Chapter 8, Problem 36P

Obtain v(t) and i(t) for t > 0 in the circuit of Fig. 8.84.

Figure 8.84

For Prob. 8.36.

To determine

Find the expression of voltage $v\left(t\right)$, and the expression of current $i\left(t\right)$ for $t>0$ in the circuit of Figure 8.84.

Answer to Problem 36P

For $t>0$, the expression of voltage $v\left(t\right)$ is, $\left[-16+\left(40\cos \left(0.6t\right)+53.33\sin \left(0.6t\right)\right)e^{-0.8t}\right]u\left(t\right)\text{V}$, and the expression of current $i\left(t\right)$ is, $\left[13.333e^{-0.8t}\sin \left(0.6t\right)\right]u\left(t\right)\text{A}$ for $t>0$.

Explanation of Solution

Given data:

Refer to Figure 8.84 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 to calculate the step response of a series $RLC$ circuit when the response of the 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,

$V_s$ is the step input voltage,

$A_1$ and $A_2$ are constants, and

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

Write 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 $\left(i_2\right)$ becomes zero.

The reduced diagram of Figure 1 is shown in Figure 2.

Refer to Figure 2, the circuit is open circuited, there is no current flow through the inductor and the voltage across the capacitor is same as the voltage source $\left(v_1\right)$.

$\begin{array}{l}{i}_L\left(0^{-}\right)=0\text{A}\\ v_C\left(0^{-}\right)=24\text{V}\end{array}$

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

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

$v\left(0\right)=v_C\left(0^{-}\right)=v_C\left(0^{+}\right)=24\text{V}$

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

$\begin{array}{c}i=10\left(1\right)\text{A}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =10\text{A}\end{array}$

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

Use source transformation to convert the current source $\left(i_2\right)$ into voltage source $\left(v_2\right)$.

Write an expression to calculate the voltage source $\left(v_2\right)$.

$v_2=i_2{R}_3$ (5)

Substitute $10\text{A}$ for $i_2$, and $4\Omega$ for $R_3$ in equation (5) to find $v_2$.

$\begin{array}{c}{v}_2=\left(10\text{A}\right)\left(4\Omega \right)\\ =40\text{V}\end{array}$

The Figure 3 is reduced as shown in Figure 4.

Refer to Figure 4, the resistors $R_1$, $R_2$ and $R_3$ are connected in series form.

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

$R=R_1+R_2+R_3$ (6)

Substitute $2\Omega$ for $R_1$, $2\Omega$ for $R_2$ and $4\Omega$ for $R_3$ in equation (6) to find $R$.

 $\begin{array}{c}R=2\Omega +2\Omega +4\Omega \\ =8\Omega \end{array}$

The Figure 4 is reduced as shown in Figure 5.

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

Substitute $8\Omega$ for $R$, and $5\text{H}$ for $L$ in equation (1) to find $\alpha$.

$\begin{array}{c}\alpha =\frac{8\Omega }{2\left(5\text{H}\right)}\\ =\frac{8\Omega }{2\left(5\Omega \cdot \text{s}\right)}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =0.8\frac{\text{Np}}{\text{s}}\end{array}$

Substitute $5\text{H}$ for $L$, and $200\text{mF}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{5\left(200\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(5\text{H}\right)\left(200\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{\sqrt{\left(5\frac{{\text{s}}^2}{\text{F}}\right)\left(200\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =1\frac{\text{rad}}{\text{s}}\end{array}$

Comparing the value of neper and natural frequency, the value of neper frequency is greater than the natural frequency $\alpha >{\omega }_0$. Therefore, the system is over damped.

Substitute $0.8$ for $\alpha$, and $1$ for ${\omega }_0$ in equation (4) to find $s_{1,2}$.

$\begin{array}{c}{\omega }_d=\sqrt{{\left(1\right)}^2-{\left(0.8\right)}^2}\\ =\sqrt{0.36}\\ =0.6\end{array}$

Substitute $0.8$ for $\alpha$, and $0.6$ for ${\omega }_d$ in equation (3) to find $v\left(t\right)$.

$v\left(t\right)=\left[V_s+\left(A_1\cos \left(0.6t\right)+A_2\sin \left(0.6t\right)\right)e^{-0.8t}\right]\text{V}$

$v\left(t\right)=\left[V_s+A_1{e}^{-0.8t}\cos \left(0.6t\right)+A_2{e}^{-0.8t}\sin \left(0.6t\right)\right]\text{V}$ (7)

At time $t\to \infty$, the circuit again reaches steady state condition, the capacitor acts like open circuit and the inductor acts like circuit. Now, the Figure 5 is reduced as shown in Figure 6.

