Chapter 14, Problem 44P

For the circuit in Fig. 14.83, find:

1. (a) the resonant frequency ω₀
2. (b) Z_in(ω₀)

Figure 14.83

To determine

Find the value of the resonant frequency $\left({\omega }_0\right)$ for the circuit in Figure 14.83.

Answer to Problem 44P

The value of the resonant frequency $\left({\omega }_0\right)$ is $2.357\frac{\text{krad}}{\text{s}}$.

Explanation of Solution

Given data:

Refer to Figure 14.83 in the textbook.

Formula used:

Write a general expression to calculate the impedance of a resistor in frequency domain.

$Z_R=R$ (1)

Here,

$R$ is the value of the resistor.

Write a general expression to calculate the impedance of an inductor in frequency domain.

$Z_L=j\omega L$ (2)

Here,

$L$ is the value of the inductor.

Write a general expression to calculate the impedance of a capacitor in frequency domain.

$Z_C=\frac{1}{j\omega C}$ (3)

Here,

$C$ is the value of the capacitor.

Calculation:

The given circuit is redrawn as Figure 1.

The Figure 1 is converted into frequency domain and drawn as Figure 2 using the equations (1), (2), and (3).

Refer to Figure 2, the resistor $R_1$ and inductor is connected in parallel form and the combination is connected in parallel with the series combination of the capacitor and resistor $R_2$.

Write the expression to calculate impedance of the circuit in Figure 2.

$\begin{array}{c}{Z}_{\text{in}}=\left(R_1\parallel j\omega L\right)\parallel \left(R_2+\frac{1}{j\omega C}\right)\\ =\left(\frac{R_1\times j\omega L}{R_1+j\omega L}\right)\parallel \left(R_2+\frac{1}{j\omega C}\right)\\ =\frac{\left(\frac{j\omega R_{1}L}{R_1+j\omega L}\right)\times \left(R_2+\frac{1}{j\omega C}\right)}{\left(\frac{j\omega R_{1}L}{R_1+j\omega L}+R_2+\frac{1}{j\omega C}\right)}\\ =\frac{\left(\frac{j\omega R_1{R}_{2}L}{R_1+j\omega L}+\frac{j\omega R_{1}L}{j\omega C\left(R_1+j\omega L\right)}\right)}{\left(\frac{\left(j\omega R_{1}L\right)\left(j\omega C\right)+R_2\left(j\omega C\right)\left(R_1+j\omega L\right)+1\left(R_1+j\omega L\right)}{j\omega C\left(R_1+j\omega L\right)}\right)}\end{array}$

Simplify the above equation to find $Z_{\text{in}}$.

$\begin{array}{c}{Z}_{\text{in}}=\frac{\left(\frac{j\omega R_1{R}_{2}L\left(j\omega C\right)+j\omega R_{1}L}{j\omega C\left(R_1+j\omega L\right)}\right)}{\left(\frac{\left(j\omega R_{1}L\right)\left(j\omega C\right)+R_2\left(j\omega C\right)\left(R_1+j\omega L\right)+1\left(R_1+j\omega L\right)}{j\omega C\left(R_1+j\omega L\right)}\right)}\\ =\frac{j^2{\omega }^2{R}_1{R}_{2}LC+j\omega R_{1}L}{j^2{\omega }^2{R}_{1}LC+j\omega R_1{R}_{2}C+j^2{\omega }^2{R}_{2}LC+R_1+j\omega L}\\ =\frac{-{\omega }^2{R}_1{R}_{2}LC+j\omega R_{1}L}{-{\omega }^2{R}_{1}LC+j\omega R_1{R}_{2}C-{\omega }^2{R}_{2}LC+R_1+j\omega L}\left\{\because j^2=1\right\}\\ =\frac{-{\omega }^2{R}_1{R}_{2}LC+j\left(\omega R_{1}L\right)}{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)+j\omega \left(L+R_1{R}_{2}C\right)}\end{array}$

Multiply and divide the above equation by the conjugate of denominator to find $Z_{\text{in}}$.

