An ideal inductor of inductance L and an ideal capacitor of capacitance C are connected in series. Initially, at t = 0 , the capacitor has a charge Q 0 and the current flowing through the inductor i

Question

An ideal inductor of inductance L and an ideal capacitor of capacitance C are connected in series. Initially, at t = 0 , the capacitor has a charge Q 0 ​ and the current flowing through the inductor is I 0 ​ . Which of the following statements is/are correct?

TestHub · Accepted Answer

For an ideal LC circuit, the angular frequency of oscillations is $ω = \frac{1}{L C}$ . So, option A is correct.

The total energy of the LC circuit is conserved. At $t = 0$ , the energy stored in the capacitor is $\frac{Q _{0}^{2}}{2 C}$ and in the inductor is $\frac{1}{2} L I_{0}^{2}$ . Thus, the total energy $E_{t o t a l} = \frac{Q _{0}^{2}}{2 C} + \frac{1}{2} L I_{0}^{2}$ . So, option D is correct. At maximum charge $Q_{ma x}$ , the current in the inductor is zero. Therefore, $E_{t o t a l} = \frac{Q _{ma x}^{2}}{2 C}$ . Equating this with the total energy, $\frac{Q _{ma x}^{2}}{2 C} = \frac{Q _{0}^{2}}{2 C} + \frac{1}{2} L I_{0}^{2}$ , which gives $Q_{ma x} = Q_{0}^{2} + L C I_{0}^{2}$ . So, option B is correct. At maximum current $I_{ma x}$ , the charge on the capacitor is zero. Therefore, $E_{t o t a l} = \frac{1}{2} L I_{ma x}^{2}$ . Equating this with the total energy, $\frac{1}{2} L I_{ma x}^{2} = \frac{Q _{0}^{2}}{2 C} + \frac{1}{2} L I_{0}^{2}$ , which gives $I_{ma x} = I_{0}^{2} + \frac{Q _{0}^{2}}{L C}$ . So, option C is correct.

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Options:(select one or more)

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