Chapter 8, Problem 46P

To determine

Design a problem to better understand the step response of a parallel $RLC$ circuit.

Explanation of Solution

Problem design:

Design a problem to understand the step response of a parallel $RLC$ circuit if the value of resistance $\left(R\right)$ is $2\text{k}\Omega$, the value of inductance $\left(L\right)$ is $8\text{mH}$, the value of capacitance $\left(C\right)$ is $5\mu \text{F}$, the value of voltage source $\left(v\right)$ is $12u\left(t\right)\text{V}$ in the Figure 8.93 in the textbook.

Formula used:

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

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

Here,

$L$ is the value of inductance.

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 parallel $RLC$ circuit when the response of the system is under damped.

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

Here,

$I_s$ is the step input current,

${\omega }_d$ 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)

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 is zero, the voltage source becomes zero. If the value of voltage source is zero, it acts as a short circuit. Now, the Figure 1 is reduced as shown in Figure 2.

Refer to Figure 2, there is no current and voltage through the circuit. Therefore, the current flows through inductor and voltage across the capacitor is zero.

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

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

The current through inductor and voltage across the 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 voltage source becomes,

$\begin{array}{c}v=12\left(1\right)\text{V}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =12\text{V}\end{array}$

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

The Figure 3 can also be drawn as shown in Figure 4.

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

$i_1=\frac{v}{R}$ (5)

Substitute $12\text{V}$ for $v$, and $2\text{k}\Omega$ for $R$ in equation (5) to find $i_1$.

$\begin{array}{c}{i}_1=\frac{12\text{V}}{2\text{k}\Omega }\\ =\frac{12\text{V}}{2\times {10}^3\Omega }\left\{\because 1\text{k}={10}^3\right\}\\ =6\times {10}^{-3}\text{A}\\ =6\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Now, the Figure 4 is reduced as shown in Figure 5.

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

Substitute $2\text{k}\Omega$ for $R$, and $5\mu \text{F}$ for $C$ in equation (1) to find $\alpha$.

$\begin{array}{c}\alpha =\frac{1}{2\left(2\text{k}\Omega \right)\left(5\mu \text{F}\right)}\\ =\frac{1}{2\left(2\times {10}^3\Omega \right)\left(5\times {10}^{-6}\text{F}\right)}\left\{\because 1\text{k}={10}^3,1\mu ={10}^{-6}\right\}\\ =\frac{1}{2\left(2\times {10}^3\Omega \right)\left(5\times {10}^{-6}\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =50\frac{\text{Np}}{\text{s}}\end{array}$

Substitute $8\text{mH}$ for $L$, and $5\mu \text{F}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(8\text{mH}\right)\left(5\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(8\times {10}^{-3}\text{H}\right)\left(5\times {10}^{-6}\text{F}\right)}}\left\{\because 1\text{k}={10}^3,1\mu ={10}^{-6}\right\}\\ =\frac{1}{\sqrt{\left(8\times {10}^{-3}\frac{{\text{s}}^2}{\text{F}}\right)\left(5\times {10}^{-6}\text{F}\right)}}\left\{\because 1\text{H}=\frac{\text{1}{\text{s}}^2}{\text{1}\text{F}}\right\}\\ =5000\frac{\text{rad}}{\text{s}}\end{array}$

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 under damped.

Substitute $50$ for $\alpha$, and $5000$ for ${\omega }_0$ in equation (4) to find ${\omega }_d$.

$\begin{array}{c}{\omega }_d=\sqrt{{\left(5000\right)}^2-{\left(50\right)}^2}\\ =5000\end{array}$

Substitute $50$ for $\alpha$, and $5000$ for ${\omega }_d$ in equation (3) to find $i\left(t\right)$.

$i\left(t\right)=\left[I_s+\left(A_1\cos \left(5000t\right)+A_2\sin \left(5000t\right)\right)e^{-50t}\right]\text{A}$ (6)

For a time $t\to \infty$, the circuit again reaches steady state, the capacitor acts like open circuit and the inductor acts like short circuit.

Now, the reduced circuit of Figure 5 is shown in Figure 6.

Refer to Figure 6, the short circuited inductor and resistor are connected in parallel. Since the full current flows through the short circuited inductor, the resistor is also shorted. Now, the reduced diagram of Figure 6 is shown in Figure 7.

