Equation of current in an oscillatory circuit. Oscillatory circuit. Free electromagnetic oscillations. Conversion of energy in an oscillatory circuit. Thompson's formula
Electrical oscillations mean periodic changes in charge, current and voltage. The simplest system in which free electrical oscillations are possible is the so-called oscillatory circuit. This is a device consisting of a capacitor and a coil connected to each other. We will assume that there is no active resistance of the coil, in which case the circuit is called ideal. When energy is imparted to this system, undamped harmonic vibrations charge on the capacitor, voltage and current.
You can impart energy to the oscillatory circuit different ways. For example, by charging a capacitor from a direct current source or exciting a current in an inductor. In the first case, the energy is possessed by the electric field between the plates of the capacitor. In the second, the energy is contained in the magnetic field of the current flowing through the circuit.
§1 Equation of oscillations in a circuit
Let us prove that when energy is imparted to the circuit, undamped harmonic oscillations will occur in it. To do this you need to get differential equation harmonic vibrations of the form.
Let's say the capacitor is charged and shorted to the coil. The capacitor will begin to discharge and current will flow through the coil. According to Kirchhoff's second law, the sum of the voltage drops along a closed circuit is equal to the sum of the emf in this circuit 
.
In our case, the voltage drop is because the circuit is ideal. The capacitor in the circuit behaves as a current source; the potential difference between the plates of the capacitor acts as the EMF, where is the charge on the capacitor and is the electrical capacitance of the capacitor. In addition, when a changing current flows through the coil, a Self-induced emf, where is the inductance of the coil, is the rate of change of current in the coil. Since the self-induction emf prevents the process of discharging the capacitor, Kirchhoff’s second law takes the form
But the current in the circuit is the discharge or charging current of the capacitor, therefore. Then
The differential equation is transformed to the form
By introducing the notation, we obtain the well-known differential equation of harmonic oscillations.
This means that the charge on the capacitor in the oscillating circuit will change according to the harmonic law
where is the maximum charge value on the capacitor, is the cyclic frequency, is the initial phase of oscillations.
Charge oscillation period 
. This expression is called the Thompson formula.
Capacitor voltage
Circuit current
We see that in addition to the charge on the capacitor, according to the harmonic law, the current in the circuit and the voltage on the capacitor will also change. The voltage oscillates in phase with the charge, and the current strength leads the charge in
phase on .
Energy electric field capacitor
Energy magnetic field current
Thus, the energies of the electric and magnetic fields also change according to the harmonic law, but with double the frequency.
Summarize
Electrical oscillations should be understood as periodic changes in charge, voltage, current, electric field energy, and magnetic field energy. These vibrations, like mechanical ones, can be either free or forced, harmonic and non-harmonic. Free harmonic electrical oscillations are possible in an ideal oscillatory circuit.
§2 Processes occurring in an oscillatory circuit
We have mathematically proven the existence of free harmonic oscillations in an oscillatory circuit. However, it remains unclear why such a process is possible. What causes oscillations in the circuit?
In the case of free mechanical vibrations, such a reason was found - this is the internal force that arises when the system is removed from the equilibrium position. This force at any moment is directed towards the equilibrium position and is proportional to the coordinate of the body (with a minus sign). Let's try to find a similar reason for the occurrence of oscillations in the oscillatory circuit.
Let the oscillations in the circuit excite by charging the capacitor and shorting it to the coil.
At the initial moment of time, the charge on the capacitor is maximum. Consequently, the voltage and energy of the electric field of the capacitor are also maximum.
There is no current in the circuit, the energy of the magnetic field of the current is zero.
First quarter of the period– capacitor discharge.
The plates of the capacitor, having different potentials, are connected by a conductor, so the capacitor begins to discharge through the coil. The charge, voltage on the capacitor and the energy of the electric field decrease.
The current that appears in the circuit increases, however, its increase is prevented by the self-induction emf that occurs in the coil. The energy of the magnetic field of the current increases.
A quarter of the period has passed- the capacitor is discharged.
The capacitor was discharged, the voltage on it became equal to zero. The energy of the electric field at this moment is also zero. According to the law of conservation of energy, it could not disappear. The energy of the capacitor field is completely converted into the energy of the magnetic field of the coil, which at this moment reaches its maximum value. Maximum current in the circuit.
