The circuit shown in the figure contains an inductor L, a capacitor , a resistor and an ideal battery. The circuit also contains two keys and . Initially, both the keys are open and there is no charge on the capacitor. [2024]

Question

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 an 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. [2024]

	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

Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [2024]

	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

Smart Achievers · Accepted Answer

(1)

$(P) When key K_{1} is closed, current in R_{0} is I_{1}$

At $t = 0$ , the circuit is as follows.

$I_{1} = 0$

$(Q) After a long time, the inductor behaves as a wire, so$

$I_{2} = \frac{V}{R} = \frac{20}{5} = 4 A$

$(R) When K_{2} is closed and K_{1} is open,$

$ω_{0} = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{25 \times 10^{- 3} \times 10 \times 10^{- 6}}} = \frac{1}{5 \times 10^{- 4}}$

$∴$ $V_{0}$ $∴$ $ω_{0} = 2 kilo-radian/s$

$(S) When K_{2} is closed and K_{1} is open,$

$\frac{1}{2} L I_{2}^{2} = \frac{1}{2} C V_{0}^{2}$

$\frac{1}{2} (25 \times 10^{- 3}) {(4)}^{2} = \frac{1}{2} (10 \times 10^{- 6}) V_{0}^{2}$

$V_{0}^{2} = 2500 \times 16$ or, $V_{0} = 50 \times 4 = 200 V$

Solution

1 P → 1; Q → 3; R → 2; S → 5

2 P → 1; Q → 2; R → 3; S → 5

3 P → 1; Q → 3; R → 2; S → 4

4 P → 2; Q → 5; R → 3; S → 4

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