How many time constants will elapse before the current in a...

For a charging RC circuit, the current at any time $t$ is given by $i = i_0 e^{-t/\tau}$, where $\tau = RC$ is the time constant and $i_0$ is the initial current.

We are given that the current decreases to half of its initial value, so $i = \dfrac{i_0}{2}$.

$\dfrac{i_0}{2} = i_0 e^{-t/\tau}$

$e^{t/\tau} = 2$

$\dfrac{t}{\tau} = \ln 2 \approx 0.693$

$t \approx 0.69 \tau$

For a discharging RC circuit, the current at any time $t$ is also given by $i = i_0 e^{-t/\tau}$.

Similarly, for the current to decrease to half of its initial value, $i = \dfrac{i_0}{2}$.

$\dfrac{i_0}{2} = i_0 e^{-t/\tau}$

$e^{t/\tau} = 2$

$\dfrac{t}{\tau} = \ln 2 \approx 0.693$

$t \approx 0.69 \tau$

Thus, in both cases, $0.69$ time constants will elapse.