JEE Advanced 2024 — Paper 1 — PHY-IV solution

JEE Advanced2024Paper 1PHY-IV

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 $ω_{0}$ .

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 $ω_{0}$ in kilo-radians/s is	(3) $4$
(S) The value of $V_{0}$ in Volt is	(4) $20$
	(5) $200$

A
P $\to$ 1; Q $\to$ 3; R $\to$ 2; S $\to$ 5
B
P $\to$ 1; Q $\to$ 2; R $\to$ 3; S $\to$ 5
C
P $\to$ 1; Q $\to$ 3; R $\to$ 2; S $\to$ 4
D
P $\to$ 2; Q $\to$ 5; R $\to$ 3; S $\to$ 4

Solution

Just after closing $K_{1}$ : The inductor opposes any sudden change in current, so initially $I_{1} = 0$ A. Hence P $\to$ 1.

Steady state with $K_{1}$ closed: The inductor behaves as a wire; the capacitor branch (with $K_{2}$ open) is open. $I_{2} = \frac{20}{R _{0}} = \frac{20}{5} = 4$ A. Hence Q $\to$ 3.

After opening $K_{1}$ and closing $K_{2}$ : The L-C loop oscillates at $ω_{0} = \frac{1}{L C _{0}} = \frac{1}{( 25 \times 1 0 ^{- 3} ) ( 10 \times 1 0 ^{- 6} )} = \frac{1}{2.5 \times 1 0 ^{- 7}}$ .

$2.5 \times 1 0^{- 7} = 5 \times 1 0^{- 4}$ , so $ω_{0} = 2000$ rad/s = $2$ krad/s. Hence R $\to$ 2.

Amplitude $V_{0}$ : Energy conservation: at the instant of switching, the inductor carries $I_{0} = 4$ A and the capacitor is uncharged. Peak charge $Q_{0} = I_{0} / ω_{0} = 4/2000 = 2 \times 1 0^{- 3}$ C. Peak voltage $V_{0} = Q_{0} / C_{0} = 2 \times 1 0^{- 3} /1 0^{- 5} = 200$ V. Hence S $\to$ 5.

$∴$ Option (A) is correct.

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