JEE Advanced2024PhysicsAlternating CurrentActual
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 . Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
Options
- AP arrow 1 ; Q arrow 3 ; R arrow 2 ; S arrow 5
- BP arrow 1 ; Q arrow 2 ; R arrow 3 ; S arrow 5
- CP arrow 1 ; Q arrow 3 ; R arrow 2 ; S arrow 4
- DP arrow 2 ; Q arrow 5 ; R arrow 3 ; S arrow 4
Correct answer
A. P arrow 1 ; Q arrow 3 ; R arrow 2 ; S arrow 5
Step-by-step solution
(P) When K₁ is closed current in R₀ is I₁ At t=0 ; circuit will be aligned & I₁=0 & P arrow(1) aligned (Q) After long time inductor behave as a wire so I₂ aligned & I₂=(20)/(5)=4 ~A & Q arrow(3) aligned (R) When K₂ is closed and K₁ open aligned & ω₀=(1)/(√(LC)) & ω₀=(1)/(√(25 × 10⁻³ × 10 × 10⁻⁶))=(1)/(5 × 10⁻⁴) & ω₀=2 × 10^3 rad / s & ω₀=2 kilo-radian / s & R arrow(2) aligned (S) Now K₂ is closed and K₁ open aligned & (1)/(2) LI₂^2=(1)/(2) CV₀^2 & 25 × 10⁻³ ×(4)^2=10 × 10⁻⁶ × V₀^2 & ~V₀^2=2500 × 16 & ~V₀=50 × 4=200 ~V & ~S arrow(5) aligned