Chapter 16, Problem 53P

In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find:

1. (a) v(0⁺), dv(0⁺)/dt
2. (b) v(t) for t ≥ 0.

To determine

Find the value of $v\left(0^{+}\right)$ and $\frac{dv\left(0^{+}\right)}{dt}$.

Answer to Problem 53P

The value of $v\left(0^{+}\right)$ is $10\text{V}$ and $\frac{dv\left(0^{+}\right)}{dt}$ is $-20\frac{\text{V}}{\text{s}}$.

Explanation of Solution

Given data:

Refer to Figure 16.76 in the textbook.

The switch is in position 1 for a long time and moved to position 2 at $t=0$.

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 ‘1’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 voltage across the resistor is same as the voltage across the capacitor which is the source voltage.

$v_C\left(0^{-}\right)=10\text{V}$.

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

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

$v\left(0\right)=v_C\left(0^{-}\right)=v_C\left(0^{+}\right)=10\text{V}$

When the switch is in position ‘2’, the Figure 1 is reduced as shown in Figure 3.

Refer to Figure 3, the capacitor, resistor and inductor are connected in parallel. For the parallel connection the voltage is same. In Figure 3, the magnitude of voltage is in opposite direction.

$v_R\left(0^{+}\right)=v_L\left(0^{+}\right)=v_C\left(0^{+}\right)$

Apply Kirchhoff’s current law for Figure 3.

$i_L\left(0^{+}\right)=i_C\left(0^{+}\right)+i_R\left(0^{+}\right)$

Substitute $0$ for $i_L\left(0^{+}\right)$ in above equation to find $i_C\left(0^{+}\right)$.

$0=i_C\left(0^{+}\right)+i_R\left(0^{+}\right)$

$i_C\left(0^{+}\right)=-i_R\left(0^{+}\right)$ (1)

Write an expression to calculate the current through resistor.

$i_R\left(0^{+}\right)=\frac{v_R\left(0^{+}\right)}{R}$

Substitute $v_C\left(0^{+}\right)$ for $v_R\left(0^{+}\right)$ in above equation to find $i_R\left(0^{+}\right)$.

$i_R\left(0^{+}\right)=\frac{v_C\left(0^{+}\right)}{R}$

Substitute $10\text{V}$ for $v_C\left(0^{+}\right)$ and $0.5\Omega$ for $R$ in above equation to find $i_R\left(0^{+}\right)$.

$\begin{array}{c}{i}_R\left(0^{+}\right)=\frac{10\text{V}}{0.5\Omega }\\ =20\text{A}\end{array}$

Substitute $20\text{A}$ for $i_R\left(0^{+}\right)$ in equation (1) to find $i_C\left(0^{+}\right)$.

$i_C\left(0^{+}\right)=-20\text{A}$

At time $t=0^{+}$, the capacitor current is calculated as follows:

$i_C\left(0^{+}\right)=C\frac{dv\left(0^{+}\right)}{dt}$

Rearrange the above equation to find $\frac{dv\left(0^{+}\right)}{dt}$.

$\frac{dv\left(0^{+}\right)}{dt}=\frac{i_C\left(0^{+}\right)}{C}$

Substitute $-20\text{A}$ for $i_C\left(0^{+}\right)$, and $1\text{F}$ for $C$ in above equation to find $\frac{dv\left(0^{+}\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0^{+}\right)}{dt}=\frac{-20\text{A}}{1\text{F}}\\ =\frac{-20\text{A}}{1\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\right)}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =-20\frac{\text{V}}{\text{s}}\end{array}$

Conclusion:

Thus, the value of $v\left(0^{+}\right)$ is $10\text{V}$ and $\frac{dv\left(0^{+}\right)}{dt}$ is $-20\frac{\text{V}}{\text{s}}$.

To determine

Find the expression of voltage $v\left(t\right)$ for $t>0$.

Answer to Problem 53P

The expression of voltage $v\left(t\right)$ for $t>0$ is $\left[11.547e^{-t}\cos \left(1.7321t+30°\right)\right]u\left(t\right)\text{V}$.

Explanation of Solution

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:

Substitute $0.5\Omega$ for $R_1$ in equation (2) to find $Z_{R_1}$.

