# 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

Verified

279.3k+ views

**Hint:**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)..

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Define absolute refractive index of a medium

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Give 10 examples for herbs , shrubs , climbers , creepers

Which planet is known as the red planet aMercury bMars class 6 social science CBSE

Which state has the longest coastline in India A Tamil class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE