Question

Obtain the resonant frequency and Q-factor of a series LCR circuit with $L{\text{ }} = {\text{ }}3.0{\text{ }}H$, $C{\text{ }} = {\text{ }}27{\text{ }}\mu F$, and $R{\text{ }} = {\text{ }}7.4{\text{ }}\Omega$. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of $2$. Suggest a suitable way.

Answer 1

In order to answer the question we will first find out the radians per Second by the formula ${\omega _r} = \dfrac{1}{{\sqrt {LC} }}$ , then we will find out the Q-Factor of the series by using the formula $Q = \dfrac{{{\omega _r}L}}{R}$ and finally we will find out the improved sharpness of the resonance of the circuit according to the given condition.

Complete answer:
The Q factor, which describes how quickly energy decays in an oscillating system, is used to define the sharpness of resonance. For a rise or decrease in damping, the sharpness of resonance increases or decreases, and as the amplitude increases, the sharpness of resonance decreases.
Inductance, $L{\text{ }} = {\text{ }}3.0{\text{ }}H$
Capacitance, $C{\text{ }} = {\text{ }}27{\text{ }}\mu F{\text{ }} = {\text{ }}27{\text{ }} \times {\text{ }}{10^{ - 6}}C$
Resistance, $R{\text{ }} = {\text{ }}7.4{\text{ }}\Omega$ At resonance, angular frequency of the source for the given LCR series circuit is given as:
${\omega _r} = \dfrac{1}{{\sqrt {LC} }}$
${\omega _r} = \dfrac{1}{{\sqrt {3 \times 27 \times {{10}^{ - 6}}} }} \\\ {\omega _r} = \dfrac{{{{10}^3}}}{9} \\\$
${\omega _r} = 111.11\,rad\,{s^{ - 1}}$
Q-Factor of the series:
$Q = \dfrac{{{\omega _r}L}}{R}$
$Q = \dfrac{{111.11 \times 3}}{{7.4}} = 45.0446$
We need to reduce $R$to half, i.e., Resistance, to increase the sharpness of the resonance by reducing its ‘full width at half limit' by a factor of $2$ without modifying
$= \dfrac{R}{2} = \dfrac{{7.4}}{2} = 3.7\Omega$
$\therefore$ For improvement in sharpness of resonance by a factor of $2$ ,Q should be doubled . To double Q with changing ${\omega _r}$, R should be reduced to half ,i.e., to $3.7\Omega$

Note: The quality factor is a ratio of resonant frequency to bandwidth, and the higher the circuit $Q,$the smaller the bandwidth, $Q{\text{ }} = {\text{ }}{{\text{f}}_r}{\text{ }}/BW$. During each phase of oscillation, it compares the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance)..