Question 17

Added on: Sep 19, 2026

MATRIX MATCHHARD

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 $\omega_0$ .

Given values from the circuit diagram: $R_0 = 5\ \Omega$ , $L = 25\ \text{mH}$ , $C_0 = 10\ \mu\text{F}$ , and Battery voltage = $20\ \text{V}$ .

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

List - I

The value of $I_2$ in Ampere is

The value of $I_1$ in Ampere is

The value of $\omega_0$ in kilo-radians/s is

The value of $V_0$ in Volt is

List-II

200

(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

Detailed Solution

1. Analysis of the first process (Closing $K_1$ ): When $K_1$ is closed, the circuit consists of the battery, $R_0$ , and $L$ in series.

Immediately after closing $K_1$ ( $t = 0$ ): The inductor opposes change in current. Since the initial current was zero, $I_1 = 0$ A. Matches: P $\to$ 1.
After a long time (steady state): The inductor acts as a short circuit ( $V_L = 0$ ). The current is determined only by the resistor: $I_2 = \frac{V}{R_0} = \frac{20}{5} = 4\ \text{A}$ Matches: Q $\to$ 3.

2. Analysis of the second process (Opening $K_1$ and closing $K_2$ ): When $K_1$ is opened and $K_2$ is closed, the battery and $R_0$ are disconnected. The inductor $L$ and capacitor $C_0$ form an LC oscillation circuit. The initial current in the inductor is $I_2 = 4$ A and the initial charge on $C_0$ is zero.

Angular frequency $\omega_0$ : $\omega_0 = \frac{1}{\sqrt{LC_0}} = \frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{25 \times 10^{-8}}} = \frac{1}{5 \times 10^{-4}} = 2000\ \text{rad/s}$ In kilo-radians/s, $\omega_0 = 2$ krad/s. Matches: R $\to$ 2.
Voltage amplitude $V_0$ : By energy conservation, the maximum energy in the inductor equals the maximum energy in the capacitor: $\frac{1}{2} L I_2^2 = \frac{1}{2} C_0 V_0^2$ $V_0 = I_2 \sqrt{\frac{L}{C_0}} = 4 \times \sqrt{\frac{25 \times 10^{-3}}{10 \times 10^{-6}}} = 4 \times \sqrt{2500} = 4 \times 50 = 200\ \text{V}$ Matches: S $\to$ 5.

Correct matching: P-1, Q-3, R-2, S-5. This corresponds to option (A).

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