In a series LCR circuit, the frequencies at which the current amplitude is $\frac{1}{\sqrt{2}}$ times the current amplitude at resonance are f₁ and f₂ (>f₁) . The frequency bandwidth of resonance which is defined as $∆\mathrm{f}={\mathrm{f}}_2-{\mathrm{f}}_1$_, expressed in terms of R and L (at the resonance frequency ${\mathrm{f}}_0\gg >\mathrm{\Delta f}$) $\mathrm{is} \frac{\mathrm{nR}}{2\mathrm{\pi L}}$ then $\mathrm{n}$ =?

In resonance At a different frequency ($\mathrm{\omega }$=2f$\mathrm{\pi }$), the current amplitude is
$\frac{{\mathrm{V}}_0}{\mathrm{Z}}=\frac{{\mathrm{V}}_0}{\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2}}$
Given $\frac{{\mathrm{V}}_0}{\left[\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2}\right]}=\frac{1}{\sqrt{2}}\frac{{\mathrm{V}}_0}{\mathrm{R}}$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2=2{\mathrm{R}}^2\\ \Rightarrow \mathrm{\omega L}-\frac{1}{\mathrm{\omega c}}=\pm \mathrm{R}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2{\mathrm{\pi f}}_1\mathrm{L}-\frac{1}{2{\mathrm{\pi f}}_1\mathrm{C}}=-\mathrm{R} ..........\left(\mathrm{i}\right)\end{array}$
and $2{\mathrm{\pi f}}_2\mathrm{L}-\frac{1}{2{\mathrm{\pi f}}_2\mathrm{C}}=\mathrm{R} ..........\left(\mathrm{ii}\right)$
Dividing (i) by f₂ and (ii) by f₁ we get
$$\begin{gathered} 2\mathrm{\pi L}\frac{{\mathrm{f}}_1}{{\mathrm{f}}_2}-\frac{1}{2{\mathrm{\pi Cf}}_1{\mathrm{f}}_2}=-\mathrm{R} .....\left(\mathrm{iii}\right) \\ 2\mathrm{\pi L}\frac{{\mathrm{f}}_2}{{\mathrm{f}}_1}-\frac{1}{2{\mathrm{\pi Cf}}_1{\mathrm{f}}_2}=\mathrm{R} .....\left(\mathrm{iv}\right) \end{gathered}$$
(iv)-(iii) gives
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{\pi L}\left(\frac{{\mathrm{f}}_2^2-{\mathrm{f}}_1^2}{{\mathrm{f}}_1{\mathrm{f}}_2}\right)=2\mathrm{R}\\ \Rightarrow \frac{\left({\mathrm{f}}_2-{\mathrm{f}}_1\right)\left({\mathrm{f}}_2+{\mathrm{f}}_1\right)}{{\mathrm{f}}_1{\mathrm{f}}_2}=\frac{\mathrm{R}}{\mathrm{\pi L}}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{f}}_1+{\mathrm{f}}_2=2{\mathrm{f}}_0\text{ and }{\mathrm{f}}_1{\mathrm{f}}_2={\mathrm{f}}_0^2\\ \therefore \mathrm{\Delta f}=\frac{\mathrm{R}}{2\mathrm{\pi L}}\end{array}$