As a result of the experiments, it was found that the amount of heat generated by the current when passing through the conductor depends on the resistance of the conductor itself, the current and the time of its passage.
This physical law was first established in 1841 by the English physicist Joule, and somewhat later (in 1844) independently by the Russian academician Emil Christianovich Lenz (1804 - 1865).
The quantitative relationships that occur when a conductor is heated by current are called the Joule-Lenz law.
The above has been established:
Since 1 cal = 0.472 kgm, then
Thus,
1 J = 0.24 cal.
Energy electric current is determined by the formula
A = I 2 × r × t J.
Since the energy of the current goes to heating, the amount of heat released by the current in the conductor is equal to:
Q= 0.24 × I 2 × r × t cal.
This formula, which expresses the Joule-Lenz law, shows and defines the law that the amount of heat in calories released by current when passing through a conductor is equal to a coefficient of 0.24 times the square of the current in amperes, resistance in ohms and time in seconds.
Video - "Joule-Lenz Law, Physics Grade 8":
Example 1. Determine how much heat a current of 6 A will release by passing through a conductor with a resistance of 2 ohms for 3 minutes.
Q= 0.24 × I 2 × r × t= 0.24 x 36 x 2 x 180 = 3110.4 cal.
The formula for the Joule-Lenz law can be written as follows:
Q= 0.24 × I × I × r × t ,
and since I × r = U, then you can write:
Q= 0.24 × I × U× t cal.
Example 2. The electric hotplate is connected to a 120V network. The current flowing along the spiral of the tile is 5A. It is required to determine how much heat the current will release in 2 hours.
Q= 0.24 × I × U× t= 0.24 × 5 × 120 × 7200 = 1,036,800 calories = 1036.8 kcal.
Video - "Heating of conductors by electric current":
E. H. Lenz summarized the experiments electromagnetic induction, expounding this generalization in the form of "Lenz's rule". In his works on the theory of electrical machines, Lenz studied the phenomenon of "armature reaction" in DC machines, proved the principle of reversibility of electrical machines. Lenz, working with Jacobi, investigated the force of attraction of electromagnets, established the dependence of the magnetic moment on the magnetizing force.
12 (24) February 1804 - 29 January (10 February) 1865 (age 60)
Lenz was a member of the St. Petersburg Academy of Sciences and the rector of St. Petersburg University.
Consider a homogeneous conductor, to the ends of which a voltage U is applied. During a time dt, a charge dq = Idt is transferred through the conductor cross-section. Since the current is the movement of the charge dq under the action electric field, then, the work of the current is
dA = Udq = IU dt (13.28)
If the resistance of the conductor is R, then using Ohm's law, we get
Power current
(13.30)
If the current passes through a fixed metal conductor, then all the work of the current goes to heating it, and, according to the law of conservation of energy,
(13.31)
Thus, using expression (13.28) and (13.31), we obtain
(13.32)
The expression is Joule-Lenz law experimentally established independently by Joule and Lenz.
§ 13.7 Ohm's and Joule-Lenz's laws in differential form.
Substituting the expression for resistance into Ohm's law, we get
(13.33)
where the value 
, the reciprocal of the resistivity, is called specific electrical conductivity
substance of the conductor. Its unit is siemens per meter (S / m).
Considering that 
- the strength of the electric field in the conductor, 
is the current density, the formula can be written as
j = γE (13.34)
Joule-Lenz law in differential form
We select in the conductor an elementary cylindrical volume dV = dSdℓ (the axis of the cylinder coincides with the direction of the current (Figure 13.9)), the resistance of which is 
... According to the Joule-Lenz law, heat will be released in this volume during the time
(13.35)
The amount of heat released per unit of time in a unit of volume is called specific thermal power of current ... She is equal
ω = ρ ∙ j 2 (13.36)
Using the differential form of Ohm's law (j = γE) and the ratio, we obtain ω = j ∙ E = γ ∙ E 2 (13.37)
Examples of problem solving
Example. The current in the conductor evenly increases fromI 0 = 0 toI max = 3A for the time τ = 6s. Determine the chargeQpassed through the conductor.
Given: I 0 = 0; I max = 3A; τ = 6s .
Find: Q.
Solution. The charge dQ passing through the cross-section of the conductor in time dt,
According to the condition of the problem, the current increases uniformly, i.e. I = kt, where the proportionality coefficient
.
