Combination of two identical capacitors, a resistor `R` and a dc voltage source of voltage `6V` is used in an experiment on a `(C-R)` circuit. It is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combinatiomn reduces to half its original voltage is `10` second. For series combination the time needed for reducing teh voltage of the fully charged series combination by half is-

Options

A. 20s
B. 10s
C. 5s
D. 2.5s

Explanation

To solve the problem, we need to determine the time required for the voltage of a fully charged series combination of two identical capacitors to reduce to half its original value, given that the time for the parallel combination is 10 seconds.

### Step-by-Step Solution:

1. **Understanding the Time Constant for Parallel Combination**:
- For the parallel combination of capacitors, the equivalent capacitance \( C_P \) is given by:
\[ C_P = C + C = 2C \]
- The time constant \( T_P \) for the parallel combination is:
\[ T_P = R \cdot C_P = R \cdot 2C \]
- The time required for the voltage to reduce to half its original value (half-life) is given by:
\[ t_{1/2} = T_P \cdot \ln(2) = 2RC \cdot \ln(2) \]
- We know from the problem that this time is 10 seconds:
\[ 2RC \cdot \ln(2) = 10 \text{ seconds} \]

2. **Understanding the Time Constant for Series Combination**:
- For the series combination of capacitors, the equivalent capacitance \( C_S \) is given by:
\[ C_S = \frac{C \cdot C}{C + C} = \frac{C}{2} \]
- The time constant \( T_S \) for the series combination is:
\[ T_S = R \cdot C_S = R \cdot \frac{C}{2} \]
- The time required for the voltage to reduce to half its original value is:
\[ t_{1/2} = T_S \cdot \ln(2) = \left(R \cdot \frac{C}{2}\right) \cdot \ln(2) \]

3. **Finding the Relationship Between \( T_S \) and \( T_P \)**:
- We can express \( T_S \) in terms of \( T_P \):
\[ T_S = \frac{1}{4} T_P \]
- Since \( T_P = 2RC \), we have:
\[ T_S = \frac{1}{4} \cdot 2RC = \frac{RC}{2} \]

4. **Calculating the Half-Life for Series Combination**:
- Now substituting \( T_S \) into the equation for half-life:
\[ t_{1/2} = T_S \cdot \ln(2) = \left(\frac{RC}{2}\right) \cdot \ln(2) \]
- From the parallel case, we know:
\[ 2RC \cdot \ln(2) = 10 \text{ seconds} \]
- Therefore:
\[ RC \cdot \ln(2) = 5 \text{ seconds} \]
- Substituting this into the series half-life equation:
\[ t_{1/2} = \left(\frac{5}{2}\right) = 2.5 \text{ seconds} \]

### Final Answer:
The time needed for reducing the voltage of the fully charged series combination by half is **2.5 seconds**.