Average power dissipation in a pure capacitor is:

Options

A. `(1)/(2)CV^(2)`
B. `(1)/(4)CV^(2)`
C. Zero
D. `CV^(2)`

Explanation

To determine the average power dissipation in a pure capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Power Dissipation**:
In an AC circuit, the average power dissipation can be calculated using the formula:
\[ P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi \]
where \( V_{\text{rms}} \) is the root mean square voltage, \( I_{\text{rms}} \) is the root mean square current, and \( \phi \) is the phase difference between the voltage and current.

2. **Phase Difference in a Capacitor**:
In a pure capacitor, the current leads the voltage by 90 degrees. This means that the phase angle \( \phi \) is 90 degrees.

3. **Calculating Cosine of the Phase Angle**:
The cosine of 90 degrees is:
\[ \cos(90^\circ) = 0 \]

4. **Substituting into the Power Formula**:
Substituting \( \cos(90^\circ) = 0 \) into the power dissipation formula:
\[ P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot 0 = 0 \]

5. **Conclusion**:
Therefore, the average power dissipation in a pure capacitor is:
\[ P_{\text{avg}} = 0 \]

### Final Answer:
The average power dissipation in a pure capacitor is **0**.