Chapter 16, Problem 26P

To determine

Find the expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 16.49.

Answer to Problem 26P

The expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 16.49 is $2.582e^{-5t}\cos \left(19.365t-14.48°\right)u\left(t\right)\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 16.49 in the textbook.

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 a DC circuit, at steady state condition when the switch is in position A at time $t=0^{-}$ the capacitor acts like open circuit and the inductor acts like short circuit.

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

Refer to Figure 2, the short circuited inductor is connected in parallel with resistors $R_1$ and $R_2$. The full current flows through the short circuited inductor so the resistor $R_1$ and $R_2$ also short circuited.

Now, the Figure 2 is reduced as shown in Figure 3.

Refer to Figure 3, the current flow through the inductor is same as the value of current source $\left(i_s\right)$.

$i_L\left(0^{-}\right)=2.5\text{A}$

Refer to Figure 3, there is no capacitor placed in a circuit. Therefore, the voltage across the capacitor is zero.

$v_C\left(0^{-}\right)=0$

The current through inductor and voltage across capacitor is always continuous so that,

$i\left(0\right)=i_L\left(0^{-}\right)=i_L\left(0^{+}\right)=2.5\text{A}$

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

For time $t>0$ the switch is in position B and the Figure 1 is reduced as shown in Figure 4.

Substitute $10\Omega$ for $R_2$ in equation (1) to find $Z_{R_2}$.

$Z_{R_2}=10$

Substitute $0.25\text{H}$ for $L$ in equation (2) to find $Z_L$.

$\begin{array}{c}{Z}_L=s\left(0.25\text{H}\right)\\ =0.25s\end{array}$

Substitute $10\text{mF}$ for $C$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{s\left(10\text{mF}\right)}\\ =\frac{1}{s\left(10\times {10}^{-3}\text{F}\right)}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{100}{s}\end{array}$

Using element transformation methods with initial conditions convert the Figure 4 into s-domain.

Apply Kirchhoff’s current law for the circuit shown in Figure 5.

$\begin{array}{l}\frac{V\left(s\right)}{\left(\frac{100}{s}\right)}+\frac{V\left(s\right)}{10}+\frac{V\left(s\right)}{0.25s}+\frac{i\left(0\right)}{s}=0\\ \frac{sV\left(s\right)}{100}+\frac{V\left(s\right)}{10}+\frac{V\left(s\right)}{0.25s}=-\frac{i\left(0\right)}{s}\\ V\left(s\right)\left[\frac{s}{100}+\frac{1}{10}+\frac{1}{0.25s}\right]=-\frac{i\left(0\right)}{s}\\ V\left(s\right)\left[\frac{s^2+10s+400}{100s}\right]=-\frac{i\left(0\right)}{s}\end{array}$

Substitute $2.5$ for $i\left(0\right)$ in above equation.

$V\left(s\right)\left[\frac{s^2+10s+400}{100s}\right]=-\frac{2.5}{s}$

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

$V\left(s\right)=-\frac{2.5}{s}\left(\frac{100s}{s^2+10s+400}\right)$

$V\left(s\right)=\frac{-250}{s^2+10s+400}$ . (4)

From the equation (4), the characteristic equation is

$s^2+10s+400=0$ (5)

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

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (6)

Comparing the equation (5) with the equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=10\\ c=400\end{array}$

Substitute $1$ for $a$, $10$ for $b$, and $400$ for $c$ in equation (6) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-10\pm \sqrt{{\left(10\right)}^2-4\left(1\right)\left(400\right)}}{2\left(1\right)}\\ =\frac{-10\pm \sqrt{10-1600}}{2\left(1\right)}\\ =\frac{-10\pm \sqrt{-1500}}{2}\\ =\frac{-10\pm j38.73}{2}\end{array}$

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

$\begin{array}{c}{s}_{1,2}=\frac{-10+j38.73}{2},\frac{-10-j38.73}{2}\\ =-5+j19.365,-5-j19.365\end{array}$

Substitute the roots of characteristic equation in equation (4) to find $V\left(s\right)$.

$\begin{array}{c}V\left(s\right)=\frac{-250}{\left(s-\left(-5+j19.365\right)\right)\left(\left(s-\left(-5-j19.365\right)\right)\right)}\\ =\frac{-12}{\left(s-\left(-5+j19.365\right)\right)\left(s-\left(-5-j19.365\right)\right)}\end{array}$

Take partial fraction for above equation.

