Chapter 16, Problem 65P

For the RLC circuit shown in Fig. 16.88, find the complete response if v(0) = 100 V when the switch is closed.

To determine

Find the expression of voltage response $v\left(t\right)$ for the $RLC$ circuit shown in Figure 16.88.

Answer to Problem 65P

The expression of voltage response $v\left(t\right)$ for the $RLC$ circuit shown in Figure 16.88 is $\left[110.1e^{-3t}+192t{e}^{-3t}-10.1\cos \left(4t\right)+34.58\sin \left(4t\right)\right]u\left(t\right)\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 16.88 in the textbook.

The value of initial voltage across the capacitor $v\left(0\right)$ is $100\text{V}$.

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

$Z_R=R$ (1)

Here,

$R$ is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

$Z_L=sL$ (2)

Here,

$L$ is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

$Z_C=\frac{1}{sC}$ (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

For time $t>0$, the switch is closed and the Figure 1 is reduced as shown in Figure 2.

Apply Laplace transform for $v_s\left(t\right)$ in Figure 2 to find $V_s\left(s\right)$.

$\begin{array}{c}{V}_s\left(s\right)=100\left(\frac{s}{s^2+4^2}\right)\left\{\because \mathcal{L}\left(\cos \left(at\right)u\left(t\right)\right)=\frac{s}{s^2+a^2}\right\}\\ =\frac{100s}{s^2+16}\end{array}$

Substitute $6\Omega$ for $R$ in equation (1) to find $Z_R$.

$Z_R=6\Omega$

Substitute $1\text{H}$ for $L$ in equation (2) to find $Z_L$.

$\begin{array}{c}{Z}_L=s\left(1\text{H}\right)\\ =s\end{array}$

Substitute $\frac{1}{9}\text{F}$ for $C$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{s\left(\frac{1}{9}\text{F}\right)}\\ =\frac{9}{s}\end{array}$

Using element transformation methods with initial conditions convert the Figure 2 into s-domain.

Apply Kirchhoff’s voltage law for the circuit shown in Figure 3.

$\begin{array}{l}\frac{-100s}{s^2+16}+6I\left(s\right)+sI\left(s\right)+\frac{9}{s}I\left(s\right)+\frac{v\left(0\right)}{s}=0\\ 6I\left(s\right)+sI\left(s\right)+\frac{9}{s}I\left(s\right)=\frac{100s}{s^2+16}-\frac{v\left(0\right)}{s}\\ I\left(s\right)\left[6+s+\frac{9}{s}\right]=\frac{100s}{s^2+16}-\frac{v\left(0\right)}{s}\\ I\left(s\right)\left[\frac{6s+s^2+9}{s}\right]=\frac{100s}{s^2+16}-\frac{v\left(0\right)}{s}\end{array}$

Substitute $100$ for $v\left(0\right)$ in above equation.

$\begin{array}{l}I\left(s\right)\left[\frac{6s+s^2+9}{s}\right]=\frac{100s}{s^2+16}-\frac{100}{s}\\ I\left(s\right)\left[\frac{s^2+6s+9}{s}\right]=\frac{100s^2-100s^2-1600}{s\left(s^2+16\right)}\\ I\left(s\right)\left[\frac{s^2+6s+9}{s}\right]=\frac{-1600}{s\left(s^2+16\right)}\end{array}$

Simplify the above equation to find $I\left(s\right)$.

$\begin{array}{c}I\left(s\right)=\frac{-1600}{s\left(s^2+16\right)}\left(\frac{s}{s^2+6s+9}\right)\\ =\frac{-1600}{\left(s^2+16\right)\left(s^2+3s+3s+9\right)}\\ =\frac{-1600}{\left(s^2+16\right)\left(s\left(s+3\right)+3\left(s+3\right)\right)}\\ =\frac{-1600}{\left(s^2+16\right){\left(s+3\right)}^2}\end{array}$

Refer to Figure 3, the voltage $V\left(s\right)$ is calculated as follows:

$V\left(s\right)=\frac{9}{s}I\left(s\right)+\frac{v\left(0\right)}{s}$

Substitute $100$ for $v\left(0\right)$, and $\frac{-1600}{\left(s^2+16\right){\left(s+3\right)}^2}$ for $I\left(s\right)$ in above equation to find $V\left(s\right)$.

$V\left(s\right)=\frac{9}{s}\left(\frac{-1600}{\left(s^2+16\right){\left(s+3\right)}^2}\right)+\frac{100}{s}$

$V\left(s\right)=\frac{-14400}{s\left(s^2+16\right){\left(s+3\right)}^2}+\frac{100}{s}$ (4)

Assume,

$V_1\left(s\right)=\frac{-14400}{s\left(s^2+16\right){\left(s+3\right)}^2}$ (5)

Substitute equation (5) in equation (4).

