Chapter 16, Problem 51P

In the circuit of Fig. 16.74, find i(t) for t > 0.

To determine

Find the expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 16.74.

Answer to Problem 51P

The expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 16.74 is $\left[-12+41.17e^{-15.125t}\cos \left(4.608t-73.06°\right)\right]u\left(t\right)\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 16.74 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 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 switch is open and there is no source connected to the circuit. Therefore, the voltage across the capacitor and current through inductor are zero.

$\begin{array}{l}{i}_L\left(0^{-}\right)=0\text{A}\\ v_C\left(0^{-}\right)=0\text{V}\end{array}$

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

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

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

For time $t>0$, the switch is closed and the Figure 1 is drawn as shown in Figure 3.

Substitute $6\Omega$ for $R_1$ in equation (1) to find $Z_{R_1}$.

$Z_{R_1}=6\Omega$

Substitute $4\Omega$ for $R_2$ in equation (1) to find $Z_{R_2}$.

$Z_{R_2}=4\Omega$

Substitute $\frac{1}{4}\text{H}$ for $L$ in equation (2) to find $Z_L$.

$\begin{array}{c}{Z}_L=s\left(\frac{1}{4}\text{H}\right)\\ =\frac{1}{4}s\end{array}$

Substitute $\frac{1}{25}\text{F}$ for $C$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{s\left(\frac{1}{25\text{F}}\right)}\\ =\frac{25}{s}\end{array}$

Refer to Figure 3, the circuit for $t>0$ is drawn. The source voltage $v$ is also written as follows:

$\begin{array}{c}v=v_s\\ =120\text{V}\\ =120\left(1\right)\\ =120u\left(t\right)\text{V}\left\{u\left(t\right)=1\text{for}t>0\right\}\end{array}$

Apply Laplace transform for above to find $V_s\left(s\right)$.

$\begin{array}{c}{V}_s\left(s\right)=120\left(\frac{1}{s}\right)\left\{\because \mathcal{L}\left(u\left(t\right)\right)=\frac{1}{s}\right\}\\ =\frac{120}{s}\end{array}$

Convert the Figure 3 into s-domain.

Apply nodal analysis at node $V\left(s\right)$ in Figure 4.

$\begin{array}{l}\frac{V\left(s\right)-\frac{120}{s}}{6+\frac{1}{4}s}+\frac{V\left(s\right)}{\left(\frac{25}{s}\right)}+\frac{V\left(s\right)}{4}=0\\ \frac{V\left(s\right)}{\left(6+\frac{1}{4}s\right)}-\frac{\frac{120}{s}}{\left(6+\frac{1}{4}s\right)}+\frac{V\left(s\right)}{\left(\frac{25}{s}\right)}+\frac{V\left(s\right)}{4}=0\\ \frac{V\left(s\right)}{\left(\frac{s+24}{4}\right)}-\frac{\frac{120}{s}}{\left(\frac{s+24}{4}\right)}+\frac{sV\left(s\right)}{25}+\frac{V\left(s\right)}{4}=0\\ \frac{4V\left(s\right)}{s+24}-\frac{4\left(120\right)}{s\left(s+24\right)}+\frac{sV\left(s\right)}{25}+\frac{V\left(s\right)}{4}=0\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}\frac{4V\left(s\right)}{s+24}+\frac{sV\left(s\right)}{25}+\frac{V\left(s\right)}{4}=\frac{480}{s\left(s+24\right)}\\ V\left(s\right)\left[\frac{4}{s+24}+\frac{s}{25}+\frac{1}{4}\right]=\frac{480}{s\left(s+24\right)}\\ V\left(s\right)\left[\frac{4\left(25\right)\left(4\right)+4s\left(s+24\right)+25\left(s+24\right)}{100\left(s+24\right)}\right]=\frac{480}{s\left(s+24\right)}\\ V\left(s\right)\left[\frac{400+4s^2+96s+25s+600}{100\left(s+24\right)}\right]=\frac{480}{s\left(s+24\right)}\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}V\left(s\right)\left[\frac{4s^2+121s+1000}{100\left(s+24\right)}\right]=\frac{480}{s\left(s+24\right)}\\ V\left(s\right)\left[\frac{4\left(s^2+30.25s+250\right)}{100\left(s+24\right)}\right]=\frac{480}{s\left(s+24\right)}\\ V\left(s\right)\left[\frac{\left(s^2+30.25s+250\right)}{25\left(s+24\right)}\right]=\frac{480}{s\left(s+24\right)}\end{array}$

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

$V\left(s\right)=\frac{480}{s\left(s+24\right)}\left(\frac{25\left(s+24\right)}{s^2+30.25s+250}\right)$

$V\left(s\right)=\frac{12000}{s\left(s^2+30.25s+250\right)}$ (4)

From the above equation, the characteristic equation is

$s^2+30.25s+250=0$ (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 equation (5) with the equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=30.25\\ c=250\end{array}$

