Two capacitors `C_1 and C_2 (C_1 gt C_2)` are charged separtately to same potential. Now they are allowed to discharge through similar resistors. Initial rate of discharging will be

Options

A. more for `C_1`
B. more for `C_2`
C. same for both
D. cannot say

Explanation

To solve the problem, we need to analyze the initial rate of discharge of two capacitors, \( C_1 \) and \( C_2 \), which are charged to the same potential and then discharged through similar resistors.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two capacitors \( C_1 \) and \( C_2 \) where \( C_1 > C_2 \). Both capacitors are charged to the same potential \( V_0 \) and are then allowed to discharge through identical resistors.

2. **Formula for Current During Discharge**:
The current \( I \) flowing from a capacitor during discharge can be expressed as:
\[ I(t) = \frac{V_0}{R} e^{-\frac{t}{RC}} \]
where \( R \) is the resistance and \( C \) is the capacitance of the capacitor.

3. **Initial Rate of Discharge**:
We need to find the initial rate of discharge, which corresponds to the current at \( t = 0 \). Substituting \( t = 0 \) into the equation gives:
\[ I(0) = \frac{V_0}{R} e^{0} = \frac{V_0}{R} \]
This means that at \( t = 0 \), the current is simply \( \frac{V_0}{R} \).

4. **Calculating for Both Capacitors**:
- For capacitor \( C_1 \):
\[ I_1(0) = \frac{V_0}{R} \]
- For capacitor \( C_2 \):
\[ I_2(0) = \frac{V_0}{R} \]

5. **Conclusion**:
Since both capacitors are discharged through the same resistor and have the same initial voltage, the initial rate of discharge (current) will be the same for both capacitors:
\[ I_1(0) = I_2(0) = \frac{V_0}{R} \]

### Final Answer:
The initial rate of discharge for both capacitors is the same.