Chapter 8, Problem 41P

For the network in Fig. 8.89, find i(t) for t > 0.

Figure 8.89

For Prob. 8.41.

To determine

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

Answer to Problem 41P

The expression of current $i\left(t\right)$ for $t>0$ is $\left[8.7e^{-2t}\sin \left(4.583t\right)\right]u\left(t\right)\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 8.89 in the textbook.

Formula used:

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

$\alpha =\frac{R}{2L}$ (1)

Here,

$R$ is the value of resistance, and

$L$ is the value of inductance.

Write an expression to calculate the natural frequency for a series $RLC$ circuit.

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

Here,

$C$ is the value of capacitance.

The three types of responses for a series $RLC$ 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 for the step response of a series $RLC$ circuit when the response of system is under damped.

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

Here,

$A_1$ and $A_2$ are constants, and

${\omega }_d$ is the damped natural frequency.

Write a an expression to calculate the damped natural frequency.

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

Write an expression to calculate the value of step input.

$u\left(t\right)=\{\begin{array}{l}0t<0\\ 1t>0\end{array}$

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. The value of step input for $t<0$ is zero.

Since the value of step input for $t<0$ is zero, the current source $\left(i\right)$ becomes zero. The reduced diagram of Figure 1 is shown in Figure 2.

Refer to Figure 2, the circuit is open circuited. Therefore, the voltage across the capacitor is same as the source voltage $\left(v_2\right)$ and there is no current flow through the inductor.

$v_C\left(0^{-}\right)=20\text{V}$

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

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

$\begin{array}{l}v\left(0\right)=v_C\left(0^{-}\right)=v_C\left(0^{+}\right)=20\text{V}\\ i\left(0\right)=i_L\left(0^{-}\right)=i_L\left(0^{+}\right)=0\text{A}\end{array}$

For $t>0$, the value of step input is $1$. Therefore, the current source becomes,

$\begin{array}{c}i=40\left(1\right)\text{A}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =40\text{A}\end{array}$

Now, the Figure 1 is reduced as shown in Figure 3.

Use source transformation to convert the current source $\left(i\right)$ into voltage source $\left(v_1\right)$.

Write an expression to calculate the voltage source $\left(v_1\right)$.

$v_1=i{R}_1$ (5)

Substitute $40\text{A}$ for $i$, and $1\Omega$ for $R_1$ in equation (5) to find $v_1$.

$\begin{array}{c}{v}_1=\left(40\text{A}\right)\left(1\Omega \right)\\ =40\text{V}\end{array}$

The Figure 3 is reduced as shown in Figure 4.

Refer to Figure 4, the voltage source $v_1$ and $v_2$ are in series form, hence the sources are added. The voltage for a step input is calculated as follows,

$V_s=-v_1+v_2$ (6)

Substitute $40\text{V}$ for $v_1$, and $20\text{V}$ for $v_2$ in equation (6) to find $V_s$.

$\begin{array}{c}{V}_s=-40\text{V}+20\text{V}\\ =-20\text{V}\end{array}$

Refer to Figure 4, the resistors $R_1$, and $R_2$ are connected in series form.

Write an expression to calculate the equivalent resistance for series connected resistors.

$R=R_1+R_2$ (7)

Substitute $1\Omega$ for $R_1$, and $3\Omega$ for $R_2$ in equation (7) to find $R$.

$\begin{array}{c}R=1\Omega +3\Omega \\ =4\Omega \end{array}$

The Figure 4 is reduced as shown in Figure 5.

Refer to Figure 5, the circuit shows the step response of a series $RLC$ circuit.

Substitute $4\Omega$ for $R$, and $1\text{H}$ for $L$ in equation (1) to find $\alpha$.

$\begin{array}{c}\alpha =\frac{4\Omega }{2\left(1\text{H}\right)}\\ =\frac{4\Omega }{2\left(1\Omega \cdot \text{s}\right)}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =2\frac{\text{Np}}{\text{s}}\end{array}$

Substitute $1\text{H}$ for $L$, and $40\text{mF}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(1\text{H}\right)\left(40\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(1\text{H}\right)\left(40\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{\sqrt{\left(1\frac{{\text{s}}^2}{\text{F}}\right)\left(40\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =5\frac{\text{rad}}{\text{s}}\end{array}$

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

Substitute $2$ for $\alpha$, and $5$ for ${\omega }_0$ in equation (4) to find ${\omega }_d$.

