Determine the heat released at the resistance. Joule–Lenz law

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The amount of heat released per unit time in the section of the circuit under consideration is proportional to the product of the square of the current in this section and the resistance of the section

Joule Lenz's law in integral form in thin wires:

If the current strength changes over time, the conductor is stationary and there are no chemical transformations in it, then heat is generated in the conductor.

- The power of heat released per unit volume of the medium during the flow of electric current is proportional to the product of the electric current density and the electric field value

The conversion of electrical energy into heat is widely used in electric furnaces and various electric heating devices. The same effect in electrical machines and devices leads to involuntary energy expenditure (energy loss and decreased efficiency). Heat, by causing these devices to heat up, limits their load; When overloaded, the temperature rise may damage the insulation or shorten the service life of the installation.

In the formula we used:

Quantity of heat

Current work

Conductor voltage

Current strength in the conductor

Time interval

Romanova_1 / kursachi / Kursovik Romanova / EXAMPLE / 13 Thermal calculation. Thermal power on a resistor formula

Problem on the topic “Laws of direct current”. The problem may be of interest to 10th grade students and graduates to prepare for the Unified State Exam. By the way, this kind of problem was on the Unified State Exam in part 1 with a slightly different question (it was necessary to find the ratio of the amounts of heat released by the resistors).

Which resistor will generate the greatest (smallest) amount of heat? R1 = R4 = 4 Ohm, R2 = 3 Ohm, R3 = 2 Ohm. Give a solution. To answer the question of the problem, it is necessary to compare the amount of heat generated by each of their resistors. To do this, we will use the formula of the Joule-Lenz law. That is, the main task will be to determine the current strength (or comparison) flowing through each resistor.

According to the laws of series connection, the current flowing through resistors R1 and R2, and R3 and R4 is the same. To determine the current strength in the upper and lower branches, we use the law of parallel connection, according to which the voltage on these branches is the same. Describing the voltage on the lower and upper branches according to Ohm’s law for a section of the circuit, we have: Substituting the numerical values of the resistances of the resistors, we obtain: That is, we obtain the relationship between the currents flowing in the upper and lower branches: Having determined the current strength through each of these resistors, we determine the amount of heat released on each of the resistors. Comparing the numerical coefficients, we come to the conclusion that the maximum amount of heat will be released on the fourth resistor, and the minimum amount of heat will be released on the second.

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Thermal power - calculation formula

Owners of private houses, apartments or any other objects have to deal with thermal engineering calculations. This is the basis of the fundamentals of building design.

Understanding the essence of these calculations in official papers is not as difficult as it seems.

You can also learn to perform calculations for yourself in order to decide what kind of insulation to use, how thick it should be, what power the boiler should be purchased and whether the available radiators are enough for a given area.

The answers to these and many other questions can be found if you understand what thermal power is. Formula, definition and scope of application - read the article.

What is thermal design?

Simply put, a thermal calculation helps to find out exactly how much heat a building stores and loses, and how much energy heating must produce to maintain comfortable conditions in the home.

When assessing heat loss and the degree of heat supply, the following factors are taken into account:

What kind of object is it: how many floors does it have, the presence of corner rooms, is it residential or industrial, etc.
How many people will “live” in the building.
An important detail is the glazing area. And the dimensions of the roof, walls, floors, doors, ceiling heights, etc.
What is the duration of the heating season, the climatic characteristics of the region.
According to SNiPs, the temperature standards that should be in the premises are determined.
Thickness of walls, ceilings, selected thermal insulators and their properties.

Other conditions and features may be taken into account, for example, for production facilities, working days and weekends, the power and type of ventilation, the orientation of housing to the cardinal points, etc. are considered.

Why do you need a thermal calculation?

How did builders of the past manage to do without thermal calculations?

The surviving merchant houses show that everything was done simply with reserves: smaller windows, thicker walls. It turned out to be warm, but not economically profitable.

Thermal engineering calculations allow us to build in the most optimal way. No more or less materials are taken, but exactly as much as needed. The dimensions of the building and the costs of its construction are reduced.

Calculating the dew point allows you to build in such a way that materials do not deteriorate for as long as possible.

To determine the required boiler power, you also cannot do without calculations. Its total power consists of energy costs for heating rooms, heating hot water for household needs, and the ability to cover heat loss from ventilation and air conditioning. A power reserve is added for periods of peak cold weather.