Then we can write
By integrating (1) and substituting the expression for k, we find the required charge passed through the conductor:
Answer : Q = 9 Cl .
Example. On an iron guide (ρ = 7.87 g / cm 3 , М = 56 ∙ 10 -3 kg / mol) with a cross sectionS= 0.5 mm 2 current flowsI= 0.1 A. Determine the average speed of the ordered (directed) movement of electrons, assuming that the number of free electrons per unit volume of the conductor is equal to the number of atomsn"per unit volume of the conductor
Given: ρ = 7.87 g / cm 3, = 7.87 ∙ 10 3 kg / m 3; M = 56 ∙ 10 -3 kg / mol; I = 0.1A; S = 0.5 mm 2 = 0.5 10 -6 m 2.
Find:
.
Solution ... Conductor current density
j = ne 
,
where 
is the average speed of the ordered movement of electrons in a conductor; n is the concentration of electrons (the number of electrons per unit volume); e = 1.6 ∙ 10 -19 C is the electron charge.
According to the condition of the problem,
(2)
(take into account that 
, where is the mass of the conductor; M is its molar mass; N A = 6.02 ∙ 10 23 mol -1 - Avogadro's constant; 
- iron density).
Taking into account formula (2) and the fact that the current density 
, expression (1) can be written as
,
Whence the sought speed of the ordered motion of electrons
Answer:
= 14.8 μm / s.
Example. Resistance of homogeneous wireR= 36 ohms. Determine how many equal segments the wire was cut into, if after their parallel connection the resistance turned out to beR 1 = 1 Ohm.
Given R= 36 Ohm;R 1 = 1 ohm.
Find: N.
Solution. An uncut wire can be thought of as N series-connected resistances. Then
where r is the resistance of each segment.
In case of parallel connection of N lengths of wires
or 
(2)
From expressions (1) and (2) we find the required number of segments
Answer: N = 6
Example. Determine the current density in a copper wire with a length of ℓ = 100 m, if the potential difference at its ends is φ 1 -φ 2 = 10V. Copper resistivity ρ = 17 nOhm ∙ m.
Given ℓ = 100 m; φ 1 -φ 2 = 10V; ρ = 17 nΩ ∙ m = 1.7 ∙ 10 -8 Ohm ∙ m.
Find: j.
Solution. According to Ohm's law in differential form,
where 
- specific electrical conductivity of the conductor; 
- the strength of the electric field inside a homogeneous conductor, expressed through the potential difference at the ends of the conductor and its length.
Substituting the written formulas into expression (1), we find the required current density
Answer: j = 5.88 MA / m 2.
Example. Current flows through the incandescent lampI= 1A, Temperature of tungsten filament diameterd 1 = 0.2 mm is equal to 2000 ° C. The current is supplied by copper wires with a cross-sectionS 2 = 5mm 2 ... Determine the strength of the electrostatic field: 1) in tungsten; 2) in copper. Resistivity of tungsten at 0 ° С ρ 0 = 55 nΩ ∙ m, its temperature coefficient of resistance α 1 = 0.0045 hail -1 , the resistivity of copper ρ 2 = 17nOhm ∙ m.
Given: I= 1A;d 1 = 0.2 mm = 2 ∙ 10 -4 m; T = 2000 ° C;S 2 = 5mm 2 =5∙10 -6 m 2 ; ρ 0 = 55 nOhm ∙ m = 5.5 ∙ 10 -8 Ohm ∙ m: α 1 = 0.0045 ° C -1 ; ρ 2 = 17nOhm ∙ m = 1.7 ∙ 10 -8 Ohm ∙ m.
Find: E 1; E 2.
Solution. According to Ohm's law in differential form, the current density
(1)
where 
- specific electrical conductivity of the conductor; E is the strength of the electric field.
The resistivity of tungsten changes linearly with temperature:
ρ = ρ 0 (1 + αt). (2)
Current density in tungsten
(3)
Substituting expression (2) and (3) into formula (1), we find the required strength of the electrostatic field in tungsten
.
Electrostatic field strength in copper
(take into account that 
).
Answer: 1) E 1 = 17.5 V / m; 2) E 2 = 3.4 mV / m.
Example. Through conductor resistanceR= 10 Ohm a current flows, the current increases linearly. Quantity of heatQreleased in the conductor during τ = 10 s is equal to 300 J. Determine the chargeq, passed during this time along the conductor, if in the initialmThe moment of time the current in the conductor is zero.
