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(D) Given: Capacitance $C = 20 \mu F = 20 	imes 10^{-6} 	ext{ F}$, Voltage $V = 35 	ext{ V}$, Inductance $L = 200 	ext{ mH} = 200 	imes 10^{-3} \

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$0.35$

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(D) Given: Capacitance $C = 20 \mu F = 20 	imes 10^{-6} 	ext{ F}$, Voltage $V = 35 	ext{ V}$, Inductance $L = 200 	ext{ mH} = 200 	imes 10^{-3} 	ext{ H}$.In an $LC$ circuit, the energy stored in the capacitor is converted into the energy stored in the inductor.The maximum energy in the capacitor is $U_E = \frac{1}{2} C V^2$.The maximum energy in the inductor is $U_B = \frac{1}{2} L I_{	ext{max}}^2$.By conservation of energy, $U_E = U_B$, so $\frac{1}{2} C V^2 = \frac{1}{2} L I_{	ext{max}}^2$.Solving for $I_{	ext{max}}$: $I_{	ext{max}} = V \sqrt{\frac{C}{L}}$.Substituting the values: $I_{	ext{max}} = 35 	imes \sqrt{\frac{20 	imes 10^{-6}}{200 	imes 10^{-3}}} = 35 	imes \sqrt{\frac{1}{10000}} = 35 	imes \frac{1}{100} = 0.35 	ext{ A}$.

$A$ capacitor of capacity $20 \mu F$ is charged to a potential of $35 \text{ V}$. The battery is then disconnected. $A$ pure inductor coil of $200 \text{ mH}$ is connected across the capacitor so that $LC$ oscillations are set up. The maximum current in the coil is: (in $\text{ A}$)

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