he 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.

List-I	List-II
(P) The value of 𝐼1 in Ampere is	(1) 0
(Q) The value of 𝐼2 in Ampere is	(2) 2
(R) The value of 𝜔0 in kilo-radians/s is	(3) 4
(S) The value of 𝑉0 in Volt is	(4) 20
	(5) 200

Correct Option: P → 1; Q → 3; R → 2; S → 5

To solve this problem, we analyze the circuit configurations step-by-step using the values given in the circuit diagram:
- Battery Voltage, $V=20\text{V}$
- Resistor, $R_0=5\Omega$
- Inductor, $L=25\text{mH}$
- Capacitor, $C_0=10\mu \text{F}$

Step 1: Finding $I_1$ (Immediately after closing ${\text{K}}_1$)
Initially, both switches ${\text{K}}_1$ and ${\text{K}}_2$ are open, and there is no current in the inductor. At $t=0$, switch ${\text{K}}_1$ is closed while ${\text{K}}_2$ remains open. An inductor opposes any sudden change in current. Since the current through the inductor was $0$ just before closing the switch, it must remain $0$ immediately after closing the switch:
$I_L\left(0^{+}\right)=0$
Because the capacitor branch is open, the current flowing through $R_0$ is the same as the current through the inductor. Thus, the initial current $I_1$ is:
$I_1=0\text{A}$
This corresponds to P → 1.

Step 2: Finding $I_2$ (After a long time in steady state)
After a long time, the circuit reaches a steady state. In a DC circuit, an inductor acts as a short circuit (zero resistance) at steady state. The current flows from the battery, through the resistor $R_0$, and through the shorted inductor $L$ back to the battery. The steady-state current $I_2$ is:
$I_2=\frac{V}{R_0}=\frac{20}{5}=4\text{A}$
This corresponds to Q → 3.

Step 3: Finding the angular frequency ${\omega }_0$
When ${\text{K}}_2$ is closed and ${\text{K}}_1$ is opened simultaneously, the battery and the resistor $R_0$ are disconnected from the loop containing the inductor $L$ and the capacitor $C_0$. This forms a closed, parallel LC oscillating circuit. The angular frequency of oscillation ${\omega }_0$ is:
${\omega }_0=\frac{1}{\sqrt{L{C}_0}}$
Substituting the given values:
${\omega }_0=\frac{1}{\sqrt{25\times {10}^{-3}\times 10\times {10}^{-6}}}=\frac{1}{\sqrt{250\times {10}^{-9}}}=\frac{1}{\sqrt{25\times {10}^{-8}}}=\frac{1}{5\times {10}^{-4}}=2000\text{rad/s}=2\text{kilo-rad/s}$
This corresponds to R → 2.

Step 4: Finding the amplitude of the voltage oscillation $V_0$
At the instant ${\text{K}}_1$ is opened and ${\text{K}}_2$ is closed, the current flowing in the inductor is $I_2=4\text{A}$ and the charge on the capacitor is zero. The total energy stored in the LC circuit is equal to the initial magnetic energy of the inductor:
$E=\frac{1}{2}L{I}_2^2$
During the oscillation, the maximum voltage across the capacitor $V_0$ occurs when all the electromagnetic energy is stored in the electric field of the capacitor:
$\frac{1}{2}{C}_0{V}_0^2=\frac{1}{2}L{I}_2^2$
Solving for $V_0$:
$V_0=I_2\sqrt{\frac{L}{C_0}}$
Substituting the values:
$V_0=4\times \sqrt{\frac{25\times {10}^{-3}}{10\times {10}^{-6}}}=4\times \sqrt{2500}=4\times 50=200\text{V}$
This corresponds to S → 5.