Chapter 5, Problem 5.68HP

To determine

The current $i_L$ through the inductor for time $t>0$ .

Answer to Problem 5.68HP

The expression for the current flowing through the inductor is $\left[3.33\times {10}^{-3}\text{A}+e^{-0.42t}\left(\begin{array}{l}\left(-2.774\times {10}^{-3}\right)\cos \left(707.1\text{rad}/\text{s}t\right)\\ +2.36\sin \left(707.1\text{rad}/\text{s}t\right)\end{array}\right)\right]$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

 

The conversion from $\text{mF}$ into $\text{F}$ is given by

  $\text{1}\text{mF}=1\times {10}^{-3}\text{F}$

The conversion from $2\text{mF}$ into $\text{F}$ is given by,

  $2\text{mF}=2\times {10}^{-3}\text{F}$

The conversion from $1\text{mH}$ into $\text{H}$ is given by,

  $1\text{mH}=1\times {10}^{-3}\text{H}$

The conversion from $\text{k}\Omega$ into $\Omega$ is given by,

  $\text{1}\text{k}\Omega ={10}^3\Omega$

The conversion from $\text{3}\text{k}\Omega$ into $\Omega$ is given by,

  $3\text{k}\Omega =3\times {10}^3\Omega$

For time $t=0^{-}$ the switch is closed circuit reaches state reaches steady state, the capacitor acts as open circuit and the inductor as short circuit. Mark the values and redraw the circuit.

The required diagram is shown in Figure 2

 

From above, the expression for the initial voltage across the capacitor is given by,

  $v_C\left(0^{-}\right)=V_S\left(\frac{R}{R+R_S}\right)$

Substitute $2\text{V}$ for $V_S$, $600\Omega$ for $R_S$ and $3\text{k}\Omega$ for $R$ in the above equation.

  $\begin{array}{c}{v}_C\left(0^{-}\right)=\left(2\text{V}\right)\left(\frac{3\text{k}\Omega }{3\text{k}\Omega +600\Omega }\right)\\ =\left(2\text{V}\right)\left(\frac{3\times {10}^3\Omega }{3\times {10}^3\Omega +600\Omega }\right)\\ =1.67\text{V}\end{array}$

The expression for the voltage across the capacitor for time $t=0^{+}$ is given by,

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

Substitute $1.67\text{V}$ for $v_C\left(0^{-}\right)$ in the above equation.

  $v_C\left(0^{+}\right)=1.67\text{V}$

The expression for the current flowing through the inductor for time $t=0^{-}$ is given by,

  $i_L\left(0^{-}\right)=\frac{V_S}{R+R_S}$

Substitute $2\text{V}$ for $V_S$, $600\Omega$ for $R_S$ and $3\text{k}\Omega$ for $R$ in the above equation.

  $\begin{array}{c}{i}_L\left(0^{-}\right)=\frac{2\text{V}}{3\times {10}^3\Omega +600\Omega }\\ =0.556\times {10}^3\text{A}\end{array}$

The inductor opposes sudden change in the current, thus the current $i_L\left(0^{+}\right)$ is given by,

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

Substitute $0.556\times {10}^3\text{A}$ for $i_L\left(0^{-}\right)$ in the above equation.

  $i_L\left(0^{+}\right)=0.556\times {10}^3\text{A}$

Close the switch and redraw the circuit for time $t=0$ .

The required diagram is shown in Figure 3

 

The expression for the voltage across the inductor is given by,

  $v_C\left(t\right)=L\frac{d{i}_L\left(t\right)}{dt}$

Apply KVL to the top node.

  $\frac{v_C\left(t\right)-V_S}{R_S}+V\frac{d{v}_C\left(t\right)}{dt}+i_L\left(t\right)=0$

Substitute $L\frac{d{i}_L\left(t\right)}{dt}$ for $v_C\left(t\right)$ in the above equation.

  $\begin{array}{l}\frac{L\frac{d{i}_L\left(t\right)}{dt}-V_S}{R_S}+V\frac{d\left[L\frac{d{i}_L\left(t\right)}{dt}\right]}{dt}+i_L\left(t\right)=0\\ LC\frac{d^2{i}_L\left(t\right)}{d{t}^2}+\frac{1}{R_S}\frac{d{i}_L\left(t\right)}{dt}+i_L\left(t\right)=\frac{V_S}{R_S}\end{array}$   .......... (1)

The standard second order equation for the differential equation.

