A capacitor with a capacitance of 8.0 is wired to a battery of emf...

The current in an RC circuit during charging is given by the equation:

$I(t) = I_0 e^{-t/\tau}$

where $I_0 = \dfrac{V}{R}$ is the initial current and $\tau = RC$ is the time constant.

At exactly one time constant, $t = \tau$, the current becomes:

$I(\tau) = I_0 e^{-1} = \dfrac{I_0}{e}$

The initial current $I_0$ is:

$I_0 = \dfrac{6.0}{24} = 0.25 \text{ A}$

Substituting $I_0$ and the value of $e \approx 2.718$ into the equation for current:

$I(\tau) = \dfrac{0.25}{2.718} \approx 0.09 \text{ A}$