Chapter 8, Problem 62P

Find the response v_R(t) for t > 0 in the circuit of Fig. 8.107. Let R = 8 Ω, L = 2 H, and C = 125 mF.

Figure 8.107

To determine

Find the response of voltage across the resistor $v_R\left(t\right)$ for $t>0$ in the circuit of Figure 8.107.

Answer to Problem 62P

The response of voltage across the resistor $v_R\left(t\right)$ for $t>0$ is $\left(80-320t{e}^{-2t}\right)u\left(t\right)\text{V}$.

Explanation of Solution

Given data:

The value of resistance $\left(R\right)$ is $8\Omega$.

The value of inductance $\left(L\right)$ is $2\text{H}$.

The value of capacitance $\left(C\right)$ is $125\text{mF}$.

Refer to Figure 8.107 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 voltage across capacitor for the step response of a series $RLC$ circuit when the response of system is critically damped.

$v_C\left(t\right)=\left[V_s+\left(A_{1}t+A_2\right)e^{-\alpha t}\right]\text{V}$ (3)

Here,

$A_1$ and $A_2$ are constants, and

$V_s$ is the step input voltage.

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 $\left(i_1\right)$ 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,

$\begin{array}{c}{i}_1=10\left(1\right)\text{A}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =10\text{A}\end{array}$

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

Refer to Figure 3, the voltage across the capacitor is calculated as follows,

$v\left(t\right)=\left(10\text{A}-i\left(t\right)\right)\left(8\Omega \right)$ (4)

Use source transformation to convert the current source $\left(i_1\right)$ into voltage source $\left(v_1\right)$.

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

$v_1=i_{1}R$ (5)

Substitute $10\text{A}$ for $i_1$, and $8\Omega$ for $R$ in equation (5) to find $v_1$.

$\begin{array}{c}{v}_1=\left(10\text{A}\right)\left(8\Omega \right)\\ =80\text{V}\end{array}$

The Figure 3 is reduced as shown in Figure 4.

Refer to Figure 4, the circuit shows a step response of a series $RLC$ circuit.

Substitute $8\Omega$ for $R$, and $2\text{H}$ for $L$ in equation (1) to find $\alpha$.

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

Substitute $2\text{H}$ for $L$, and $125\text{mF}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(2\text{H}\right)\left(125\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(2\text{H}\right)\left(125\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{\sqrt{\left(2\frac{{\text{s}}^2}{\text{F}}\right)\left(125\times {10}^{-3}\text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =2\frac{\text{rad}}{\text{s}}\end{array}$

Comparing the value of neper and natural frequency, the value of neper frequency is equal to the natural frequency $\alpha ={\omega }_0$. Therefore, the system is critically damped.

Substitute $2$ for $\alpha$ in equation (3) to find $v_C\left(t\right)$.

$v_C\left(t\right)=\left[V_s+\left(A_{1}t+A_2\right)e^{-2t}\right]\text{V}$

$v_C\left(t\right)\text{=}\left[V_s+A_{1}t{e}^{-2t}+A_2{e}^{-2t}\right]\text{V}$ (6)

For time $t\to \infty$, the circuit again reaches a steady state, the capacitor acts like open circuit and the inductor acts like short circuit. Now, the Figure 4 is reduced as shown in Figure 5.

Refer to Figure 5, the circuit is open circuited so there is no current flow through inductor and voltage across the capacitor is same as the value of voltage source $\left(v_1\right)$.

$v_C\left(\infty \right)=80\text{V}$

For a step input,

$V_s=v_C\left(\infty \right)=v\left(\infty \right)$ (7)

Substitute $80\text{V}$ for $v_C\left(\infty \right)$ in equation (7) to find $V_s$.

$V_s=80\text{V}$

Substitute $80\text{V}$ for $V_s$ in equation (6) to find $v_C\left(t\right)$.

$v_C\left(t\right)\text{=}\left[80+A_{1}t{e}^{-2t}+A_2{e}^{-2t}\right]\text{V}$ (8)

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

$\begin{array}{c}{v}_C\left(0\right)\text{=}\left[80+A_1\left(0\right)e^{-2\left(0\right)}+A_2{e}^{-2\left(0\right)}\right]\text{V}\\ =\left[80+A_1\left(0\right)\left(1\right)+A_2\left(1\right)\right]\text{V}\left\{\because e^0=1\right\}\end{array}$

$v_C\left(0\right)=\left[80+A_2\right]\text{V}$ (9)

Substitute $0\text{V}$ for $v_C\left(0\right)$ in equation (9) to find $A_2$.

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

Simplify the above equation to find $A_2$.

$A_2=-80$

Substitute $-80$ for $A_2$ in equation (8) to find $v\left(t\right)$.

$v_C\left(t\right)\text{=}\left[80+A_{1}t{e}^{-2t}-80e^{-2t}\right]\text{V}$ (10)

Differentiate equation (10) with respect to $t$ to find $\frac{d{v}_C\left(t\right)}{dt}$.

