Chapter 14, Problem 37P

Rework Prob. 14.25 if the elements are connected in parallel.

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 37P

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 $2\text{k}\Omega$, $\left(1.4212+j53.3\right)\Omega$, $\left(8.85+j132.74\right)\Omega$, $\left(8.85-j132.74\right)\Omega$ and $\left(1.4212-j53.3\right)\Omega$ respectively.

Explanation of Solution

Given data:

In a parallel RLC network,

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 admittance at resonance of parallel RLC circuit.

$Y\left({\omega }_0\right)=\frac{1}{R}$        (2)

Here,

$R$ is the value of the resistor.

Write the expression to calculate the admittance of the parallel RLC circuit.

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

Write the expression that shows the general relationship between admittance and impedance.

$Z\left({\omega }_0\right)=\frac{1}{Y\left({\omega }_0\right)}$        (4)

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 $Y\left({\omega }_0\right)$.

$Y\left({\omega }_0\right)=\frac{1}{2\text{k}\Omega }$

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

$Z\left({\omega }_0\right)=\frac{1}{Y\left({\omega }_0\right)}$

Substitute $\frac{1}{2\text{k}\Omega }$ for $Y\left({\omega }_0\right)$ in above equation to find $Z\left({\omega }_0\right)$.

$\begin{array}{c}Z\left({\omega }_0\right)=\frac{1}{\left(\frac{1}{2\text{k}\Omega }\right)}\\ =2\text{k}\Omega \end{array}$

(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 $Y\left(\frac{{\omega }_0}{4}\right)$.

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

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 $Y\left(\frac{{\omega }_0}{4}\right)$.

$\begin{array}{c}Y\left(\frac{{\omega }_0}{4}\right)=\frac{1}{\left(2\text{k}\Omega \right)}+\frac{4}{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)}+\frac{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)}{4}\\ =\left(\begin{array}{l}\frac{1}{\left(2\times {10}^3\Omega \right)}+\left(\frac{4\left(-j\right)}{\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)}\right)\\ +\left(\frac{j\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)}{4}\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-\left(\frac{4j}{\left(200\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)\right)}\right)+\left(\frac{j\left(5\times {10}^{-3}\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }\right)}{4}\right)\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-j0.02{\Omega }^{-1}+j\left(1.25\times {10}^{-3}\right){\Omega }^{-1}\right)\end{array}$

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

$\begin{array}{c}Y\left(\frac{{\omega }_0}{4}\right)=\left(\left(0.5\times {10}^{-3}\right){\Omega }^{-1}-j\left(18.75\times {10}^{-3}\right){\Omega }^{-1}\right)\\ =\left(0.5-j18.75\right)\text{m}{\Omega }^{-1}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

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

$Z\left(\frac{{\omega }_0}{4}\right)=\frac{1}{Y\left(\frac{{\omega }_0}{4}\right)}$

Substitute $\left(0.5-j18.75\right)\text{m}{\Omega }^{-1}$ for $Y\left(\frac{{\omega }_0}{4}\right)$ in above equation to find $Z\left(\frac{{\omega }_0}{4}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{4}\right)=\frac{1}{\left(0.5-j18.75\right)\text{m}{\Omega }^{-1}}\\ =\frac{1}{\left(0.5-j18.75\right)\times {\text{10}}^{-3}}\Omega \left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(1.4212+j53.3\right)\Omega \end{array}$

(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 $Y\left(\frac{{\omega }_0}{2}\right)$.

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

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 $Y\left(\frac{{\omega }_0}{2}\right)$.

$\begin{array}{c}Y\left(\frac{{\omega }_0}{2}\right)=\frac{1}{\left(2\text{k}\Omega \right)}+\frac{2}{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)}+\frac{j\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)}{2}\\ =\left(\begin{array}{l}\frac{1}{\left(2\times {10}^3\Omega \right)}+\left(\frac{2\left(-j\right)}{\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)}\right)\\ +\left(\frac{j\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)}{2}\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-\left(\frac{2j}{\left(200\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)\right)}\right)+\left(\frac{j\left(5\times {10}^{-3}\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }\right)}{2}\right)\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-j0.01{\Omega }^{-1}+j\left(2.5\times {10}^{-3}\right){\Omega }^{-1}\right)\end{array}$

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

$\begin{array}{c}Y\left(\frac{{\omega }_0}{2}\right)=\left(\left(0.5\times {10}^{-3}\right){\Omega }^{-1}-j\left(7.5\times {10}^{-3}\right){\Omega }^{-1}\right)\\ =\left(0.5-j7.5\right)\text{m}{\Omega }^{-1}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Use equation (4) to find $Z\left(\frac{{\omega }_0}{2}\right)$.

$Z\left(\frac{{\omega }_0}{2}\right)=\frac{1}{Y\left(\frac{{\omega }_0}{2}\right)}$

Substitute $\left(0.5-j7.5\right)\text{m}{\Omega }^{-1}$ for $Y\left(\frac{{\omega }_0}{2}\right)$ in above equation to find $Z\left(\frac{{\omega }_0}{2}\right)$.

