A capacitor of capacitance 500,mu F is charge | vedclass

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(A) The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.Substituting the given values,$U = \frac{1}{2} 	imes (500 	imes 10^{-6} \,

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(A) The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.Substituting the given values,$U = \frac{1}{2} 	imes (500 	imes 10^{-6} \, F) 	imes (100 \, V)^2 = \frac{1}{2} 	imes 500 	imes 10^{-6} 	imes 10^4 = 2.5 \, J$.In an $LC$ circuit,the total energy is conserved. The maximum current $I_{max}$ occurs when the entire energy of the capacitor is transferred to the inductor.Thus,$\frac{1}{2} LI_{max}^2 = U_{total}$.$\frac{1}{2} 	imes (50 	imes 10^{-3} \, H) 	imes I_{max}^2 = 2.5 \, J$.$I_{max}^2 = \frac{2.5 	imes 2}{50 	imes 10^{-3}} = \frac{5}{0.05} = 100$.$I_{max} = \sqrt{100} = 10 \, A$.

$A$ capacitor of capacitance $500\,\mu F$ is charged completely using a dc supply of $100\,V$. It is now connected to an inductor of inductance $50\,mH$ to form an $LC$ circuit. The maximum current in $LC$ circuit will be $.........A$.

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