$V\left(s\right)=\frac{-250}{\left[\begin{array}{l}\left(s-\left(-5+j19.365\right)\right)\\ \left(\left(s-\left(-5-j19.365\right)\right)\right)\end{array}\right]}=\left[\begin{array}{l}\frac{A}{\left(\left(s-\left(-5+j19.365\right)\right)\right)}\\ +\frac{B}{\left(\left(s-\left(-5-j19.365\right)\right)\right)}\end{array}\right]$ (7)

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

$\frac{-250}{\left[\begin{array}{l}\left(s-\left(-5+j19.365\right)\right)\\ \left(\left(s-\left(-5-j19.365\right)\right)\right)\end{array}\right]}=\frac{A\left(s-\left(-5-j19.365\right)\right)+B\left(s-\left(-5+j19.365\right)\right)}{\left(s-\left(-5+j19.365\right)\right)\left(s-\left(-5-j19.365\right)\right)}$

Simplify the above equation as follows:

$-250=A\left(s-\left(-5-j19.365\right)\right)+B\left(s-\left(-5+j19.365\right)\right)$ (8)

Substitute $-5+j19.365$ for $s$ in equation (8) to find $A$.

$\begin{array}{l}-250=A\left(\left(-5+j19.365\right)-\left(-5-j19.365\right)\right)+B\left(\left(-5+j19.365\right)-\left(-5+j19.365\right)\right)\\ -250=A\left(j38.73\right)+B\left(0\right)\\ A\left(j38.73\right)=-250\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{-250}{\left(j38.73\right)}\\ =6.45\angle 90°\end{array}$

Substitute $-5-j19.365$ for $s$ in equation (8) to find $B$.

$\begin{array}{l}-250=A\left(\left(-5-j19.365\right)-\left(-5-j19.365\right)\right)+B\left(\left(-5-j19.365\right)-\left(-0.5+j1.9365\right)\right)\\ -12=A\left(0\right)+B\left(-j38.73\right)\\ B\left(-j38.73\right)=-250\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{-250}{\left(-j38.73\right)}\\ =6.45\angle -90°\end{array}$

Substitute $6.45\angle 90°$ for $A$, and $6.45\angle -90°$ for $B$ in equation (7) to find $V\left(s\right)$.

$\begin{array}{c}V\left(s\right)=\frac{6.45\angle -90°}{\left(s-\left(-5+j19.365\right)\right)}+\frac{6.45\angle 90°}{\left(s-\left(-5-j19.365\right)\right)}\\ =\frac{6.45\left(\cos \left(90°\right)+\sin \left(90°\right)\right)}{\left(s-\left(-5+j19.365\right)\right)}+\frac{6.45\left(\cos \left(-90\right)°+\sin \left(-90\right)°\right)}{\left(s-\left(-5-j19.365\right)\right)}\\ =\frac{6.45e^{j\left(90°\right)}}{\left(s-\left(-5+j19.365\right)\right)}+\frac{6.45e^{j\left(-90°\right)}}{\left(s-\left(-5-j19.365\right)\right)}\left\{\because e^{i\theta }=\cos \theta +i\sin \theta \right\}\end{array}$

Refer to Figure 5, the current $I\left(s\right)$ is calculated as follows:

$I\left(s\right)=\frac{V\left(s\right)}{0.25s}+\frac{i\left(0\right)}{s}$

Substitute $\frac{6.45e^{j\left(90°\right)}}{\left(s-\left(-5+j19.365\right)\right)}+\frac{6.45e^{j\left(-90°\right)}}{\left(s-\left(-5-j19.365\right)\right)}$ for $V\left(s\right)$, and $2.5$ for $i\left(0\right)$ in above equation to find $I\left(s\right)$.

$\begin{array}{c}I\left(s\right)=\frac{\left[\frac{6.45e^{j\left(90°\right)}}{\left(s-\left(-5+j19.365\right)\right)}+\frac{6.45e^{j\left(-90°\right)}}{\left(s-\left(-5-j19.365\right)\right)}\right]}{0.25s}+\frac{2.5}{s}\\ =\left[\frac{6.45e^{j\left(90°\right)}}{0.25s\left(s-\left(-5+j19.365\right)\right)}+\frac{6.45e^{j\left(-90°\right)}}{0.25s\left(s-\left(-5-j19.365\right)\right)}\right]+\frac{2.5}{s}\end{array}$

$I\left(s\right)=\left[\frac{25.8e^{j\left(90°\right)}}{s\left(s-\left(-5+j19.365\right)\right)}+\frac{25.8e^{j\left(-90°\right)}}{s\left(s-\left(-5-j19.365\right)\right)}\right]+\frac{2.5}{s}$ (9)

Assume,

$I_1\left(s\right)=\frac{25.8e^{j\left(90°\right)}}{s\left(s-\left(-5+j19.365\right)\right)}$ (10)

$I_2\left(s\right)=\frac{25.8e^{j\left(-90°\right)}}{s\left(s-\left(-5-j19.365\right)\right)}$ (11)

Substitute equation (10) and (11) in equation (9).

