Question 313: AC fundamentals

Question

A series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit containing a resistance ' $R$ ' has angular frequency ' $\omega$ '. At resonance the voltage across resistance and inductor are ' $V_R$ ' and ' $\mathrm{V}_{\mathrm{L}}$ ' respectively, then value of capacitance will be

Accepted Answer

$$\begin{aligned}
& 	ext { At resonance, } \mathrm{C}=\frac{1}{\omega^2 \mathrm{~L}} \
& \mathrm{~L}=\frac{\mathrm{V}_{\mathrm{L}}}{\mathrm{I} \omega} 	ext { and } \mathrm{I}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \
	herefore \quad \mathrm{~L} & =\frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{~V}_{\mathrm{R}} \omega} \
	herefore \quad & \mathrm{C}=\frac{\mathrm{V}_{\mathrm{R}} \omega}{\mathrm{~V}_{\mathrm{L}} \mathrm{R} \omega^2}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{~V}_{\mathrm{L}} \mathrm{R} \omega}
\end{aligned}$$

Answer

$\frac{V_R}{V_L \omega R}$

Answer

$\frac{V_L}{V_R R \omega^2}$

Answer

$\frac{V_R}{V_L R \omega^2}$

Answer

$\frac{V_L R}{V_R \omega}$

AC fundamentals

A series \mathrm{L}-\mathrm{C}-\mathrm{R} circuit containing a resistance ' R ' has angular frequency ' \omega '. At resonance the voltage across resistance and inductor are ' V_R ' and ' \mathrm{V}_{\mathrm{L}} ' respectively, then value of capacitance will be

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