It would seem that at this moment the current in the circuit should stop, because the cause of the current – the electric field – has disappeared. However, the disappearance of the current is again prevented by the self-induction emf in the coil. Now it will support the decreasing current, and it will continue to flow in the same direction, charging the capacitor. The second quarter of the period begins.
Second quarter of the period
– recharging the capacitor.
The current, supported by the self-induction emf, continues to flow in the same direction, gradually decreasing. This current charges the capacitor in opposite polarity. The charge and voltage on the capacitor increase.
The energy of the magnetic field of the current, decreasing, turns into the energy of the electric field of the capacitor.
The second quarter of the period has passed - the capacitor has recharged.
The capacitor recharges as long as current exists. Therefore, at the moment when the current stops, the charge and voltage on the capacitor take on the maximum value.
The energy of the magnetic field at this moment was completely converted into the energy of the electric field of the capacitor.
The situation in the circuit at this moment is equivalent to the original one. The processes in the circuit will repeat, but in the opposite direction. One complete oscillation in the circuit, lasting for a period, will end when the system returns to its original state, that is, when the capacitor is recharged in its original polarity.
It is easy to see that the cause of oscillations in the circuit is the phenomenon of self-induction. The self-induction EMF prevents the current from changing: it prevents it from instantly increasing and instantly disappearing.
By the way, it would not be amiss to compare the expressions for calculating the quasi-elastic force in a mechanical oscillatory system and the self-induction emf in the circuit:
Previously, differential equations were obtained for mechanical and electrical oscillatory systems:
Despite the fundamental differences in the physical processes of mechanical and electrical oscillatory systems, the mathematical identity of the equations describing the processes in these systems is clearly visible. We should talk about this in more detail.
§3 Analogy between electrical and mechanical vibrations
A careful analysis of differential equations for a spring pendulum and an oscillatory circuit, as well as formulas connecting quantities characterizing processes in these systems, allows us to identify which quantities behave the same (Table 2).
| Spring pendulum | Oscillatory circuit |
| Body Coordinate() | Charge on capacitor () |
| Body speed | Current strength in the circuit |
| Potential energy of an elastically deformed spring | Electric field energy of a capacitor |
| Kinetic energy of cargo | Magnetic field energy of a current coil |
| The reciprocal of the spring stiffness | Capacitor capacity |
| Cargo weight | Coil inductance |
Elastic force  | Self-induction emf equal to the voltage across the capacitor  |
table 2
What is important is not just the formal similarity between the quantities describing the processes of oscillation of the pendulum and the processes in the circuit. The processes themselves are identical!
The extreme positions of the pendulum are equivalent to the state of the circuit when the charge on the capacitor is maximum.
The equilibrium position of the pendulum is equivalent to the state of the circuit when the capacitor is discharged. At this moment, the elastic force becomes zero, and there is no voltage on the capacitor in the circuit. The speed of the pendulum and the current in the circuit are maximum. The potential energy of elastic deformation of the spring and the energy of the electric field of the capacitor are equal to zero. The energy of the system consists of the kinetic energy of the load or the energy of the magnetic field of the current.
The discharge of a capacitor proceeds similarly to the movement of a pendulum from its extreme position to its equilibrium position. The process of recharging the capacitor is identical to the process of removing the load from the equilibrium position to the extreme position.
Total energy of the oscillatory system 
or 
remains unchanged over time.
A similar analogy can be traced not only between a spring pendulum and an oscillatory circuit. Universal laws of free vibrations of any nature! These patterns, illustrated by the example of two oscillatory systems (a spring pendulum and an oscillatory circuit), are not only possible, but must see in the oscillations of any system.
In principle, it is possible to solve the problem of any oscillatory process by replacing it with pendulum oscillations. To do this, it is enough to competently construct an equivalent mechanical system, solve a mechanical problem and replace quantities in the final result. For example, you need to find the period of oscillation in a circuit containing a capacitor and two coils connected in parallel.
The oscillating circuit contains one capacitor and two coils. Since the coil behaves like the weight of a spring pendulum, and the capacitor like a spring, the equivalent mechanical system must contain one spring and two weights. The problem is how the weights are attached to the spring. Two cases are possible: one end of the spring is fixed, and one weight is attached to the free end, the second is on the first, or the weights are attached to different ends springs.
When coils of different inductances are connected in parallel, different currents flow through them. Consequently, the speeds of loads in an identical mechanical system must also be different. Obviously, this is only possible in the second case.
