When a capacitor is connected to a voltage source, it takes a certain length of time for the capacitor to become fully charged. If a high resistance is connected in series with the capacitor, the time for charging is increased.

For any given circuit containing capacitance and resistance only, the time in seconds required to charge the capacitor to 63.2 percent of its full charge is called the time constant for that circuit.

This same time constant applies when the capacitor is discharged through the same resistance and is the time required for the capacitor to lose 63.2 percent of its charge.

The charging and discharging of a capacitor in terms of time constants is illustrated in the graph of Fig. 3.8. It will be noted that it takes six time constants to charge the capacitor to 99.8 percent of full charge.

The discharge curve is the exact reverse of the charge curve. When the capacitor is short circuited, it will lose 63.2 percent of its charge in one time constant and almost 99.8 percent of its charge in six time constants.

To determine the length of a time constant in seconds for any particular capacitor-resistance circuit, it is necessary to multiply the capacitance in microfarads by the resistance in megohorns, that is,

T = CR

As an example of how the time constant may be used in determining the performance of a capacitance-resistance circuit, we shall assume that a 20-pf capacitor is connected in series with a 10,000-ohm resistor and that 110 volts is applied to the circuit at intervals off sec.

The time constant is equal to 20 x 0.01 or 0.2 sec. (Note that 10,000 ohms is equal to 0.01 megohm.)
The time interval is given as sec, hence, the number of time constants is 2.5.

If we examine a time constant chart or graph, we find that the voltage at 2.5 time constants will be approximately 92 percent of full voltage. Applying this to our problem, 92 percent of 110 volts is approximately 101 volts. Thus, we find that the capacitor in this problem will charge to approximately 10 1 volts.


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