In a series LCR circuit, at resonance, power factor is …….. .

Options

A. `V^2/([omegaL-1/(omegaC)])`
B. `I^2Lomega`
C. `I^2R`
D. `V^2/(Comega`

Explanation

To solve the question regarding the power factor in a series LCR circuit at resonance, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a series LCR circuit, which consists of an inductor (L), a capacitor (C), and a resistor (R) connected in series with an AC voltage source.

2. **Power Factor Definition**:
- The power factor (PF) is defined as the cosine of the phase angle (φ) between the voltage and the current in the circuit. Mathematically, it is given by:
\[ \text{Power Factor (PF)} = \cos \phi = \frac{R}{Z} \]
where \( R \) is the resistance and \( Z \) is the impedance of the circuit.

3. **Impedance in LCR Circuit**:
- The total impedance \( Z \) in a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance.

4. **Condition at Resonance**:
- At resonance, the inductive reactance equals the capacitive reactance:
\[ X_L = X_C \]
This means that \( X_L - X_C = 0 \).

5. **Calculating Impedance at Resonance**:
- Substituting \( X_L = X_C \) into the impedance formula:
\[ Z = \sqrt{R^2 + 0^2} = R \]
Therefore, at resonance, the impedance \( Z \) is equal to the resistance \( R \).

6. **Calculating Power Factor at Resonance**:
- Now substituting \( Z = R \) back into the power factor formula:
\[ \cos \phi = \frac{R}{R} = 1 \]
Hence, the power factor at resonance is:
\[ \text{Power Factor} = 1 \]

### Final Answer:
The power factor in a series LCR circuit at resonance is **1**.

---