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In an LCR series circuit, if the angular frequency is gradually in creased then match the following columns
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The correct answer is:
$(\mathrm{A})-(\mathrm{iv}),(\mathrm{B})-(\mathrm{i}),(\mathrm{C})-(\mathrm{ii}),(\mathrm{D})-(\mathrm{iii})$
$\begin{aligned} & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}, \text { so } \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\omega} \\ & \therefore(\mathrm{A})-(\mathrm{iv}) \\ & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}, \text { so } \mathrm{X}_{\mathrm{L}} \propto \omega \\ & \therefore(\mathrm{B})-(\mathrm{i})\end{aligned}$
And $\mathrm{R}$ is not a function of $\omega$
$\therefore(\mathrm{C})-(\mathrm{ii})$
Impedence $\mathrm{Z}$ is the minimum at resonance frequency.
$\therefore(\mathrm{D})-(\mathrm{iii})$
And $\mathrm{R}$ is not a function of $\omega$
$\therefore(\mathrm{C})-(\mathrm{ii})$
Impedence $\mathrm{Z}$ is the minimum at resonance frequency.
$\therefore(\mathrm{D})-(\mathrm{iii})$
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