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An $LC$ circuit contains a $20\,\,mH$ inductor and a $50\,\,\mu F$ capacitor with an initial charge of $10 \,\,mC.$ The resistance of the circuit is negligible. Let the instant the circuit is closed be $t = 0.$ At what time is the energy stored completely magnetic $t=$ .......... $m/s$
$ 0$
$1.57$
$3.14$
$6.28$
Solution
Given, $\quad \mathrm{L}=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{\,H}$
$\mathrm{C}=50 \mu \mathrm{F}=50 \times 10^{-6} \mathrm{\,F}$
For $LC$ circuit the frequency,
$\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}} \quad \text { or } \mathrm{T}=2 \pi \sqrt{\mathrm{LC}} \quad\left(\because \mathrm{T}=\frac{1}{\mathrm{f}}\right)$
At time $\mathrm{t}=\frac{\mathrm{T}}{4},$ energy stored is completely magnetic,
The time, $\quad t=\frac{T}{4}$
$\mathrm{t}=\frac{2 \pi \sqrt{\mathrm{LC}}}{4}$
or $\mathrm{t}=\frac{2 \pi \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}{4}$
or $\mathrm{t}=\frac{3.14 \sqrt{10^{-6}}}{2}$ or $\mathrm{t}=\frac{3.14 \times 10^{-3}}{2}$
or $\mathrm{t}=1.57 \times 10^{-3} \mathrm{s}=1.57 \mathrm{\,ms}$
