary in order to select the most optimal cable. If you choose the wrong conductor, the power supply system will very quickly fail or will not start at all. To avoid possible errors, it is recommended to use an online voltage loss calculator. The data obtained using the calculator will ensure stable and safe operation of lines and networks.

Causes of energy loss during electricity transmission

Significant losses occur as a result of excessive dispersion. Due to excess heat, the cable can become very hot, especially under heavy loads and incorrect calculations of electricity losses. Excessive heat causes damage to the insulation, creating a real threat to the health and life of people.

Electricity losses often occur due to too long cable lines, with a high load power. In case of prolonged use, electricity costs increase significantly. Incorrect calculations can cause equipment malfunctions, for example, security alarms. Voltage loss in the cable becomes important when the equipment power supply is low voltage DC or AC, rated from 12 to 48V.

How to calculate voltage loss

An online voltage loss calculator will help you avoid possible problems. The source data table contains data on the length of the cable, its cross-section and the material from which it is made. For calculations, information about load power, voltage and current will be required. In addition, the power factor and temperature characteristics of the cable are taken into account. After pressing the button, data appears on energy losses as a percentage, indicators of conductor resistance, reactive power and voltage experienced by the load.

The basic calculation formula is the following: ΔU=IхRL, in which ΔU means the voltage loss on the settlement line, I is the consumed current, determined primarily by the consumer parameters. RL reflects the resistance of the cable, depending on its length and cross-sectional area. It is the latter value that plays a decisive role in the loss of power in wires and cables.

Opportunities for reducing losses

The main way to reduce losses in a cable is to increase its cross-sectional area. In addition, you can reduce the length of the conductor and reduce the load. However, the last two methods cannot always be used due to technical reasons. Therefore, in many cases, the only option is to reduce the cable resistance by increasing the cross-section.

A significant disadvantage of a large cross-section is considered to be a noticeable increase in material costs. The difference becomes noticeable when cable systems stretch over long distances. Therefore, at the design stage, you must immediately select a cable with the required cross-section, for which you will need to calculate the power loss using a calculator. This program is of great importance when drawing up projects for electrical installation work, since manual calculations take a lot of time, and in the online calculator mode the calculation takes literally a few seconds.