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: (a) 2.0 A (b) 3.2 A (c) 1.5 A (d) 1.2 A

Answer: (c). Imax = V√(C/L) Imax = 12√100×10⁻⁶6.4×10⁻³ = 12√(1/64) = 12/8 So the answer is (c) 1.5 A.

A capacitor of capacitance 100 is charged to a potential of 12 V and… | JEE Main 2024 (29 Jan, Shift 1)

Q10JEE Main 2024 · 29 JAN · SHIFT 1Easy

LC Oscillations

A capacitor of capacitance $100\,\mu\text{F}$ is charged to a potential of $12\text{ V}$ and connected to a $6.4\text{ mH}$ inductor to produce oscillations. The maximum current in the circuit would be:

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Answer: (c)

$I_{max} = V\sqrt{\frac{C}{L}}$
$I_{max} = 12\sqrt{\frac{100\times10^{-6}}{6.4\times10^{-3}}} = 12\sqrt{\frac{1}{64}} = \frac{12}{8}$
So the answer is (c) $1.5\text{ A}$ .

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