7.6. Capacitors
7.6.3. Change in electrical capacity capacitor and capacitor bank
The capacitance of a capacitor can be changed by increasing or decreasing the distance between its plates, replacing the dielectric in the space between them, etc. In this case, it turns out that the determining factor is whether the capacitor is disconnected or connected to the voltage source.
Permanent hearing loss is very short signals, but also less loud, but exposed for a long time. A noise level of 85 dB for 8 hours is considered safe. 88 dB for 4 hours, 91 dB for 2 hours and 100 dB for 15 minutes.
Decibel. The enormous range of sound intensities we encounter every day makes it impossible to use a linear scale to compare the volume of different sounds. The perceived auditory sensation is approximately proportional to the logarithm of the arousal. This is a large block, so a unit ten times smaller is used - the decibel. The difference in intensity received by our hearing can be up to 130 dB. Power levels, voltages, pressure currents and acoustic velocities can also be specified in decibels.
If the capacitor (or capacitor bank):
- connected to a voltage source, then the potential difference (voltage) between the plates of the capacitor remains unchanged and equal to the voltage at the poles of the source:
U = const;
- disconnected from the voltage source, the charge on the capacitor plates remains unchanged:
Q = const.
The logarithmic measure tells us that a 3 decibel increase in some value on an absolute scale gives us twice the change in that value, and a 20 dB difference gives us an absolute 100 times the change in value. A person begins to hear sound only from a certain sound pressure level. Based on the study, statistical curves of our sensitivity to hearing were constructed depending on the frequency of sounds. The sound pressure level at which the ear begins to hear sound varies for different frequencies.
The most sensitive hearing is for a tone around 4 kHz. For tones with different frequencies, the threshold of sensitivity at which we begin to hear something is different. The graph shows that to listen to a tone such as 100 Hz, the sound pressure level must be 35 dB higher than for a 1 kHz tone, for which the sound pressure level is 0 dB. The same is true at the upper limit of audibility, where the auditory sensation changes into pain. The bandwidth heard by humans can be divided into three parts.
When connected to each other covers of the same name two charged capacitors take place parallel connection.
U = Q total C total,
where Qtot is the charge of the capacitor bank; Ctot is the electrical capacity of the battery;
Ctot = C 1 + C 2,
where C 1 is the electrical capacity of the first capacitor; C 2 - electrical capacity of the second capacitor;
Acoustic vibrations from 16 Hz to 300 Hz are defined as low tones. Many people have difficulty distinguishing sounds, but most of us can easily cope with this. Musicians or those with hearing ability can even determine the musical interval, i.e. the distance between them. And few have the ability to instantly recognize the pitch of a sound. If they are musically educated, they know that, for example, this rare skill is called absolute pitch. This is an innate trait and cannot be acquired.
Some musicians can determine pitch, but most often it is a skill that applies to the instrument they play. A violinist can determine the pitch of a violin, but will have difficulty determining the tone of a piano. We don't have to worry about missing a feature. Let's listen to music for pleasure and our musical sensitivity will develop over time. Beethoven had been losing his hearing since the age of 27, but despite his disability, he wrote several brilliant works.
- total charge
Qtot = Q 1 + Q 2,
When connected to each other different linings two charged capacitors takes place (as in the case of connecting plates of the same name) parallel connection.
The parameters of such a capacitor bank are calculated as follows:
- capacitor bank voltage
U = Q total C total,
We only need the joy of listening. To obtain the appropriate acoustic atmosphere of a room, experts create them in accordance with the principles of acoustics. Already at the design stage he receives correct form, and then uses such materials and finishing elements that allow him to achieve the desired effects. Few of us have the opportunity to build or seriously recreate a room and use expensive systems to improve the room's acoustics. In most cases there is no need, because many of the rooms in which we listen to music have good acoustics, so it is enough to adjust the speakers and a small adjustment in the furniture to satisfy the effects.
