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A capacitor of capacity ' $\mathrm{C}$ ' is charged to a potential ' $\mathrm{V}$ '. It is connected in parallel to an inductor of inductance ' $L$ '. The maximum current that will flow in the circuit is
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Verified Answer
The correct answer is:
$\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$
The charge oscillates according to the equations
$$
\begin{aligned}
& \mathrm{q}=\mathrm{q}_0 \cos \omega \mathrm{t} \\
& \mathrm{i}=-\frac{\mathrm{dq}}{\mathrm{dt}}=\omega \mathrm{q}_0 \sin \omega \mathrm{t}=\mathrm{i}_0 \sin \omega \mathrm{t} \\
& \text { where } \mathrm{i}_0=\omega \mathrm{q}_0, \omega=\frac{1}{\sqrt{\mathrm{LC}}}, \mathrm{q}_0=C V \\
& \therefore \mathrm{i}_0=\frac{1}{\sqrt{\mathrm{LC}}} \cdot \mathrm{CV}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{q}=\mathrm{q}_0 \cos \omega \mathrm{t} \\
& \mathrm{i}=-\frac{\mathrm{dq}}{\mathrm{dt}}=\omega \mathrm{q}_0 \sin \omega \mathrm{t}=\mathrm{i}_0 \sin \omega \mathrm{t} \\
& \text { where } \mathrm{i}_0=\omega \mathrm{q}_0, \omega=\frac{1}{\sqrt{\mathrm{LC}}}, \mathrm{q}_0=C V \\
& \therefore \mathrm{i}_0=\frac{1}{\sqrt{\mathrm{LC}}} \cdot \mathrm{CV}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}
\end{aligned}
$$
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