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PhysicsEMI/ACLC oscillationMedium2 minPYQ_2024

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A capacitor of capacitance $100 μF$ 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 :

Options:

Answer:

Solution:

By energy conservation, it follows that

$\frac{1}{2} C V^{2} = \frac{1}{2} L I_{\max}^{2} . . . (1)$

Equation (1) implies that

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

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:LC oscillation

⏱ 2mℹ️ Source: PYQ_2024

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