$\begin{array}{c}{\omega }_d=\sqrt{{\left(5\right)}^2-{\left(2\right)}^2}\\ =\sqrt{21}\\ =4.583\end{array}$

Substitute $-20\text{V}$ for $V_s$, $2$ for $\alpha$, and $4.583$ for ${\omega }_d$  in equation (3) to find $v\left(t\right)$.

$v\left(t\right)=\left[-20+\left(A_1\cos \left(4.583t\right)+A_2\sin \left(4.583t\right)\right)e^{-2t}\right]\text{V}$

$v\left(t\right)=\left[-20+A_1{e}^{-2t}\cos \left(4.583t\right)+A_2{e}^{-2t}\sin \left(4.583t\right)\right]\text{V}$ (8)

Substitute $0$ for $t$ in equation (8) to find $v\left(0\right)$.

$\begin{array}{c}v\left(0\right)=\left[-20+A_1{e}^{-2\left(0\right)}\cos \left(4.583\left(0\right)\right)+A_2{e}^{-2\left(0\right)}\sin \left(4.583\left(0\right)\right)\right]\text{V}\\ \text{=}\left[-20+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[-20+A_1\left(1\right)\left(1\right)+A_2\left(1\right)\left(0\right)\right]\text{V}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$$v\left(0\right)=\left[-20+A_1\right]\text{V}$ (9)

Substitute $20\text{V}$ for $v\left(0\right)$ in equation (9) to find $A_1$.

$\begin{array}{l}20\text{V}=\left[-20+A_1\right]\text{V}\\ -20+A_1=20\end{array}$

Simplify the above equation to find $A_1$.

$\begin{array}{c}{A}_1=20+20\\ =40\end{array}$

Substitute $40$ for $A_1$ in equation (8) to find $v\left(t\right)$.

$v\left(t\right)=\left[-20+40e^{-2t}\cos \left(4.583t\right)+A_2{e}^{-2t}\sin \left(4.583t\right)\right]\text{V}$ (10)

Differentiate equation (10) with respect to $t$.

$\frac{dv\left(t\right)}{dt}\text{=}\left[\begin{array}{l}0+40e^{-2t}\left(-2\right)\cos \left(4.583t\right)+40e^{-2t}\left(-\sin \left(4.583t\right)\right)\left(4.583\right)\\ +A_2{e}^{-2t}\left(-2\right)\sin \left(4.583t\right)+A_2{e}^{-2t}\left(\cos \left(4.583t\right)\left(4.583\right)\right)\end{array}\right]\frac{\text{V}}{\text{s}}$

$\frac{dv\left(t\right)}{dt}\text{=}\left[\begin{array}{l}-80e^{-2t}\cos \left(4.583t\right)-183.32e^{-2t}\sin \left(4.583t\right)\\ -2A_2{e}^{-2t}\sin \left(4.583t\right)+4.583A_2{e}^{-2t}\cos \left(4.583t\right)\end{array}\right]\frac{\text{V}}{\text{s}}$ (11)

Substitute $0$ for $t$ in equation (11) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}\text{=}\left[\begin{array}{l}-80e^{-2\left(0\right)}\cos \left(4.583\left(0\right)\right)-183.32e^{-2\left(0\right)}\sin \left(4.583\left(0\right)\right)\\ -2A_2{e}^{-2\left(0\right)}\sin \left(4.583\left(0\right)\right)+4A_2{e}^{-2\left(0\right)}\cos \left(4.583\left(0\right)\right)\end{array}\right]\frac{\text{V}}{\text{s}}\\ =\left[\begin{array}{l}-80\left(1\right)\cos \left(0\right)-183.32\left(1\right)\sin \left(0\right)\\ -2A_2\left(1\right)\sin \left(0\right)+4.583A_2\left(1\right)\cos \left(0\right)\end{array}\right]\frac{\text{V}}{\text{s}}\left\{\because e^0=1\right\}\\ =\left[\begin{array}{l}-80\left(1\right)\left(1\right)-183.32\left(1\right)\left(0\right)\\ -2A_2\left(1\right)\left(0\right)+4.583A_2\left(1\right)\left(1\right)\end{array}\right]\frac{\text{V}}{\text{s}}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$$\frac{dv\left(0\right)}{dt}=\left[-80+4.583A_2\right]\frac{\text{V}}{\text{s}}$ (12)

For a series $RLC$ circuit, the current through resistor, inductor and capacitor are same.

