Four capacitors with capacitances. Find the total capacitance of the capacitor bank. Series connection of capacitors. Software packages for the study of electrical circuits

« Physics - Grade 10 "

"Electric capacity" is the last topic of the "Electrostatics" section. When solving problems on this topic, all the information obtained in the study of electrostatics may be required: the law of conservation of electric charge, the concepts of field strength and potential, information about the behavior of conductors in an electrostatic field, about the field strength in dielectrics, about the law of conservation of energy in relation to electrostatic phenomena. The main formula for solving problems for electrical capacity is the formula (14.22).

Task 1.

The capacitance of a capacitor connected to a constant voltage source U \u003d 1000 V is equal to C 1 \u003d 5 pF. The distance between its plates was reduced by n = 3 times. Determine the change in charge on the capacitor plates and the energy of the electric field.

Decision.

According to formula (14.22), the charge of the capacitor is q = CU. Hence the change in charge Δq - (C 2 - C)U \u003d (nC 1 - C 1)U \u003d (n - 1) C 1 U \u003d 10 -8 C.

Change in electric field energy

Task 2.

Capacitor charge q = 3 10 -8 C. Capacitor capacitance C \u003d 10 pF. Determine the speed that an electron acquires when flying in a capacitor from one plate to another. The initial speed of the electron is zero. Specific charge of an electron

Decision.

The initial kinetic energy of an electron is equal to zero, and the final one is equal to Apply the law of conservation of energy where A is the work of the electric field of the capacitor:

Hence,

Finally

Task 3.

Four capacitors with capacities C 1 \u003d C 2 \u003d\u003d 1 μF, C 3 \u003d 3 μF, C 4 \u003d 2 μF are connected, as shown in Figure 14.46. Voltage U = 140 V is applied to points A and B. Determine the charge q1 and the voltage U1 on each of the capacitors.

To determine the charge and voltage, first of all, we find the capacity of the capacitor bank. The equivalent capacitance of the second and third capacitors C 2.3 \u003d C 2 + C 3 and the equivalent capacity of the entire capacitor bank, which is three capacitors connected in series with capacities C 1, C 2.3, C 4, we find from the ratio

1 / Cequiv \u003d 1 / C 1 + 1 / C 2.3 + 1 / C 4, Seq \u003d (4/7) 10 -6 F.

The charges on these capacitors are the same:

q 1 \u003d q 2.3 \u003d q 4 \u003d Seq \u003d 8 10 -5 Cl.

Therefore, the charge of the first capacitor q 1 = 8 10 -5 C, and the potential difference between its plates, or voltage, U 1 = q 1 / C 1 = 80 V.

For the fourth capacitor, we similarly have q 4 \u003d 8 10 -5 C, U 4 \u003d q 4 / C 4 \u003d 40 V.

Let's find the voltage on the second and third capacitors: U 2 \u003d U 3 \u003d q 2.3 / C 2.3 \u003d 20 V.

Thus, on the second capacitor, the charge q 2 = C 2 U 2 = 2 10-5 C, and on the third capacitor q 3 = C 3 U 3 = 6 10 -5 C. Note that q 2,3 = q 2 + g 3 .

Task 4.

Determine the equivalent electrical capacitance in the circuit shown in Figure (14.47 a), if the capacitances of the capacitors are known.

Decision.

Often when solving problems in which it is required to determine the equivalent electrical capacitance, the connection of capacitors is not obvious. In this case, if it is possible to determine the points of the circuit at which the potentials are equal, then we can connect these points or exclude the capacitors connected to these points, since they cannot accumulate charge (Δφ \u003d 0) and, therefore, do not play a role in the distribution of charges .

