A capacitor of 1,mu F initially charged to 10 | vedclass

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(B) In an $LC$ circuit,the total energy is conserved.Initially,the energy is stored in the capacitor as electrostatic potential energy: $U_E = \frac{1

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$1$

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(B) In an $LC$ circuit,the total energy is conserved.Initially,the energy is stored in the capacitor as electrostatic potential energy: $U_E = \frac{1}{2} CV_0^2$.When the current reaches its maximum value $I_0$,all energy is stored in the inductor as magnetic potential energy: $U_B = \frac{1}{2} LI_0^2$.By the principle of conservation of energy: $\frac{1}{2} CV_0^2 = \frac{1}{2} LI_0^2$.Solving for $I_0$: $I_0 = V_0 \sqrt{\frac{C}{L}}$.Given $C = 1 	imes 10^{-6} \, F$,$V_0 = 10 \, V$,and $L = 0.1 	imes 10^{-3} \, H = 10^{-4} \, H$.Substituting the values: $I_0 = 10 	imes \sqrt{\frac{10^{-6}}{10^{-4}}} = 10 	imes \sqrt{10^{-2}} = 10 	imes 0.1 = 1 \, A$.

$A$ capacitor of $1\,\mu F$ initially charged to $10\,V$ is connected across an ideal inductor of $0.1\,mH$. The maximum current in the circuit is.....$A$

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