$i=i_R=i_L=i_C$

Write an expression to calculate the current through capacitor.

$i_C\left(t\right)=C\frac{dv\left(t\right)}{dt}$ (13)

Substitute $i_L$ for $i_C$ in equation (13) to find $\frac{dv\left(t\right)}{dt}$.

$i_L\left(t\right)=C\frac{dv\left(t\right)}{dt}$

$\frac{dv\left(t\right)}{dt}=\frac{i_L\left(t\right)}{C}$ (14)

Substitute $0$ for $t$ in equation (14) to find $\frac{dv\left(0\right)}{dt}$.

$\frac{dv\left(0\right)}{dt}=\frac{i_L\left(0\right)}{C}$ (15)

Substitute $0\text{A}$ for $i_L\left(0\right)$, and $40\text{mF}$ for $C$ in equation (15) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\frac{0\text{A}}{40\text{mF}}\\ =\frac{0\text{A}}{40\times {10}^{-3}\text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{0\text{A}}{40\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =0\frac{\text{V}}{\text{s}}\end{array}$

Substitute $0\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(0\right)}{dt}$ in equation (12) to find $A_2$.

$\begin{array}{l}0\frac{\text{V}}{\text{s}}=\left[-80+4.583A_2\right]\frac{\text{V}}{\text{s}}\\ -80+4.583A_2=0\\ 4.583A_2=80\end{array}$

Simplify the above equation to find $A_2$.

$\begin{array}{c}{A}_2=\frac{80}{4.583}\\ =17.456\end{array}$

Substitute $17.456$ for $A_2$ in equation (10) to find $v\left(t\right)$.

$v\left(t\right)=\left[-20+40e^{-2t}\cos \left(4.583t\right)+17.456e^{-2t}\sin \left(4.583t\right)\right]\text{V}$

Substitute $17.456$ for $A_2$ in equation (11) to find $\frac{dv\left(t\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(t\right)}{dt}\text{=}\left[\begin{array}{l}-80e^{-2t}\cos \left(4.583t\right)-183.32e^{-2t}\sin \left(4.583t\right)\\ -2\left(17.456\right)e^{-2t}\sin \left(4.583t\right)+4.583\left(17.456\right)e^{-2t}\cos \left(4t\right)\end{array}\right]\frac{\text{V}}{\text{s}}\\ =\left[\begin{array}{l}-80e^{-2t}\cos \left(4.583t\right)-183.32e^{-2t}\sin \left(4.583t\right)\\ -34.912e^{-2t}\sin \left(4.583t\right)+80e^{-2t}\cos \left(4.583t\right)\end{array}\right]\frac{\text{V}}{\text{s}}\\ =\left[-218.232e^{-2t}\sin \left(4.583t\right)\right]\frac{\text{V}}{\text{s}}\end{array}$

Refer to Figure 5, the direction of current through capacitor is in opposite direction.

Therefore,

+

$i_C\left(t\right)=-i\left(t\right)$

Substitute $-i\left(t\right)$ for $i_C\left(t\right)$ in equation (13) to find $i\left(t\right)$.

$-i\left(t\right)=C\frac{dv\left(t\right)}{dt}$

$i\left(t\right)=-C\frac{dv\left(t\right)}{dt}$ (16)

Substitute $\left[-218.232e^{-2t}\sin \left(4.583t\right)\right]\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(t\right)}{dt}$ , and $40\text{mF}$ for $C$ in equation (16) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=-\left(40\text{mF}\right)\left[-218.232e^{-2t}\sin \left(4.583t\right)\right]\frac{\text{V}}{\text{s}}\\ =-\left(40\times {10}^{-3}\right)\left[-218.232e^{-2t}\sin \left(4.583t\right)\right]\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =-\left(40\times {10}^{-3}\right)\left[-218.232e^{-2t}\sin \left(4.583t\right)\right]\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =8.7e^{-2t}\sin \left(4.583t\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1}\text{F}\cdot 1\text{V}}{\text{1}\text{s}}\right\}\end{array}$

$i\left(t\right)=\left[8.7e^{-2t}\sin \left(4.583t\right)\right]u\left(t\right)\text{A}\left\{\because \text{For step input }u\left(t\right)=1\text{for}t>0\right\}$

Conclusion:

Thus, the expression of current $i\left(t\right)$ for $t>0$ is $\left[8.7e^{-2t}\sin \left(4.583t\right)\right]u\left(t\right)\text{A}$.