Chapter 14, Problem 25P

A series RLC network has R = 2 kΩ, L = 40 mH, and C = 1 μF. Calculate the impedance at resonance and at one-fourth, one-half, twice, and four times the resonant frequency.

To determine

Find the value of the impedance at resonance and at one-fourth, one-half, twice and four times the resonant frequency.

Answer to Problem 25P

The value of the impedance at resonance $Z\left({\omega }_0\right)$, at one-fourth $Z\left(\frac{{\omega }_0}{4}\right)$, one-half $Z\left(\frac{{\omega }_0}{2}\right)$, twice $Z\left(2{\omega }_0\right)$ and four times $Z\left(4{\omega }_0\right)$ of the resonant frequency is $2\text{k}\Omega$, $\left(2-j0.75\right)\text{k}\Omega$, $\left(2-j0.3\right)\text{k}\Omega$, $\left(2+j0.3\right)\text{k}\Omega$ and $\left(2+j0.75\right)\text{k}\Omega$ respectively.

Explanation of Solution

Given data:

The value of the resistor $\left(R\right)$ is $2\text{k}\Omega$.

The value of the inductor $\left(L\right)$ is $40\text{mH}$.

The value of the capacitor $\left(C\right)$ is $1\mu \text{F}$.

Formula used:

Write the expression to calculate the resonant frequency.

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

Here,

$L$ is the value of the inductor, and

$C$ is the value of the capacitor.

Write the expression to calculate the impedance at resonance of series RLC circuit.

$Z\left({\omega }_0\right)=R$        (2)

Here,

$R$ is the value of the resistor.

Write the expression to calculate the impedance of the series RLC circuit.

$Z\left({\omega }_0\right)=R+j{\omega }_{0}L+\frac{1}{j{\omega }_{0}C}$        (3)

Calculation:

Substitute $40\text{mH}$ for $L$ and $1\mu \text{F}$ for $C$ in equation (1) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(40\text{mH}\right)\left(1\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(40\times {10}^{-3}\right)\left(1\times {10}^{-6}\right)\text{H}\cdot \text{F}}}\left\{\because 1\text{m}={10}^{-3},1\mu ={10}^{-6}\right\}\\ =\frac{1}{\sqrt{\left(40\times {10}^{-3}\right)\left(1\times {10}^{-6}\right)\frac{{\text{s}}^2}{\text{F}}\cdot \text{F}}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =\frac{1}{\sqrt{\left(4\times {10}^{-8}\right){\text{s}}^2}}\end{array}$

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

$\begin{array}{c}{\omega }_0=5\times {10}^3\frac{\text{rad}}{\text{s}}\\ =5\frac{\text{krad}}{\text{s}}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

(1) Impedance at resonance:

Substitute $2\text{k}\Omega$ for $R$ in equation (2) to find $Z\left({\omega }_0\right)$.

$Z\left({\omega }_0\right)=2\text{k}\Omega$

(2) Impedance at one-fourth of the resonant frequency:

Here, the resonant frequency $\left({\omega }_0\right)$ is $\frac{{\omega }_0}{4}$.

Substitute $\frac{{\omega }_0}{4}$ for ${\omega }_0$ in equation (3) to find $Z\left(\frac{{\omega }_0}{4}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{4}\right)=R+j\left(\frac{{\omega }_0}{4}\right)L+\frac{1}{j\left(\frac{{\omega }_0}{4}\right)C}\\ =R+\frac{j{\omega }_{0}L}{4}+\frac{4}{j{\omega }_{0}C}\end{array}$

Substitute $5\frac{\text{krad}}{\text{s}}$ for ${\omega }_0$, $2\text{k}\Omega$ for $R$, $40\text{mH}$ for $L$ and $1\mu \text{F}$ for $C$ in above equation to find $Z\left(\frac{{\omega }_0}{4}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{4}\right)=\left(2\text{k}\Omega \right)+\left(\frac{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)}{4}\right)+\left(\frac{4}{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)}\right)\\ =\left(\begin{array}{l}\left(2\times {10}^3\Omega \right)+\left(\frac{j\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)}{4}\right)\\ +\left(\frac{4\left(-j\right)}{\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)}\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(2\times {10}^3\Omega \right)+j50\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)-j\left(\frac{4}{5\times {10}^{-3}\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\right)\\ =\left(2\times {10}^3\Omega \right)+j50\Omega -j800\Omega \end{array}$

Simplify the above equation to find $Z\left(\frac{{\omega }_0}{4}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{4}\right)=\left(2\times {10}^3\Omega \right)-j750\Omega \\ =\left(2\times {10}^3\Omega \right)-\left(j750\times {10}^{-3}\times {10}^3\right)\Omega \\ =2\text{k}\Omega -j0.75\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\\ =\left(2-j0.75\right)\text{k}\Omega \end{array}$

(3) Impedance at one-half of the resonant frequency:

Here, the resonant frequency $\left({\omega }_0\right)$ is $\frac{{\omega }_0}{2}$.