Refer to Figure 6, the voltage across the capacitor is the negative value of voltage source $\left(v\right)$.

$v_C\left(\infty \right)=-16\text{V}$

For a step input,

$V_s=v_C\left(\infty \right)=v\left(\infty \right)$ (8)

Substitute $-16\text{V}$ for $v_C\left(\infty \right)$ in equation (8) to find $V_s$.

$V_s=-16\text{V}$

Substitute $-16\text{V}$ for $V_s$ in equation (7) to find $v\left(t\right)$.

$v\left(t\right)=\left[-16+A_1{e}^{-0.8t}\cos \left(0.6t\right)+A_2{e}^{-0.8t}\sin \left(0.6t\right)\right]\text{V}$ (9)

Substitute $0$ for $t$ in equation (9) to find $v\left(0\right)$.

$\begin{array}{c}v\left(0\right)=\left[-16+A_1{e}^{-0.8\left(0\right)}\cos \left(0.6\left(0\right)\right)+A_2{e}^{-0.8\left(0\right)}\sin \left(0.6\left(0\right)\right)\right]\text{V}\\ \text{=}\left[-16+A_1\left(1\right)\cos \left(0\right)+A_2\left(1\right)\sin \left(0\right)\right]\text{V}\left\{\because e^0=1\right\}\\ \text{=}\left[-16+A_1\left(1\right)\left(1\right)+A_2\left(1\right)\left(0\right)\right]\text{V}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$

$v\left(0\right)=\left[-16+A_1\right]\text{V}$ (10)

Substitute $24\text{V}$ for $v\left(0\right)$ in equation (10) to find $A_1$.

$\begin{array}{l}24\text{V}=\left[-16+A_1\right]\text{V}\\ -16+A_1=24\end{array}$

Simplify the above equation to find $A_1$.

$\begin{array}{c}{A}_1=24+16\\ =40\end{array}$

Substitute $40$ for $A_1$ in equation (9) to find $v\left(t\right)$.

$v\left(t\right)=\left[-16+40e^{-0.8t}\cos \left(0.6t\right)+A_2{e}^{-0.8t}\sin \left(0.6t\right)\right]\text{V}$ (11)

Differentiate equation (11) with respect to $t$ to find $\frac{dv\left(t\right)}{dt}$.

$\frac{dv\left(t\right)}{dt}=\left[\begin{array}{l}0+40e^{-0.8t}\left(-0.8\right)\cos \left(0.6t\right)+40e^{-0.8t}\left(-\sin \left(0.6t\right)\right)\left(0.6\right)\\ +A_2{e}^{-0.8t}\left(-0.8\right)\sin \left(0.6t\right)+A_2{e}^{-0.8t}\left(\cos \left(0.6t\right)\right)\left(0.6\right)\end{array}\right]\frac{\text{V}}{\text{s}}$

$\frac{dv\left(t\right)}{dt}=\left[\begin{array}{l}-32e^{-0.8t}\cos \left(0.6t\right)-24e^{-0.8t}\sin \left(0.6t\right)\\ -0.8A_2{e}^{-0.8t}\sin \left(0.6t\right)+0.6A_2{e}^{-0.8t}\cos \left(0.6t\right)\end{array}\right]\frac{\text{V}}{\text{s}}$ (12)

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

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\left[\begin{array}{l}-32e^{-0.8\left(0\right)}\cos \left(0.6\left(0\right)\right)-40e^{-0.8\left(0\right)}\sin \left(0.6\left(0\right)\right)\\ -0.8A_2{e}^{-0.8\left(0\right)}\sin \left(0.6\left(0\right)\right)+0.6A_2{e}^{-0.8\left(0\right)}\cos \left(0.6\left(0\right)\right)\end{array}\right]\frac{\text{V}}{\text{s}}\\ =\left[\begin{array}{l}-32\left(1\right)\cos \left(0\right)-40\left(1\right)\sin \left(0\right)\\ -0.8A_2\left(1\right)\sin \left(0\right)+0.6A_2\left(1\right)\cos \left(0\right)\end{array}\right]\frac{\text{V}}{\text{s}}\left\{\because e^0=1\right\}\\ =\left[\begin{array}{l}-32\left(1\right)\left(1\right)-40\left(1\right)\left(0\right)\\ -0.8A_2\left(1\right)\left(0\right)+0.6A_2\left(1\right)\left(1\right)\end{array}\right]\frac{\text{V}}{\text{s}}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$

$\frac{dv\left(0\right)}{dt}=\left[-32+0.6A_2\right]\frac{\text{V}}{\text{s}}$ . (13)

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

$i\left(t\right)=i_R\left(t\right)=i_L\left(t\right)=i_C\left(t\right)$

Write an expression to calculate the current through capacitor.

