Series connection of capacitors. Find the total capacity of the capacitor bank. Series connection of capacitors Calculating the total battery capacity

« Physics - Grade 10 "

"Electric capacity" is the last topic of the section "Electrostatics". 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 on electrical capacity is the formula (14.22).

Objective 1.

The electrical capacity of a capacitor connected to a constant voltage source U = 1000 V is equal to C 1 = 5 pF. The distance between its plates was reduced by n = 3 times. Determine the change in the charge on the plates of the capacitor and the energy of the 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 = (nC 1 - C 1) U = (n - 1) C 1 U = 10 -8 Cl.

Changing the energy of the electric field

Objective 2.

Capacitor charge q = 3 10 -8 Cl. The capacity of the capacitor is C = 10 pF. Determine the speed that an electron acquires as it travels in a capacitor from one plate to another. The initial velocity of the electron is zero. Specific charge of an electron

Solution.

The initial kinetic energy of the 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

Objective 3.

Four capacitors with capacities C 1 = C 2 = = 1 μF, C 3 = 3 μF, C 4 = 2 μF are connected, as shown in Figure 14.46. A 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 capacity of the second and third capacitors C 2,3 = 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, will be found from the ratio

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

The charges on these capacitors are the same:

q 1 = q 2,3 = q 4 = Ceq = 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, similarly, we have q 4 = 8 10 -5 C, U 4 = q 4 / C 4 = 40 V.

Let's find the voltage across the second and third capacitors: U 2 = U 3 = q 2,3 / C 2,3 = 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 capacity, 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 you can connect these points or exclude the capacitors connected to these points, since they cannot accumulate charge (Δφ = 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 capacities of the corresponding capacitors, the potentials of points A and B are equal. Therefore, you can, for example, connect points A and B. The circuit 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 eq will be determined by the formula 1 / C eq = 1 / 2C 1 + 1 / 2C 2, whence

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

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

Task 5.

Energy of a flat air condenser W 1 = 2 10 -7 J. Determine the energy of the capacitor after filling it with a dielectric with a dielectric constant ε = 2, if:

1) the capacitor is disconnected from the power supply;

2) the capacitor is connected to a power source.

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 С 2 = εС 1.

Four capacitors, capacitors of which 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, the voltage at the terminals of which is U = 10 V, then the energy W3 of the electrostatic field of the capacitor C3 is equal to ... 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 μJ.

Time elapsed: 3 minutes. assessment of the problem: 6 out of 10 points.

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

Two resistors, the resistances of which are R1 = 0.64 Ohm and R2 = 2.56 Ohm, are connected in series for the first time, and in parallel for the second, and after connection are alternately connected to a direct current source. In both cases, the powers released in the outer sections of the circuit are the same. If the current strength at short circuit of this source Ik = 15 A, then the maximum useful power Рmax of the source is equal to ... 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 home page test):

Time elapsed: 6.5 minutes. assessment of the problem: 8 out of 10 points.

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

Electrical 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 electrical capacitance of a solitary conducting sphere of radius R, located in an infinite medium with a dielectric constant,

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

Electric capacity of a flat capacitor:

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

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

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

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 a dielectric constant):

Electric capacity C of capacitors connected in series:

- in general:

where n is the number of capacitors;

- in the case of two capacitors:

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

Electric capacity of parallel-connected capacitors:

- in general: .


– surface charge density, C / m 2.

Capacitor electric field energy:

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

Examples of problem solving

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

Solution... Capacitor capacity by definition
where Q is the charge on the capacitor plates; U is the potential difference of the plates. Replacing in this equality the total potential difference U with the sum U 1 + U 2 voltages on the layers of dielectrics, 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 layers of the dielectric, respectively; D is the dielectric displacement of the field in dielectrics. Multiplying the numerator and denominator of equality (4.2) by 0 and taking into account that D = , we finally get:

(4.3)

Making calculations using formula (4.3), we find:

Example 2. Two identical flat capacitors are connected in parallel and charged to a voltage of 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.

Solution... When capacitors are connected in parallel, their total capacity will be:

C baht = C 1 + C 2 = 2C; (C 1 = C 2 = C).

