JEE Advanced
Physics
Alternating Current
2023
JEE Advanced 2023 (Paper 1)
JEE Advanced Physics Question (2023) — Solution
Question
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 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
Options
- A. P → 2 ,   Q → 3 ,   R → 5 ,   S → 1
- B. P → 3 ,   Q → 1 ,   R → 4 ,   S → 2
- C. P → 4 ,   Q → 5 ,   R → 3 ,   S → 1
- D. P → 4 ,   Q → 2 ,   R → 1 ,   S → 5
Answer
B. P → 3 ,   Q → 1 ,   R → 4 ,   S → 2
Step-by-step solution
Resonant angular frequency is given by, 1 L C = 10 5 1 50 × 10 - 3 C = 10 5 ⇒ C = 2 × 10 - 9   F Given: V = 45 sin ω t . Therefore, V 0 = 45 . Now, I 0 = V 0 R = 45 R                               . . . ii Inductive reactance, X L = ω L = 8 × 10 4 × 50 × 10 - 3 = 4000   Ω . and capacitive reactance, X C = 1 ω C = 1 8 × 10 4 × 2 × 10 - 9 = 6250   Ω . For new current amplitude, we can write 0 . 05 I 0 = 45 R 2 + X L - X C 2 ⇒ 0 . 05 I 0 = 45 R 2 + 6250 - 4000 2 ⇒ 0 . 05 × 45 R = 45 R 2 + 6250 - 4000 2 ⇒ R 2 + 6250 - 4000 2 = R 2 0 . 05 2 ⇒ R 2 + 2250 2 = 400 R 2 ⇒ R = 2250 399 = 112 . 67   Ω Where X L 0 = X C 0 are at resonant frequencies On solving, ⇒ I 0 = 45 R ≃ 400   mA Quality factor Q = X L R ≃ 44 . 44 Now, Q = ω 0 ∆ ω ⇒ ∆ ω ≃ 2250   rad   s - 1 . Peak power = 45 × 400 1000   W = 18   W Therefore, P → 3 ,   Q → 1 ,   R → 4 ,   S → 2 .
Practice more on Quantrex App →
Related: Physics — Alternating Current · All PYQ Banks