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1 Training minimum in physics PHYSICS Topic The law of conservation of energy in electrical circuits QUESTIONS We consider electrical circuits that may contain batteries, resistors, capacitors and inductors Formulas for the energy of a capacitor and an inductor Formulate the law of conservation of energy for electrical circuit How is battery life determined? When is it positive? When is it negative? 4 What electrical elements produce heat? 5 Formulate the Joule-Lenz Law 6 How is the heat Q, released on the resistor by the resistance for any time, if the current I t flows through it? 7 What formula determines the rate of change in the energy of a capacitor? 8 What formula determines the rate of change in the energy of an inductor? TASKS All kinds of tasks for a class 5 circuit Fig. Task In the circuit shown in Fig., all elements can be considered ideal. The parameters of the elements are shown in Fig. Before closing the key, there was no current in the circuit. immediately after opening the key?) What work will the source do during the entire time of the experiment?) How much heat will be released in the circuit for the entire time of the experiment? 4) How much heat will be released in the circuit during time t? Task In the electrical circuit shown in Fig., all elements can be considered ideal Before the key was closed, there was no current in the circuit Key K is closed for some time and then opened. , while the key was open) the amount of heat Q was released in the circuit Find the time Problem In the electrical circuit shown in Fig, all elements can be considered ideal Before the key was closed, there was no current in the circuit Key K was closed for a while and then opened It turned out that during while the key was closed, and during the time that the key was open, the circuit stood out equal amounts heat How much charge flowed through the source during the time that the switch was closed? How much heat was released in the circuit during the entire time of the experiment?
2 Task 4 In the electrical circuit shown in the figure, all elements are ideal, key K is open Coil inductance, resistor resistance, battery EMF Key K is closed In the first seconds after key K is closed, the battery has done work 5% less than the work it has done for the next seconds) Determine the time) How much heat will be released in the circuit in time 4 after the key K is closed? Task 5 In the electrical circuit shown in Fig., all elements can be considered ideal. The parameters of the elements are shown in the figure. Before the key was closed, there was no current in the circuit. Key K is closed for a while, and then opened. It turned out that after opening the key, two times more heat than when the key is closed Find the ratio of the charge flowing through the source when the key is closed to the charge flowing through the resistor after the key is opened Task 6 In the electrical circuit shown in the figure, all elements can be considered ideal The parameters of the elements are shown in the figure the circuit was missing Key K is closed for some time, and then opened It turned out that the charge flowing through the coil with the key closed is 4 times greater than the charge flowing through the coil after opening the key Determine the time Find the ratio of heat released in the circuit after opening the key, to the heat released in the circuit when the key is closed Task 7 An electric circuit consists of an ideal battery with an EMF, an inductor, a capacitor with a capacitance C and a resistor with an unknown resistance (figure on the right) The key K is closed for a while and then opened During the time that the key was closed, a charge flowed through the resistor q) How much heat was released in the circuit during the time the key was closed?) How much heat was released in the circuit after the key was opened? Schemes - classes Problem 8 In the electrical circuit shown in the figure on the left, all elements are ideal The capacitor is initially not charged, the key K is open The key K is closed and then opened at the moment when the voltage across the capacitor becomes equal It is known that while the key K was closed, through resistor with resistance leaked charge 6 C How much heat was released in the circuit while the key K was closed? Problem 9 How much heat will be released on the resistor in the circuit shown in the figure on the right, after moving the key K from position to position? Neglect the internal resistance of the battery Problem In the electrical circuit shown in the figure on the left, all elements are ideal The capacitor is initially charged to voltage, the key K is open The key K is closed) Determine the change in the energy of the capacitor) Determine the work that the battery will do? What state will the battery be in?) How much heat will be released in the circuit? 4) What is the highest rate of change in the energy of the capacitor (the highest in absolute value)?
3 Task In the electrical circuit shown in the figure on the right, at the initial moment the key K is closed After the key is opened, the amount of heat is released on the resistor Q) How much heat will be released on the resistor?) What is the battery EMF? The resistances and the inductance of the coil are known Neglect the internal resistance of the battery Task In the circuit shown in the left figure, with the key K open, the capacitor C is charged to voltage U, and the capacitor C to voltage U The key K is closed) What will be the current in the circuit immediately after closing the key K (indicate the direction)?) Determine the rate of change in the energy of the capacitor with capacitance C immediately after closing the key K?) Determine the magnitude and sign of the charge of the left plate of the capacitor with capacitance C in the steady state? 4) What charge will flow through the resistor with resistance (indicate the direction)? 5) Find the change in the energy of a capacitor with a capacity C? 6) How much heat will be released in the circuit? 7) How much heat will be released on the resistor resistance? Task In the circuit shown in the figure on the right, a capacitor with a capacitance C is charged to a voltage U, and a capacitor with a capacitance C to a voltage U (fig on the right) Similar charged plates are connected by a resistor with a resistance Key K is closed for a while and then opened) Find the current in the circuit immediately after closing the key K (indicate the direction)) How much heat was released in the circuit if, at the moment of opening the key K, the current in the circuit was half the initial one? Task 4 In the circuit shown in the left figure, all elements are ideal At the initial moment of time, the keys K and K are open, the capacitors are not charged The keys close at the same time) Find the initial current through each of the batteries) Determine the charges of the capacitors in the steady state) Find the total operation of the batteries 4 ) How much heat will be released in the whole circuit after the keys are closed? Assume that Task 5 The electrical circuit consists of a battery with an EMF and an internal resistance r, a capacitor with a capacitance C and a resistor with a resistance of 5r. The key K is closed and then opened at the moment when the currents through the capacitor and the resistor are equal in magnitude) What instantaneous power does the source develop immediately before the key is opened?) How much heat will be released in the circuit after the key is opened?
