While working on a physics project at school physics, lab, you require a 4 `mu`F capacitor in a circuit across a potential difference of 1 k V. Unfortunately, 4 `mu`F capacitors are out or stock in your lab but 2 `mu`F capacitors which can withstand a potential difference of 400 V are available in plenty . if you decide to use the 2 `mu`F capacitors in place of `4 mu` F capacitor minimum number of capacitors required are

Options

A. 16
B. 18
C. 20
D. 12

Answer

To solve the problem of how many 2 µF capacitors are needed to replace a 4 µF capacitor while ensuring they can withstand a potential difference of 1000 V, we can follow these steps:

### Step 1: Determine the Required Voltage Rating
The 4 µF capacitor needs to withstand a potential difference of 1000 V. Since the 2 µF capacitors can only withstand 400 V each, we need to connect them in series to increase the voltage rating.

### Step 2: Calculate the Number of Capacitors in Series
To find out how many 2 µF capacitors are needed in series to withstand at least 1000 V, we can set up the equation:
\[ \text{Total Voltage} = n \times V_{\text{max}} \]
Where:
- \( n \) = number of capacitors in series
- \( V_{\text{max}} = 400 \, V \)

We need:
\[ n \times 400 \, V \geq 1000 \, V \]
Solving for \( n \):
\[ n \geq \frac{1000 \, V}{400 \, V} = 2.5 \]
Since \( n \) must be a whole number, we round up to the next whole number:
\[ n = 3 \]
Thus, we need at least 3 capacitors in series to withstand the required voltage.

### Step 3: Calculate the Equivalent Capacitance
When capacitors are connected in series, the equivalent capacitance \( C_{\text{eq}} \) is given by:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
For three 2 µF capacitors:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]
Thus:
\[ C_{\text{eq}} = \frac{2}{3} \, \mu F \]

### Step 4: Determine How Many Sets of Series Capacitors are Needed
Now, we need to create an equivalent capacitance of 4 µF using the equivalent capacitance of the series connection. We can connect these series groups in parallel.

Let \( x \) be the number of series connections needed:
\[ C_{\text{total}} = C_{\text{eq}} \times x = 4 \, \mu F \]
Substituting \( C_{\text{eq}} \):
\[ \frac{2}{3} \, \mu F \times x = 4 \, \mu F \]
Solving for \( x \):
\[ x = 4 \, \mu F \times \frac{3}{2} = 6 \]

### Conclusion
Therefore, the minimum number of 2 µF capacitors required is:
\[ \text{Total Capacitors} = 3 \, \text{(in series)} \times 6 \, \text{(sets)} = 18 \]

### Final Answer
The minimum number of 2 µF capacitors required is **18**.

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