$V\left(s\right)=V_1\left(s\right)+\frac{100}{s}$ (6)

Take partial fraction for equation (5).

$V_1\left(s\right)=\frac{-14400}{s{\left(s+3\right)}^2\left(s^2+16\right)}=\frac{A}{s}+\frac{B}{s+3}+\frac{C}{{\left(s+3\right)}^2}+\frac{Ds+E}{\left(s^2+16\right)}$ . (7)

The equation (7) can also be written as follows:

$\frac{-14400}{s{\left(s+3\right)}^2\left(s^2+16\right)}=\frac{A{\left(s+3\right)}^2\left(s^2+16\right)+Bs\left(s+3\right)\left(s^2+16\right)+Cs\left(s^2+16\right)+\left(Ds+E\right)s{\left(s+3\right)}^2}{s{\left(s+3\right)}^2\left(s^2+16\right)}$

Simplify the above equation as follows:

$\begin{array}{l}-14400=A{\left(s+3\right)}^2\left(s^2+16\right)+Bs\left(s+3\right)\left(s^2+16\right)+Cs\left(s^2+16\right)+\left(Ds+E\right)s{\left(s+3\right)}^2\\ -14400=\left[\begin{array}{l}A\left(s^2+6s+9\right)\left(s^2+16\right)+\left(B{s}^2+3Bs\right)\left(s^2+16\right)\\ +\left(C{s}^3+16Cs\right)+\left(D{s}^2+Es\right)\left(s^2+6s+9\right)\end{array}\right]\left\{\because {\left(a+b\right)}^2=a^2+2ab+b^2\right\}\\ -14400=\left[\begin{array}{l}A{s}^4+16A{s}^2+6A{s}^3+96As+9A{s}^2+144A+B{s}^4+16B{s}^2+3B{s}^3+48Bs\\ +C{s}^3+16Cs+D{s}^4+6D{s}^3+9D{s}^2+E{s}^3+6E{s}^2+9Es\end{array}\right]\\ -14400=\left[\begin{array}{l}A{s}^4+25A{s}^2+6A{s}^3+96As++144A+B{s}^4+16B{s}^2+3B{s}^3+48Bs\\ +C{s}^3+16Cs+D{s}^4+6D{s}^3+9D{s}^2+E{s}^3+6E{s}^2+9Es\end{array}\right]\end{array}$

Simplify the above equation as follows:

$-14400=\left[\begin{array}{l}\left(A+B+D\right)s^4+\left(6A+3B+C+6D+E\right)s^3\\ +\left(25A+16B+9D+6E\right)s^2\\ +\left(96A+48B+16C+9E\right)s+144A\end{array}\right]$ (8)

Equate the co-efficient of constant term in equation (8) to find $A$.

$144A=-14400$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{-14400}{144}\\ =-100\end{array}$

Equate the co-efficient of $s^4$ term in equation (8).

$A+B+D=0$

Simplify the above equation to find $B$.

$B=-D-A$

Substitute $-100$ for $A$ in above equation to find $B$.

$B=-D+100$ (9)

Equate the co-efficient of $s^3$ term in equation (8).

$6A+3B+C+6D+E=0$

Substitute $-100$ for $A$, and $-D+100$ for $B$ in above equation.

$\begin{array}{l}6\left(-100\right)+3\left(-D+100\right)+C+6D+E=0\\ -600-3D+300+C+6D+E=0\\ 3D+C+E-300=0\end{array}$

$3D+C+E=300$ . (10)

Equate the co-efficient of $s^2$ term in equation (8).

$25A+16B+9D+6E=0$

Substitute $-100$ for $A$, and $-D+100$ for $B$ in above equation.

$\begin{array}{l}25\left(-100\right)+16\left(-D+100\right)+9D+6E=0\\ -2500-16D+1600+9D+6E=0\\ -7D+6E=900\\ 6E=900+7D\end{array}$

Simplify the above equation to find $E$.

$E=\frac{900+7D}{6}$ (11)

Equate the co-efficient of $s$ term in equation (8).

$96A+48B+16C+9E=0$

Substitute $-100$ for $A$, and $-D+100$ for $B$ in above equation.

$\begin{array}{l}96\left(-100\right)+48\left(-D+100\right)+16C+9E=0\\ -9600-48D+4800+16C+9E=0\\ -48D+16C+9E-4800=0\end{array}$

$-48D+16C+9E=4800$ (12)

Substitute the equation (11) in equation (10).

$\begin{array}{l}3D+C+\frac{900+7D}{6}=300\\ \frac{18D+6C+900+7D}{6}=300\\ 25D+6C+900=300\left(6\right)\\ 25D+6C=1800-900\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}25D+6C=900\\ 6C=900-25D\end{array}$

Simplify the above equation to find $C$.

