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 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.

	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 \(\omega _ 0\) in kilo-radians/s is	(3)	4
(S)	The value of \(V_0\) in Volt is	(4)	20
		(5)	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

Answer 1

Correct option is (A) P → 1; Q → 3; R → 2; S → 5 

(P) When \(K_1\) is closed current in \(R_0\) is \(I_1\)

At t = 0; circuit will be

\(I_1 = 0\)

\(P \rightarrow (1)\)

(Q) After long time inductor behave as a wire so \(I_2\)

\(I_2 = \frac {20}{5} = 4A\)

\(Q \rightarrow (3)\)

(R) When \(K_2\) is closed and \(K_1\) open

\(\omega _0 =\frac {1}{\sqrt {LC}}\)

\(\omega _0 =\frac {1}{\sqrt {25 \times 10 ^{-3}\times 10 \times 10^{-6}}} = \frac {1}{5 \times 10 ^{-4}}\)

\(\omega _ 0 = 2 \times 10 ^3 \,rad/s\)

\(\omega _0 =\) 2 kilo-radian/s

\(R \rightarrow (2)\)

(S) Now \(K_2\) is closed and \(K_1\) open

\(\frac {1}{2} L I ^2_2 = \frac {1}{2} CV ^2 _ 0\)

\(25 \times 10 ^{-3} \times (4)^2 = 10 \times 10 ^{-6} \times V ^2_0\)

\(V^2_0 = 2500 \times 16\)

\(V_0 = 50 \times 4 = 200 V\)

\(S \rightarrow (5)\)