Battery charge q 1 = C bat U 0 = 2CU 0.

When the distance between the capacitor plates 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 = C ’baht U = 3CU.

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

Tasks

401. Find the electrical capacity C of a solitary metal ball of radius R = 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 ground, taking it for a ball with a radius of R = 6400 km. (Answer: 180 pF).

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

405. Determine the electrical capacity 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  = 7). (Answer: 6.2 nF).

406. Five capacitors of the same capacity are connected in series to form a battery. A static voltmeter is connected in parallel to one of the capacitors, the capacity of which is half the capacity of each capacitor. The voltmeter reads 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 electrical capacity of the capacitor (dielectric constant of mica  = 7, ebonite  = 3). (Answer:

408. N spherical drops of radius r are charged to the same potential φ 0. All drops merge into one large one. Determine the potential and charge density at the surface of the large droplet. (Answer: ).

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

410. A slab of paraffin with a thickness of d = 1 cm was inserted into a flat condenser, which adheres tightly to its plates. How much should the distance between the plates be increased to get the same capacity? (Dielectric constant of paraffin  = 2). (Answer: 0.5 cm).

411. The capacitor consists of two concentric spheres. The radius R 1 of the inner sphere is 10 cm, of the outer R 2 = 10.2 cm. The space between the spheres is filled with paraffin. The inner sphere is imparted with a charge of Q = 5 μC. Determine the potential difference U between the spheres. (Dielectric constant of paraffin = 2). (Answer: 4, 41 kV).

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

413. Two capacitors with electrical capacities C 1 = 3 μF and C 2 = 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. The capacitor with electrical capacity C1 = 0.2 μF was charged to a potential difference U 1 = 320 V. After it was connected in parallel with the second capacitor charged to a potential difference U 2 = 450 V, the voltage U across it changed to 400 V. Calculate the capacity C 2 of the second capacitor. (Answer:
).

415. A capacitor with an electrical capacity of C 1 = 0.6 μF was charged to a potential difference U 1 = 300 V and connected to the second capacitor with an electrical capacity of C 2 = 0.4 μF, charged to a potential difference U 2 = 150 V. Find the charge ΔQ flowed 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 cm 2. Dielectric - glass ( = 7). What is the thickness of the glass? (Answer: 2, 32 mm).

417. The capacitors are connected as shown in fig. 4.2. The capacitances of the capacitors: C 1 = 0.2 μF, C 2 = 0.1 μF, C 3 = 0.3 μF, C 4 = 0.4 μF. Determine the electrical capacity C of the capacitor bank. (Answer: 0.21 μF).

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 electrical capacity C of the capacitor bank. (Answer: 20 pF).

419. The capacitors are connected as shown in fig. 4.4. The electrical capacitances of the capacitors: C 1 = 2 μF, C 2 = 2 μF, C 3 = 3 μF, C 4 = 1 μF. The potential difference on the plates of the fourth capacitor U 4 = 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 electrical capacity C of the capacitor bank. (Answer: 2 pF. Indication. Prove that if C 1 / C 2 = C 3 / C 4, then φ A = φ B, and, therefore, the capacity C 5 does not matter in determining the total capacity of the circuit).

421. A flat capacitor, between the plates of which there is a dielectric plate of permeability , is connected to the accumulator. The capacitor charge is Q 0. What charge ΔQ will pass through the battery when the record 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 is S = 100 cm 2, the distance between them is d = 1 mm. (Answer:
).

423. On the plates of the flat capacitor, the charge is uniformly distributed with the surface density σ = 0.2 µC / m 2. The distance d between the plates is 1 mm. How much will the potential difference across its plates change with an increase in the distance d between the plates to 3 mm? (Answer: 22.6V).

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 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. What amount of heat Q will be released during the discharge of a flat capacitor, if the potential difference U between the plates is 15 kV, the distance d = 1 mm, the dielectric is mica and the area S of each plate is 300 cm 2. (Answer:

427. The force F-attraction between the plates of a flat air condenser 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 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, by removing the plates from each other, increase the distance between them to d 2 = 3.5 cm. (Answer: 2.5 J / m 3).