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

Substitute $i\left(t\right)$ for $i_C\left(t\right)$ in equation (14) to find $i\left(t\right)$.

$i\left(t\right)=C\frac{dv\left(t\right)}{dt}$ (15)

Rearrange the equation (15) to find $\frac{dv\left(t\right)}{dt}$.

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

Substitute $0$ for $t$ in equation (16) to find $\frac{dv\left(0\right)}{dt}$.

$\frac{dv\left(0\right)}{dt}=\frac{i\left(0\right)}{C}$ (17)

Substitute $0\text{A}$ for $i\left(0\right)$, and $200\text{mF}$ for $C$ in equation (17) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\frac{0\text{A}}{200\text{mF}}\\ =\frac{0\text{A}}{200\times {10}^{-3}\text{F}}\\ =\frac{0\text{A}}{200\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =0\frac{\text{V}}{\text{s}}\end{array}$

Substitute $0\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(0\right)}{dt}$ in equation (13) to find $A_2$.

$\begin{array}{l}0\frac{\text{V}}{\text{s}}=\left[-32+0.6A_2\right]\frac{\text{V}}{\text{s}}\\ -32+0.6A_2=0\\ 0.6A_2=32\end{array}$

Simplify the above equation to find $A_2$.

$\begin{array}{c}{A}_2=\frac{32}{0.6}\\ =53.33\end{array}$

Substitute $53.33$ for $A_2$ in equation (11) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left[-16+40e^{-0.8t}\cos \left(0.6t\right)+53.33e^{-0.8t}\sin \left(0.6t\right)\right]\text{V}\\ \text{=}\left[-16+\left(40\cos \left(0.6t\right)+53.33\sin \left(0.6t\right)\right)e^{-0.8t}\right]u\left(t\right)\text{V}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\end{array}$

Substitute $53.33$ for $A_2$ in equation (12) to find $\frac{dv\left(t\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(t\right)}{dt}=\left[\begin{array}{l}-32e^{-0.8t}\cos \left(0.6t\right)-24e^{-0.8t}\sin \left(0.6t\right)\\ -0.8\left(53.33\right)e^{-0.8t}\sin \left(0.6t\right)+0.6\left(53.33\right)e^{-0.8t}\cos \left(0.6t\right)\end{array}\right]\frac{\text{V}}{\text{s}}\\ =\left[\begin{array}{l}-32e^{-0.8t}\cos \left(0.6t\right)-24e^{-0.8t}\sin \left(0.6t\right)\\ -42.664e^{-0.8t}\sin \left(0.6t\right)+32e^{-0.8t}\cos \left(0.6t\right)\end{array}\right]\frac{\text{V}}{\text{s}}\\ =\left[-66.664e^{-0.8t}\sin \left(0.6t\right)\right]\frac{\text{V}}{\text{s}}\end{array}$

Substitute $\left[-66.664e^{-0.8t}\sin \left(0.6t\right)\right]\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(t\right)}{dt}$, and $200\text{mF}$ for $C$ in equation (15) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(200\text{mF}\right)\left[-66.664e^{-0.8t}\sin \left(0.6t\right)\right]\frac{\text{V}}{\text{s}}\\ =\left(200\times {10}^{-3}\right)\left[-66.664e^{-0.8t}\sin \left(0.6t\right)\right]\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =13.333e^{-0.8t}\sin \left(0.6t\right)\frac{\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\cdot \text{V}}{\text{s}}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =\left[13.333e^{-0.8t}\sin \left(0.6t\right)\right]u\left(t\right)\text{A}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\end{array}$

Conclusion:

Thus, the expression of voltage $v\left(t\right)$ is $\left[-16+\left(40\cos \left(0.6t\right)+53.33\sin \left(0.6t\right)\right)e^{-0.8t}\right]u\left(t\right)\text{V}$, and the expression of current $i\left(t\right)$ is $\left[13.333e^{-0.8t}\sin \left(0.6t\right)\right]u\left(t\right)\text{A}$ for $t>0$.