How many `6muF, 200V` condensers are needed to make a condenser of `18muF, 600V?`

Options

A. 9
B. 18
C. 3
D. 27

Explanation

To solve the problem of how many `6μF, 200V` capacitors are needed to create a capacitor of `18μF, 600V`, we can break it down into steps:

### Step 1: Determine the configuration for voltage
To achieve a higher voltage rating of `600V`, we need to connect capacitors in series. The voltage rating of capacitors in series adds up.

For `200V` capacitors in series:
- If we connect `n` capacitors in series, the total voltage rating becomes `n * 200V`.

To achieve `600V`, we need:
\[ n \times 200V = 600V \]
\[ n = \frac{600V}{200V} = 3 \]

So, we need **3 capacitors in series** to achieve `600V`.

### Step 2: Determine the equivalent capacitance for capacitors in series
The formula for the equivalent capacitance \( C_{eq} \) of capacitors in series is:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
For `3` capacitors of `6μF`:
\[ \frac{1}{C_{eq}} = \frac{1}{6μF} + \frac{1}{6μF} + \frac{1}{6μF} = \frac{3}{6μF} \]
\[ C_{eq} = \frac{6μF}{3} = 2μF \]

So, **3 capacitors in series give us an equivalent capacitance of `2μF`** at `600V`.

### Step 3: Determine how many sets of capacitors are needed for `18μF`
Now, we need to achieve a total capacitance of `18μF`. Since we have `2μF` from the series combination, we can connect these combinations in parallel.

To find out how many `2μF` capacitors are needed to make `18μF`:
\[ n \times 2μF = 18μF \]
\[ n = \frac{18μF}{2μF} = 9 \]

So, we need **9 sets of `3 capacitors in series`** to achieve `18μF`.

### Step 4: Calculate the total number of capacitors needed
Since each set consists of `3` capacitors, the total number of capacitors required is:
\[ \text{Total capacitors} = 9 \times 3 = 27 \]

Thus, **27 capacitors of `6μF, 200V` are needed** to create a capacitor of `18μF, 600V`.

### Final Answer
The total number of `6μF, 200V` capacitors needed is **27**.

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