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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 :

  1. A \(3.2 \mathrm{~A}\)
  2. B  \(1.5 \mathrm{~A}\)
  3. C \(2.0 \mathrm{~A}\)
  4. D  \(1.2 \mathrm{~A}\)
Verified Solution

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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