Chapter 8, Problem 45P

In the circuit of Fig. 8.92, find v(t) and i(t) for t > 0.

Figure 8.92

For Prob. 8.45.

To determine

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

Answer to Problem 45P

For $t>0$, the expression of voltage $v\left(t\right)$ is $7.5594e^{-0.5t}\sin \left(1.3229t\right)\text{V}$, and the expression of current $i\left(t\right)$ is $\left[6-\left[\left(5\cos \left(1.3229t\right)+1.8898\sin \left(1.3229t\right)\right)e^{-0.5t}\right]\right]\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 8.92 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.

The value of current source is calculated as follows for $t<0$,

$\begin{array}{c}{I}_s=\left[1+5u\left(t\right)\right]\text{A}\\ \text{=}\left[\text{1+5}\left(0\right)\right]\text{A}\left\{\because u\left(t\right)=0\text{for}t<0\right\}\\ \text{=1}\text{A}\end{array}$

Now the Figure 1 is reduced as shown in Figure 2.

Refer to Figure 2, the circuit is short circuited and full current flows through the inductor.

$i_L\left(0^{-}\right)=1\text{A}$

Refer to Figure 2, the short circuited inductor and open circuited capacitor are connected in parallel. Since the full current flows through the short circuit path there is no voltage across the capacitor.

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

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

$i\left(0\right)=i_L\left(0^{-}\right)=i_L\left(0^{+}\right)=1\text{A}$

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

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

$\begin{array}{c}{I}_s=\left[1+5\left(1\right)\right]\text{A}\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =6\text{A}\end{array}$

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

Refer to Figure 3, the circuit shows the step response of a parallel $RLC$ circuit.

Substitute $2\Omega$ for $R$, and $0.5\text{F}$ for $C$ in equation (1) to find $\alpha$.

$\begin{array}{c}\alpha =\frac{1}{2\left(2\Omega \right)\left(0.5\text{F}\right)}\\ =\frac{1}{2\left(2\Omega \right)\left(0.5\frac{\text{s}}{\Omega }\right)}\\ =0.5\frac{\text{Np}}{\text{s}}\end{array}$

Substitute $1\text{H}$ for $L$, and $0.5\text{F}$ for $C$ in equation (2) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(1\text{H}\right)\left(0.5\text{F}\right)}}\\ =\frac{1}{\sqrt{\left(1\frac{{\text{s}}^2}{\text{F}}\right)\left(0.5\text{F}\right)}}\\ =1.414\frac{\text{rad}}{\text{s}}\end{array}$

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 $0.5$ for $\alpha$, and $1.414$ for ${\omega }_0$ in equation (4) to find ${\omega }_d$.

$\begin{array}{c}{\omega }_d=\sqrt{{\left(1.414\right)}^2-{\left(0.5\right)}^2}\\ =1.3229\end{array}$

Substitute $6$ for $I_s$, $0.5$ for $\alpha$, and $1.3229$ for ${\omega }_d$ in equation (3) to find $i\left(t\right)$.

$i\left(t\right)=\left[6+\left(A_1\cos \left(1.3229t\right)+A_2\sin \left(1.3229t\right)\right)e^{-0.5t}\right]\text{A}$

$i\left(t\right)=\left[6+A_1{e}^{-0.5t}\cos \left(1.3229t\right)+A_2{e}^{-0.5t}\sin \left(1.3229t\right)\right]\text{A}$ (5)

Substitute $0$ for $t$ in equation (5) to find $i\left(0\right)$.

$\begin{array}{c}i\left(0\right)=\left[6+A_1{e}^{-0.5\left(0\right)}\cos \left(1.3229\left(0\right)\right)+A_2{e}^{-0.5\left(0\right)}\sin \left(1.3229\left(0\right)\right)\right]\text{A}\\ \text{=}\left[6+A_1\left(1\right)\cos \left(0\right)+A_2\left(1\right)\sin \left(0\right)\right]\text{A}\left\{\because e^0=1\right\}\\ =\left[6+A_1\left(1\right)\left(1\right)+A_2\left(1\right)\left(0\right)\right]\text{A}\left\{\because \cos 0°=1,\sin 0°=0\right\}\end{array}$

$i\left(0\right)=\left[6+A_1\right]\text{A}$ (6)

Substitute $1\text{A}$ for $i\left(0\right)$ in equation (6) to find $A_1$.