Given: R= 10 Ohm; τ = 10s;Q= 300J;I 0 =0.
Find: q.
Solution.
From the condition of the uniformity of the increase in the current strength (at I 0 = 0) it follows that I = kt, where k is the proportionality coefficient. Considering that 
, we can write
dq = Idt = ktdt. (1)
We integrate expression (1), then
(2)
To find the coefficient k, we write down the Joule-Lenz law for an infinitely small time interval dt:
By integrating this expression from 0 to, we get the amount of heat specified in the condition of the problem:
,
Where do we find k:
.
(3)
Substituting formula (3) into expression (2), we determine the required charge
Answer: q = 15 Cl .
Example. Determine the density of the electric current in the copper wire (specific resistance ρ = 17nOhm ∙ m), if the specific thermal power of the current ω = 1.7J / (m 3 ∙ s) ..
Given: ρ = 17nOhm ∙ m = 17 ∙ 10 -9 Ohm ∙ m; ω = 1.7J / (m 3 ∙ s).
Find: j.
Solution. According to the Joule-Lenz and Ohm laws in differential form,
(1)
,
(2)
where γ and ρ are the specific and resistance of the conductor, respectively. From law (2) we obtain that E = ρj. Substituting this expression in (1), we find the required current density:
.
Answer : j = 10 kA / m 3.
Example. Determine the internal resistance of the current source, if in the external circuit at amperageI 1 = 4A, the power P develops 1 = 10 W, and at amperageI 2 = 6A - power P 2 = 12 W.
Given: I 1 = 4A; R 1 = 10 W;I 2 = 6A; R 2 = 12 W.
Find: r.
Solution. The power developed by the current
and 
(1)
where R 1 and R 2 are the resistances of the external circuit.
According to Ohm's law for a closed circuit,
;
,
where ε is the EMF of the source. Solving these two equations for r, we get
(2)
Answer : r = 0.25 ohm.
Example ... Into a circuit consisting of an EMF source and a resistor withR= 10 Ohm, turn on the voltmeter, first in parallel, and then in series with the resistor, and the voltmeter readings are the same. Determine Internal ResistancerEMF source, if the resistance of the voltmeterR V = 500 ohms.
Given: R= 10 Ohm;R V = 500 Ohm;U 1 = U 2 .
Find: r.
R 
solution.
According to the condition of the problem, the voltmeter is connected once to the resistor in parallel (Fig. A), the second is connected in series (Fig. B), and its readings are the same.
Mathematically, it can be expressed in the following form:
where w- the power of heat release per unit volume, - the density of the electric current, - the strength of the electric field, σ - conductivity of the medium.
The law can also be formulated in integral form for the case of currents flowing in thin wires:
In mathematical form, this law has the form
where dQ- the amount of heat released over a period of time dt, I- current strength, R- resistance, Q- the total amount of heat released over a period of time from t 1 before t 2... In the case of constant amperage and resistance:
Practical value
Reduced energy losses
When transmitting electricity, the thermal effect of the current is undesirable, since it leads to energy losses. Since the transmitted power is linearly dependent on both voltage and amperage, and the heating power is quadratically dependent on amperage, it is advantageous to increase the voltage before transferring electricity, thereby lowering the amperage. However, an increase in voltage reduces the electrical safety of power lines.
To apply high voltage in the circuit, the load resistance must be increased to maintain the same power at the payload. Lead wires and load are connected in series. Wire resistance () can be considered constant. But the load resistance () grows when choosing a higher voltage in the network. The ratio of load resistance and wire resistance also increases. When the resistances are connected in series (wire - load - wire), the distribution of the allocated power () is proportional to the resistance of the connected resistances.
The current in the network is constant for all resistances. Therefore, the relation
And for in each case are constants. Therefore, the power released on the wires is inversely proportional to the load resistance, that is, it decreases with increasing voltage, since 
... Whence it follows that. In each specific case, the value is constant, therefore, the heat generated on the wire is inversely proportional to the square of the voltage at the consumer.
Choice of wires for circuits
The heat generated by the conductor with current is, to one degree or another, released in environment... In the event that the current strength in the selected conductor exceeds a certain limit permissible value, such a strong heating is possible that the conductor can provoke a fire in nearby objects or melt itself. As a rule, when assembling electrical circuits, it is enough to follow the accepted normative documents, which regulate, in particular, the choice of the conductor cross-section.
