Chapter 8, Problem 38E

The switch in Fig. 8.70 is moved from A to B at t = 0 after being at A for a long time. This places the two capacitors in series, thus allowing equal and opposite dc voltages to be trapped on the capacitors. (a) Determine v₁(0⁻), v₂(0⁻), and v_R(0⁻). (b) Find v₁(0⁺), v₂(0⁺), and v_R(0⁺). (c) Determine the time constant of v_R(t). (d) Find v_R(t), t > 0. (e) Find i(t). (f) Find v₁(t) and v₂(t) from i(t) and the initial values. (g) Show that the stored energy at t = ∞ plus the total energy dissipated in the 20 kΩ resistor is equal to the energy stored in the capacitors at t = 0.

■ FIGURE 8.70

To determine

Find the value of $v_1\left(0^{-}\right)$, $v_2\left(0^{-}\right)$ and $v_R\left(0^{-}\right)$.

Answer to Problem 38E

The value of $v_1\left(0^{-}\right)$ is $100\text{V}$, $v_2\left(0^{-}\right)$ is $0\text{V}$ and $v_R\left(0^{-}\right)$ is $0\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 8.70 in the textbook.

The switch is moved from $A$ to $B$ at time $t=0$ and allows equal and opposite dc voltage to be trapped on the capacitors.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Refer to Figure 1, the voltage across the capacitor $\left(C_1\right)$ is $v_1$, the voltage across the resistor $\left(R_2\right)$ is $v_R$ and the voltage across the capacitor $\left(C_2\right)$ is $v_2$.

For a DC circuit at steady state condition, when the switch is in position $A$ for time $t<0$, the capacitor acts like open circuit.

Now, the Figure 1 is reduced as shown in Figure 2.

Refer to Figure 2, the voltage source $\left(v\right)$ is connected in parallel with the capacitor $\left(C_1\right)$. For the parallel connection, the voltage is same.

The voltage across the capacitor $\left(C_1\right)$ is calculated as follows:

$v_1\left(0^{-}\right)=100\text{V}$

Refer to Figure 2, the resistor $\left(R_2\right)$ and capacitor $\left(C_2\right)$ is not placed in a circuit. Therefore, the voltage across the resistor $\left(R_2\right)$ and capacitor $\left(C_2\right)$ is equal to zero.

$v_R\left(0^{-}\right)=0\text{V}$

$v_2\left(0^{-}\right)=0\text{V}$

The voltage across the capacitor is always continuous so that,

$v_1\left(0\right)=v_1\left(0^{-}\right)=v_1\left(0^{+}\right)=100\text{V}$

$v_2\left(0\right)=v_2\left(0^{-}\right)=v_2\left(0^{+}\right)=0\text{V}$

Conclusion:

Thus, the value of $v_1\left(0^{-}\right)$ is $100\text{V}$, $v_2\left(0^{-}\right)$ is $0\text{V}$ and $v_R\left(0^{-}\right)$ is $0\text{V}$.

To determine

Find the value of $v_1\left(0^{+}\right)$, $v_2\left(0^{+}\right)$ and $v_R\left(0^{+}\right)$.

Answer to Problem 38E

The value of $v_1\left(0^{+}\right)$ is $100\text{V}$, $v_2\left(0^{+}\right)$ is $0\text{V}$ and $v_R\left(0^{+}\right)$ is $100\text{V}$.

Explanation of Solution

Calculation:

For time $t>0$, the switch is moved to position $B$ and the Figure 1 is reduced as shown in Figure 3.

Refer to part (a),

$v_1\left(0^{+}\right)=100\text{V}$

$v_2\left(0^{+}\right)=0\text{V}$

Apply Kirchhoff’s voltage for the circuit shown in Figure 3 for time $t=0^{+}$.

$-v_1\left(0^{+}\right)+v_R\left(0^{+}\right)+v_2\left(0^{+}\right)=0$

Substitute $100\text{V}$ for $v_1\left(0^{+}\right)$, and $0\text{V}$ for $v_2\left(0^{+}\right)$ in above equation to find $v_R\left(0^{+}\right)$.

