Question

A capacitor and an inductor are connected in series across an ac source of voltage of variable frequency. The frequency is increased continuously. The nature of the circuit before and after the resonance will be:

• A. inductive only
• B. capacitive only
• C. capacitive and inductive respectively
• D. inductive and capacitive respectively

Hint:

In a series LC circuit, the nature of the circuit depends on the frequency relative to the resonance frequency. Below resonance, the circuit is capacitive (\( X_C>X_L \)); above resonance, it is inductive (\( X_L>X_C \)).

Answer 1

The Correct Option is C

Step 1: Understand the impedance of a series LC circuit.
In a series LC circuit, the total impedance \( Z \) is given by: \[ Z = j \left( X_L - X_C \right) \] where \( X_L = \omega L \) (inductive reactance) and \( X_C = \frac{1}{\omega C} \) (capacitive reactance), and \( \omega = 2\pi f \) is the angular frequency. Step 2: Determine the resonance condition.
Resonance occurs when the inductive and capacitive reactances are equal: \[ X_L = X_C \quad \Rightarrow \quad \omega L = \frac{1}{\omega C} \] \[ \omega^2 = \frac{1}{LC} \quad \Rightarrow \quad \omega = \frac{1}{\sqrt{LC}} \] \[ f = \frac{1}{2\pi \sqrt{LC}} \] Step 3: Analyze the circuit before resonance.
Before resonance, the frequency \( f \) is less than the resonance frequency \( f_r \), so \( \omega<\omega_r \): \[ X_L = \omega L<\frac{1}{\omega C} = X_C \] \[ X_L - X_C<0 \quad \Rightarrow \quad \text{the net reactance is negative (capacitive)} \] Thus, the circuit is capacitive before resonance. Step 4: Analyze the circuit after resonance.
After resonance, the frequency \( f \) is greater than the resonance frequency \( f_r \), so \( \omega>\omega_r \): \[ X_L = \omega L>\frac{1}{\omega C} = X_C \] \[ X_L - X_C>0 \quad \Rightarrow \quad \text{the net reactance is positive (inductive)} \] Thus, the circuit is inductive after resonance. Step 5: Match with the options.
Before resonance: capacitive; after resonance: inductive. This matches option (C).