Question

In an oscillating LC circuit, the maximum charge on the capacitor is ' \(Q\) '. When the energy is stored equally between the electric and magnetic fields, the instantaneous charge on the capacitor ' \(q\) ' is

• A. \(Q\)
• B. \(\frac{Q}{2}\)
• C. \(\frac{Q}{\sqrt{2}}\)
• D. \(\frac{Q}{\sqrt{3}}\)

Hint:

LC energy: Equal energy $\rightarrow$ charge reduces by $\sqrt{2}$

Answer 1

The Correct Option is C

Concept: Total energy in LC circuit: \[ U = \frac{Q^2}{2C} \] At any instant: \[ U = \frac{q^2}{2C} + \frac{1}{2}LI^2 \]

Step 1: Equal energy condition. \[ \frac{q^2}{2C} = \frac{1}{2}LI^2 = \frac{U}{2} \]

Step 2: Substitute total energy. \[ \frac{q^2}{2C} = \frac{1}{2} \cdot \frac{Q^2}{2C} \]

Step 3: Solve. \[ q^2 = \frac{Q^2}{2} \Rightarrow q = \frac{Q}{\sqrt{2}} \]

Step 4: Conclusion.
$q = \frac{Q}{\sqrt{2}}$ Final Answer: Option (C)