4 Task 6 In the electrical circuit shown in the figure on the left, all elements are ideal. The key K is initially open, there are no currents in the circuit. The key K is closed. each of the coils during this time Task 7 An electrical circuit consists of an inductance coil, a resistor with a resistance, a battery with an EMF and an unknown internal resistance (Fig. *) Key K is closed for a while and then opened the amount of heat Q, and after opening the key, the amount Q was released in the circuit) Find the current through the coil at the moment the key was opened) Find the charge that flowed through the coil during the time the key was closed Task 8 An electric circuit consists of an inductance coil, a resistance resistor, a battery with EMF and unknown internal resistance (figure on the left) Key K is closed for a while, and then opened During the time that the key was closed, charge q flowed through the source, and energy W was stored in the coil) Find the amount of heat released in the circuit while the key was closed) What charge flowed through the coil when the key was closed? Task 9 In the electrical circuit shown in the figure on the right, the key K is closed The key K is opened After that, the battery with EMF has done work A, and the amount of heat released in the circuit is Q) Find the capacitance of the capacitor C) Find the inductance of the EMF coil of the batteries and the resistance of the resistors consider as given Assume that the Task The electric circuit consists of an ideal battery with an EMF, a flat capacitor and a resistor with a resistance A dielectric plate is inserted into the capacitor parallel to the plates, occupying half the volume of the capacitor (left figure) The dielectric constant of the dielectric is equal to The capacity of the air capacitor is C The plate is quickly removed) Which mechanical work A mech should be done in order to quickly remove the plate from the capacitor?) How much heat Q will be released in the circuit after the plate is removed? Task An electric circuit consists of an ideal battery with an EMF, a flat capacitor and a resistor with a resistance. A conductive plate is inserted parallel to the plates in the capacitor, occupying half the volume of the capacitor (rice on the right). The capacity of the air capacitor is C. The plate is quickly removed) remove the plate from the capacitor?) How much heat Q will be released in the circuit after the plate is removed?
5 Capacitor energy: W C CU qu q C ANSWERS QUESTIONS I ФI Ф Coil energy: W, where Ф magnetic flux penetrating the coil Work A B of all batteries included in the circuit goes to the release of heat Q in the electrical circuit and to the change W of the energy of this circuit: AB Q W The energy of the circuit is equal to the sum of the energies of all capacitors and all inductors AB q *, where q * modulus of the charge flowing through the battery Battery operation is positive (a “+” sign is put) if the battery is in operating mode, and negative (a sign is put) if the battery is in a recharge state 4 Only on resistors 5 If a direct current I flows through a resistance resistor, then the amount of heat released U over time is equal to Q I U I, where U I U t 6 Q I t t t U t I tt, where the summation is carried out over all small time intervals t over the time interval W t U t I t P t, where the “+” sign is put, if the capacitor is charging and the sign is set if 7 C C C C the capacitor is discharging 8 W t U t I t, where U t t I t I t PROBLEM) t) t Task t) t t 4) t Task Task Q 4)) 4) 4C) 6 4) Task 4 Task 5 8)) Q4 5 5 Task 6 Task 7) 8 Q) 4 q Q) Q q) Q C Task 8 Task 9 4 C 9 C Task C, the battery will be in a recharge state) C q C , the highest rate of change in the energy of the capacitor will be at the moment immediately after the switch is closed
6 Q) Q Q) Task Task U) (counterclockwise) U) (minus sign indicates that the energy of the capacitor decreases in this moment time)) 4 CU 4) 9 CU (counterclockwise) 4 5) 45 CU 6) 7 8 CU 7) 9 4 CU) U) CU Task Task 4) I and I 7 5) qc C, qc C and q C C 6 74) AB C) Q C 6 Problem 5 5)) 7r 98 C Problem 6 Q 9 q 4 8 and Q q 4 Q))) q W) Q Q Q q W A 8) C) Q A 9 4)) Amech Amech 8 C) C) Q 8 Q C C Problem 7 Problem 8 Problem 9 Problem Problem Compiled by: MA Penkin, lecturer at FZPTSH at Moscow Institute of Physics and Technology
IV Yakovlev Materials on physics MathUs.ru Quantity of heat. Capacitor This worksheet deals with problems for calculating the amount of heat that is released in circuits consisting of resistors and capacitors.
IV Yakovlev Materials on physics MathUs.ru Quantity of heat. Coil This worksheet deals with problems for calculating the amount of heat that is released in circuits consisting of resistors and coils.
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1.4. CLASSIFICATION OF ELECTRIC CIRCUITS
Depending on the current for which the electrical circuit is intended, it is respectively called: "Direct current electrical circuit", "Changing current electrical circuit", "Sinusoidal current electrical circuit", "Non-sinusoidal electrical circuit".
Similarly, the elements of circuits are also called - direct current machines, alternating current machines, sources electrical energy(IEE) direct current, IEE alternating current.
The elements of the circuits and the circuits made up of them are also subdivided according to the type of current-voltage characteristic (CVC). This means the dependence of their voltage on current U = f (I)
Circuit elements whose I–V characteristics are linear (Fig. 3, a) are called linear elements, and, accordingly, electrical circuits are called linear.
 |
An electrical circuit containing at least one element with a non-linear CVC (Fig. 3, b) is called non-linear.
Electric circuits of direct and alternating current are also distinguished by the method of connecting their elements - into unbranched and branched.
Finally, electrical circuits are divided according to the number of electrical energy sources - with one or several IEE.
There are active and passive circuits, sections and elements of circuits.
Active circuits are called electrical circuits containing sources of electrical energy, passive - electrical circuits that do not contain sources of electrical energy.
For the operation of an electrical circuit, the presence of active elements, i.e., energy sources, is necessary.
The simplest passive elements of an electrical circuit are resistance, inductance, and capacitance. With a certain degree of approximation, they replace the real elements of the circuit - a resistor, an inductive coil and a capacitor, respectively.
