Chapter 14, Problem 60E

To determine

The power dissipated by the $1\Omega$ resistor.

Answer to Problem 60E

The power dissipated by the $1\Omega$ resistor is $\underset{\_}{{\left\{\left[4.66e^{-t}-5.66e^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}-0.15e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}\right]u\left(t\right)\right\}}^2\text{W}}$.

Explanation of Solution

Given data:

The required diagram is shown in Figure 1.

Calculation:

The conversion of $\text{mF}$ into $\text{F}$ is done as,

$1\text{mF}=1\times {10}^{-3}\text{F}$

Hence, the conversion of $200\text{mF}$ into $\text{F}$ is,

$\begin{array}{c}200\text{mF}=200\times {10}^{-3}\text{F}\\ =0.2\text{F}\end{array}$

The $s$ domain expression for conductor is written as,

$\begin{array}{c}\frac{1}{Cs}=\frac{1}{0.2s}\\ =\frac{5}{s}\end{array}$

The Laplace transform of $e^{-4t}u\left(t\right)$ is written as,

$L\left[4e^{-t}u\left(t\right)\right]=\frac{4}{s+1}$

The Laplace transform of $5u\left(t\right)$ is written as,

$L\left[5u\left(t\right)\right]=\frac{5}{s}$

The required diagram for the $s$ domain circuit is drawn as shown in figure 2.

Apply nodal analysis at node 1 and the expression is written as,

$\begin{array}{l}\frac{V_1\left(s\right)}{1}+\frac{V_1\left(s\right)-V_2\left(s\right)}{\frac{5}{s}}=\frac{4}{s+1}\\ \frac{V_1\left(s\right)}{1}+0.2s\left[V_1\left(s\right)-V_2\left(s\right)\right]=\frac{4}{s+1}\\ V_1\left(s\right)\left(1+0.2s\right)-0.2s{V}_2\left(s\right)=\frac{4}{s+1}\end{array}$ (1)

Apply nodal analysis at node 2 and the expression is written as,

$\begin{array}{l}0.2s\left[V_1\left(s\right)-V_2\left(s\right)\right]+\frac{V_2\left(s\right)}{2s}+\frac{5}{s}=0\\ -0.2s{V}_1\left(s\right)+\left(0.2s+\frac{1}{2s}\right)V_2\left(s\right)=-\frac{5}{s}\end{array}$ (2)

Solve equation (1) and (2) by Cramer’s rule and it is written as,

$\begin{array}{c}{V}_1\left(s\right)=\frac{\left|\begin{array}{cc}\frac{4}{s+1}& -0.2s\\ \frac{-5}{s}& 0.2s+\frac{1}{2s}\end{array}\right|}{\left|\begin{array}{cc}1+0.2s& -0.2s\\ -0.2s& 0.2s+\frac{1}{2s}\end{array}\right|}\\ =\frac{\left(\frac{4}{s+1}\right)\left(0.2s+\frac{1}{2s}\right)-\left(-0.2s\right)\left(\frac{-5}{s}\right)}{\left(1+0.2s\right)\left(0.2s+\frac{1}{2s}\right)-\left(-0.2s\right)\left(-0.2s\right)}\\ =\frac{\left(\frac{4}{s+1}\right)\left(\frac{0.4s^2+1}{2s}\right)-1}{\left(0.04s^2+0.1+0.2s+\frac{1}{2s}\right)-0.04s^2}\\ =\frac{\left(\frac{1.6s^2+4-2s^2-2s}{2s\left(s+1\right)}\right)}{\left(\frac{0.08s^3+0.4s^2+0.2s+1-0.08s^3}{s}\right)}\end{array}$

Further simplify the above expression.

$\begin{array}{c}{V}_1\left(s\right)=\frac{\left(\frac{-0.4s^2+4-2s}{2s\left(s+1\right)}\right)}{\left(\frac{0.4s^2+0.2s+1}{s}\right)}\\ =\frac{-\left(s^2-10+5s\right)}{\left(s+1\right)\left(s^2+\frac{1}{2}s+\frac{5}{2}\right)}\end{array}$

Further simplify the above expression by taking partial fractions.