Electric heating devices
If the current strength is the same throughout electrical circuit, then in any selected area, the more heat will be generated, the higher the resistance of this area.
By deliberately increasing the resistance of a section of the circuit, it is possible to achieve a localized release of heat in this section. This principle is followed by electric heaters... They use a heating element- a conductor with high resistance. An increase in resistance is achieved (jointly or separately) by choosing an alloy with high resistivity (for example, nichrome, constantan), increasing the length of the conductor and reducing its cross section. Lead wires have a normal low resistance and therefore heating is usually imperceptible.
Fuses
To protect electrical circuits from the flow of excessively high currents, a piece of conductor with special characteristics is used. This is a conductor of relatively small cross-section and made of such an alloy that, at permissible currents, the heating of the conductor does not overheat it, and with excessively large overheating of the conductor is so significant that the conductor melts and opens the circuit.
see also
Notes (edit)
Links
- Effective physics. Joule-Lenz Law Web Archive Copy
- http://elib.ispu.ru/library/physics/tom2/2_3.html Joule-Lenz Law
- http://eltok.edunet.uz/dglens.htm DC Laws. Joule-Lenz law
- http://slovari.yandex.ru/dict/bse/article/00023/23600.htm TSB. Joule-Lenz law
- http://e-science.ru/physics/theory/?t=27 Joule-Lenz Law
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See what the "Joule-Lenz Law" is in other dictionaries:
- (named after the English physicist James Joule and the Russian physicist Emilia Lenz, who simultaneously, but independently of each other, discovered it in 1840) the law giving quantification thermal action of electric current. When current flows through ... ... Wikipedia
JOLE-LENTZ'S LAW- the law that determines the thermal effect of an electric current; according to this law, the amount of heat Q released in a conductor when a direct electric current passes through it is equal to the product of the square of the current I, resistance ... ... Big Polytechnic Encyclopedia
Joule-Lenz law- - [Ya.N. Luginsky, M.S.Fezi Zhilinskaya, Y.S.Kabirov. English Russian Dictionary of Electrical Engineering and Electric Power Engineering, Moscow, 1999] Subjects of electrical engineering, basic concepts EN Joule Lenz s lawJoule s law ... Technical translator's guide
Joule-Lenz law
Joule-Lenz law- Joule o dėsnis statusas T sritis automatika atitikmenys: angl. Joule s law vok. Joulesches Gesetz, n rus. Joule Lenz's law, m pranc. loi de Joule, f ryšiai: sinonimas - Džaulio dėsnis… Automatikos terminų žodynas
joule's law- Džaulio dėsnis statusas T sritis fizika atitikmenys: angl. Joule law vok. Joule Lentzsches Gesetz, n; Joulesches Gesetz, n rus. Joule's law, m; Joule Lenz's law, m pranc. loi de Joule, f ... Fizikos terminų žodynas
Joule-Lenz law- the amount of heat Q released per unit of time in a section of an electric circuit with resistance R when a direct current I flows through it is equal to Q = RI2. The law was established in 1841 by J.P. Joule (1818 1889) and confirmed in 1842 by exact ... ... Concepts of modern natural science. Glossary of basic terms
Determines the amount of heat Q released in a conductor with resistance L during time t when current I passes through it: Q = aI2Rt. Coeff. proportionality and depends on the choice of unit. measurements: if I is measured in amperes, R in ohms, t in seconds, then ... ... Physical encyclopedia
Hello. The Joule-Lenz law is unlikely when you need it, but it is included in basic course electrical engineering, and therefore now I will tell you about this law.
The Joule-Lenz law was discovered by two great scientists independently of each other: in 1841, James Prescot Joule, an English scientist who made a great contribution to the development of thermodynamics 
and in 1842 Emiliy Khristianovich Lenz, a Russian scientist of German origin, who made a great contribution to electrical engineering. Since the discovery of both scientists took place almost simultaneously and independently of each other, it was decided to call the law a double name, or rather surnames. 
Remember when, and not only him, I said that electric current heats up the conductors through which it flows. Joule and Lenz found a formula that can be used to calculate the amount of heat generated.
