Let us consider a homogeneous conductor, to the ends of which a voltage U is applied. During the time dt, a charge dq = Idt is transferred through the cross section of the conductor. Since current represents the movement of charge dq under the influence electric field, then the work of the current is equal to

dA=Udq =IU dt (13.28)

If the conductor resistance is R, then, using Ohm’s law, we get

Current power

(13.30)

If a current passes through a stationary metal conductor, then all the work done by the current goes to heat 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 represents 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 is the value , the reciprocal of resistivity, is called electrical conductivity conductor substances. Its unit is siemens per meter (S/m).

Considering that
- electric field strength in the conductor,
- current density, the formula can be written as

j = γE (13.34)

Joule-Lenz law in differential form

Let us select an elementary cylindrical volume dV=dSdℓ in the conductor (the axis of the cylinder coincides with the direction of the current (Fig. 13.9)), the resistance of which
. According to the Joule-Lenz law, heat will be released in this volume over time

(13.35)

The amount of heat released per unit time per unit volume is called specific thermal current power . It is equal

ω= ρ∙j 2 (13.36)

Using the differential form of Ohm's law (j = γE) and the relation, we obtain ω= j∙E=γ∙E 2 (13.37)

Examples of problem solving

Example. The current strength in the conductor increases uniformly fromI 0 =0 toI max =3A for time τ=6s. Determine the chargeQ, passed through the conductor.

Given: I 0 =0; I max =3A; τ=6s .

Find: Q.

Solution. Charge dQ passing through the cross section of the conductor in time dt,

According to the conditions of the problem, the current strength increases uniformly, i.e. I=kt, where proportionality coefficient

.

Then we can write

By integrating (1) and substituting the expression for k, we find the required charge passing through the conductor:

Answer : Q=9 Cl .

Example. By iron conductor (ρ =7.87 g/cm 3 , M=56∙10 -3 kg/mol) cross sectionS=0.5 mm 2 current flowsI=0.1 A. determine the average speed of the ordered (directional) 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 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 . Current density in the conductor

j=ne ,

Where - average speed of ordered movement of electrons in a conductor; n - electron concentration (number of electrons per unit volume); e=1.6∙10 -19 C – electron charge.

According to the problem conditions,

(2)

(take into account that
, where is the mass of the conductor; M – its molar mass; N A = 6.02∙10 23 mol -1 – Avogadro’s constant;
- density of iron).

Taking into account formula (2) and the fact that the current density
, expression (1) can be written as

,

Where does the desired speed of ordered electron motion come from?

Answer: =14.8 µm/s.

Example. Homogeneous wire resistanceR=36 Ohm. Determine how many equal pieces the wire was cut into if, after connecting them in parallel, the resistance turned out to be equalR 1 =1Ohm.

Given R=36 Ohm;R 1 =1 Ohm.

Find: N.

Solution. An uncut wire can be represented as N resistances connected in series. Then

where r is the resistance of each segment.

In the case of parallel connection of N wire sections

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 nOhm∙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 electric field strength 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 desired current density

Answer: j=5.88 MA/m2.

Example. Current flows through an incandescent lampI=1A, Temperature of tungsten filament diameterd 1 =0.2 mm is equal to 2000ºС. The current is supplied by copper wires with a cross-sectionS 2 =5mm 2 . Determine the electrostatic field strength: 1) in tungsten; 2) in copper. Tungsten resistivity at 0ºС ρ 0 =55 nOhm∙ m, its temperature coefficient of resistance α 1 =0.0045 deg -1 , resistivity of copper ρ 2 =17nOhm∙m.

Given: I=1A;d 1 =0.2 mm=2∙10 -4 m; T= 2000ºС;S 2 =5mm 2 =5∙10 -6 m 2 ; ρ 0 =55 nOhm∙m= 5.5∙10 -8 Ohm∙m: α 1 =0.0045ºС -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 – electric field strength.