Refer to Figure 7, the current flows through the inductor is same as the current source $\left(i_1\right)$.

$i_L\left(\infty \right)=6\text{mA}$

For a step input,

$I_s=i_L\left(\infty \right)=i\left(\infty \right)$ (7)

Substitute $6\text{mA}$ for $i_L\left(\infty \right)$ in equation (7) to find $I_s$.

$I_s=6\text{mA}$

Substitute $6\text{mA}$ for $I_s$ in equation (6) to find $i\left(t\right)$.

$i\left(t\right)=\left[6\text{mA}+\left(A_1\cos \left(5000t\right)+A_2\sin \left(5000t\right)\right)e^{-50t}\text{A}\right]$

$i\left(t\right)=\left[6\text{mA}+A_1{e}^{-50t}\cos \left(5000t\right)\text{A}+A_2{e}^{-50t}\sin \left(5000t\right)\text{A}\right]$ (8)

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

$\begin{array}{c}i\left(0\right)=\left[6\text{mA}+A_1{e}^{-50\left(0\right)}\cos \left(5000\left(0\right)\right)\text{A}+A_2{e}^{-50\left(0\right)}\sin \left(5000\left(0\right)\right)\text{A}\right]\\ =\left[\left(6\times {10}^{-3}\right)\text{A}+A_1\left(1\right)\cos \left(0\right)\text{A}+A_2\left(1\right)\sin \left(0\right)\text{A}\right]\left\{\because e^0=1,1\text{m}={10}^{-3}\right\}\\ =\left[\left(6\times {10}^{-3}\right)\text{A}+A_1\left(1\right)\left(1\right)\text{A}+A_2\left(1\right)\left(0\right)\text{A}\right]\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$$i\left(0\right)=\left[\left(6\times {10}^{-3}\right)+A_1\right]\text{A}$ (9)

Substitute $0\text{A}$ for $i\left(0\right)$ in equation (9) to find $A_1$.

$\begin{array}{l}0\text{A}=\left[\left(6\times {10}^{-3}\right)+A_1\right]\text{A}\\ \left(6\times {10}^{-3}\right)+A_1=0\end{array}$

Simplify the above equation to find $A_1$.

$A_1=-6\times {10}^{-3}$

Differentiate equation (8) with respect to $t$ to find $\frac{di\left(t\right)}{dt}$.

$\frac{di\left(t\right)}{dt}=\left[\begin{array}{l}0+A_1{e}^{-50t}\left(-50\right)\cos \left(5000t\right)+A_1{e}^{-50t}\left(-\sin \left(5000t\right)\right)\left(5000\right)\\ +A_2{e}^{-50t}\left(-50\right)\sin \left(5000t\right)+A_2{e}^{-50t}\left(\cos \left(5000t\right)\right)\left(5000\right)\end{array}\right]\frac{\text{A}}{\text{s}}$

$\frac{di\left(t\right)}{dt}=\left[\begin{array}{l}-50A_1{e}^{-50t}\cos \left(5000t\right)-5000A_1{e}^{-50t}\sin \left(5000t\right)\\ -50A_2{e}^{-50t}\sin \left(5000t\right)+5000A_2{e}^{-50t}\cos \left(5000t\right)\end{array}\right]\frac{\text{A}}{\text{s}}$ (10)

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

$\begin{array}{c}\frac{di\left(0\right)}{dt}=\left[\begin{array}{l}-50A_1{e}^{-50\left(0\right)}\cos \left(5000\left(0\right)\right)-5000A_1{e}^{-50\left(0\right)}\sin \left(5000\left(0\right)\right)\\ -50A_2{e}^{-50\left(0\right)}\sin \left(5000\left(0\right)\right)+5000A_2{e}^{-50\left(0\right)}\cos \left(5000\left(0\right)\right)\end{array}\right]\frac{\text{A}}{\text{s}}\\ =\left[\begin{array}{l}-50A_1\left(1\right)\cos \left(0\right)-5000A_1\left(1\right)\sin \left(0\right)\\ -50A_2\left(1\right)\sin \left(0\right)+5000A_2\left(1\right)\cos \left(0\right)\end{array}\right]\frac{\text{A}}{\text{s}}\left\{\because e^0=1\right\}\\ =\left[\begin{array}{l}-50A_1\left(1\right)\left(1\right)-5000A_1\left(1\right)\left(0\right)\\ -50A_2\left(1\right)\left(0\right)+5000A_2\left(1\right)\left(1\right)\end{array}\right]\frac{\text{A}}{\text{s}}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$

$\frac{di\left(0\right)}{dt}=\left[-50A_1+5000A_2\right]\frac{\text{A}}{\text{s}}$ (11)

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

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

Write an expression to calculate the voltage across inductor.

