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
\(\frac {1}{2} L I ^2_2 = \frac {1}{2} CV ^2 _ 0\)
\(25 \times 10 ^{-3} \times (4)^2 = 10 \times 10 ^{-6} \times V ^2_0\)
\(V^2_0 = 2500 \times 16\)
\(V_0 = 50 \times 4 = 200 V\)
\(S \rightarrow (5)\)