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 determining the minimum number of capacitors, each with a capacitance of \(8 \mu F\) and a voltage rating of \(250 V\), required to create a composite capacitor of \(16 \mu F\) and \(1000 V\), we can follow these steps:

### Step 1: Determine the number of capacitors needed in series for voltage
Each capacitor has a voltage rating of \(250 V\). To achieve a total voltage of \(1000 V\), we need to connect capacitors in series. The total voltage across capacitors in series is given by:

\[ V_{total} = N \times V_{individual} \]

Where:
- \(V_{total} = 1000 V\)
- \(V_{individual} = 250 V\)

Setting up the equation:

\[ 1000 = N \times 250 \]

Now, solve for \(N\):

\[ N = \frac{1000}{250} = 4 \]

Thus, we need **4 capacitors in series** to achieve \(1000 V\).

### Step 2: Determine the equivalent capacitance of capacitors in series
When capacitors are connected in series, the equivalent capacitance \(C_{eq}\) is calculated using the formula:

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

For \(N = 4\) capacitors of \(8 \mu F\):

\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]

Thus,

\[ C_{eq} = 2 \mu F \]

### Step 3: Determine the number of rows needed in parallel for total capacitance
We need a total capacitance of \(16 \mu F\). Since we have \(2 \mu F\) from one row of capacitors in series, we can find the number of rows \(M\) needed in parallel:

\[ C_{total} = M \times C_{eq} \]

Setting up the equation:

\[ 16 = M \times 2 \]

Solving for \(M\):

\[ M = \frac{16}{2} = 8 \]

Thus, we need **8 rows of capacitors** in parallel.

### Step 4: Calculate the total number of capacitors
The total number of capacitors \(T\) is given by the product of the number of capacitors in each series (which is \(N = 4\)) and the number of rows (which is \(M = 8\)):

\[ T = M \times N = 8 \times 4 = 32 \]

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

---