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Calculator for calculating voltage losses in an electric cable. Features of calculating voltage loss in the main line Formula for calculating voltage loss in a cable

How to correctly and accurately calculate the cable cross-section based on voltage loss? Very often, when designing power supply networks, a competent calculation of cable losses is required. An accurate result is important for choosing a material with the required cross-sectional area of the core. If the cable is chosen incorrectly, this will entail multiple material costs, because the system will quickly fail and cease to function. Thanks to assistant sites, where there is a ready-made program for calculating the cable cross-section and losses on it, this can be done easily and quickly.

How to use an online calculator?

In the finished table you need to enter data according to the selected cable material, system load power, network voltage, cable temperature and method of laying it. Then click the “calculate” button and get the finished result.
This calculation of voltage losses in a line can be safely used in work, if you do not take into account the resistance of the cable line under certain conditions:

When specifying the power factor, cosine phi is equal to one.
DC network lines.
AC network with a frequency of 50 Hz made of conductors with cross-sections up to 25.0–95.0.

The results obtained must be used according to each individual case, taking into account all the errors of cable and wire products.

Be sure to fill in all values!

Calculation of power loss in a cable using the school formula

You can obtain the necessary data as follows, using the following combination of indicators for calculations: ΔU=I·RL (voltage loss in the line = current consumption * cable resistance).

Why do you need to calculate the voltage loss in the cable?

Excessive energy dissipation in a cable can lead to significant power losses, excessive heating of the cable and damage to the insulation. This is dangerous for the lives of people and animals. With a significant length of the line, this will affect the cost of light, which will also adversely affect the financial condition of the owner of the premises.

In addition, uncontrolled voltage losses in the cable can cause the failure of many electrical appliances, as well as their complete destruction. Very often, residents use smaller cable sections than needed (in order to save money), which soon causes a short circuit. And the future costs of replacing or repairing electrical wiring do not cover the wallets of “thrifty” users. That is why it is so important to choose the correct cross-section of cables for the wires being laid. Any electrical installation in a residential building should be started only after a thorough calculation of cable losses. It is important to remember that electricity does not give a second chance, and therefore everything must be done correctly and efficiently from the beginning.

Ways to reduce power losses in cables

Losses can be reduced in several ways:

increasing the cross-sectional area of the cable;
decreasing the length of the material;
load reduction.

Often the last two points are more difficult, and therefore you have to do this by increasing the cross-sectional area of the electric cable core. This will help reduce resistance. This option has several costly aspects. Firstly, the cost of using such material for multi-kilometer systems is very significant, and therefore it is necessary to choose a cable of the correct cross-section in order to reduce the threshold of power loss in the cable.

Online calculation of voltage losses allows you to do this in a few seconds, taking into account all additional characteristics. For those who want to double-check the result manually, there is a physical and mathematical formula for calculating voltage losses in a cable. Of course, these are excellent assistants for every electrical network designer.

Table for calculating wire cross-section by power

Cable cross-section, mm 2

Open wiring

Gasket in channels

Aluminum

power, kWt

Video on the correct choice of wire cross-section and typical mistakes

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:

Incorrect operation of volatile installations, equipment, lighting devices.
Failure of electrical appliances to operate when the input voltage is reduced, equipment failure.
Reduced acceleration of the torque of electric motors at starting current, loss of energy taken into account, shutdown of motors when overheated.
Uneven distribution of current load between consumers at the beginning of the line and at the remote end of a long wire.
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:

The conductor carrying the current overheats and creates capacitance as part of the reactance.
Under the influence of transformations occurring on the windings of transformers, reactors, and other circuit elements, the power of electricity becomes inductive.
As a result, the resistive resistance of the metal cores is converted into active resistance of each phase of the electrical circuit.
The cable is connected to a current load with a total (complex) resistance along each current-carrying core.
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.
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. For installation, a copper conductor with a cross-section of 1.5 square meters was selected. mm, although for 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		Lighting	Nutrition		Lighting
		Mode	Start		Mode	Start
Copper	Aluminum	Cosine of phase angle = 0.8	Cosine of phase angle = 0.35	Cosine of phase angle = 1	Cosine of phase angle = 0.8	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 make your work easier 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:

The online calculator is designed to quickly perform calculations based on initial data.
You need to enter the following quantities into the calculator - current (alternating, direct), conductor (copper, aluminum), line length, cable cross-section.
Be sure to enter parameters for the number of phases, power, network voltage, power factor, line operating temperature.
After entering the initial data, the program determines the voltage drop along the cable line with maximum accuracy.
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:

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.
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.
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.

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.

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.

When designing electrical wiring, it is necessary to make accurate calculations of the voltage loss in the cable. This prevents the surface of the wires from becoming too hot during operation. Thanks to these measures, it is possible to avoid short circuits and premature breakdown of household appliances.

In addition, the formula allows you to correctly select the diameter of the wire cross-section, which is suitable for different types of electrical installation work. A wrong choice can cause a breakdown of the entire system. Online calculation helps to make the task easier.

How to calculate voltage loss?

The online calculator allows you to correctly calculate the necessary parameters, which will further reduce the occurrence of various kinds of troubles. To independently calculate the loss of electrical voltage, use the following formula:

U =(P*ro+Q*xo)*L/U nom:

P is active power. It is measured in W;
Q – reactive power. Unit of measurement var;
ro – acts as active resistance (Ohm);
xo – reactance (m);
U nom is the rated voltage (V). It is indicated in the technical data sheet of the device.

According to the rules for the design of electrical installations (PUE), the acceptable norm for possible voltage deviations is considered to be:

in power circuits it can be no higher than +/- 6%;
in living space and beyond up to +/- 5%;
at manufacturing enterprises from +/- 5% to -2%.

Electrical voltage losses from the transformer installation to the living space should not exceed +/- 10%.

During the design process, it is recommended to make the load uniform on the three-phase line. The permissible norm is 0.5 kV. During installation work, electric motors must be connected to linear conductors. The lighting line will be between phase and neutral. As a result of this, the load is correctly distributed between the conductors.

When calculating the voltage loss in a cable, the given current or power values are taken as a basis. On an extended electrical line, inductive reactance is taken into account.

How to reduce losses?

One of the ways to reduce voltage loss in a conductor is to increase its cross-section. In addition, it is recommended to reduce its length and distance from the destination. In some cases, these methods cannot always be used for technical reasons. In most cases, reducing the resistance allows the operation of the line to normalize.

The main disadvantage of a large cable cross-sectional area is the significant material costs during use. That is why correct calculation and selection of the required diameter allows you to get rid of this trouble. The online calculator is used for projects with high-voltage lines. Here the program helps to correctly calculate the exact parameters for the electrical circuit.

The main causes of voltage loss

Large electrical voltage losses occur due to excessive energy dissipation. As a result of this, the surface of the cable becomes very hot, thereby causing deformation of the insulating layer. This phenomenon is common on high-voltage lines where heavy loads occur.