A 60,mu F capacitor is charged to 100,V. This | vedclass

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(D) Given: Capacitance $C = 60\,\mu F = 60 	imes 10^{-6}\,F$,Voltage $V = 100\,V$,Inductance $L = 15\,mH = 15 	imes 10^{-3}\,H$.In an $LC$ circuit,t

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$2\sqrt{10}\,A$

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(D) Given: Capacitance $C = 60\,\mu F = 60 	imes 10^{-6}\,F$,Voltage $V = 100\,V$,Inductance $L = 15\,mH = 15 	imes 10^{-3}\,H$.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 $U_{max} = \frac{1}{2} C V^2$.The maximum energy in the inductor is $U_{max} = \frac{1}{2} L I_{max}^2$.Equating the two: $\frac{1}{2} L I_{max}^2 = \frac{1}{2} C V^2$.$I_{max} = V \sqrt{\frac{C}{L}}$.Substituting the values: $I_{max} = 100 	imes \sqrt{\frac{60 	imes 10^{-6}}{15 	imes 10^{-3}}} = 100 	imes \sqrt{4 	imes 10^{-3}} = 100 	imes 0.0632 = 6.32\,A$.Wait,calculating $I_{max} = 100 	imes \sqrt{4 	imes 10^{-3}} = 100 	imes 2 	imes 10^{-1.5} = 200 	imes 0.0316 = 6.32\,A$. Let's re-evaluate: $\sqrt{\frac{60 	imes 10^{-6}}{15 	imes 10^{-3}}} = \sqrt{4 	imes 10^{-3}} = \sqrt{0.004} = 0.0632$. Actually,$I_{max} = 100 	imes \sqrt{4 	imes 10^{-3}} = 100 	imes 0.0632 = 6.32\,A$. Re-checking the provided option $2\sqrt{10}$: $2 	imes 3.162 = 6.324\,A$. Thus,$2\sqrt{10}$ is correct.

$A$ $60\,\mu F$ capacitor is charged to $100\,V$. This charged capacitor is connected across a $15\,mH$ coil,so that $LC$ oscillations occur. The maximum current in the coil is:

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