In a discharging R – C circuit, the electric current is given by i...

The time constant of the circuit is given by $\tau = RC$.

Substituting the given values:

$\tau = (6.00 \times 10^5) \times (0.500 \times 10^{-6}) = 0.30 \text{ s}$

The current at time $t$ is given by:

$i = i_0 e^{-t/RC}$

Substituting $i_0 = 2.00 \text{ A}$ and $t = 0.31 \text{ s}$:

$i = 2.00 e^{-0.31/0.30} = 2.00 e^{-31/30}$

This can be rewritten as:

$i = 2.00 e^{-1} e^{-1/30}$

Using the approximation $e^{-x} \approx 1 - x$ for small $x$, we get $e^{-1/30} \approx 1 - \dfrac{1}{30} = \dfrac{29}{30}$.

Therefore, the current is approximately:

$i \approx \dfrac{2.00}{e} \left(\dfrac{29}{30}\right) = \dfrac{58}{30e} = \dfrac{5.8}{3e} \text{ A}$