A capacitor of capacitance `4 muF` is fully charged to a potential difference of 12 V and then charging battery is disconnected. Charged capacitor is now connected across an inductor of self-inductance of 1.2 mH. The current in the circuit at a time when the potential difference across the capacitor is 6 V is

Options

A. 0.3 A
B. 0.6 A
C. 0.9 A
D. 1.2 A

Explanation

To solve the problem step-by-step, we need to find the current in the circuit when the potential difference across the capacitor is 6 V. Here’s how we can do that:

### Step 1: Calculate the initial and final energy in the capacitor
The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]

- Initial energy when the capacitor is fully charged (V1 = 12 V):
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \, \text{F} \times (12 \, \text{V})^2 \]
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times 144 = 288 \times 10^{-6} \, \text{J} = 2.88 \times 10^{-4} \, \text{J} \]

- Final energy when the potential difference across the capacitor is 6 V (V2 = 6 V):
\[ U_2 = \frac{1}{2} \times 4 \times 10^{-6} \, \text{F} \times (6 \, \text{V})^2 \]
\[ U_2 = \frac{1}{2} \times 4 \times 10^{-6} \times 36 = 72 \times 10^{-6} \, \text{J} = 7.2 \times 10^{-5} \, \text{J} \]

### Step 2: Calculate the change in energy of the capacitor
The change in energy (ΔU) as the capacitor discharges from 12 V to 6 V is:
\[ \Delta U = U_1 - U_2 \]
\[ \Delta U = 2.88 \times 10^{-4} \, \text{J} - 7.2 \times 10^{-5} \, \text{J} = 2.16 \times 10^{-4} \, \text{J} \]

### Step 3: Relate the change in energy to the energy stored in the inductor
The energy stored in the inductor (U_L) is given by:
\[ U_L = \frac{1}{2} L I^2 \]
where L is the inductance and I is the current.

Setting the change in energy equal to the energy stored in the inductor:
\[ \Delta U = U_L \]
\[ 2.16 \times 10^{-4} = \frac{1}{2} \times 1.2 \times 10^{-3} \times I^2 \]

### Step 4: Solve for the current I
Rearranging the equation gives:
\[ I^2 = \frac{2 \times 2.16 \times 10^{-4}}{1.2 \times 10^{-3}} \]
\[ I^2 = \frac{4.32 \times 10^{-4}}{1.2 \times 10^{-3}} = 0.36 \]
\[ I = \sqrt{0.36} = 0.6 \, \text{A} \]

### Final Answer
The current in the circuit when the potential difference across the capacitor is 6 V is:
\[ I = 0.6 \, \text{A} \]