A $16 \mu \mathrm{~F}$ capacitor is charged to a 20 told potential the battery is then disconnected

Question

A $16 \mu \mathrm{~F}$ capacitor is charged to a 20 told potential the battery is then disconnected and pure 40 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is\

Professor PC Thomas Classes · Answer

Electrical energy = Magnetic energy\ $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{\mathrm{LI}^{2}}{2} 	herefore \frac{\mathrm{C}^{2} \mathrm{~V}^{2}}{\mathrm{C}}=\mathrm{LI}^{2}$\ $\mathrm{I}^{2}=\frac{\mathrm{CV}^{2}}{\mathrm{~L}}=\frac{16 	imes 10^{-6} 	imes 20 	imes 20}{40 	imes 10^{-3}}$\ $\mathrm{I}^{2}=160 	imes 10^{-3}=0.160$ or $\mathrm{I}=0.4 \mathrm{~A}$\