Two capacitors `C_(1) = 2 mF and C_(2) = 8 mF` are seperately charged from the same battery. These two capacitors are then allowed to discharge separately through resistors of same resistance with a switch in series which are kept open initially. Both switches are closed at t = 0.

Options

A. Current at t = 0 will be zero in both the capacitors.
B. At t = 0 current in `C_(1)` will be 4 times the current in `C_(2)`.
C. At t = 0 current in both the capacitors will be the same
D. Capacitor `C_(1)` loses 25% of its initial charge in lesser time than capacitor `C_(2)`.

Explanation

To solve the problem step by step, we will analyze the discharge of the capacitors and the relationships between their charge, current, and time constants.

### Step 1: Understand the Initial Conditions
- We have two capacitors:
- \( C_1 = 2 \, \text{mF} = 2 \times 10^{-3} \, \text{F} \)
- \( C_2 = 8 \, \text{mF} = 8 \times 10^{-3} \, \text{F} \)
- Both capacitors are charged from the same battery, so they have the same initial voltage \( V \).

### Step 2: Calculate the Initial Charge on Each Capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \cdot V \]
- For \( C_1 \):
\[ Q_1 = C_1 \cdot V = 2 \times 10^{-3} \cdot V \]
- For \( C_2 \):
\[ Q_2 = C_2 \cdot V = 8 \times 10^{-3} \cdot V \]

### Step 3: Write the Discharge Equation
The charge on a capacitor discharging through a resistor is given by:
\[ Q(t) = Q_0 e^{-\frac{t}{RC}} \]
Where:
- \( Q_0 \) is the initial charge.
- \( R \) is the resistance.
- \( C \) is the capacitance.

### Step 4: Analyze the Current Flow
The current \( I \) through the capacitor can be expressed as:
\[ I(t) = -\frac{dQ}{dt} = \frac{Q_0}{R} e^{-\frac{t}{RC}} \]
At \( t = 0 \):
- The initial current \( I_0 \) for both capacitors is:
\[ I_0 = \frac{V}{R} \]
Since both capacitors are charged to the same voltage and have the same resistance, the initial current will be the same for both capacitors.

### Step 5: Compare Time Constants
The time constant \( \tau \) for a capacitor is given by:
\[ \tau = R \cdot C \]
- For \( C_1 \):
\[ \tau_1 = R \cdot C_1 = R \cdot 2 \times 10^{-3} \]
- For \( C_2 \):
\[ \tau_2 = R \cdot C_2 = R \cdot 8 \times 10^{-3} \]

Since \( C_1 < C_2 \), we have \( \tau_1 < \tau_2 \). This means \( C_1 \) will discharge faster than \( C_2 \).

### Step 6: Determine Charge Loss
To find out how long it takes for \( C_1 \) to lose 25% of its charge:
- The final charge after losing 25%:
\[ Q_f = Q_1 - 0.25 Q_1 = 0.75 Q_1 \]
Using the discharge equation:
\[ 0.75 Q_1 = Q_1 e^{-\frac{t}{\tau_1}} \]
This simplifies to:
\[ 0.75 = e^{-\frac{t}{\tau_1}} \]
Taking the natural logarithm:
\[ -\frac{t}{\tau_1} = \ln(0.75) \]
Thus:
\[ t = -\tau_1 \ln(0.75) \]

### Conclusion
Since \( \tau_1 < \tau_2 \), \( C_1 \) will lose 25% of its charge in a shorter time than \( C_2 \).