In the diagram shown in Figure (14.47, a), there is no obvious parallel or series connection of capacitors, since in the general case φ A ≠ φ B in and different voltages are applied to capacitors C1 and C2. However, we note that due to the symmetry and equality of the capacitances of the corresponding capacitors, the potentials of points A and B are equal. Therefore, it is possible, for example, to connect points A and B. The scheme is converted to the form shown in Figure (14.47, b). Then the capacitors C1, as well as the capacitors C2, will be connected in parallel and C equiv will be determined by the formula 1 / C equiv \u003d 1/2C 1 + 1/2C 2, from where

You can also simply ignore the presence of the capacitor C3 in the circuit, since the charge on it is zero. Then the scheme is converted to the form shown in Figure (14.47, c). Capacitors C1 and C2 are connected in series, so

Equivalent capacitors with C "equiv are connected in parallel, so we finally get the same expression for the equivalent capacitance:

Task 5.

The energy of a flat air capacitor W 1 \u003d 2 10 -7 J. Determine the energy of the capacitor after filling it with a dielectric with a permittivity ε \u003d 2, if:

1) the capacitor is disconnected from the power supply;

2) the capacitor is connected to the power supply.

Decision.

1) Since the capacitor is disconnected from the power source, its charge q 0 remains constant. The energy of the capacitor before filling it with a dielectric after filling where C 2 = εC 1.

"Electric capacity" is the last topic of the "Electrostatics" section. When solving problems on this topic, all the information obtained in the study of electrostatics may be required: the conservation law electric charge, the concepts of field strength and potential, information about the behavior of conductors in an electrostatic field, about the field strength in dielectrics, about the law of conservation of energy in relation to electrostatic phenomena. The main formula for solving problems for electrical capacity is the formula (14.22).

Task 1. Capacitance of a capacitor connected to a source constant voltage U \u003d 1000 V, equal to C 1 \u003d 5 pF. The distance between its plates was reduced by n = 3 times. Determine the charge change on the capacitor and energy plates electric field.

Solution. According to formula (14.22), the charge of the capacitor is q = CU. Hence the change in charge Δq - (C 2 - C)U \u003d (nC 1 - C 1)U \u003d (n - 1) C 1 U \u003d 10 -8 C.

Task 2. Capacitor charge q = 3 10 -8 C. Capacitor capacitance C \u003d 10 pF. Determine the speed that an electron acquires when flying in a capacitor from one plate to another. The initial speed of the electron is zero. Specific charge of an electron

Solution. The initial kinetic energy of an electron is equal to zero, and the final one is equal to Apply the law of conservation of energy where A is the work of the electric field of the capacitor:

Hence,

Finally

Task 3. Four capacitors with capacities C 1 \u003d C 2 \u003d\u003d 1 μF, C 3 \u003d 3 μF, C 4 \u003d 2 μF are connected, as shown in Figure 14.46. Voltage U = 140 V is applied to points A and B. Determine the charge q1 and the voltage U1 on each of the capacitors.

Solution. To determine the charge and voltage, first of all, we find the capacity of the capacitor bank. The equivalent capacitance of the second and third capacitors C 2.3 \u003d C 2 + C 3, and the equivalent capacity of the entire capacitor bank, which is three series-connected capacitors with capacities C 1, C 2.3, C 4, we find from the relation

1 / Cequiv \u003d 1 / C 1 + 1 / C 2.3 + 1 / C 4, Seq \u003d (4/7) 10 -6 F.

The charges on these capacitors are the same:

q 1 \u003d q 2.3 \u003d q 4 \u003d Seq \u003d 8 10 -5 Cl.

Therefore, the charge of the first capacitor q 1 = 8 10 -5 C, and the potential difference between its plates, or voltage, U 1 = q 1 / C 1 = 80 V.

For the fourth capacitor, we similarly have q 4 \u003d 8 10 -5 C, U 4 \u003d q 4 / C 4 \u003d 40 V.

Let's find the voltage on the second and third capacitors: U 2 \u003d U 3 \u003d q 2.3 / C 2.3 \u003d 20 V.