where Qtot is the charge of the capacitor bank; Ctotal - battery capacity;
- electrical capacity of a capacitor bank
Ctot = C 1 + C 2,
where C 1 is the electrical capacity of the first capacitor; C 2 - electrical capacity of the second capacitor;
- total charge
Q total = |Q 1 − Q 2 |,
where Q 1 is the initial charge of the first capacitor, Q 1 = C 1 U 1 ; U 1 - voltage (potential difference) between the plates of the first capacitor before connection; Q 2 - initial charge of the second capacitor, Q 2 = C 2 U 2 ; U 2 - voltage (potential difference) between the plates of the second capacitor before connection.
To improve the acoustics of “difficult” rooms or when we want to achieve excellent results, we can apply appropriate “acoustic structures”, that is, elements that, depending on their design, are designed to absorb or dissipate the sound wave. Some designs combine both functions at the same time.
There are many companies that produce Acustic systems. They use their own solutions and different names companies of the same system. Some of these names have been adopted and function among audiophiles as common name. The simplest way of absorption is high tones, the most complex. Medium enough for the slightly “fluffy” surface of our sofa, carpet or even big flower. To absorb the lowest tones, we must use a very thick layer of absorbent material or even special bass traps.
Example 17. Two capacitors of the same electrical capacity are charged to a potential difference of 120 and 240 V, respectively, and then connected by similarly charged plates. What will be the potential difference between the plates of the capacitors after the indicated connection?
Solution . Before connecting the capacitor plates of the same name, each of them had a charge:
The acoustic system uses two absorbing materials: mineral wool different varieties and open porous polyurethane sponge. The rock wool material is framed in a different frame size and covered with sound permeable canvas. The surface may be flat or convex, which further distracts the sound wave. On the other hand, polyurethane sponge is cut into different sizes with different surface, for example, with pyramids, waves and faults. The varied surface further diffuses sound.
As we can see, this is not complex designs, we can easily make them ourselves, at least with mineral wool. Attenuation of low tones Sound waves below 300 Hz are absorbed to a small extent by elements of indoor equipment. To absorb it, you need to use a special "bass trap" design called an English "bass trap". Typically the resonator has a design and appearance subwoofer enclosures with one or more bass reflex holes tuned to the attenuation frequency.
- first capacitor -
- second capacitor -
When connecting plates of the same name, we obtain a parallel connection of capacitors. The potential difference between the plates of a capacitor bank is determined by the formula
U = Q total C total,
The total charge of a battery of two capacitors obtained by connecting their plates of the same name is determined by the sum of the charges of each of them:
It is placed in the corners of the room behind the listener, where there are arrows of low-frequency standing waves. This is the only design that can accurately combat very low, specific frequency sounds. Helmholtz resonators are relatively easy to make at home. Dimensions and settings are carried out according to the same principle as when building loudspeakers or a subwoofer. Calculate a Bass Trap page or look online. Plate absorbers - the structure resembles a fairly flat box with a certain capacity, in which one wall is a membrane.
Qtot = Q 1 + Q 2,
U = Q tot C tot = Q 1 + Q 2 2 C = C U 1 + C U 2 2 C = U 1 + U 2 2.
Let's calculate:
U = 120 + 240 2 = 180 V.
The potential difference between the capacitor plates after this connection will be 180 V.
Example 18. Two identical flat capacitors are charged to a potential difference of 200 and 300 V. Determine the potential difference between the plates of the capacitors after connecting their opposite plates.
Depending on the weight and surface of the membrane, the capacity of the box and the type of filling, different frequencies may be suppressed. Unlike Helmholtz resonators, they operate relatively broadband and cannot be tuned to a specific frequency. They are placed on the wall behind the speakers or behind the listener. Stream "sinks". It is a kind of box or tube that is open at both ends. The holes are blinded in different thicknesses with materials with open pores. Interior the box is filled with damping material.