In a real circuit, not only a resistor or a rheostat as devices designed to use their electrical resistances has electrical resistance, but also any conductor, coil, capacitor, winding of any electromagnetic element, etc. But common property of all devices with electrical resistance is the irreversible conversion of electrical energy into thermal energy. Indeed, it is known from the course of physics that at a current i in a resistor with a resistance r, during the time dt, in accordance with the Joule-Lenz law, energy is released
dw = ri 2 dt,
or we can say that power is consumed in this resistor
p = dw/dt = ri 2 = ui,
Where u- voltage at the resistor terminals.
The thermal energy released in the resistance is usefully used or dissipated in space: But since the conversion of electrical energy into thermal energy in a passive element is irreversible, in the equivalent circuit, in all cases when it is necessary to take into account the irreversible energy conversion, resistance is turned on. In a real device, such as an electromagnet, electrical energy can be converted into mechanical energy (armature attraction), but in an equivalent circuit this device is replaced by a resistance in which an equivalent amount of thermal energy is released. And when analyzing the circuit, we are already indifferent to what is actually the consumer of energy: an electromagnet or an electric stove.
A value equal to the ratio of the constant voltage in the section of the passive electrical circuit to the direct current in it in the absence of e. d.s., called electrical resistance to direct current. It differs from AC resistance, which is determined by dividing the active power of a passive electrical circuit by the square of the effective current. The fact is that with alternating current, due to the surface effect, the essence of which is the displacement of alternating current from the central parts to the periphery of the conductor section, the resistance of the conductor increases and the more, the greater the frequency of the alternating current, the diameter of the conductor and its electrical and magnetic conductivity material. In other words, in the general case, the conductor always has more resistance to alternating current than to direct current. In AC circuits, resistance is called active. Chains characterized only electrical resistances their elements are called resistive .
Inductance L, measured in Henry (G), characterizes the property of a section of a circuit or coil to accumulate the energy of a magnetic field. In a real circuit, not only inductive coils, as circuit elements designed to use their inductance, have inductance, but also wires, capacitor leads, and rheostats. However, for the sake of simplicity, in many cases it is assumed that all the energy of the magnetic field is concentrated only in the coils.
With increasing current in the coil, the energy of the magnetic field is stored, which can be defined asw m \u003d L i 2 / 2 .
Capacitance C, measured in farads (F), characterizes the ability of a circuit section or capacitor to store energy electric floor I. In a real circuit, electrical capacitance exists not only in capacitors, as elements designed specifically to use their capacitance, but also between conductors, between turns of coils (interturn capacitance), between a wire and ground or the frame of an electrical device. However, in equivalent circuits, it is assumed that only capacitors have capacitance.
The energy of the electric field stored in the capacitor with increasing voltage is 
.
Thus, the parameters of the electrical circuit characterize the properties of the elements to absorb energy from the electrical circuit and convert it into other types of energy (irreversible processes), as well as create their own electric or magnetic fields in which energy can accumulate and, under certain conditions, return to the electrical circuit. The elements of the DC electric circuit are characterized by only one parameter - resistance. Resistance determines the property of an element to absorb energy from an electrical circuit and convert it into other forms of energy.
1.5. DC ELECTRIC CIRCUIT. OHM'S LAW
In the presence of electric current in conductors, moving free electrons collide with ions crystal lattice, experience resistance to their movement. This resistance is quantified by the amount of resistance.
| Rice. 4 |
Consider an electrical circuit (Fig. 4), which shows the IEE (highlighted by dashed lines) with emf on the left. E and internal resistance r, and on the right is an external circuit - a consumer of electrical energy R. To determine the quantitative characteristics of this resistance, we use Ohm's law for a section of the circuit.
Under the influence of e. d.s. in the circuit (Fig. 4) a current arises, the value of which can be determined by the formula:
I = U/R (1.6)
This expression is Ohm's law for a circuit section: the current strength in a circuit section is proportional to the voltage applied to this section.
From received find expressions R = U / I and U = I R.
It should be noted that the above expressions are valid provided that R is a constant value, i.e. for a linear circuit characterized by the dependence I = (l / R)U (the current linearly depends on the voltage and the slope angle φ of the straight line in Fig. 3, a is equal to φ = arctan(1/R)). An important conclusion follows from this: Ohm's law is valid for linear circuits when R = const.
The unit of resistance is the resistance of such a section of the circuit, in which a current of one ampere is set at a voltage of one volt:
1 ohm = 1 V/1A.
The larger units of resistance are kiloohm (kΩ): 1 kΩ = ohm and meg (mΩ): 1 mΩ = ohm.
In general R = ρ L/S, where ρ - resistivity of a conductor with a cross-sectional area S and length l.
However, in real circuits, the voltage U determined not only by the magnitude of the emf, but also depends on the magnitude of the current and resistance r IEE, since any energy source has internal resistance.
Consider now a complete closed circuit (Fig. 4). According to Ohm's law, we obtain for the outer section of the chain U=IR and for internal U 0=I r. A since the e.f.s. equal to the sum of the voltages in individual sections of the circuit, then
E = U + U 0 = IR + Ir
. (1.7)
Expression (1. 7) is Ohm's law for the entire circuit: the current strength in the circuit is directly proportional to the emf. source.
From expression E=U+ follows that U = E - Ir, i.e. in the presence of current in the circuit, the voltage at its terminals is less than the emf. source by the voltage drop across the internal resistance r source.
It is possible to measure voltages (with a voltmeter) in various parts of the circuit only when the circuit is closed. emf the same is measured between the source terminals with an open circuit, i.e. at idle, when I current in the circuit zero in this case E = U.
1.6. METHODS OF CONNECTING RESISTANCES
When calculating circuits, one has to deal with various consumer connection schemes. In the case of a circuit with a single source, a mixed connection is often obtained, which is a combination of parallel and series connections known from the course of physics. The task of calculating such a circuit is to determine, with known resistances of consumers, the currents flowing through them, the voltages, powers on them and the power of the entire circuit (all consumers).
A connection in which the same current passes through all sections is called a series connection of circuit sections. Any closed path that passes through several sections is called an electrical circuit loop. For example, the circuit shown in Fig. 4 is single-loop.
Consider the various ways of connecting resistances in more detail.