Thus, on the second capacitor, the charge q 2 = C 2 U 2 = 2 10-5 C, and on the third capacitor q 3 = C 3 U 3 = 6 10 -5 C. Note that q 2,3 = q 2 + g 3 .

Task 4. Determine the equivalent electrical capacitance in the circuit shown in Figure (14.47 a), if the capacitances of the capacitors are known.

Solution. Often, when solving problems in which it is required to determine the equivalent electric capacitance, the connection of capacitors is not obvious. In this case, if it is possible to determine the points of the circuit at which the potentials are equal, then we can connect these points or exclude the capacitors connected to these points, since they cannot accumulate charge (Δφ \u003d 0) and, therefore, do not play a role in the distribution of charges .

In the diagram shown in Figure (14.47, a), there is no obvious parallel or series connection of capacitors, since in the general case φ A ≠ φ B in and different voltages are applied to capacitors C1 and C2. However, we note that due to the symmetry and equality of the capacitances of the corresponding capacitors, the potentials of points A and B are equal. Therefore, it is possible, for example, to connect points A and B. The scheme is converted to the form shown in Figure (14.47, b). Then the capacitors C1, as well as the capacitors C2, will be connected in parallel and C equiv will be determined by the formula 1 / C equiv \u003d 1/2C 1 + 1/2C 2, from where

You can also simply ignore the presence of the capacitor C3 in the circuit, since the charge on it is zero. Then the scheme is converted to the form shown in Figure (14.47, c). Capacitors C1 and C2 are connected in series, so

Equivalent capacitors with C "equiv are connected in parallel, so we finally get the same expression for the equivalent capacitance:

Task 5. The energy of a flat air capacitor W 1 \u003d 2 10 -7 J. Determine the energy of the capacitor after filling it with a dielectric with a permittivity ε \u003d 2, if:

1) the capacitor is disconnected from the power supply;

2) the capacitor is connected to the power supply.

Solution 1) Since the capacitor is disconnected from the power source, its charge q 0 remains constant. The energy of the capacitor before filling it with a dielectric after filling where C 2 = εC 1.

Tasks for independent solution

1. The potential difference between the plates of a capacitor with a capacity of 0.1 μF has changed by 175 V. Determine the change in the charge of the capacitor.

2. An electron flies into the space between the plates of a flat capacitor with a speed of 2-10 7 m / s, directed parallel to the capacitor plates. How far will the electron move towards the positively charged plate during its movement inside the capacitor if the length of the capacitor is 0.05 m and the potential difference between the plates is 200 V? The distance between the capacitor plates is 0.02 m. The ratio of the electron charge modulus to its mass is 1.76 10 11 C/kg.

3. A flat capacitor was charged using a current source with a voltage of U \u003d 200 V. Then the capacitor was disconnected from this current source. What will be the voltage U 1 between the plates if the distance between them is increased from the initial d \u003d 0.2 mm to d 1 \u003d 0.7 mm?

4. Determine the capacity of the air spherical capacitor. Radii of spheres R 1 and R 2 .

5. A metal plate with a thickness of d 0 is inserted into a flat air condenser. The charge on the capacitor plates q. The capacitor is disconnected from the source. The distance between the plates d, the area of ​​the plates S. Determine the change in the capacitance of the capacitor and the energy of its electric field.

Review the material in chapter 14 according to the following plan.

1. Write down the basic concepts and physical quantities and give them a definition.

2. Formulate the laws and write down the basic formulas.

3. Indicate the units of physical quantities and their expression in terms of the basic SI units.

4. Describe the main experiments confirming the validity of laws.

A series connection of capacitors is a battery formed by a chain of capacitors. There is no branching, the output of one element is connected to the input of the next.

