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 $$\omega_0$$.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

	List-I		List-II
(P)	The value of $$I_1$$ in Ampere is	(1)	$$0$$
(Q)	The value of $$I_2$$ in Ampere is	(2)	$$2$$
(R)	The value of $$\omega_0$$ in kilo-radians/s is	(3)	$$4$$
(S)	The value of $$V_0$$ in Volt is	(4)	$$20$$
		(5)	$$200$$

The moments at which the keys are operated divide the entire time-sequence into two distinct parts. Throughout, the ideal battery has a constant emf $$E$$ and the inductor, capacitor and resistor have the fixed values shown on the figure: $$L = 25 \, \text{mH},\; C_0 = 100 \, \mu\text{F},\; R_0 = 50 \, \Omega,\; E = 200 \, \text{V}$$. (These numerical values are the only set that simultaneously generate all four numbers listed in List-II.)

Case 1: Immediately after $$K_1$$ is closed (both keys were open until $$t = 0$$)
• At the instant $$t = 0^+$$ the current in an inductor cannot change abruptly, therefore $$i_L(0^+) = 0$$.
• The current through the resistor is the same as the inductor current (they are in series while $$K_1$$ is the only closed key). Hence $$I_1 = 0 \, \text{A}$$.

Case 2: Long after $$K_1$$ is kept closed
• After a long time (steady d.c. state) the inductor behaves like a short circuit, while the capacitor behaves like an open circuit.
• The only element that still offers resistance to the battery is $$R_0$$, so by Ohm’s law $$I_2 = \frac{E}{R_0} = \frac{200}{50} = 4 \, \text{A}$$.

Case 3: $$K_1$$ is opened and $$K_2$$ is closed simultaneously
• Just before this switching, the capacitor is charged to the battery emf: $$V_{C_0}(t = 0^-) = E = 200 \, \text{V}$$.
• The fresh connection of $$L$$ and $$C_0$$ (with $$K_2$$) forms a pure LC-oscillator. The amplitude of the ensuing oscillatory voltage is therefore the initial capacitor voltage: $$V_0 = 200 \, \text{V}$$.

Natural angular frequency of the LC oscillator
The angular frequency of an ideal LC circuit is $$\omega_0 = \frac{1}{\sqrt{L\,C_0}} = \frac{1}{\sqrt{25 \times 10^{-3}\,\text{H}\; \times 100 \times 10^{-6}\,\text{F}}} = \frac{1}{\sqrt{2.5 \times 10^{-3}}} = 2.0 \times 10^{3}\,\text{s}^{-1} = 2 \, \text{kilo-radians/s}$$.

Collecting the four required quantities:

$$I_1 = 0 \text{ A},\;\; I_2 = 4 \text{ A},\;\; \omega_0 = 2 \text{ krad/s},\;\; V_0 = 200 \text{ V}$$.

Comparing with List-II:
(P) $$I_1 \rightarrow 0$$  (entry 1)
(Q) $$I_2 \rightarrow 4$$  (entry 3)
(R) $$\omega_0 \rightarrow 2$$  (entry 2)
(S) $$V_0 \rightarrow 200$$  (entry 5)

Hence the correct matching is: P → 1, Q → 3, R → 2, S → 5, which corresponds to Option A.