Chapter 21, Problem 1P

Find the resonant ${\text{ω}}_s$ and f_s for the series circuit with the following parameters:

a. $R\text{ = 10Ω, }L\text{ = 1H, }C\text{ = 16}\mu \text{F}$

b. $R\text{ = 300Ω, }L\text{ = 0}\text{.51H, }C\text{ = 0}\text{.16}\mu \text{F}$

c. $R\text{ = 20Ω, }L\text{ = 0}\text{.27mH, }C\text{ = 7}\text{.5}\mu \text{F}$

To determine

(a)

The resonant frequency ${\omega }_s$ and $f_s$ for a given series circuit.

Answer to Problem 1P

The value of resonant is ${\omega }_s=250\text{rad/sec}$ and $f_s=39.79\text{Hz}$.

Explanation of Solution

Given:

The series circuit has following parameters

  $\begin{array}{l}R=10\Omega \\ L=1\text{H}\\ C=16\text{μF}\end{array}$

Formula used:${\omega }_s$ is calculated by

  ${\omega }_s=\frac{1}{\sqrt{LC}}$

Resonant frequency $f_s$ is calculated by

  $f_s=\frac{1}{2\pi \sqrt{LC}}$

Calculation:

With the known values, ${\omega }_s$ is calculated as

  $\begin{array}{c}{\omega }_s=\frac{1}{\sqrt{LC}}\\ =\frac{1}{\sqrt{1\times (16\times {10}^{-6})}}\\ =\frac{1}{4\times {10}^{-3}}\\ =0.25\times {10}^3\\ {\omega }_s=250\text{rad/sec}\end{array}$

Resonant frequency $f_s$ is calculated as

  $\begin{array}{c}{f}_s=\frac{1}{2\pi \sqrt{LC}}\\ =\frac{1}{2\pi \sqrt{1(16\times {10}^{-6})}}\\ =.03979\times {10}^3\\ f_s=39.79\text{Hz}\end{array}$

Conclusion:

Therefore, the value of resonant is ${\omega }_s=250\text{rad/sec}$ and $f_s=39.79\text{Hz}$

To determine

(b)

The resonant frequency ${\omega }_s$ and $f_s$ for a given series circuit.

Answer to Problem 1P

The value of resonant is ${\omega }_s=3500\text{rad/sec}$ and $f_s=557Hz$

Explanation of Solution

Given:

The series circuit has following parameters

  $\begin{array}{l}R=300\Omega \\ L=0.51\text{H}\\ C=0.16\text{μF}\end{array}$

Formula used:${\omega }_s$ is calculated by

  ${\omega }_s=\frac{1}{\sqrt{LC}}$

Resonant frequency $f_s$ is calculated by

  $f_s=\frac{1}{2\pi \sqrt{LC}}$

Calculation:

With the known values, ${\omega }_s$ is calculated as

  $\begin{array}{c}{\omega }_s=\frac{1}{\sqrt{LC}}\\ =\frac{1}{\sqrt{(0.51)(0.16\times {10}^{-6})}}\\ =3.5\times {10}^3\\ {\omega }_s=3500\text{rad/sec}\end{array}$

Resonant frequency $f_s$ is calculated as

  $\begin{array}{c}{f}_s=\frac{1}{2\pi \sqrt{LC}}\\ =\frac{1}{2\pi \sqrt{(0.51)(0.16\times {10}^{-6})}}\\ =.557\times {10}^3\\ f_s=557\text{Hz}\end{array}$

Conclusion:

Therefore, the value of resonant is ${\omega }_s=3500\text{rad/sec}$ and $f_s=557\text{Hz}$

To determine

(c)

The resonant frequency ${\omega }_s$ and $f_s$ for a given series circuit.

Answer to Problem 1P

The value of resonant is ${\omega }_s=22.22\times {10}^3\text{rad/sec}$ and $f_s=3.536\text{kHz}$

Explanation of Solution

Given:

The given circuit has parameters

  $\begin{array}{l}R=20\Omega \\ L=0.27mH\\ C=7.5\mu F\end{array}$

Formula used:${\omega }_s$ is calculated by

  ${\omega }_s=\frac{1}{\sqrt{LC}}$

Resonant frequency $f_s$ is calculated by

  $f_s=\frac{1}{2\pi \sqrt{LC}}$

Calculation:

With the known values, ${\omega }_s$ is calculated as

  $\begin{array}{l}{\omega }_s=\frac{1}{\sqrt{LC}}\\ =\frac{1}{\sqrt{(0.27\times {10}^{-3})(7.5\times {10}^{-6})}}\\ =22.22\times {10}^3\\ {\omega }_s=22.22\times {10}^3\text{rad/sec}\end{array}$

Resonant frequency $f_s$ is calculated as

  $\begin{array}{l}{f}_s=\frac{1}{2\pi \sqrt{LC}}\\ =\frac{1}{2\pi \sqrt{(0.27\times {10}^{-3})(7.5\times {10}^{-6})}}\\ =3.536\times {10}^3\\ f_s=3.536kHz\end{array}$

Conclusion:

Therefore, the value of resonant is ${\omega }_s=22.22\times {10}^3\text{rad/sec}$ and $f_s=3.536\text{kHz}$.