21.6 Dc circuits containing resistors and capacitors  (Page 2/10)

Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. [link] (b) shows a graph of capacitor voltage versus time ( $t$ ) starting when the switch is closed at $t=0$ . The voltage approaches emf asymptotically, since the closer it gets to emf the less current flows. The equation for voltage versus time when charging a capacitor $C$ through a resistor $R$ , derived using calculus, is

where $V$ is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the natural logarithm. Note that the units of $\text{RC}$ are seconds. We define

where $\tau$ (the Greek letter tau) is called the time constant for an $\text{RC}$ circuit. As noted before, a small resistance $R$ allows the capacitor to charge faster. This is reasonable, since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor $C$ , the less time needed to charge it. Both factors are contained in $\tau =\text{RC}$ .

More quantitatively, consider what happens when $t=\tau =\text{RC}$ . Then the voltage on the capacitor is

This means that in the time $\tau =\text{RC}$ , the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time $\tau$ . It is a characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time, $\tau$ . In just a few multiples of the time constant $\tau$ , then, the final value is very nearly achieved, as the graph in [link] (b) illustrates.

Discharging a capacitor

Discharging a capacitor through a resistor proceeds in a similar fashion, as [link] illustrates. Initially, the current is $I_0=\frac{V_0}{R}$ , driven by the initial voltage $V_0$ on the capacitor. As the voltage decreases, the current and hence the rate of discharge decreases, implying another exponential formula for $V$ . Using calculus, the voltage $V$ on a capacitor $C$ being discharged through a resistor $R$ is found to be

The graph in [link] (b) is an example of this exponential decay. Again, the time constant is $\tau =\text{RC}$ . A small resistance $R$ allows the capacitor to discharge in a small time, since the current is larger. Similarly, a small capacitance requires less time to discharge, since less charge is stored. In the first time interval $\tau =\text{RC}$ after the switch is closed, the voltage falls to 0.368 of its initial value, since $V=V_0\cdot e^{-1}=0\text{.}\text{368}{V}_0$ .