JEE Advanced 2024 · Paper 1 · Physics Q17

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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 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 $R_0$ is found to be $I_1$ . After a long time, the current attains a steady state value $I_2$ . Thereafter, K₂ is closed and simultaneously K₁ 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.

$Circuit with two vertical side wires (left and right) and three horizontal branches between them. Top branch: key K₂ (open switch) in series with capacitor $C_0 = 10\,\mu\mathrm{F}$. Middle branch: inductor $L = 25\,\mathrm{mH}$, directly across the two side wires, so the inductor is in parallel with the K₂-C₀ branch. Below the middle branch, the left side wire contains resistor $R_0 = 5\,\Omega$ and the right side wire contains key K₁ (open switch). Bottom branch: ideal battery of 20 V joining the lower ends of the two side wires. With K₁ closed, the battery, R₀, the inductor and K₁ form one loop; with K₂ closed and K₁ open, the inductor and capacitor form an LC loop and the battery branch is disconnected.$

List-I

P
The value of $I_1$ in Ampere is
Q
The value of $I_2$ in Ampere is
R
The value of $\omega_0$ in kilo-radians/s is
S
The value of $V_0$ in Volt is

List-II

1
$0$
2
$2$
3
$4$
4
$20$
5
$200$

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 and worked solution

Official answer: A

With $K_1$ closed, $R_0$ and $L$ are in series with the battery.

(P) Just after closing, the inductor current is still 0, so $I_1 = 0$ . → (1)

(Q) In steady state the inductor is a short: $I_2 = \frac{20}{5} = 4\,\mathrm{A}$ . → (3)

Then $K_1$ opens and $K_2$ closes: $L$ and $C_0$ form an LC loop starting with current $4\,\mathrm{A}$ .

(R) $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{25\times10^{-3}\times10\times10^{-6}}} = 2000\,\mathrm{rad/s} = 2\ \text{krad/s}$ . → (2)

(S) $\frac12 LI_2^2 = \frac12 C V_0^2 \implies V_0 = I_2\sqrt{L/C} = 4\times50 = 200\,\mathrm{V}$ . → (5)

Option (A).

Topics:LC oscillations LR circuits

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