One application of $L C R$ series ac circuit is to high pass or low pass filters, which filter out either the low or high frequency components of a signal. A high pass filter is shown in figure, where the output voltage is taken across the $L R$ combination, where $L R$ combination represents an inductance ‘L’ and resistance ‘R’
Find the ratio $V_{o u t} / V_{s}$ as a function of the angular frequency $ω$ of the source

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$\sqrt{\frac{R^{2} + {(\frac{1}{ω C})}^{2}}{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

$\frac{R}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

$\sqrt{\frac{R^{2} + {(ω L)}^{2}}{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

answer is B.

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Detailed Solution

The voltage across the resistor and the inductor are $90^{0}$ out of phase, hence
$V_{o u t} = \sqrt{V_{R}^{2} + V_{L}^{2}}$
$X_{L} = ω L, X_{C} = \frac{1}{ω C}$
Hence $X_{L} = ω L, X_{C} = \frac{1}{ω C}$
$I = \frac{V_{s}}{Z}$ and $V_{o u t} = I \sqrt{R^{2} + X_{L}^{2}}$
$= V_{s} \sqrt{\frac{R^{2} + ω^{2} L^{2}}{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$
$\frac{V_{o u t}}{V_{s}} = \sqrt{\frac{R^{2} + ω^{2} L^{2}}{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$