$\begin{array}{c}Z\left(\frac{{\omega }_0}{2}\right)=\frac{1}{\left(0.5-j7.5\right)\text{m}{\Omega }^{-1}}\\ =\frac{1}{\left(0.5-j7.5\right)\times {\text{10}}^{-3}}\Omega \left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(8.85+j132.74\right)\Omega \end{array}$

(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 $Y\left(2{\omega }_0\right)$.

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

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 $Y\left(2{\omega }_0\right)$.

$\begin{array}{c}Y\left(2{\omega }_0\right)=\frac{1}{\left(2\text{k}\Omega \right)}+\frac{1}{j\left(2\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)}+j\left(2\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)\\ =\left(\begin{array}{l}\frac{1}{\left(2\times {10}^3\Omega \right)}+\left(\frac{\left(-j\right)}{\left(2\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)}\right)\\ +\left(j\left(2\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-\left(\frac{j}{\left(400\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)\right)}\right)+\left(j\left(10\times {10}^{-3}\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }\right)\right)\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-j\left(2.5\times {10}^{-3}\right){\Omega }^{-1}+j\left(10\times {10}^{-3}\right){\Omega }^{-1}\right)\end{array}$

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

$\begin{array}{c}Y\left(2{\omega }_0\right)=\left(\left(0.5\times {10}^{-3}\right){\Omega }^{-1}+j\left(7.5\times {10}^{-3}\right){\Omega }^{-1}\right)\\ =\left(0.5+j7.5\right)\text{m}{\Omega }^{-1}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Use equation (4) to find $Z\left(2{\omega }_0\right)$.

$Z\left(2{\omega }_0\right)=\frac{1}{Y\left(2{\omega }_0\right)}$

Substitute $\left(0.5+j7.5\right)\text{m}{\Omega }^{-1}$ for $Y\left(2{\omega }_0\right)$ in above equation to find $Z\left(2{\omega }_0\right)$.

$\begin{array}{c}Z\left(2{\omega }_0\right)=\frac{1}{\left(0.5+j7.5\right)\text{m}{\Omega }^{-1}}\\ =\frac{1}{\left(0.5+j7.5\right)\times {\text{10}}^{-3}}\Omega \left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(8.85-j132.74\right)\Omega \end{array}$

(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 $Y\left(4{\omega }_0\right)$.

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

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 $Y\left(4{\omega }_0\right)$.

$\begin{array}{c}Y\left(4{\omega }_0\right)=\frac{1}{\left(2\text{k}\Omega \right)}+\frac{1}{j\left(4\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(40\text{mH}\right)}+j\left(4\right)\left(5\frac{\text{krad}}{\text{s}}\right)\left(1\mu \text{F}\right)\\ =\left(\begin{array}{l}\frac{1}{\left(2\times {10}^3\Omega \right)}+\left(\frac{\left(-j\right)}{\left(4\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(40\times {10}^{-3}\text{H}\right)}\right)\\ +\left(j\left(4\right)\left(5\times {10}^3\frac{\text{1}}{\text{s}}\right)\left(1\times {10}^{-6}\text{F}\right)\right)\end{array}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-\left(\frac{j}{\left(800\frac{\text{1}}{\text{s}}\cdot \left(\Omega \cdot \text{s}\right)\right)}\right)+\left(j\left(20\times {10}^{-3}\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }\right)\right)\right)\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\left(\left(5\times {10}^{-4}\right){\Omega }^{-1}-j\left(1.25\times {10}^{-3}\right){\Omega }^{-1}+j\left(20\times {10}^{-3}\right){\Omega }^{-1}\right)\end{array}$

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

$\begin{array}{c}Y\left(4{\omega }_0\right)=\left(\left(0.5\times {10}^{-3}\right){\Omega }^{-1}+j\left(18.75\times {10}^{-3}\right){\Omega }^{-1}\right)\\ =\left(0.5+j18.75\right)\text{m}{\Omega }^{-1}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

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

$Z\left(4{\omega }_0\right)=\frac{1}{Y\left(4{\omega }_0\right)}$

Substitute $\left(0.5+j18.75\right)\text{m}{\Omega }^{-1}$ for $Y\left(4{\omega }_0\right)$ in above equation to find $Z\left(4{\omega }_0\right)$.

$\begin{array}{c}Z\left(4{\omega }_0\right)=\frac{1}{\left(0.5+j18.75\right)\text{m}{\Omega }^{-1}}\\ =\frac{1}{\left(0.5+j18.75\right)\times {\text{10}}^{-3}}\Omega \left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(1.4212-j53.3\right)\Omega \end{array}$

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 $2\text{k}\Omega$, $\left(1.4212+j53.3\right)\Omega$, $\left(8.85+j132.74\right)\Omega$, $\left(8.85-j132.74\right)\Omega$ and $\left(1.4212-j53.3\right)\Omega$ respectively.