An inductance of `2H` carries a current of `2A`. To prevent sparking when the circuit is broken a capacitor of `4muF` is connected across the inductance. The voltage rating of the capacitor is of the order of

Options

A. `10^3V`
B. `10V`
C. `10^5V`
D. `10^6V`

Explanation

To find the voltage rating of the capacitor connected across an inductor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Stored in the Inductor**:
The energy stored in an inductor (E) can be expressed using the formula:
\[ E = \frac{1}{2} L I^2 \]
where \(L\) is the inductance and \(I\) is the current.

2. **Substitute the Given Values**:
Given that the inductance \(L = 2 \, H\) and the current \(I = 2 \, A\), we can substitute these values into the energy formula:
\[ E = \frac{1}{2} \times 2 \times (2)^2 \]
\[ E = \frac{1}{2} \times 2 \times 4 = 4 \, J \]

3. **Energy Transfer to the Capacitor**:
When the circuit is broken, this energy will be transferred to the capacitor. The energy stored in a capacitor (E) is given by:
\[ E = \frac{1}{2} C V^2 \]
where \(C\) is the capacitance and \(V\) is the voltage across the capacitor.

4. **Set the Energies Equal**:
Since the energy in the inductor will be equal to the energy stored in the capacitor:
\[ \frac{1}{2} C V^2 = 4 \]

5. **Substitute the Capacitance Value**:
Given that the capacitance \(C = 4 \, \mu F = 4 \times 10^{-6} \, F\), we can substitute this into the equation:
\[ \frac{1}{2} \times (4 \times 10^{-6}) \times V^2 = 4 \]

6. **Solve for Voltage \(V\)**:
Rearranging the equation gives:
\[ V^2 = \frac{4 \times 2}{4 \times 10^{-6}} = \frac{8}{4 \times 10^{-6}} = 2 \times 10^{6} \]
Taking the square root of both sides:
\[ V = \sqrt{2 \times 10^{6}} = \sqrt{2} \times 10^{3} \]

7. **Calculate the Voltage**:
Approximating \(\sqrt{2} \approx 1.414\):
\[ V \approx 1.414 \times 10^{3} \approx 2 \times 10^{3} \, V \]

8. **Determine the Order of Voltage**:
The voltage rating of the capacitor is therefore of the order of \(10^{3} \, V\).

### Final Answer:
The voltage rating of the capacitor is of the order of \(10^{3} \, V\).