A capacitor of capacitance 2 mu F is charged | vedclass

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(A) Given: Capacitance $C = 2 	imes 10^{-6} \, F$,Inductance $L = 0.6 	imes 10^{-3} \, H$,Initial voltage $V_0 = 12 \, V$,Instantaneous voltage $V =

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$0.6$

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(A) Given: Capacitance $C = 2 	imes 10^{-6} \, F$,Inductance $L = 0.6 	imes 10^{-3} \, H$,Initial voltage $V_0 = 12 \, V$,Instantaneous voltage $V = 6 \, V$.Using the principle of conservation of energy in an $LC$ circuit:Total energy $E = \frac{1}{2} C V_0^2 = \frac{1}{2} C V^2 + \frac{1}{2} L I^2$.Substituting the values:$\frac{1}{2} 	imes (2 	imes 10^{-6}) 	imes (12)^2 = \frac{1}{2} 	imes (2 	imes 10^{-6}) 	imes (6)^2 + \frac{1}{2} 	imes (0.6 	imes 10^{-3}) 	imes I^2$.$(2 	imes 10^{-6}) 	imes 144 = (2 	imes 10^{-6}) 	imes 36 + (0.6 	imes 10^{-3}) 	imes I^2$.$288 	imes 10^{-6} = 72 	imes 10^{-6} + 0.6 	imes 10^{-3} 	imes I^2$.$216 	imes 10^{-6} = 0.6 	imes 10^{-3} 	imes I^2$.$I^2 = \frac{216 	imes 10^{-6}}{0.6 	imes 10^{-3}} = 360 	imes 10^{-3} = 0.36$.$I = \sqrt{0.36} = 0.6 \, A$.

$A$ capacitor of capacitance $2 \mu F$ is charged to a potential difference of $12 \, V$. It is then connected across an inductor of inductance $0.6 \, mH$. The current in the circuit when the potential difference across the capacitor is $6 \, V$ is :

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