We have already found the period of this oscillatory system. It is equal 
. Replacing the masses of the loads with the inductance of the coils, and the reciprocal of the spring stiffness with the capacitance of the capacitor, we obtain 
.
§4 Oscillating circuit with a direct current source
Consider an oscillatory circuit containing a direct current source. Let the capacitor be initially uncharged. What will happen in the system after key K is closed? Will oscillations be observed in this case and what is their frequency and amplitude?
Obviously, after closing the key, the capacitor will begin to charge. We write down Kirchhoff's second law:
The current in the circuit is the charging current of the capacitor, therefore. Then . The differential equation is transformed to the form
*We solve the equation by changing variables.
Let's denote . We differentiate twice and, taking into account the fact that , we obtain . The differential equation takes the form
This is a differential equation of harmonic oscillations, its solution is the function
where is the cyclic frequency, integration constants and are found from the initial conditions.
The charge on the capacitor changes according to the law
Immediately after the key is closed, the charge on the capacitor is zero and there is no current in the circuit 
. Taking into account the initial conditions, we obtain a system of equations:
Solving the system, we obtain and . After the key is closed, the charge on the capacitor changes according to the law.
It is easy to see that harmonic oscillations occur in the circuit. The presence of a direct current source in the circuit did not affect the oscillation frequency; it remained equal. The “equilibrium position” has changed - at the moment when the current in the circuit is maximum, the capacitor is charged. The amplitude of charge oscillations on the capacitor is equal to Cε.
The same result can be obtained more simply by using an analogy between oscillations in a circuit and oscillations of a spring pendulum. DC source is equivalent to DC force field, in which a spring pendulum is placed, for example, a gravitational field. The absence of charge on the capacitor at the moment the circuit is closed is identical to the absence of spring deformation at the moment the pendulum is brought into oscillatory motion.
In a constant force field, the period of oscillation of a spring pendulum does not change. The oscillation period in the circuit behaves the same way - it remains unchanged when a direct current source is introduced into the circuit.
In the equilibrium position, when the speed of the load is maximum, the spring is deformed:
When the current in the oscillating circuit is maximum 
. Kirchhoff's second law will be written as follows
At this moment, the charge on the capacitor is equal to The same result could be obtained based on expression (*) by making the replacement
§5 Examples of problem solving
Problem 1 Law of energy conservation
L= 0.5 µH and a capacitor with a capacity WITH= 20 pF electrical oscillations occur. What is the maximum voltage across the capacitor if the current amplitude in the circuit is 1 mA? The active resistance of the coil is negligible.
Solution:
(1)
2 At the moment when the voltage on the capacitor is maximum (maximum charge on the capacitor), there is no current in the circuit. The total energy of the system consists only of the energy of the electric field of the capacitor
(2)
3 At the moment when the current in the circuit is maximum, the capacitor is completely discharged. The total energy of the system consists only of the energy of the magnetic field of the coil
(3)
4 Based on expressions (1), (2), (3) we obtain the equality 
. The maximum voltage across the capacitor is
Problem 2 Law of energy conservation
In an oscillating circuit consisting of an inductive coil L and a capacitor with a capacity WITH, electrical oscillations occur with a period T = 1 μs. Maximum charge value 
. What is the current in the circuit at the moment when the charge on the capacitor is equal to ? The active resistance of the coil is negligible.
Solution:
1 Since the active resistance of the coil can be neglected, the total energy of the system, consisting of the energy of the electric field of the capacitor and the energy of the magnetic field of the coil, remains unchanged over time:
(1)
2 At the moment when the charge on the capacitor is maximum, there is no current in the circuit. The total energy of the system consists only of the energy of the electric field of the capacitor
(2)
3 Based on (1) and (2) we obtain the equality 
. The current in the circuit is 
.
4 The period of oscillation in the circuit is determined by Thomson's formula. From here. Then for the current in the circuit we obtain
Problem 3 Oscillatory circuit with two parallel connected capacitors
In an oscillating circuit consisting of an inductive coil L and a capacitor with a capacity WITH, electrical oscillations occur with charge amplitude . At the moment when the charge on the capacitor is maximum, switch K is closed. What will be the period of oscillation in the circuit after closing the key? What is the amplitude of the current in the circuit after the switch is closed? Neglect the ohmic resistance of the circuit.