The operation is relatively broadband and cannot be tuned to a specific frequency. They are placed in the corners of the room behind the listener. Scattering Scattering is the reduction or elimination of reflections of sound waves from large surfaces. Without proper dissipation, the listener will receive very clear individual reflections, sometimes even causing echoes. Diffusing elements are located on the side walls, and even the ceiling, instead of the first reflections, as well as behind the columns, sometimes in the corners of the room. The most famous still is the egg stamp, located on the wall or ceiling.
Solution . Before connecting opposite plates of capacitors, each of them had a charge:
- first capacitor -
Q 1 = C 1 U 1 = CU 1,
where C 1 is the electrical capacity of the first capacitor, C 1 = C; U 1 - potential difference between the plates of the first capacitor;
- second capacitor -
Q 2 = C 2 U 2 = CU 2,
where C 2 is the electrical capacity of the second capacitor, C 2 = C; U 2 is the potential difference between the plates of the second capacitor.
More professional ones are built from sponge with a varied surface, and also in the form of a half-cylinder, which is excellent at interrupting the continuity of the sound wave. They are also based on wooden structure, filled with appropriate damping material and covered with canvas, and external shape reflects and distracts the wave. These are relatively simple and inexpensive acoustic elements. Will this do anything?
Professional research uses many types of systems, both dissipative and damping. Does it make sense to use them in your home? Yes, if we use them correctly. If the room has properly arranged furniture and other objects, their use may be unnecessary, or only those that we lack to achieve certain effects are sufficient. But if we have almost Empty room with obvious for a long time reverb, with booming bass, use political systems - the only way improve acoustics.
When connecting opposite plates, we obtain a parallel connection of capacitors. The potential difference between the plates of a capacitor bank is determined by the formula
U = Q total C total,
where Q total is the total battery charge; C total - the total electrical capacity of the battery.
The total charge of a battery of two capacitors obtained by connecting their opposite plates is determined by the modulus of the charge difference of each of them:
The impact of using acoustic elements is very easily audible and can also be measured using appropriate instruments. Measurements show a significant reduction in reverberation time and a small reduction in volume due to the lower proportion of reflected waves. When using bass traps, the quality of low tones improves, and the characteristic rumbling resulting from wave interference disappears. Room attenuation causes a significant reduction in background noise, resulting in a significantly improved listening environment.
Q total = |Q 1 − Q 2 |,
and the total electrical capacity of a battery of two identical capacitors connected in parallel is
Ctot = C1 + C2 = 2C.
Therefore, the potential difference between the battery plates is determined by the expression
U = Q tot C tot = |
Let's calculate:
Q 1 − Q 2 |
2 C = | C U 1 − C U 2 | 2 C = |
U 1 − U 2 |
2. U = | 200 − 300 |
They do not occupy the entire height of the room, but only a belt about 1 m wide at the height of our ears. The bright, smaller quarter tones in the corners are "cushions" of damping material in the corners under the ceiling. Many users of audio kits have experienced an unpleasant surprise when, while listening to music, the speakers, usually high-frequency speakers, suddenly “burn out.” This happens to young people who like to listen to loud music with muted bass and high frequencies. Speakers overheating. Speakers for home use have different characteristics loads than professional speakers.
Solution . When a metal plate is placed in a flat capacitor as shown in the figure, the free electrons in the metal are redistributed:
- the plane facing the positively charged capacitor plate receives an excess of electrons and is charged with a negative charge q 1 = −q;
- the plane facing the negatively charged capacitor plate has a lack of electrons and is charged with a positive charge q 2 = +q.
As a result of charge redistribution, the plate remains neutral:
They are not suitable for long, loud games. Deactivation is a high amplitude and high power distorted signal generated by an amplifier. If the source supplies more voltage to the amplifier input than the manufacturer recommends, even with slight overclocking, the sound from the speakers will be very distorted. Each amplifier gives full power at a certain position of the potentiometer knob. However, we must remember that the potentiometer is not a regulator of the power of the amplifier, but a regulator of the volume of the music being listened to.