When gasifying a facility, coordination with services is required. The annual gas consumption for heating and the total power of heat sources in gigacalories are calculated.

Calculations are needed when selecting heating system elements. The system of pipes and radiators is calculated - you can find out what their length and surface area should be. The loss of power when turning the pipeline, at joints and passing through fittings is taken into account.

When calculating thermal energy costs, knowledge of how to convert Gcal to KW and vice versa can be useful. The following article discusses this topic in detail with calculation examples.

A complete calculation of a warm water floor is given in this example.

Did you know that the number of sections of heating radiators is not taken out of thin air? Too little of them will lead to the fact that the house will be cold, and too much will create heat and lead to excessive dryness of the air. The link http://microklimat.pro/sistemy-otopleniya/raschet-sistem-otopleniya/kolichestva-sekcij-radiatorov.html provides examples of correct calculation of radiators.

Calculation of thermal power: formula

Let's look at the formula and give examples of how to make calculations for buildings with different dissipation coefficients.

Vx(delta)TxK= kcal/h (thermal power), where:

The first indicator “V” is the volume of the calculated premises;
Delta "T" - temperature difference - is the value that shows how many degrees warmer inside the room than outside;
“K” is the dissipation coefficient (it is also called the “heat transmittance coefficient”). The value is taken from the table. Typically the figure ranges from 4 to 0.6.

Approximate dissipation coefficient values for simplified calculations

If it is an uninsulated metal profile or board, then “K” will be = 3 – 4 units.
Single brickwork and minimal insulation - “K” = from 2 to 3.
Two brick wall, standard ceiling, windows and
doors – “K” = from 1 to 2.
The warmest option. Double-glazed windows, brick walls with double insulation, etc. - “K” = 0.6 – 0.9.

A more accurate calculation can be made by calculating the exact dimensions of the surfaces of the house in m2 that differ in properties (windows, doors, etc.), making calculations for them separately and adding up the resulting indicators.

Example of thermal power calculation

Let’s take a certain room of 80 m2 with a ceiling height of 2.5 m and calculate the power of the boiler we will need to heat it.

First, we calculate the cubic capacity: 80 x 2.5 = 200 m3. Our house is insulated, but not enough - the dissipation coefficient is 1.2.

Frosts can be down to -40 °C, but indoors you want to have a comfortable +22 degrees, the temperature difference (delta “T”) is 62 °C.

We substitute numbers into the heat loss power formula and multiply:

200 x 62 x 1.2 = 14880 kcal/hour.

We convert the resulting kilocalories into kilowatts using a converter:

1 kW = 860 kcal;
14880 kcal = 17302.3 W.

We round up with a margin, and we understand that in the most severe frost of -40 degrees we will need 18 kW of energy per hour.

Multiply the perimeter of the house by the height of the walls:

(8 + 10) x 2 x 2.5 = 90 m2 of wall surface + 80 m2 of ceiling = 170 m2 of surface in contact with the cold. The heat loss we calculated above amounted to 18 kW/h, dividing the surface of the house by the estimated energy consumed, we find that 1 m2 loses approximately 0.1 kW or 100 W every hour at an outdoor temperature of -40 °C, and indoor temperature +22 °C .

These data can become the basis for calculating the required thickness of insulation on the walls.

Let's give another example of a calculation; it is more complicated in some aspects, but more accurate.

Formula:

Q = S x (delta)T / R:

Q – the desired value of heat loss at home in W;
S – area of cooling surfaces in m2;
T – temperature difference in degrees Celsius;
R – thermal resistance of the material (m2 x K/W) (Square meters multiplied by Kelvin and divided by Watt).

So, to find “Q” of the same house as in the example above, let’s calculate the area of its surfaces “S” (we will not count the floor and windows).

“S” in our case = 170 m2, of which 80 m2 is the ceiling and 90 m2 is the walls;
T = 62 °C;
R – thermal resistance.

We look for “R” using the thermal resistance table or formula. The formula for calculating the thermal conductivity coefficient is as follows:

R= H/ K.T. (N – material thickness in meters, K.T. – thermal conductivity coefficient).

In this case, our house has walls made of two bricks covered with foam plastic 10 cm thick. The ceiling is covered with sawdust 30 cm thick.

The heating system of a private home must be designed taking into account energy savings. Read the calculation of the heating system of a private house, as well as recommendations for choosing boilers and radiators, carefully.