$\begin{array}{c}{V}_1\left(s\right)=\frac{\frac{14}{3}}{\left(s+1\right)}-\frac{\frac{17}{3}s}{\left(s^2+\frac{1}{2}s+\frac{5}{2}\right)}-\frac{\frac{5}{3}}{\left(s^2+\frac{1}{2}s+\frac{5}{2}\right)}\\ =\frac{14}{3\left(s+1\right)}-\frac{17s}{3\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}-\frac{5}{3\left({\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right)}\\ =\frac{14}{3\left(s+1\right)}-\frac{17}{3}\left[\begin{array}{l}\frac{\left(s+\frac{1}{4}\right)}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\\ -\frac{1}{4}\frac{1}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\end{array}\right]-\frac{5}{3\left({\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right)}\end{array}$

The Laplace transform of $\left(\sin at\right)u\left(t\right)$ is written as,

$L\left[\left(\sin at\right)u\left(t\right)\right]=\frac{a}{s^2+a^2}$

The Laplace transform of $\left(\cos at\right)u\left(t\right)$ is written as,

$L\left[\left(\cos at\right)u\left(t\right)\right]=\frac{s}{s^2+a^2}$

The properties for Laplace transform are written as,

$L\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=L\left\{f_1\left(t\right)\right\}+L\left\{f_2\left(t\right)\right\}$

$L^{-1}\left\{kF\left(s\right)\right\}=k{L}^{-1}\left\{F\left(s\right)\right\}$

The inverse Laplace of the given function is written as,

$v_{\text{1}}\left(t\right)=L^{-1}\left\{V_1\left(s\right)\right\}$ (3)

Substitute $\left[\frac{14}{3\left(s+1\right)}-\frac{17}{3}\left[\begin{array}{l}\frac{\left(s+\frac{1}{4}\right)}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\\ -\frac{1}{4}\frac{1}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\end{array}\right]-\frac{5}{3\left({\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right)}\right]$ for $V_1\left(s\right)$ in equation (3).

$\begin{array}{c}{v}_{\text{1}}\left(t\right)=L^{-1}\left\{\frac{14}{3\left(s+1\right)}-\frac{17}{3}\left[\begin{array}{l}\frac{\left(s+\frac{1}{4}\right)}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\\ -\frac{1}{4}\frac{1}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\end{array}\right]-\frac{5}{3\left({\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right)}\right\}\\ =\left[\begin{array}{l}\frac{14}{3}{L}^{-1}\frac{1}{\left(s+1\right)}-\frac{17}{3}\left[\begin{array}{l}{L}^{-1}\left(\frac{\left(s+\frac{1}{4}\right)}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\right)\\ -\frac{17}{3}{L}^{-1}\left(-\frac{1}{4}\frac{1}{\left[{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right]}\right)\end{array}\right]\\ -\frac{5}{3}\left\{-\frac{1}{\left({\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{39}}{4}\right)}^2\right)}\right\}\end{array}\right]\\ =\frac{14}{3}{e}^{-t}u\left(t\right)-\frac{17}{3}\left[\begin{array}{l}{e}^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}u\left(t\right)-\\ 0.16e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}u\left(t\right)\end{array}\right]-1.06e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}u\left(t\right)\\ =\left[4.66e^{-t}-5.66e^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}-0.15e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}\right]u\left(t\right)\text{V}\end{array}$

The power dissipated in $1\Omega$ resistor is written as,

$p\left(t\right)=\frac{v_1^2\left(t\right)}{R}$

Here,

$p\left(t\right)$ is the power dissipated in $1\Omega$ resistor,

$R$ is the resistor.

Substitute $\left\{\left[4.66e^{-t}-5.66e^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}-0.15e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}\right]u\left(t\right)\right\}\text{V}$ for $v_{\text{1}}\left(t\right)$ and $1\Omega$ for $R$ in the above formula.

$\begin{array}{c}p\left(t\right)=\frac{{\left(\left\{\left[4.66e^{-t}-5.66e^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}-0.15e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}\right]u\left(t\right)\right\}\text{V}\right)}^2}{1\Omega }\\ ={\left\{\left[4.66e^{-t}-5.66e^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}-0.15e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}\right]u\left(t\right)\right\}}^2\text{W}\end{array}$

Conclusion:

Therefore, the power dissipated by the $1\Omega$ resistor is $\underset{\_}{{\left\{\left[4.66e^{-t}-5.66e^{-\frac{1}{4}t}\cos \frac{\sqrt{39}t}{4}-0.15e^{-\frac{1}{4}t}\sin \frac{\sqrt{39}t}{4}\right]u\left(t\right)\right\}}^2\text{W}}$.