Question

In an $L-C$ oscillation circuit, maximum charge on capacitor can be $Q_0$. If at any instant the electrical energy is $\frac{3}{4}th$ of the magnetic energy, then the charge on the capacitor at that instant is

• A. $\frac{2Q_0}{3}$
• B. $\sqrt{\frac{3}{7}}{Q}_0$
• C. $\frac{3}{7}{Q}_0$
• D. $\sqrt{\frac{2}{3}}{Q}_0$

Answer 1

The correct option is B $\sqrt{\frac{3}{7}}{Q}_0$

Total energy stored in the circuit is

$E_0=\frac{Q_0^2}{2C}$

Let at any instant the charge on the capacitor is $Q$.

From energy conservation principle,

$\text{Magnetic energy + Electrical energy = Total energy}$

$\frac{4}{3}\times \left(\frac{Q^2}{2C}\right)+\frac{Q^2}{2C}=\frac{Q_0^2}{2C}$

$[\because \text{Magnetic energy}=\frac{4}{3}\times \text{Electrical energy}$

$\Rightarrow \frac{Q^2}{2C}[\frac{4}{3}+1]=\frac{Q_0^2}{2C}$

$\Rightarrow Q^2=\frac{3}{7}{Q}_0^2$

$\therefore Q=\sqrt{\frac{3}{7}}{Q}_0$

Hence, $\left(B\right)$ is correct answer.