Physical processes in serial connection

When capacitors are connected in series, the charge of each is equivalent. Due to the natural principle of balance. Only the extreme plates are connected to the source, the others are charged by redistribution of charges between them. Using equality, we find:

q = q1 = q2 = U1 C1 = U2 C2, from which we write:

The voltages between the capacitors are distributed inversely with the nominal capacitances. In sum, both make up the voltage of the mains. When discharging, the design is able to give a charge q, regardless of how many capacitors are connected in series. We find the battery capacity from the formula:

C = q/u = q/(U1 + U2), substituting the expressions above, reducing to a common denominator:

1/C = 1/C1 + 1/C2.

Calculating total battery capacity

When capacitors are connected in series, the values ​​\u200b\u200binverse to the nominal capacities are added to the battery. Bringing the last expression to a common denominator, inverting the fractions, we get:

C \u003d C1C2 / (C1 + C2).

The expression is used to find the battery capacity. If there are more than two capacitors, the formula becomes more complicated. To find the answer, the denominations are multiplied among themselves, the numerator of the fraction comes out. Pairwise products of two denominations are put in the denominator, sorting through the combinations. In practice, it is sometimes more convenient to calculate using reciprocals. The result is to divide the unit.

Connection of series capacitors

The formula is greatly simplified if the battery ratings are the same. You just need to divide the number by the total number of elements, getting the resulting value. The voltage will be distributed evenly, therefore, it is enough to divide the nominal value of the supply network equally into the total number. When powered by a 12 volt battery, 4 containers, 3 volts will drop on each.

One simplification will be made for the case when the denominations are equal, one capacitance is included as a variable to adjust the result. Then the maximum voltage of each element can be approximately found by dividing the source voltage by the amount reduced by one. The result will be obtained, obviously having a certain margin. Concerning variable capacity, the requirements are much more stringent. Ideally, the operating value overlaps the source voltage.

The need for a serial connection

At first glance, the idea of ​​connecting capacitors in series with a battery seems to make no sense. The first advantage is obvious: the requirements for the maximum voltage of the plates are reduced. More operating voltage, more expensive product. In a similar way, the world is seen by a radio amateur who has several low-voltage capacitors on his hands, who wants to use iron integral part high voltage circuit.

Calculating the acting stresses by the element using the above formulas, one can easily solve the problem. Let's look at an example to make it clearer:

Let a battery with a voltage of 12 volts, three capacities with nominal values ​​of 1, 2 and 4 nF be installed. Find the voltage when the battery cells are connected in series.

To find three unknowns, take the trouble to write an equal number of equations. It is known from the course of higher mathematics. The result will look like this:

1. U1 + U2 + U3 = 12;
2. U1/U2 = 2/1 = 2, from which we write: U1 = 2U2;
3. U2/U3 = 4/2 = 2, where you can see: U2 = 2U

It is not difficult to notice, we substitute the last two expressions for the first one, expressing 12 volts through the voltage of the third capacitor. You will get the following:

4U3 + 2U3 + U3 = 12, from where we find the voltage of the third capacitor is 12/7 = 1.714 volts, U2 - 3.43 volts, U1 - 6.86 volts. The sum of the numbers gives 12, each less than the voltage of the supply battery. Moreover, the greater the difference, the smaller the face value of the neighbors. From this rule it follows: in a series connection, low capacitance capacitors have a higher operating voltage. For definiteness, we will find the value of the compiled battery, at the same time we will illustrate the formula, since it is described above purely verbally:

C \u003d C1C2C3 / (C1C2 + C2C3 + C1C3) \u003d 8 / (2 + 8 + 4) \u003d 8/14 \u003d 571 pF.

The resulting value is less than each capacitor that makes up the series connection. From the rule it can be seen: the maximum influence on the total capacity has a smaller one. Therefore, if it is necessary to adjust the full rating of the battery, it must be variable capacitor. Otherwise, turning the screw will not have much effect on the final result.

We see another pitfall: after adjustment, the voltage distribution across the capacitors will change. Calculate extreme cases so that the voltage does not exceed the operating value for the components of the battery.

