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 solve the problem of when the voltage leads the current in a series LCR circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding 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 relationship between the inductive reactance (X_L) and the capacitive reactance (X_C).

2. **Voltage Relationships**: The voltage across the inductor (V_L) leads the current (I) by 90 degrees, while the voltage across the capacitor (V_C) lags the current by 90 degrees. For the voltage to lead the current, the voltage across the inductor must be greater than the voltage across the capacitor:
\[ V_L > V_C \]

3. **Expressing Voltages**: The voltages across the inductor and capacitor can be expressed in terms of current (I):
\[ V_L = I \cdot X_L \quad \text{and} \quad V_C = I \cdot X_C \]

4. **Setting Up the Inequality**: From the condition \( V_L > V_C \), we can substitute the expressions for the voltages:
\[ I \cdot X_L > I \cdot X_C \]
Since the current (I) is common and non-zero, we can divide both sides by I:
\[ X_L > X_C \]

5. **Substituting Reactances**: The inductive reactance \( X_L \) and capacitive reactance \( X_C \) are given by:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]
Thus, we can rewrite the inequality:
\[ \omega L > \frac{1}{\omega C} \]

6. **Rearranging the Inequality**: Multiplying both sides by \( \omega \) (assuming \( \omega > 0 \)):
\[ \omega^2 L > \frac{1}{C} \]

7. **Final Form**: Rearranging gives us:
\[ \omega^2 > \frac{1}{LC} \]
Taking the square root of both sides, we find:
\[ \omega > \frac{1}{\sqrt{LC}} \]

8. **Resonant Frequency**: The resonant angular frequency \( \omega_0 \) is defined as:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
Therefore, we conclude that:
\[ \omega > \omega_0 \]

### Conclusion:
In a series LCR circuit, the voltage leads the current when the angular frequency \( \omega \) is greater than the resonant angular frequency \( \omega_0 \).