A `4mu F` capacitor is connected to a battery of emf 24V. Through a resistance of `5 M Omega` and a switch which is kept open initially. Internal resistance of the battery is negligible. Switch is closed at t=0. Potential difference across capacitor and resistor at t=0 are respectively.

Options

A. 12V, 12V
B. 0V, 24V
C. 24V, 0V
D. 6V, 18V

Explanation

To solve the problem, we need to determine the potential difference across the capacitor and the resistor at the moment the switch is closed (t=0).

### Step-by-Step Solution:

1. **Understanding the Circuit Configuration**:
- We have a capacitor (C = 4 µF) connected to a battery (emf = 24 V) through a resistor (R = 5 MΩ). The switch is initially open.

2. **Closing the Switch at t=0**:
- When the switch is closed at t=0, the circuit is completed, and current starts to flow.

3. **Initial Condition of the Capacitor**:
- At t=0, the capacitor is uncharged. Therefore, the initial voltage across the capacitor (V_C) is 0 V. This is because a capacitor initially behaves like a short circuit when it has no charge.

4. **Applying Kirchhoff’s Voltage Law**:
- According to Kirchhoff’s Voltage Law, the sum of the potential differences in a closed loop must equal the emf supplied by the battery.
- The total voltage provided by the battery is 24 V. Since the capacitor has 0 V across it at t=0, all the voltage from the battery appears across the resistor.

5. **Calculating the Voltage Across the Resistor**:
- Since the voltage across the capacitor is 0 V, the voltage across the resistor (V_R) must be equal to the emf of the battery:
\[ V_R = V_{battery} - V_C = 24 V - 0 V = 24 V \]

6. **Final Answer**:
- Therefore, at t=0, the potential difference across the capacitor is 0 V and across the resistor is 24 V.

### Summary of Results:
- Voltage across the capacitor (V_C) = 0 V
- Voltage across the resistor (V_R) = 24 V