JEE Advanced 2023 — Paper 1 — PHY-IV solution

JEE Advanced2023Paper 1PHY-IV

A series LCR circuit is connected to a $45 sin (ω t)$ Volt source. The resonant angular frequency of the circuit is $1 0^{5} rad s^{- 1}$ and current amplitude at resonance is $I_{0}$ . When the angular frequency of the source is $ω = 8 \times 1 0^{4} rad s^{- 1}$ , 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) 44.4
(Q) The quality factor of the circuit	(2) 18
(R) The bandwidth of the circuit in rad s $^{- 1}$	(3) 400
(S) The peak power dissipated at resonance in Watt	(4) 2250
	(5) 500

A
P $\to$ 2, Q $\to$ 3, R $\to$ 5, S $\to$ 1
B
P $\to$ 3, Q $\to$ 1, R $\to$ 4, S $\to$ 2
C
P $\to$ 4, Q $\to$ 5, R $\to$ 3, S $\to$ 1
D
P $\to$ 4, Q $\to$ 2, R $\to$ 1, S $\to$ 5

Solution

Capacitance. Resonance condition $ω_{0} = 1/ L C$ gives $C = \frac{1}{L ω _{0}^{2}} = \frac{1}{( 5 \times 1 0 ^{- 2} ) ( 1 0 ^{10} )} = 2 \times 1 0^{- 9}$ F.

Off-resonance impedance. At $ω = 0.8 ω_{0} = 8 \times 1 0^{4} rad s^{- 1}$ : $X_{L} = ω L = 4000 Ω$ , $X_{C} = 1/ (ω C) = 6250 Ω$ . The current amplitude $0.05 I_{0}$ implies $Z = 20 R$ , so $400 R^{2} = R^{2} + (X_{C} - X_{L})^{2} = R^{2} + (2250)^{2}$ . Hence $399 R^{2} = 225 0^{2}$ , giving $R \approx 112.6 Ω$ ( $\approx 900/8$ ).

(P) Peak current at resonance. $I_{0} = V_{0} / R = 45/ (900/8) = 0.4$ A $= 400$ mA. Match $\to$ (3).

(Q) Quality factor. $Q = \frac{1}{R} \frac{L}{C} = \frac{8}{900} \frac{5 \times 1 0 ^{- 2}}{2 \times 1 0 ^{- 9}} = \frac{8}{900} \cdot 5000 \approx 44.4$ . Match $\to$ (1).

(R) Bandwidth. $Δ ω = ω_{0} / Q = 1 0^{5} /44.4 \approx 2250$ rad/s. Match $\to$ (4).

(S) Peak power at resonance. $P_{p e ak} = I_{0}^{2} R = (0.4)^{2} \cdot (900/8) = 0.16 \cdot 112.5 = 18$ W. Match $\to$ (2).

Mapping: P $\to$ 3, Q $\to$ 1, R $\to$ 4, S $\to$ 2. Answer: (B).

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