The average power dissipated in a pure inductor `L carrying an alternating current of rms value I is .

Options

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

Explanation

To solve the problem of finding the average power dissipated in a pure inductor carrying an alternating current of RMS value \( I \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Power in AC Circuits**:
In alternating current (AC) circuits, the average power (\( P \)) dissipated can be calculated using 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, and \( \cos \phi \) is the power factor.

2. **Identify the Characteristics of a Pure Inductor**:
In a pure inductor:
- The resistance \( R = 0 \).
- The voltage leads the current by \( 90^\circ \), which means the phase angle \( \phi = 90^\circ \).

3. **Calculate the Power Factor**:
The power factor is defined as:
\[ \cos \phi = \frac{R}{Z} \]
Since \( R = 0 \) for a pure inductor, we have:
\[ \cos \phi = \frac{0}{Z} = 0 \]

4. **Substitute the Power Factor into the Power Formula**:
Now, substituting \( \cos \phi = 0 \) into the power formula:
\[ P = I_{\text{rms}} \times V_{\text{rms}} \times 0 = 0 \]

5. **Conclusion**:
Therefore, the average power dissipated in a pure inductor carrying an alternating current of RMS value \( I \) is:
\[ P_{\text{average}} = 0 \]

### Final Answer:
The average power dissipated in a pure inductor is \( 0 \).