So, initially, the formula looked like this:
The unit of measurement for this formula was calories and the coefficient k, which is equal to 0.24, was "responsible" for this, that is, the formula for obtaining data in calories looks like this: 
But since in the SI measurement system, in view of the large number of measured quantities and the avoidance of confusion, the designation joule was adopted, the formula has changed somewhat. k became equal to one, and therefore the coefficient was no longer written in the formula and it began to look like this:
Here: Q - the amount of heat generated, measured in Joules (designation in the SI system - J);
I is the current measured in Amperes, A;
R - resistance, measured in Ohms, Ohm;
t is the time measured in seconds, s;
and U is the voltage, measured in volts, V.
Look carefully to see if one part of this formula reminds you of anything? And more specifically? But this is power, or rather the power formula from Ohm's law. And to be honest, I have never seen such a representation of the Joule-Lenz law on the Internet:
Now we recall the mnemonic table and get at least three formula expressions of the Joule-Lenz law, depending on what quantities we know: 
It would seem that everything is very simple, but it seems to us only when we already know this law, and then both great scientists discovered it not theoretically, but experimentally and then were able to substantiate it theoretically.
Where can this Joule-Lenz law come in handy?
In electrical engineering, there is the concept of a long-term permissible current flowing through wires. This is the current that the wire is able to withstand long time(that is, infinitely long), without destroying the wire (and insulation, if any, because the wire can be without insulation). Of course, you can now take the data from the PUE (Electrical Installation Rules), but you received this data solely on the basis of the Joule-Lenz law.
In electrical engineering, fuses are also used. Their main quality is reliability of actuation. For this, a conductor of a certain cross section is used. Knowing the melting point of such a conductor, you can calculate the amount of heat that is necessary for the conductor to melt from flowing through it. large values current, and by calculating the current, you can calculate the resistance that such a conductor must have. In general, as you already understood, using the Joule-Lenz law, you can calculate the cross-section or resistance (values are interdependent) of the conductor for the fuse.
Also, remember, we talked about. There, using the example of a light bulb, I told the paradox that a more powerful lamp in series connection shines weaker. And probably remember why: the voltage drop across the resistance is the stronger, the lower the resistance. And since power is, and the voltage drops very much, it turns out that a large resistance will emit a large number of heat, that is, the current will have to work harder to overcome the large resistance. And the amount of heat that the current will release in this case can be calculated using the Joule-Lenz law. If we take a series connection of resistances, then it is better to use an expression through the square of the current, that is, the original form of the formula:
And for parallel connection of resistances, since the current in parallel branches depends on the resistance, while the voltage on each parallel branch is the same, then the formula is best represented in terms of voltage:
You all use examples of the Joule-Lenz law in Everyday life- first of all, these are all kinds of heating devices. As a rule, they use nichrome wire and the thickness (cross-section) and length of the conductor are selected taking into account that prolonged thermal exposure does not lead to rapid destruction of the wire. In exactly the same way, they achieve the glow of a tungsten filament in an incandescent lamp. The same law determines the degree of possible heating of almost any electrical and electronic device.
In general, despite the apparent simplicity, the Joule-Lenz law plays a very huge role in our life. This law gave a great impetus for theoretical calculations: the release of heat by currents, the calculation of the specific temperature of the arc, conductor and any other electrically conductive material, the loss of electrical power in thermal equivalent, etc.
You may ask how to convert Joules to Watts and that's pretty frequent question in the Internet. Although the question is somewhat wrong, as you read on, you will understand why. The answer is quite simple: 1 J = 0.000278 Watt * hour, while 1 Watt * hour = 3600 Joules. Let me remind you that the instantaneous power consumption is measured in Watts, that is, it is directly used while the circuit is turned on. And Joule determines the work of an electric current, that is, the power of the current over a period of time. Remember, in Ohm's Law I gave an allegorical situation. Current is money, voltage is a store, resistance is a sense of proportion and money, power is the amount of products that you can carry (take away) at one time, but how far, how quickly and how many times you can take them away is work ... That is, it is impossible to compare work and power, but it can be expressed in units that are more understandable to us: watts and hours.
I think that now it will not be difficult for you to apply the Joule-Lenz law in practice and theory, if necessary, and even make the conversion of Joules to Watts and vice versa. And thanks to the understanding that the Joule-Lenz law is the product of electrical power and time, you can more easily remember it, and even if you suddenly forgot the basic formula, then remembering only Ohm's law you can again get the Joule-Lenz law. And on this I say goodbye to you.
