Question

What is Bandwidth? Write its value in the LCR circuit.

Answer 1

Hint: In order to solve this question we need to understand Capacitance, inductor and resistance. Capacitor is an electrical device that is used to store energy in form of charges, an inductor is an electrical device which is used to store energy in the form of a magnetic field and resistance is as an electrical device which is used to derive current in circuit. So when the inductor, capacitor and resistance are connected in series with AC Voltage then it is known as LCR circuit. LCR circuit initially shows inductive behavior and later it shows capacitive behavior.

Complete step by step answer:
Bandwidth in an LCR circuit is defined as the difference between two half frequencies. Half frequencies are those frequencies for which the power dissipation is half the maximum value. To find the bandwidth of an LCR circuit, we need to find the resonant frequency and quality factor.Consider an inductor “L”, Capacitor “C” and Resistance “R” connected in series with each other.Let “Z” be the impedance in circuit, then we know Z can be written as;
$Z=[R^2+({X_L}^2-{X_C}^2){]}^{{\displaystyle \frac{1}{2}}}$
Here, $\omega$ is the angular frequency in the circuit.

Resonance is the condition when the impedance of circuit is minimum and for this, Inductive reactance $X_L$ and capacitive reactance $X_C$ must become equal with each other.We know,
$X_L=\omega L$ and $X_C={\displaystyle \frac{1}{\omega C}}$
Equating both we get, $X_L=X_C$
$\omega L={\displaystyle \frac{1}{\omega C}}$
$\Rightarrow {\omega }^2={\displaystyle \frac{1}{LC}}$
$\Rightarrow \omega ={\displaystyle \frac{1}{\sqrt{LC}}}$
So the resonant frequency is, $f={\displaystyle \frac{\omega }{2\pi }}$
$f={\displaystyle \frac{1}{2\pi \sqrt{LC}}}$
And the Quality factor is a quantity which shows the quality of circuit,
$Q={\displaystyle \frac{X_L}{R}}$
$\Rightarrow Q={\displaystyle \frac{\omega L}{R}}$

Bandwidth is mathematically defined as ratio of resonant frequency and quality factor, denoted by
$\mathrm{\Delta }f={\displaystyle \frac{f}{Q}}$
Putting values we get,
$$\mathrm{\Delta }f={\displaystyle \frac{{\displaystyle \frac{1}{2\pi \sqrt{LC}}}}{{\displaystyle \frac{\omega L}{R}}}}$$
Putting the value of $\omega$ and solving we get,
$$\mathrm{\Delta }f={\displaystyle \frac{R}{2\pi \sqrt{LC}({\displaystyle \frac{1}{\sqrt{LC}}}L)}}$$
$$\therefore \mathrm{\Delta }f={\displaystyle \frac{R}{2\pi L}}$$

Note: It should be remembered that circuit behave as inductor first because initially uncharged inductor behave as closed circuit and initially uncharged capacitor behave as open circuit, so initially circuit behave as inductor and later o after resonance circuit behave as capacitor because the voltage across capacitor is building up.