Chapter 16, Problem 17P

If i_s(t) = 7.5e^–2t u(t) A in the circuit shown in Fig. 16.40, find the value of i_o(t).

To determine

Find the expression of current $i_o\left(t\right)$ in the circuit of Figure 16.40.

Answer to Problem 17P

The expression of current $i_o\left(t\right)$ in the circuit of Figure 16.40 is $7.5\left[e^{-2t}-0.7559e^{-0.5t}\sin \left(1.3229t\right)\right]u\left(t\right)\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 16.40 in the textbook.

The value of source current is,

$i_s\left(t\right)=7.5e^{-2t}u\left(t\right)\text{A}$ (1)

Formula used:

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

$Z_R=R$ (2)

Here,

$R$ is the value of resistance.

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

$Z_L=sL$ (3)

Here,

$L$ is the value of inductance.

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

$Z_C=\frac{1}{sC}$ (4)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Substitute $2\Omega$ for $R$ in equation (2) to find $Z_R$.

$Z_R=2\Omega$

Substitute $1\text{H}$ for $L$ in equation (3) to find $Z_L$.

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

Substitute $0.5\text{F}$ for $C$ in equation (4) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{s\left(0.5\text{F}\right)}\\ =\frac{2}{s}\end{array}$

Apply Laplace transform for equation (1) to find $I_s\left(s\right)$.

$I_s\left(s\right)=\frac{7.5}{s+2}\left\{\because \mathcal{L}\left(e^{-at}u\left(t\right)\right)=\frac{1}{s+a}\right\}$

Convert the Figure 1 into s-domain as shown in Figure 2.

Apply nodal analysis at node $V\left(s\right)$ in the circuit shown in Figure 2.

$\begin{array}{l}-\frac{7.5}{s+2}+\frac{V\left(s\right)}{s}+\frac{V\left(s\right)}{2}+\frac{V\left(s\right)}{\left(\frac{2}{s}\right)}=0\\ \frac{V\left(s\right)}{s}+\frac{V\left(s\right)}{2}+\frac{sV\left(s\right)}{2}=\frac{7.5}{s+2}\\ V\left(s\right)\left[\frac{1}{s}+\frac{1}{2}+\frac{s}{2}\right]=\frac{7.5}{s+2}\\ V\left(s\right)\left[\frac{2+s+s^2}{2s}\right]=\frac{7.5}{s+2}\end{array}$

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

$V\left(s\right)=\left(\frac{7.5}{s+2}\right)\left(\frac{2s}{s^2+s+2}\right)$

$V\left(s\right)=\frac{15s}{\left(s+2\right)\left(s^2+s+2\right)}$ (5)

From the above equation , the characteristic equation is

$s^2+s+2=0$ (6)

Write a general expression to calculate the roots of quadratic equation $\left(a{s}^2+bs+c=0\right)$.

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (7)

Comparing the equation (6) with the equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=1\\ c=2\end{array}$

Substitute $1$ for $a$, $1$ for $b$, and $2$ for $c$ in equation (7) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}\\ =\frac{-1\pm \sqrt{1-8}}{2\left(1\right)}\\ =\frac{-1\pm \sqrt{-7}}{2}\\ =\frac{-1\pm j2.6457}{2}\end{array}$

Simplify the above equation to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-1\pm j2.6457}{2},\frac{-1\pm j2.6457}{2}\\ =-0.5+j1.3229,-0.5-j1.3229\end{array}$

Substitute the roots of characteristic equation in equation (5) to find $V\left(s\right)$.

$\begin{array}{c}V\left(s\right)=\frac{15s}{\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\left(\left(s-\left(-0.5-j1.3229\right)\right)\right)}\\ =\frac{15s}{\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\left(s-\left(-0.5-j1.3229\right)\right)}\end{array}$

Refer to Figure 2, the current through the capacitor is calculated as follows:

$\begin{array}{c}{I}_o\left(s\right)=\frac{V\left(s\right)}{\left(\frac{2}{s}\right)}\\ =\frac{sV\left(s\right)}{2}\end{array}$

Substitute $\frac{15s}{\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\left(s-\left(-0.5-j1.3229\right)\right)}$ for $V\left(s\right)$ in above equation to find $I_o\left(s\right)$.

$\begin{array}{c}{I}_o\left(s\right)=\frac{s\left[\frac{15s}{\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\left(s-\left(-0.5-j1.3229\right)\right)}\right]}{2}\\ =\frac{7.5s^2}{\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\left(s-\left(-0.5-j1.3229\right)\right)}\end{array}$

Take partial fraction for above equation.

