Capacitance Help (page 3)
Capacitance impedes the flow of ac charge carriers by temporarily storing the energy as an electrical field. This energy is given back later. Capacitance generally is not important in pure dc circuits, but it can have significance in circuits where dc is pulsating and not steady. Capacitance, like inductance, can appear when it is not wanted or intended. Capacitive effects become more evident as the frequency increases.
The Property Of Capacitance
Imagine two huge flat sheets of metal that are excellent electrical conductors. Suppose that they are each the size of the state of Nebraska and are placed one over the other, separated uniformly by a few centimeters of air. If these two sheets of metal are connected to the terminals of a battery, they will become charged, one positively and the other negatively. This will take a little while because the sheets are so big.
If the plates were small, they would both become charged almost instantly, attaining a potential difference equal to the voltage of the battery. However, because the plates are gigantic, it takes awhile for the negative plate to “fill up” with electrons, and it also takes some time for the positive plate to get electrons “sucked out.“
Ultimately, the potential difference between the two plates becomes equal to the battery voltage, and an electrical field exists in the space between the plates. This electrical field is small at first; the plates don’t charge right away. However, the field increases over a period of time, depending on how large the plates are, as well as on how far apart they are. Energy is stored in this electrical field. Capacitance is a manifestation of the ability of the plates, and of the space between them, to store this energy. In formulas, capacitance is symbolized by the italicized uppercase letter C .
It is out of the question to make a capacitor of the preceding dimensions. However, two sheets or strips of foil can be placed one atop the other, separated by a thin, nonconducting sheet such as paper, and then the whole assembly can be rolled up to get a large effective surface area. When this is done, the electrical flux becomes great enough that the device exhibits significant capacitance. Two sets of several plates can be meshed together, with air in between them, and the resulting capacitance is significant at high ac frequencies.
In a capacitor, the electrical flux concentration is multiplied when a dielectric of a certain type is placed between the plates. Some plastics work well for this purpose. The dielectric increases the effective surface area of the plates so that a physically small component can be made to have a large capacitance. Capacitance is directly proportional to the surface area of the conducting plates or sheets. Capacitance is inversely proportional to the separation between conducting sheets; the closer the sheets are to each other, the greater is the capacitance. The capacitance also depends on the dielectric constant of the material between the plates. This is the electrostatic equivalent of magnetic permeability. A vacuum has a dielectric constant of 1. Dry air is about the same as a vacuum. Some substances have high dielectric constants that multiply the effective capacitance many times.
In theory, if the dielectric constant of a material is x , then placing that material between the plates of a capacitor will increase the capacitance by a factor of x compared with the capacitance when there is only dry air or a vacuum between the plates. In practice, this is true only if the dielectric is 100 percent efficient—if it does not turn any of the energy contained in the electrical field into heat. It is also true only if all the electrical lines of flux between the plates are forced to pass through the dielectric material. These are ideal scenarios, and while they can never be attained absolutely, many manufactured capacitors come close.
The Unit Of Capacitance
When a battery is connected between the plates of a capacitor, it takes some time before the electrical field reaches its full intensity. The voltage builds up at a rate that depends on the capacitance. The greater the capacitance, the slower is the rate of change of voltage in the plates.
The unit of capacitance is an expression of the ratio between the amount of current flowing and the rate of voltage change across the plates of a capacitor. A capacitance of 1 farad , abbreviated F, represents a current flow of 1 ampere (1 A) while there is a potential-difference increase or decrease of 1 volt per second (1 V/s). A capacitance of 1 F also results in 1 V of potential difference for an electric charge of 1 coulomb (1 C).
The farad is a huge unit of capacitance. You’ll almost never see a capacitor with a value of 1 F. Commonly employed units of capacitance are the microfarad (μF) and the picofarad (pF). A capacitance of 1 μF represents 0.000001 (10 −6 ) F, and 1 pF is 0.000001 μF, or 10 −12 F.
Capacitors In Series
With capacitors, there is rarely any significant mutual interaction. At very high ac frequencies, however, interelectrode capacitance can sometimes be a problem for engineers. This effect, which shows up as an inherent tiny capacitance between wires that run near and parallel to each other, is almost always undesirable in practical circuits.
Capacitors in series add together like resistors or inductors in parallel. If you connect two capacitors of the same value in series, the result is half the capacitance of either component alone. In general, if there are several capacitors in series, the composite value is less than that of any of the single components. It is important that you always use the same size units when determining the capacitance of any combination. Don’t mix microfarads with picofarads. The answer you get will be in whichever size units you use for the individual components.
Suppose that you have several capacitors with values C 1 , C 2 , C 3 ,..., C n connected in series (Fig. 15-6). You can find the reciprocal of the total capacitance 1/ C using the following formula:
1/ C = 1/ C 1 + 1/ C 2 + 1/ C 3 + ... + 1/C n
The total capacitance C is found by taking the reciprocal of the number you get for 1/ C .
Fig. 15-6 . Capacitances in series add like resistances or inductances in parallel.
Capacitors In Series Practice Problems
Two capacitors, with values of C 1 = 0.10 μF and C 2 = 0.050 μF, are connected in series. What is the total capacitance?
Using the preceding formula, first find the reciprocals of the values. They are 1/ C 1 = 10 μF −1 and 1/ C 2 = 20 μF −1 . (“Reciprocal microfarads” don’t have any practical meaning, but using them can help us remember that we must take the reciprocal of the sum of the numbers before we come up with capacitance.) Then
1/ C = 10 μF −1 + 20 μF −1 = 30 μF −1
C = 1/30 μF −1 = 0.033 μF
Two capacitors with values of 0.0010 μF and 100 pF are connected in series. What is the total capacitance?
Convert to the same size units. A value of 100 pF represents 0.000100 μF. Then you can say that C 1 = 0.0010 μF and C 2 = 0.000100 μF. The reciprocals are 1/ C 1 = 1000 μF −1 and 1/ C 2 = 10,000 μF −1 . Therefore:
1/ C = 1000 μF −1 + 10,000 μF −1 = 11,000 μF −1
C = 0.000091 μF
This number is a little awkward, and you might rather say that it is 91 pF.
In the preceding problem, you can choose picofarads to work with rather than microfarads. In either case, there is some tricky decimal placement involved. It’s important to double-check calculations when numbers get like this. Calculators will take care of the decimal placement problem, sometimes using exponent notation and sometimes not, but a calculator can only work with what you put into it. If you enter a wrong number, you will get a wrong answer, and if you miss a digit, you’ll be off by a factor of 10 (an order of magnitude).
Capacitors In Parallel
Capacitances in parallel add like resistances or inductances in series (Fig. 15-7). That is, the total capacitance is the sum of the individual component values. Again, you need to be sure that you use the same size units all the way through.
Capacitors In Parallel Practice Problem
Three capacitors are set up in parallel, having values of C 1 = 0.100 μF, C 2 = 0.0100 μF, and C 3 = 0.00100 μF. What is C , the total capacitance?
Add them up: C = 0.100 μF + 0.0100 μF + μF 0.00100 = 0.11100 μF. Because the values are given to three significant figures, the final answer should be stated as C = 0.111 μF.
Practice problems of these concepts can be found at: Alternating Current Practice Test
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