Q144 / 150 · MHT CET 2024 (02 May Shift 1) · Physics

In the given circuit, when $\mathrm{S}_1$ is closed, the capacitor gets fully charged. Now $\mathrm{S}_1$ is open and $\mathrm{S}_2$ is closed. Then Figure

  • A
    there is no exchange of energy between L and C .
  • B
    the current in the circuit is in the same direction.
  • Correct answer
    C
    the instantaneous current in the circuit may be $\mathrm{V}\left(\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}\right)$.
  • D
    the energy stored in the circuit is purely in the form of magnetic energy.

Solution

Maximum energy in capacitor $=$ Maximum energy in inductor $\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}^2$ $\begin{array}{ll}\therefore & I^2=\frac{C}{L} V^2 \\ \therefore & I=V \sqrt{\frac{C}{L}}\end{array}$

Hence, the correct answer is option: 'C'.

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