Minimum number of capacitors each of `8 muF` and 250 V used to make a composite capacitor of `16 muF` and 1000 V are

Options

A. 8
B. 32
C. 16
D. 24

Explanation

To solve the problem of finding the minimum number of capacitors required to create a composite capacitor of 16 µF and 1000 V using capacitors of 8 µF and 250 V, we can follow these steps:

### Step 1: Determine the Voltage Requirement
We need to create a composite capacitor that can handle 1000 V. Each capacitor can handle a maximum of 250 V. To find out how many capacitors we need in series to achieve the required voltage, we can use the formula:

\[ \text{Number of capacitors in series} = \frac{\text{Total Voltage}}{\text{Voltage rating of one capacitor}} = \frac{1000 \text{ V}}{250 \text{ V}} = 4 \]

**Hint:** To increase the voltage rating of capacitors, connect them in series. The total voltage rating is the sum of the individual voltage ratings.

### Step 2: Determine the Capacitance Requirement
Next, we need to achieve a total capacitance of 16 µF. When capacitors are connected in series, the equivalent capacitance (C_eq) can be calculated using the formula:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \]

For 'n' identical capacitors of capacitance C, this simplifies to:

\[ C_{eq} = \frac{C}{n} \]

In our case, we have 4 capacitors of 8 µF in series:

\[ C_{eq} = \frac{8 \text{ µF}}{4} = 2 \text{ µF} \]

**Hint:** When capacitors are connected in series, the equivalent capacitance is less than the smallest capacitance in the series.

### Step 3: Calculate the Number of Rows Needed
Now, we need to find out how many such series combinations (rows) are required to achieve a total capacitance of 16 µF. We can use the formula for total capacitance in parallel:

\[ C_{total} = C_{eq} \times \text{Number of Rows} \]

Let 'm' be the number of rows. We want:

\[ 16 \text{ µF} = 2 \text{ µF} \times m \]

Solving for m gives:

\[ m = \frac{16 \text{ µF}}{2 \text{ µF}} = 8 \]

**Hint:** To increase the total capacitance, connect the equivalent capacitors in parallel.

### Step 4: Calculate the Total Number of Capacitors
Now, we know that we need 4 capacitors in each row and we have 8 rows. Therefore, the total number of capacitors required is:

\[ \text{Total Capacitors} = \text{Number of capacitors per row} \times \text{Number of rows} = 4 \times 8 = 32 \]

### Final Answer
The minimum number of capacitors required is **32 capacitors**.

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