Solution:
1 Closing the key leads to the appearance of another capacitor in the circuit, connected in parallel to the first. The total capacitance of two parallel-connected capacitors is equal to .
The period of oscillations in the circuit depends only on its parameters and does not depend on how the oscillations were excited in the system and what energy was imparted to the system for this. According to Thomson's formula.
2 To find the current amplitude, let’s find out what processes occur in the circuit after the switch is closed.
The second capacitor was connected at the moment when the charge on the first capacitor was maximum, therefore, there was no current in the circuit.
The loop capacitor should begin to discharge. The discharge current, having reached the node, should be divided into two parts. However, in the branch with the coil, a self-induction EMF arises, which prevents the discharge current from increasing. For this reason, the entire discharge current will flow into the branch with the capacitor, the ohmic resistance of which is zero. The current will stop as soon as the voltages on the capacitors are equal, and the initial charge on the capacitor will be redistributed between the two capacitors. The time of charge redistribution between two capacitors is negligible due to the absence of ohmic resistance in the branches with capacitors. During this time, the current in the branch with the coil will not have time to arise. Fluctuations in new system will continue after the redistribution of charge between the capacitors.
It is important to understand that in the process of redistributing charge between two capacitors, the energy of the system is not conserved! Before the key was closed, one capacitor, a circuit one, had energy:
After charge redistribution, the capacitor bank has energy:
It is easy to see that the energy of the system has decreased!
3 We find the new current amplitude using the law of conservation of energy. During the oscillation process, the energy of the capacitor bank is converted into the energy of the magnetic field of the current:
Please note that the law of conservation of energy begins to “work” only after the redistribution of charge between the capacitors is completed.
Problem 4 Oscillatory circuit with two capacitors connected in series
The oscillatory circuit consists of a coil of inductance L and two series-connected capacitors C and 4C. A capacitor of capacity C is charged to voltage, a capacitor of capacity 4C is not charged. After the key is closed, oscillations begin in the circuit. What is the period of these oscillations? Determine the current amplitude, maximum and minimum voltage values on each capacitor.
Solution:
1 At the moment when the current in the circuit is maximum, there is no self-inductive emf in the coil 
. We write down Kirchhoff’s second law for this moment
We see that at the moment when the current in the circuit is maximum, the capacitors are charged to the same voltage, but in the opposite polarity:
2 Before closing the switch, the total energy of the system consisted only of the energy of the electric field of capacitor C:
At the moment when the current in the circuit is maximum, the energy of the system is the sum of the energy of the magnetic field of the current and the energy of two capacitors charged to the same voltage:
According to the law of conservation of energy
To find the voltage on the capacitors, we will use the law of conservation of charge - the charge of the lower plate of capacitor C is partially transferred to the upper plate of capacitor 4C:
We substitute the found voltage value into the law of conservation of energy and find the amplitude of the current in the circuit:
3 Let us find the limits within which the voltage on the capacitors changes during oscillations.
It is clear that at the moment the circuit was closed, there was a maximum voltage on capacitor C. Capacitor 4C was not charged, therefore.
After the key is closed, capacitor C begins to discharge, and capacitor with capacity 4C begins to charge. The process of discharging the first and charging the second capacitor ends as soon as the current in the circuit stops. This will happen after half the period. According to the laws of conservation of energy and electric charge:
Solving the system, we find:
.
The minus sign means that after half a cycle the capacitor C is charged in the opposite polarity to the original one.
Problem 5 Oscillatory circuit with two coils connected in series
The oscillatory circuit consists of a capacitor with capacitance C and two inductance coils L 1 And L 2. At the moment when the current in the circuit has reached its maximum value, an iron core is quickly introduced into the first coil (compared to the oscillation period), which leads to an increase in its inductance by μ times. What is the voltage amplitude during further oscillations in the circuit?
Solution:
1 When the core is quickly inserted into the coil, the magnetic flux must be maintained (the phenomenon electromagnetic induction). Therefore, a rapid change in the inductance of one of the coils will lead to rapid change current in the circuit.
