Correct Option: (C) capacitive and inductive respectively
Step 1 (Governing Law & Impedance Relation): [0.25 Mark]
The impedance $Z$ of an $LCR$ series circuit (or $LC$ series circuit) is given by:
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$Where the inductive reactance $X_L = \omega L = 2\pi f L$ and capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.
Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]
- Before Resonance ($f < f_r$): As frequency $f$ is low, $X_C = \frac{1}{2\pi f C}$ is large and $X_L = 2\pi f L$ is small. Thus, $X_C > X_L$. The net reactance $(X_L - X_C)$ is negative, making the circuit capacitive.
- At Resonance ($f = f_r$): $X_L = X_C$, so $Z = R$ (minimum impedance).
- After Resonance ($f > f_r$): As frequency $f$ increases, $X_L$ becomes larger and $X_C$ becomes smaller. Thus, $X_L > X_C$. The net reactance $(X_L - X_C)$ is positive, making the circuit inductive.
Step 3 (Final Concluded Answer): [0.25 Mark]
Since the circuit behaves as capacitive when $f < f_r$ and inductive when $f > f_r$, the correct sequence is capacitive and inductive respectively. Therefore, option (C) is correct.
💡 EL10 Concept Insight: Think of the capacitor as a "slow" component that hates high frequencies and the inductor as a "fast" component that hates low frequencies; before resonance, the capacitor dominates the circuit's behavior, but after resonance, the inductor takes over.