$\begin{array}{c}{Z}_{\text{in}}=\left(\begin{array}{l}\left(\frac{-{\omega }^2{R}_1{R}_{2}LC+j\left(\omega R_{1}L\right)}{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)+j\omega \left(L+R_1{R}_{2}C\right)}\right)\\ \times \left(\frac{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)-j\omega \left(L+R_1{R}_{2}C\right)}{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)-j\omega \left(L+R_1{R}_{2}C\right)}\right)\end{array}\right)\\ \\ =\frac{\left(\begin{array}{l}\left(-{\omega }^2{R}_1{R}_{2}LC+j\omega R_{1}L\right)\\ \left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC-j\omega L-j\omega R_1{R}_{2}C\right)\end{array}\right)}{{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)}^2-{\left(j\omega \left(L+R_1{R}_{2}C\right)\right)}^2}\left\{\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right\}\\ \\ =\frac{\left(\begin{array}{l}-{\omega }^2{R}_1{}^2{R}_{2}LC+{\omega }^4{R}_1{}^2{R}_2{L}^2{C}^2+{\omega }^4{R}_1{R}_2{}^2{L}^2{C}^2\\ +j{\omega }^3{R}_1{R}_2{L}^{2}C+j{\omega }^3{R}_1{}^2{R}_2{}^{2}L{C}^2+j\omega R_1{}^{2}L\\ -j{\omega }^3{R}_1{}^2{L}^{2}C-j{\omega }^3{R}_1{R}_2{L}^{2}C-j^2{\omega }^2{R}_1{L}^2-j^2{\omega }^2{R}_1{}^2{R}_{2}LC\end{array}\right)}{{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)}^2-{\left(j\omega \left(L+R_1{R}_{2}C\right)\right)}^2}\\ \\ =\frac{\left(\begin{array}{l}-{\omega }^2{R}_1{}^2{R}_{2}LC+{\omega }^4{R}_1{}^2{R}_2{L}^2{C}^2+{\omega }^4{R}_1{R}_2{}^2{L}^2{C}^2\\ +j{\omega }^3{R}_1{R}_2{L}^{2}C+j{\omega }^3{R}_1{}^2{R}_2{}^{2}L{C}^2+j\omega R_1{}^{2}L\\ -j{\omega }^3{R}_1{}^2{L}^{2}C-j{\omega }^3{R}_1{R}_2{L}^{2}C-\left(-1\right){\omega }^2{R}_1{L}^2-\left(-1\right){\omega }^2{R}_1{}^2{R}_{2}LC\end{array}\right)}{{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)}^2-\left(-1\right){\left(\omega \left(L+R_1{R}_{2}C\right)\right)}^2}\left\{\because j^2=-1\right\}\end{array}$

Simplify the above equation to find $Z_{\text{in}}$.

$Z_{\text{in}}=\frac{\left(\begin{array}{l}-{\omega }^2{R}_1{}^2{R}_{2}LC+{\omega }^4{R}_1{}^2{R}_2{L}^2{C}^2+{\omega }^4{R}_1{R}_2{}^2{L}^2{C}^2\\ +j{\omega }^3{R}_1{R}_2{L}^{2}C+j{\omega }^3{R}_1{}^2{R}_2{}^{2}L{C}^2+j\omega R_1{}^{2}L\\ -j{\omega }^3{R}_1{}^2{L}^{2}C-j{\omega }^3{R}_1{R}_2{L}^{2}C+{\omega }^2{R}_1{L}^2+{\omega }^2{R}_1{}^2{R}_{2}LC\end{array}\right)}{{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)}^2+{\left(\omega \left(L+R_1{R}_{2}C\right)\right)}^2}$

$Z_{\text{in}}=\frac{\left(\begin{array}{l}\left({\omega }^4{R}_1{R}_2{L}^2{C}^2\left(R_1+R_2\right)+{\omega }^2{R}_1{L}^2\right)\\ +j\left({\omega }^3{R}_1{R}_{2}LC\left(L+R_1{R}_{2}C\right)+\omega R_{1}L\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)\right)\end{array}\right)}{{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)}^2+{\left(\omega \left(L+R_1{R}_{2}C\right)\right)}^2}$ (4)

Equate the imaginary part to zero in equation (4).