$I_o\left(s\right)=\left[\frac{7.5s^2}{\left[\begin{array}{l}\left(s+2\right)\\ \left(s-\left(-0.5+j1.3229\right)\right)\\ \left(s-\left(-0.5-j1.3229\right)\right)\end{array}\right]}\right]=\left[\begin{array}{l}\frac{A}{s+2}\\ +\frac{B}{\left(s-\left(-0.5+j1.3229\right)\right)}\\ +\frac{C}{\left(s-\left(-0.5-j1.3229\right)\right)}\end{array}\right]$ (8)

The equation (8) can be written as follows:

$\frac{7.5s^2}{\left[\begin{array}{l}\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\\ \left(s-\left(-0.5-j1.3229\right)\right)\end{array}\right]}=\left[\frac{\begin{array}{l}A\left(s-\left(-0.5+j1.3229\right)\right)\\ \left(s-\left(-0.5-j1.3229\right)\right)\\ +B\left(s+2\right)\left(s-\left(-0.5-j1.3229\right)\right)\\ +C\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\end{array}}{\left[\begin{array}{l}\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\\ \left(s-\left(-0.5-j1.3229\right)\right)\end{array}\right]}\right]$

Simplify the above equation as follows:

$7.5s^2=\left[\begin{array}{l}A\left(s-\left(-0.5+j1.3229\right)\right)\left(s-\left(-0.5-j1.3229\right)\right)\\ +B\left(s+2\right)\left(s-\left(-0.5-j1.3229\right)\right)\\ +C\left(s+2\right)\left(s-\left(-0.5+j1.3229\right)\right)\end{array}\right]$ . (9)

Substitute $-2$ for $s$ in equation (9) to find $A$.

$\begin{array}{l}7.5{\left(-2\right)}^2=\left[\begin{array}{l}A\left(-2-\left(-0.5+j1.3229\right)\right)\left(-2-\left(-0.5-j1.3229\right)\right)\\ +B\left(-2+2\right)\left(-2-\left(-0.5-j1.3229\right)\right)\\ +C\left(-2+2\right)\left(-2-\left(-0.5+j1.3229\right)\right)\end{array}\right]\\ 30=\left[\begin{array}{l}A\left(-2-\left(-0.5+j1.3229\right)\right)\left(-2-\left(-0.5-j1.3229\right)\right)\\ +B\left(0\right)\left(-2-\left(-0.5-j1.3229\right)\right)\\ +C\left(0\right)\left(-2-\left(-0.5+j1.3229\right)\right)\end{array}\right]\\ 30=\left[\begin{array}{l}A\left[\begin{array}{l}4+2\left(\left(-0.5-j1.3229\right)\right)+2\left(-0.5+j1.3229\right)\\ +\left(-0.5+j1.3229\right)\left(-0.5-j1.3229\right)\end{array}\right]\\ +0+0\end{array}\right]\\ 30=\left[\begin{array}{l}A\left[\begin{array}{l}4+-1-j2.6458-1+j2.6458\\ +{\left(-0.5\right)}^2-{\left(j1.3229\right)}^2\end{array}\right]\\ +0+0\end{array}\right]\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}30=A\left[2+0.25-j^{2}1.75\right]\\ 30=A\left(2.25-\left(-1\right)1.75\right)\left\{\because j^2=-1\right\}\\ 30=A\left(2.25+1.75\right)\\ 30=A\left(4\right)\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{30}{4}\\ =7.5\end{array}$

Substitute $-0.5+j1.3229$ for $s$ in equation (9) to find $B$.

$\begin{array}{l}7.5{\left(-0.5+j1.3229\right)}^2=\left[\begin{array}{l}A\left(\left(-0.5+j1.3229\right)-\left(-0.5+j1.3229\right)\right)\\ \left(\left(-0.5+j1.3229\right)-\left(-0.5-j1.3229\right)\right)\\ +B\left(-0.5+j1.3229+2\right)\left(\left(-0.5+j1.3229\right)-\left(-0.5-j1.3229\right)\right)\\ +C\left(\left(-0.5+j1.3229\right)+2\right)\left(\left(-0.5+j1.3229\right)-\left(-0.5+j1.3229\right)\right)\end{array}\right]\\ 7.5\left(-1.5-j1.3229\right)=\left[\begin{array}{l}A\left(0\right)+B\left(1.5+j1.3229\right)\left(j2.646\right)\\ +C\left(0\right)\end{array}\right]\\ -11.25-j9.92175=B\left(-3.5+j3.969\right)\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{-11.25-j9.92175}{\left(-3.5+j3.969\right)}\\ =2.8346\angle 90°\end{array}$

Substitute $-0.5-j1.3229$ for $s$ in equation (8) to find $C$.

