The circuit shown in the figure contains an inductor $L$, a capacitor $C_0$, a resistor $R_0$ and an ideal…

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 $\mathrm{K}_1$ and $\mathrm{K}_2$. 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, $\mathrm{K}_2$ is closed and simultaneously $\mathrm{K}_1$ 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.
  1. $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$
  2. $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5$
  3. $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$
  4. $\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4$

Solution

(P) When $K_1$ is closed current in $R_0$ is $I_1$ At $\mathrm{t}=0$; circuit will be $\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1)\end{aligned}$ (Q) After long time inductor behave as a wire so $\mathrm{I}_2$ $\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{~A} \\ & \mathrm{Q} \rightarrow(3)\end{aligned}$ (R) When $\mathrm{K}_2$ is closed and $\mathrm{K}_1$ open $\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \text { kilo-radian } / \mathrm{s} \\ & \mathrm{R} \rightarrow(2)\end{aligned}$ (S) Now $\mathrm{K}_2$ is closed and $\mathrm{K}_1$ open $\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{~V}_0^2=2500 \times 16 \\ & \mathrm{~V}_0=50 \times 4=200 \mathrm{~V} \\ & \mathrm{~S} \rightarrow(5)\end{aligned}$ !

Asked in: JEE Advanced 2024 (Paper 1)

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