$-100\text{V}+v_R\left(0^{+}\right)+0\text{V}=0$

$v_R\left(0^{+}\right)=100\text{V}$

Conclusion:

Thus, the value of $v_1\left(0^{+}\right)$ is $100\text{V}$, $v_2\left(0^{+}\right)$ is $0\text{V}$ and $v_R\left(0^{+}\right)$ is $100\text{V}$.

To determine

Find the value of the time constant.

Answer to Problem 38E

The value of time constant $\left(\tau \right)$ is $80\text{ms}$.

Explanation of Solution

Formula used:

Write a general expression to calculate the time constant.

$\tau =RC$        (1)

Here,

$R$ is the value of resistance, and

$C$ is the value of capacitance.

Calculation:

Refer to Figure 3, the capacitors $C_1$ and $C_2$ are connected in series.

The equivalent capacitance $\left(C\right)$ is calculated as follows:

$\begin{array}{c}C=\frac{\left(20\mu \text{F}\right)\left(5\mu \text{F}\right)}{20\mu \text{F}+5\mu \text{F}}\\ =\frac{\left(20\mu \text{F}\right)\left(5\mu \text{F}\right)}{25\mu \text{F}}\\ =4\mu \text{F}\end{array}$

Now, the Figure 3 is reduced as shown in Figure 4.

Refer to Figure 4, it shows the $RC$ circuit.

Use equation (1) to find $\tau$ with a resistance of $R_2$.

$\tau =R_{2}C$

Substitute $20\text{k}\Omega$ for $R_2$, and $4\mu \text{F}$ for $C$ in above equation to find $\tau$.

$\begin{array}{c}\tau =\left(20\text{k}\Omega \right)\left(4\mu \text{F}\right)\\ =\left(20\times {10}^3\Omega \right)\left(4\times {10}^{-6}\frac{\text{s}}{\Omega }\right)\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega },1\text{k}={10}^3,1\mu ={10}^{-6}\right\}\\ =80\times {10}^{-3}\text{s}\\ =80\text{ms}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the value of time constant $\left(\tau \right)$ is $80\text{ms}$.

To determine

Find the expression of voltage $v_R\left(t\right)$ for time $t>0$.

Answer to Problem 38E

The expression of voltage $v_R\left(t\right)$ for time $t>0$ is $100e^{-\frac{t}{80\text{ms}}}\text{V}$.

Explanation of Solution

Formula used:

Write a general expression to calculate the voltage response of an $RC$ circuit.

$v_C\left(t\right)=v_C\left(0^{+}\right)e^{-\frac{t}{\tau }}$        (2)

Here,

$v_R\left(0\right)$ is the initial voltage across the resistor $R_2$.

Refer to Figure 3, the resistor $\left(R_2\right)$ and capacitance $\left(C\right)$ is connected in parallel. For the parallel connection, the voltage is same.

$v_R\left(t\right)=v_C\left(t\right)$

Substitute $v_R\left(t\right)$ for $v_C\left(t\right)$ in equation (2) to find $v_R\left(t\right)$.

$v_R\left(t\right)=v_R\left(0^{+}\right)e^{-\frac{t}{\tau }}$

Substitute $100\text{V}$ for $v_R\left(0^{+}\right)$, and $80\text{ms}$ for $\tau$ in above equation to find $v_R\left(t\right)$.

$v_R\left(t\right)=100e^{-\frac{t}{80\text{ms}}}\text{V}$

Conclusion:

Thus, the expression of voltage $v_R\left(t\right)$ for time $t>0$ is $100e^{-\frac{t}{80\text{ms}}}\text{V}$.

To determine

Find the expression of current $i\left(t\right)$.

Answer to Problem 38E

The expression of current $i\left(t\right)$ is $5e^{-\frac{t}{80\text{ms}}}\text{mA}$.

Explanation of Solution

Calculation:

Refer to Figure 3, the capacitor $\left(C_1\right)$, resistor $\left(R_2\right)$ and the capacitor $\left(C_2\right)$ are connected in series. For the series connection, the current is same.

