An ideal capacitor of capacitance 0.2, mu F i | vedclass

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Question

(A) Given: Capacitance $C = 0.2\, \mu F = 0.2 	imes 10^{-6}\, F$.Inductance $L = 0.5\, mH = 0.5 	imes 10^{-3}\, H$.Initial potential difference $V_0

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$0.17$

Vedclass · Answer

(A) Given: Capacitance $C = 0.2\, \mu F = 0.2 	imes 10^{-6}\, F$.Inductance $L = 0.5\, mH = 0.5 	imes 10^{-3}\, H$.Initial potential difference $V_0 = 10\, V$.Potential difference at time $t$ is $V = 5\, V$.By the law of conservation of energy in an $LC$ circuit,the total energy remains constant:$\frac{1}{2} C V_0^2 = \frac{1}{2} C V^2 + \frac{1}{2} L I^2$Substituting the values:$\frac{1}{2} 	imes (0.2 	imes 10^{-6}) 	imes (10)^2 = \frac{1}{2} 	imes (0.2 	imes 10^{-6}) 	imes (5)^2 + \frac{1}{2} 	imes (0.5 	imes 10^{-3}) 	imes I^2$$(0.2 	imes 10^{-6}) 	imes 100 = (0.2 	imes 10^{-6}) 	imes 25 + (0.5 	imes 10^{-3}) 	imes I^2$$20 	imes 10^{-6} = 5 	imes 10^{-6} + (0.5 	imes 10^{-3}) 	imes I^2$$15 	imes 10^{-6} = (0.5 	imes 10^{-3}) 	imes I^2$$I^2 = \frac{15 	imes 10^{-6}}{0.5 	imes 10^{-3}} = 30 	imes 10^{-3} = 0.03$$I = \sqrt{0.03} = \sqrt{3 	imes 10^{-2}} = \sqrt{3} 	imes 10^{-1} \approx 1.732 	imes 0.1 = 0.1732\, A$.Thus,the current is approximately $0.17\, A$.

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