The resistivity of tungsten changes with temperature according to a linear law:

ρ=ρ 0 (1+αt). (2)

Current density in tungsten

(3)

Substituting expressions (2) and (3) into formula (1), we find the desired electrostatic field strength 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 a conductor with resistanceR=10 Ohm current flows, the current increases linearly. Quantity of heatQ, released in the conductor during the time τ = 10 s, is equal to 300 J. Determine the chargeq, passed through the conductor during this time, if at the initialmmoment in 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 uniform increase in 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)

Let's integrate expression (1), then

(2)

To find the coefficient k, we write the Joule-Lenz law for an infinitely small period of time dt:

By integrating this expression from 0 to, we obtain the amount of heat specified in the problem statement:

,

Where do we find k:

. (3)

Substituting formula (3) into expression (2), we determine the required charge

Answer: q=15 Cl .

Example. Determine the electric current density in a copper wire (specific resistance ρ=17 nOhm∙m), if the specific thermal power of the current ω=1.7 J/(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 into (1), we find the required current density:

.

Answer : j=10 kA/m 3 .

Example. Determine the internal resistance of the current source if the current in the external circuit isI 1 =4A power P develops 1 =10 W, and at current strengthI 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 is

And
(1)

where R 1 and R 2 are the external circuit resistances.

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 . In a circuit consisting of an EMF source and a resistor with a resistanceR=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 Ohm.

Given: R=10 Ohm;R V =500 Ohm;U 1 = U 2 .

Find: r.

R decision. According to the conditions of the problem, the voltmeter is connected once to the resistor in parallel (Fig. a), the second time in series (Fig. b), and its readings are the same.

A physical law that evaluates the thermal effect of electric current. The Joule-Lenz law was discovered in 1841 by James Joule and in 1842, completely independently, by Emilius Lenz.

as we already know, when free electrons move along a conductor, it must overcome the resistance of the material. During this movement of charges, constant collisions of atoms and molecules of the substance occur. In this case, the energy of movement and resistance is converted into heat. Its dependence on current was first described by two independent scientists, James Joule and Emil Lenz. That is why the law received a double name.

Definition, the amount of heat released per unit time in a specific section of an electrical circuit is directly proportional to the product of the square of the current in a given section and its resistance.

Mathematically, the formula can be written as follows:

Q = а×I 2 ×R×t

Where Q– amount of heat generated, A– heat coefficient (usually it is taken equal to 1 and is not taken into account), I– current strength, R– material resistance, t– time of current flow through the conductor. If the heat coefficient a = 1, That Q measured in joules. If a = 0.24, That Q measured in small calories.

Any conductor always heats up if current flows through it. But overheating of conductors is very dangerous, because it can damage not only electronic equipment, but also cause a fire. For example, in the event of a short circuit, the overheating of the conductor material is enormous. Therefore, to protect against short circuits and large overheating in electronic circuits special radio components are added - fuses. For their manufacture, a material is used that quickly melts and de-energizes the supply circuit when the current reaches maximum values. Fuses must be selected depending on the cross-sectional area of ​​the conductor.

The Joule-Lenz law is relevant for both direct and alternating current. According to it, many different heating devices work. After all, the thinner the conductor, the greater the current passes through it over a longer period of time, the more more quantity heat will be released as a result of this.

I hope you remember to remember that current depends on voltage. The question arises, why does the laptop not heat up as much as an iron? Because at the base there is a spiral wire made of steel, which has low resistance. Plus there is a steel sole, so the iron heats up to high temperatures, and we can iron with it.

And it has a voltage stabilizer that reduces 220 volts to 19 volts. Plus, the resistance of all circuits and components is quite high. Additionally for cooling there is a cooler and copper thermal radiators.

The work of the Joule-Lenz law is clearly visible in practice. Most famous example its application is an ordinary incandescent lamp or in which the filament glows due to the passage of a high voltage current through it.

Based on the Joule-Lenz law, and works, where the creation of a welded joint is accomplished by heating the metal, due to the current passing through it and deforming the parts being welded by compression.

Electric arc welding also works on the physical principles of the Joule-Lenz law. To carry out welding work, the electrodes are heated to such a state that a welding arc. Effect voltaic arc discovered by the Russian scientist V.V. Petrov, using the Joule-Lenz principle.

