In a charging RC circuit, determine the number of time constants...

The energy stored in a capacitor at any time $t$ is given by $U = \dfrac{q^2}{2C}$, where $q$ is the charge on the capacitor at time $t$.

The maximum (equilibrium) energy stored in the capacitor is $U_0 = \dfrac{q_0^2}{2C}$, where $q_0$ is the maximum charge.

Given that the energy attains half of its equilibrium value:

$U = \dfrac{U_0}{2}$

$\dfrac{q^2}{2C} = \dfrac{1}{2} \dfrac{q_0^2}{2C}$

$q = \dfrac{q_0}{\sqrt{2}}$

For a charging RC circuit, the charge as a function of time is:

$q = q_0(1 - e^{-t/\tau})$

Substituting $q = \dfrac{q_0}{\sqrt{2}}$:

$\dfrac{q_0}{\sqrt{2}} = q_0(1 - e^{-t/\tau})$

$1 - e^{-t/\tau} = \dfrac{1}{\sqrt{2}}$

$e^{-t/\tau} = 1 - \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}-1}{\sqrt{2}}$

$e^{t/\tau} = \dfrac{\sqrt{2}}{\sqrt{2}-1} = \sqrt{2}(\sqrt{2}+1) = 2 + \sqrt{2}$

Taking the natural logarithm on both sides:

$\dfrac{t}{\tau} = \ln(2 + \sqrt{2})$

Since $\sqrt{2} \approx 1.414$:

$\dfrac{t}{\tau} = \ln(3.414) \approx 1.228 \approx 1.23$

Thus, the number of time constants that will elapse is approximately $1.23$.