2 During the time the core was introduced into the coil, the charge on the capacitor did not have time to change; it remained uncharged (the core was introduced at the moment when the current in the circuit was maximum). After a quarter of the period, the energy of the magnetic field of the current will transform into the energy of a charged capacitor:
We substitute the current value into the resulting expression I and find the voltage amplitude on the capacitor:
Problem 6 Oscillatory circuit with two parallel-connected coils
Inductors L 1 and L 2 are connected through switches K1 and K2 to a capacitor with capacitance C. At the initial moment, both switches are open, and the capacitor is charged to a potential difference. First, switch K1 is closed and, when the voltage on the capacitor becomes zero, K2 is closed. Determine the maximum voltage on the capacitor after closing K2. Neglect the coil resistances.
Solution:
1 When switch K2 is open, oscillations occur in the circuit consisting of a capacitor and the first coil. By the time K2 closes, the energy of the capacitor has transferred into the energy of the magnetic field of the current in the first coil:
2 After closing K2, there are two coils connected in parallel in the oscillating circuit.
The current in the first coil cannot stop due to the phenomenon of self-induction. At the node it is divided: one part of the current goes to the second coil, and the other charges the capacitor.
3 The voltage across the capacitor will be maximum when the current stops I, charging capacitor. Obviously, at this moment the currents in the coils will be equal.
:The pendulum moves forward at the speed of the center of mass 
and oscillates relative to the center of mass.
The elastic force is maximum at the moment of maximum deformation of the spring. Obviously, at this moment the relative speed of the loads becomes zero, and relative to the table the weights move at the speed of the center of mass. We write down the law of conservation of energy:
Solving the system, we find
We make a replacement
and we obtain the previously found value for the maximum voltage 
§6 Tasks for independent solution
Exercise 1 Calculation of the period and frequency of natural oscillations
1 The oscillatory circuit includes a variable inductance coil that varies within L 1= 0.5 µH to L 2= 10 µH, and a capacitor whose capacitance can vary from C 1= 10 pF to
C 2=500 pF. What frequency range can be covered by tuning this circuit?
2 How many times will the frequency of natural oscillations in the circuit change if its inductance is increased by 10 times and its capacitance is reduced by 2.5 times?
3 An oscillating circuit with a 1 µF capacitor is tuned to a frequency of 400 Hz. If you connect a second capacitor in parallel to it, then the oscillation frequency in the circuit becomes equal to 200 Hz. Determine the capacitance of the second capacitor.
4 The oscillating circuit consists of a coil and a capacitor. How many times will the frequency of natural oscillations in the circuit change if a second capacitor is connected in series to the circuit, the capacitance of which is 3 times less than the capacitance of the first?
5 Determine the period of oscillation of the circuit, which includes a coil (without a core) of length V= 50 cm m cross-sectional area
S= 3 cm 2, having N= 1000 turns, and capacitor capacity WITH= 0.5 µF.
6 The oscillatory circuit includes an inductor L= 1.0 µH and an air capacitor whose plate area S= 100 cm 2. The circuit is tuned to a frequency of 30 MHz. Determine the distance between the plates. The active resistance of the circuit is negligible.
Charge the capacitor from the battery and connect it to the coil. In the contour we created, the electromagnetic vibrations(Fig. 46). The discharge current of the capacitor, passing through the coil, creates a magnetic field around it. This means that during the discharge of a capacitor, the energy of its electric field transforms into the energy of the magnetic field of the coil, just as when a pendulum or string oscillates, potential energy transforms into kinetic energy.
As the capacitor discharges, the voltage across its plates drops and the current in the circuit increases, and by the time the capacitor is completely discharged, the current will be maximum (current amplitude). But even after the end of the capacitor discharge, the current will not stop - the decreasing magnetic field of the coil will maintain the movement of charges, and they will again begin to accumulate on the capacitor plates. In this case, the current in the circuit decreases, and the voltage across the capacitor increases. This process of the reverse transition of the energy of the magnetic field of the coil into the energy of the electric field of the capacitor is somewhat reminiscent of what happens when the pendulum, having passed the midpoint, rises upward.
By the time the current in the circuit stops and the magnetic field of the coil disappears, the capacitor will be charged to the maximum (amplitude) voltage of reverse polarity. The latter means that on the plate where there were previously positive charges, there will now be negative ones, and vice versa. Therefore, when the discharge of the capacitor begins again (and this will happen immediately after it is fully charged), a current in the opposite direction will flow in the circuit.