How and how to insulate a wooden house from the inside, you will learn by reading this information. Choice of insulation and insulation technology.

From the table of thermal conductivity coefficients (measured by W / (m2 x K) Watt divided by the product of a square meter by Kelvin). We find the values for each material, they will be:

brick - 0.67;
polystyrene foam – 0.037;
sawdust – 0.065.

Substitute the data into the formula (R= H/ K.T.):

R (ceiling 30 cm thick) = 0.3 / 0.065 = 4.6 (m2 x K) / W;
R (brick wall 50 cm) = 0.5 / 0.67 = 0.7 (m2 x K) / W;
R (foam 10 cm) = 0.1 / 0.037 = 2.7 (m2 x K) / W;
R (wall) = R (brick) + R (foam) = 0.7 + 2.7 = 3.4 (m2 x K) / W.

Now we can start calculating heat loss “Q”:

Q for ceiling = 80 x 62 / 4.6 = 1078.2 W.
Q walls = 90 x 62 / 3.4 = 1641.1 W.
All that remains is to add 1078.2 + 1641.1 and convert to kW, it turns out (if you round right away) 2.7 kW of energy in 1 hour.

You can notice how big the difference was in the first and second cases, although the volume of the houses and the temperature outside the window in the first and second cases were exactly the same.

It's all about the degree of fatigue of the houses (although, of course, the data could have been different if we had calculated the floors and windows).

Conclusion

The given formulas and examples show that when making thermal calculations it is very important to take into account as many factors as possible that influence heat loss. This includes ventilation, window area, degree of fatigue, etc.

And the approach, when 1 kW of boiler power is taken per 10 m2 of a house, is too approximate to seriously rely on it.

Video on the topic

microclimat.pro

13 Thermal calculation

10. Thermal calculation.

The design of the IC must be such that the heat released during its operation does not lead to failure of the elements due to overheating under the most unfavorable operating conditions. The main heat-generating elements include, first of all, resistors, active elements and components. The power dissipated by capacitors and inductors is small. Film switching of the IC, due to the low electrical resistance and high thermal conductivity of metal films, helps remove heat from the hottest elements and equalize the temperature of the GIS board and the semiconductor IC chip.

Rice. 10.1. Option for mounting the board on the case.

Thermal calculation of resistors.

Let's calculate the thermal resistance of the resistor using formula (10.1)

п = 0.03 [W/cm °С] - thermal conductivity coefficient of the substrate material;

δп = 0.06 cm – board thickness.

RT=0.06/0.03=2 cm2∙°С/W

Let's calculate the temperature of film resistors using the formula

PR – power released on the resistor;

SR – area occupied by the resistor on the board;

P0 – total power allocated by all components of the microcircuit;

Sp – board area.

PR = 0.43 mW – power released by the resistor;

SR = 0.426mm2 – area occupied by the resistor;

Sn = 80 mm2 – board area;

RT = 2 cm2∙°С/W – thermal resistance of the resistor;

Tam.av = 40С – maximum ambient temperature;

T = 125С = maximum permissible temperature of film resistors.

TR=(0.43∙10-3∙200)/0.426+(24.82∙10-3∙200)/80+40=40.26 С<125 С

The temperature of the remaining resistors is calculated similarly using the MathCad program. The calculation results are presented in Table 10.1

Table. 10.1

Resistor

The table shows that for all film resistors the specified thermal conditions are observed.

Thermal calculation for a hanging element.

Thermal resistance will be calculated using the formula:

k = 0.003 [W/cm °C] - thermal conductivity coefficient of the glue;

δк1 = 0.01 cm – glue thickness.

Rt=(0.06/0.03)+(0.01/0.003)=5.33 cm2∙°С/W

Let's calculate the temperature of the hinged element using the formula:

Calculation of transistor KT202A, VT14

Pne = 2.6 mW – power released by the transistor;

Sne = 0.49 mm2 – area occupied by the transistor;

P0 = 24.82 mW – power released by all components of the board;

Sn = 80 mm2 – board area;

Т0С = 40С – maximum ambient temperature;

T = 85С = maximum permissible temperature of the transistor.

Tne=(2.6∙10-3∙533)/0.49+(24.82∙10-3∙533)/80+40=42.99С<85С

Therefore, the specified thermal regime is observed.

