An ideal capacitor of capacitance `0.2muF` is charged to a potential difference of `10V`. The charging battery is than disconnected. The capacitor is then connected to an ideal inductor of self inductance `0.5mH`. The current at a time when the potential difference across the capacitor is `5V` is :

Options

A. Option1 `0.34 A`
B. Option2 `0.17 A`
C. Option3 `0.25 A`
D. Option4 `0.15A`

Answer

To solve the problem step by step, we will use the principle of conservation of energy in an LC circuit.

### Step 1: Calculate the initial energy stored in the capacitor
The initial energy stored in the capacitor when it is fully charged can be calculated using the formula:
\[ U_i = \frac{1}{2} C V_1^2 \]
Where:
- \(C = 0.2 \mu F = 0.2 \times 10^{-6} F\)
- \(V_1 = 10 V\)

Substituting the values:
\[ U_i = \frac{1}{2} \times (0.2 \times 10^{-6}) \times (10)^2 \]
\[ U_i = \frac{1}{2} \times (0.2 \times 10^{-6}) \times 100 = 0.01 \times 10^{-6} = 0.01 \times 10^{-3} = 1 \times 10^{-5} J \]

### Step 2: Set up the energy conservation equation
When the capacitor is connected to the inductor, the total energy is conserved. The energy at the point when the potential difference across the capacitor is \(5V\) can be expressed as:
\[ U_f = \frac{1}{2} C V_2^2 + \frac{1}{2} L I^2 \]
Where:
- \(V_2 = 5 V\)
- \(I\) is the current we want to find.

### Step 3: Substitute values into the energy conservation equation
Setting the initial energy equal to the final energy:
\[ \frac{1}{2} C V_1^2 = \frac{1}{2} C V_2^2 + \frac{1}{2} L I^2 \]
This simplifies to:
\[ \frac{1}{2} C (V_1^2 - V_2^2) = \frac{1}{2} L I^2 \]

### Step 4: Rearranging to find \(I^2\)
Rearranging gives:
\[ I^2 = \frac{C (V_1^2 - V_2^2)}{L} \]

### Step 5: Substitute the known values
Substituting \(C\), \(V_1\), \(V_2\), and \(L\):
- \(C = 0.2 \times 10^{-6} F\)
- \(V_1 = 10 V\)
- \(V_2 = 5 V\)
- \(L = 0.5 \times 10^{-3} H\)

Thus:
\[ I^2 = \frac{0.2 \times 10^{-6} (10^2 - 5^2)}{0.5 \times 10^{-3}} \]
\[ = \frac{0.2 \times 10^{-6} (100 - 25)}{0.5 \times 10^{-3}} \]
\[ = \frac{0.2 \times 10^{-6} \times 75}{0.5 \times 10^{-3}} \]
\[ = \frac{15 \times 10^{-6}}{0.5 \times 10^{-3}} = 30 \times 10^{-3} = 0.03 \]

### Step 6: Calculate \(I\)
Taking the square root gives:
\[ I = \sqrt{0.03} \approx 0.1732 \, A \]

### Final Answer
The current at the time when the potential difference across the capacitor is \(5V\) is approximately:
\[ \boxed{0.1732 \, A} \]