$I\left(s\right)=I_1\left(s\right)+I_2\left(s\right)+\frac{2.5}{s}$ (12)

Take partial fraction for equation (10).

$I_1\left(s\right)=\frac{25.8e^{j\left(90°\right)}}{s\left(s-\left(-5+j19.365\right)\right)}=\frac{C}{s}+\frac{D}{\left(s-\left(-5+j19.365\right)\right)}$ (13)

The equation (13) can also be written as follows:

$\frac{25.8e^{j\left(90°\right)}}{s\left(s-\left(-5+j19.365\right)\right)}=\frac{C\left(s-\left(-5+j19.365\right)\right)+Ds}{s\left(s-\left(-5+j19.365\right)\right)}$

Simplify the above equation as follows:

$25.8e^{j\left(90°\right)}=C\left(s-\left(-5+j19.365\right)\right)+Ds$ (14)

Substitute $0$ for $s$ in equation (14) to find $C$.

$\begin{array}{l}25.8e^{j\left(90°\right)}=C\left(0-\left(-5+j19.365\right)\right)+D\left(0\right)\\ C\left(5-j19.365\right)=25.8e^{j\left(90°\right)}\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{25.8e^{j\left(90°\right)}}{\left(5-j19.365\right)}\\ =1.291\angle 165.52°\end{array}$

Substitute $-5+j19.365$ for $s$ in equation (14) to find $D$.

$\begin{array}{l}25.8e^{j\left(90°\right)}=C\left(\left(-5+j19.365\right)-\left(-5+j19.365\right)\right)+D\left(-5+j19.365\right)\\ D\left(-5+j19.365\right)=25.8e^{j\left(90°\right)}\end{array}$

Simplify the above equation to find $D$.

$\begin{array}{c}D=\frac{25.8e^{j\left(90°\right)}}{\left(-5+j19.365\right)}\\ =1.291\angle -14.48°\end{array}$

Substitute $1.291\angle 165.52°$ for $C$, and $1.291\angle -14.48°$ for $D$ in equation (13) to find $I_1\left(s\right)$.

$I_1\left(s\right)=\frac{1.291\angle 165.52°}{s}+\frac{1.291\angle -14.48°}{\left(s-\left(-5+j19.365\right)\right)}$

Take partial fraction for equation (11).

$I_2\left(s\right)=\frac{25.8e^{j\left(-90°\right)}}{s\left(s-\left(-5-j19.365\right)\right)}=\frac{E}{s}+\frac{F}{\left(s-\left(-5-j19.365\right)\right)}$ (15)

The equation (13) can also be written as follows:

$\frac{25.8e^{j\left(-90°\right)}}{s\left(s-\left(-5-j19.365\right)\right)}=\frac{E\left(s-\left(-5-j19.365\right)\right)+Fs}{s\left(s-\left(-5-j19.365\right)\right)}$

Simplify the above equation as follows:

$25.8e^{j\left(-90°\right)}=E\left(s-\left(-5-j19.365\right)\right)+Fs$ (16)

Substitute $0$ for $s$ in equation (16) to find $E$.

$\begin{array}{l}25.8e^{j\left(-90°\right)}=E\left(0-\left(-5-j19.365\right)\right)+F\left(0\right)\\ E\left(5+j19.365\right)=25.8e^{j\left(-90°\right)}\end{array}$

Simplify the above equation to find $E$.

$\begin{array}{c}E=\frac{25.8e^{j\left(-90°\right)}}{\left(5+j19.365\right)}\\ =1.291\angle -165.52°\end{array}$

Substitute $-5-j19.365$ for $s$ in equation (16) to find $F$.

$\begin{array}{l}25.8e^{j\left(-90°\right)}=E\left(\left(-5-j19.365\right)-\left(-5-j19.365\right)\right)+F\left(-5-j19.365\right)\\ F\left(-5-j19.365\right)=25.8e^{j\left(-90°\right)}\end{array}$

Simplify the above equation to find $F$.