1.6.1 Series connection of resistances
If two or more resistors are connected, as shown in Fig. 5, one after the other without branching and the same current passes through them, then such a connection is called series.
| Rice. 5 |
According to Ohm's law, you can determine the voltage in individual sections of the circuit (resistance)
U 1 =IR 1 ; U 2 =IR2 ; U 3 =IR 3 .
Since the current in all sections has the same value, the voltages in the sections are proportional to their resistances, i.e.
U 1 /U 2 = R 1 /R 2 ; U 2 /U 3 = R 2 /R 3 .
The capacities of individual sections are respectively equal
P 1 = U 1 I;P 2 = U 2 I;P 3 = U 3 I.
And the power of the entire circuit, equal to the sum of the powers of individual sections, is defined as
P =P 1 +P 2 +P 3 =U 1 I+U 2 I+U 3 I= (U 1 +U 2 +U 3)I=UI,
whence it follows that the voltage at the circuit terminals U equal to the sum of the stresses in the individual sections
U=U 1 +U 2 + U 3 .
Dividing the right and left sides of the last equation by the current, we get
R=R 1 +R 2 +R 3 .
Here R = U/I- the resistance of the entire circuit, or, as it is often called, the equivalent resistance of the circuit, i.e. such an equivalent resistance, replacing which all the resistances of the circuit (R 1 ,R 2 , R 3) with a constant voltage at its terminals, we get the same current value.
1.6.2. Parallel connection of resistances
| Rice. 6 |
A parallel connection of resistances is a connection (Fig. 6), in which one terminal of each of the resistances is connected to one point in the electrical circuit, and the other terminal of each of the same resistances is connected to another point in the electrical circuit. So between two points the electrical circuit will include several resistances. forming parallel branches.
Since in this case the voltage on all branches will be the same, the currents in the branches may be different, depending on the values of the individual resistances. These currents can be determined by Ohm's law:
Voltages between branching points (A and B Fig.6)
Therefore, both incandescent lamps and motors designed to operate at a certain (rated) voltage are always connected in parallel.
The law of conservation of energy in capacitor circuits Task 1 A Q 0 W A kmech ist Option 1 When the key K2 is open, the key K1 is closed and after the end of the transients it is opened. After that, the key K2 is closed. Solution. According to the law of conservation of energy, the change in energy in a capacitor is determined by the ratio mechA work mechanical forces is zero, since there is no movement inside the capacitors. source the work of the current source is zero, since when the key K2 is closed, the key K1 is open, the current source is turned off. Q is the amount of heat that is released during the movement of charges. W W kn The initial and final energies of the capacitors correspond respectively to the open and closed key K2. For the initial state (capacitors are charged from the current source): Q Q W W kk 0 kn kk For the final state (only capacitor C2 and capacitor C3 parallel to it remain in the circuit.). Capacitor charges are conserved because the circuit is open. q 23 2 Ec W kk 2 q 23 2 C 23 2 2 E c 4 2 (c 2) c 2 3 2 E c 2 Q E c Option 2. 2 3 2 E c 1 3 2 E c 2 c C o q o W kn 2) c 2 c Ec 2 1 () C C C 6 (c c 3 c C C C c 6 3 2 1 q q q 2 E C 1 3 2 C U 2 s E o 2 2 cE 2 2 o 2 o kn ist Q A kk ist kkkn When open key K2, the key K1 is closed and after the end of the transient processes, the key K2 is closed. Solution. In this case, the key K2 is closed under voltage, the current source remains connected constantly, participates in the recharging of the capacitors, therefore, does the work. The law of conservation of energy in this case takes the form: - W W Q W W A c C C 3 2 ok 3 Ec E C ok 2 2 C E 3 E c ok 2 2 Current source operation: E q E q A (source ok We substitute the energies of the capacitors in the ratio for Q and get the answer. E (3 Ec 2 Ec) q on) 2 E c 2 c 3 c W kk 2 Q E c 2 2 E c E c 2 E c 3 2 1 3 The same answer in the first and second options is not a pattern, but an accident. Task 2 In the initial state for the scheme of Fig. 2 С1=2С, С2=3С, emf. the current source is equal to E. In a flat air capacitor C1, with the help of an external force, the plates were very quickly pushed apart, increasing the distance between the plates by 2 times. What amount of heat will be released in the circuit in the subsequent transient process? Solution. When the plate moves quickly against the Coulomb force, the charge of the plates is conserved, the Coulomb force does negative work, and the external force does positive work. The second plate moves in the field of the first plate, the electric capacitance of the first capacitor decreases by 2 times. A fur F k dЕ q 1 2 q d q n 1 2 S 0 2 n d 2 q d 1 n 2 S 0 2 q 1 n 2 C n For the initial state ( before the start of movement): C 0 n 1 n С C 2 C C 2 1 n q 0 n q 1 n q 2 n 2 3 c c 3 2 c c Ec 6 5 6 5 s A mech 2 2 36 E c 25 2 0.72 2 E c W kn 2 6 sE 5 2 0.6 2 E c , so for the fact that if the sum of the voltages on C1 and C2 remained equal to E, the charge would go into the current source, which means that the current source would do negative work. For the final state: 0 3 cE 2 4 C E k 2 3 4 3 8 Energy conservation law W W Q Q W W AA Task 3 kkn mech kkn ist mech ist AA cE Ec 6 5 Ec) 9 20 2 Ec 0.45 2 E c 2 0.375 cE 2 (0.375 0.6 0.72 0.45) E c 2 0.495 E c 2 С2=С, emf the current source is equal to E. In a flat air capacitor C1, with the help of an external force, the plates were very quickly moved, reducing the distance between the plates by 2 times. What amount of heat will be released in the circuit in the subsequent transient process? Solution. For the initial state: the first capacitor is increased by 2 times. In this case, for the constant potential difference on the current source, a larger charge is required, therefore, in the subsequent transient process, the charge will flow from the current source, and