The average power dissipation in a pure capacitance in `AC` circuit is

Options

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

Answer

To find the average power dissipation in a pure capacitance in an AC circuit, we can follow these steps:

### Step 1: Understand the formula for average power
The average power \( P \) in an AC circuit is given by the formula:
\[ P = I_{\text{rms}} \times V_{\text{rms}} \times \cos \phi \]
where:
- \( I_{\text{rms}} \) is the root mean square current,
- \( V_{\text{rms}} \) is the root mean square voltage,
- \( \phi \) is the phase difference between the current and voltage.

### Step 2: Determine the phase angle for a pure capacitive circuit
In a pure capacitive circuit, the current leads the voltage by \( 90^\circ \). Therefore, the phase angle \( \phi \) is:
\[ \phi = 90^\circ \]

### Step 3: Calculate the cosine of the phase angle
Now, we need to find \( \cos \phi \):
\[ \cos 90^\circ = 0 \]

### Step 4: Substitute the values into the power formula
Substituting \( \cos \phi \) into the power formula, we get:
\[ P = I_{\text{rms}} \times V_{\text{rms}} \times \cos 90^\circ = I_{\text{rms}} \times V_{\text{rms}} \times 0 \]

### Step 5: Conclude the average power dissipation
Since the product is zero, we conclude that:
\[ P = 0 \]

Thus, the average power dissipation in a pure capacitance in an AC circuit is \( 0 \).

### Final Answer:
The average power dissipation in a pure capacitance in an AC circuit is \( 0 \).

---