The resonant frequency of a series LCR circuit is ' $\mathrm{f}_R$ '. The circuit is connected to a…
The resonant frequency of a series LCR circuit is ' $\mathrm{f}_R$ '. The circuit is connected to a sinusoidally alternating e.m.f. of frequency ' $2 \mathrm{f}_{\mathrm{R}}$ '. The inductive reactance becomes. ' $\mathrm{X}_{\mathrm{L}_1}$ ' and capacitive reactance becomes ' $\mathrm{X}_{\mathrm{C}_1}$ ' after changing the frequency. ' $\mathrm{X}_{C_1}$ ' is equal to
$2 \mathrm{X}_{\mathrm{L}_1}$
$\frac{1}{4} \mathrm{X}_{\mathrm{L}_1}$
$\frac{1}{2} \mathrm{X}_{\mathrm{L}_1}$
$\mathrm{x}_{\mathrm{L}_1}$
Solution
$X_C=\frac{1}{\omega^2}$ and $X_L=(\omega L)$
We know, $\omega=2 \pi f \Rightarrow f \propto \omega$ if $f$ is doubled
$\therefore \mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{X}_{\mathrm{C}}}{2}$ and $\mathrm{X}_{\mathrm{L}_1}=2 \mathrm{X}_{\mathrm{L}}$
We know, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$
$\Rightarrow 2 \mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{XL}_1}{2} \Rightarrow\left(\mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{X}_{\mathrm{L}_1}}{4}\right)$