the current source will perform positive work. 2 s sE) qsE c ok 3 s 2 3 C ok sE 2 C c 1 to 2 (3 AE sE source 2 3 sE 2 W kk A mech 2 q 1 n 2 S on d n 2 2 q 1 n 4 Cc 1 2 2 E s 4 2 sE 4 WА kk kn Task 4 A ist сЕ mech 2 нd 1 2 cЕ 1.5 2 сЕ 2 0.25 cЕ 2 0.25 cЕ 2 1 01 02 0 Solution This problem with nonzero initial conditions and its peculiarity is that when the key K is closed, the total charge of the right plate of the capacitor C1 and the left plate of the capacitor C2 is not zero: (according to the law of conservation of electric charge) for any subsequent transients... Since the circuit is connected to a current source, when the key K is closed, the charges of the capacitors (right plates) will change and will be equal after the transient q1 and q2, and the voltages U1 and U2. These charges and voltages must comply with the law of conservation of charge and the ratio of voltages in series consonant inclusion. We get a system of two equations. If the capacitor C2 were turned on oppositely (by polarity), then the signs of both q2 and U2 would change to the opposite. 1 U U q q 2 1 2 E q 0 q q 1 2 C C 1 q q 1 2 2 E q 0 q C 1 2 (q 1 q C EC C 0 2) 1 1 We find the charges of capacitors. q 1 q 2 EC C q C 1 0 EC C U C U C C 2 02 1 2 EC C q C 2 0 EC C U C U C C 2 01 2 1 1 2 C C 2 1 2 C C 2 1 1 1 2 01 2 1 C C 2 2 01 C C q p , i.e. 0 1 1 2 1 q p or 0 It is clear from the relations that situations are possible when capacitors can be recharged to opposite polarities as a result of a transient process. Work of the current source (for the positive pole): East q 2 1 2 q 2 q 2 q 02 You can show that EC C U C C 1 01 1 2 2 02 C 1 Q Q Q Q Q Q Q Q Q v Q8 2 1 2 2 - U C 2 02 EC C U C C U C C 1 01 1 2 - 02 2 C C 1 1 2 2 2 C U 2 2 For the final state: W k 2 q 2 2 C 2 2 q 1 2 C 1 2 C U about 2 about . Let us determine the value of the released heat at the following numerical values: C1=c, C2=3s, E= 8 V, U01 =4 V, U02 =2 V. q 0 q 1 q 4 8 2 3 2 c c c 3 2 c 11 c c c 3 c 2 c 4 c 3 3 c c c 4 c 14 2 c 3 c q 2 8 c 8 c 3 s 4 s c 2 3 s 15 2 s 3 2 s Wc n W k 2 s 16 2 11 (2 8 1.5 s c) 3 4 s 2 2 s 12 s A source Q W W As source Task 5. 15 s (2 2) 2 3 s 121 s 8 75 s 8 24.5 s 14 s 24.5 s 12 s 1.5 2 1 E U U , therefore charges will not flow either from the source or to the source Solution. 1. Heat is released only when there is a redistribution of charges, i.e. current flows. When the key is opened, this can only happen from a current source. The potential difference between points A and B does not change in this case, since ABU (charges can flow if the potential of the positive pole of the current source is not equal to the potential of t.B, and the potential of the negative pole of the source is not equal to the potential of t.A). This means that the charges of the capacitors will not change, the work of the current source is zero, therefore, heat will not be released when the key is opened. 2. The invariance of the charges of capacitors can be proved using the law of conservation of charge for the midpoint of the circuit. For the initial state: 3 1 C C C 1 3 EC C 3 C C C 3 1 1 2 2 q 23 (C C U U ) 23 23 2 3 q 3 n C U 3 23 the left plate of the capacitor C3 from the midpoint, then its negative charge q3n also leaves with it. Therefore, according to the charge conservation law for the midpoint, we obtain: q 1 q 2 q 3 n 1 EC C 3 C C C 3 1 2 Solving this equation together with the equation for voltages in series connection U U 1 2 E q q 2 1 C C 1 2 E , it is possible to determine q1 and q2 the charges of capacitors that have been established after the transient process. We get q 1 ) EC С С 3 С С С 1 3 (1 2 2 , the value of which is equal to q1н, which means that there will be no redistribution of charges when the key is opened.
2.12.1 Third-party source of electromagnetic field and electric current in the electrical circuit.
☻ Third party source is such integral part electrical circuit, without which the electric current in the circuit is not possible. This divides the electrical circuit into two parts, one of which is capable of conducting current, but does not excite it, and the other “third-party” conducts current and excites it. Under the action of the EMF of a third-party source, not only an electric current is excited in the circuit, but also an electromagnetic field, and both are accompanied by the transfer of energy from the source to the circuit.
2.12.2 EMF source and current source.
☻ A third-party source, depending on its internal resistance, can be a source of EMF 
or current source 
EMF source: 
,
does not depend on 
.
Current source: 
,
does not depend on 
.
Thus, any source that can withstand a stable voltage in the circuit when the current changes in it can be considered as an EMF source. This also applies to sources of stable voltage in electrical networks. Obviously the conditions 
or 
for real third-party sources should be considered as idealized approximations, convenient for the analysis and calculation of electrical circuits. So at 
the interaction of a third-party source with the chain is determined by simple equalities
,
,
.
Electromagnetic field in an electrical circuit.
☻ Third party sources are either energy storage devices or energy generators. The transfer of energy by sources to the circuit occurs only through the electromagnetic field, which is excited by the source in all elements of the circuit, regardless of their technical features and application value, as well as the combination of physical properties in each of them. It is the electromagnetic field that is the primary factor that determines the distribution of the source energy over the circuit elements and determines the physical processes in them, including the electric current.
2.12.4 Resistance in DC and AC circuits.
Fig 2.12.4
Generalized schemes of single-circuit circuits of direct and alternating current.
☻ In simple single-circuit DC and AC circuits, the dependence of current on the EMF of the source can be expressed by similar formulas
,
.
