Q184 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
In an ideal LC circuit, if the energy stored in the inductor is equal to energy stored in a capacitor, then the magnitude of charge Q stored in capacitor is ( $Q_0$ is the maximum charge)
- Correct answerA$\dfrac{Q_0}{\sqrt{2}}$
- B$\dfrac{Q_0}{2}$
- C$\sqrt{2}Q_0$
- D$Q_0$
Solution
Total energy in an ideal LC circuit is conserved and is equal to the maximum energy stored in the capacitor.
$E_{total} = \dfrac{Q_0^2}{2C}$
At any instant, the total energy is the sum of the energy stored in the capacitor ($U_C$) and the inductor ($U_L$).
$E_{total} = U_C + U_L$
Given that the energy stored in the inductor is equal to the energy stored in the capacitor ($U_L = U_C$):
$E_{total} = 2U_C$
Substituting the expressions for energy:
$\dfrac{Q_0^2}{2C} = 2 \left( \dfrac{Q^2}{2C} \right)$
$Q^2 = \dfrac{Q_0^2}{2}$
$Q = \dfrac{Q_0}{\sqrt{2}}$
Answer: $\dfrac{Q_0}{\sqrt{2}}$
$E_{total} = \dfrac{Q_0^2}{2C}$
At any instant, the total energy is the sum of the energy stored in the capacitor ($U_C$) and the inductor ($U_L$).
$E_{total} = U_C + U_L$
Given that the energy stored in the inductor is equal to the energy stored in the capacitor ($U_L = U_C$):
$E_{total} = 2U_C$
Substituting the expressions for energy:
$\dfrac{Q_0^2}{2C} = 2 \left( \dfrac{Q^2}{2C} \right)$
$Q^2 = \dfrac{Q_0^2}{2}$
$Q = \dfrac{Q_0}{\sqrt{2}}$
Answer: $\dfrac{Q_0}{\sqrt{2}}$
Hence, the correct answer is option: 'A'.