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JEE Advanced 2023 · Paper 1

JEE Advanced 2023 Physics question on Electromagnetic Induction and AC

A series LCR circuit is connected to a 45sin⁡(ωt)45\sin(\omega t) Volt source. The resonant angular frequency of the circuit is 10510^5 rad s−1^{-1} and current amplitude at resonance is I0I_0. When the angular frequency of the source is ω=8×104\omega = 8 \times 10^4 rad s−1^{-1}, the current amplitude in the circuit is 0.05 I00.05\, I_0. If L=50L = 50 mH, match each entry in List-I with an appropriate value from List-II and choose the correct option.
List-I
(P) I0I_0 in mA
(Q) The quality factor of the circuit
(R) The bandwidth of the circuit in rad s−1^{-1}
(S) The peak power dissipated at resonance in Watt
List-II
(1) 44.4
(2) 18
(3) 400
(4) 2250
(5) 500
  1. A.
    P→2,Q→3,R→5,S→1P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1
  2. B.
    P→3,Q→1,R→4,S→2P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2
  3. C.
    P→4,Q→5,R→3,S→1P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1
  4. D.
    P→4,Q→2,R→1,S→5P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5
Show answer

Correct option: B

[JEE Advanced 2023 · Paper 1]

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