This makes it possible to present the circuits themselves with similar schemes, as shown in Fig. 2.12.4.
It is important to emphasize that in an alternating current circuit the value 
means no active circuit resistance 
, but the impedance of the circuit, which exceeds the active resistance for the reason that the inductive and capacitive elements of the circuit provide additional reactance to the alternating current, so that
,
,
.
Reactances 
And 
determined by the frequency of the alternating current 
, inductance 
inductive elements (coils) and capacitance 
capacitive elements (capacitors).
2.12.5 Phase shift
☻ Circuit elements with reactances cause a special electromagnetic phenomenon in the alternating current circuit - a phase shift between EMF and current
,
,
Where 
- phase shift, the possible values of which are determined by the equation
.
The absence of a phase shift is possible in two cases, when 
or when there are no capacitive and inductive elements in the circuit. The phase shift makes it difficult to output the source power to the electrical circuit.
2.12.6 The energy of the electromagnetic field in the circuit elements.
☻ The energy of the electromagnetic field in each element of the circuit consists of the energy of the electric field and the energy of the magnetic field
.
However, a chain element can be designed in such a way that for it one of the terms of this sum will be dominant, and the other - not essential. So at characteristic frequencies of alternating current in the capacitor 
, and in the coil, on the contrary, 
. Therefore, we can assume that the capacitor is the energy storage of the electric field, and the coil is the energy storage of the magnetic field and for them, respectively
,
,
where it is taken into account that for the capacitor 
, and for the coil 
. Two coils in one circuit can be inductively independent or inductively coupled through their common magnetic field. In the latter case, the energy of the magnetic fields of the coils is supplemented by the energy of their magnetic interaction
,
,
.
Mutual induction coefficient 
depends on the degree of inductive coupling between the coils, in particular on their relative position. Inductive coupling may be insignificant or absent completely, then 
.
A characteristic element of an electrical circuit is a resistor with a resistance 
. For him, the energy of the electromagnetic field 
, because 
. Since the energy of the electric field in the resistor 
experiences an irreversible conversion into thermal energy, then for the resistor
,
where is the amount of heat 
corresponds to the Joule-Lenz law.
A special element of an electrical circuit is its electromechanical element, capable of performing mechanical work when an electric current passes through it. An electric current in such an element excites a force or moment of force, under the action of which linear or angular displacements of the element itself or its parts relative to each other occur. These mechanical phenomena associated with electric current are accompanied by the transformation of the energy of the electromagnetic field in the element into its mechanical energy, so that
where is the work 
expressed according to its mechanical definition.
2.12.7 The law of conservation and transformation of energy in an electrical circuit.
☻ A third-party source is not only a source of EMF, but also a source of energy in an electrical circuit. During 
from the source, energy enters the circuit, equal to the work of the EMF of the source
Where 
- the power of the source, or, what is also, the intensity of energy supply from the source to the circuit. The source energy is converted into circuits into other types of energy. So in a single circuit 
with a mechanical element, the operation of the source is accompanied by a change in the energy of the electromagnetic field in all elements of the circuit in full accordance with the energy balance
This equation for the circuit under consideration expresses the laws of conservation of energy. It follows from it
.
After appropriate substitutions, the power balance equation can be represented as
.
This equation in a generalized form expresses the law of conservation of energy in an electrical circuit based on the concept of power.
Law
Kirchhoff
☻ After differentiation and reduction of the current, Kirchhoff's law follows from the presented law of conservation of energy
where in a closed circuit the listed voltages on the circuit elements mean
,
,
,
,
.
2.12.9 Application of the law of conservation of energy for the calculation of the electrical circuit.
☻ The given equations of the law of conservation of energy and Kirchhoff's law apply only to quasi-stationary currents, in which the circuit is not a source of electromagnetic field radiation. The equation of the law of conservation of energy allows in a simple and visual form to analyze the operation of numerous single-circuit electrical circuits, both AC and DC.
Setting constants 
zero individually or in combination, you can calculate different options for electrical circuits, including when 
And 
. Some options for calculating such circuits are discussed below.
2.12.10 Chain 
at 
☻ A single-circuit circuit in which through a resistor 
the capacitor is charged from a source with a constant emf ( 
). Accepted: 
,
,
, and 
at 
. Under such conditions, the law of conservation of energy for a given circuit can be written in the following equivalent versions
,
,
.
From the solution of the last equation follows:
,
.
2.12.11 Chain 
at 
☻ A single-circuit circuit in which a source of constant EMF ( 
) is closed to the elements 
And 
. Accepted: 
,
,
, and 
at 
. Under such conditions, the law of conservation of energy for a given circuit can be represented in the following equivalent versions
,
,
.
From the solution of the last equation it follows
.
2.12.12 Chain 
at 
And 
☻ A single-circuit circuit without an EMF source and without a resistor, in which a charged capacitor 
closes on an inductive element 
. Accepted: 
,
,
,
,
, as well as at 
And 
. Under such conditions, the law of conservation of energy for a given circuit, taking into account the fact that 
,
,
.
The last equation corresponds to free undamped oscillations. It follows from his decision
,
,
,
,
.
This circuit is an oscillatory circuit.
2.12.13 ChainRLCat
☻ A single-circuit circuit without an EMF source in which a charged capacitor WITH closes on the circuit elements R and L. Accepted: 
,
, as well as at 
And 
. Under such conditions, the law of conservation of energy for a given circuit is lawful, taking into account the fact that 
, can be written as follows
,
,
.
The last equation corresponds to free damped oscillations. It follows from his decision
,
,
,
,
.
This circuit is an oscillatory circuit with a dissipative element - a resistor, due to which the total energy of the electromagnetic field decreases during oscillations.
2.12.14 ChainRLCat 
☻ Single circuit RCL is an oscillatory circuit with a dissipative element. A variable emf acts in the circuit
and excites forced oscillations in it, including resonance.
Accepted: 
. Under these conditions, the energy conservation law can be written in several equivalent versions.
