An electric circuit requires a total capacitance of 2mF across a potential of 1000 V. Large number of 1mF capacitances are available each of which would breakdown if the potential is more than 350 V. How many capacitances are required to make the circuit

Options

A. 24
B. 20
C. 18
D. 12

Explanation

To solve the problem, we need to determine how many 1 mF capacitors are required to achieve a total capacitance of 2 mF while ensuring that the voltage across each capacitor does not exceed 350 V. Here's the step-by-step solution:

### Step 1: Determine the configuration of capacitors
Since the total voltage required is 1000 V and each capacitor can only withstand a maximum of 350 V, we will need to connect the capacitors in series. In a series configuration, the total voltage is the sum of the voltages across each capacitor.

**Hint:** Remember that in series, the voltage across each capacitor adds up.

### Step 2: Calculate the number of capacitors in series
To find out how many capacitors are needed in series to handle 1000 V, we can use the formula:
\[ \text{Number of capacitors in series} = \frac{\text{Total voltage}}{\text{Voltage rating per capacitor}} \]
Substituting the values:
\[ \text{Number of capacitors in series} = \frac{1000 \text{ V}}{350 \text{ V}} \approx 2.857 \]
Since we cannot have a fraction of a capacitor, we round up to 3 capacitors in series.

**Hint:** Always round up when dealing with capacitors in series to ensure you do not exceed the voltage limit.

### Step 3: Calculate the equivalent capacitance of capacitors in series
The formula for the equivalent capacitance \( C_{eq} \) of capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
For 3 capacitors of 1 mF each:
\[ \frac{1}{C_{eq}} = \frac{1}{1 \text{ mF}} + \frac{1}{1 \text{ mF}} + \frac{1}{1 \text{ mF}} = \frac{3}{1 \text{ mF}} \]
Thus,
\[ C_{eq} = \frac{1}{3} \text{ mF} \approx 0.333 \text{ mF} \]

**Hint:** The equivalent capacitance of capacitors in series is always less than the smallest individual capacitor.

### Step 4: Determine how many series groups are needed to achieve 2 mF
Now, we need to achieve a total capacitance of 2 mF. Since each series group of 3 capacitors gives us 0.333 mF, we can determine how many such groups are needed:
\[ \text{Number of series groups} = \frac{2 \text{ mF}}{0.333 \text{ mF}} \approx 6 \]

**Hint:** When combining capacitors, ensure to check how many groups are needed to reach the desired capacitance.

### Step 5: Calculate the total number of capacitors required
Since each series group consists of 3 capacitors, the total number of capacitors required is:
\[ \text{Total capacitors} = \text{Number of series groups} \times \text{Capacitors per group} \]
\[ \text{Total capacitors} = 6 \times 3 = 18 \]

**Hint:** Multiply the number of series groups by the number of capacitors in each group to find the total.

### Final Answer
Thus, the total number of 1 mF capacitors required to achieve a total capacitance of 2 mF across a potential of 1000 V is **18 capacitors**.