A 60,mu F capacitor is charged to 50, V. This | vedclass

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(D) In $LC$ oscillations,the energy stored in the capacitor is converted into magnetic energy in the inductor.By the law of conservation of energy,the

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$10$

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(D) In $LC$ oscillations,the energy stored in the capacitor is converted into magnetic energy in the inductor.By the law of conservation of energy,the maximum electrostatic energy equals the maximum magnetic energy:$\frac{1}{2} C V^2 = \frac{1}{2} L I_{\max}^2$$I_{\max} = V \sqrt{\frac{C}{L}}$Given: $C = 60 	imes 10^{-6} \, F$,$V = 50 \, V$,$L = 1.5 	imes 10^{-3} \, H$.Substituting the values:$I_{\max} = 50 	imes \sqrt{\frac{60 	imes 10^{-6}}{1.5 	imes 10^{-3}}}$$I_{\max} = 50 	imes \sqrt{\frac{60 	imes 10^{-3}}{1.5}}$$I_{\max} = 50 	imes \sqrt{40 	imes 10^{-3}} = 50 	imes \sqrt{0.04}$$I_{\max} = 50 	imes 0.2 = 10 \, A$.

$A$ $60\,\mu F$ capacitor is charged to $50\, V$. This charged capacitor is connected across a $1.5\, mH$ coil,so that $LC$ oscillations occur. The max current in the coil is......$A$

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