,
,
,
From the solution of the last equation it follows that current oscillations in the circuit are forced and occur with the frequency of the effective EMF
, but with a phase shift with respect to it, so that
,
Where 
is the phase shift, the value of which is determined by the equation
.
The power supplied to the circuit from the source is variable
The average value of this power over one period of oscillation is determined by the expression
.
Fig 2.12.14
Addiction Resonance
Thus, the power output from the source to the circuit is determined by the phase shift. Obviously, in its absence, the indicated power becomes maximum and this corresponds to resonance in the circuit. It is achieved because the resistance of the circuit in the absence of a phase shift takes on a minimum value equal only to the active resistance.
.
It follows from this that the conditions are satisfied at resonance.
,
,
,
Where 
is the resonant frequency.
With forced oscillations of the current, its amplitude depends on the frequency
.
The resonant value of the amplitude is achieved in the absence of a phase shift, when 
And 
. Then
,
On fig. 2.12.14 shows the resonance curve 
with forced oscillations in the RLC circuit.
2.12.15 Mechanical energy in electrical circuits
☻ Mechanical energy is excited by special electromechanical circuit elements, which, when an electric current passes through them, perform mechanical work. These can be electric motors, electromagnetic vibrators, etc. Electric current in these elements excites forces or moments of forces, under the action of which linear, angular or oscillatory movements occur, while the electromechanical element becomes a carrier of mechanical energy
Options for the technical implementation of electromechanical elements are almost limitless. But either way, the same thing happens. physical phenomenon– conversion of electromagnetic field energy into mechanical energy
.
It is important to emphasize that this transformation takes place under the conditions of an electric circuit and with the unconditional fulfillment of the law of conservation of energy. It should be noted that the electromechanical element of the circuit, for any purpose and technical design, is an energy storage of the electromagnetic field 
. It accumulates on the internal capacitive or inductive parts of the electromechanical element, between which mechanical interaction is initiated. In this case, the mechanical power of the electromechanical element of the circuit is not determined by the energy 
, and the time derivative of it, i.e. the intensity of its change R within the element itself
.
Thus, in the case of a simple circuit, when a third-party EMF source is closed only to an electromechanical element, the law of conservation of energy is represented in the form
,
,
where the inevitable irreversible thermal power losses of the third-party source are taken into account. In the case of a more complex circuit, in which there are additional energy storage devices of the electromagnetic field W , the energy conservation law is written as
.
Given that 
And 
, the last equation can be written as
.
In a simple chain 
and then
.
A more rigorous approach requires taking into account friction processes, which further reduce the useful mechanical power of the electromechanical circuit element.
By special agreement with the editorial board and the editors of the journal "Kvant"
The law of conservation of energy determines in the most general form the energy balance for all kinds of changes in any system. Let's write it like this:
Where A ex - work done on the system under consideration by external forces, Δ W- change in the energy of the system, Q- the amount of heat released in the system. Let's agree that if A ext > 0, then positive work is done on the system, and if A external< 0, положительную работу совершает система; если ΔW> 0, then the energy of the system increases, and if Δ W < 0, энергия уменьшается; наконец, если Q> 0, then heat is released in the system, and if Q < 0, тепло системой поглощается.
In this article, we will look at how the law of conservation of energy "works" in electrostatics. In general, an electrostatic system contains interacting charges in an electric field.
Let us consider each term in equation (1) separately.
Let's start with energy. The energy of interaction of charges is expressed through the characteristics of the electric field of this system of charges. So, for example, the energy of a charged capacitor with a capacity C given famous expression
(2)
Where q- lining charge, U- tension between them. Recall that a capacitor is a system of two conductors (plates, plates), which has the following property: if a charge is transferred from one plate to another q(i.e., charge one plate with a charge + q, and the other q), then all field lines of the field created in this way will begin on one (positively charged) plate and end on the other. The field of a capacitor exists only inside it.
The energy of a charged capacitor can also be represented as the energy of a field localized in the space between the plates with an energy density where E- field strength. In essence, it is this fact that gives grounds to speak of the field as an object that really exists - this object has an energy density. But we must remember that this is just an equivalent way of determining the energy of the interaction of charges (which we now call the energy of the electric field). Thus, we can calculate the energy of the capacitor both by formulas (2) and by the formula
(3)
Where V- volume of the condenser. The last formula is easy to use, of course, only in the case of a homogeneous field, but the representation of energy in this form is very clear, and therefore convenient.
Of course, in addition to the energy of interaction of charges (the energy of the electric field), the energy of the system can include both the kinetic energy of charged bodies and their potential energy in the field of gravity, and the energy of springs attached to bodies, etc.
Now about work external forces. In addition to normal mechanical work A fur (for example, by pushing the capacitor plates apart), for an electrical system, we can talk about the work of an external electric field. For example, about the operation of a battery charging or recharging a capacitor. The task of the battery is to create a fixed potential difference inherent in this source between the bodies to which it is attached. She does this in the only possible way - she takes the charge from one body and transfers it to another. The source never creates charges, only moves them. The total charge of the system is conserved in this case - this is one of the cornerstone laws of nature.
In sources of different designs, the electric field necessary to move charges is created by various "mechanisms". In batteries and accumulators - these are electrochemical reactions, in dynamos - electromagnetic induction. It is essential that for the chosen system of charges (charged bodies) this field is external, external. When through a source with EMF charge Δ flows from the negative pole to the positive pole q, external forces do work
Moreover, if Δ q> 0, then A baht > 0 - the battery is running low; if Δ q < 0, то A baht< 0 - the battery is charged and chemical (or magnetic) energy accumulates in it.
Finally, about the release of heat. We only note that this is Joule heat, i.e. heat associated with the flow of current through a resistance.
Now let's discuss some specific problems.
Task 1. Two identical flat plate capacitors C each attached to two identical batteries with emf. At some point, one capacitor is disconnected from the battery while the other is left connected. Then slowly separate the plates of both capacitors, reducing the capacitance of each in n once. What mechanical work is done in each case?
