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. calculate the total capacitance of the circuit. express your answer with the appropriate units. 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 time - constant formula

The time - constant $\tau$ of an RC circuit is given by $\tau = RC$. We know $\tau = 14.0\ \mu s=14.0\times 10^{- 6}\ s$ and $R = 20.0\ k\Omega=20.0\times10^{3}\ \Omega$. We can solve for $C$.
$C=\frac{\tau}{R}$

Step2: Calculate capacitance

Substitute the values of $\tau$ and $R$ into the formula:
$C=\frac{14.0\times 10^{-6}\ s}{20.0\times 10^{3}\ \Omega}=7.00\times 10^{-10}\ F = 0.700\ nF$

Step3: Recall charging formula for capacitor voltage

The voltage across a charging capacitor in an RC - circuit is given by $V = \mathcal{E}(1 - e^{-\frac{t}{RC}})$. We know $V = 14.0\ V$, $\mathcal{E}=31.0\ V$, $R = 20.0\times 10^{3}\ \Omega$ and $C = 7.00\times 10^{-10}\ F$. First, rewrite the formula for $t$:
$1 - e^{-\frac{t}{RC}}=\frac{V}{\mathcal{E}}$
$e^{-\frac{t}{RC}}=1-\frac{V}{\mathcal{E}}$
$-\frac{t}{RC}=\ln(1 - \frac{V}{\mathcal{E}})$
$t=-RC\ln(1 - \frac{V}{\mathcal{E}})$

Step4: Calculate time $t$

Substitute the values:
$t=-(20.0\times 10^{3}\ \Omega)\times(7.00\times 10^{-10}\ F)\times\ln(1-\frac{14.0\ V}{31.0\ V})$
$t=-(20.0\times 10^{3}\ \Omega)\times(7.00\times 10^{-10}\ F)\times\ln(\frac{31.0 - 14.0}{31.0})=-(20.0\times 10^{3}\ \Omega)\times(7.00\times 10^{-10}\ F)\times\ln(\frac{17}{31.0})$
$t = 5.97\times 10^{-6}\ s=5.97\ \mu s$

Answer:

Part A:
$C = 0.700\ nF$
Part B:
$t = 5.97\ \mu s$