Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at t=0 as shown in figure.The charges on the capacitor at a time `t=CR` are, respectively,

A. VC,VC
B. VC/e,VC
C. VC,VC/e
D. VC/e,VC/e

Options

A. VC,VC
B. VC/e,VC
C. VC,VC/e
D. VC/e,VC/e

Answer

To solve the problem, we need to analyze the behavior of two identical capacitors A and B, which are charged to the same potential V and connected in two different circuits at time t=0.

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- Capacitor A is connected in a circuit with a diode in forward bias, meaning the positive terminal of the capacitor is connected to the anode of the diode.
- Capacitor B is connected in a circuit with a diode in reverse bias, meaning the positive terminal of the capacitor is connected to the cathode of the diode.

2. **Initial Charge on Capacitors**:
- Both capacitors are initially charged to the same potential V. Therefore, the initial charge (Q) on each capacitor is given by:
\[ Q = C \times V \]
where C is the capacitance.

3. **Behavior of Capacitor A (Forward Biased)**:
- In the forward bias condition, the diode allows current to flow. The voltage across the capacitor will decrease over time as it discharges through the resistor R.
- The voltage across the capacitor as a function of time (t) is given by:
\[ V(t) = V_0 \cdot e^{-t/(RC)} \]
where \( V_0 = V \).

4. **Calculate Voltage at t = CR**:
- Substitute \( t = CR \) into the voltage equation:
\[ V(CR) = V \cdot e^{-CR/(RC)} = V \cdot e^{-1} = \frac{V}{e} \]
- The charge on capacitor A at this time is:
\[ Q_A = C \cdot V(CR) = C \cdot \frac{V}{e} = \frac{CV}{e} \]

5. **Behavior of Capacitor B (Reverse Biased)**:
- In the reverse bias condition, the diode blocks current flow, effectively making the circuit open. Therefore, the charge on capacitor B remains unchanged.
- The charge on capacitor B at time t = CR is still:
\[ Q_B = C \cdot V = CV \]

6. **Final Charges on Capacitors**:
- At time \( t = CR \):
- Charge on Capacitor A: \( Q_A = \frac{CV}{e} \)
- Charge on Capacitor B: \( Q_B = CV \)

### Conclusion:
The charges on the capacitors at time \( t = CR \) are:
- Capacitor A: \( \frac{CV}{e} \)
- Capacitor B: \( CV \)

Thus, the correct answer is **C. \( CV, \frac{CV}{e} \)**.