JEE Advanced 2024 · Paper 1

JEE Advanced 2024 Physics question on Electromagnetic Induction and AC

The circuit shown in the figure contains an inductor LL, a capacitor C0C_0, a resistor R0R_0 and an ideal battery. The circuit also contains two keys K1K_1 and K2K_2. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K1K_1 is closed and immediately after this the current in R0R_0 is found to be I1I_1. After a long time, the current attains a steady state value I2I_2. Thereafter, K2K_2 is closed and simultaneously K1K_1 is opened and the voltage across C0C_0 oscillates with amplitude V0V_0 and angular frequency ω0\omega_0. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
  1. A.
    P→1;Q→3;R→2;S→5P \to 1; Q \to 3; R \to 2; S \to 5
  2. B.
    P→1;Q→2;R→3;S→5P \to 1; Q \to 2; R \to 3; S \to 5
  3. C.
    P→1;Q→3;R→2;S→4P \to 1; Q \to 3; R \to 2; S \to 4
  4. D.
    P→2;Q→5;R→3;S→4P \to 2; Q \to 5; R \to 3; S \to 4
Show answer

Correct option: A

[JEE Advanced 2024 · Paper 1]

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