Chapter 8, Problem 24P

The switch in Fig. 8.77 moves from position A to position B at t = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make-before-break switch). Determine i(t) for t > 0.

Figure 8.77

For Prob. 8.24.

To determine

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

Answer to Problem 24P

The expression of current $i\left(t\right)$ for $t>0$ is $e^{-5t}\left(4\cos \left(19.365t\right)+1.033\sin \left(19.365t\right)\right)\text{A}$.

Explanation of Solution

Formula used:

Write an expression to calculate the neper frequency for a parallel $RLC$ circuit.

$\alpha =\frac{1}{2RC}$ (1)

Here,

$R$ is the value of resistance, and

$C$ is the value of capacitance.

Write an expression to calculate the neper frequency for a parallel  $RLC$ circuit.

${\omega }_0=\frac{1}{\sqrt{LC}}$ (2)

Here,

$L$ is the value of inductance.

The three types of responses of the parallel  $RLC$ circuit are,

1. i. When $\alpha >{\omega }_0$ , the system is overdamped,
2. ii. When $\alpha ={\omega }_0$ , the system is critically damped, and
3. iii. When $\alpha <{\omega }_0$ , the system is under damped.

Write a general expression to calculate the voltage response of parallel  $RLC$ circuit when the system is underdamped.

$v\left(t\right)=\left(A_1\cos {\omega }_{d}t+A_2\sin {\omega }_{d}t\right)e^{-\alpha t}\text{V}$ (3)

Here,

${\omega }_d$ is the damped natural frequency, and

$A_1$ and $A_2$ are constants.

Write an expression to calculate the damped natural frequency.

${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$ (4)

Calculation:

The given circuit is redrawn as shown in Figure 1.

For a DC circuit, at steady state condition when time $t=0^{-}$ the switch is at position $A$ and 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, resistors $R$ and $R_1$ are connected in parallel.

Since the inductor is short circuit, the total current flows through the inductor.

The Figure 2 is reduced as shown in Figure 3.

Refer to Figure 3, the current through inductor is,

$i_L\left(0^{-}\right)=\text{4}\text{A}$

The voltage across the capacitor is,

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

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

$\begin{array}{l}v\left(0\right)=v_C\left(0^{-}\right)=v_C\left(0^{+}\right)=0\text{V}\\ i\left(0\right)=i_L\left(0^{-}\right)=i_L\left(0^{+}\right)=\text{4}\text{A}\end{array}$

For time $t>0$, the switch is at position $B$ and the circuit becomes source free parallel $RLC$ circuit as shown in Figure 4.

Substitute $10\Omega$ for $R$, and $10\text{mF}$ for $C$ in equation (1) to find $\alpha$.

$\begin{array}{c}\alpha =\frac{1}{2\left(10\Omega \right)\left(10\text{mF}\right)}\\ =\frac{1}{2\left(10\Omega \right)\left(10\times {10}^{-3}\text{F}\right)}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{2\left(10\Omega \right)\left(10\times {10}^{-3}\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =5\frac{\text{Np}}{\text{s}}\end{array}$

Substitute $0.25\text{H}$ for $L$, and $10\text{mF}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(0.25\text{H}\right)\left(10\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(0.25\text{H}\right)\left(10\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{\sqrt{\left(0.25\frac{{\text{s}}^2}{\text{F}}\right)\left(10\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{H}=\frac{\text{1}{\text{s}}^2}{\text{1}\text{F}}\right\}\\ =20\frac{\text{rad}}{\text{s}}\end{array}$

Comparing the value of neper and natural frequency, the value of neper frequency is less than natural frequency $\left(\alpha <{\omega }_0\right)$. Therefore, the system is underdamped.

Substitute $5$ for $\alpha$, and $20$ for ${\omega }_0$ in equation (4) to find ${\omega }_d$.

$\begin{array}{c}{\omega }_d=\sqrt{{\left(20\right)}^2-{\left(5\right)}^2}\\ =19.365\end{array}$

Substitute $5$ for $\alpha$, and $19.365$ for ${\omega }_d$ in equation (3) to find $v\left(t\right)$.