Power given by the amplifier in this moment, depends on the type of music. Listening loudly to quiet parts of a song does not cause the amplifier to produce high power, but energetic, powerful music can easily cause overload, i.e. give more, instant power loudspeakers and therefore a bloated speaker. Paradoxically, the weaker the amplifier, the easier it is to overload it. If we connect speakers, for example, with a 100 W amplifier with a power of 30 W, then to achieve greater volume we “overclock” the potentiometer to the maximum.
Q = q 1 + q 2 = −q + q = 0.
The redistribution of charge in the metal plate leads to the formation of a bank of two capacitors:
- the positively charged capacitor plate and the negatively charged plane of the metal plate have charges of the same magnitude and opposite sign; they can be considered as a capacitor with electrical capacitance
C 1 = ε 0 S d 1 ,
where ε 0 is the electrical constant, ε 0 = 8.85 ⋅ 10 −12 C 2 /(N ⋅ m 2); S is the area of the capacitor plate; d 1 is the distance between the positively charged capacitor plate and the negatively charged plane of the metal plate;
- the negatively charged capacitor plate and the positively charged plane of the metal plate also have charges of the same sign of the same magnitude; they can be considered as a capacitor with electrical capacitance
C 2 = ε 0 S d 2 ,
where d 2 is the distance between the negatively charged capacitor plate and the positively charged plane of the metal plate.
Both capacitors have equal charges and form a series connection. Electric capacity of a battery of two capacitors at serial connection is determined by the formula
1 C total = 1 C 1 + 1 C 2, or C total = C 1 C 2 C 1 + C 2.
With a symmetrical arrangement of the plate in the space between the plates of the capacitor (d 1 = d 2 = d), the electrical capacitances of the capacitors are the same:
C 1 = C 2 = ε 0 S d ,
the total electrical capacity of the battery is given by the expression
Ctot = C 1 C 2 C 1 + C 2 = C 2 = ε 0 S 2 d,
where d = (d 0 − a )/2; d 0 - the distance between the plates of the capacitor before the insertion of the plate; a is the thickness of the metal plate.
Potential difference between battery plates
U = Q total C total = 2 d q ε 0 S = q (d 0 − a) ε 0 S ,
where Qtot is the charge of a battery of series-connected capacitors, Qtot = q.
The initial potential difference is determined by the formula
U 0 = Q 0 C 0 = Q 0 d 0 ε 0 S ,
where Q 0 is the charge of the capacitor before inserting the plate, Q 0 = q (the capacitor is disconnected from the voltage source); C 0 is the electrical capacity of the capacitor before inserting the plate.
The ratio of the potential difference before and after the introduction of the metal plate is determined by the expression
U U 0 = d 0 − a d 0 .
From here we find the required potential difference
U = U 0 d 0 − a d 0 .
Taking into account d 0 = 3a, the expression takes the form:
U = U 0 3 a − a 3 a = 2 3 U 0 .
Let's calculate:
U = 2 3 ⋅ 180 = 120 V.
As a result of introducing a metal plate into the capacitor, the potential difference between its plates decreased and amounted to 120 V.
Example 20. A flat-plate air capacitor is charged to 240 V and disconnected from the voltage source. It is vertically immersed in some liquid with a dielectric constant of 2.00 per one-third of the volume. Find the potential difference that will be established between the plates of the capacitor.
Solution . When a flat air capacitor is partially immersed in a liquid dielectric, as shown in the figure, free electrons on its plates are redistributed in such a way that:
- part of the capacitor plates immersed in the dielectric has a charge q 1;
- the part of the capacitor plates remaining in the air has a charge q 2.
As a result of charge redistribution over the area of the capacitor plates, a charge is established on its plates:
Qtot = q 1 + q 2.