The temperature of the remaining transistors is calculated similarly using the MathCad program. The calculation results are presented in Table 10.2

Table 10.2

Transistor

The table shows that for all transistors the specified thermal conditions are observed. Consequently, the thermal conditions for the entire circuit are met.

studfiles.net

Thermal power of electric current and its practical application

The reason for heating a conductor lies in the fact that the energy of electrons moving in it (in other words, current energy) during the sequential collision of particles with ions of the molecular lattice of a metal element is converted into a warm type of energy, or Q, and this is how the concept of “thermal power” is formed.

The work of a current is measured using the International System of Units (SI) using joules (J), and current power is defined as “watt” (W). Departing from the system in practice, they can also use non-system units that measure the work of the current. Among them are watt-hour (W × h), kilowatt-hour (abbreviated kW × h). For example, 1 W × h denotes the work of a current with a specific power of 1 watt and a duration of one hour.

If electrons move along a stationary metal conductor, in this case all the useful work of the generated current is distributed to heat the metal structure, and, based on the provisions of the law of conservation of energy, this can be described by the formula Q=A=IUt=I2Rt=(U2/R)* t. Such relationships accurately express the well-known Joule-Lenz law. Historically, it was first determined empirically by the scientist D. Joule in the mid-19th century, and at the same time, independently of him, by another scientist - E. Lenz. Thermal power found practical application in technical design with the invention of an ordinary incandescent lamp in 1873 by Russian engineer A. Ladygin.

Thermal current power is used in a number of electrical appliances and industrial installations, namely, in thermal measuring instruments, heating-type electric stoves, electric welding and inventory equipment, household appliances using the electric heating effect are very common - boilers, soldering irons, kettles, irons.

The thermal effect also finds itself in the food industry. With a high proportion of use, the possibility of electric contact heating is used, which guarantees thermal output. It is determined by the fact that the current and its thermal power, influencing the food product, which has a certain degree of resistance, causes uniform heating in it. An example can be given of how sausages are produced: through a special dispenser, minced meat enters metal molds, the walls of which also serve as electrodes. Here, constant uniformity of heating is ensured over the entire area and volume of the product, the set temperature is maintained, the optimal biological value of the food product is maintained, along with these factors, the duration of technological work and energy consumption remain the least.

The specific thermal power of the electric current (ω), in other words, the amount of heat released per unit volume for a certain unit of time, is calculated as follows. The elementary cylindrical volume of a conductor (dV), with a cross-sectional conductor cross-section dS, length dl parallel to the direction of current, and resistance are composed by the equations R=p(dl/dS), dV=dSdl.

According to the definitions of the Joule-Lenz law, in the allotted time (dt) in the volume we have taken, a level of heat will be released equal to dQ=I2Rdt=p(dl/dS)(jdS)2dt=pj2dVdt. In this case, ω=(dQ)/(dVdt)=pj2 and, applying Ohm’s law here to establish the current density j=γE and the ratio p=1/γ, we immediately obtain the expression ω=jE= γE2. It gives the concept of the Joule-Lenz law in differential form.

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Embedder page » Thermal calculations

All electronic components generate heat, so the ability to calculate radiators so as not to overshoot by a couple of orders of magnitude is very useful for any electronics engineer.

Thermal calculations are very simple and have a lot in common with electronic circuit calculations. Here's a look at a common thermal design problem I just encountered

Task

You need to select a radiator for a 5-volt linear stabilizer, which is powered by a maximum of 12 volts and produces 0.5A. The maximum released power is (12-5)*0.5 = 3.5W

Dive into theory

In order not to create entities, people scratched the pumpkin and realized that heat is very similar to electric current, and for thermal calculations you can use the usual Ohm's law, only

Voltage (U) is replaced by temperature (T)

Current (I) is replaced by power (P)

Resistance is replaced by thermal resistance. Normal resistance has the dimension Volt/Ampere, and thermal resistance has the dimension °C/Watt

As a result, Ohm's law is replaced by its thermal analogue:

A small note - in order to indicate that what is meant is thermal (and not electrical) resistance, add the letter theta to the letter R: I don’t have such a letter on my keyboard, and I’m too lazy to copy from the symbol table, so I’ll just use the letter R.

Let's continue

Heat is generated in the stabilizer crystal, and our goal is to prevent it from overheating (to prevent overheating of the crystal, and not the case, this is important!).

