A 1 mu F capacitor is charged to 50 V and is | vedclass

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(A) The energy stored in a charged capacitor is given by $U = \frac{1}{2} CV^2$. When the capacitor discharges through an inductor,the energy is trans

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$0.5$

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(A) The energy stored in a charged capacitor is given by $U = \frac{1}{2} CV^2$. When the capacitor discharges through an inductor,the energy is transferred from the electric field of the capacitor to the magnetic field of the inductor. At the moment of maximum current $(I_0)$,all the energy is stored in the inductor as magnetic energy,given by $U = \frac{1}{2} LI_0^2$. By the law of conservation of energy: $\frac{1}{2} CV^2 = \frac{1}{2} LI_0^2$. Solving for $I_0$: $I_0^2 = \frac{CV^2}{L}$. Given: $C = 1 \mu F = 10^{-6} \ F$,$V = 50 \ V$,$L = 10 \ mH = 10 	imes 10^{-3} \ H = 10^{-2} \ H$. Substituting the values: $I_0^2 = \frac{10^{-6} 	imes (50)^2}{10^{-2}} = \frac{10^{-6} 	imes 2500}{10^{-2}} = 2500 	imes 10^{-4} = 0.25$. Therefore,$I_0 = \sqrt{0.25} = 0.5 \ A$.

$A$ $1 \mu F$ capacitor is charged to $50 \ V$ and is then discharged through a $10 \ mH$ inductor of negligible resistance. The maximum current in the inductor is (in $A$)

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