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 0. [figure] Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
- P → 1; Q → 3; R → 2; S → 5
- P → 1; Q → 2; R → 3; S → 5
- P → 1; Q → 3; R → 2; S → 4
- P → 2; Q → 5; R → 3; S → 4
Answer
(1) P → 1; Q → 3; R → 2; S → 5
Solution
With K1 closed, K2 open: series loop battery–R0–L–K1.
Just after closing, inductor current cannot jump ⇒ I1=0 → P → 1.
Steady state: inductor acts as a plain wire ⇒ I2= 205=4 A → Q → 3.
Then L and C0 form an isolated LC loop with initial current 4 A: 0= 1 LC0= 1 25 10-3 10 10-6=2000 rad s-1=2 krad/s → R → 2.
Energy: 12LI22= 12C0V02 ⇒ V0=I2 LC0=4 2500=200 V → S → 5.
Answer: (A)