Answer :

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.


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