Question

The figure shows variation of R,XL and XC with frequency ' f ' in a series LCR circuit. Then for what frequency point is the circuit capacitive?

• A. B
• B. D
• C. A
• D. C

Answer 1

The Correct Option is C

In a series LCR circuit:

• \( X_L = \omega L \) increases with frequency \( f \)
• \( X_C = \frac{1}{\omega C} \) decreases with frequency \( f \)

The circuit is:

• Inductive if \( X_L > X_C \)
• Capacitive if \( X_C > X_L \)
• At resonance when \( X_L = X_C \)

Step 1: Analyze the Graph

From the graph:

• The point where \( X_L = X_C \) is marked as point C (resonance).
• To the left of point C, \( X_C > X_L \), so the circuit behaves as capacitive.

Step 2: Identify the Correct Point

Among the marked points, point A is to the left of resonance (C), meaning the circuit is capacitive at point A.

Conclusion:

The circuit is capacitive at frequency point \({\text{A}} \), so the correct answer is (C).

Answer 2

In a series LCR circuit, the impedance \( Z \) is given by the equation: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where \( X_L \) is the inductive reactance, \( X_C \) is the capacitive reactance, and \( R \) is the resistance.
When \( X_L > X_C \), the circuit behaves inductively.
When \( X_C > X_L \), the circuit behaves capacitively.
At the resonance point, \( X_L = X_C \), and the circuit becomes purely resistive.

Looking at the given graph:
At low frequencies, \( X_C \) dominates and the circuit behaves capacitively.
As frequency increases, \( X_L \) increases and \( X_C \) decreases. The circuit becomes inductive as \( X_L \) exceeds \( X_C \).

Now, from the graph:
At point A, \( X_C \) is greater than \( X_L \), meaning the circuit is capacitive.
At points B, C, and D, \( X_L \) is greater than \( X_C \), indicating an inductive circuit.

Thus, the circuit is capacitive at point A.