MM0704, Lesson 1
charged after five time constants. In other words, when t = 5TCs, the transient action is complete. One steady state has
been altered and is now settled into another steady state: transience. From this, you can see that the value of R and C
determines the time required to charge the capacitor. Remember that the product of R and C is referred to as a time
constant, and this product is a measure of time as follows:
R () X C (fd) = time constants (sec),
R (M) X C (fd) = time constants (sec),
R () X C (fd) = time constants (sec).
The last equation above is the one most commonly used in electronics.
Discharging a Capacitor Through a Resistor.
If a charged capacitor
is placed across
1-15A, the capacitor will begin to discharge. During discharge, the capacitor voltage and resistor voltage is the same
(EC = ER). At the first instant, the voltage is 100 V and the current through meter A is 10 mamps. After 1 TC, (R x C
= 103 x .1 = 1,000 sec) the voltage is 36.8 V and the current is 3.68 mamps (63.2 percent change). See figure 1-15B.
After 2 TCs, 2,000 sec, the voltage is 13.5 V and the current is 1.35 mamps (another 63.2 percent change, 86.5
percent total change). See figure 1-15C. After 3 TCs, 3,000 sec, the voltage is 4.9 V and the current is 0.49 mamps.
After 4 TCs, E = 1.8; after 5 TCs, E = 0.0, and for all practical purposes, the capacitor is discharged and the current is
Universal Time Constant Graph. Thus far,
it is possible
to find the resistor and capacitor voltage and the current
series RC circuit with direct current applied for any whole number of time constants. Suppose, however, you want to
know this information for a fractional number of time constants. For example, use the circuit shown in figure 1-16 and
find the EC, ER, and I after 1,500 sec (1.5 TC).
You know that EC is 63.2 V after 1,000 sec (1 TC) and 86.5 V after 2,000 sec (2 TCs). Remembering that the
capacitor does not charge at a linear rate,