A 4 mu F capacitor is charged to 10 V. The b | vedclass

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(A) The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.Given $C = 4 \mu F = 4 	imes 10^{-6} \ F$ and $V = 10 \ V$,the energy is $

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$0.2$

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(A) The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.Given $C = 4 \mu F = 4 	imes 10^{-6} \ F$ and $V = 10 \ V$,the energy is $U = \frac{1}{2} 	imes 4 	imes 10^{-6} 	imes (10)^2 = 2 	imes 10^{-4} \ J$.When the capacitor is connected to the inductor,the total energy is conserved and oscillates between the electric field of the capacitor and the magnetic field of the inductor.The maximum energy in the inductor is given by $U_{max} = \frac{1}{2} L I_{max}^2$.Equating the energies: $\frac{1}{2} L I_{max}^2 = \frac{1}{2} C V^2$.$I_{max}^2 = \frac{C V^2}{L} = \frac{4 	imes 10^{-6} 	imes 100}{10 	imes 10^{-3}} = \frac{4 	imes 10^{-4}}{10^{-2}} = 4 	imes 10^{-2} = 0.04$.Therefore,$I_{max} = \sqrt{0.04} = 0.2 \ A$.

$A$ $4 \mu F$ capacitor is charged to $10 \ V$. The battery is then disconnected and a pure $10 \ mH$ coil is connected across the capacitor so that $LC$ oscillations are set up. The maximum current in the coil is (in $A$)

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