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JEE Main 2018 (15 Apr, Shift 1) · LC Oscillations

LC Oscillations question from JEE Main 2018 (15 Apr, Shift 1)

Q19JEE Main 2018 · 15 APR · SHIFT 1Medium

LC Oscillations

An ideal capacitor of capacitance 0.2 μF0.2\,\mu F is charged to a potential difference of 10 V.10\ V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH.0.5\ mH. The current at a time when the potential difference across the capacitor is 5 V,5\ V, is:

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Answer: (c)

  1. Energy conservation: 12C(V12−V22)=12LI2\frac{1}{2}C(V_1^2 - V_2^2) = \frac{1}{2}LI^2
  2. I=0.2×10−6(100−25)0.5×10−3=15×10−65×10−4=0.03I = \sqrt{\frac{0.2\times10^{-6}(100 - 25)}{0.5\times10^{-3}}} = \sqrt{\frac{15\times10^{-6}}{5\times10^{-4}}} = \sqrt{0.03}
  3. I≈0.17 AI \approx 0.17\text{ A}
  4. So the answer is (c) 0.17 A0.17\text{ A}.
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