Consider the following circuit. By keeping $S_1$ closed, the capacitor is fully charged and then $S_1$ is…

Consider the following circuit. By keeping $S_1$ closed, the capacitor is fully charged and then $S_1$ is opened and $S_2$ is closed, then
  1. At time $t=0$, the energy stored in the circuit is purely in the form of magnetic energy.
  2. At $t\gt0$, there is no exchange of energy between L and C .
  3. At any time $t\gt0$, the current in the circuit is in the same direction.
  4. At any time $\mathrm{t}\gt0$, the instantaneous current in the circuit may be $\mathrm{V} \sqrt{\frac{\mathrm{c}}{\mathrm{L}}}$

Solution

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

Asked in: MHT CET 2024 (09 May Shift 2)

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