An unchanged capacitor is connected to a battery. On charging the capacitor
(A) All the energy supplied is stored in the capacitor
(B) half the energy supplied is dissipated in form if heat in connecting wire
(D) the energy stored depends upon the time for which the capcitor is charged

Options

A. `A,C`
B. `B,C`
C. `A,D`
D. `C,D`

Explanation

To solve the problem, we need to analyze the statements regarding the charging of a capacitor connected to a battery. Let's break down the statements one by one.

### Step-by-Step Solution:

1. **Understanding the Energy Supplied by the Battery**:
- When a capacitor of capacitance \( C \) is connected to a battery with voltage \( V \), the total charge \( Q \) stored in the capacitor is given by:
\[ Q = C \cdot V \]
- The energy supplied by the battery when charging the capacitor is:
\[ \text{Energy supplied} = Q \cdot V = C \cdot V \cdot V = C V^2 \]

2. **Energy Stored in the Capacitor**:
- The energy stored in the capacitor when fully charged is given by:
\[ \text{Energy stored} = \frac{1}{2} C V^2 \]

3. **Comparison of Supplied Energy and Stored Energy**:
- From the above equations, we see that the energy supplied by the battery is \( C V^2 \) and the energy stored in the capacitor is \( \frac{1}{2} C V^2 \).
- This indicates that not all the energy supplied is stored in the capacitor; in fact, half of it is stored.

4. **Heat Dissipation**:
- The difference between the energy supplied and the energy stored is the energy that is dissipated as heat in the connecting wires:
\[ \text{Energy dissipated} = \text{Energy supplied} - \text{Energy stored} = C V^2 - \frac{1}{2} C V^2 = \frac{1}{2} C V^2 \]
- Thus, half of the energy supplied by the battery is dissipated as heat.

5. **Dependence on Charging Time**:
- The energy stored in the capacitor is given by \( \frac{1}{2} C V^2 \), which is dependent on the capacitance \( C \) and the voltage \( V \), but not on the time for which the capacitor is charged.
- In an ideal scenario with zero resistance, the capacitor charges instantaneously, meaning the energy stored does not depend on the charging time.

### Conclusion:
- **Statement A**: Incorrect - Not all energy is stored in the capacitor.
- **Statement B**: Correct - Half of the energy supplied is dissipated as heat.
- **Statement C**: Incorrect - The energy stored does not depend on the time for which the capacitor is charged.

Therefore, the correct answer is **Statement B**.