Correct ANSWER is (b) 2 A

The explanation is: Given: V2 = 10 V; V1 = 20 V; C = 5 μF; Inductance (I) = 0.5 mH

Initial charge on the capacitor (q1) = C × V1 = 5 × 10^-6 × 20 ……………….1

q1 = 10^-4 C …………..B

The instantaneous charge on the capacitor as the capacitor discharges through the inductor ➔ q2

q2 = q1cos (ωt)➔ \(\frac {q_2}{q_2}\) = cos (ωt) ………………..A

Also, q2 = C × V2 = 5 × 10^-6 × 10 …………………….2

q2 = 0.5 × 10^-4 C

From 1 and 2 ➔ \(\frac {q_2}{q_2} = \frac {V_2}{V_1}\) ➔ \(\frac {q_2}{q_2}\) = 0.5 = \(\frac {1}{2}\)

From equation A, we can equate as follows ➔ cos (ωt) = \(\frac {1}{2}\)

ωt = \(\frac {\pi }{2}\) rad ………………..3

For an LC circuit ➔ ω=\(\frac {1}{\sqrt {LC}}\)

ω=20000\(\frac {rad}{s}\) …………………….4

The current through the circuit is given as:

Current (I)=-\(\frac {dq}{dt}\)

Charge decreases with RESPECT to time, so, \(\frac {dq}{dt}\) obtained will be negative and this is why we add a negative sign to MAKE a current positive.

Current = q1 ω sinωt

Considering 3, 4 and B

Current = 10^-4 × 20000 × sin (\(\frac {\pi }{2}\))

Current = 2 A[Sin(\(\frac {\pi }{2}\)) = 1]

Therefore, the current flowing through the circuit is 2 A.