Long cable lines are characterized by significant resistance, which makes adjustments to the operation of the network. Depending on the brand of cable and other parameters, the resistance value will also differ. And the amount of voltage on the cable line is directly proportional to this resistance.

Using an online calculator, calculating voltage losses in a cable comes down to the following steps:

• Indicate the cable length in meters and the material of the current-carrying conductors in the appropriate boxes;
• Conductor cross-section in mm²;
• The amount of electricity consumed in amperes or watts (place the indicator next to the power or current, depending on what parameter you know and what value you will indicate);
• Enter the voltage value in the network;
• Enter the power factor cosφ;
• Specify the cable temperature;

After you have entered the above data into the fields of the calculator, click the “calculate” button and in the corresponding columns you will receive the calculation result - the amount of voltage loss in the cable ΔU in%, the resistance of the wire itself R pr in Ohm, reactive power Q pr in VAR and voltage at load U n.

To calculate these values, the entire system, including cable and load, is replaced with an equivalent one, which can be represented as follows:

As you can see in the figure, depending on the type of power supply to the load (single-phase or three-phase), the resistance of the cable line will have a series or parallel connection with respect to the load. Calculation in the calculator is carried out using the following formulas:

• ΔU – voltage loss;
• U L – linear voltage;
• U Ф – phase voltage;
• I – current flowing in the line;
• Z K – cable line impedance;
• R K – active resistance of the cable line;
• X K – cable line reactance.

Of these, U L, U F, I, are specified at the data entry stage. To determine the total resistance Z K, the arithmetic addition of its active R K and reactive X K components is performed. Active and reactive resistance is determined by the formulas:

R K = (ρ * l) / S

R K – active resistance of the cable line, where

ρ is the resistivity for the corresponding metal (copper or aluminum), but the value of the resistivity of the material is not constant and can vary depending on the temperature, which is why, to bring it to real conditions, recalculation is performed in relation to temperature:

ρ t = ρ 20 *

• a is the coefficient of temperature change in the resistivity of the material.
• ρ 20 – specific resistance of the material at a temperature of +20ºС.
• t is the actual temperature of the conductor at a given time.
• l – cable line length (if the load is single-phase and the cable has two cores, then both of them are connected in series and the length must be multiplied by 2)
• S – cross-sectional area of ​​the conductor.

Reactive power is determined by the following formula: Q = S*sin φ, where

Where S is the apparent power, which can be defined as the product of the current in the circuit and the input voltage of the source or as the ratio of active power to power factor.

To calculate the voltage per load, the following calculations are made: U H = U - ΔU, where

• Where U N is the magnitude of the voltage applied to the load;
• U – voltage at the input to the cable line
• ΔU – voltage drop in the cable line.

Main switchboard 2.2. Indications of phase voltages after the first section of the cable line

Backup power supply parameters:

• Maximum power of diesel power plant – 600 kW,
• Cable line – 3 cables AVBbShv 4x240, connected in parallel,
• Cable line length – 250 m.

Based on these parameters, we can clearly conclude that the capacity of the diesel power plant and the backup cable line, taking into account the voltage drop, will be enough for no more than half of the maximum load requirements, which is completely unacceptable.

Therefore, it makes no sense to monitor the quality of food through the diesel power station.

Download file

In conclusion - as promised, a good book on calculating voltage loss and voltage loss in a cable. It will be very interesting to everyone who is interested in this article. Nowadays such books are no longer written.

/ Brochure from the Electrician's Library. Provides instructions and calculations necessary for selecting cross-sections of wires and cables up to 1000 V. Useful for those who are interested in primary sources., zip, 1.57 MB, downloaded: 385 times./

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.

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.

So, today on the agenda is the question - how to calculate the cross-section of a wire based on the permissible voltage loss.

And of course, a program for electricians will help us with this, which is called “Electrician”.

For those who do not know why to make calculations based on voltage loss, let me remind you that with a large length of wire, a voltage drop occurs in this section and very little can “reach” the load if the wire cross-section is chosen incorrectly.

Typically organizations that do major renovation of apartments, be sure to look at the condition of the electrical wiring and, in general, all electrical equipment and, when making repairs, replace dilapidated and outdated wires, automatic devices, etc.

In this case, it is necessary to correctly select the cross-section of the new wiring not only according to the heating conditions, but also according to the permissible voltage loss.

Let's imagine this situation. You have to renovate your apartment, or if you have a house, then at home.

You are renovating the electrical wiring in your house and decide to run a separate outlet wire into the room. But this room is far away and the length of the wire is about 30 meters to the last socket.

You know that you will never plug anything powerful into the sockets, the most you can plug in is iron, TV, computer which in total runs no more 3kW and current at this power I=P/U=3000/220=13.64 A or if we round then 14 amps.

According to the PUE, a copper cross-section of 1.5 sq. mm is suitable for such current. True, the wire insulation will be about 60 degrees Celsius at a room temperature of +25, but the rules allow the following load:

Now let’s see what the “Electrician” program will tell us in our case, we will find out how many volts are “lost” on a 30m wire and how many “reach” the outlet.

So, open the “Electrician” program and we are interested in the button called “Losses”, click on it:

This window opens, where you need to put a dot on “Voltage loss”:

In the next window that opens, click on the “Cable lines and other wires” button:

Well, in the next window we indicate the necessary parameters, listed from top to bottom:

Find - Losses in %

Conductor material- copper

Asked by:

3- Power P, kW

4- Allowable losses,% (in our example this value is not important, you can also set 4):

Next, you need to select the inductive reactance, you don’t need to bother too much here, just click on the “Select Xo” button and in the window that opens, click on the value “Cable with vinyl or polyvinyl chloride insulation”:

Next, we enter the value of the cosine phi, I set it to 0.85 since we do not have a purely active load, and we enter the following value - the wire length is 30 m:

That’s all, now you can find out the result, to do this, click on the “Calculation” button:

And now we see the result - as much as 10 volts of voltage are “lost” on a section of copper wire with a cross-section of 1.5 sq. mm and a length of 30 meters!

That is, with a 3 kW load on there will no longer be 220 volts, but only 210. For fun, you can calculate how many volts are “lost” if the wire has a cross-section of 2.5 sq. mm:

As you can see, it’s already less, the voltage drop over a section 30 m long will be only 6 volts.

You can also find out vice versa - what cross-section of the wire is needed if you know the required value of voltage loss, to do this, at the top of the window you need to put a dot on “Section in mm square.” and enter the required values ​​- I circled them in red in the picture:

In this way, using the “Electrician” program, you can determine not only the value of the voltage drop on the electrical wiring, but also find out the required cross-section for the correct selection of wires when installing electrical wiring.

I hope this information will help you and come in handy more than once.

I will be glad to see your comments, if you have any technical questions, please ask them on the forum, that’s where I answer questions - .

Subscribe to my video channel on YouTube !

Watch many more home electrical videos!