The periodically repeated exchange of energy between the capacitor and the coil represents electromagnetic oscillations in the circuit. During these oscillations, an alternating current flows in the circuit (that is, not only the magnitude, but also the direction of the current changes), and an alternating voltage acts on the capacitor (that is, not only the voltage magnitude changes, but also the polarity of the charges accumulating on the plates). One of the directions of current voltage is conventionally called positive, and the opposite direction is called negative.
By observing changes in voltage or current, you can build a graph of electromagnetic oscillations in the circuit (Fig. 46), just as we built a graph of mechanical oscillations of a pendulum (). On a graph, positive current or voltage values are plotted above the horizontal axis, and negative currents or voltages are plotted below this axis. That half of the period when the current flows in the positive direction is often called the positive half-cycle of the current, and the other half - the negative half-cycle of the current. We can also talk about positive and negative half-cycles of voltage.
I would like to emphasize once again that we use the words “positive” and “negative” completely conditionally, only to distinguish two opposite directions of current.
The electromagnetic oscillations we have become familiar with are called free or natural oscillations. They occur whenever we transfer a certain amount of energy to the circuit, and then allow the capacitor and coil to freely exchange this energy. The frequency of free oscillation (that is, the frequency of the alternating voltage and current in the circuit) depends on how quickly the capacitor and coil can store and release energy. This, in turn, depends on the inductance Lk and capacitance Ck of the circuit, just as the frequency of vibration of a string depends on its mass and elasticity. The greater the inductance L of the coil, the more time it takes to create a magnetic field in it, and the longer this magnetic field can maintain current in the circuit. The larger the capacitance C of the capacitor, the longer it will take to discharge and the longer it will take for this capacitor to recharge. Thus, the more Lk and Ck of the circuit, the slower the electromagnetic oscillations occur in it, the lower their frequency. The dependence of the frequency f o of free oscillations on L to and C to the circuit is expressed by a simple formula, which is one of the basic formulas of radio engineering:
The meaning of this formula is extremely simple: in order to increase the frequency of natural oscillations f 0, you need to reduce the inductance L k or the capacitance C k of the circuit; to reduce f 0, inductance and capacitance must be increased (Figure 47).
From the formula for frequency one can easily derive (we have already done this with the formula of Ohm’s law) calculation formulas for determining one of the parameters of the circuit L k or C k at a given frequency f0 and a known second parameter. Formulas convenient for practical calculations are given on sheets 73, 74 and 75.
Free electromagnetic oscillations These are periodic changes in the charge on the capacitor, the current in the coil, as well as electric and magnetic fields in the oscillatory circuit that occur under the influence of internal forces.
Continuous electromagnetic oscillations
To excite electromagnetic oscillations it is used oscillatory circuit , consisting of an inductor L connected in series and a capacitor with capacitance C (Fig. 17.1).
Let's consider an ideal circuit, i.e. a circuit whose ohmic resistance is zero (R=0). To excite oscillations in this circuit, it is necessary either to impart a certain charge to the capacitor plates, or to excite a current in the inductor. Let at the initial moment of time the capacitor be charged to a potential difference U (Fig. (Fig. 17.2, a); therefore, it has potential energy 
.At this moment in time, the current in the coil I = 0 .
This state of the oscillatory circuit is similar to the state of a mathematical pendulum, deflected by an angle α (Fig. 17.3, a). At this time, the current in the coil is I=0. After connecting a charged capacitor to the coil, under the influence of the electric field created by the charges on the capacitor, free electrons in the circuit will begin to move from the negatively charged plate of the capacitor to the positively charged one. The capacitor will begin to discharge, and an increasing current will appear in the circuit. The alternating magnetic field of this current will generate an electric vortex. This electric field will be directed opposite to the current and therefore will not allow it to immediately reach its maximum value. The current will increase gradually. When the force in the circuit reaches its maximum, the charge on the capacitor and the voltage between the plates are zero. This will happen after a quarter of the period t = π/4. At the same time, the energy e 
electric field transforms into magnetic field energyW e =1/2C U 2 0. At this moment, there will be so many electrons transferred to it on the positively charged plate of the capacitor that their negative charge completely neutralizes the positive charge of the ions present there. The current in the circuit will begin to decrease and the induction of the magnetic field it creates will begin to decrease. The changing magnetic field will again generate an electric vortex, which this time will be directed in the same direction as the current. The current supported by this field will flow in the same direction and gradually recharge the capacitor. However, as charge accumulates on the capacitor, its own electric field will increasingly inhibit the movement of electrons, and the current strength in the circuit will become less and less. When the current drops to zero, the capacitor will be completely overcharged.