$v\left(t\right)=\left(A_1\cos \left(19.365t\right)+A_2\sin \left(19.365t\right)\right)e^{-5t}\text{V}$

$v\left(t\right)\text{=}\left[A_1{e}^{-5t}\cos \left(19.365t\right)+A_2{e}^{-5t}\sin \left(19.365t\right)\right]\text{V}$ (5)

Substitute $0$ for $t$ in equation (5) to find $v\left(0\right)$.

$\begin{array}{c}v\left(0\right)\text{=}\left[A_1{e}^{-5\left(0\right)}\cos \left(19.365\left(0\right)\right)+A_2{e}^{-5\left(0\right)}\sin \left(19.365\left(0\right)\right)\right]\text{V}\\ \text{=}\left[A_1\left(1\right)\cos \left(0\right)+A_2\left(1\right)\sin \left(0\right)\right]\text{V}\left\{\because e^0=1\right\}\\ =\left[A_1\left(1\right)\left(1\right)+A_2\left(1\right)\left(0\right)\right]\text{V}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$

$v\left(0\right)=A_1\text{V}$ (6)

Substitute $0\text{V}$ for $v\left(0\right)$ in equation (6) to find $A_1$.

$0\text{V}=A_1\text{V}$

$A_1=0$

Substitute $0$ for $A_1$ in equation (5) to find $v\left(t\right)$.

$v\left(t\right)\text{=}\left[\left(0\right)e^{-5t}\cos \left(19.365t\right)+A_2{e}^{-5t}\sin \left(19.365t\right)\right]\text{V}$ 

$v\left(t\right)\text{=}{A}_2{e}^{-5t}\sin \left(19.365t\right)\text{V}$ (7)

Differentiate equation (7) with respect to $t$.

$\frac{dv\left(t\right)}{dt}=\left[A_2{e}^{-5t}\left(-5\right)\sin \left(19.365t\right)+A_2{e}^{-5t}\left(\cos \left(19.365t\right)\right)\left(19.365\right)\right]\frac{\text{V}}{\text{s}}$

$\frac{dv\left(t\right)}{dt}=\left[-5A_2{e}^{-5t}\sin \left(19.365t\right)+19.365A_2{e}^{-5t}\left(\cos \left(19.365t\right)\right)\right]\frac{\text{V}}{\text{s}}$ (8)

Substitute $0$ for $t$ in equation (8) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\left[-5A_2{e}^{-5\left(0\right)}\sin \left(19.365\left(0\right)\right)+19.365A_2{e}^{-5\left(0\right)}\left(\cos \left(19.365\left(0\right)\right)\right)\right]\frac{\text{V}}{\text{s}}\\ =\left[-5A_2\left(1\right)\sin \left(0\right)+19.365A_2\left(1\right)\left(\cos \left(0\right)\right)\right]\frac{\text{V}}{\text{s}}\left\{\because e^0=1\right\}\\ =\left[-5A_2\left(1\right)\left(0\right)+19.365A_2\left(1\right)\left(1\right)\right]\frac{\text{V}}{\text{s}}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$$\frac{dv\left(0\right)}{dt}=19.365A_2\frac{\text{V}}{\text{s}}$ (9)

Apply Kirchhoff’s current law for Figure 3.

$\frac{v\left(t\right)}{R}+i_L\left(t\right)+C\frac{dv\left(t\right)}{dt}=0$

$\frac{dv\left(t\right)}{dt}=-\frac{1}{C}\left(\frac{v\left(t\right)}{R}+i_L\left(t\right)\right)$ (10)

Substitute $0$ for $t$ in equation (10) to find $\frac{dv\left(0\right)}{dt}$.

$\frac{dv\left(0\right)}{dt}=-\frac{1}{C}\left(\frac{v\left(0\right)}{R}+i_L\left(0\right)\right)$ (11)

Substitute $0\text{V}$ for $v\left(0\right)$, $4\text{A}$ for $i_L\left(0\right)$, $10\text{mF}$ for $C$ and $10\Omega$ for $R$ in equation (11) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=-\frac{1}{10\text{mF}}\left(0\text{V}+4\text{A}\right)\\ =-\frac{4\text{A}}{\left(10\times {10}^{-3}\text{F}\right)}\\ =-\frac{4\text{A}}{\left(10\times {10}^{-3}\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\right)}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =-400\frac{\text{V}}{\text{s}}\end{array}$

Substitute $-400\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(0\right)}{dt}$ in equation (9) to find $A_2$.

