A circuit has a self inductance of 1H and carries a current of 2A. To prevent sparking when the circuit is switched off, a capacitor which can withstand 400V is used. The least capacitance of the capacitor connected across the switch must be equal to

Options

A. `50muF`
B. `25muF`
C. `100muF`
D. `12.5muF`

Answer

To solve the problem, we need to determine the least capacitance of the capacitor that can prevent sparking when the circuit is switched off. Here's the step-by-step solution:

### Step 1: Identify the energy stored in the inductor
The energy (E) stored in an inductor can be calculated using the formula:
\[ E = \frac{1}{2} L I^2 \]
where:
- \(L\) is the self-inductance (1 H),
- \(I\) is the current (2 A).

### Step 2: Calculate the energy stored in the inductor
Substituting the values into the formula:
\[ E = \frac{1}{2} \times 1 \times (2)^2 = \frac{1}{2} \times 1 \times 4 = 2 \text{ Joules} \]

### Step 3: Identify the energy stored in the capacitor
The energy stored in a capacitor can be calculated using the formula:
\[ E = \frac{1}{2} C V^2 \]
where:
- \(C\) is the capacitance,
- \(V\) is the voltage (400 V).

### Step 4: Set the energy of the inductor equal to the energy of the capacitor
To prevent sparking, the energy stored in the capacitor must equal the energy released by the inductor:
\[ \frac{1}{2} C V^2 = 2 \]

### Step 5: Solve for capacitance \(C\)
Rearranging the equation to solve for \(C\):
\[ C = \frac{2 \times 2}{V^2} \]
Substituting \(V = 400\) V:
\[ C = \frac{2 \times 2}{(400)^2} = \frac{4}{160000} = \frac{1}{40000} \text{ Farads} \]

### Step 6: Convert Farads to microfarads
To convert Farads to microfarads, we multiply by \(10^6\):
\[ C = \frac{1}{40000} \times 10^6 = 25 \text{ microfarads} \]

### Final Answer
The least capacitance of the capacitor connected across the switch must be equal to:
\[ \boxed{25 \text{ microfarads}} \]

---