$\begin{array}{l}1\text{A}=\left[6+A_1\right]\text{A}\\ 6+A_1=1\end{array}$

Simplify the above equation to find $A_1$.

$\begin{array}{c}{A}_1=1-6\\ =-5\end{array}$

Substitute $-5$ for $A_1$ in equation (5) to find $i\left(t\right)$.

$i\left(t\right)=\left[6-5e^{-0.5t}\cos \left(1.3229t\right)+A_2{e}^{-0.5t}\sin \left(1.3229t\right)\right]\text{A}$ (7)

Differentiate equation (7) with respect to $t$.

$\frac{di\left(t\right)}{dt}=\left[\begin{array}{l}0-5e^{-0.5t}\left(-0.5\right)\cos \left(1.3229t\right)-5e^{-0.5t}\left(-\sin \left(1.3229t\right)\right)\left(1.3229\right)\\ +A_2{e}^{-0.5t}\left(-0.5\right)\sin \left(1.3229t\right)+A_2{e}^{-0.5t}\left(\cos \left(1.3229t\right)\right)\left(1.3229\right)\end{array}\right]\frac{\text{A}}{\text{s}}$

$\frac{di\left(t\right)}{dt}=\left[\begin{array}{l}2.5e^{-0.5t}\cos \left(1.3229t\right)+6.6145e^{-0.5t}\sin \left(1.3229t\right)\\ -0.5A_2{e}^{-0.5t}\sin \left(1.3229t\right)+1.3229A_2{e}^{-0.5t}\cos \left(1.3229t\right)\end{array}\right]\frac{\text{A}}{\text{s}}$ (8)

Substitute $0$ for $t$ in equation (8) to find $\frac{di\left(0\right)}{dt}$.

$\begin{array}{c}\frac{di\left(0\right)}{dt}=\left[\begin{array}{l}2.5e^{-0.5\left(0\right)}\cos \left(1.3229\left(0\right)\right)+6.6145e^{-0.5\left(0\right)}\sin \left(1.3229\left(0\right)\right)\\ -0.5A_2{e}^{-0.5\left(0\right)}\sin \left(1.3229\left(0\right)\right)+1.3229A_2{e}^{-0.5\left(0\right)}\cos \left(1.3229\left(0\right)\right)\end{array}\right]\frac{\text{A}}{\text{s}}\\ =\left[\begin{array}{l}2.5\left(1\right)\cos \left(0\right)+6.6145\left(1\right)\sin \left(0\right)\\ -0.5A_2\left(1\right)\sin \left(0\right)+1.3229A_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}2.5\left(1\right)\left(1\right)+6.6145\left(1\right)\left(0\right)\\ -0.5A_2\left(1\right)\left(0\right)+1.3229A_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[2.5+1.3229A_2\right]\frac{\text{A}}{\text{s}}$ (9)

For the parallel $RLC$ connection the voltage is same for resistor, inductor and capacitor.

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

Write an expression to calculate the voltage across the inductor.

$v_L\left(t\right)=L\frac{di\left(t\right)}{dt}$ (10)

Rearrange equation (10) to find $\frac{di\left(t\right)}{dt}$.

$\frac{di\left(t\right)}{dt}=\frac{v_L\left(t\right)}{L}$ (11)

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

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

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

 $\frac{di\left(0\right)}{dt}=\frac{v\left(0\right)}{L}$                                                                                                       (13)

Substitute $1\text{H}$ for $L$, and $0\text{V}$ for $v\left(0\right)$ in equation (13) to find $\frac{di\left(0\right)}{dt}$.