$\begin{array}{l}-400\frac{\text{V}}{\text{s}}=19.365A_2\frac{\text{V}}{\text{s}}\\ 19.365A_2=-400\end{array}$

Simplify the above equation to find $A_2$.

$\begin{array}{c}{A}_2=\frac{-400}{19.365}\\ =-20.66\end{array}$

Substitute $-20.66$ for $A_2$ in equation (7) to find $v\left(t\right)$.

$v\left(t\right)\text{=}-20.66e^{-5t}\sin \left(19.365t\right)\text{V}$

For the parallel $RLC$ circuit, the voltage across resistor, inductor and capacitor are same.

$v\left(t\right)=v_R=v_L=v_C$

Refer to Figure 4, the current $i\left(t\right)$ is mentioned through the inductor.

Write an expression to calculate the current through inductor.

$i\left(t\right)=\frac{1}{L}{\displaystyle \int v\left(t\right)}dt$ (12)

Substitute $-20.66e^{-5t}\sin \left(19.365t\right)\text{V}$ for $v\left(t\right)$, and $0.25\text{H}$ for $L$ in equation (12) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\frac{1}{0.25\text{H}}{\displaystyle \int -20.66e^{-5t}\sin \left(19.365t\right)\text{V}}dt\\ =\frac{-20.66}{0.25}{\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt\frac{\text{V}}{\text{H}}\end{array}$

$i\left(t\right)=-82.64{\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt\frac{\text{V}}{\text{H}}$ (13)

Assume,

$I_1={\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt$

Substitute $I_1$ for ${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt$ in equation (13) to find $i\left(t\right)$.

$i\left(t\right)=-82.64I_1\frac{\text{V}}{\text{H}}$ (14)

Use integration by parts,

Assume

$\begin{array}{l}u=e^{-5t}\\ \frac{du}{dt}=e^{-5t}\left(-5\right)\\ du=-5e^{-5t}dt\end{array}$

$dv=\sin \left(19.365t\right)$

Simplify the above equation to find $v$.

$\begin{array}{c}v={\displaystyle \int \sin \left(19.365t\right)dt}\\ =\frac{-\cos \left(19.365t\right)}{19.365}\\ =-0.0516\cos \left(19.365t\right)\end{array}$

Write an expression to calculate the integration by parts.

${\displaystyle \int udv}=uv-{\displaystyle \int vdu}$ (15)

Substitute $e^{-5t}$ for $u$, $-5e^{-5t}dt$ for $du$, $\sin \left(19.365t\right)$ for $dv$, and $-0.0516\cos \left(19.365t\right)$ for $v$ in equation (15).

${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=\left(e^{-5t}\right)\left(-0.0516\cos \left(19.365t\right)\right)-{\displaystyle \int \left(-0.0516\cos \left(19.365t\right)\right)\left(-5e^{-5t}dt\right)}$

${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=-0.0516e^{-5t}\cos \left(19.365t\right)-0.258{\displaystyle \int e^{-5t}}\cos \left(19.365t\right)dt$ (16)

Assume,

$I_2={\displaystyle \int e^{-5t}\cos \left(19.365t\right)}dt$

Substitute $I_2$ for ${\displaystyle \int e^{-5t}\cos \left(19.365t\right)}dt$ in equation (16).

${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=-0.0516e^{-5t}\cos \left(19.365t\right)-0.258I_2$ (17)

Assume,

$\begin{array}{l}u=e^{-5t}\\ \frac{du}{dt}=e^{-5t}\left(-5\right)\\ du=-5e^{-5t}dt\end{array}$

$dv=\cos \left(19.365t\right)$

Simplify the above equation to find $v$.

$\begin{array}{c}v={\displaystyle \int \cos \left(19.365t\right)dt}\\ =\frac{\sin \left(19.365t\right)}{19.365}\\ =0.0516\sin \left(19.365t\right)\end{array}$

Substitute $e^{-5t}$ for $u$, $-5e^{-5t}dt$ for $du$, $\cos \left(19.365t\right)$ for $dv$, and $0.0516\sin \left(19.365t\right)$ for $v$ in equation (15).

