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 the equivalent capacitance \( C_{eq} \) of capacitors in series is given by:

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

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

### Step 1: Write down 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
\[ \frac{1}{C_{eq}} = \frac{1}{1 \, \text{pF}} + \frac{1}{2 \, \text{pF}} + \frac{1}{4 \, \text{pF}} \]

### Step 3: Calculate each term
Calculating each term gives:
\[ \frac{1}{1 \, \text{pF}} = 1 \]
\[ \frac{1}{2 \, \text{pF}} = 0.5 \]
\[ \frac{1}{4 \, \text{pF}} = 0.25 \]

### Step 4: Sum the terms
Now, add these values together:
\[ \frac{1}{C_{eq}} = 1 + 0.5 + 0.25 = 1.75 \]

### Step 5: Take the reciprocal to find \( C_{eq} \)
Now, take the reciprocal to find \( C_{eq} \):
\[ C_{eq} = \frac{1}{1.75} \, \text{pF} \]

### Step 6: Simplify the fraction
To simplify \( \frac{1}{1.75} \):
\[ C_{eq} = \frac{1}{\frac{7}{4}} = \frac{4}{7} \, \text{pF} \]

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