Question

How many time constants will elapse before the current in a charging RC circuit drops to half of its initial value? Answer the same question for a discharging RC circuit.

Answer 1

The growth of charge across a capacitor,
$$\begin{gathered} q=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right) \\ q=\frac{Q}{2} \\ \Rightarrow \frac{Q}{2}=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right) \\ \Rightarrow \frac{1}{2}=\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right) \\ \Rightarrow e^{-\frac{t}{RC}}=\frac{1}{2} \\ \Rightarrow \frac{t}{R\mathrm{C}}=\ln 2 \\ \text{L}\mathrm{et}t=nRC \\ \Rightarrow \frac{nRC}{RC}=0.69 \\ \Rightarrow n=0.69 \end{gathered}$$

The decay of charge across a capacitor,
$$\begin{gathered} q=Q{\mathrm{e}}^{-\frac{t}{RC}} \\ q=\frac{Q}{2} \\ \Rightarrow \frac{Q}{2}=Q{\mathrm{e}}^{-\frac{t}{RC}} \\ \Rightarrow \frac{1}{2}={\mathrm{e}}^{-\frac{t}{RC}} \\ \text{L}\mathrm{et}\mathit{}t\mathit{=}nRC \\ \Rightarrow \frac{nRC}{RC}=\ln 2 \\ \Rightarrow n=0.69 \end{gathered}$$