Suppose the initi

Given: Capacitance C = 30 uF = 30 × 10 ^-6F

Inductance L = 27 mH = 20 × 10 ^-6H

The charge on capacitor, Q = 6 mC = 6 × 10^-3C

The diagram is given :

The total energy stored in the circuit can be calculated as followed:

E = 1/2 Q²C

E = 1/2 × (6 × 10^-3C)² × 30 × 10^-6F

On calculating, we get

⇒ E = 0.6 J

At later time, total energy will be same i.e. E = 0.6 J as energy is shared between inductor and capacitor.