The current $i\left(t\right)$ is calculated as follows:

$i\left(t\right)=\frac{v_R\left(t\right)}{R_2}$

Substitute $100e^{-\frac{t}{80\text{ms}}}\text{V}$ for $v_R\left(t\right)$, and $20\text{k}\Omega$ for $R_2$ in above equation to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\frac{100e^{-\frac{t}{80\text{ms}}}\text{V}}{20\text{k}\Omega }\\ =\frac{100e^{-\frac{t}{80\text{ms}}}\text{V}}{20\times {10}^3\Omega }\\ =5\times {10}^{-3}{e}^{-\frac{t}{80\text{ms}}}\text{A}\\ =5e^{-\frac{t}{80\text{ms}}}\text{mA}\end{array}$

Conclusion:

Thus, the expression of current $i\left(t\right)$ is $5e^{-\frac{t}{80\text{ms}}}\text{mA}$.

To determine

Find the value of voltage $v_1\left(t\right)$ and $v_2\left(t\right)$ from $i\left(t\right)$ and the initial values.

Answer to Problem 38E

The value of voltage $v_1\left(t\right)$ is $\left[100-20\left(1-e^{-\frac{t}{80\text{ms}}}\right)\right]\text{V}$ and $v_2\left(t\right)$ is $80\left(1-e^{-\frac{t}{80\text{ms}}}\right)\text{V}$.

Explanation of Solution

Formula used:

Write a general expression to calculate the voltage across the capacitor.

$v\left(t\right)=\frac{1}{C}{\displaystyle \int idt}+v\left(0\right)$        (3)

Calculation:

From the given data, the capacitor $\left(C_1\right)$ and $\left(C_2\right)$ has equal and opposite dc voltages.

Use equation (1) to find the voltage $v_1\left(t\right)$ across the capacitor $\left(C_1\right)$.

$v_1\left(t\right)=\left[-\frac{1}{C_1}{\displaystyle \int idt}\right]+v_1\left(0\right)$

Substitute $20\mu \text{F}$ for $C_1$, $5e^{-\frac{t}{80\text{ms}}}\text{mA}$ for $i\left(t\right)$, and $100\text{V}$ for $v_1\left(0\right)$ in above equation to find $v_1\left(t\right)$.

$\begin{array}{c}{v}_1\left(t\right)=\left[-\frac{1}{20\mu \text{F}}{\displaystyle {\int }_0^{t}5e^{-\frac{t}{80\text{ms}}}\text{mA}dt}\right]+100\text{V}\\ \text{=}\left[-\frac{1}{20\times {10}^{-6}\text{F}}{\displaystyle {\int }_0^{t}5\times {10}^{-3}{e}^{-\frac{t}{80\times {10}^{-3}\text{s}}}\text{A}dt}\right]+100\text{V}\left\{\begin{array}{l}\because 1\text{m}={10}^{-3}\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left[-\frac{5\times {10}^{-3}}{20\times {10}^{-6}\text{F}}\left({\left[\frac{e^{-\frac{t}{80\times {10}^{-3}\text{s}}}}{\left(-\frac{1}{80\times {10}^{-3}}\right)}\right]}_0^t\text{A}\cdot \text{s}\right)\right]+100\text{V}\\ \text{=}\left[-\frac{5\times {10}^{-3}\left(80\times {10}^{-3}\right)}{20\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}}\left({\left[e^{-\frac{t}{80\times {10}^{-3}\text{s}}}\right]}_0^t\text{A}\cdot \text{s}\right)\right]+100\text{V}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\end{array}$

Simplify the above equation to find $v_1\left(t\right)$.

$\begin{array}{c}{v}_1\left(t\right)=\left[-\frac{5\times {10}^{-3}\left(80\times {10}^{-3}\right)\left[-e^{-\frac{t}{80\times {10}^{-3}\text{s}}}+e^{-\frac{0}{80\times {10}^{-3}\text{s}}}\right]}{20\times {10}^{-6}}\text{V}\right]+100\text{V}\\ =-20\left(-e^{-\frac{t}{80\times {10}^{-3}\text{s}}}+1\right)\text{V}+100\text{V}\left\{e^0=1\right\}\\ =\left[100-20\left(1-e^{-\frac{t}{80\text{ms}}}\right)\right]\text{V}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Use equation (1) to find the voltage $v_2\left(t\right)$ across the capacitor $\left(C_2\right)$.