Software packages for the study of electrical circuits

Apart from online calculators There are more powerful tools for calculating the series connection of capacitors. A big disadvantage of publicly available tools is due to the reluctance of sites to check the program code, which means they contain errors. It is bad if one capacitance fails, broken by the testing process of an incorrectly assembled circuit. Not the only downside. Sometimes the schemes are much more complex, it is impossible to understand in a complex way.

Some devices have filters. high frequency, using a capacitor, included in cascades. Then, in addition to the circuit through the resistor to the ground, a series connection of capacitors is formed on the circuit. Usually do not apply the formula shown above. It is generally accepted that each filter stage exists separately, the result of the signal passage is described by the amplitude-frequency characteristic. A graph showing how much the spectral component of the signal will cut off at the output.

Those wishing to carry out approximate calculations are advised to familiarize themselves with the software package personal computer Electronics Workbench. The design is made according to English standards, take the trouble to take into account the nuance: the designation of resistors on the electrical circuit with a broken zigzag. Denominations, names of elements will be stated in a foreign manner. It does not interfere with the use of a shell that provides the operator with a mountain of power sources of various kinds.

And most importantly - Electronics Workbench will allow you to set control points on each, in real time to see the voltage, current, spectrum, waveform. The project should be supplemented with an ammeter, voltmeter, and other similar devices.

With the help of such a software package, simulate the situation, see how much the voltage drops on the battery cell. Saves you from cumbersome calculations, greatly speeding up the circuit design process. Errors are excluded at the same time. It becomes easy and simple to add, remove capacitors with an immediate evaluation of the result.

Working example

The screenshot shows the Electronics Workbench 5.12 desktop with the assembled electric circuit series connection of capacitors. Each 1uF capacitance, the same elements are taken for demonstration purposes. So that everyone can easily check the correctness.

Series capacitor bank

Let's look at the source first. Alternating voltage with a frequency of 60 Hz. The developer's country has a different standard than Russian ones. It is recommended to right click the source, visit properties, set:

1. Frequency (frequency) 50 Hz instead of 60 Hz.
2. The effective voltage value (voltage) is 220 volts instead of 120.
3. Phase (phase - imitation of reactivity) to take according to your needs.

For letter-eaters, it will be useful to look through the properties of the circuit elements. The source is free to set the voltage tolerance in percent. It is enough to add one 1 kΩ resistor, the circuit becomes a high-pass filter. It is recommended not to simplify the steps. Put the ground sign correctly, make sure: the circuit is completely trivial. Otherwise, the results will make you break your head for a long time.

Electric capacitance of a solitary conductor or capacitor:

where Q is the charge imparted to the conductor (capacitor);  is the change in potential caused by this charge.

The electric capacitance of a solitary conducting sphere of radius R, located in an infinite medium with permittivity ,

.

If the sphere is hollow and filled with a dielectric, then its electric capacitance does not change from this.

Electric capacitance of a flat capacitor:

where S is the area of ​​the plates (of each plate); d is the distance between them;  is the permittivity of the dielectric filling the space between the plates.

The electric capacitance of a flat capacitor filled with n layers of dielectric with a thickness of d i each with dielectric permittivities  i (layered capacitor),

The electric capacitance of a spherical capacitor (two concentric spheres with radii R 1 and R 2, the space between which is filled with a dielectric with a permittivity)

The electric capacitance of a cylindrical capacitor (two coaxial cylinders with a length l and radii R 1 and R 2, the space between which is filled with a dielectric with permittivity):

Capacitance C of series-connected capacitors:

- in general:

where n is the number of capacitors;

– in the case of two capacitors:

- in the case of n identical capacitors with electric capacity C 1 each

Capacitance of capacitors connected in parallel:

- in general: .


– surface charge density, C/m 2 .

Capacitor electric field energy:

The volume energy density of the electric field in a linear isotropic medium with a relative permittivity  is as follows:

.

