For an LCR series circuit with an aac source of angular frequency `omega`.

Options

A. circuit will be capacitive if `omegagt(1)/(sqrt(LC))`
B. circuit will be inductive if `omega=(1)/(sqrt(LC))`
C. Power factor of circuit will be unity if capacitive reactance equals inductive reactance
D. circuit will be leading voltage if `omegagt(1)/(sqrt(LC))`

Explanation

To solve the problem regarding the LCR series circuit with an AC source of angular frequency \( \omega \), we will analyze each of the given options step by step.

### Step 1: Analyze the condition for a capacitive circuit
- The circuit will be capacitive if the capacitive reactance \( X_C \) is greater than the inductive reactance \( X_L \).
- Mathematically, this can be expressed as:
\[ X_C > X_L \implies \frac{1}{\omega C} > \omega L \]
- Rearranging this inequality gives:
\[ \omega < \frac{1}{\sqrt{LC}} \]
- Therefore, the first option is correct: **The circuit will be capacitive if \( \omega < \frac{1}{\sqrt{LC}} \)**.

### Step 2: Analyze the condition for an inductive circuit
- The circuit will be inductive if the inductive reactance \( X_L \) is greater than the capacitive reactance \( X_C \).
- This can be expressed as:
\[ X_L > X_C \implies \omega L > \frac{1}{\omega C} \]
- Rearranging gives:
\[ \omega > \frac{1}{\sqrt{LC}} \]
- Therefore, the second option is correct: **The circuit will be inductive if \( \omega > \frac{1}{\sqrt{LC}} \)**.

### Step 3: Analyze the condition for power factor
- The power factor \( \cos \phi \) is defined as:
\[ \cos \phi = \frac{R}{Z} \]
- The power factor is unity when the impedance \( Z \) is equal to the resistance \( R \), which occurs when:
\[ Z = \sqrt{R^2 + (X_C - X_L)^2} = R \]
- Squaring both sides gives:
\[ R^2 + (X_C - X_L)^2 = R^2 \implies (X_C - X_L)^2 = 0 \implies X_C = X_L \]
- Therefore, the third option is correct: **The power factor of a circuit will be unity if \( X_C = X_L \)**.

### Step 4: Analyze the condition for leading voltage
- The circuit will have a leading voltage if the capacitive reactance \( X_C \) is greater than the inductive reactance \( X_L \).
- This is the same condition we analyzed in Step 1:
\[ X_C > X_L \implies \frac{1}{\omega C} > \omega L \implies \omega < \frac{1}{\sqrt{LC}} \]
- Therefore, the fourth option is incorrect: **The circuit will be leading if \( \omega < \frac{1}{\sqrt{LC}} \)**, not \( \omega > \frac{1}{\sqrt{LC}} \).

### Conclusion
After analyzing all four options, we find that:
- **Correct Options**: A, B, and C
- **Incorrect Option**: D

### Final Answer
The correct answer is that options A, B, and C are correct.

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