Physics · Alternating Current

JEE Main 2024 — 29 January, Shift 1 — Question 46

A capacitor of capacitance $100 \mu \mathrm{~F}$ 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 :

Option A:
3.2 A
Option B:
1.5 A
Correct
Option C:
2.0 A
Option D:
1.2 A

Answer: B

Step-by-step solution

By energy conservation $\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \mathrm{LI}_{\max }^{2}$

$I_{\text {max }}=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V}$

$=\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12$

$=\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Alternating Current
Topic: L-C Circuits oscillations and Quality Factor, Resonance