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Electrical Oscillations

A 2μ F capaci...

Question

A $2 μ F$ capacitor is initially charged to $20$ Volts and then shorted across a $8 μ H$ inductor. The maximum value of the current in the circuit (in $A$ ) is

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Solution

For $L - C$ circuit,
Maximum current is given by
$I_{m} = ω Q_{0} = \frac{C V_{0}}{\sqrt{L C}} = \frac{2 μ F . 20 V}{\sqrt{(8 μ H) (2 μ F)}}$
$= 10.0 A$

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