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

430. The capacitance C of a flat capacitor is 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 needs to be done to remove the dielectric from the capacitor? Friction is negligible. (Answer: 0.18 J).

Objective 1. Electric capacity of the capacitor connected to the source constant voltage U = 1000 V, equal to C 1 = 5 pF. The distance between its plates was reduced by n = 3 times. Determine the change in charge on the plates of the capacitor and energy 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 = (nC 1 - C 1) U = (n - 1) C 1 U = 10 -8 Cl.

Objective 2. Capacitor charge q = 3 10 -8 Cl. The capacity of the capacitor is C = 10 pF. Determine the speed that an electron acquires as it travels in a capacitor from one plate to another. The initial velocity of the electron is zero. Specific charge of an electron

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

Hence,

Finally

Objective 3. Four capacitors with capacities C 1 = C 2 = = 1 μF, C 3 = 3 μF, C 4 = 2 μF are connected, as shown in Figure 14.46. A 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, we first of all find the capacity of the capacitor bank. The equivalent capacity of the second and third capacitors C 2,3 = 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, will be found from the ratio

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

The charges on these capacitors are the same:

q 1 = q 2,3 = q 4 = Ceq = 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, similarly, we have q 4 = 8 10 -5 C, U 4 = q 4 / C 4 = 40 V.

Let's find the voltage across the second and third capacitors: U 2 = U 3 = q 2,3 / C 2,3 = 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 capacity, 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 you can connect these points or exclude the capacitors connected to these points, since they cannot accumulate charge (Δφ = 0) and, therefore, do not play a role in the distribution of charges ...

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

Task 5. Energy of a flat air condenser W 1 = 2 10 -7 J. Determine the energy of the capacitor after filling it with a dielectric with a dielectric constant ε = 2, if:

1) the capacitor is disconnected from the power supply;

2) the capacitor is connected to a power source.

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 С 2 = εС 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 capacitor charge.

2. An electron flies into the space between the plates of a flat capacitor at a speed of 2-10 7 m / s, directed parallel to the plates of the capacitor. How far towards the positively charged plate will the electron be displaced 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 plates of the capacitor is 0.02 m. The ratio of the modulus of the electron charge to its mass is 1.76 10 11 C / kg.

3. The flat capacitor was charged using a current source with a voltage of U = 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 = 0.2 mm to d 1 = 0.7 mm?

4. Determine the capacity of the air spherical condenser. The radii of the spheres R 1 and R 2.

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

Sample assignments for the exam

C1. A small ball with a charge q = 4 10 -7 C and a mass of 3 g, suspended on a weightless thread with a coefficient of elasticity of 100 N / m, is located between the vertical plates of the air condenser (see Fig.). The distance between the capacitor plates is 5 cm. What is the potential difference between the capacitor plates if the filament elongation is 0.5 mm?

C2. An electron flies into a flat capacitor L = 5 cm long at an angle a = 15 ° to the plates. The energy of the electron is W = 2.4 10 -16 J. The distance between the plates is d = 1 cm. Determine the potential difference between the plates of the capacitor U, at which the electron at the exit from the capacitor will move parallel to the plates. Electron charge q e = 1.6 10 -19 C.

C3. Capacitors, electrical capacity which are 2 μF and 10 μF, are charged to a voltage of 5 V each, and then the "plus" of one of them is connected to the "minus" of the other and the free leads are connected with a 1000 Ohm resistor. Determine the amount of heat that will be generated in the resistor.

Review Chapter 14 as follows

1. Write down the basic concepts and physical quantities and define them.

2. Formulate laws and write down basic formulas.

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

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

Series connection of capacitors - 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 equal. Due to the natural principle of balance. Only the extreme plates are connected to the source, the others are charged by redistributing charges between them. Using equality, we find:

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

The voltages between the capacitors are distributed in inverse proportion to the nominal capacitances. Both add up to the mains voltage. When discharging, the structure is capable of giving off 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, resulting in a common denominator:

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

Calculating the total battery capacity

When capacitors are connected in series in a battery, values are added that are inverse to the nominal capacitances. Bringing the last expression to a common denominator, turning the fractions, we get:

C = C1C2 / (C1 + C2).

