A series LCR circuit is connected to a 45 sin (ω t) Volt source. The resonant angular frequency of the circuit is 10 5 rad s -1 and current amplitude at resonance is I 0 . When the angular frequency of the source is ω=8 × 10 4 rad s -1 , the current amplitude in the circuit is 0.05 I 0 . If L =50 mH , match each entry in List-I with an appropriate value from List-II and choose the correct option LIST-I LIST-II --- --- P) I 0 in mA & (1) 1) 44.4 Q) The quality factor of the circuit 2) 18 R) The bandwidth of the circuit in rad s s -1 3) 400 S) The peak power dissipated at resonance in Watt 4) 2250 5) 500

P rightarrow 2, Q rightarrow 3, R rightarrow 5, S rightarrow 1

P rightarrow 4, Q rightarrow 5, R rightarrow 3, S rightarrow 1

P rightarrow 4, Q rightarrow 2, R rightarrow 1, S rightarrow 5

Physics · Alternating Current

JEE Advanced 2023 — Paper 1 — Question 32

A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^{5} \mathrm{rad} \mathrm{s}^{-1}$ and current amplitude at resonance is $\mathrm{I}_{0}$ . When the angular frequency of the source is $\omega=8 \times$ $10^{4} \mathrm{rad} \mathrm{s}^{-1}$ , the current amplitude in the circuit is $0.05 \mathrm{I}_{0}$ . If $\mathrm{L}=50 \mathrm{mH}$ , match each entry in List-I with an appropriate value from List-II and choose the correct option

LIST-I	LIST-II
P) $I_{0}$ in mA & $(1)$	1) 44.4
Q) The quality factor of the circuit	2) 18
R) The bandwidth of the circuit in rad s s $^{-1}$	3) 400
S) The peak power dissipated at resonance in Watt	4) 2250
	5) 500

Option A:
$\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 1$
Option B:
$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 2$
Correct
Option C:
$\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 1$
Option D:
$\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 2, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 5$

Answer: B

Step-by-step solution

$\mathrm{v}=45 \sin (\omega \mathrm{t})$ $\omega_{\mathrm{r}}=10^{5} \mathrm{rad} / \mathrm{s}$ $\omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{LC}}}$ $10^{5}=\frac{1}{\sqrt{50 \times 10^{-3} \times \mathrm{C}}}$ $\mathrm{C}=2 \times 10^{-9} \mathrm{~F}$ $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=4000 \Omega$ $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=6250 \Omega$ $\mathrm{X}=\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}$ $0.25 \mathrm{I}_{0}=\frac{45}{\mathrm{Z}}$ $\mathrm{R}=0.05 \mathrm{Z}$ $\mathrm{R}=0.05 \times \sqrt{2250^{2}+\mathrm{R}^{2}}$ $\mathrm{R}=112.6 \Omega$ $\mathrm{I}_{0}=\frac{45}{\mathrm{R}}=400 \mathrm{~mA}$ $\mathrm{Q}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}=44.4$ Bandwidth $=\frac{\mathrm{R}}{\mathrm{L}}=2250 \mathrm{rad} / \mathrm{s}$ $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=18 \mathrm{~W}$ (p) $\rightarrow 3$ , (q) $\rightarrow 1$ , (r) $\rightarrow 4$ , (s) $\rightarrow 2$

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor