Question 2.17: Discharge of a capacitor...
Discharge of a capacitor
Step-by-Step
Consider the circuit of Figure 2.28. Before the switch is closed, the capacitor has a voltage V across it. Suppose the switch is closed at lime t = 0. A current then flows in the circuit and the voltage, v, across the capacitor decays with time. The voltage across the capacitor is given by
v= \begin{cases}V & t<0 \\ V \mathrm{e}^{-t /(R C)} & t \geqslant 0\end{cases}The quantity RC is known as the time constant of the circuit and is usually denoted by π. So
v= \begin{cases}V & t<0 \\ V \mathrm{e}^{-t / \tau} & t \geqslant 0\end{cases}If π is small, then the capacitor voltage decays more quickly than if π is large. This is illustrated in Figure 2.29.
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