$v_L\left(t\right)=L\frac{di\left(t\right)}{dt}$ (12)

Substitute $v\left(t\right)$ for $v_L\left(t\right)$ in equation (12) to find $v\left(t\right)$.

$v\left(t\right)=L\frac{di\left(t\right)}{dt}$ (13)

Rearrange equation (13) to find $\frac{di\left(t\right)}{dt}$.

$\frac{di\left(t\right)}{dt}=\frac{v\left(t\right)}{L}$ (14)

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

$\frac{di\left(0\right)}{dt}=\frac{v\left(0\right)}{L}$ (15)

Substitute $0\text{V}$ for $v\left(0\right)$, and $8\text{mH}$ for $L$ in equation (15) to find $\frac{di\left(0\right)}{dt}$.

$\begin{array}{c}\frac{di\left(0\right)}{dt}=\frac{0\text{V}}{8\text{mH}}\\ =\frac{0\text{V}}{8\times {10}^{-3}\text{H}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{0\text{V}}{8\times {10}^{-3}\left(\frac{\text{V}\cdot \text{s}}{\text{A}}\right)}\left\{\because 1\text{H}=\frac{\text{1}\text{V}\cdot 1\text{s}}{\text{1}\text{A}}\right\}\\ =0\frac{\text{A}}{\text{s}}\end{array}$

Substitute $0\frac{\text{A}}{\text{s}}$ for $\frac{di\left(0\right)}{dt}$ in equation (11) to find $A_2$.

$\begin{array}{l}0\frac{\text{A}}{\text{s}}=\left[-50A_1+5000A_2\right]\frac{\text{A}}{\text{s}}\\ -50A_1+5000A_2=0\\ 5000A_2=50A_1\end{array}$

Simplify the above equation to find $A_2$.

$A_2=\frac{50A_1}{5000}$ (16)

Substitute $-6\times {10}^{-3}$ for $A_1$ in equation (16) to find $A_2$.

$\begin{array}{c}{A}_2=\frac{50\left(-6\times {10}^{-3}\right)}{5000}\\ =-0.06\times {10}^{-3}\end{array}$

Substitute $-6\times {10}^{-3}$ for $A_1$, and $-0.06\times {10}^{-3}$ for $A_2$ in equation (8) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left[\begin{array}{l}6\text{mA}+\left(-6\times {10}^{-3}\right)e^{-50t}\cos \left(5000t\right)\text{A}\\ +\left(-0.06\times {10}^{-3}\right)e^{-50t}\sin \left(5000t\right)\text{A}\end{array}\right]\\ =\left[\begin{array}{l}6\times {10}^{-3}+\left(-6\times {10}^{-3}\right)e^{-50t}\cos \left(5000t\right)\\ +\left(-0.06\times {10}^{-3}\right)e^{-50t}\sin \left(5000t\right)\end{array}\right]\text{A}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left[6-6e^{-50t}\cos \left(5000t\right)-0.06e^{-50t}\sin \left(5000t\right)\right]\times {10}^{-3}\text{A}\\ \text{=}\left[6-\left(6\cos \left(5000t\right)+0.06\sin \left(5000t\right)\right)e^{-50t}\right]u\left(t\right)\text{mA}\left\{\because 1\text{m}={10}^{-3},u\left(t\right)=1\text{for}t>0\right\}\end{array}$

Therefore, the expression of current $i\left(t\right)$ for $t>0$ is $\left[6-\left(6\cos \left(5000t\right)+0.06\sin \left(5000t\right)\right)e^{-50t}\right]u\left(t\right)\text{mA}$.

Conclusion:

Thus, thus a problem to make the better understand of a step response of a parallel $RLC$ circuit is designed.