$C=\frac{900-25D}{6}$ (13)

Substitute equation (11) in equation (12).

$\begin{array}{l}-48D+16C+9\left(\frac{900+7D}{6}\right)=4800\\ -48D+16C+3\left(\frac{900+7D}{2}\right)=4800\\ \frac{-96D+32C+2700+21D}{2}=4800\\ -75D+32C+2700=4800\left(2\right)\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}-75D+32C=9600-2700\\ -75D+32C=6900\end{array}$

Substitute the equation (13) in above equation to find $D$.

$\begin{array}{l}-75D+32\left(\frac{900-25D}{6}\right)=6900\\ -75D+16\left(\frac{900-25D}{3}\right)=6900\\ \frac{-225D+14400-400D}{3}=6900\\ -625D+14400=6900\left(3\right)\\ -625D=20700-14400\end{array}$

Simplify the above equation to find $D$.

$\begin{array}{c}D=\frac{6300}{-625}\\ =-10.1\end{array}$

Substitute $-10.1$ for $D$ in equation (13) to find $C$.

$\begin{array}{c}C=\frac{900-25\left(-10.1\right)}{6}\\ =192\end{array}$

Substitute $-10.1$ for $D$ in equation (11) to find $E$.

$\begin{array}{c}E=\frac{900+7\left(-10.1\right)}{6}\\ =138.3\end{array}$

Substitute $-10.1$ for $D$ in equation (9) to find $B$.

$\begin{array}{c}B=-\left(-10.1\right)+100\\ =110.1\end{array}$

Substitute $-100$ for $A$, $110.1$ for $B$, $192$ for $C$, $-10.1$ for $D$, and $138.3$ for $E$ in equation (7) to find $V_1\left(s\right)$.

$V_1\left(s\right)=\frac{-100}{s}+\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}+\frac{-10.1s+138.3}{\left(s^2+16\right)}$

Substitute $\frac{-100}{s}+\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}+\frac{-10.1s+138.3}{\left(s^2+16\right)}$ for $V_1\left(s\right)$ in equation (6) to find $V\left(s\right)$.

$\begin{array}{c}V\left(s\right)=\frac{-100}{s}+\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}+\frac{-10.1s+138.3}{\left(s^2+16\right)}+\frac{100}{s}\\ =\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}+\frac{-10.1s+138.3}{\left(s^2+16\right)}\\ =\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}-\frac{10.1s}{\left(s^2+16\right)}+\frac{138.3}{\left(s^2+16\right)}\\ =\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}-\frac{10.1s}{\left(s^2+16\right)}+\frac{138.3\left(4\right)}{4\left(s^2+16\right)}\end{array}$

Simplify the above equation as follows:

$\begin{array}{c}V\left(s\right)=\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}-\frac{10.1s}{\left(s^2+16\right)}+\frac{\left(34.58\right)\left(4\right)}{\left(s^2+16\right)}\\ =\frac{110.1}{s+3}+\frac{192}{{\left(s+3\right)}^2}-\frac{10.1s}{\left(s^2+4^2\right)}+\frac{\left(34.58\right)\left(4\right)}{\left(s^2+4^2\right)}\end{array}$

Take inverse Laplace transform for above equation to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left[\begin{array}{l}110.1e^{-3t}u\left(t\right)+192t{e}^{-3t}u\left(t\right)\\ -10.1\cos \left(4t\right)u\left(t\right)+34.58\sin \left(4t\right)u\left(t\right)\end{array}\right]\left\{\begin{array}{l}\because {\mathcal{L}}^{-1}\left(\frac{1}{s+a}\right)=e^{-at}u\left(t\right),\\ {\mathcal{L}}^{-1}\left(\frac{1}{{\left(s+a\right)}^n}\right)=\frac{t^{n-1}}{\left(n-1\right)!}{e}^{-at}u\left(t\right)\\ {\mathcal{L}}^{-1}\left(\frac{s}{s^2+a^2}\right)=\cos \left(at\right)u\left(t\right),\\ {\mathcal{L}}^{-1}\left(\frac{a}{s^2+a^2}\right)=\sin \left(at\right)u\left(t\right)\end{array}\right\}\\ =\left[110.1e^{-3t}+192t{e}^{-3t}-10.1\cos \left(4t\right)+34.58\sin \left(4t\right)\right]u\left(t\right)\text{V}\end{array}$

Conclusion:

Thus, the expression of voltage response $v\left(t\right)$ for the $RLC$ circuit shown in Figure 16.88 is $\left[110.1e^{-3t}+192t{e}^{-3t}-10.1\cos \left(4t\right)+34.58\sin \left(4t\right)\right]u\left(t\right)\text{V}$.