A capacitor of capacitance 100 is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be

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Solution

Q.
A capacitor of capacitance 100 $μ F$ is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be [2024]

1 3.2 A

2 1.5 A

3 2.0 A

4 1.2 A

Ans.
(2)

From energy conservation, $\frac{1}{2} C V^{2} = \frac{1}{2} L i_{\max^{2}}$

$I_{\max} = \sqrt{\frac{C}{L}} V = \sqrt{\frac{100 \times 10^{- 6}}{6.4 \times 10^{- 3}}} \times 12 = \frac{12}{8} = \frac{3}{2} = 1.5 A$

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