  $\frac{1}{{\omega }_n^2}\frac{d^{2}x\left(t\right)}{d{t}^2}+\frac{2\varsigma }{{\omega }_n}\frac{dx\left(t\right)}{dt}+x\left(t\right)=K_{S}f\left(t\right)$

From above and from equation (1), the angular frequency is derived as,

  $\begin{array}{l}\frac{1}{{\omega }_n^2}=LC\\ {\omega }_n=\frac{1}{\sqrt{LC}}\end{array}$

Substitute $1\times {10}^{-3}\text{H}$ for $L$ and $2\times {10}^{-3}\text{F}$ for $C$ in the above equation.

  $\begin{array}{c}{\omega }_n=\frac{1}{\sqrt{\left(1\times {10}^{-3}\text{H}\right)\left(2\times {10}^{-3}\text{F}\right)}}\\ =707.1\text{rad}/\text{s}\end{array}$

The expression for the damping coefficient is given by,

  $\varsigma =\frac{{\omega }_{n}L}{2R_S}$

Substitute $707.1\text{rad}/\text{s}$ for ${\omega }_n$, $1\times {10}^{-3}\text{H}$ for $L$ and $600\Omega$ for $R_S$ in the above equation.

  $\begin{array}{c}\varsigma =\frac{\left(707.1\text{rad}/\text{s}\right)\left(1\times {10}^{-3}\text{H}\right)}{2\left(600\Omega \right)}\\ =5.89\times {10}^{-4}\end{array}$

The value of $\varsigma$ is less then $1$, thus the circuit is under damped.

The expression for the output response equation of the inductor current is given by,

  $i_L\left(t\right)=i_{L,F}\left(t\right)+i_{L,N}\left(t\right)$   .......... (2)

The expression for the forced response of the system is given by,

  $i_{L,F}\left(t\right)=\frac{V_S}{R_S}$

Substitute $600\Omega$ for $R_S$ and $2\text{V}$ for $V_S$ in the above equation.

  $\begin{array}{c}{i}_{L,F}\left(t\right)=\frac{2\text{V}}{600\Omega }\\ =3.33\times {10}^{-3}\text{A}\end{array}$

The expression for the natural response for the system is give by,

  $i_{L,N}=e^{-\varsigma {\omega }_{n}t}\left({\alpha }_1\cos \left({\omega }_{d}t\right)+{\alpha }_2\sin \left({\omega }_{d}t\right)\right)$

Substitute $e^{-\varsigma {\omega }_{n}t}\left({\alpha }_1\cos \left({\omega }_{d}t\right)+{\alpha }_2\sin \left({\omega }_{d}t\right)\right)$ for $i_{L,N}$ and $3.33\times {10}^{-3}\text{A}$ for $i_{L,F}\left(t\right)$ in equation (3).

  $i_L\left(t\right)=3.33\times {10}^{-3}\text{A}+e^{-\varsigma {\omega }_{n}t}\left({\alpha }_1\cos \left({\omega }_{d}t\right)+{\alpha }_2\sin \left({\omega }_{d}t\right)\right)$   .......... (4)

The expression to calculate the damping frequency of the circuit is given by,

  ${\omega }_d={\omega }_n\sqrt{1-{\varsigma }^2}$

Substitute $707.1\text{rad}/\text{s}$ for ${\omega }_n$ and $5.89\times {10}^{-4}$ for $\varsigma$ in the above equation.

  $\begin{array}{c}{\omega }_d=\left(707.1\text{rad}/\text{s}\right)\sqrt{1-{\left(5.89\times {10}^{-4}\right)}^2}\\ =707.1\text{rad}/\text{s}\end{array}$

Substitute $707.1\text{rad}/\text{s}$ for ${\omega }_d$, $707.1\text{rad}/\text{s}$ for ${\omega }_n$ and $5.89\times {10}^{-4}$ for $\varsigma$ in equation (4).

  $\begin{array}{c}{i}_L\left(t\right)=\left[3.33\times {10}^{-3}\text{A}+e^{-\left(5.89\times {10}^{-4}\right)\left(707.1\text{rad}/\text{s}\right)t}\left(\begin{array}{l}{\alpha }_1\cos \left(707.1\text{rad}/\text{s}t\right)\\ +{\alpha }_2\sin \left(707.1\text{rad}/\text{s}t\right)\end{array}\right)\right]\\ =\left[3.33\times {10}^{-3}\text{A}+e^{-0.42t}\left(\begin{array}{l}{\alpha }_1\cos \left(707.1\text{rad}/\text{s}t\right)\\ +{\alpha }_2\sin \left(707.1\text{rad}/\text{s}t\right)\end{array}\right)\right]\end{array}$   .......... (5)

Substitute $0$ for $t$ in the above equation.