$v_2\left(t\right)=\left[\frac{1}{C_2}{\displaystyle \int idt}\right]+v_2\left(0\right)$

Substitute $5\mu \text{F}$ for $C_2$, $5e^{-\frac{t}{80\text{ms}}}\text{mA}$ for $i\left(t\right)$, and $0\text{V}$ for $v_2\left(0\right)$ in above equation to find $v_2\left(t\right)$.

$\begin{array}{c}{v}_2\left(t\right)=\left[\frac{1}{5\mu \text{F}}{\displaystyle {\int }_0^{t}5e^{-\frac{t}{80\text{ms}}}\text{mA}dt}\right]+0\text{V}\\ \text{=}\left[\frac{1}{5\times {10}^{-6}\text{F}}{\displaystyle {\int }_0^{t}5\times {10}^{-3}{e}^{-\frac{t}{80\times {10}^{-3}\text{s}}}\text{A}dt}\right]\left\{\begin{array}{l}\because 1\text{m}={10}^{-3}\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left[\frac{5\times {10}^{-3}}{5\times {10}^{-6}\text{F}}\left({\left[\frac{e^{-\frac{t}{80\times {10}^{-3}\text{s}}}}{\left(-\frac{1}{80\times {10}^{-3}}\right)}\right]}_0^t\text{A}\cdot \text{s}\right)\right]\\ \text{=}\left[\frac{5\times {10}^{-3}\left(80\times {10}^{-3}\right)}{5\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}}\left({\left[-e^{-\frac{t}{80\times {10}^{-3}\text{s}}}\right]}_0^t\text{A}\cdot \text{s}\right)\right]\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\end{array}$

Simplify the above equation to find $v_1\left(t\right)$.

$\begin{array}{c}{v}_2\left(t\right)=\left[\frac{5\times {10}^{-3}\left(80\times {10}^{-3}\right)\left[-e^{-\frac{t}{80\times {10}^{-3}\text{s}}}+e^{-\frac{0}{80\times {10}^{-3}\text{s}}}\right]}{5\times {10}^{-6}}\text{V}\right]\\ =80\left(-e^{-\frac{t}{80\times {10}^{-3}\text{s}}}+1\right)\text{V}\left\{e^0=1\right\}\\ =80\left(1-e^{-\frac{t}{80\text{ms}}}\right)\text{V}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the value of voltage $v_1\left(t\right)$ is $\left[100-20\left(1-e^{-\frac{t}{80\text{ms}}}\right)\right]\text{V}$ and $v_2\left(t\right)$ is $80\left(1-e^{-\frac{t}{80\text{ms}}}\right)\text{V}$.

To determine

Show that the stored energy at $t=\infty$ plus the total energy dissipated in $20\text{k}\Omega$ resistor is equal to the energy stored in the capacitors at $t=0$.

Explanation of Solution

Formula used:

Write a general expression to calculate the energy stored in a capacitor.

$w=\frac{1}{2}C{v}^2$        (4)

Write a general expression to calculate the energy stored in a resistor.

$w_R={\displaystyle \int i^2\left(t\right)Rdt}$        (5)

Here,

$R$ is the value of resistance.

Calculation:

Refer to Figure 1, there are two capacitors placed in a circuit. Therefore, the total energy stored in a capacitor is,

$w=w_1+w_2$        (6)

Use equation (1) to find $w_1$ for capacitor $C_1$.

$w_1=\frac{1}{2}{C}_1{v}_1^2\left(t\right)$        (7)

Use equation (1) to find $w_2$ for capacitor $C_2$.

$w_2=\frac{1}{2}{C}_2{v}_2^2\left(t\right)$        (8)

Substitute equation (7) and (8) in (6).

$w=\frac{1}{2}{C}_1{v}_1^2\left(t\right)+\frac{1}{2}{C}_2{v}_2^2\left(t\right)$        (9)

Substitute $0$ for $t$ in equation (9) to find $w\left(0\right)$.

$w\left(0\right)=\frac{1}{2}{C}_1{\left(v_1\left(0\right)\right)}^2+\frac{1}{2}{C}_2{\left(v_2\left(0\right)\right)}^2$

Substitute $20\mu \text{F}$ for $C_1$, $5\mu \text{F}$ for $C_2$, $100\text{V}$ for $v_1\left(0\right)$, and $0\text{V}$ for $v_2\left(0\right)$ in above equation to find $w\left(0\right)$.

