Minimum number of capacitors of `2muF` capacitance each required to obtain a capacitor of `5muF` will be

Options

A. Three
B. Four
C. Five
D. Six

Explanation

To find the minimum number of capacitors of \(2 \mu F\) each required to obtain a capacitor of \(5 \mu F\), we can approach the problem step by step.

### Step 1: Understand the Capacitance in Series and Parallel
- When capacitors are connected in **parallel**, the total capacitance \(C_{total}\) is the sum of the individual capacitances:
\[ C_{total} = C_1 + C_2 + C_3 + \ldots \]
- When capacitors are connected in **series**, the total capacitance \(C_{total}\) is given by:
\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots \]

### Step 2: Determine the Required Combination
We need to achieve a total capacitance of \(5 \mu F\) using \(2 \mu F\) capacitors.

### Step 3: Finding Combinations
1. **Parallel Combination**:
- If we connect \(n\) capacitors in parallel, the total capacitance is:
\[ C_{total} = n \times 2 \mu F \]
- To achieve \(5 \mu F\) using only parallel combinations:
\[ n \times 2 = 5 \implies n = \frac{5}{2} = 2.5 \]
- Since \(n\) must be an integer, we cannot achieve \(5 \mu F\) with only parallel combinations.

2. **Series Combination**:
- If we connect \(m\) capacitors in series, the total capacitance is:
\[ \frac{1}{C_{total}} = \frac{m}{2} \implies C_{total} = \frac{2}{m} \mu F \]
- To achieve \(1 \mu F\) (which we can use in combination with parallel capacitors):
\[ \frac{2}{m} = 1 \implies m = 2 \]

### Step 4: Combining Series and Parallel
To achieve \(5 \mu F\):
- We can use \(2\) capacitors in series to get \(1 \mu F\).
- Now, we need \(4 \mu F\) more to reach \(5 \mu F\).
- To achieve \(4 \mu F\), we can connect \(2\) capacitors in parallel:
\[ C_{parallel} = 2 + 2 = 4 \mu F \]

### Step 5: Total Capacitors Used
- For \(1 \mu F\) (2 capacitors in series): **2 capacitors**
- For \(4 \mu F\) (2 capacitors in parallel): **2 capacitors**
- Total capacitors used:
\[ 2 + 2 = 4 \]

Thus, the minimum number of \(2 \mu F\) capacitors required to obtain a total capacitance of \(5 \mu F\) is **4 capacitors**.

### Final Answer
The minimum number of capacitors required is **4**.