${\displaystyle \int e^{-5t}\cos \left(19.365t\right)}dt=\left(e^{-5t}\right)\left(0.0516\sin \left(19.365t\right)\right)-{\displaystyle \int \left(0.0516\sin \left(19.365t\right)\right)\left(-5e^{-5t}dt\right)}$

${\displaystyle \int e^{-5t}\cos \left(19.365t\right)}dt=0.0516e^{-5t}\sin \left(19.365t\right)+0.258{\displaystyle \int e^{-5t}}\sin \left(19.365t\right)dt$ (18)

Substitute $I_2$ for ${\displaystyle \int e^{-5t}\cos \left(19.365t\right)}dt$ in equation (18).

$I_2=0.0516e^{-5t}\sin \left(19.365t\right)+0.258{\displaystyle \int e^{-5t}}\sin \left(19.365t\right)dt$

Substitute $0.0516e^{-5t}\sin \left(19.365t\right)+0.258{\displaystyle \int e^{-5t}}\sin \left(19.365t\right)dt$ for $I_2$ in equation (17).

$\begin{array}{l}{\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=\left[\begin{array}{l}-0.0516e^{-5t}\cos \left(19.365t\right)\\ -0.258\left(0.0516e^{-5t}\sin \left(19.365t\right)+0.258{\displaystyle \int e^{-5t}}\sin \left(19.365t\right)dt\right)\end{array}\right]\\ {\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=\left[\begin{array}{l}-0.0516e^{-5t}\cos \left(19.365t\right)-0.0133e^{-5t}\sin \left(19.365t\right)\\ -0.06656{\displaystyle \int e^{-5t}}\sin \left(19.365t\right)dt\end{array}\right]\\ {\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt+0.06656{\displaystyle \int e^{-5t}}\sin \left(19.365t\right)dt=\left[\begin{array}{l}-0.0516e^{-5t}\cos \left(19.365t\right)\\ -0.0133e^{-5t}\sin \left(19.365t\right)\end{array}\right]\\ {\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt\left(1+0.06656\right)=\left[-0.0516e^{-5t}\cos \left(19.365t\right)-0.0133e^{-5t}\sin \left(19.365t\right)\right]\end{array}$

Simplify the above equation to find ${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt$.

$\begin{array}{l}{\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt\left(1.06656\right)=-0.0516e^{-5t}\cos \left(19.365t\right)-0.0133e^{-5t}\sin \left(19.365t\right)\\ {\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=\frac{-0.0516e^{-5t}\cos \left(19.365t\right)-0.0133e^{-5t}\sin \left(19.365t\right)}{\left(1.06656\right)}\text{s}\end{array}$

${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt=-0.0484e^{-5t}\cos \left(19.365t\right)-0.0125e^{-5t}\sin \left(19.365t\right)\text{s}$ (19)

Substitute $I_1$ for ${\displaystyle \int e^{-5t}\sin \left(19.365t\right)}dt$ in equation (19).

$I_1=-0.0484e^{-5t}\cos \left(19.365t\right)-0.0125e^{-5t}\sin \left(19.365t\right)\text{s}$

Substitute $-0.0484e^{-5t}\cos \left(19.365t\right)-0.0125e^{-5t}\sin \left(19.365t\right)\text{s}$ for $I_1$ in equation (14) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=-82.64\left(-0.0484e^{-5t}\cos \left(19.365t\right)-0.0125e^{-5t}\sin \left(19.365t\right)\text{s}\right)\frac{\text{V}}{\text{H}}\\ =\left(4e^{-5t}\cos \left(19.365t\right)+1.033e^{-5t}\sin \left(19.365t\right)\right)\frac{\text{V}\cdot \text{s}}{\left(\frac{\text{V}\cdot \text{s}}{\text{A}}\right)}\left\{\because 1\text{H}=\frac{\text{1}\text{V}\cdot 1\text{s}}{\text{1}\text{A}}\right\}\\ =\left(4e^{-5t}\cos \left(19.365t\right)+1.033e^{-5t}\sin \left(19.365t\right)\right)\text{A}\\ =e^{-5t}\left(4\cos \left(19.365t\right)+1.033\sin \left(19.365t\right)\right)\text{A}\end{array}$

Conclusion:

Thus, the expression of current $i\left(t\right)$ for $t>0$ is,

$e^{-5t}\left(4\cos \left(19.365t\right)+1.033\sin \left(19.365t\right)\right)\text{A}$.