In a series LCR circuit, the phase difference between the voltage and the current is `45^(@)`. Then the power factor will be

Options

A. `0.607`
B. `0.707`
C. `0.808`
D. 1

Explanation

To solve the problem, we need to find the power factor of a series LCR circuit given that the phase difference between the voltage and the current is \( 45^\circ \).

### Step-by-Step Solution:

1. **Understand the Concept of Power Factor**:
The power factor (PF) in an AC circuit is defined as the cosine of the phase angle (\( \phi \)) between the voltage and the current. Mathematically, it is expressed as:
\[ \text{Power Factor (PF)} = \cos(\phi) \]

2. **Identify the Given Information**:
From the problem, we know:
- The phase difference (\( \phi \)) is \( 45^\circ \).

3. **Substitute the Given Phase Angle into the Power Factor Formula**:
We substitute \( \phi \) into the power factor formula:
\[ \text{PF} = \cos(45^\circ) \]

4. **Calculate \( \cos(45^\circ) \)**:
The value of \( \cos(45^\circ) \) is a well-known trigonometric value:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \]

5. **Convert \( \frac{1}{\sqrt{2}} \) to Decimal Form**:
To express the power factor in decimal form:
\[ \frac{1}{\sqrt{2}} \approx 0.707 \]

6. **Conclusion**:
Therefore, the power factor of the circuit is:
\[ \text{Power Factor} = 0.707 \]

### Final Answer:
The power factor is \( 0.707 \).

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