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(B) By the principle of conservation of energy in an $LC$ circuit,the maximum electrostatic energy stored in the capacitor is equal to the maximum mag

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) By the principle of conservation of energy in an $LC$ circuit,the maximum electrostatic energy stored in the capacitor is equal to the maximum magnetic energy stored in the inductor.$\frac{1}{2} CV^2 = \frac{1}{2} LI_{\max}^2$Rearranging the formula to solve for the maximum current $I_{\max}$:$I_{\max} = V \sqrt{\frac{C}{L}}$Given values:$C = 100 \ \mu F = 100 	imes 10^{-6} \ F = 10^{-4} \ F$$L = 6.4 \ mH = 6.4 	imes 10^{-3} \ H$$V = 12 \ V$Substituting the values:$I_{\max} = 12 	imes \sqrt{\frac{100 	imes 10^{-6}}{6.4 	imes 10^{-3}}}$$I_{\max} = 12 	imes \sqrt{\frac{10^{-4}}{6.4 	imes 10^{-3}}} = 12 	imes \sqrt{\frac{10^{-1}}{6.4}} = 12 	imes \sqrt{\frac{0.1}{6.4}} = 12 	imes \sqrt{\frac{1}{64}}$$I_{\max} = 12 	imes \frac{1}{8} = 1.5 \ A$

$A$ capacitor of capacitance $100 \ \mu F$ is charged to a potential of $12 \ V$ and connected to a $6.4 \ mH$ inductor to produce oscillations. The maximum current in the circuit would be: (in $A$)

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