$\frac{d{v}_C\left(t\right)}{dt}\text{=}\left[0+A_{1}t{e}^{-2t}\left(-2\right)+A_1\left(1\right)e^{-2t}-80e^{-2t}\left(-2\right)\right]\frac{\text{V}}{\text{s}}$

$\frac{d{v}_C\left(t\right)}{dt}=\left[-2A_{1}t{e}^{-2t}+A_1{e}^{-2t}+160e^{-2t}\right]\frac{\text{V}}{\text{s}}$ (11)

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

$\begin{array}{c}\frac{d{v}_C\left(0\right)}{dt}=\left[-2A_1\left(0\right)e^{-2\left(0\right)}+A_1{e}^{-2\left(0\right)}+160e^{-2\left(0\right)}\right]\frac{\text{V}}{\text{s}}\\ =\left[-2A_1\left(0\right)\left(1\right)+A_1\left(1\right)+160\left(1\right)\right]\frac{\text{V}}{\text{s}}\left\{\because e^0=1\right\}\end{array}$

$\frac{d{v}_C\left(0\right)}{dt}=\left[A_1+160\right]\frac{\text{V}}{\text{s}}$ (12)

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.

$i_C\left(t\right)=C\frac{d{v}_C\left(t\right)}{dt}$ (13)

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

$i\left(t\right)=C\frac{d{v}_C\left(t\right)}{dt}$ (14)

Rearrange the equation (14) to find $\frac{dv\left(t\right)}{dt}$.

$\frac{d{v}_C\left(t\right)}{dt}=\frac{i\left(t\right)}{C}$ (15)

Substitute $0$ for $t$ in equation (15) to find $\frac{d{v}_C\left(0\right)}{dt}$.

$\frac{d{v}_C\left(0\right)}{dt}=\frac{i\left(0\right)}{C}$ (16)

Substitute $0\text{A}$ for $i\left(0\right)$, and $125\text{mF}$ for $C$ in equation (16) to find $\frac{d{v}_C\left(0\right)}{dt}$.

$\begin{array}{c}\frac{d{v}_C\left(0\right)}{dt}=\frac{0\text{A}}{125\text{mF}}\\ =\frac{0\text{A}}{125\times {10}^{-3}\text{F}}\\ =\frac{0\text{A}}{125\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =0\frac{\text{V}}{\text{s}}\end{array}$

Substitute $0\frac{\text{V}}{\text{s}}$ for $\frac{d{v}_C\left(0\right)}{dt}$ in equation (12) to find $A_1$.

$\begin{array}{l}0\frac{\text{V}}{\text{s}}=\left[A_1+160\right]\frac{\text{V}}{\text{s}}\\ A_1+160=0\end{array}$

Simplify the above equation to find $A_1$.

$A_1=-160$

Substitute $-160$ for $A_1$ in equation (10) to find $v_C\left(t\right)$.

$\begin{array}{c}{v}_C\left(t\right)\text{=}\left[80-160t{e}^{-2t}-80e^{-2t}\right]\text{V}\\ \text{=}80\left[1-\left(2t+1\right)e^{-2t}\right]u\left(t\right)\text{V}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\end{array}$

Substitute $-160$ for $A_1$ in equation (10) to find $\frac{d{v}_C\left(t\right)}{dt}$.

$\begin{array}{c}\frac{d{v}_C\left(t\right)}{dt}=\left[-2\left(-160\right)t{e}^{-2t}+\left(-160\right)e^{-2t}+160e^{-2t}\right]\frac{\text{V}}{\text{s}}\\ =\left[320t{e}^{-2t}-160e^{-2t}+160e^{-2t}\right]\frac{\text{V}}{\text{s}}\\ =\left[320t{e}^{-2t}\right]\frac{\text{V}}{\text{s}}\end{array}$

Substitute $\left[320t{e}^{-2t}\right]\frac{\text{V}}{\text{s}}$ for $\frac{d{v}_C\left(t\right)}{dt}$, and $125\text{mF}$ for $C$ in equation (14) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(125\text{mF}\right)\left[320t{e}^{-2t}\right]\frac{\text{V}}{\text{s}}\\ =\left(125\times {10}^{-3}\right)\left[320t{e}^{-2t}\right]\frac{\text{F}\cdot \text{V}}{\text{s}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =40t{e}^{-2t}\frac{\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\cdot \text{V}}{\text{s}}\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =40t{e}^{-2t}\text{A}\end{array}$

Substitute $40t{e}^{-2t}\text{A}$ for $i\left(t\right)$ in equation (4) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left(\left(10\text{A}\right)-\left(40t{e}^{-2t}\text{A}\right)\right)\left(8\Omega \right)\\ =\left(10-40t{e}^{-2t}\right)\left(8\right)\text{A}\cdot \Omega \\ =\left(80-320t{e}^{-2t}\right)u\left(t\right)\text{V}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\end{array}$

Refer to Figure 3, the voltage across resistor is mentioned as $v\left(t\right)$.

Therefore,

$v_R\left(t\right)=v\left(t\right)$ (17)

Substitute $\left(80-320t{e}^{-2t}\right)u\left(t\right)\text{V}$ for $v\left(t\right)$ in equation (17) to find $v_R\left(t\right)$.

$v_R\left(t\right)=\left(80-320t{e}^{-2t}\right)u\left(t\right)\text{V}$

Conclusion:

Thus, the response of voltage across resistor $v_R\left(t\right)$ for $t>0$ is $\left(80-320t{e}^{-2t}\right)u\left(t\right)\text{V}$.