Chapter 14, Problem 63E

(a)

To determine

The value of $v_2$ at $t=1\text{ms}$.

Answer to Problem 63E

The voltage $v_2$ at $t=1\text{ms}$ is $-5.132\times {10}^{-5}\text{V}$.

Explanation of Solution

Given data:

The resistive component of the circuit is of $100\Omega$.

The capacitive components of the circuit are $600\mu \text{F}$ and $\text{500}\mu \text{F}$.

The values of voltage dependent current sources are $5v_2$ and $3v_2$.

The inductive component of the circuit is of $2\text{mH}$.

The input voltage of the circuit is $14tu\left(t\right)\text{V}$.

The time $t$ is $1\text{ms}$.

The given diagram is shown in Figure 1.

Calculation:

Let the resistance $R=100\Omega$, capacitance $C_1=600\text{μF}$, capacitance $C_2=500\text{μF}$ and inductance be $L=2\text{mH}$.

The conversion of $\mu \text{F}$ into $\text{F}$ is given by,

$\text{1}\mu \text{F}={10}^{-6}\text{F}$

The conversion of $600\mu \text{F}$ into $\text{F}$ is given by,

$600\mu \text{F}=6\text{00}\times {10}^{-6}\text{F}$

The conversion of $500\mu \text{F}$ into $\text{F}$ is given by,

$500\mu \text{F}=5\text{00}\times {10}^{-6}\text{F}$

The conversion from $\text{mH}$ to $\text{H}$ is given by,

$\text{1}\text{mH}={10}^{-3}\text{H}$

The conversion from $\text{mH}$ to $\text{H}$ is given by,

$2\text{mH}=2\times {10}^{-3}\text{H}$

The voltage source $14tu\left(t\right)\text{V}$ in $s$ domain is written as,

$L\left(14tu\left(t\right)\right)=\frac{14}{s^2}$

The Laplace transform of capacitance is given by,

$L\left(C\right)=\frac{1}{Cs}$

The Laplace transform of $6\text{00}\times {10}^{-6}\text{F}$ is given as,

$L\left(6\text{00}\times {10}^{-6}\right)=\frac{1}{\left(6\text{00}\times {10}^{-6}\right)s}$

The Laplace transform of $5\text{00}\times {10}^{-6}\text{F}$ is given as,

$L\left(5\text{00}\times {10}^{-6}\right)=\frac{1}{\left(\text{500}\times {10}^{-6}\right)s}$

The Laplace transform of inductor is given by,

$L\left(L\right)=Ls$

The Laplace transform of $2\times {10}^{-3}\text{H}$ is given as,

$\begin{array}{c}L\left(2\times {10}^{-3}\text{H}\right)=\left(2\times {10}^{-3}\text{H}\right)s\\ =0.002s\end{array}$

Mark the nodes apply mesh analysis to the circuit and redraw the circuit in $s$ domain.

The required diagram is shown in Figure 2.

The output voltage $v_2$ equals to the voltage drop across the $0.002s$ inductor and the current flowing through the mesh is $i_2$.

The value of node voltage $v_2$ is given as,

$v_2=\left(0.002s\right)i_2$        (1)

The current source $3v_2$ is in between loop 1 and loop 2 so it forms super mesh.

The super mesh equation is given as,

$i_2-i_1=3v_2$

Substitute $0.002s{i}_2$  for $v_2$ in the above equation.

$\begin{array}{l}{i}_2-i_1=3\left(0.002s{i}_2\right)\\ i_1=\left(1-0.006s\right)i_2\end{array}$        (2)

The current flowing through the loop 3 is given by,

$i_3=-5v_2$

Substitute $\left(0.002s\right)i_2$ for $v_2$ in the above equation.

$\begin{array}{l}{i}_3=-5\left(0.002s\right)i_2\\ i_3=-0.01s{i}_2\end{array}$        (3)

Apply Kirchhoff’s voltage law at the super mesh.

$\begin{array}{l}\frac{14}{s^2}+100i_1+\frac{1}{6\times {10}^{-4}s}{i}_1+\frac{2000}{s}\left(i_2-i_3\right)+0.002s{i}_2=0\\ i_1\left(100+\frac{1}{6\times {10}^{-4}s}\right)+i_2\left(\frac{2000+0.002s^2}{s}\right)-\frac{2000}{s}{i}_3=\frac{14}{s^2}\end{array}$

Substitute $-0.01s{i}_2$ for $i_3$ and $\left(1-0.006s\right)i_2$ for $i_1$ in the above equation.

$\begin{array}{l}\left(1-0.006s\right)i_2\left(100+\frac{1}{6\times {10}^{-4}s}\right)+i_2\left(\frac{2000+0.002s^2}{s}\right)-\frac{2000}{s}\left(-0.01s{i}_2\right)=\frac{14}{s^2}\\ i_2\left[100+\frac{1}{6\times {10}^{-4}s}-0.6s-10+\frac{2000+0.002s^2}{s}-20\right]=\frac{14}{s^2}\\ i_2\left[\frac{600s\times {10}^{-4}-3.6s^2-60s+12000+0.0012s^2+120s}{s}-20\right]=\frac{14}{s^2}\\ i_2\left[\frac{-3.588s^2+660s+22000}{s}\right]=\frac{14}{s}\end{array}$

Solve further as,

$i_2=\left[\frac{14}{s\left(-0.598s^2+110s+3666.67\right)}\right]$

Substitute $\frac{14}{s\left(-0.598s^2+110s+3666.67\right)}$ for $i_2$ in equation (1).