The area of the capacitor plates when partially immersed in a liquid dielectric is divided into two parts:
- the part immersed in the dielectric has an area S 1 ; the corresponding part of the capacitor can be considered as a separate capacitor with electrical capacitance
C 1 = ε 0 ε S 1 d,
where ε 0 is the electrical constant, ε 0 = 8.85 ⋅ 10 −12 C 2 /(N ⋅ m 2); ε is the dielectric constant of the capacitor; d is the distance between the capacitor plates;
- the part remaining in the air has an area S 2 ; the corresponding part of the capacitor can be considered as a separate capacitor with electrical capacitance
C 2 = ε 0 S 2 d.
Both capacitors have the same potential difference between the plates and form a parallel connection. Electric capacity of a battery of two capacitors at parallel connection is determined by the formula
C total = C 1 + C 2 = ε 0 ε S 1 d + ε 0 S 2 d = ε 0 d (ε S 1 + S 2),
and the charge on the battery plates is
Q total = C total U = ε 0 d (ε S 1 + S 2) U,
where U is the potential difference between the battery plates.
The electrical capacity of a capacitor before it is immersed in a dielectric is determined by the expression
C 0 = ε 0 S 0 d ,
and the charge on its plates is
Q 0 = C 0 U 0 = ε 0 S 0 d U 0 ,
where U 0 is the potential difference between the plates of the capacitor before the introduction of the plate; S 0 - lining area.
The capacitor is disconnected from the voltage source, so its charge does not change after partial immersion in the dielectric:
Q 0 = Q total,
or, explicitly,
ε 0 S 0 d U 0 = ε 0 d (ε S 1 + S 2) U .
After simplification we have:
S 0 U 0 = (εS 1 + S 2)U .
It follows that the desired potential difference is determined by the expression
U = U 0 S 0 ε S 1 + S 2 .
Taking into account the fact that part of the capacitor plates is immersed in the dielectric, i.e.
S 1 = ηS 0 , S 2 = S 0 − S 1 = S 0 − ηS 0 = S 0 (1 − η), η = 1 3 ,
U = U 0 S 0 ε η S 0 + S 0 (1 − η) = U 0 ε η + 1 − η .
From here we find the required potential difference:
U = 240 2.00 ⋅ 1 3 + 1 − 1 3 = 180 V.
“Problems on electric current” - Basic formulas. Tasks. Work formula electric current...Tasks of the first level. Electricity. Current strength. Purpose of the lesson: Terminological dictation. Quiz. 2. There are two lamps with a power of 60 W and 100 W, designed for a voltage of 220V. Voltage. Physics lesson: generalization on the topic “Electricity”.
“Application of capacitors” - Rig radio station plateau. Lapel microphone. LED clusters and modules, flexible LED strips. Condenser microphone. A person has the capacity of a sphere with a radius of 30 cm. More information about the supplied products can be found on the website www.e-neon.ru. Current rectifier circuit. Capacitor CTEALTG STC - 1001.
Charges are separated by performing mechanical work. Solar batteries are made from photocells. Used in solar powered, light sensors, calculators, video cameras. Fixing the material. First electric battery appeared in 1799. Sealed small-sized batteries (SMA).
“Electric current” - Electrolytic action of current. Separation of the victim from the live part that is energized. Biological effects current Factors influencing the outcome of electric shock. Be careful when handling electricity! Mechanical action of current. General electrical injuries.
"Work and Current Power" - James Watt. Units of work. Power units. Learn to determine the power and operation of current. i=P/u. A=P*t. Work of electric current. James Joule. Calculate the energy consumed (1 kWh costs 1.37 rubles). Work and power of electric current. The power of an electric current is the work done by the current per unit of time.
“Alternating currents” - The Hopkinson brothers developed the theory of electromagnetic circuits. Power frequency value alternating current due to technical and economic considerations. AC voltage converted to DC by a semiconductor rectifier. In 1848, the French mechanic G. Ruhmkorff invented the induction coil.