The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances 1 pF, 2pF and 4pF is

Options

A. `4/7` pF
B. 1 pF
C. 2 pF
D. `7/4` pE

Answer

To find the minimum value of effective capacitance that can be obtained by combining three capacitors of capacitances 1 pF, 2 pF, and 4 pF, we need to connect them in series. The formula for calculating the equivalent capacitance \( C_{eq} \) for capacitors in series is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Where:
- \( C_1 = 1 \, \text{pF} \)
- \( C_2 = 2 \, \text{pF} \)
- \( C_3 = 4 \, \text{pF} \)

### Step 1: Write the formula for capacitors in series
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

### Step 2: Substitute the values of the capacitors
\[ \frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} \]

### Step 3: Calculate the individual fractions
\[ \frac{1}{C_{eq}} = 1 + 0.5 + 0.25 \]

### Step 4: Find a common denominator and add the fractions
The common denominator for 1, 2, and 4 is 4. Thus, we can rewrite the fractions:
\[ \frac{1}{C_{eq}} = \frac{4}{4} + \frac{2}{4} + \frac{1}{4} = \frac{4 + 2 + 1}{4} = \frac{7}{4} \]

### Step 5: Take the reciprocal to find \( C_{eq} \)
\[ C_{eq} = \frac{4}{7} \, \text{pF} \]

### Conclusion
The minimum value of effective capacitance that can be obtained by combining the three capacitors in series is:
\[ C_{eq} = \frac{4}{7} \, \text{pF} \]