$Z_{R_1}=0.5\Omega$

Substitute $0.25\text{H}$ for $L$ in equation (3) to find $Z_L$.

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

Substitute $1\text{F}$ for $C$ in equation (4) to find $Z_C$.

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

Using element transformation methods with initial conditions convert the Figure 3 into s-domain.

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

$\begin{array}{l}\frac{V\left(s\right)}{0.25s}+\frac{V\left(s\right)}{0.5}+\frac{V\left(s\right)-\frac{v\left(0\right)}{s}}{\left(\frac{1}{s}\right)}=0\\ \frac{V\left(s\right)}{0.25s}+\frac{V\left(s\right)}{0.5}+sV\left(s\right)-\left(s\right)\frac{v\left(0\right)}{s}=0\\ \frac{V\left(s\right)}{0.25s}+\frac{V\left(s\right)}{0.5}+sV\left(s\right)-v\left(0\right)=0\\ V\left(s\right)\left[\frac{1}{0.25s}+\frac{1}{0.5}+s\right]=v\left(0\right)\end{array}$

Substitute $10$ for $v\left(0\right)$ in above equation.

$\begin{array}{l}V\left(s\right)\left[\frac{1}{0.25s}+\frac{1}{0.5}+s\right]=10\\ V\left(s\right)\left[\frac{2+s+0.5s^2}{0.5s}\right]=10\\ V\left(s\right)\left[\frac{0.5s^2+s+2}{0.5s}\right]=10\\ V\left(s\right)\left[\frac{0.5\left(s^2+2s+4\right)}{0.5s}\right]=10\end{array}$

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

$V\left(s\right)=\frac{10s}{s^2+2s+4}$ (5)

From the above equation , the characteristic equation is

$s^2+2s+4=0$ (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 $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=2\\ c=4\end{array}$

Substitute $1$ for $a$, $2$ for $b$, and $4$ for $c$ in equation (7) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\\ =\frac{-2\pm \sqrt{4-16}}{2\left(1\right)}\\ =\frac{-2\pm \sqrt{-12}}{2}\\ =\frac{-2\pm j3.4642}{2}\end{array}$

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

$\begin{array}{c}{s}_{1,2}=\frac{-2+j3.4642}{2},\frac{-2-j3.4642}{2}\\ =-1+j1.7321,-1-j1.7321\end{array}$

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

$\begin{array}{c}V\left(s\right)=\frac{10s}{s\left(s-\left(-1+j1.7321\right)\right)\left(\left(s-\left(-1-j1.7321\right)\right)\right)}\\ =\frac{10s}{s\left(s-\left(-1+j1.7321\right)\right)\left(s-\left(-1-j1.7321\right)\right)}\end{array}$

Take partial fraction for above equation.

$V\left(s\right)=\frac{10s}{\left(s-\left(-1+j1.7321\right)\right)\left(s-\left(-1-j1.7321\right)\right)}=\left[\begin{array}{l}\frac{A}{\left(s-\left(-1+j1.7321\right)\right)}\\ +\frac{B}{\left(s-\left(-1-j1.7321\right)\right)}\end{array}\right]$ (8)

The equation (8) can also be written as follows:

$\frac{10s}{\left(s-\left(-1+j1.7321\right)\right)\left(s-\left(-1-j1.7321\right)\right)}=\left[\frac{\begin{array}{l}A\left(s-\left(-1-j1.7321\right)\right)\\ +B\left(s-\left(-1+j1.7321\right)\right)\end{array}}{\left[\begin{array}{l}\left(s-\left(-1+j1.7321\right)\right)\\ \left(s-\left(-1-j1.7321\right)\right)\end{array}\right]}\right]$

Simplify the above equation as follows:

$10s=A\left(s-\left(-1-j1.7321\right)\right)+B\left(s-\left(-1+j1.7321\right)\right)$ . (9)

Substitute $-1+j1.732$ for $s$ in equation (9) to find $A$.

$\begin{array}{l}10\left(-1+j1.7321\right)=A\left(\left(-1+j1.7321\right)-\left(-1-j1.7321\right)\right)+B\left(\left(-1+j1.7321\right)-\left(-1+j1.7321\right)\right)\\ -10+j17.321=A\left(j3.4642\right)+B\left(0\right)\\ A\left(j3.4642\right)=-10+j17.321\end{array}$

Simplify the above equation to find $A$.

