If we take two metal plates, separate them with a dielectric (insulator), and apply a DC voltage between the plates, current will not be able to cross the dielectric. However, a surplus of electrons will build up on the plate connected to the negative terminal of the voltage source, and shortage of electrons will occur on the plate connected to the positive terminal. The voltage source will try to force electrons into one plate (negative terminal) and draw them out of the other (positive terminal).
At some point in time, these plates will become completely saturated; no further electrons can be forced into the negative plate, and no more electrons can be drawn from the positive plate. At this point, the plates have an electrical potential equal to that of the voltage source. In fact, the plates now act like a second voltage source, one in parallel with the first but with the opposite polarity. Fig. 1 shows the equivalent circuit. Obviously, since these opposing voltages are equal, they cancel each other out, and no current can flow between the voltage source and the plates in either direction. The plates are said to be charged.
What happens if the voltage source is removed from the circuit? The answer is that the plates will remain charged because there is no place for the electrons on the negative plate to go. Similarly, there is no place for the positive plate to draw electrons from. In effect, the voltage is stored by the plates. [ILLUSTRATION FOR FIGURE 2 OMITTED].
Replacing the missing voltage source with a resistor, as shown in Fig. 3, provides a current path for the excess electrons stored on the negative plate to flow to the positively charged plate. This current flow will continue until both plates are returned to an electrically neutral state. This is called discharging the plates.
Such a device as noted above (two conductive plates separated by a dielectric) is called a capacitor. It's used to store electrical energy. (Note: At one time a capacitor was known as a condenser, but this term has fallen from use.)
A capacitor can't hold a charge indefinitely. Even air can conduct some current, so the charge will slowly seep off into the air. There will also be some leakage through the dielectric. All other factors being equal, the lower the internal leakage, the better the capacitor.
Alternating current and the capacitor
What happens when we apply an alternating current to our capacitor? During the first part of the cycle, as the source voltage increases from zero, it will charge the plates of the capacitor similar to when a DC voltage is applied, and the polarity of the charge capacitor opposes that of the source voltage.
The capacitor may or may not be completely charged by the time the applied voltage passes its peak and starts to decrease again. This will depend on the size of the plates, how much voltage is applied, and the frequency of the AC signal. In either case, as the applied voltage decreases, a point will be reached when it is less than the charge stored in the capacitor. This will allow the capacitor to start discharging through the AC voltage source.
The capacitor may or may not be completely discharged when the AC voltage reverses polarity, but since the source polarity is the same as the capacitor polarity the voltages aid, quickly discharging the capacitor the rest of the way, then charging it with the opposite polarity from the original charge. When the AC voltage source reverses direction, the capacitor is discharged again, and the entire process is repeated with the next cycle of the AC waveform.
Example of capacitor in an AC circuit
Let's look at Fig. 5. If the voltage source is DC, the lamp will not light because the DC current can't flow through the circuit; it's blocked by the dielectric. In effect, the current "sees" the capacitor as an open circuit.
If this same circuit has an AC voltage source, the lamp will light, indicating that AC current is flowing through the circuit. What's happening here? Remembering our prior discussion of AC voltage application, we know that the process of charging, discharging, and recharging a capacitor from an AC voltage source results in the same effect as if the current was actually flowing through the capacitor itself. Moreover, if we decrease the frequency of the AC voltage source, the lamp will dim; if we increase the frequency, the lamp will burn brighter. Thus, a capacitor lets more current flow as the frequency of the source voltage is increased.
As we've seen, AC current can flow through a circuit with a capacitance. The apparent resistance of a capacitor in an AC circuit is less than its DC resistance. This apparent AC resistance is called capacitive reactance, and its value decreases as the applied frequency increases. A capacitive reactance slows down voltage more than it does current, so the voltage lags the current by 90 degrees (assuming a purely capacitive circuit).
Before we can talk about the equation for calculating capacitive reactance, we should know how capacitance is determined. The basic unit of capacitance is the farad (F). If 1A of current flows when the applied voltages changes at a rate of 1V per sec, we have 1F of capacitance.
With the above in mind, capacitive reactance is calculated by the following equation:
[X.sub.c] = 1 [divided by] 2[Pi]FC
where F is the frequency in hertz and C is the capacitance in farads.
Note that if the voltage source is DC, the applied frequency is zero. Thus the denominator in the equation above is zero. Any number divided by zero is unsolvable; it equals infinity. An infinite resistance in a circuit, of course, acts like an open or incomplete circuit.