$\begin{array}{c}F=\frac{25.8e^{j\left(-90°\right)}}{\left(-5-j19.365\right)}\\ =1.291\angle 14.48°\end{array}$

Substitute $1.291\angle -165.52°$ for $E$, and $1.291\angle 14.48°$ for $F$ in equation (15) to find $I_2\left(s\right)$.

$I_2\left(s\right)=\frac{1.291\angle -165.52°}{s}+\frac{1.291\angle 14.48°}{\left(s-\left(-5-j19.365\right)\right)}$

Substitute $\frac{1.291\angle 165.52°}{s}+\frac{1.291\angle -14.48°}{\left(s-\left(-5+j19.365\right)\right)}$ for $I_1\left(s\right)$, and $\frac{1.291\angle -165.52°}{s}+\frac{1.291\angle 14.48°}{\left(s-\left(-5-j19.365\right)\right)}$ for $I_2\left(s\right)$ in equation (12) to find $I\left(s\right)$.

$\begin{array}{c}I\left(s\right)=\left[\begin{array}{c}\frac{1.291\angle 165.52°}{s}+\frac{1.291\angle -14.48°}{\left(s-\left(-5+j19.365\right)\right)}+\frac{1.291\angle -165.52°}{s}\\ +\frac{1.291\angle 14.48°}{\left(s-\left(-5-j19.365\right)\right)}+\frac{2.5}{s}\end{array}\right]\\ =\frac{-2.5}{s}+\frac{1.291\angle -14.48°}{\left(s-\left(-5+j19.365\right)\right)}+\frac{1.291\angle 14.48°}{\left(s-\left(-5-j19.365\right)\right)}+\frac{2.5}{s}\\ =\frac{1.291\angle -14.48°}{\left(s-\left(-5+j19.365\right)\right)}+\frac{1.291\angle 14.48°}{\left(s-\left(-5-j19.365\right)\right)}\\ =\frac{1.291\left(\cos \left(-14.48°\right)+\sin \left(-14.48°\right)\right)}{\left(s-\left(-5+j19.365\right)\right)}+\frac{1.291\left(\cos \left(14.48\right)°+\sin \left(14.48\right)°\right)}{\left(s-\left(-5-j19.365\right)\right)}\end{array}$$I\left(s\right)=\frac{1.291e^{j\left(-14.48°\right)}}{\left(s-\left(-5+j19.365\right)\right)}+\frac{1.291e^{j\left(14.48°\right)}}{\left(s-\left(-5-j19.365\right)\right)}\left\{\because e^{i\theta }=\cos \theta +i\sin \theta \right\}$

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

$\begin{array}{c}i\left(t\right)=\left[\begin{array}{c}1.291e^{j\left(-14.48°\right)}{e}^{\left(-5+j19.365\right)t}u\left(t\right)\\ +1.291e^{j\left(14.48°\right)}{e}^{\left(-5-j19.365\right)t}u\left(t\right)\end{array}\right]\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s+a}\right)=e^{-at}u\left(t\right)\right\}\\ =\left[1.291e^{\left(-5t+j19.365t-j14.48°\right)}+1.291e^{\left(-5t-j19.365t+j14.48°\right)}\right]u\left(t\right)\left\{\because e^a\cdot e^b=e^{a+b}\right\}\\ =\left[1.291e^{-5t}{e}^{j\left(19.365t-14.48°\right)}+1.291e^{-5t}{e}^{-j\left(1.9365t-14.48°\right)}\right]u\left(t\right)\end{array}$

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

$\begin{array}{c}i\left(t\right)=\left[1.291e^{-5t}{e}^{j\left(19.365t-14.48°\right)}+1.291e^{-5t}{e}^{-j\left(1.9365t-14.48°\right)}\right]\\ =1.291e^{-5t}\left[e^{j\left(19.365t-14.48°\right)}+e^{-j\left(1.9365t-14.48°\right)}\right]u\left(t\right)\\ =1.291e^{-5t}\left[2\cos \left(19.365t-14.48°\right)\right]u\left(t\right)\left\{\because \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}\right\}\\ =2.582e^{-5t}\cos \left(19.365t-14.48°\right)u\left(t\right)\text{A}\end{array}$

Conclusion:

Thus, the expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 16.49 is $2.582e^{-5t}\cos \left(19.365t-14.48°\right)u\left(t\right)\text{A}$.