Examples of problem solving

Example 1 Determine the electric capacitance of a flat capacitor with two layers of dielectrics: porcelain with a thickness of d 1 \u003d 2 mm and ebonite with a thickness of d 2 \u003d 1.5 mm, if the area S of the plates is 100 cm 2.

Decision. Capacitor capacitance by definition
where Q is the charge on the capacitor plates; U is the potential difference between the plates. Replacing in this equality the total potential difference U by the sum U 1 + U 2 of the voltages on the dielectric layers, we get:

(4.1)

Taking into account that Q=S, equality (4.1) can be rewritten as:

(4.2)

where  is the surface charge density on the plates; E 1 and E 2 are the field strengths in the first and second dielectric layers, respectively; D is the dielectric field displacement in dielectrics. Multiplying the numerator and denominator of equality (4.2) by  0 and taking into account that D=, we finally get:

(4.3)

Having made calculations according to the formula (4.3), we find:

.

Example 2 Two identical flat capacitors are connected in parallel and charged to a voltage U 0 = 480 V. After disconnecting from the current source, the distance between the plates of one of the capacitors was halved. What will be the voltage U on the capacitors.

Decision. When capacitors are connected in parallel, their total capacitance will be:

C baht \u003d C 1 + C 2 \u003d 2C; (C 1 \u003d C 2 \u003d C).

Battery charge q 1 \u003d C baht U 0 \u003d 2CU 0.

When the distance between the plates of the capacitor is halved, its electrical capacity will double (according to the formula
) and becomes C’ = 2C, then their total capacity is C’ baht = 2C+C= 3C.

The charge will become q 2 \u003d C ' baht U \u003d 3CU.

According to the law of conservation of electric charge q 1 \u003d q 2, since the capacitor bank is disconnected from the source. Therefore, 2CU 0 = 3CU, whence
AT.

Tasks

401. Find the capacitance C of a solitary metal ball with a radius R \u003d 1 cm. (Answer: 1.11 pF).

402. Determine the charges on each of the capacitors in the circuit shown in fig. 4.1, if C 1 = 2 μF, C 2 = 4 μF, C 3 = 6 μF,  = 18 V. (Answer: Q 1 = 30 μC; Q 2 = 12 μC; Q 1 = 18 μC).

403. Determine the electrical capacity From the earth, taking it for a ball with a radius R = 6400 km. (Answer: 180 pF).

404. A ball of radius R 1 = 6 cm is charged to a potential φ 1 = 300 V, and a ball of radius R 2 = 4 cm is charged to a potential φ 2 = 500 V. Determine the potential φ of the balls after they are connected by a metal conductor. Ignore the capacitance of the connecting conductor. (Answer:
).

405. Determine the capacitance C of a flat mica capacitor, the area S of the plates of which is 100 cm 2, and the distance between them is 0.1 mm (dielectric constant of mica  \u003d 7). (Answer: 6.2 nF).

406. Five capacitors of the same capacity are connected in series into a battery. A static voltmeter is connected in parallel to one of the capacitors, the capacitance of which is half the capacitance of each capacitor. The voltmeter shows 500 V. What is the potential difference across the entire battery? (Answer: 3500 V).

407. The distance d between the plates of a flat capacitor is 1.33 mm, the area S of the plates is 20 cm 2. In the space between the capacitor plates there are two layers of dielectrics: mica with a thickness of d 1 = 0.7 mm and ebonite with a thickness of d 2 = 0.3 mm. Determine the capacitance of the capacitor (dielectric constant of mica  = 7, ebonite  = 3). (Answer:

408. N ball drops of radius r are charged to the same potential φ 0 . All drops merge into one big one. Determine the potential and charge density on the surface of a large drop. (Answer: ).

409. Two concentric metal spheres with radii R 1 = 2 cm and R 2 = 2.1 cm form a spherical capacitor. Determine its electric capacitance C if the space between the spheres is filled with paraffin (paraffin permittivity  = 2). (Answer:
).

