A series $L-C-R$ circuit containing a resistance ' $R$ ' has angular frequency ' $\omega$ '. At resonance…
A series $L-C-R$ circuit containing a resistance ' $R$ ' has angular frequency ' $\omega$ '. At resonance the voltage across resistance and inductor are ' $\mathrm{V}_{\mathrm{R}}$ ' and ' $\mathrm{V}_{\mathrm{L}}$ ' respectively, then value of capacitance will be
$\frac{V_R}{V_L \omega R}$
$\frac{V_L}{V_R R \omega^2}$
$\frac{V_R}{V_L R \omega^2}$
$\frac{V_L R}{V_R \omega}$
Solution
$\begin{aligned} & \text { At resonance, } C=\frac{1}{\omega^2 L} \\ & L=\frac{V_L}{I \omega} \text { and } I=\frac{V_R}{R} \\ \therefore \quad & L=\frac{V_L R}{V_R \omega} \\ \therefore \quad & C=\frac{V_R \omega}{V_L R \omega^2}=\frac{V_R}{V_L R \omega}\end{aligned}$