$\begin{array}{l}7.5{\left(-0.5-j1.3229\right)}^2=\left[\begin{array}{l}A\left(\left(-0.5-j1.3229\right)-\left(-0.5+j1.3229\right)\right)\\ \left(\left(-0.5-j1.3229\right)-\left(-0.5-j1.3229\right)\right)\\ +B\left(-0.5-j1.3229+2\right)\left(\left(-0.5-j1.3229\right)-\left(-0.5-j1.3229\right)\right)\\ +C\left(\left(-0.5-j1.3229\right)+2\right)\left(\left(-0.5-j1.3229\right)-\left(-0.5+j1.3229\right)\right)\end{array}\right]\\ 7.5\left(-1.5+j1.3229\right)=\left[\begin{array}{l}A\left(0\right)+B\left(0\right)\\ +C\left(1.5-j1.3229\right)\left(-j2.646\right)\end{array}\right]\\ -11.25+j9.92175=C\left(-3.5-j3.969\right)\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{-11.25+j9.92175}{\left(-3.5-j3.969\right)}\\ =2.8346\angle -90°\end{array}$

Substitute $7.5$ for $A$, $2.8346\angle 90°$ for $B$, and $2.8346\angle -90°$ for $C$ in equation (8) to find $I_o\left(s\right)$.

$\begin{array}{c}{I}_o\left(s\right)=\left[\frac{7.5}{s+2}+\frac{2.8346\angle 90°}{\left(s-\left(-0.5+j1.3229\right)\right)}+\frac{2.8346\angle -90°}{\left(s-\left(-0.5-j1.3229\right)\right)}\right]\\ =\left[\frac{7.5}{s+2}+\frac{2.8346\left(\cos \left(90°\right)+j\sin \left(90°\right)\right)}{\left(s-\left(-0.5+j1.3229\right)\right)}+\frac{2.8346\left(\cos \left(-90°\right)+j\sin \left(-90°\right)\right)}{\left(s-\left(-0.5-j1.3229\right)\right)}\right]\\ =\left[\frac{7.5}{s+2}+\frac{2.8346e^{j90°}}{\left(s-\left(-0.5+j1.3229\right)\right)}+\frac{2.8346e^{-j90°}}{\left(s-\left(-0.5-j1.3229\right)\right)}\right]\left\{\because e^{i\theta }=\cos \theta +i\sin \theta \right\}\end{array}$

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

$\begin{array}{c}{i}_o\left(t\right)=\left[\begin{array}{c}7.5e^{-2t}u\left(t\right)+2.8346e^{j90°}{e}^{\left(-0.5+j1.3229\right)t}u\left(t\right)\\ +2.8346e^{-j90°}{e}^{\left(-0.5-j1.3229\right)t}u\left(t\right)\end{array}\right]\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s-a}\right)=e^{at}u\left(t\right)\right\}\\ =\left[\begin{array}{l}7.5e^{-2t}+2.8346e^{\left(-0.5t+j1.3229t+j90°\right)}\\ +2.8346e^{\left(-0.5t-j1.3229t-j90°\right)}\end{array}\right]u\left(t\right)\left\{\because e^a\cdot e^b=e^{a+b}\right\}\\ =\left[7.5e^{-2t}+2.8346e^{-0.5t}{e}^{j1.3229t+j90°}+2.8346e^{-0.5t}{e}^{-j1.3229t-j90°}\right]u\left(t\right)\\ =\left[7.5e^{-2t}+2.8346e^{-0.5t}{e}^{j\left(1.3229t+90°\right)}+2.8346e^{-0.5t}{e}^{-j\left(1.3229t+j90°\right)}\right]u\left(t\right)\end{array}$

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

$\begin{array}{c}{i}_o\left(t\right)=\left[7.5e^{-2t}+2.8346e^{-0.5t}{e}^{j\left(1.3229t+90°\right)}+2.8346e^{-0.5t}{e}^{-j\left(1.3229t+j90°\right)}\right]u\left(t\right)\\ =\left[7.5e^{-2t}+2.8346e^{-0.5t}\left[e^{j\left(1.3229t+90°\right)}+e^{-j\left(1.3229t+j90°\right)}\right]\right]u\left(t\right)\\ =\left[7.5e^{-2t}+2.8346e^{-0.5t}\left[2\cos \left(1.3229t+90°\right)\right]\right]u\left(t\right)\left\{\because \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}\right\}\\ =\left[7.5e^{-2t}+5.6692e^{-0.5t}\cos \left(1.3229t+90°\right)\right]u\left(t\right)\text{A}\end{array}$

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

$\begin{array}{c}{i}_o\left(t\right)=\left[7.5e^{-2t}+5.6692e^{-0.5t}\left(-\sin \left(1.3229t\right)\right)\right]u\left(t\right)\text{A}\left\{\because \cos \left(90°+\theta \right)=-\sin \theta \right\}\\ =7.5\left[e^{-2t}-0.7559e^{-0.5t}\sin \left(1.3229t\right)\right]u\left(t\right)\text{A}\end{array}$

Conclusion:

Thus, the expression of current $i_o\left(t\right)$ in the circuit of Figure 16.40 is $7.5\left[e^{-2t}-0.7559e^{-0.5t}\sin \left(1.3229t\right)\right]u\left(t\right)\text{A}$.