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If maximum energy is stored in capacitor at $t=0$, then find the time after which current in the circuit will be maximum
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Verified Answer
The correct answer is:
$\frac{\pi}{4} \mathrm{~ms}$
In $L-C$ oscillations, total time in one complete cycle is, $T=2 \pi \sqrt{L C}$
$\begin{aligned} & T=2 \pi \sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}} \\ & \Rightarrow T=2 \pi \sqrt{25 \times 10^{-8}} \\ & =\pi \times 10^{-3} \mathrm{sec}\end{aligned}$
After time $T / 4$ capacitor is fully discharged and current in circuit will be maximum, then $\frac{T}{4}=\frac{\pi \times 10^{-3}}{4}=\frac{\pi}{4} \mathrm{~ms}$
$\begin{aligned} & T=2 \pi \sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}} \\ & \Rightarrow T=2 \pi \sqrt{25 \times 10^{-8}} \\ & =\pi \times 10^{-3} \mathrm{sec}\end{aligned}$
After time $T / 4$ capacitor is fully discharged and current in circuit will be maximum, then $\frac{T}{4}=\frac{\pi \times 10^{-3}}{4}=\frac{\pi}{4} \mathrm{~ms}$
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