Substitute $1$ for $a$, $30.25$ for $b$, and $250$ for $c$ in equation (6) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-30.25\pm \sqrt{{\left(30.25\right)}^2-4\left(1\right)\left(250\right)}}{2\left(1\right)}\\ =\frac{-30.25\pm \sqrt{915.0625-1000}}{2\left(1\right)}\\ =\frac{-30.25\pm \sqrt{-84.9375}}{2}\\ =\frac{-30.25\pm j9.216}{2}\end{array}$

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

$\begin{array}{c}{s}_{1,2}=\frac{-30.25+j9.216}{2},\frac{-30.25-j9.216}{2}\\ =-15.25+j4.608,-15.25-j4.608\end{array}$

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

$\begin{array}{c}V\left(s\right)=\frac{12000}{s\left(s-\left(-15.125+j4.608\right)\right)\left(\left(s-\left(-15.125-j4.608\right)\right)\right)}\\ =\frac{12000}{s\left(s-\left(-15.125+j4.608\right)\right)\left(s-\left(-15.125-j4.608\right)\right)}\end{array}$

Take partial fraction for above equation.

$V\left(s\right)=\frac{12000}{\left[\begin{array}{l}s\left(s-\left(-15.125+j4.608\right)\right)\\ \left(s-\left(-15.125-j4.608\right)\right)\end{array}\right]}$

$V\left(s\right)=\left[\begin{array}{l}\frac{A}{s}\\ +\frac{B}{\left(s-\left(-15.125+j4.608\right)\right)}\\ +\frac{C}{\left(s-\left(-15.125-j4.608\right)\right)}\end{array}\right]$ (7)

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

$\frac{12000}{\left[\begin{array}{l}s\left(s-\left(-15.125+j4.608\right)\right)\\ \left(s-\left(-15.125-j4.608\right)\right)\end{array}\right]}=\left[\frac{\begin{array}{l}A\left(s-\left(-15.125+j4.608\right)\right)\left(s-\left(-15.125-j4.608\right)\right)\\ +Bs\left(s-\left(-15.125-j4.608\right)\right)\\ +Cs\left(s-\left(-15.125+j4.608\right)\right)\end{array}}{\left[\begin{array}{l}s\left(s-\left(-15.125+j4.608\right)\right)\\ \left(s-\left(-15.125-j4.608\right)\right)\end{array}\right]}\right]$

Simplify the above equation as follows:

$12000=\left[\begin{array}{l}A\left(s-\left(-15.125+j4.608\right)\right)\left(s-\left(-15.125-j4.608\right)\right)\\ +Bs\left(s-\left(-15.125-j4.608\right)\right)\\ +Cs\left(s-\left(-15.125+j4.608\right)\right)\end{array}\right]$ . (8)

Substitute $0$ for $s$ in equation (8) to find $A$.

$\begin{array}{l}12000=\left[\begin{array}{l}A\left(0-\left(-15.125+j4.608\right)\right)\left(0-\left(-15.125-j4.608\right)\right)\\ +B\left(0\right)\left(0-\left(-15.125-j4.608\right)\right)\\ +C\left(0\right)\left(0-\left(-15.125+j4.608\right)\right)\end{array}\right]\\ 12000=\left[A\left(-\left(-15.125+j4.608\right)\right)\left(-\left(-15.125-j4.608\right)\right)+0+0\right]\\ 12000=\left[A\left(-15.125+j4.608\right)\left(-15.125-j4.608\right)\right]\\ 12000=A\left({\left(-15.125\right)}^2-{\left(j4.608\right)}^2\right)\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}12000=A\left(228.765-j^{2}21.234\right)\\ 12000=A\left(228.765-\left(-1\right)21.234\right)\left\{\because j^2=-1\right\}\\ 12000=A\left(228.765+21.234\right)\\ 12000=A\left(250\right)\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{12000}{250}\\ =48\end{array}$

Substitute $-15.125+j4.608$ for $s$ in equation (8) to find $B$.

$\begin{array}{c}12000=\left[\begin{array}{l}A\left(\left(-15.125+j4.608\right)-\left(-15.125+j4.608\right)\right)\\ \left(\left(-15.125+j4.608\right)-\left(-15.125-j4.608\right)\right)\\ +B\left(-15.125+j4.608\right)\\ \left(\left(-15.125+j4.608\right)-\left(-15.125-j4.608\right)\right)\\ +C\left(-15.125+j4.608\right)\\ \left(\left(-15.125+j4.608\right)-\left(-15.125+j4.608\right)\right)\end{array}\right]\\ 12000=\left[\begin{array}{l}A\left(0\right)\\ +B\left(-15.125+j4.608\right)\left(j9.216\right)\\ +C\left(0\right)\end{array}\right]\\ 12000=\left[0+B\left(-15.125+j4.608\right)\left(j9.216\right)+0\right]\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{12000}{\left(j9.216\right)\left(-15.125+j4.608\right)}\\ =-24+j78.7\\ =82.35\angle 106.94°\end{array}$

Substitute $-15.125-j4.608$ for $s$ in equation (8) to find $C$.