To what temperature the crystal can be heated is written in the datasheet:

Usually, the limiting temperature of the crystal is called Tj (j = junction = junction - the temperature-sensitive insides of microcircuits mainly consist of pn junctions. We can assume that the temperature of the junctions is equal to the temperature of the crystal)

Without radiator

The thermal diagram looks very simple:

Especially for cases of using a case without a radiator, the thermal resistance crystal-atmosphere (Rj-a) is written in the datasheets (you already know what j is, a = ambient = environment)

Note that the temperature of the “ground” is not zero, but equal to the ambient air temperature (Ta). The air temperature depends on the conditions in which the radiator is located. If it is in the open air, then you can set Ta = 40 °C, but if in a closed box, then the temperature can be much higher!

We write Ohm's thermal law: Tj = P*Rj-a + Ta. We substitute P = 3.5, Rj-a = 65, we get Tj = 227.5 + 40 = 267.5 °C. A bit much, though!

We hook the radiator

The thermal circuit of our example with a stabilizer on a radiator becomes like this:

Rj-c – resistance from the crystal to the case heat sink (c = case = case). Given in the datasheet. In our case – 5 °C/W – from the datasheet
Rc-r – body-radiator resistance. It's not that simple. This resistance depends on what is between the case and the heatsink. For example, a silicone gasket has a thermal conductivity coefficient of 1-2 W/(m*°C), and KPT-8 paste – 0.75 W/(m*°C). Thermal resistance can be obtained from the thermal conductivity coefficient using the formula:
R = gasket thickness/(thermal conductivity coefficient * area of one side of the gasket)
Often Rc-r can be ignored altogether. For example, in our case (we use a TO220 case, with KPT-8 paste, the average depth of the paste taken from the ceiling is 0.05mm). Total, Rc-r = 0.5 °C/W. With a power of 3.5W, the temperature difference between the stabilizer housing and the radiator is 1.75 degrees. That's not a lot. For our example, let's take Rc-r = 2 °C/W

Rr-a is the thermal resistance between the radiator and the atmosphere. It is determined by the geometry of the radiator, the presence of airflow, and a bunch of other factors. This parameter is much easier to measure than to calculate (see at the end of the article). For example - Rr-c = 12.5 °C/W

Ta = 40°C - here we figured that the atmospheric temperature is rarely higher, you can take 50 degrees, to be sure.

We substitute all this data into Ohm's law, and we get Tj = 3.5*(5+2+12.5) + 40 = 108.25 °C

This is significantly less than the limit of 150 °C. Such a radiator can be used. In this case, the radiator housing will heat up to Tc = 3.5 * 12.5 + 40 = 83.75 °C. This temperature can already soften some plastics, so you need to be careful.

Radiator-atmosphere resistance measurement.

Most likely, you already have a bunch of radiators lying around that can be used. Thermal resistance is very easy to measure. This requires resistance and a power source.

We sculpt resistance onto the radiator using thermal paste:

We connect the power source and set the voltage so that some power is released at the resistance. It is better, of course, to heat the radiator with the power that it will dissipate in the final device (and in the position in which it will be located, this is important!). I usually leave this structure for half an hour so that it warms up well.

Once the temperature has been measured, the thermal resistance can be calculated

Rr-a = (T-Ta)/P. For example, my radiator heated up to 81 degrees, and the air temperature was 31 degrees. thus Rr-a = 50/4 = 12.5 °C/W.

Estimation of radiator area

An ancient radio amateur's handbook provided a graph by which you can estimate the radiator area. Here he is:

It's very easy to work with. We select the overheating that we want to achieve and see what area corresponds to the required power for such overheating.

For example, with a power of 4W and an overheat of 20 degrees, you will need a 250cm^2 radiator. This graph overestimates the area and does not take into account a bunch of factors such as forced airflow, fin geometry, etc.

bsvi.ru

The Joule-Lenz law is a law of physics that defines a quantitative measure of the thermal effect of electric current. This law was formulated in 1841 by the English scientist D. Joule and completely separately from him in 1842 by the famous Russian physicist E. Lenz. Therefore, it received its double name - the Joule-Lenz law.

Law definition and formula

The verbal formulation has the following form: the power of heat generated in a conductor when flowing through it is proportional to the product of the electric field density value and the intensity value.

