In an LCR series circuit, if the angular frequency is gradually increased then match the following columns $…

In an LCR series circuit, if the angular frequency is gradually increased then match the following columns $ \begin{array}{|l|l|l|l|} \hline & \text{Column-I} & & \text{Column-II} \\ \hline \text { (A) } & \text { Capacitive reactance } & \text { (i) } & \text { Will continuously increase } \\ \hline \text { (B) } & \text { Inductive reactance } & \text { (ii) } & \text { Will remain constant } \\ \hline \text { (C) } & \text { Resistance } & \text { (iii) } & \begin{array}{l} \text { Will first decrease and then } \\ \text { increase } \end{array} \\ \hline \text { (D) } & \text { Total impedance } & \text { (iv) } & \text { Will continuously decrease } \\ \hline \end{array} $
  1. $(\mathrm{A})-(\mathrm{iv}),(\mathrm{B})-(\mathrm{i}),(\mathrm{C})-(\mathrm{ii}),(\mathrm{D})-(\mathrm{iii})$
  2. (A) - (i), (B) - (iii), (C) - (iv), (D) - (ii)
  3. $(\mathrm{A})-(\mathrm{ii}),(\mathrm{B})-(\mathrm{iii}),(\mathrm{C})-(\mathrm{i}),(\mathrm{D})-(\mathrm{iv})$
  4. $(A)-(i),(B)-(i v),(C)-(i i),(D)-(i i i)$

Solution

$\begin{aligned} & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}, \text { so } \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\omega} \\ & \therefore(\mathrm{A})-(\mathrm{iv}) \\ & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}, \text { so } \mathrm{X}_{\mathrm{L}} \propto \omega \\ & \therefore(\mathrm{B})-(\mathrm{i})\end{aligned}$ And $\mathrm{R}$ is not a function of $\omega$ $\therefore(\mathrm{C})-(\mathrm{ii})$ Impedence $\mathrm{Z}$ is the minimum at resonance frequency. $\therefore(\mathrm{D})-(\mathrm{iii})$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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