A series LCR circuit is connected to a 45 (ω t) Volt source. The resonant angular frequency of the circuit is 10^5 \,{rad}\,{s}^{-1} and current amplitude at resonance is I₀. When the angular frequency of the source is ω = 8 × 10^4 rad s^{-1}, the current amplitude in the circuit is 0.05 I₀. 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) I₀ in {mA} (1) 44.4 (Q) The quality factor of the circuit (2) 18 (R) The bandwidth of the circuit in rad s (3) 400 (S) The peak power dissipated at resonance in Watt (4) 2250 (5) 500 [JEE Adv 2023 P1]

Question

Examsnet · Accepted Answer

Correct Answer is : P→3,Q→1,R→4,S→2P arrow 3, Q arrow 1, R arrow 4, S arrow 2P→3,Q→1,R→4,S→2

List-I	List-II
(P) $I_0$ in $\text{mA}$	(1) 44.4
(Q) The quality factor of the circuit	(2) 18
(R) The bandwidth of the circuit in rad s	(3) 400
(S) The peak power dissipated at resonance in Watt	(4) 2250
	(5) 500

JEE Advanced 2023 Paper 1 Solutions