  $\begin{array}{c}{i}_L\left(0\right)=\left[3.33\times {10}^{-3}\text{A}+e^{-0.42\left(0\right)}\left(\begin{array}{l}{\alpha }_1\cos \left(707.1\text{rad}/\text{s}\left(0\right)\right)\\ +{\alpha }_2\sin \left(707.1\text{rad}/\text{s}\left(0\right)\right)\end{array}\right)\right]\\ =3.33\times {10}^{-3}+{\alpha }_1\end{array}$

Substitute $0.556\times {10}^{-3}\text{A}$ for $i_L\left(0\right)$ in the above equation.

  $\begin{array}{l}0.556\times {10}^{-3}\text{A}=3.33\times {10}^{-3}+{\alpha }_1\\ {\alpha }_1=-2.774\times {10}^{-3}\end{array}$

The differentiation of equation (5) with respect to $dt$ is given by,

  $\begin{array}{l}\frac{d{i}_L\left(0\right)}{dt}=\left[\begin{array}{l}{e}^{-0.42t}\left[-707.1{\alpha }_1\sin \left(707.1t\right)+707.1{\alpha }_2\cos \left(707.1t\right)\right]-\\ 0.42e^{-0.42t}\left[{\alpha }_1\cos \left(707.1t\right)+{\alpha }_2\sin \left(707.1t\right)\right]\end{array}\right]\\ \frac{v_C\left(t\right)}{L}=\left[\begin{array}{l}{e}^{-0.42t}\left[-707.1{\alpha }_1\sin \left(707.1t\right)+707.1{\alpha }_2\cos \left(707.1t\right)\right]-\\ 0.42e^{-0.42t}\left[{\alpha }_1\cos \left(707.1t\right)+{\alpha }_2\sin \left(707.1t\right)\right]\end{array}\right]\end{array}$

Substitute $1\times {10}^{-3}\text{H}$ for $L$, $0$ for $t$ and $1.67\text{V}$ for $v_C\left(0\right)$ in the above equation.

  $\begin{array}{c}\frac{1.67\text{V}}{1\times {10}^{-3}\text{H}}=\left[\begin{array}{l}{e}^{-0.42\left(0\right)}\left[-707.1{\alpha }_1\sin \left(707.1\left(0\right)\right)+707.1{\alpha }_2\cos \left(707.1\left(0\right)\right)\right]-\\ 0.42e^{-0.42\left(0\right)}\left[{\alpha }_1\cos \left(707.1\left(0\right)\right)+{\alpha }_2\sin \left(707.1\left(0\right)\right)\right]\end{array}\right]\\ =707.1{\alpha }_2-0.42{\alpha }_1\end{array}$

Substitute $-2.774\times {10}^{-3}$ for ${\alpha }_1$ in the above equation.

  $\begin{array}{l}\frac{1.67\text{V}}{1\times {10}^{-3}\text{H}}=07.1{\alpha }_2-0.42\left(-2.774\times {10}^{-3}\right)\\ 1670=707.1{\alpha }_2+1.165\times {10}^{-3}\\ {\alpha }_2=2.36\end{array}$

Substitute $-2.774\times {10}^{-3}$ for ${\alpha }_1$ and $2.36$ for ${\alpha }_2$ in equation (5).

  $i_L\left(t\right)=\left[3.33\times {10}^{-3}\text{A}+e^{-0.42t}\left(\begin{array}{l}\left(-2.774\times {10}^{-3}\right)\cos \left(707.1\text{rad}/\text{s}t\right)\\ +2.36\sin \left(707.1\text{rad}/\text{s}t\right)\end{array}\right)\right]$

Conclusion:

Therefore, the expression for the current flowing through the inductor is $\left[3.33\times {10}^{-3}\text{A}+e^{-0.42t}\left(\begin{array}{l}\left(-2.774\times {10}^{-3}\right)\cos \left(707.1\text{rad}/\text{s}t\right)\\ +2.36\sin \left(707.1\text{rad}/\text{s}t\right)\end{array}\right)\right]$ .