A series RLC circuit is activated by an ac source of voltage V_S … - Vidyadip

MCQ

A series $RLC$ circuit is activated by an $ac$ source of voltage $V_S$ volt and variable angular frequency $(\omega )$ as shown in the circuit. $V_{RL}$ and $V_C$ are the potential drops across $RL$ and $C$ respectively. Select the correct statement

A
At low frequency limit, both $V_{RL}$ and $V_C$ are proportional to $\omega $
✓
At high frequency limit, $V_{RL}$ approaches $V_S$ but $V_C$ is proportional to $\frac{1}{{{\omega ^2}}}$
C
At high frequency limit, both $V_{RL}$ and $V_C$ are proportional to $\frac{1}{{{\omega ^2}}}$
D
At low frequency limit, $V_{RL}$ is proportional to $\frac{1}{\omega }$ , whereas $V_C$ approaches $V_S$

✓

Answer

Correct option: B.

At high frequency limit, $V_{RL}$ approaches $V_S$ but $V_C$ is proportional to $\frac{1}{{{\omega ^2}}}$

b
$\mathrm{V}_{\mathrm{RL}}=\mathrm{i}_{0} \times \sqrt{\omega^{2} \mathrm{L}^{2}+\mathrm{R}^{2}}$

$ - \frac{{{{\rm{V}}_0}}}{{\sqrt {{{\left( {\omega {\rm{L}} - \frac{1}{{\omega {\rm{C}}}}} \right)}^2} + {{\rm{R}}^2}} }} \times \sqrt {{\omega ^2}{{\rm{L}}^2} + {{\rm{R}}^2}} $

$\omega = 0,\quad {{\rm{V}}_{{\rm{RL}}}} \propto {{\rm{V}}_0} \times \omega {\rm{C}}$

$\mathrm{V}_{\mathrm{c}}=\mathrm{i}_{0} \times \frac{1}{\omega \mathrm{C}}$

$\omega = \infty \quad {{\rm{V}}_{{\rm{RL}}}} = {{\rm{V}}_0}$

$ = \frac{{{V_0}}}{{\sqrt {{{\left( {{\omega ^2}{\rm{LC}} - 1} \right)}^2} + {\omega ^2}{{\rm{R}}^2}{{\rm{C}}^2}} }}$

$\omega=0$

$\mathrm{V}_{\mathrm{c}}=\frac{\mathrm{V}_{0}}{\mathrm{\omega}} \sqrt{\mathrm{LC}}$

$\omega=\infty$

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