$\begin{array}{c}w\left(0\right)=\frac{1}{2}\left(20\mu \text{F}\right){\left(100\text{V}\right)}^2+\frac{1}{2}\left(5\mu \text{F}\right){\left(0\text{V}\right)}^2\\ =\frac{1}{2}\left(20\times {10}^{-6}\text{F}\right)\left(10000{\text{V}}^2\right)+0\left\{\because 1\mu ={10}^{-6}\right\}\\ =\frac{1}{2}\left(20\times {10}^{-6}\frac{\text{C}}{\text{V}}\right)\left(10000{\text{V}}^2\right)\left\{\because 1\text{F}=\frac{\text{1}\text{C}}{\text{1}\text{V}}\right\}\\ =0.1\text{C}\cdot \text{V}\end{array}$

Simplify the above equation to find $w\left(0\right)$.

$\begin{array}{c}w\left(0\right)=0.1\times {10}^3\times {10}^{-3}\text{J}\left\{\because 1\text{J}=1\text{C}\cdot \text{1}\text{V}\right\}\\ =100\text{mJ}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Refer to part (f),

$v_1\left(t\right)=\left[100-20\left(1-e^{-\frac{t}{80\text{ms}}}\right)\right]\text{V}$

Substitute $\infty$ for $t$ in above equation to find $v_1\left(\infty \right)$.

$\begin{array}{c}{v}_1\left(\infty \right)=\left[100-20\left(1-e^{-\frac{\infty }{80\text{ms}}}\right)\right]\text{V}\\ \text{=}\left[100-20\left(1-e^{-\infty }\right)\right]\text{V}\\ \text{=}\left[100-20\left(1-0\right)\right]\text{V}\left\{\because e^{-\infty }=0\right\}\\ =80\text{V}\end{array}$

Refer to part (f),

$v_2\left(t\right)=80\left(1-e^{-\frac{t}{80\text{ms}}}\right)\text{V}$

Substitute $\infty$ for $t$ in above equation to find $v_2\left(\infty \right)$.

$\begin{array}{c}{v}_2\left(\infty \right)=80\left(1-e^{-\frac{\infty }{80\text{ms}}}\right)\text{V}\\ =80\left(1-e^{-\infty }\right)\text{V}\\ =80\left(1-0\right)\text{V}\left\{\because e^{-\infty }=0\right\}\\ =80\text{V}\end{array}$

Substitute $\infty$ for $t$ in equation (9) to find $w\left(\infty \right)$.

$w\left(\infty \right)=\frac{1}{2}{C}_1{\left(v_1\left(\infty \right)\right)}^2+\frac{1}{2}{C}_2{\left(v_2\left(\infty \right)\right)}^2$

Substitute $20\mu \text{F}$ for $C_1$, $5\mu \text{F}$ for $C_2$, $80\text{V}$ for $v_1\left(\infty \right)$, and $80\text{V}$ for $v_2\left(\infty \right)$ in above equation to find $w\left(\infty \right)$.

$\begin{array}{c}w\left(\infty \right)=\frac{1}{2}\left(20\mu \text{F}\right){\left(80\text{V}\right)}^2+\frac{1}{2}\left(5\mu \text{F}\right){\left(80\text{V}\right)}^2\\ =\frac{1}{2}{\left(80\text{V}\right)}^2\left[20\mu \text{F+}5\mu \text{F}\right]\\ =\frac{1}{2}{\left(80\text{V}\right)}^2\left[25\mu \text{F}\right]\\ =\frac{1}{2}\left(6400{\text{V}}^2\right)\left(25\times {10}^{-6}\text{F}\right)\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Simplify the above equation to find $w\left(0\right)$.

$\begin{array}{c}w\left(\infty \right)=\frac{1}{2}\left(25\times {10}^{-6}\frac{\text{C}}{\text{V}}\right)\left(6400{\text{V}}^2\right)\left\{\because 1\text{F}=\frac{\text{1}\text{C}}{\text{1}\text{V}}\right\}\\ =0.08\text{C}\cdot \text{V}\\ =0.08\times {10}^3\times {10}^{-3}\text{J}\left\{\because 1\text{J}=1\text{C}\cdot \text{1}\text{V}\right\}\\ =80\text{mJ}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $20\text{k}\Omega$ for $R$, and $5e^{-\frac{t}{80\text{ms}}}\text{mA}$ for $i\left(t\right)$ in equation (5) to find $w_R$.