The system states shown in Fig. 17.2 and 17.3, correspond to successive moments in time T
= 0;
;
;
And T.
The self-inductive emf arising in the circuit is equal to the voltage on the capacitor plates: ε = U
And 
Believing 
, we get
(17.1)
Formula (17.1) is similar to the differential equation of harmonic vibration considered in mechanics; his decision will be
q = q max sin(ω 0 t+φ 0) (17.2)
where q max is the largest (initial) charge on the capacitor plates, ω 0 is the circular frequency of the circuit’s natural oscillations, φ 0 is the initial phase.
According to the accepted notation, 
where
(17.3)
Expression (17.3) is called Thomson's formula and shows that when R=0, the period of electromagnetic oscillations arising in the circuit is determined only by the values of inductance L and capacitance C.
According to the harmonic law, not only the charge on the capacitor plates changes, but also the voltage and current in the circuit:
where U m and I m are the amplitudes of voltage and current.
From expressions (17.2), (17.4), (17.5) it follows that the oscillations of charge (voltage) and current in the circuit are phase shifted by π/2. Consequently, the current reaches its maximum value at those moments in time when the charge (voltage) on the capacitor plates is zero, and vice versa.
When a capacitor is charged, an electric field appears between its plates, the energy of which
or 
When a capacitor is discharged onto an inductor, a magnetic field arises in it, the energy of which
In an ideal circuit, the maximum energy of the electric field is equal to the maximum energy of the magnetic field:
The energy of a charged capacitor periodically changes over time according to the law
or 
Considering that 
, we get
The energy of the magnetic field of the solenoid changes with time according to the law
(17.6)
Considering that I m =q m ω 0, we obtain
(17.7)
The total energy of the electromagnetic field of the oscillatory circuit is equal to
W =W e +W m = (17.8)
In an ideal circuit, the total energy is conserved and the electromagnetic oscillations are undamped.
Damped electromagnetic oscillations
A real oscillatory circuit has ohmic resistance, so the oscillations in it are damped. In relation to this circuit, we write Ohm's law for the complete circuit in the form
(17.9)
Transforming this equality:
and making the replacement:
And 
,where β-damping coefficient we get
(10.17) - this is differential equation of damped electromagnetic oscillations
.
The process of free oscillations in such a circuit no longer obeys the harmonic law. For each period of oscillation, part of the electromagnetic energy stored in the circuit is converted into Joule heat, and the oscillations become fading(Fig. 17.5). For small attenuations ω ≈ ω 0, the solution to the differential equation will be an equation of the form
(17.11)
Damped oscillations in an electrical circuit are similar to damped mechanical oscillations of a load on a spring in the presence of viscous friction.
The logarithmic damping decrement is equal to
(17.12)
Time interval 
during which the amplitude of oscillations decreases by e ≈ 2.7 times is called decay time
.
Quality factor Q of the oscillatory system determined by the formula:
(17.13)
For an RLC circuit, the quality factor Q is expressed by the formula
(17.14)
The quality factor of electrical circuits used in radio engineering is usually on the order of several tens or even hundreds.
1. Oscillatory circuit.
2 Oscillatory circuit equation
3. Free vibrations in the circuit
4. Free damped oscillations in the circuit
5. Forced electrical oscillations.
6. Resonance in a series circuit
7. Resonance in a parallel circuit
8. Alternating current
1. 5.1. Oscillatory circuit.
Let us find out how electrical oscillations arise and are maintained in an oscillatory circuit.
Let first The upper plate of the capacitor is positively charged ,and the lower one is negative(Fig. 11.1, A).
In this case, all the energy of the oscillatory circuit is concentrated in the capacitor.
Let's close the key TO.. The capacitor will begin to discharge, and through the coil L current will flow. The electrical energy of the capacitor will begin to convert into magnetic energy of the coil. This process will end when the capacitor is completely discharged and the current in the circuit reaches its maximum (Fig. 11.1, b).
From this moment the current, without changing direction, will begin to decrease. However, it will not stop immediately - it will be supported by e. d.s. self-induction. The current will recharge the capacitor, and an electric field will arise, tending to weaken the current. Finally, the current will stop and the charge on the capacitor will reach its maximum.
