A capacitor of capacitance $100 \mu \mathrm{F}$ is charged to a potential of $12 \mathrm{~V}$ and connected to a $6.4 \mathrm{~mH}$ inductor to produce oscillations. The maximum current in the circuit would be :
Solution
<p>By energy conservation</p>
<p>$$\begin{aligned}
& \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\text {max }}^2 \\
& \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \\
& =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12 \\
& =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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