Three capacitors `C_(a) t c_(b) lt C_(c)` are connected in series. Their resultant capacitance will be

Options

A. equivalent capacitance greater than `C_c`
B. equivalent capacitance less than `C_c` but greater then `C_a`
C. equivalent capacitance less than `C_c`
D. equivalent capacitance is infinite

Answer

To find the resultant capacitance of three capacitors \( C_a \), \( C_b \), and \( C_c \) connected in series, we can follow these steps:

### Step 1: Write the formula for capacitance in series
For capacitors connected in series, the formula for the resultant capacitance \( C_{eff} \) is given by:
\[ \frac{1}{C_{eff}} = \frac{1}{C_a} + \frac{1}{C_b} + \frac{1}{C_c} \]

### Step 2: Substitute the values
Substituting the values of the capacitors into the formula:
\[ \frac{1}{C_{eff}} = \frac{1}{C_a} + \frac{1}{C_b} + \frac{1}{C_c} \]

### Step 3: Analyze the inequalities
Given that \( C_a < C_b < C_c \), we can infer the following:
- \( \frac{1}{C_a} > \frac{1}{C_b} > \frac{1}{C_c} \)

### Step 4: Establish bounds for \( C_{eff} \)
From the inequalities, we can derive:
- Since \( \frac{1}{C_a} \) is the largest, we can say:
\[ \frac{1}{C_{eff}} < \frac{3}{C_a} \]
This implies:
\[ C_{eff} > \frac{C_a}{3} \]

- Similarly, since \( \frac{1}{C_c} \) is the smallest:
\[ \frac{1}{C_{eff}} > \frac{3}{C_c} \]
This implies:
\[ C_{eff} < \frac{C_c}{3} \]

### Step 5: Combine the inequalities
Combining these results, we find:
\[ \frac{C_a}{3} < C_{eff} < \frac{C_c}{3} \]

### Step 6: Conclusion
Since \( C_a < C_b < C_c \), we can conclude that the effective capacitance \( C_{eff} \) is bounded by \( C_a \) and \( C_c \). Therefore, the resultant capacitance will be less than \( C_c \) and greater than \( C_a \).

### Final Result
Thus, the resultant capacitance \( C_{eff} \) of the three capacitors in series is:
\[ C_{a}/3 < C_{eff} < C_{c}/3 \]