In series LCR circuit voltage leads the current. When
(Given that `omega_(0)` = resonant angular frequency)

Options

A. `omega lt omega_(0)`
B. `omega = omega_(0)`
C. `omega gt omega_(0)`
D. None of these

Answer

To determine the condition under which the voltage leads the current in a series LCR circuit, we need to analyze the relationship between the angular frequency (ω) and the resonant angular frequency (ω₀).

### Step-by-Step Solution:

1. **Understand the Circuit**: In a series LCR circuit, we have an inductor (L), a capacitor (C), and a resistor (R) connected in series. The behavior of the circuit is influenced by the frequency of the applied voltage.

2. **Resonant Frequency**: The resonant angular frequency (ω₀) is given by the formula:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
At this frequency, the inductive reactance (X_L = ωL) and capacitive reactance (X_C = \frac{1}{ωC}) are equal, leading to a purely resistive circuit.

3. **Phase Relationship**: In an LCR circuit:
- If the applied frequency (ω) is less than the resonant frequency (ω₀), the circuit is capacitive, and the current leads the voltage.
- If the applied frequency (ω) is greater than the resonant frequency (ω₀), the circuit is inductive, and the voltage leads the current.

4. **Condition for Voltage Leading Current**: Therefore, for the voltage to lead the current, the condition must be:
\[ \omega > \omega_0 \]

5. **Conclusion**: Hence, the final answer is that in a series LCR circuit, the voltage leads the current when the applied frequency (ω) is greater than the resonant frequency (ω₀).

### Final Answer:
Voltage leads the current in a series LCR circuit when:
\[ \omega > \omega_0 \]