If the process of changing the charge on the capacitor is carried out slowly all the time, no heat will be released. Indeed, if through a resistor with a resistance R charge leaked Δ q during t, then the amount of heat released on the resistor during this time
For sufficiently large t quantity of heat Q can be arbitrarily small.
In the first case, the charge on the plates is fixed (the battery is off), equal to Mechanical work is determined by the change in the energy of the capacitor:
In the second case, the potential difference across the capacitor is fixed and the battery is working, therefore
Charge flows through the battery
This charge is less than zero, which means that the battery is charging and its work
The field energy in the capacitor decreases:
Thus,
The battery is charged by the work of pushing the plates apart and by the energy of the capacitor.
Note that the words about pushing the plates apart do not play a significant role. The same result will be with any other changes leading to a decrease in capacitance in n once.
Task 2. In the circuit shown in the figure, find the amount of heat released in each resistor after the key is closed. Capacitor C 1 charged to voltage U 1, and a capacitor with a capacity C 2 - up to voltage U 2. Resistors R 1 and R 2 .
The energy conservation law (1) for this system has the form
The initial energy of the capacitors is
To determine the energy in the final state, we use the fact that the total charge of the capacitors cannot change. He is equal 
(for cases when the capacitors were connected by like or oppositely charged plates, respectively). After the key is closed, this charge charges a capacitor with a capacity C 1 + C 2 (capacitors with capacities C 1 and C 2 connected in parallel). Thus,
And
As it should be, in both cases heat is released - there are Joule losses. It is remarkable that the released amount of heat does not depend on the resistance of the circuit - at low resistances, large currents flow and vice versa.
Now let's find how the amount of heat Q distributed among the resistors. Through resistance R 1 and R 2 at each moment of the recharging process, the same currents flow, which means that at each moment the powers released on the resistances are equal
And 
Hence,
Besides, . Therefore, finally
Task 3. In the circuit in Figure 2, a capacitor with a capacity C charged to voltage U. How much chemical energy will be stored in the battery with EMF after the key is closed? How much heat will be released in the resistor?
The initial charge on the capacitor . After the recharge is completed, the charge on the capacitor will become equal to . The charge flowing through the battery in the case when a negatively charged capacitor plate is connected to the minus of the battery will be equal to
Otherwise, and at the same time, the battery will be discharged (Δ q> 0). And in the first case, the battery is being charged (Δ q < 0), и количество химической энергии, запасенной в аккумуляторе после замыкания ключа, равно работе батареи:
Now we write the law of conservation of energy (1) -
- and find the released amount of heat:
Task 4. A flat capacitor is placed in an external uniform field with a strength perpendicular to the plates. On plates with an area S charges are distributed + q And - q. Distance between plates d. What is the minimum work that must be done to swap the plates? Position parallel to the field? Take it out of the field?
Work will be minimal when the process is carried out very slowly - no heat is generated. Then, according to the law of conservation of energy,
To find Δ W, we use formula (3). The field between the plates is a superposition of the field this flat capacitor -
– and external field .
When changing the plates, the field changes to -, and the field outside does not change, i.e., the change in the energy of the system is associated with a change in its density between the capacitor plates:
If the directions of the vectors and were the same, then the energy density between the plates decreased after changing the places of the plates, and if the directions were opposite, then the energy density increased. Thus, in the first case, the capacitor wants to turn around on its own and must be held ( A < 0), а во втором случае
When the capacitor plates are parallel to the field and perpendicular to each other. The field energy inside the capacitor in this case is equal to 
. Then
When the capacitor was taken out of the field, in the place where it was, the field became, and in itself now the field, i.e. Δ W And A min are the same as in the previous case.
Task 5. Capacitor with a capacity WITH without a dielectric charged with a charge q. How much heat will be released in the capacitor if it is filled with a substance with a permittivity ε? The same, but the capacitor is connected to a battery with an emf.
When pouring the dielectric, the capacitance of the capacitor increased by ε times.
In the first case, the charge on the plates is fixed, there are no external forces, and the energy conservation law (1) has the form
Heat is released due to a decrease in the energy of interaction of charges.
In the second case, there is battery operation and the voltage across the capacitor is fixed:
Exercises
1. Two identical flat capacitors with a capacity WITH each connected in parallel and charged to a voltage U. The plates of one of the capacitors are slowly parted over a long distance. What is the work being done?
2. Two capacitors, each with a capacity WITH, charged to voltage U and connected through a resistor (Fig. 4). The plates of one of the capacitors are quickly moved apart, so that the distance between them doubles, and the charge on the plates does not change during their movement. How much heat will be released in the resistor?
3. A flat plate air capacitor is connected to an EMF battery. Plate area S, the distance between them d. The condenser contains a metal plate with a thickness d 1 parallel to the plates (Fig. 5). What is the minimum work required to remove the plate from the condenser?
4. Large thin conductive plate area S and thickness d placed in a uniform electric field with a strength perpendicular to the surface of the plate. What amount of heat will be released in the plate if the field is turned off instantly? What is the minimum work required to remove the plate from the field?
5. One of the plates of a flat capacitor is suspended on a spring (Fig. 6). Area of each plate S, the distance between them at the initial moment d. Capacitor on a short time connected to the battery, and it charged up to voltage U. What should be the minimum stiffness of the spring so that the plates do not touch? Ignore the displacement of the plates during charging.
Answers.
1. (the entire charge is on the capacitor, the plates of which were not moved apart).
2. (at the first moment after the separation of the plates, a capacitor with a capacity of WITH with tension U and a capacitor WITH/2 with voltage 2 U).
3. 
(the minimum work to remove the plate is equal to the difference between the change in the energy of the capacitor and the work of the battery).
4. 
(immediately after the external field is turned off, there is a field of polarization charges in the plate, the strength of which is equal to E; removing the plate from the field is equivalent to creating a field with strength E in the volume of the plate).
5. (the result is obtained from the law of conservation of energy 
and from the equilibrium condition of the plate ).