Except mathematical formula, this law also has a differential form. Let us assume that a current flows through a stationary conductor and all its work is spent only on heating. Then, according to the law of conservation of energy, we obtain the following mathematical expression.

In the 19th century, independently of each other, the Englishman J. Joule and the Russian E. H. Lenz studied the heating of conductors by electric current and experimentally established a pattern: the amount of heat released in a current-carrying conductor is directly proportional to the square of the current, the resistance of the conductor and the time it takes for the current to pass.
Later it was found that this statement is true for any conductors: solid, liquid, gaseous. Therefore, the open pattern is called Joule-Lenz law:

The figure shows an installation diagram with which you can experimentally verify the Joule-Lenz law. By dividing the current by the voltage, the resistance is calculated using the formula R=U/I. The thermometer measures the increase in water temperature. By formulas Q=I2Rt And Q=cmDt° calculate the amounts of heat that, according to the results of the experiment, should coincide.
For those who are more deeply interested in physics, we specifically note that the Joule-Lenz law can be obtained not only experimentally, but also derived theoretically. Let's do it.


The resulting formula A=I2Rt is similar to the formula of the Joule-Lenz law, but on the left side it is the work of the current, and not the amount of heat. What gives us the right to consider these quantities equal? Let's write it down first law of thermodynamics(see § 6-h) and express the work from it:
DU = Q + A, therefore A =DU-Q.
Let's remember that DU- this is a change in the internal energy of a conductor heated by current; Q- the amount of heat given off by the conductor (this is indicated by the “-” sign in front); A- work done on the conductor. Let's find out what kind of work this is.
The conductor itself is motionless, but electrons move inside it, constantly colliding with ions of the crystal lattice and transferring part of their energy to them. kinetic energy. To prevent the flow of electrons from weakening, work is constantly done on them by the forces of the electric field created by the source of electricity. Therefore, A is the work done by the electric field forces to move electrons inside the conductor.
Let us now discuss the quantity DU(change in internal energy) applied to a conductor in which current begins to flow.
The conductor will gradually heat up, which means it internal energy will increase. As it heats up, the difference between the temperatures of the conductor and the environment will increase. According to Newton's law (see § 6-k), the heat transfer power of the conductor will increase. After some time, this will cause the temperature of the conductor to stop increasing. From now on the internal energy of the conductor will cease to change, that is, the value DU will become equal to zero.
Then the first law of thermodynamics for this state will be: A = -Q. That is If the internal energy of the conductor does not change, then the work done by the current is completely converted into heat. Using this conclusion, we write all three formulas for calculating the work of the current in a different form:

For now we will consider these formulas to be equal. Later we will discuss that the right formula is always valid (that's why it is called the law), and the two left ones are true only under certain conditions, which we will formulate when studying physics in high school.

Simultaneously, but independently of each other, who discovered it in 1840) is a law that gives a quantitative assessment of the thermal effect of electric current.

When current flows through a conductor, electrical energy is converted into thermal energy, and the amount of heat released will be equal to the work of electrical forces:

Q = W

Joule-Lenz law: the amount of heat generated in a conductor is directly proportional to the square of the current, the resistance of the conductor and the time of its passage.

Practical significance

Reduced energy loss

When transmitting electricity, the thermal effect of current is undesirable, since it leads to energy losses. Since the transmitted power depends linearly on both voltage and current, and the heating power depends quadratically on the current, it is advantageous to increase the voltage before transmitting electricity, thereby reducing the current. Increasing voltage reduces the electrical safety of power lines. If high voltage is used in the circuit, in order to maintain the same consumer power, the consumer resistance will have to be increased (quadratic dependence. 10V, 1 Ohm = 20V, 4 Ohm). The supply wires and the consumer are connected in series. Wire resistance ( R w) constant. But consumer resistance ( R c) increases when choosing a higher voltage in the network. The ratio of consumer resistance to wire resistance also increases. When resistors are connected in series (wire - consumer - wire), the distribution of released power ( Q) is proportional to the resistance of the connected resistors. ; ; ; The current in the network is constant for all resistances. Therefore we have the relation Q c / Q w = R c / R w ; Q c And R w these are constants (for each specific task). Let's determine that . Consequently, the power released on the wires is inversely proportional to the consumer resistance, that is, it decreases with increasing voltage. because . (Q c- constant); Let's combine the last two formulas and deduce that ; for each specific task it is a constant. Consequently, the heat generated on the wire is inversely proportional to the square of the voltage at the consumer. The current flows evenly.

