The resultant capacitance of n condensers of capacitances `C_(1)C_(2)...........C_(n)` connected in series is given by

Options

A. `C_(s)=1/C_(1)+1/C_(2)+.........+1/C_(n)`
B. `1/C_(s)=1/C_(1)+.........+1/C_(n)`
C. `C_(s)=C_(1)+C_(2)+.........+C_(n)`
D. `C_(s)=C_(1)-C_(2)-.........-C_(n)`

Answer

To find the resultant capacitance of \( n \) capacitors \( C_1, C_2, \ldots, C_n \) connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Series Connection**:
- When capacitors are connected in series, the total voltage across the combination is the sum of the voltages across each capacitor.

2. **Voltage Across Each Capacitor**:
- For a capacitor, the relationship between charge \( Q \), voltage \( V \), and capacitance \( C \) is given by:
\[ V = \frac{Q}{C} \]
- Therefore, for each capacitor \( C_i \) (where \( i = 1, 2, \ldots, n \)), the voltage across it can be expressed as:
\[ V_i = \frac{Q}{C_i} \]

3. **Total Voltage**:
- The total voltage \( V \) across the series combination of capacitors is:
\[ V = V_1 + V_2 + V_3 + \ldots + V_n \]
- Substituting the expressions for \( V_i \):
\[ V = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \ldots + \frac{Q}{C_n} \]

4. **Expressing Total Voltage in Terms of Resultant Capacitance**:
- Let \( C_s \) be the resultant capacitance of the series combination. The total voltage can also be expressed as:
\[ V = \frac{Q}{C_s} \]

5. **Setting the Equations Equal**:
- Since both expressions represent the total voltage \( V \), we can set them equal to each other:
\[ \frac{Q}{C_s} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \ldots + \frac{Q}{C_n} \]

6. **Cancelling \( Q \)**:
- Assuming \( Q \neq 0 \), we can divide both sides by \( Q \):
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots + \frac{1}{C_n} \]

7. **Final Result**:
- Thus, the formula for the resultant capacitance \( C_s \) of \( n \) capacitors in series is:
\[ \frac{1}{C_s} = \sum_{i=1}^{n} \frac{1}{C_i} \]