Question

question 3 of 3 current attempt in progress what multiple of the time - constant τ gives the time taken by an initially uncharged capacitor in an rc - series circuit to be charged to 85.1% of its final charge? number units attempts: 0 of 3 used save for later etextbook and media submit answer view policies

Explanation:

Step1: Recall charging formula for RC - circuit

The charge on a capacitor in an RC - circuit as a function of time is given by $q(t)=q_{max}(1 - e^{-t/RC})$, where $q_{max}$ is the final charge. We want to find the time $t$ when $q = 0.851q_{max}$.

Step2: Substitute into the formula

Substitute $q = 0.851q_{max}$ into $q(t)=q_{max}(1 - e^{-t/RC})$:
\[

$$\begin{align*} 0.851q_{max}&=q_{max}(1 - e^{-t/RC})\\ 0.851&=1 - e^{-t/RC}\\ e^{-t/RC}&=1 - 0.851\\ e^{-t/RC}&=0.149 \end{align*}$$

\]

Step3: Take the natural - logarithm of both sides

Taking the natural - logarithm of both sides of $e^{-t/RC}=0.149$, we get $-\frac{t}{RC}=\ln(0.149)$.
Since $\ln(0.149)\approx - 1.904$, then $t=-RC\times\ln(0.149)\approx1.904RC$.
The time constant $\tau = RC$. So the time $t$ is approximately $1.90\tau$.

Answer:

$1.90$ (assuming the units of time are in terms of the time - constant $\tau = RC$)