The expression is used to find the capacity of a battery. 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 denominators are put in the denominator, sorting out the combinations. In practice, it is sometimes more convenient to calculate in terms of reciprocal values. The result is to divide by one.

Series connection of 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 rating of the supply network equally by the total number. When powered by a 12 volt battery, 4 capacities, 3 volts will drop on each.

We will make one simplification for the case when the ratings are equal, one capacitor is switched on with a variable to adjust the result. Then the maximum voltage of each element can be approximately found by dividing the voltage of the source by the amount reduced by one. The result will be a result that obviously has a certain margin. Concerning variable capacity, the requirements are much more stringent. Ideally, the operating value will overlap the source voltage.

Serial connection required

At first glance, the idea of connecting capacitors with a battery in a sequential manner seems meaningless. The first advantage is obvious: the requirements for the maximum voltage of the plates are reduced. The higher the working voltage, the more expensive the product. Similarly, the world is seen by a radio amateur who has several low-voltage capacitors on his hands, who wants to use iron part of high voltage circuit.

Calculating the effective stresses of the element using the above formulas, one can easily solve the problem posed. Let's consider an example for greater clarity:

Let a battery with a voltage of 12 volts, three capacities with denominations of 1, 2 and 4 nF be installed. Let's find the voltage when the cells are connected in series by the battery.

To find the three unknowns, take the trouble to draw up an equal number of equations. Known from the course of higher mathematics. The result will look like this:

U1 + U2 + U3 = 12;
U1 / U2 = 2/1 = 2, whence we write down: U1 = 2U2;
U2 / U3 = 4/2 = 2, whence you can see: U2 = 2U

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

4U3 + 2U3 + U3 = 12, whence 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 lower the denomination of the neighbors. From this rule follows: in series connection, low-capacity capacitors have a higher operating voltage. For definiteness, we will find the nominal value of the compiled battery, at the same time we will illustrate the formula, since it is described above purely verbally:

С = С1С2С3 / (С1С2 + С2С3 + С1С3) = 8 / (2 + 8 + 4) = 8/14 = 571 pF.

The resulting rating is less than each capacitor in the series connection. The rule shows that the maximum impact on the total capacity is less. Therefore, if it is necessary to adjust the total battery rating, it should be variable capacitor... Otherwise, turning the screw will have little effect on the final result.

We see another pitfall: after adjustment, the voltage distribution across the capacitors will change. Consider extreme cases so that the voltage does not exceed the operating value for the cells that make up the battery.

Electrical Circuit Research Software Packages

In addition to online calculators more powerful tools are available for calculating the series connection of capacitors. A big minus of publicly available tools is explained by the unwillingness of sites to check the program code, which means they contain errors. It is bad if one container fails, broken by the process of testing an incorrectly assembled circuit. Not the only drawback. 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 stages. Then, in the diagram, in addition to a short circuit through a resistor to ground, a series connection of capacitors is formed. Usually, the formula shown above is not applied. It is generally accepted that each stage of the filter 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 be cut at the output.

Those wishing to carry out approximate calculations are advised to familiarize themselves with the software package personal computer Electronics Workbench. The construct 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 spelled out in a foreign manner. It does not interfere with the use of the shell, which provides the operator with a mountain of power sources of various kinds.

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

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

Working example

Screenshot shows Electronics Workbench 5.12 desktop with assembled electrical circuit series connection of capacitors. Each with a capacity of 1 μF, identical elements taken for demonstration purposes. So that everyone can easily check the correctness.

Series capacitor bank

Let us first pay attention to the source. Alternating voltage with a frequency of 60 Hz. In the country of the developer, there is a different standard than the Russian ones. Recommended right click mouse click source, visit properties, expose:

The frequency is 50 Hz instead of 60 Hz.
The effective voltage is 220 volts instead of 120.
Take the phase (phase - imitation of reactivity) according to your needs.

It will be useful for literalists to flip through the properties of the chain elements. At the source, you are free to set the voltage tolerance as a percentage. It is enough to add one 1K resistor and the circuit becomes a high-pass filter. It is recommended not to oversimplify the actions. Put the grounding sign correctly, make sure: the scheme is completely trivial. Otherwise, the results will make you break your head for a long time.