$\begin{array}{c}{w}_R={\displaystyle \underset{0}{\overset{t}{\int }}{\left(5e^{-\frac{t}{80\text{ms}}}\text{mA}\right)}^2\left(20\text{k}\Omega \right)dt}\\ =\left(20\text{k}\Omega \right){\displaystyle \underset{0}{\overset{t}{\int }}{\left(5e^{-\frac{t}{80\text{ms}}}\text{mA}\right)}^{2}dt}\\ =\left(20\times {10}^3\Omega \right){\displaystyle \underset{0}{\overset{t}{\int }}{\left(5\times {10}^{-3}{e}^{-\frac{t}{80\times {10}^{-3}}}\text{A}\right)}^{2}dt}\left\{\because 1\text{k}={10}^3\right\}\\ =\left(20\times {10}^3\Omega \right){\left(5\times {10}^{-3}\right)}^2{\displaystyle \underset{0}{\overset{t}{\int }}{e}^{-\frac{2t}{80\times {10}^{-3}}}{\text{A}}^{2}dt}\end{array}$

Simplify the above equation to find $w_R$.

$\begin{array}{c}{w}_R=\left(20\times {10}^3\Omega \right){\left(5\times {10}^{-3}\right)}^2{\displaystyle \underset{0}{\overset{t}{\int }}{e}^{-\frac{t}{40\times {10}^{-3}}}{\text{A}}^{2}dt}\\ =\left(20\times {10}^3\Omega \right){\left(5\times {10}^{-3}\right)}^2\frac{{\left[-e^{-\frac{t}{40\times {10}^{-3}}}\right]}_0^t}{\frac{1}{40\times {10}^{-3}}}{\text{A}}^2\cdot \text{s}\\ \text{=}\left(20\times {10}^3\right)\left(25\times {10}^{-6}\right)\left(40\times {10}^{-3}\right)\left[-e^{-\frac{t}{40\times {10}^{-3}}}+e^{-\frac{0}{40\times {10}^{-3}}}\right]{\text{A}}^2\cdot \text{H}\text{}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =0.02\left[1-e^{-\frac{t}{40\times {10}^{-3}}}\right]{\text{A}}^2\cdot \frac{\text{J}}{{\text{A}}^2}\left\{\because 1\text{H}=\frac{\text{1}\text{J}}{\text{1}{\text{A}}^2}\right\}\end{array}$

Simplify the above equation to find $w_R$.

$\begin{array}{c}{w}_R=0.02\left[1-e^{-\frac{t}{40\times {10}^{-3}}}\right]\text{J}\\ =0.02\times {10}^3\times {10}^{-3}\left[1-e^{-\frac{t}{40\times {10}^{-3}}}\right]\text{J}\\ =20\left[1-e^{-\frac{t}{40\times {10}^{-3}}}\right]\text{mJ}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $\infty$ for $t$ in above equation to find $w_R\left(\infty \right)$.

$\begin{array}{c}{w}_R\left(\infty \right)=20\left[1-e^{-\frac{\infty }{40\times {10}^{-3}}}\right]\text{mJ}\\ =20\left[1-e^{-\infty }\right]\text{mJ}\\ =20\left[1-0\right]\text{mJ}\text{}\left\{\because e^{-\infty }=0\right\}\\ =\text{20}\text{mJ}\end{array}$

Add $w_R\left(\infty \right)$ and $w\left(\infty \right)$.

$w_R\left(\infty \right)+w\left(\infty \right)=10\text{0}\text{mJ}$

Therefore,

$w_R\left(\infty \right)+w\left(\infty \right)=w\left(0\right)=100\text{mJ}$

Therefore, the stored energy at $t=\infty$ plus the total energy dissipated in $20\text{k}\Omega$ resistor is equal to the energy stored in the capacitors at $t=0$.

Conclusion:

Thus, the stored energy at $t=\infty$ plus the total energy dissipated in $20\text{k}\Omega$ resistor is equal to the energy stored in the capacitors at $t=0$ is showed.