Question

in (figure 1), the total resistance is 20.0 kω, and the batterys emf is 31.0 v. the time - constant is measured to be 14.0 μs. part b calculate the time it takes for the voltage across the capacitor to reach 14.0 v after the switch is closed. express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Recall RC - charging formula

The voltage across a capacitor in an RC - circuit during charging is given by $V = V_0(1 - e^{-\frac{t}{RC}})$, where $V$ is the voltage across the capacitor at time $t$, $V_0$ is the emf of the battery, and $RC$ is the time - constant $\tau$. We know that $\tau=14.0\ \mu s$ and $V_0 = 31.0\ V$, and $V = 14.0\ V$.

Step2: Rearrange the formula for $t$

First, rewrite the formula as $\frac{V}{V_0}=1 - e^{-\frac{t}{\tau}}$. Then, $e^{-\frac{t}{\tau}}=1-\frac{V}{V_0}$. Taking the natural logarithm of both sides, we get $-\frac{t}{\tau}=\ln(1 - \frac{V}{V_0})$, and $t=-\tau\ln(1 - \frac{V}{V_0})$.

Step3: Substitute the values

Substitute $\tau = 14.0\ \mu s$, $V = 14.0\ V$, and $V_0 = 31.0\ V$ into the formula.
$1-\frac{V}{V_0}=1-\frac{14.0}{31.0}\approx1 - 0.4516 = 0.5484$.
$\ln(1 - \frac{V}{V_0})=\ln(0.5484)\approx - 0.604$.
$t=-14.0\times(- 0.604)\ \mu s\approx8.46\ \mu s$.

Answer:

$t = 8.46\ \mu s$