$\begin{array}{l}\frac{\left({\omega }^3{R}_1{R}_{2}LC\left(L+R_1{R}_{2}C\right)+\omega R_{1}L\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)\right)}{{\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)}^2+{\left(\omega \left(L+R_1{R}_{2}C\right)\right)}^2}=0\\ \omega R_{1}L\left({\omega }^2{R}_{2}C\left(L+R_1{R}_{2}C\right)+\left(R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)\right)=0\\ \left({\omega }^2{R}_{2}CL+{\omega }^2{R}_1{R}_2{}^2{C}^2+R_1-{\omega }^2{R}_{1}LC-{\omega }^2{R}_{2}LC\right)=0\\ {\omega }^2{R}_1{R}_2{}^2{C}^2+R_1-{\omega }^2{R}_{1}LC=0\end{array}$

Simplify the above equation.

$\begin{array}{l}{R}_1\left({\omega }^2{R}_2{}^2{C}^2+1-{\omega }^{2}LC\right)=0\\ {\omega }^2{R}_2{}^2{C}^2+1-{\omega }^{2}LC=0\\ {\omega }^{2}LC-{\omega }^2{R}_2{}^2{C}^2=1\\ {\omega }^2\left(LC-R_2{}^2{C}^2\right)=1\end{array}$

Rearrange the above equation to find ${\omega }_0{}^2$.

${\omega }_0{}^2=\frac{1}{\left(LC-R_2{}^2{C}^2\right)}$

Take square root on both sides of the above equation to find ${\omega }_0$.

$\sqrt{{\omega }_0{}^2}=\sqrt{\frac{1}{\left(LC-R_2{}^2{C}^2\right)}}$

${\omega }_0=\frac{1}{\sqrt{\left(LC-R_2{}^2{C}^2\right)}}$ (5)

Substitute $20\text{mH}$ for $L$, $9\mu \text{F}$ for $C$ and $0.1\Omega$ for $R_2$ in equation (5) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(\left(20\text{mH}\right)\left(9\mu \text{F}\right)-{\left(0.1\Omega \right)}^2{\left(9\mu \text{F}\right)}^2\right)}}\\ =\frac{1}{\sqrt{\left(\left(20\times {10}^{-3}\right)\left(9\times {10}^{-6}\right)\text{H}\cdot \text{F}-{\left(0.1\right)}^2{\left(9\times {10}^{-6}\right)}^2{\Omega }^2\cdot {\text{F}}^2\right)}}\left\{\because 1\text{m}={10}^{-3},1\text{μ}={10}^{-6}\right\}\\ =\frac{1}{\sqrt{\left(\left(1.8\times {10}^{-7}\right)\left(\frac{{\text{s}}^{\text{2}}}{\text{F}}\right)\cdot \text{F}-\left(8.1\times {10}^{-13}\right){\left(\frac{\text{s}}{\text{F}}\right)}^2\cdot {\text{F}}^2\right)}}\left\{\because 1\text{H}=\frac{\text{1}{\text{s}}^{\text{2}}}{\text{1}\text{F}},1\Omega =\frac{\text{1}\text{s}}{\text{1}\text{F}}\right\}\\ =\frac{1}{\sqrt{\left(\left(1.8\times {10}^{-7}\right){\text{s}}^{\text{2}}-\left(8.1\times {10}^{-13}\right){\text{s}}^{\text{2}}\right)}}\end{array}$

Simplify the above equation to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(1.7999\times {10}^{-7}\right){\text{s}}^{\text{2}}}}\\ =2.357\times {10}^3\frac{\text{rad}}{\text{s}}\\ =2.357\frac{\text{krad}}{\text{s}}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

Conclusion:

Thus, the value of the resonant frequency $\left({\omega }_0\right)$ is $2.357\frac{\text{krad}}{\text{s}}$.

To determine

Find the value of the input impedance $Z_{\text{in}}\left({\omega }_0\right)$.

Answer to Problem 44P

The value of the input impedance $Z_{\text{in}}\left({\omega }_0\right)$ is $1\Omega$.

