A capacitor of capacity `2muF` is charged to a potential difference of 12V. It is then connected across an inductor of inductance `6muH`. At an instant when potential difference across the capacitor is 6V, what is the current(in A)?

Answer

To solve the problem, we need to find the current flowing through the inductor at the moment when the potential difference across the capacitor is 6V. We will use the principle of conservation of energy in an LC circuit.

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance of the capacitor, \( C = 2 \mu F = 2 \times 10^{-6} F \)
- Initial potential difference across the capacitor, \( V_0 = 12 V \)
- Inductance of the inductor, \( L = 6 \mu H = 6 \times 10^{-6} H \)
- Potential difference across the capacitor at the instant of interest, \( V = 6 V \)

2. **Calculate the initial energy stored in the capacitor:**
The energy stored in the capacitor when it is charged to \( V_0 \) is given by the formula:
\[ U_C = \frac{1}{2} C V_0^2 \]
Substituting the values:
\[ U_C = \frac{1}{2} \times (2 \times 10^{-6}) \times (12)^2 \]
\[ U_C = \frac{1}{2} \times (2 \times 10^{-6}) \times 144 = 1.44 \times 10^{-4} J \]

3. **Calculate the energy stored in the capacitor when the voltage is 6V:**
The energy stored in the capacitor at this moment is:
\[ U_C' = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U_C' = \frac{1}{2} \times (2 \times 10^{-6}) \times (6)^2 \]
\[ U_C' = \frac{1}{2} \times (2 \times 10^{-6}) \times 36 = 3.6 \times 10^{-5} J \]

4. **Calculate the energy stored in the inductor:**
The total energy in the circuit must remain constant, so we can find the energy stored in the inductor \( U_L \) at this instant:
\[ U_L = U_C - U_C' = 1.44 \times 10^{-4} J - 3.6 \times 10^{-5} J \]
\[ U_L = 1.08 \times 10^{-4} J \]

5. **Relate the energy in the inductor to the current:**
The energy stored in the inductor is given by:
\[ U_L = \frac{1}{2} L I^2 \]
Setting this equal to the energy we calculated:
\[ 1.08 \times 10^{-4} = \frac{1}{2} \times (6 \times 10^{-6}) \times I^2 \]

6. **Solve for the current \( I \):**
Rearranging the equation:
\[ I^2 = \frac{2 \times 1.08 \times 10^{-4}}{6 \times 10^{-6}} \]
\[ I^2 = \frac{2.16 \times 10^{-4}}{6 \times 10^{-6}} = 36 \]
\[ I = \sqrt{36} = 6 A \]

### Final Answer:
The current flowing through the inductor at the instant when the potential difference across the capacitor is 6V is **6 A**.