Mathematically, the Joule-Lenz law is expressed as follows:

ω = j E = ϭ E²,

where ω is the amount of heat released in units. volume;

E and j are the intensity and density, respectively, of the electric fields;

σ is the conductivity of the medium.

Physical meaning of the Joule–Lenz law

The law can be explained as follows: current flowing through a conductor represents the movement of an electric charge under the influence. Thus, the electric field does some work. This work is spent on heating the conductor.

In other words, energy transforms into another quality – heat.

But excessive heating of current-carrying conductors and electrical equipment should not be allowed, as this can lead to damage. Severe overheating of the wires is dangerous when quite large currents can flow through the conductors.

In integral form for thin conductors Joule–Lenz law sounds like this: the amount of heat that is released per unit time in the section of the circuit under consideration is determined as the product of the square of the current strength and the resistance of the section.

Mathematically, this formulation is expressed as follows:

Q = ∫ k I² R t,

in this case Q is the amount of heat released;

I – current value;

R - active resistance of conductors;

t – exposure time.

The value of the parameter k is usually called the thermal equivalent of work. The value of this parameter is determined depending on the bit depth of the units in which the values used in the formula are measured.

The Joule-Lenz law is quite general in nature, since it does not depend on the nature of the forces generating the current.

From practice, it can be argued that it is valid both for electrolytes and conductors and semiconductors.

Application area

There are a huge number of areas of application of Joule Lenz’s law in everyday life. For example, a tungsten filament in an incandescent lamp, an arc in electric welding, a heating filament in an electric heater, and many others. etc. This is the most widely accepted physical law in everyday life.

Content:

The famous Russian physicist Lenz and the English physicist Joule, conducting experiments to study the thermal effects of electric current, independently derived the Joule-Lenz law. This law reflects the relationship between the amount of heat generated in a conductor and the electric current passing through this conductor over a certain period of time.

Properties of electric current

When electric current passes through a metal conductor, its electrons constantly collide with various foreign particles. These can be ordinary neutral molecules or molecules that have lost electrons. In the process of moving, an electron can split off another electron from a neutral molecule. As a result, its kinetic energy is lost, and instead of a molecule, a positive ion is formed. In other cases, an electron, on the contrary, combines with a positive ion and forms a neutral molecule.

In the process of collisions of electrons and molecules, energy is consumed, which is subsequently converted into heat. The expenditure of a certain amount of energy is associated with all movements during which resistance has to be overcome. At this time, the work spent on overcoming friction resistance is converted into thermal energy.

Joule Lenz law formula and definition

According to Lenz's Joule law, an electric current passing through a conductor is accompanied by an amount of heat directly proportional to the square of the current and the resistance, as well as the time of flow of this current through the conductor.

In the form of a formula, the Joule-Lenz law is expressed as follows: Q = I 2 Rt, in which Q displays the amount of heat released, I - , R - conductor resistance, t - time period. The value "k" represents the thermal equivalent of work and is used in cases where the amount of heat is measured in calories, current in , resistance in Ohms, and time in seconds. The numerical value of k is 0.24, which corresponds to a current of 1 ampere, which, with a conductor resistance of 1 Ohm, releases an amount of heat equal to 0.24 kcal within 1 second. Therefore, to calculate the amount of heat released in calories, the formula Q = 0.24I 2 Rt is used.

When using the SI system of units, the amount of heat is measured in joules, so the value of “k”, in relation to the Joule-Lenz law, will be equal to 1, and the formula will look like: Q = I 2 Rt. According to I = U/R. If this current value is substituted into the basic formula, it will take the following form: Q = (U 2 /R)t.

Basic formula Q = I 2 Rt is very convenient to use when calculating the amount of heat that is released in the case of a series connection. The current strength in all conductors will be the same. When several conductors are connected in series at once, each of them will release so much heat that will be proportional to the resistance of the conductor. If three identical wires made of copper, iron and nickel are connected in series, then the maximum amount of heat will be released by the latter. This is due to the highest resistivity of nickel and the stronger heating of this wire.

When the same conductors are connected in parallel, the value of the electric current in each of them will be different, and the voltage at the ends will be the same. In this case, the formula Q = (U 2 /R)t is more suitable for calculations. The amount of heat generated by a conductor will be inversely proportional to its conductivity. Thus, the Joule-Lenz law is widely used to calculate electrical lighting installations, various heating and heating devices, as well as other devices related to the conversion of electrical energy into heat.