$\begin{array}{c}12000=\left[\begin{array}{l}A\left(\left(-15.125-j4.608\right)-\left(-15.125+j4.608\right)\right)\\ \left(\left(-15.125-j4.608\right)-\left(-15.125-j4.608\right)\right)\\ +B\left(-15.125-j4.608\right)\\ \left(\left(-15.125-j4.608\right)-\left(-15.125-j4.608\right)\right)\\ +C\left(-15.125-j4.608\right)\\ \left(\left(-15.125-j4.608\right)-\left(-15.125+j4.608\right)\right)\end{array}\right]\\ 12000=\left[A\left(0\right)+B\left(0\right)+C\left(-15.125-j4.608\right)\left(-j9.216\right)\right]\end{array}$

Simplify the above equation to find $C$.

$\begin{array}{c}C=\frac{12000}{\left(-j9.216\right)\left(-15.125-j4.608\right)}\\ =-24-j7.87\\ =82.35\angle -106.94°\end{array}$

Substitute $48$ for $A$, $82.35\angle 106.94°$ for $B$, and $82.35\angle -106.94°$ for $C$ in equation (7) to find $V\left(s\right)$.

$V\left(s\right)=\left[\frac{48}{s}+\frac{82.35\angle 106.94°}{\left(s-\left(-15.125+j4.608\right)\right)}+\frac{82.35\angle -106.94°}{\left(s-\left(-15.125-j4.608\right)\right)}\right]$

Refer to Figure 3, the current through resistor $R_2$ is calculated as follows:

$I\left(s\right)=-\frac{V\left(s\right)}{4}$

Substitute $\left[\frac{48}{s}+\frac{82.35\angle 106.94°}{\left(s-\left(-15.125+j4.608\right)\right)}+\frac{82.35\angle -106.94°}{\left(s-\left(-15.125-j4.608\right)\right)}\right]$ for $V\left(s\right)$ in above equation to find $I\left(s\right)$.

$\begin{array}{c}I\left(s\right)=\frac{\left[\frac{48}{s}+\frac{82.35\angle 106.94°}{\left(s-\left(-15.125+j4.608\right)\right)}+\frac{82.35\angle -106.94°}{\left(s-\left(-15.125-j4.608\right)\right)}\right]}{\left(-4\right)}\\ =\left[\frac{-12}{s}+\frac{20.587\angle -73.06°}{\left(s-\left(-15.125+j4.608\right)\right)}+\frac{20.587\angle 73.06°}{\left(s-\left(-15.125-j4.608\right)\right)}\right]\\ =\left[\begin{array}{l}\frac{-12}{s}+\frac{20.587\left(\cos \left(-73.06°\right)+j\sin \left(-73.06°\right)\right)}{\left(s-\left(-15.125+j4.608\right)\right)}\\ +\frac{20.587\left(\cos \left(73.06°\right)+j\sin \left(73.06°\right)\right)}{\left(s-\left(-15.125-j4.608\right)\right)}\end{array}\right]\\ =\left[\begin{array}{l}\frac{-12}{s}+\frac{20.587e^{-j73.06°}}{\left(s-\left(-15.125+j4.608\right)\right)}\\ +\frac{20.587e^{j73.06°}}{\left(s-\left(-15.125-j4.608\right)\right)}\end{array}\right]\left\{\because e^{i\theta }=\cos \theta +i\sin \theta \right\}\end{array}$

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

$\begin{array}{c}i\left(t\right)=\left[\begin{array}{c}-12u\left(t\right)+20.587e^{-j73.06°}{e}^{\left(-15.125+j4.608\right)t}u\left(t\right)\\ +20.587e^{j73.06°}{e}^{\left(-15.125-j4.608\right)t}u\left(t\right)\end{array}\right]\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s-a}\right)=e^{at}u\left(t\right)\right\}\\ =\left[\begin{array}{l}-12+20.587e^{\left(-15.125t+j4.608t-j73.06°\right)}\\ +20.587e^{\left(-15.125t-j4.608t+j73.06°\right)}\end{array}\right]u\left(t\right)\left\{\because e^a\cdot e^b=e^{a+b}\right\}\\ =\left[-12+20.587e^{-15.125t}{e}^{j4.608t-j73.06°}+20.587e^{-15.125t}{e}^{-j4.608t+j73.06°}\right]u\left(t\right)\\ =\left[-12+20.587e^{-15.125t}{e}^{j\left(4.608t-73.06°\right)}+20.587e^{-15.125t}{e}^{-j\left(4.608t-j73.06°\right)}\right]u\left(t\right)\end{array}$

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

$\begin{array}{c}i\left(t\right)=\left[-12+20.587e^{-15.125t}\left[e^{j\left(4.608t-73.06°\right)}+e^{-j\left(4.608t-j73.06°\right)}\right]\right]u\left(t\right)\\ =\left[-12+20.587e^{-15.125t}\left[2\cos \left(4.608t-73.06°\right)\right]\right]u\left(t\right)\left\{\because \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}\right\}\\ =\left[-12+41.17e^{-15.125t}\cos \left(4.608t-73.06°\right)\right]u\left(t\right)\text{A}\end{array}$

Conclusion:

Thus, the expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 16.74 is $\left[-12+41.17e^{-15.125t}\cos \left(4.608t-73.06°\right)\right]u\left(t\right)\text{A}$.