410. A paraffin tile d = 1 cm thick is pushed into a flat capacitor, which fits snugly against its plates. By how much should the distance between the plates be increased to get the same capacitance? (Dielectric constant of paraffin = 2). (Answer: 0.5 cm).

411. A condenser consists of two concentric spheres. The radius R 1 of the inner sphere is 10 cm, the outer one R 2 = 10.2 cm. The gap between the spheres is filled with paraffin. The inner sphere is given a charge Q = 5 μC. Determine the potential difference U between the spheres. (Dielectric constant of paraffin= 2). (Answer: 4.41 kV).

412. To an air capacitor charged to a potential difference U = 600 V and disconnected from a voltage source, a second uncharged capacitor of the same size and shape, but with a dielectric (porcelain), was connected in parallel. Determine the dielectric constant ε of porcelain if, after connecting the second capacitor, the potential difference decreased to U 1 = 100 V. (Answer: 5).

413. Two capacitors with electrical capacities C 1 \u003d 3 μF and C 2 \u003d 6 μF are interconnected and connected to a battery with an emf equal to 120 V. Determine the charges Q 1 and Q 2 of the capacitors and the potential difference U 1 and U 2 between their plates, if the capacitors are connected: 1) in parallel; 2) sequentially. (Answer: 360 µC; 720 µC; 120 V).

414. A capacitor with an electrical capacity C1 = 0.2 μF was charged to a potential difference U 1 = 320 V. After it was connected in parallel with a second capacitor charged to a potential difference U 2 = 450 V, the voltage U on it changed to 400 V. Calculate the capacitance C 2 of the second capacitor. (Answer:
).

415. A capacitor with an electrical capacity C 1 = 0.6 μF was charged to a potential difference U 1 = 300 V and connected to a second capacitor with an electrical capacity C 2 = 0.4 μF, charged to a potential difference U 2 = 150 V. Find the charge ΔQ flowing from the plates the first capacitor to the second. (Answer:
).

416. Three identical flat capacitors are connected in series. The capacitance C of such a capacitor bank is 80 pF. The area S of each plate is 100 cm2. Dielectric - glass (= 7). What is the thickness of the glass? (Answer: 2.32 mm).

417. Capacitors are connected as shown in fig. 4.2. Electric capacitance of capacitors: C 1 \u003d 0.2 μF, C 2 \u003d 0.1 μF, C 3 \u003d 0.3 μF, C 4 \u003d 0.4 μF. Determine the electrical capacity C of the capacitor bank. (Answer: 0.21uF).

418. Capacitors with electrical capacities C 1 = 10 nF, C 2 = 40 nF, C 3 = 2 nF, C 4 = 30 nF are connected as shown in fig. 4.3. Determine the capacitance C of the capacitor bank. (Answer: 20 pF).

419. Capacitors are connected as shown in fig. 4.4. The capacitance of capacitors: C 1 \u003d 2 μF, C 2 \u003d 2 μF, C 3 \u003d 3 μF, C 4 \u003d 1 μF. The potential difference on the plates of the fourth capacitor U 4 \u003d 100 V. Find the charges and potential differences on the plates of each capacitor, as well as the total charge and potential difference of the capacitor bank. (Answer: 200 μC; 120 μC; 120 μC; 100 μC; 110 V; 60 V; 40 V; 220 μC; 210 V).

420. Capacitors with electrical capacities C 1 = 1 pF, C 2 = 2 pF, C 3 = 2 pF, C 4 = 4 pF, C 5 = 3 pF are connected as shown in fig. 4.5. Determine the capacitance C of the capacitor bank. (Answer: 2 pF. Instruction. Prove that if C 1 /C 2 \u003d C 3 /C 4, then φ A \u003d φ B, and, therefore, the capacity of C 5 does not matter when determining the total capacity of the circuit).