The transfer of electricity during the movement of current into another energy occurs at the molecular level. During this process, the temperature of the conductor rises by a certain amount. describes this phenomenon interaction of atoms and ions of a current conductor with current electrons. 
Electricity properties
While moving along a metal conductor, electrons collide with a lot of randomly located foreign particles. Periodically, as a result of contact, new electrons are released from the neutral molecule. A positive ion is formed from a molecule, and disappears in an electron kinetic energy... Sometimes there is a second option - the formation of a neutral-type molecule due to the combination of a positive ion and an electron.
All these processes are accompanied by the expenditure of a certain amount of energy, which is further converted into heat. Overcoming resistance in the course of all these movements determines the expenditure of energy and the transformation of the work necessary for this into heat.
The R parameters are identical to those of the standard resistance. To one degree or another, a certain amount of energy is converted into heat when current passes through any conductor. It is this transformation that is considered by the Joule-Lenz law.
Formula and its components
The transition into the internal energy of the conductor of the results of the work of the current is confirmed by numerous experiments. After the accumulation of the critical volume, the surplus of energy is returned to the surrounding bodies with the heating of the conductor.
The classical calculation formula for this phenomenon:
We take Q to denote the amount of heat released and substitute it for A. Now, in the resulting expression Q = U * I * t, we replace U = IR and derive the classical Joule-Lenz formula: 
In daisy-chain diagrams, this basic formula will be the most convenient method for calculations. In this case, in all conductors, the current strength always remains the same. The amount of heat released is proportional to the resistance of each of the available conductors.
But with a parallel connection, the voltage at the ends will be the same, and the nominal value of the electric current in each element is significantly different. It can be argued that there is inverse proportion between the amount of heat and the conductivity of a single conductor. Here the formula becomes more appropriate:
Q = (U2 / R) t
Practical examples of the phenomenon of thermal action of current
Many researchers and scientists have studied the characteristics of the flow of electricity. But the most impressive results were obtained by the Russian scientist Emiliy Khristianovich Lenz and the Englishman James Joule. Independently of each other, a law was formulated, with the help of which an assessment was made of the heat obtained in the process of the action of electricity on the conductor. The final expression was named after its authors.
A few examples can help you understand the nature and characteristics heat exposure current.
Heating devices
The heating function in the construction of such devices is performed by a metal spiral. If it is necessary to heat water, it is important to maintain a balance between the parameters of network energy and heat exchange. The installation of the spiral is carried out in isolation.
The tasks of minimizing energy losses are solved in various ways. One of the options is to increase the voltage, but this is fraught with a decrease in the level of operational safety of the lines.
The method of selection of wires is also used, the heat loss in which depends on the properties various metals and alloys. The production of spirals is carried out from materials intended for work with high loads.
Incandescent lamp
The discovery of the Joule-Lenz law contributed to the rapid progress of electrical engineering. The example of its use for lighting elements remains especially indicative.
A tungsten thread is drawn inside such a light bulb. The whole process is based on high resistivity and the refractoriness of this metal.
The transformation of energy into heat causes the effect of heating and glowing of the spiral. The downside is always the consumption of the main volume of energy for heating, and the glow itself is performed due to its small part.
For a more accurate understanding of this process, a concept is introduced such as the coefficient useful action, by which the efficiency of the workflow is determined.
Electric arc
In this case, we are talking about a powerful light source and a method of welding metal structures.
The principle of such a process is the connection of a current source of high power and minimum voltage to a pair of coal rods, followed by contact of these elements.
Household fuses
When using electrical circuits, special devices are used. The main element in such fuses will be a low-melting wire. It is screwed into a porcelain case that fits into the chuck.
Being part of a common circuit, such a conductor melts and opens the network with a sharp increase in heat release.
Physics Grade 8: Joule-Lenz Law
A detailed study of the passage of electricity through a conductor and the heating that occurs during this is set out in school curriculum... Practical examples show all the nuances that affect the magnitude of the thermal effect of the current.
The plan for conducting a training session is usually structured as follows:
- Necessary experiments to demonstrate the dependence of the volume of heat on resistance and current strength.
- A detailed study of the Joule-Lenz law, its basic formula and the meaning of all its components.
- Historical facts, excluding the likelihood of plagiarism on the part of both authors.
- Summing up the general results of the lesson.
- Practical application for performing calculations.
- Solving problems based on the information received.
The material is fixed during homework to assess the amount of heat generated during the flow of current through a conductor with the indicated parameters.
