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(B) The impedance of the series $RLC$ circuit is $Z = \sqrt{R^2 + (\omega L - 1/(\omega C))^2}$. The current in the circuit is $I = V_S / Z$.For $V_{R

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The impedance of the series $RLC$ circuit is $Z = \sqrt{R^2 + (\omega L - 1/(\omega C))^2}$. The current in the circuit is $I = V_S / Z$.For $V_{RL}$ (voltage across $RL$): $V_{RL} = I 	imes \sqrt{R^2 + (\omega L)^2} = V_S 	imes \frac{\sqrt{R^2 + \omega^2 L^2}}{\sqrt{R^2 + (\omega L - 1/(\omega C))^2}}$.For $V_C$ (voltage across $C$): $V_C = I 	imes (1/(\omega C)) = V_S 	imes \frac{1/(\omega C)}{\sqrt{R^2 + (\omega L - 1/(\omega C))^2}} = V_S 	imes \frac{1}{\sqrt{(\omega RC)^2 + (\omega^2 LC - 1)^2}}$.$1$. At low frequency limit $(\omega 	o 0)$:$V_{RL} \approx V_S 	imes \frac{R}{1/(\omega C)} = V_S \omega RC \propto \omega$.$V_C \approx V_S 	imes \frac{1/(\omega C)}{1/(\omega C)} = V_S$. Thus,$V_C$ approaches $V_S$.$2$. At high frequency limit $(\omega 	o \infty)$:$V_{RL} \approx V_S 	imes \frac{\omega L}{\omega L} = V_S$. Thus,$V_{RL}$ approaches $V_S$.$V_C \approx V_S 	imes \frac{1/(\omega C)}{\omega L} = \frac{V_S}{\omega^2 LC} \propto 1/\omega^2$.Comparing these with the options,option $B$ is correct.

$A$ series $RLC$ circuit is connected to an $ac$ source of voltage $V_S$ and variable angular frequency $\omega$,as shown in the figure. $V_{RL}$ and $V_C$ are the potential drops across the $RL$ combination and the capacitor $C$,respectively. Select the correct statement.

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