About the charge of the capacitor.

We'll close the chain. In the chain, the charge current of the capacitor goes. This means that from the left side of the capacitor, some of the electrons will go into the wire, and the same number of electrons will go from the wire to the right plate. Both plates will be charged with opposite charges of the same size.

Between the plates in the dielectric there will be an electric field.

Now let's break the chain. The capacitor remains charged. We will shorten the wire of its electrode. The capacitor will instantly discharge. This means that the excess of electrons will leave the wire on the right side of the wire, and a lack of electrons will enter the left plate from the wire. On both plates of electrons will be the same, the capacitor will be discharged.

To what voltage is the capacitor charged?

It is charged to the voltage that is applied to it from the power source.

Resistance of the capacitor.

We'll close the chain. The capacitor began to be charged and immediately became a source of current, voltage, ED C .. The figure shows that the EDS of the capacitor is directed against the current source charging it.

The counteraction of the electromotive force of a charged capacitor by the charge of this capacitor is called the capacitive resistance.

All the energy expended by the current source to overcome the capacitive resistance is converted into energy electric field   capacitor. When the capacitor is discharged, the entire energy of the electric field will return back to the circuit in the form of energy electric current. Thus, the capacitive resistance is reactive, i.e. not causing irreversible energy losses.

Why does the direct current do not pass through the capacitor, but does the alternating current pass?

Turn on the DC circuit. The lamp will flash and go out, why? Because in the circuit passed the current charge of the capacitor. As soon as the capacitor is charged up to the battery voltage, the current in the circuit will stop.

Now close the chain alternating current. In the first quarter of the period, the voltage on the generator rises from 0 to a maximum. In the circuit there is a current charge of the capacitor. In the second quarter of the period, the voltage on the generator decreases to zero. The condenser is discharged through the generator. After this, the capacitor is recharged and discharged. Thus, the charge and discharge currents of the capacitor are in the circuit. The light bulb will light constantly.

In a circuit with a capacitor, the current passes through the entire closed circuit, including the capacitor's dielectric. In the charging capacitor an electric field is formed which polarizes the dielectric. Polarization is the rotation of electrons in atoms in elongated orbits.

Simultaneous polarization of a huge number of atoms forms a current, called bias current.   Thus, in the wires there is a current in a dielectric of the same magnitude.

The capacitive capacitance of a capacitor is determined by the formula

Considering the graph, we conclude: the current in the circuit with a purely capacitive resistance is ahead of the voltage by 90 °.

The question arises how the current in the circuit can outrun the voltage on the generator? In the circuit there is a current from two current sources alternately, from the generator and from the condenser. When the voltage across the generator is zero, the current in the circuit is maximal. This is the discharge current of the capacitor.

About a real capacitor

A real capacitor has two impedances: active and capacitive.They should be considered included sequentially.

The voltage applied by the generator to the active resistance and the current flowing along the active resistance coincide in phase.

The voltage applied by the generator to the capacitive resistance and the current flowing along the capacitive resistance are shifted in phase by 90 0. The resulting voltage applied by the generator to the capacitor can be determined by the parallelogram rule.

At the active resistance, the voltage U act and current I coincide in phase. On the capacitive resistance, the voltage U c lags behind the current I by 90 0. The resulting voltage applied by the generator to the capacitor is determined by the parallelogram rule. This resultant voltage lags the current I by some angle φ always less than 90 °.

Determination of the resulting resistance of the capacitor

The resulting resistance of the capacitor can not be found by summing the values ​​of its active and capacitive resistances. This is done by formula

With an alternating voltage on a real capacitor, in addition to the bias current, there are small conduction currents through the thickness of the dielectric (volumetric current) and over the surface (surface current). The conduction currents and the polarization of the dielectric accompany energy losses.

Thus, in a real capacitor, along with a change in the energy of the electric field (this characterizes reactive power Q ) because of the imperfection of the dielectric there is an irreversible process of transformation electric power   in heat, the speed of which is expressed active power P . Therefore, in the replacement circuit, the real capacitor must be represented by the active and reactive elements.

The division of a real capacitor into two elements is a calculated technique, since it is impossible to distinguish them constructively. However, the same scheme of substitution has a real chain of two elements, one of which is characterized only by the active power P (Q = 0), the other by the reactive (capacitive) power Q (P = 0).

The circuit of replacement of the condenser with parallel connection of elements

A real capacitor (with losses) can be represented by an equivalent circuit parallel connection active G   and capacitiveB   from conductivity (Figure 13.15), with the active conductivity being determined by the capacitor losses G = P / U c 2 , and capacitance - the design of the capacitor. Suppose that the conductivities G and B c for such a chain are known, and the voltage has equation

u = Umsinωt.

