Chapter 25, Problem 116P

(a)

To determine

The sketch for the graph of the voltage across the capacitor $C$ and the current in $R_2$ between $t=0$ at $t=10.0\text{s}$ .

Answer to Problem 116P

The required sketch is shown in Figure 2.

Explanation of Solution

Given:

The resistance $R_1$ is $2\text{M}\Omega$ .

The resistance $R_2$ is $5.00\text{M}\Omega$ .

The capacitance $C$ is $1.00\text{μF}$ .

The time $t$ at which switch is close is $0$ .

The time $t$ at which switch is close is $2.00\text{s}$ .

The given diagram is shown in Figure 1

  

Figure 1

Formula:

The expression for the value of the current $I_1$ is given by,

  $\begin{array}{l}{I}_1=I_2+I_3\\ I_3=I_1-I_2\end{array}$

The expression to determine the formula for the resistance $R_1$ is given by,

  $\begin{array}{l}\epsilon -R_1{I}_1-\frac{Q}{C}=0\\ I_1=\frac{\epsilon -\frac{Q}{C}}{R_1}\end{array}$

The loop rule for the loop with $R_2$ is given by,

  $\begin{array}{l}\frac{Q}{C}-R_2{I}_2=0\\ R_2\frac{d{I}_2}{dt}=\frac{I_3}{C}\\ R_2\frac{d{I}_2}{dt}=\frac{I_1-I_2}{C}\\ R_2\frac{d{I}_2}{dt}=\frac{\frac{\epsilon -\frac{Q}{C}}{R_1}-I_2}{C}\end{array}$

Solve further as,

  $\frac{d{I}_2}{dt}=\frac{\epsilon }{R_1{R}_{2}C}-\left(\frac{R_1+R_2}{R_1{R}_{2}C}\right)I_2$

The general for the solution is given by,

  $\begin{array}{l}{I}_2\left(t\right)=a+b{e}^{\frac{-t}{\tau }}\\ \frac{d{I}_2}{dt}=\frac{d}{dt}\left[a+b{e}^{\frac{-t}{\tau }}\right]\\ \frac{d{I}_2}{dt}=-\frac{b}{\tau }{e}^{\frac{-t}{\tau }}\end{array}$

The expression for $b$ when time $t=0$ is given by,

  $b=-a$

The expression to determine the voltage $V_C\left(t\right)$ is given by,

  $V_C\left(t\right)=I_2\left(t\right)R_2$

Calculation:

The expression for the constant $\tau$ is calculated as,

  $\begin{array}{l}\frac{d{I}_2}{dt}=\frac{\epsilon }{R_1{R}_{2}C}-\left(\frac{R_1+R_2}{R_1{R}_{2}C}\right)I_2\\ -\frac{b}{\tau }{e}^{\frac{-t}{\tau }}=\frac{\epsilon }{R_1{R}_{2}C}-\left(\frac{R_1+R_2}{R_1{R}_{2}C}\right)\left(a+b{e}^{\frac{-t}{\tau }}\right)\\ \tau =\frac{R_1{R}_{2}C}{R_1+R_2}\\ a=\frac{\epsilon }{R_1+R_2}\end{array}$

The value of $b$ is evaluated as,

  $\begin{array}{c}b=-a\\ =-\frac{\epsilon }{R_1+R_2}\end{array}$

The value of $\tau$ is evaluated as,

  $\begin{array}{c}\tau =\frac{R_1{R}_{2}C}{R_1+R_2}\\ =\frac{\left(5\text{M}\Omega \right)\left(2\text{M}\Omega \right)\left(1\text{μF}\right)}{\left(5\text{M}\Omega \right)+\left(2\text{M}\Omega \right)}\\ =1.43\text{s}\end{array}$

The expression for the current $I_2\left(t\right)$ is evaluated as,

  $\begin{array}{c}{I}_2\left(t\right)=a+b{e}^{\frac{-t}{\tau }}\\ =\frac{\epsilon }{R_1+R_2}\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\\ =\frac{10\text{V}}{2\text{M}\Omega +5\text{M}\Omega }\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\\ =1.43\text{μA}\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\end{array}$

The expression for the voltage $V_C\left(t\right)$ is evaluated as,

  $\begin{array}{c}{V}_C\left(t\right)=I_2\left(t\right)R_2\frac{\pi }{4}\\ =\left(5\text{M}\Omega \right)\left(1.43\text{μA}\right)\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\\ =7.15\text{V}\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\end{array}$

The plot for the voltage across capacitor as a function of time is shown in Figure 2

  

Figure 2

Conclusion:

Therefore, the required sketch is shown in Figure 2.

(b)

To determine

The sketch for the graph of the voltage across the capacitor $C$ and the current in $R_2$ between $t=2$ at $t=10.0\text{s}$ .

Answer to Problem 116P

The value of the voltage at $t=2\text{s}$ is $5.38\text{V}$ and at $t=8\text{s}$ is $1.62\text{V}$ .

Explanation of Solution

Given:

The given time $t$ is $2\sec$ and $8\sec$ .

Formula:

The expression for the voltage $V_C\left(t\right)$ is given by,

  $V_C\left(t\right)=7.15\text{V}\left(1-e^{\frac{-t}{1.43\text{s}}}\right)$

Calculation:

The value of the voltage at time $t=2\text{s}$ is calculated as,

  $\begin{array}{c}{V}_C\left(2\right)=7.15\text{V}\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\\ =7.15\text{V}\left(1-e^{\frac{-2}{1.43\text{s}}}\right)\\ =5.38\text{V}\end{array}$

The value of the voltage at time $t=8\text{s}$ is calculated as,

  $\begin{array}{c}{V}_C\left(8\right)=7.15\text{V}\left(1-e^{\frac{-t}{1.43\text{s}}}\right)\\ =7.15\text{V}\left(1-e^{\frac{-8}{1.43\text{s}}}\right)\\ =1.62\text{V}\end{array}$

Conclusion:

Therefore, the value of the voltage at $t=2\text{s}$ is $5.38\text{V}$ and at $t=8\text{s}$ is $1.62\text{V}$ .