In an RC circuit, how many time constants must elapse before the...

The power delivered by the battery in an RC circuit is given by $P = VI$, where $V$ is the constant voltage of the battery and $I$ is the current.

The current in a charging RC circuit as a function of time is $I = I_0 e^{-t/\tau}$, where $I_0$ is the maximum current at $t = 0$ and $\tau$ is the time constant.

Substituting the expression for current, the power delivered by the battery is:

P = V I_0 e^{-t/\tau} = P_\max e^{-t/\tau}

We need to find the time $t$ when the power decreases to half of its maximum value, so P = \dfrac{P_\max}{2}.

\dfrac{P_\max}{2} = P_\max e^{-t/\tau}

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

Taking the natural logarithm on both sides:

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

$t = \tau \ln(2) \approx 0.693 \tau$

Thus, approximately $0.69$ time constants must elapse.