JEE Mains · Physics · STD 12 -7. Alternating current
A capacitor of capacitance \(100\ \mu \mathrm{F}\) is charged to a potential of \(12 \mathrm{~V}\) and connected to a \(6.4 \ \mathrm{mH}\) inductor to produce oscillations. The maximum current in the circuit would be :
- A \(3.2 \mathrm{~A}\)
- B \(1.5 \mathrm{~A}\)
- C \(2.0 \mathrm{~A}\)
- D \(1.2 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(1.5 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
By energy conservation \( \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\max }^2 \) \( \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \) \( =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12\) \( =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A}\)
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