Substitute $\frac{{\omega }_0}{2}$ for ${\omega }_0$ in equation (3) to find $Z\left(\frac{{\omega }_0}{2}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{2}\right)=R+j\left(\frac{{\omega }_0}{2}\right)L+\frac{1}{j\left(\frac{{\omega }_0}{2}\right)C}\\ =R+\frac{j{\omega }_{0}L}{2}+\frac{2}{j{\omega }_{0}C}\end{array}$

Substitute $5\frac{\text{krad}}{\text{s}}$ for ${\omega }_0$, $2\text{k}\Omega$ for $R$, $40\text{mH}$ for $L$ and $1\mu \text{F}$ for $C$ in above equation to find $Z\left(\frac{{\omega }_0}{2}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{2}\right)=\left(2\text{k}\Omega \right)+\left(\frac{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)}{2}\right)+\left(\frac{2}{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)}\right)\\ =\left(\begin{array}{l}\left(2\times {10}^3\Omega \right)+\left(\frac{j\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)}{2}\right)\\ +\left(\frac{2\left(-j\right)}{\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)}\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(2\times {10}^3\Omega \right)+j100\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)-j\left(\frac{2}{5\times {10}^{-3}\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\right)\\ =\left(2\times {10}^3\Omega \right)+j100\Omega -j400\Omega \end{array}$

Simplify the above equation to find $Z\left(\frac{{\omega }_0}{2}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{2}\right)=\left(2\times {10}^3\Omega \right)-j300\Omega \\ =\left(2\times {10}^3\Omega \right)-\left(j300\times {10}^{-3}\times {10}^3\right)\Omega \\ =2\text{k}\Omega -j0.3\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\\ =\left(2-j0.3\right)\text{k}\Omega \end{array}$

(4) Impedance at twice of the resonant frequency:

Here, the resonant frequency $\left({\omega }_0\right)$ is $2{\omega }_0$.

Substitute $2{\omega }_0$ for ${\omega }_0$ in equation (3) to find $Z\left(2{\omega }_0\right)$.

$\begin{array}{c}Z\left(2{\omega }_0\right)=R+j\left(2{\omega }_0\right)L+\frac{1}{j\left(2{\omega }_0\right)C}\\ =R+j2{\omega }_{0}L+\frac{1}{j2{\omega }_{0}C}\end{array}$

Substitute $5\frac{\text{krad}}{\text{s}}$ for ${\omega }_0$, $2\text{k}\Omega$ for $R$, $40\text{mH}$ for $L$ and $1\mu \text{F}$ for $C$ in above equation to find $Z\left(2{\omega }_0\right)$.

$\begin{array}{c}Z\left(2{\omega }_0\right)=\left(2\text{k}\Omega \right)+\left(j\left(2\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)\right)+\left(\frac{1}{j\left(2\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)}\right)\\ =\left(\begin{array}{l}\left(2\times {10}^3\Omega \right)+\left(j\left(2\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)\right)\\ +\left(\frac{\left(-j\right)}{\left(2\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)}\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(2\times {10}^3\Omega \right)+j400\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)-j\left(\frac{1}{0.01\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\right)\\ =\left(2\times {10}^3\Omega \right)+j400\Omega -j100\Omega \end{array}$

Simplify the above equation to find $Z\left(2{\omega }_0\right)$.

$\begin{array}{c}Z\left(2{\omega }_0\right)=\left(2\times {10}^3\Omega \right)+j300\Omega \\ =\left(2\times {10}^3\Omega \right)+\left(j300\times {10}^{-3}\times {10}^3\right)\Omega \\ =2\text{k}\Omega +j0.3\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\\ =\left(2+j0.3\right)\text{k}\Omega \end{array}$

(5) Impedance at four times of the resonant frequency:

Here, the resonant frequency $\left({\omega }_0\right)$ is $4{\omega }_0$.

Substitute $4{\omega }_0$ for ${\omega }_0$ in equation (3) to find $Z\left(4{\omega }_0\right)$.

$\begin{array}{c}Z\left(4{\omega }_0\right)=R+j\left(4{\omega }_0\right)L+\frac{1}{j\left(4{\omega }_0\right)C}\\ =R+j4{\omega }_{0}L+\frac{1}{j4{\omega }_{0}C}\end{array}$

Substitute $5\frac{\text{krad}}{\text{s}}$ for ${\omega }_0$, $2\text{k}\Omega$ for $R$, $40\text{mH}$ for $L$ and $1\mu \text{F}$ for $C$ in above equation to find $Z\left(4{\omega }_0\right)$.

$\begin{array}{c}Z\left(4{\omega }_0\right)=\left(2\text{k}\Omega \right)+\left(j\left(4\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)\right)+\left(\frac{1}{j\left(4\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)}\right)\\ =\left(\begin{array}{l}\left(2\times {10}^3\Omega \right)+\left(j\left(4\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)\right)\\ +\left(\frac{\left(-j\right)}{\left(4\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)}\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(2\times {10}^3\Omega \right)+j800\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)-j\left(\frac{1}{0.02\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\right)\\ =\left(2\times {10}^3\Omega \right)+j800\Omega -j50\Omega \end{array}$

Simplify the above equation to find $Z\left(4{\omega }_0\right)$.

$\begin{array}{c}Z\left(4{\omega }_0\right)=\left(2\times {10}^3\Omega \right)+j750\Omega \\ =\left(2\times {10}^3\Omega \right)+\left(j750\times {10}^{-3}\times {10}^3\right)\Omega \\ =2\text{k}\Omega +j0.75\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\\ =\left(2+j0.75\right)\text{k}\Omega \end{array}$

Conclusion:

Thus, the value of the impedance at resonance $Z\left({\omega }_0\right)$, at one-fourth $Z\left(\frac{{\omega }_0}{4}\right)$, one-half $Z\left(\frac{{\omega }_0}{2}\right)$, twice $Z\left(2{\omega }_0\right)$ and four times $Z\left(4{\omega }_0\right)$ of the resonant frequency is $2\text{k}\Omega$, $\left(2-j0.75\right)\text{k}\Omega$, $\left(2-j0.3\right)\text{k}\Omega$, $\left(2+j0.3\right)\text{k}\Omega$ and $\left(2+j0.75\right)\text{k}\Omega$ respectively.