$\begin{array}{c}{v}_2=\left(0.002s\right)\left(\frac{14}{s\left(-0.598s^2+110s+3666.67\right)}\right)\\ =\frac{-0.028s}{0.598s\left(s^2-183.95s-6131.56\right)}\\ =\frac{-0.04682}{\left(s^2-183.95s-6131.56\right)}\\ =\frac{-0.04682}{\left(s+28.818\right)\left(s-212.768\right)}\end{array}$

The above equation in partial form is written as,

$v_2\left(s\right)=\frac{A}{\left(s-212.768\right)}+\frac{B}{\left(s+28.818\right)}$        (4)

Substitute $\frac{-0.04682}{\left(s+28.818\right)\left(s-212.768\right)}$ for $v_2$ in above equation.

$\begin{array}{l}\frac{-0.04682}{\left(s+28.818\right)\left(s-212.768\right)}=\frac{A}{\left(s-212.768\right)}+\frac{B}{\left(s+28.818\right)}\\ \frac{-0.04682}{\left(s+28.818\right)\left(s-212.768\right)}=\frac{A\left(s+28.818\right)+B\left(s-212.768\right)}{\left(s+28.818\right)\left(s-212.768\right)}\\ -0.04682=A\left(s+28.818\right)+B\left(s-212.768\right)\end{array}$        (5)

Substitute $-28.818$ for $s$ in the above equation.

$\begin{array}{l}-0.04682=A\left(-28.818+28.818\right)+B\left(-28.818-212.768\right)\\ -0.04682=B\left(-28.818-212.768\right)+0\\ B=\frac{-0.04682}{\left(-28.818-212.768\right)}\\ B=0.1938\times {10}^{-3}\end{array}$

Substitute $212.768$ for $s$ in equation (5).

$\begin{array}{l}-0.04682=A\left(212.768+28.818\right)+B\left(212.768-212.768\right)\\ A=\frac{-0.04682}{\left(212.768+28.818\right)}\\ A=-0.1938\times {10}^{-3}\end{array}$

Substitute $-0.1938\times {10}^{-3}$ for $A$ and $0.1938\times {10}^{-3}$ for $B$ in equation (4).

$v_2\left(s\right)=\frac{-0.1938\times {10}^{-3}}{\left(s-212.768\right)}+\frac{0.1938\times {10}^{-3}}{\left(s+28.818\right)}$

Apply inverse Laplace transform to the above equation.

$\begin{array}{c}L\left(v_2\right)=L\left[\frac{-0.1938\times {10}^{-3}}{\left(s-212.768\right)}+\frac{0.1938\times {10}^{-3}}{\left(s+28.818\right)}\right]\\ v_2\left(t\right)=\left[-0.1938e^{212.768t}+0.1938e^{-28.818t}\right]\times {10}^{-3}\text{V}\\ =0.1938\left[e^{-28.818t}-e^{212.768t}\right]\times {10}^{-3}\text{V}\end{array}$        (5)

7

The conversion from $\text{ms}$ to $\text{s}$ is given by,

$\text{1}\text{ms}={10}^{-3}\text{s}$

The conversion of $10\text{ms}$ to $\text{s}$ is given as,

$10\text{ms}=10\times {10}^{-3}\text{s}$

Substitute $1\times {10}^{-3}$ for $t$ in equation (5).

$\begin{array}{c}{v}_2\left(1\times {10}^{-3}\text{ s}\right)=0.1938\left[-e^{212.76\left(1\times {10}^{-3}\right)}+e^{-28.818\left(1\times {10}^{-3}\right)}\right]\times {10}^{-3}\text{V}\\ =0.1938\left[-1.237+0.9722\right]\times {10}^{-3}\text{V}\\ \text{=}-5.132\times {10}^{-5}\text{V}\end{array}$

Conclusion:

Therefore he value of $v_2$ at $t=1\text{ms}$ is $-5.132\times {10}^{-5}\text{V}$.

(b)

To determine

The value of $v_2$ at $t=100\text{ms}$.

Answer to Problem 63E

The voltage $v_2$ at $t=100\text{ms}$ is $-0.3368\times {10}^6\text{V}$.

Explanation of Solution

Given data:

The value of $t=100\text{ms}$.

Calculation:

The conversion from $\text{ms}$ to $\text{s}$ is given by,

$\text{1}\text{ms}={10}^{-3}\text{s}$

The conversion of $100\text{ms}$ to $\text{s}$ is given as,

$100\text{ms}=100\times {10}^{-3}\text{s}$

Substitute $100\times {10}^{-3}\text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_2\left(100\text{ms}\right)=\left[-0.1938e^{212.76\left(100\times {10}^{-3}\right)}+0.1938e^{-28.818\left(100\times {10}^{-3}\right)}\right]\times {10}^{-3}\text{V}\\ =0.1938\left[0.056-1737.998\times {10}^6\right]\times {10}^{-3}\\ =-0.3368\times {10}^6\text{V}\end{array}$

Conclusion:

Thus, the voltage $v_2$ at $t=100\text{ms}$ is $-0.3368\times {10}^6\text{V}$.

(c)

To determine

The value of $v_2$ at $t=10\text{s}$.

Answer to Problem 63E

The voltage $v_2$ at $t=10\text{s}$ is $-\infty$.

Explanation of Solution

Given data:

The value of $t=10\text{s}$

Calculation:

Substitute $10\text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_2\left(10\text{s}\right)=\left[-0.1938e^{212.76\left(10\text{s}\right)}+0.1938e^{-28.818\left(10\text{s}\right)}\right]\times {10}^{-3}\text{V}\\ =\left[-0.1938e^{212.76\times 10}+0.1938e^{-28.818\times 10}\right]\times {10}^{-3}\text{V}\\ =0.1938\left[-\infty +0\right]\times {10}^{-3}\\ =-\infty \end{array}$

Conclusion:

Thus, the voltage $v_2$ at $t=10\text{s}$ is $-\infty$.