Explanation of Solution

Calculation:

From part a, ${\omega }_0=\omega =2.357\frac{\text{krad}}{\text{s}}$ and

$Z_{\text{in}}\left({\omega }_0\right)=\left(R_1\parallel j\omega L\right)\parallel \left(R_2+\frac{1}{j\omega C}\right)$ (6)

To find $\left(R_1\parallel j\omega L\right)$:

$\left(R_1\parallel j\omega L\right)=\left(\frac{R_1\times j\omega L}{R_1+j\omega L}\right)$

$\left(R_1\parallel j\omega L\right)=\frac{j\omega R_{1}L}{R_1+j\omega L}$ (7)

Substitute $1\Omega$ for $R_1$, $20\text{mH}$ for $L$ and $2.357\frac{\text{krad}}{\text{s}}$ for $\omega$ in equation (7) to find $\left(R_1\parallel j\omega L\right)$.

$\begin{array}{c}\left(R_1\parallel j\omega L\right)=\frac{j\left(2.357\frac{\text{krad}}{\text{s}}\right)\left(1\Omega \right)\left(20\text{mH}\right)}{1\Omega +j\left(2.357\frac{\text{krad}}{\text{s}}\right)\left(20\text{mH}\right)}\\ =\frac{j\left(2.357\times {10}^3\right)\left(1\right)\left(20\times {10}^{-3}\right)\frac{\Omega \cdot \text{H}}{\text{s}}}{1\Omega +j\left(2.357\times {10}^3\right)\left(20\times {10}^{-3}\right)\frac{\text{H}}{\text{s}}}\left\{\because 1\text{k}={10}^3,1\text{m}={10}^{-3}\right\}\end{array}$

$\begin{array}{c}\left(R_1\parallel j\omega L\right)=\frac{j47.14\frac{\Omega \cdot \Omega \cdot \text{s}}{\text{s}}}{1\Omega +j47.14\frac{\Omega \cdot \text{s}}{\text{s}}}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =\frac{j47.14{\Omega }^2}{1\Omega +j47.14\Omega }\end{array}$

To find $\left(R_2+\frac{1}{j\omega C}\right)$:

Substitute $0.1\Omega$ for $R_2$, $9\mu \text{F}$ for $C$ and $2.357\frac{\text{krad}}{\text{s}}$ for $\omega$ to find $\left(R_2+\frac{1}{j\omega C}\right)$.

$\begin{array}{c}\left(R_2+\frac{1}{j\omega C}\right)=\left(0.1\Omega \right)+\left(\frac{1}{j\left(2.357\frac{\text{krad}}{\text{s}}\right)\left(9\mu \text{F}\right)}\right)\\ =0.1\Omega +\left(\frac{1}{j\left(2.357\times {10}^3\right)\left(9\times {10}^{-6}\right)\frac{\text{F}}{\text{s}}}\right)\left\{\because 1\text{k}={10}^3,1\mu ={10}^{-6}\right\}\\ =0.1\Omega +\left(\frac{1}{j0.0212\frac{\left(\frac{\text{s}}{\Omega }\right)}{\text{s}}}\right)\left\{\because 1\text{F}=\frac{1\text{s}}{1\Omega }\right\}\\ =0.1\Omega -j47.14\Omega \end{array}$

Substitute $\frac{j47.14{\Omega }^2}{1\Omega +j47.14\Omega }$ for $\left(R_1\parallel j\omega L\right)$ and $0.1\Omega -j47.14\Omega$ for $\left(R_2+\frac{1}{j\omega C}\right)$ in equation (6) to find $Z_{\text{in}}\left({\omega }_0\right)$.

$\begin{array}{c}{Z}_{\text{in}}\left({\omega }_0\right)=\left(\frac{j47.14{\Omega }^2}{1\Omega +j47.14\Omega }\right)\parallel \left(0.1\Omega -j47.14\Omega \right)\\ =\left(0.9996\Omega +j0.0212\Omega \right)\parallel \left(0.1\Omega -j47.14\Omega \right)\\ =\frac{\left(0.9996\Omega +j0.0212\Omega \right)\times \left(0.1\Omega -j47.14\Omega \right)}{\left(0.9996\Omega +j0.0212\Omega \right)+\left(0.1\Omega -j47.14\Omega \right)}\\ =1\Omega \end{array}$

Conclusion:

Thus, the value of the input impedance $Z_{\text{in}}\left({\omega }_0\right)$ is $1\Omega$.