Question

The switch in the circuit shifts from $1$ to $2$ when $V_c>\frac{2V}{3}$and goes back to $1$ from $2$ when $V_c<\frac{V}{3}$. The voltmeter reads voltage as plotted. What is the period $T$ of the wave in terms of $R$ and $C$?

• A. $RC\ln 3$
• B. $\frac{RC}{2}\ln 3$
• C. $2RC\ln 2$
• D. $\frac{RC}{3}\ln 3$

Answer 1

The correct option is C

$2RC\ln 2$

Step 1: Given data

$\mathrm{Lower}\mathrm{limit},V=\frac{V}{3}$

$\mathrm{Upper}\mathrm{limit},V_0=\frac{2V}{3}$

Step 2: To find

We have to find the time period $\left(T\right)$

Step 3: Formula used

$\mathrm{q}={\mathrm{q}}_0{\mathrm{e}}^{\frac{-{\mathrm{t}}_2}{\mathrm{RC}}}$

Here,

$\mathrm{q}$ is the charge.

$\mathrm{R}$ is the resistance.

${\mathrm{t}}_2$ is the time period of discharging of the capacitor.

$\mathrm{C}$ is the capacitor.

Step 4: Calculate the time period $\left(T\right)$.

During the time interval $t_2$, the capacitor is discharging with the help of a resistor $R$ because the potential difference across it is increasing $(q\propto V)$.
$$\begin{gathered} \Rightarrow q=q_0{e}^{-t_2/RC} \\ \Rightarrow CV=C{V}_0{e}^{-t_2/RC} \\ \Rightarrow V=V_0{e}^{-t_2/RC}...\left(1\right) \end{gathered}$$
From the graph,
$V_0=\frac{2V}{3}$ and $V=\frac{V}{3}$

Use equation (1).

$$\begin{gathered} \frac{V}{3}=\frac{2V}{3}{e}^{-\frac{t_2}{RC}} \\ \frac{1}{2}=e^{-\frac{t_2}{RC}} \\ \ln \left(\frac{1}{2}\right)=\ln \left(e^{-\frac{t_2}{RC}}\right)\left[\mathrm{Taking}\ln \mathrm{on}\mathrm{both}\mathrm{sides}\right] \\ \ln \left(\frac{1}{2}\right)=-\frac{t_2}{RC} \\ \ln \left(1\right)-\ln \left(2\right)=-\frac{t_2}{RC} \\ 0-\ln \left(2\right)=-\frac{t_2}{RC} \\ -\ln \left(2\right)=-\frac{t_2}{RC} \\ \Rightarrow t_2=RC\ln 2 \end{gathered}$$

During the time interval $t_1$, the capacitor is charging with the help of a resistor $R$ because the potential difference across it is decreasing.

The charging equation is given as,
$$\begin{gathered} \Rightarrow q=q_0\left(1-e^{-t_1/RC}\right) \\ \Rightarrow CV=C{V}_0\left(1-e^{-t_1/RC}\right) \\ \Rightarrow V=V_0\left(1-e^{-t_1/RC}\right)...\left(2\right) \end{gathered}$$
Again from the graph,
$V_0=\frac{2V}{3}$ and $V=\frac{V}{3}$

Use equation (2).

$$\begin{gathered} \frac{V}{3}=\frac{2V}{3}\left(1-e^{-\frac{t_1}{RC}}\right) \\ \frac{1}{2}=\left(1-e^{-\frac{t_1}{RC}}\right) \\ \frac{1}{2}-1=-e^{-\frac{t_1}{RC}} \\ -\frac{1}{2}=-e^{-\frac{t_1}{RC}} \\ \frac{1}{2}=e^{-\frac{t_1}{RC}} \\ \ln \left(\frac{1}{2}\right)=\ln \left(e^{-\frac{t_1}{RC}}\right)\left[\mathrm{Taking}\ln \mathrm{on}\mathrm{both}\mathrm{sides}\right] \\ \ln \left(\frac{1}{2}\right)=-\frac{t_1}{RC} \\ \ln \left(1\right)-\ln \left(2\right)=-\frac{t_1}{RC} \\ 0-\ln \left(2\right)=-\frac{t_1}{RC} \\ -\ln \left(2\right)=-\frac{t_1}{RC} \\ \Rightarrow t_1=RC\ln 2 \end{gathered}$$

Thus, the time period of the waveform is given as,

$$\begin{gathered} T=t_1+t_2 \\ =RC\ln 2+RC\ln 2 \\ =2RC\ln 2 \end{gathered}$$

Hence, Option (C) is correct.