$\begin{array}{c}\frac{di\left(0\right)}{dt}=\frac{0\text{V}}{1\text{H}}\\ =\frac{0\text{V}}{1\frac{\text{V}\cdot \text{s}}{\text{A}}}\left\{\because 1\text{H}=\frac{\text{1}\text{V}\cdot 1\text{s}}{\text{1}\text{A}}\right\}\\ =0\frac{\text{A}}{\text{s}}\end{array}$

Substitute $0\frac{\text{A}}{\text{s}}$ for $\frac{di\left(0\right)}{dt}$ in equation (9) to find $A_2$.

$\begin{array}{l}0\frac{\text{A}}{\text{s}}=\left[2.5+1.3229A_2\right]\frac{\text{A}}{\text{s}}\\ 2.5+1.3229A_2=0\\ 1.3229A_2=-2.5\end{array}$

Simplify the above equation to find

$\begin{array}{c}{A}_2=\frac{-2.5}{1.3229}\\ =-1.8898\end{array}$

Substitute $-1.8898$ for $A_2$ in equation (7) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left[6-5e^{-0.5t}\cos \left(1.3229t\right)-1.8898e^{-0.5t}\sin \left(1.3229t\right)\right]\text{A}\\ \text{=}\left[6-\left[\left(5\cos \left(1.3229t\right)+1.8898\sin \left(1.3229t\right)\right)e^{-0.5t}\right]\right]\text{A}\end{array}$

Substitute $-1.8898$ for $A_2$ in equation (8) to find $\frac{di\left(t\right)}{dt}$.

$\begin{array}{c}\frac{di\left(t\right)}{dt}=\left[\begin{array}{l}2.5e^{-0.5t}\cos \left(1.3229t\right)+6.6145e^{-0.5t}\sin \left(1.3229t\right)\\ -0.5\left(-1.8898\right)e^{-0.5t}\sin \left(1.3229t\right)+1.3229\left(-1.8898\right)e^{-0.5t}\cos \left(1.3229t\right)\end{array}\right]\frac{\text{A}}{\text{s}}\\ =\left[\begin{array}{l}2.5e^{-0.5t}\cos \left(1.3229t\right)+6.6145e^{-0.5t}\sin \left(1.3229t\right)\\ +0.9449e^{-0.5t}\sin \left(1.3229t\right)-2.5e^{-0.5t}\cos \left(1.3229t\right)\end{array}\right]\frac{\text{A}}{\text{s}}\\ =\left[7.5594e^{-0.5t}\sin \left(1.3229t\right)\right]\frac{\text{A}}{\text{s}}\end{array}$

Substitute $\left[7.5594e^{-0.5t}\sin \left(1.3229t\right)\right]\frac{\text{A}}{\text{s}}$ for $\frac{di\left(t\right)}{dt}$, and $1\text{H}$ for $L$ in equation (10) to find $v_L\left(t\right)$.

$\begin{array}{c}{v}_L\left(t\right)=\left(1\text{H}\right)\left[7.5594e^{-0.5t}\sin \left(1.3229t\right)\right]\frac{\text{A}}{\text{s}}\\ =\left[7.5594e^{-0.5t}\sin \left(1.3229t\right)\right]\frac{\text{A}\cdot \text{H}}{\text{s}}\\ =7.5594e^{-0.5t}\sin \left(1.3229t\right)\text{V}\left\{\because 1\text{V}=\frac{\text{1}\text{A}\cdot 1\text{H}}{\text{1}\text{s}}\right\}\end{array}$

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

$v\left(t\right)=7.5594e^{-0.5t}\sin \left(1.3229t\right)\text{V}$

Conclusion:

Thus, the expression of voltage $v\left(t\right)$ is $7.5594e^{-0.5t}\sin \left(1.3229t\right)\text{V}$, and the expression of current $i\left(t\right)$ is $\left[6-\left[\left(5\cos \left(1.3229t\right)+1.8898\sin \left(1.3229t\right)\right)e^{-0.5t}\right]\right]\text{A}$ for $t>0$.