A 2,mu F capacitor is charged to 100,V. After | vedclass

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(C) The energy stored in the capacitor is given by $U_E = \frac{1}{2}CV^2$. When the capacitor discharges through the inductor,this energy is transfer

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(C) The energy stored in the capacitor is given by $U_E = \frac{1}{2}CV^2$. When the capacitor discharges through the inductor,this energy is transferred to the magnetic field of the inductor,given by $U_B = \frac{1}{2}LI_{\max}^2$. By the law of conservation of energy,$U_E = U_B$,so $\frac{1}{2}CV^2 = \frac{1}{2}LI_{\max}^2$. Substituting the given values: $C = 2 	imes 10^{-6}\,F$,$V = 100\,V$,and $L = 20 	imes 10^{-3}\,H$. $(2 	imes 10^{-6}) 	imes (100)^2 = (20 	imes 10^{-3}) 	imes I_{\max}^2$. $(2 	imes 10^{-6}) 	imes 10^4 = 0.02 	imes I_{\max}^2$. $0.02 = 0.02 	imes I_{\max}^2$. $I_{\max}^2 = 1$,which gives $I_{\max} = 1\,A$.

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