Chapter 8, Problem 39P

Determine v(t) for t > 0 in the circuit of Fig. 8.87.

Figure 8.87

For Prob. 8.39.

To determine

Find the expression of voltage $v\left(t\right)$ for $t>0$ in the circuit of Figure 8.87.

Answer to Problem 39P

The expression of voltage $v\left(t\right)$ for $t>0$ is $\left[-60+60.2102e^{-0.167t}-0.2102e^{-47.83t}\right]\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 8.87 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 overdamped.

$v\left(t\right)=\left[V_s+A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}\right]\text{V}$ (3)

Here,

$A_1$ and $A_2$ are constants, and

$s_1$ and $s_2$ are the roots of characteristic equation.

Write a general expression to calculate the roots of characteristic equation.

$s_{1,2}=-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}$ (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 $\left(i\right)$ and voltage source $\left(v_2\right)$ becomes zero. The reduced diagram of Figure 1 is shown in Figure 2.

Refer to Figure 2, there is no current and voltage through the circuit. Therefore, the current through inductor and voltage across the capacitor is 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 $t>0$, the value of step input is $1$. Therefore, the current and voltage source becomes,

$\begin{array}{c}{v}_2=20\left(1\right)\text{V}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =20\text{V}\end{array}$

$\begin{array}{c}i=20\left(1\right)\text{A}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =20\text{A}\end{array}$

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 $20\text{A}$ for $i$, and $4\Omega$ for $R_1$ in equation (5) to find $v_1$.

$\begin{array}{c}{v}_1=\left(20\text{A}\right)\left(4\Omega \right)\\ =80\text{V}\end{array}$

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 $80\text{V}$ for $v_1$, and $20\text{V}$ for $v_2$ in equation (6) to find $V_s$.

$\begin{array}{c}{V}_s=-80\text{V}+20\text{V}\\ =-60\text{V}\end{array}$

The Figure 4 is reduced as shown in Figure 5.

Refer to Figure 5, 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.

$R=R_1+R_2+R_3$ (7)

Substitute $4\Omega$ for $R_1$, $3\Omega$ for $R_2$ and $5\Omega$ for $R_3$ in equation (7) to find $R$.

 $\begin{array}{c}R=4\Omega +3\Omega +5\Omega \\ =12\Omega \end{array}$

The Figure 5 is reduced as shown in Figure 6.

Refer to Figure 6, the circuit shows the step response of a series $RLC$ circuit.

Substitute $12\Omega$ for $R$, and $250\text{mH}$ for $L$ in equation (1) to find $\alpha$.

$\begin{array}{c}\alpha =\frac{12\Omega }{2\left(250\text{mH}\right)}\\ =\frac{12\Omega }{2\left(250\times {10}^{-3}\text{H}\right)}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{12\Omega }{2\left(250\times {10}^{-3}\Omega \cdot \text{s}\right)}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =24\frac{\text{Np}}{\text{s}}\end{array}$

Substitute $250\text{mH}$ for $L$, and $500\text{mF}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(250\text{mH}\right)\left(500\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(250\times {10}^{-3}\text{H}\right)\left(500\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{\sqrt{\left(250\times {10}^{-3}\frac{{\text{s}}^2}{\text{F}}\right)\left(500\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =2.828\frac{\text{rad}}{\text{s}}\end{array}$

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 $24$ for $\alpha$ , and $2.828$ for ${\omega }_0$ in equation (4) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=-24\pm \sqrt{{\left(24\right)}^2-{\left(2.828\right)}^2}\\ =-24\pm \sqrt{568}\\ =-24\pm 23.833\end{array}$

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

$\begin{array}{c}{s}_{1,2}=-24+23.833,-24-23.833\\ =-0.167,-47.83\end{array}$

The roots of characteristic equation are,

$\begin{array}{l}{s}_1=-0.167\\ s_2=-47.83\end{array}$

Substitute $-60\text{V}$ for $V_s$, $-0.167$ for $s_1$, and $-47.83$ for $s_2$ in equation (3) to find $v\left(t\right)$.

$v\left(t\right)=\left[-60+A_1{e}^{-0.167t}+A_2{e}^{-47.83t}\right]\text{V}$ (8)

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

$\begin{array}{c}v\left(0\right)=\left[-60+A_1{e}^{-0.167\left(0\right)}+A_2{e}^{-47.83\left(0\right)}\right]\text{V}\\ \text{=}\left[-60+A_1\left(1\right)+A_2\left(1\right)\right]\text{V}\left\{\because e^0=1\right\}\end{array}$

$v\left(0\right)\text{=}\left[-60+A_1+A_2\right]\text{V}$ (9)

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

$\begin{array}{l}0\text{V=}\left[-60+A_1+A_2\right]\text{V}\\ -60+A_1+A_2=0\\ A_1+A_2=60\end{array}$

Simplify the equation to find $A_1$.

$A_1=60-A_2$ (10)

Differentiate equation (8) with respect to $t$.

$\frac{dv\left(t\right)}{dt}=\left[0+A_1{e}^{-0.167t}\left(-0.167\right)+A_2{e}^{-47.83t}\left(-47.83\right)\right]\frac{\text{V}}{\text{s}}$

$\frac{dv\left(t\right)}{dt}=\left[-0.167A_1{e}^{-0.167t}-47.83A_2{e}^{-47.83t}\right]\frac{\text{V}}{\text{s}}$ (11)

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

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\left[-0.167A_1{e}^{-0.167\left(0\right)}-47.83A_2{e}^{-47.83\left(0\right)}\right]\frac{\text{V}}{\text{s}}\\ =\left[-0.167A_1\left(1\right)-47.83A_2\left(1\right)\right]\frac{\text{V}}{\text{s}}\left\{\because e^0=1\right\}\end{array}$

$\frac{dv\left(0\right)}{dt}=\left[-0.167A_1-47.83A_2\right]\frac{\text{V}}{\text{s}}$ (12)

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

$i_R=i_L=i_C$

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 $500\text{mF}$ for $C$ in equation (15) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\frac{0\text{A}}{100\text{mF}}\\ =\frac{0\text{A}}{100\times {10}^{-3}\text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{0\text{A}}{100\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =0\frac{\text{V}}{\text{s}}\end{array}$

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

$0\frac{\text{V}}{\text{s}}=\left[-0.167A_1-47.83A_2\right]\frac{\text{V}}{\text{s}}$

$-0.167A_1-47.83A_2=0$ (16)

Substitute $60-A_2$ for $A_1$ in equation (16) to find $A_2$.

$\begin{array}{l}-0.167\left(60-A_2\right)-47.83A_2=0\\ -10.02+0.167A_2-47.83A_2=0\\ -47.663A_2=10.02\end{array}$

Simplify the above equation to find $A_2$.

$\begin{array}{c}{A}_2=\frac{10.02}{\left(-47.663\right)}\\ =-0.2102\end{array}$

Substitute $-0.2102$ for $A_2$ in equation (10) to find $A_1$.

$\begin{array}{c}{A}_1=60-\left(-0.2102\right)\\ =60+0.2102\\ =60.2102\end{array}$

Substitute $60.2102$ for $A_1$, and $-0.2102$ for $A_2$ in equation (8) to find $v\left(t\right)$.

$v\left(t\right)=\left[-60+60.2102e^{-0.167t}-0.2102e^{-47.83t}\right]\text{V}$ for $t>0$

Conclusion:

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

$\left[-60+60.2102e^{-0.167t}-0.2102e^{-47.83t}\right]\text{V}$.