$\begin{array}{c}A=\frac{-10+j17.321}{j3.4642}\\ =5+j2.886\\ =5.773\angle 30°\end{array}$

Substitute $-1-j1.7321$ for $s$ in equation (9) to find $B$.

$\begin{array}{l}10\left(-1-j1.7321\right)=A\left(\left(-1-j1.7321\right)-\left(-1-j1.7321\right)\right)+B\left(\left(-1-j1.7321\right)-\left(-1+j1.7321\right)\right)\\ -10-j17.321=A\left(0\right)+B\left(-j3.4642\right)\\ B\left(-j3.4642\right)=-10-j17.321\end{array}$

Simplify the above equation to find $B$.

$\begin{array}{c}B=\frac{-10-j17.321}{-j3.4642}\\ =5-j2.886\\ =5.773\angle -30°\end{array}$

Substitute $5.773\angle 30°$ for $A$, and $5.773\angle -30°$ for $B$ in equation (8) to find $V\left(s\right)$.

$V\left(s\right)=\left[\frac{5.773\angle 30°}{\left(s-\left(-1+j1.7321\right)\right)}+\frac{5.773\angle -30°}{\left(s-\left(-1-j1.7321\right)\right)}\right]$

$\begin{array}{c}V\left(s\right)=\left[\frac{5.773\angle 30°}{\left(s-\left(-1+j1.7321\right)\right)}+\frac{5.773\angle -30°}{\left(s-\left(-1-j1.7321\right)\right)}\right]\\ =\left[\frac{5.773\left(\cos \left(30°\right)+j\sin \left(30°\right)\right)}{\left(s-\left(-1+j1.7321\right)\right)}+\frac{20.587\left(\cos \left(-30°\right)+j\sin \left(-30°\right)\right)}{\left(s-\left(-1-j1.7321\right)\right)}\right]\\ =\left[\frac{5.773e^{j30°}}{\left(s-\left(-1+j1.7321\right)\right)}+\frac{5.773e^{-j30°}}{\left(s-\left(-1-j1.7321\right)\right)}\right]\left\{\because e^{i\theta }=\cos \theta +i\sin \theta \right\}\end{array}$

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

$\begin{array}{c}v\left(t\right)=\left[\frac{5.773e^{j30°}}{\left(s-\left(-1+j1.7321\right)\right)}+\frac{5.773e^{-j30°}}{\left(s-\left(-1-j1.7321\right)\right)}\right]\\ =\left[5.773e^{j30°}{e}^{\left(-1+j1.7321\right)t}u\left(t\right)+5.773e^{-j30°}{e}^{\left(-1-j1.7321\right)t}u\left(t\right)\right]\left\{\because {\mathcal{L}}^{-1}\left(\frac{1}{s-a}\right)=e^{at}u\left(t\right)\right\}\\ =\left[5.773e^{\left(-1t+j41.7321t+j30°\right)}+5.773e^{\left(-1t-j1.7321t-j30°\right)}\right]u\left(t\right)\left\{\because e^a\cdot e^b=e^{a+b}\right\}\\ =\left[5.773e^{-t}{e}^{j1.7321t+j30°}+5.773e^{-t}{e}^{-j1.7321t-j30°}\right]u\left(t\right)\\ =\left[5.773e^{-t}{e}^{j\left(1.7321t+30°\right)}+5.773e^{-t}{e}^{-j\left(1.7321t+30°\right)}\right]u\left(t\right)\end{array}$

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

$\begin{array}{c}v\left(t\right)==\left[5.773e^{-t}\left[e^{j\left(1.7321t+30°\right)}+e^{-j\left(1.7321t+30°\right)}\right]\right]u\left(t\right)\\ =\left[5.773e^{-t}\left[2\cos \left(1.7321t+30°\right)\right]\right]u\left(t\right)\left\{\because \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}\right\}\\ =\left[11.547e^{-t}\cos \left(1.7321t+30°\right)\right]u\left(t\right)\text{V}\end{array}$

Conclusion:

Thus, the expression of voltage $v\left(t\right)$ for $t>0$ is $\left[11.547e^{-t}\cos \left(1.7321t+30°\right)\right]u\left(t\right)\text{V}$.