The resultant capacity of n condensers of capacitances `C_(1),C_(2)........C_(n)` connected in parallel is

Options

A. `C_(p)=C_(1)+C_(2)+........+C_(n)`
B. `C_(p)=C_(1)-C_(2)-C_(3)-........-C_(n)`
C. `1/C_(p)=1/C_(1)+1/C_(2)+........+1/C_(n)`
D. `C_(p)=1/C_(1)+1/C_(2)+........+1/C_(n)`

Answer

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

### Step-by-Step Solution:

1. **Understanding Parallel Connection**: In a parallel connection, all capacitors are connected across the same two points, which means they all have the same voltage across them.

2. **Charge on Each Capacitor**: The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

3. **Charge on Individual Capacitors**: For each capacitor connected in parallel:
- The charge on capacitor \( C_1 \) is \( Q_1 = C_1 \times V \)
- The charge on capacitor \( C_2 \) is \( Q_2 = C_2 \times V \)
- The charge on capacitor \( C_3 \) is \( Q_3 = C_3 \times V \)
- Continuing this way, the charge on capacitor \( C_n \) is \( Q_n = C_n \times V \)

4. **Total Charge in the Circuit**: The total charge \( Q \) supplied by the equivalent capacitor \( C_p \) connected to the same voltage \( V \) is:
\[ Q = C_p \times V \]

5. **Summing the Charges**: Since the total charge in the circuit is the sum of the charges on each capacitor, we can write:
\[ Q = Q_1 + Q_2 + Q_3 + \ldots + Q_n \]
Substituting the expressions for \( Q_1, Q_2, \ldots, Q_n \):
\[ Q = (C_1 \times V) + (C_2 \times V) + (C_3 \times V) + \ldots + (C_n \times V) \]

6. **Factoring Out Voltage**: We can factor out \( V \) from the right side:
\[ Q = V \times (C_1 + C_2 + C_3 + \ldots + C_n) \]

7. **Equating the Charges**: Now we can equate the two expressions for \( Q \):
\[ C_p \times V = V \times (C_1 + C_2 + C_3 + \ldots + C_n) \]

8. **Cancelling Voltage**: Since \( V \) is common and non-zero, we can cancel it from both sides:
\[ C_p = C_1 + C_2 + C_3 + \ldots + C_n \]

9. **Final Result**: Therefore, the resultant capacitance \( C_p \) of \( n \) capacitors connected in parallel is given by:
\[ C_p = C_1 + C_2 + C_3 + \ldots + C_n \]