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(B) In an $L-C$ oscillation circuit,the total time period $T$ for one complete cycle is given by:$T = 2\pi \sqrt{LC}$Given $L = 25 \, mH = 25 	imes 1

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$\frac{\pi}{4} \, ms$

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(B) In an $L-C$ oscillation circuit,the total time period $T$ for one complete cycle is given by:$T = 2\pi \sqrt{LC}$Given $L = 25 \, mH = 25 	imes 10^{-3} \, H$ and $C = 10 \, \mu F = 10 	imes 10^{-6} \, F$.Substituting the values:$T = 2\pi \sqrt{25 	imes 10^{-3} 	imes 10 	imes 10^{-6}}$$T = 2\pi \sqrt{250 	imes 10^{-9}} = 2\pi \sqrt{25 	imes 10^{-8}}$$T = 2\pi 	imes 5 	imes 10^{-4} = 10\pi 	imes 10^{-4} = \pi 	imes 10^{-3} \, s = \pi \, ms$.The energy in the capacitor is maximum at $t = 0$. The current in the circuit becomes maximum when the energy is fully transferred from the capacitor to the inductor,which occurs at $t = \frac{T}{4}$.Therefore,the time required is:$t = \frac{T}{4} = \frac{\pi \, ms}{4} = \frac{\pi}{4} \, ms$.

If maximum energy is stored in the capacitor at $t = 0$,find the time after which the current in the circuit will be maximum. Given $L = 25 \, mH$ and $C = 10 \, \mu F$.

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