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JEE Advanced Physics Alternating Current 2024 JEE Advanced 2024 (Paper 1)

JEE Advanced Physics Question (2024) — Solution

Question

The circuit shown in the figure contains an inductor L, a capacitor C_0, a resistor R_0 and an ideal battery. The circuit also contains two keys K _1 and K _2. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K_1 is closed and immediately after this the current in R_0 is found to be I_1. After a long time, the current attains a steady state value I_2. Thereafter, K _2 is closed and simultaneously K _1 is opened and the voltage across C_0 oscillates with amplitude V_0 and angular frequency _0. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

Options

  1. A. P 1 ; Q 3 ; R 2 ; S 5
  2. B. P 1 ; Q 2 ; R 3 ; S 5
  3. C. P 1 ; Q 3 ; R 2 ; S 4
  4. D. P 2 ; Q 5 ; R 3 ; S 4

Answer

A. P 1 ; Q 3 ; R 2 ; S 5

Step-by-step solution

(P) When K_1 is closed current in R_0 is I_1 At t =0; circuit will be aligned & I _1=0 \\ & P (1) aligned (Q) After long time inductor behave as a wire so I _2 aligned & I _2= 20 5 =4 ~A \\ & Q (3) aligned (R) When K _2 is closed and K _1 open aligned & _0= 1 LC \\ & _0= 1 25 10^ -3 10 10^ -6 = 1 5 10^ -4 \\ & _0=2 10^3 rad / s \\ & _0=2 kilo-radian / s \\ & R (2) aligned (S) Now K _2 is closed and K _1 open aligned & 1 2 LI _2^2= 1 2 CV _0^2 \\ & 25 10^ -3 (4)^2=10 10^ -6 V _0^2 \\ & ~V _0^2=2500 16 \\ & ~V _0=50 4=200 ~V \\ & ~S (5) aligned

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