Verified Solution • 2025–26 Session moderate

QJEEM24Ph5b3a: JEE Main 2024 Physics - Class 12 JEEM24Ph5b3a Solution: A capacitor of capacitance is charged to a potential of and…

A capacitor of capacitance 100μF100 \mu \mathrm{F} is charged to a potential of 12 V12 \mathrm{~V} and connected to a 6.4 mH6.4 \mathrm{~mH} inductor to produce oscillations. The maximum current in the circuit would be :

Multiple Choice Options

  • A 2.0 A
  • B 3.2 A
  • C 1.5 A
  • D 1.2 A

Step-by-Step Verified Solution

By energy conservation

amp;12CV2=12LImax 2amp;Imax⁡=CLVamp;=100×10−66.4×10−3×12amp;=128=32=1.5 A\begin{aligned}& \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\text {max }}^2 \\& \mathrm{I}_{\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}\end{aligned}

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