Power dissipated in pure inductance will be-

Options

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

Answer

To determine the power dissipated in a pure inductance, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit**: In a pure inductive AC circuit, we have an alternating current (AC) source connected to an inductor (L).

2. **Power Formula**: The average power (P) in an AC circuit can be calculated using the formula:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi \]
where:
- \( V_{\text{rms}} \) is the root mean square voltage,
- \( I_{\text{rms}} \) is the root mean square current,
- \( \phi \) is the phase difference between the voltage and current.

3. **Phase Difference in Pure Inductance**: In a pure inductive circuit, the voltage leads the current by 90 degrees. Therefore, the phase difference \( \phi \) is 90 degrees.

4. **Calculate Cosine of the Phase Angle**: We need to find \( \cos(90^\circ) \):
\[ \cos(90^\circ) = 0 \]

5. **Substituting into the Power Formula**: Now, substitute \( \cos(90^\circ) \) into the power formula:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \times 0 \]

6. **Conclusion**: Since any number multiplied by zero is zero, we find that:
\[ P = 0 \]
Thus, the power dissipated in a pure inductance is zero.

### Final Answer:
The power dissipated in pure inductance will be **0 watts**.

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