A charged capacitor is connected to an inductor . At a particular instant, energy stored in inductor is 8/9 times of initial energy stored In capacitor . What is the ratio of initial charge on capacitor to that with instantaneous charge on capacitor ?

Explanation

To solve the problem, we need to analyze the energy stored in the capacitor and the inductor at different instances.

### Step-by-Step Solution:

1. **Initial Energy in the Capacitor:**
The initial energy stored in the capacitor (E₀) can be expressed as:
\[ E_0 = \frac{Q_i^2}{2C} \]
where \( Q_i \) is the initial charge on the capacitor and \( C \) is the capacitance.

2. **Energy in the Inductor at a Particular Instant:**
At a particular instant, it is given that the energy stored in the inductor (E_L) is:
\[ E_L = \frac{8}{9} E_0 \]

3. **Energy Conservation:**
Since the total energy in the system is conserved, the energy in the capacitor at that instant (E_C) can be calculated as:
\[ E_C = E_0 - E_L = E_0 - \frac{8}{9} E_0 = \frac{1}{9} E_0 \]

4. **Energy in the Capacitor at Instant:**
The energy stored in the capacitor at that instant can also be expressed in terms of the instantaneous charge \( Q_f \):
\[ E_C = \frac{Q_f^2}{2C} \]
Setting this equal to the expression we derived for \( E_C \):
\[ \frac{Q_f^2}{2C} = \frac{1}{9} E_0 \]

5. **Substituting Initial Energy:**
Substitute \( E_0 = \frac{Q_i^2}{2C} \) into the equation:
\[ \frac{Q_f^2}{2C} = \frac{1}{9} \left(\frac{Q_i^2}{2C}\right) \]

6. **Simplifying the Equation:**
Cancel \( \frac{1}{2C} \) from both sides:
\[ Q_f^2 = \frac{1}{9} Q_i^2 \]

7. **Finding the Ratio of Charges:**
Taking the square root of both sides gives:
\[ \frac{Q_i}{Q_f} = 3 \]

### Final Answer:
The ratio of the initial charge on the capacitor to the instantaneous charge on the capacitor is:
\[ \frac{Q_i}{Q_f} = 3 \]