From this moment the capacitor will begin to discharge again, the current will flow in the opposite direction, etc. - the process will be repeated
In the circuit in the absence of resistance conductors will be carried out strictly periodic oscillations. During the process, the charge on the plates of the capacitor, the voltage across it and the current through the coil periodically change.
Oscillations are accompanied by mutual transformations of the energy of electric and magnetic fields.
If the resistance of the conductors 
, then in addition to the described process, the conversion of electromagnetic energy into Joule heat will occur.
Circuit conductor resistanceR usually calledactive resistance.
1.5.2. Oscillatory circuit equation
Let us find the equation of oscillations in a circuit containing a series-connected capacitor WITH, inductor L,
active resistance R
and external variable e. d.s. 
(Fig. 1.5.1).
Let's choose the positive direction of traversing the circuit, for example clockwise.
Let's denote through q the charge of that plate of the capacitor, the direction from which to the other plate coincides with the selected positive direction of bypassing the circuit.
Then the current in the circuit is determined as 
(1)
Therefore, if I > Oh, that's it dq > 0, and vice versa (sign I matches the sign dq).
According to Ohm's law for a section of the circuit 1 R.L.2
.
(2),
Where 
- uh. d.s. self-induction.
In our case 
(sign q
must match the sign of the difference 
, because C > 0).
Therefore, equation (2) can be rewritten as 
or taking into account (1) as 
That's what it is oscillatory circuit equation
- linear differential inhomogeneous equation of the second order with constant coefficients. Finding with this equation q(t),
we can easily calculate the voltage across the capacitor 
and current strength I- according to formula (1).
The equation of the oscillatory circuit can be given a different form:
(5)
where the notation is introduced
.
(6)
Size 
- called natural frequency contour,
β - attenuation coefficient.
If ξ = 0, then the oscillations are usually called free.
- At R = Oh they will undamped,
- at R ≠0 - damped.
Consider the following oscillatory circuit. We will assume that its resistance R is so small that it can be neglected.
The total electromagnetic energy of the oscillatory circuit at any time will be equal to the sum of the energy of the capacitor and the energy of the magnetic field of the current. The following formula will be used to calculate it:
W = L*i^2/2 + q^2/(2*C).
The total electromagnetic energy will not change over time, since there is no energy loss through the resistance. Although its components will change, they will always add up to the same number. This is ensured by the law of conservation of energy.
From this we can obtain equations describing free oscillations in an electric oscillatory circuit. The equation will look like this:
q"’ = -(1/(L*C))*q.
The same equation, up to notation, is obtained when describing mechanical vibrations. Considering the analogy between these types of oscillations, we can write down a formula describing electromagnetic oscillations.
Frequency and period of electromagnetic oscillations
But first, let's look at the frequency and period of electromagnetic oscillations. The value of the frequency of natural vibrations can again be obtained from an analogy with mechanical vibrations. The coefficient k/m will be equal to the square of the natural oscillation frequency.
Therefore, in our case the square frequencies free oscillations will be equal to 1/(L*C)
ω0 = 1/√(L*C).
From here period free vibrations:
T = 2*pi/ω0 = 2*pi*√(L*C).
This formula is called Thompson's formulas. It follows from this that the oscillation period increases with increasing capacitance of the capacitor or inductance of the coil. These conclusions are logical, since with an increase in capacitance, the time spent charging the capacitor increases, and with an increase in inductance, the current strength in the circuit will increase more slowly, due to self-induction.
Charge Oscillation Equation capacitor is described by the following formula:
q = qm*cos(ω0*t), where qm is the amplitude of oscillations of the capacitor charge.
The current strength in the oscillatory circuit circuit will also perform harmonic oscillations:
I = q’= Im*cos(ω0*t+pi/2).
Here Im is the amplitude of current fluctuations. Note that between the oscillations of charge and current strength there is a difference in vases equal to pi/2.
The figure below shows graphs of these fluctuations.
Again, by analogy with mechanical vibrations, where fluctuations in the speed of a body are ahead of fluctuations in the coordinates of this body by pi/2.
In real conditions, the resistance of the oscillatory circuit cannot be neglected, and therefore the oscillations will be damped.
With a very high resistance R, oscillations may not begin at all. In this case, the energy of the capacitor is released in the form of heat at the resistance.