It is required to determine the currents in the circuit and the power. Investigation of a circuit with an active resistance and a circuit with a capacitance showed that at a sinusoidal voltage the currents in them are also sinusoidal. With the parallel connection of the branches G and B c, according to the first Kirchhoff law, the total current i is equal to the sum of the currents in the branches with active and capacitive conductances:

i = i G + i c, (13.30)

Taking into account that the current   i G   The phase coincides with the voltage, and the current   i c ahead of the voltage by a quarter period, the equation of the total current can be written in the following form:

Vector diagram of currents in a circuit with a capacitor

To determine the effective value of the total current I by the vector addition method, we construct a vector diagram according to equation

I = I G + I C

The effective values ​​of the current components:

I G = GU (13.31)

I C = B C U (13.32)

The vector in the vector diagram is the vector of the voltage U   (Figure 13.16, a), its direction coincides with the positive direction of the axis, from which the phase angles are measured (the initial phase of the voltage φ a =0). Vector I   G   coincides in direction with the vector U, and vector I C is directed perpendicular to the vector U with a positive angle. From the vector diagram it is seen that the total voltage vector lags behind the total current vector by an angle φ , the value of which is greater than zero, but less than 90 °. Vector I   is a hypotenuse of a right triangle whose legs are the constituent vectors I G and I C:

  At a voltage u = Um sinωt   According to the vector diagram, the current equation

i = I m sin (ωt + φ )

Condenser conductivity triangle

The sides of the current triangles, expressed in current units, are divided by the voltage U. We obtain a similar triangle of conductivity (Figure 13.16, b), whose legs are the active G = I G / U and capacitive In c = I c / U   conductivity, and hypotenuse - the total conductance of the chain Y = I / U . From the conductivity triangle

The relation between the effective values ​​of the voltage and current is expressed by the formulas

I = UY

U = I / Y (13.35)

From the triangles of currents and conductivities determine the quantities

cosφ = I G / I = G / Y; sin φ = I c / I = B c / Y; tg φ = I C / I G = B c / G. (13.36)

Power circuit with capacitor

Expression of the instantaneous power of a real capacitor

p = ui = Um sinωt * I m sin (ωt + φ)

coincides with the expression of the instantaneous power of the coil. Reasoning similar to those made when considering the instantaneous power curve (see Fig.13.11) can be carried out for a real capacitor based on the graph in Fig. 13.17. The values ​​of the active, reactive and total powers are expressed by the same formulas as those obtained for the coil [ (13.19) - (13.22)]. This is not difficult to show if the sides of the current triangle, expressed in current units, are multiplied by the voltage U. As a result of multiplication, we obtain a similar triangle of powers (Figure 13.16, c), whose legs are powers; active

  P = UI G = UIcosφ

reactive

Q = UI C = UIsinφ

complete

The circuit of replacement of the condenser with a serial connection of elements

The real capacitor, as well as on the design diagram, can be represented by a series connection of two sections: with active R   and capacitiveX   from   resistances. In Fig. 13.18, and such a scheme is shown in comparison with the scheme of parallel the connection of active and capacitive conductivities (Fig. 13, 18, 6). All the conclusions and formulas obtained for the coil remain valid for the capacitor, provided that the inductive resistance is replaced by a capacitive one. Condensers used in practice have relatively low energy losses. Therefore, in substitution schemes they are most often represented only by the reactive part, i.e., capacity C   The parts of the chain, individual elements   - the resistor R and the capacitor C, have such a replacement circuit, as shown in Fig. 13.18, a. If you are interested, read which apply in the industry.

We now assume that the portion of the chain contains the capacitor of the capacitance C, whereby the resistance and inductance of the section can be neglected, and let us see by what law the voltage at the ends of the section will change in this case. Denote the voltage between points a   and b   across u   and we will assume the charge of the capacitor q   and current i   Positive if they correspond to Fig. Then

and, consequently,

If the current in the circuit changes according to the law

then the charge of the condenser is

.

Integral constant q   0 here denotes an arbitrary constant charge of the capacitor, not connected with the current oscillations, and therefore we set. Consequently,

. (2)

Comparing (1) and (2), we see that with sinusoidal oscillations of the current in the circuit, the voltage across the capacitor also varies according to the cosine law. However, the voltage fluctuations on the capacitor lag behind the current oscillations by p / 2. The changes in the current and voltage in time are shown graphically in Fig. The result obtained has a simple physical meaning. The voltage at the capacitor at any time is determined by the existing capacitor charge. But this charge was formed by a current that had previously occurred in an earlier stage of oscillations. Therefore, the voltage oscillations are delayed relative to the current oscillations.

Formula (2) shows that the voltage amplitude on the capacitor is

Comparing this expression with Ohm's law for the chain section with direct current   (), we see that the quantity

plays the role of the resistance of the circuit, it was called capacitive resistance. The capacitive resistance depends on the frequency w, and at high frequencies even small capacitances can represent a very small resistance for alternating current. It is important to note that the capacitive resistance determines the relationship between the amplitude, not the instantaneous values ​​of the current and voltage.

Instantaneous AC power

changes with time according to a sinusoidal law with a doubled frequency. During the time from 0 to T/ 4 the power is positive, and in the next quarter of the period the current and voltage have opposite signs and the power becomes negative. Since the average value for the oscillation period is zero, the average AC power on the capacitor.

The current in the circuit with the capacitor can flow only if the voltage applied to it changes, and the current flowing through the circuit during the charge and discharge of the capacitor will be the greater, the larger the capacity of the capacitor and the faster the EMF changes.
  The capacitor included in the alternating current circuit affects the strength of the current flowing through the circuit, that is, it behaves as a resistance. The value of the capacitive resistance is the smaller, the larger the capacitance and the higher the frequency of the alternating current. Conversely, the capacitor's resistance to alternating current increases with decreasing capacitance and decreasing frequency.

X C = 1 / (2πƒC)

where Xc is the reactance of the capacitor, f is the frequency, and C is the capacitance.

To calculate the reactance of the capacitor, fill out the following form:

Calculation of the capacity for reactance:

Capacity calculation: C = 1 / (2πƒX C)

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