421. A flat capacitor, between the plates of which there is a plate of dielectric permittivity , is attached to the battery. The charge of the capacitor is equal to Q 0. What charge ΔQ will pass through the battery when the plate is removed? (Answer:
).

422. A flat air capacitor is charged to a potential difference U = 1000 V. With what force F are its plates attracted to one another? The area of ​​the plates S= 100 cm 2, the distance between them d= 1 mm. (Answer:
).

423. On the plates of a flat capacitor, a charge is uniformly distributed with a surface density σ \u003d 0.2 μC / m 2. The distance d between the plates is 1 mm. By how much will the potential difference across its plates change as the distance d between the plates increases to 3 mm? (Answer: 22.6 V).

424. The distance d between the plates of a flat capacitor is 2 cm, the potential difference is U = 6 kV. The charge Q of each plate is 10 nC. Calculate the energy W of the capacitor field and the force F of the mutual attraction of the plates. (Answer: 30 µJ).

425. Determine the charges of the capacitors Q 1, Q 2, Q 3 in the circuit, the parameters of which are shown in fig. 4.6.

426. How much heat Q will be released when a flat capacitor is discharged if the potential difference U between the plates is 15 kV, the distance d \u003d 1 mm, the dielectric is mica and the area S of each plate is 300 cm 2. (Answer:

427. The force F of attraction between the plates of a flat air capacitor is 50 mN. The area S of each plate is 200 cm 2 . Find the energy density w capacitor fields. (Answer: 0.209 J).

428. A flat air condenser consists of two round plates with a radius r = 10 cm each. The distance d 1 between the plates is 1 cm. The capacitor was charged to a potential difference U = 1.2 kV and disconnected from the current source. What work A needs to be done in order to remove the plates from each other and increase the distance between them to d 2 \u003d 3.5 cm. (Answer: 2.5 J / m 3).

429. Capacitors with electrical capacities C 1 = 1 μF, C 2 = 2 μF C 3 = 3 μF are included in a circuit with a voltage U = 1.1 kV. Determine the energy of each capacitor in the following cases: 1) their sequential connection; 2) parallel connection. (Answer: 50 µJ).

430. Electric capacitance From a flat capacitor is equal to 111 pF. The dielectric is porcelain. The capacitor was charged to a potential difference U= 600 V and disconnected from the voltage source. What work A must be done to remove the dielectric from the capacitor? Friction is negligible. (Answer: 0.18 J).

Four capacitors, the capacitances of which are C1 = 1.0 μF, C2 = 4.0 μF, C3 = 2.0 μF and C4 = 3.0 μF, are connected to a battery (see Fig.). If the battery is connected to a source whose terminal voltage is U = 10 V, then the energy W3 of the electrostatic field of the capacitor C3 is ... μJ.

To determine the energy W3 of the electrostatic field of the capacitor C3, it is necessary to know the charge accumulated by this capacitor. Capacitors C3 and C4 are connected in series with each other and in parallel with capacitors C1 and C2 connected in series. Total capacitance:

Correct answer: 36 mJ.

1. elapsed time: 3 minutes. task score: 6 out of 10 points.

2. task level: 3 (basic). subjective difficulty: 6 out of 10 points.

Two resistors, the resistances of which R1 = 0.64 ohms and R2 = 2.56 ohms, are connected in series for the first time, and in parallel for the second time, and after the connection they are connected in turn to a direct current source. In both cases, the power released in the outer sections of the circuit is the same. If the current strength during a short circuit of this source Ik \u003d 15 A, then the maximum useful power Pmax of the source is ... W.

The maximum useful power of the source is achieved when the external resistance of the circuit is equal to the internal resistance of the source and is equal to:

The maximum useful power Pmax of the source is 72 W.

Correct answer: 72 W.

Notes (details on the main page of the test):

1. elapsed time: 6.5 minutes. task score: 8 out of 10 points.

2. task level: 4 (profile). subjective difficulty: 7 out of 10 points.