Choosing wires for circuits

The heat generated by a current-carrying conductor is released to varying degrees in environment. If the current strength in the selected conductor exceeds a certain maximum permissible value, such strong heating is possible that the conductor can cause a fire in objects near it or melt itself. As a rule, when assembling electrical circuits, it is enough to follow the accepted regulatory documents, which regulate, in particular, the choice of conductor cross-section.

Electric heating devices

If the current strength is the same throughout the entire electrical circuit, then in any selected section the more heat will be generated, the higher the resistance of this section.

By deliberately increasing the resistance of a section of a circuit, localized heat generation can be achieved in that section. They work on this principle electric heating devices. They use a heating element- conductor with high resistance. Increased 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. The lead wires have a generally low resistance and therefore their heating is usually unnoticeable.

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 with a relatively small cross-section and made of such an alloy that, at permissible currents, heating the conductor does not overheat it, but at excessively high currents, the overheating of the conductor is so significant that the conductor melts and opens the circuit.

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The Joule-Lenz law determines the amount of heat released in a conductor with resistance during a time t when an electric current passes through it.

Q = a*I*2R*t, where
Q - amount of heat released (in Joules)
a - proportionality coefficient
I - current strength (in Amperes)
R - Conductor resistance (in Ohms)
t - Travel time (in seconds)

The Joule-Lenz law explains that electric current is a charge that moves under the influence of an electric field. In this case, the field does work, and the current has power and energy is released. When this energy passes through a stationary metal conductor, it becomes thermal energy, since it is aimed at heating the conductor.

In differential form, the Joule-Lenz law is expressed as the volumetric thermal power density of the current in the conductor will be equal to the product of the electrical conductivity and the square of the electric field strength.

Application of the Joule-Lenz law

Incandescent lamps were invented in 1873 by the Russian engineer Lodygin. In incandescent lamps, as in electric heating devices, the Joule-Lenz law applies. They use a heating element, which is a high-resistance conductor. Due to this element, it is possible to achieve localized heat release in the area. Heat generation will appear with increasing resistance, increasing the length of the conductor, or choosing a specific alloy.

One of the areas of application of the Joule-Lenz law is to reduce energy losses.
The thermal effect of current leads to energy loss. When transmitting electricity, the transmitted power depends linearly on voltage and current, and the heating power depends on the current quadratically, so if you increase the voltage while lowering the current before supplying electricity, it will be more profitable. But an increase in voltage leads to a decrease in electrical safety. To increase the level of electrical safety, the load resistance is increased according to the increase in voltage in the network.

Also, the Joule-Lenz law affects the choice of wires for circuits. If not correct selection wires, strong heating of the conductor, as well as it, is possible. This occurs when the current exceeds the maximum permissible values ​​and too much energy is released. When selecting the correct wires, you should follow the regulatory documents.

Sources:

• Physical encyclopedia

There is a directly proportional relationship between current and voltage, described by Ohm's law. This law determines the relationship between current, voltage and resistance in a section of an electrical circuit.

Instructions

Remember current and voltage.
- Electric current is an ordered flow of charged particles (electrons). For quantitative determination, the value I is used, called current strength.
- Voltage U is the potential difference at the ends of a section of an electrical circuit. It is this difference that causes the electrons to move, like a fluid flowing.

Current strength is measured in amperes. In electrical circuits, the current strength is determined by an ammeter. The unit of voltage is , you can measure the voltage in a circuit using a voltmeter. Assemble a simple electrical circuit from a current source, a resistor, an ammeter and a voltmeter.

When a circuit is closed and current flows through it, record the instrument readings. Change the voltage at the ends of the resistance. You will see that the ammeter reading will increase as the voltage increases and vice versa. This experience demonstrates a directly proportional relationship between current and voltage.