JEE Advanced Previous Year Questions (2018 - 2025): Alternating Current

2024

Q1: The circuit shown in the figure contains an inductor L, a capacitor C₀, a resistor R₀, and an ideal battery. The circuit also contains two keys K₁ and K₂. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key K₁ is closed and immediately after this the current in R₀ is found to be I₁. After a long time, the current attains a steady state value I₂. Thereafter, K₂ is closed and simultaneously K₁ is opened and the voltage across C₀ oscillates with amplitude V₀ and angular frequency ω₀.
(a) P → 1; Q → 3; R → 2; S → 5
(b) P → 1; Q → 2; R → 3; S → 5
(c) P → 1; Q → 3; R → 2; S → 4
(d) P → 2; Q → 5; R → 3; S → 4     [JEE Advanced 2024 Paper 1]
Ans: (a)
(P) When K₁ $�1$ is closed current in R₀ $�0$ is I₁
At t = 0 $t=0$; circuit will be
I₁ = 0
P → (1)
(Q) After long time inductor behave as a wire so  I₂
I₂ = 20 / 5 = 4 A
Q → (3)
(R) When K₂ is closed and K₁ open

(S) Now $�2$K₂ is closed and $�1$K₁ open

12 L I₂² = 12 C V₀²
25 × 10^-3 × (4)² = 10 × 10^-6 × V₀²
V₀² = 2500 × 16
V₀ = 50 × 4 = 200 V
S → (5)

2023

Q1: A series LCR circuit is connected to a 45sin⁡(ωt) Volt source. The resonant angular frequency of the circuit is   10⁵ rad s^-1 and current amplitude at resonance is I₀. When the angular frequency of the source is   ω = 8 × 10⁴ rad s^-1, the current amplitude in the circuit is 0.05I₀. If  L = 50 mH, match each entry in List-I with an appropriate value from List-II and choose the correct option.

At, ω = 8 × 10⁴ rad s^-1 

From (i) and (ii), we get

= 400 mA