Difficulty: Medium
Correct Answer: series RC circuit
Explanation:
Introduction / Context:
Series RLC circuits display different qualitative behaviors depending on frequency relative to resonance. Recognizing whether the circuit behaves more like RL or RC helps predict current phase, impedance magnitude, and filter characteristics without exhaustive mathematics.
Given Data / Assumptions:
Concept / Approach:
Reactances: X_L = ωL (increases with frequency) and X_C = 1/(ωC) (decreases with frequency). Below resonance, ω is small, so X_L is relatively small, and X_C is relatively large. In a series sum X_total = R + j(X_L − X_C), the capacitive magnitude tends to dominate (X_C > X_L), yielding a net capacitive reactance. Thus, the circuit looks like a series RC (capacitive) network from a phase perspective (current leads voltage).
Step-by-Step Solution:
1) Write reactances: X_L = ωL; X_C = 1/(ωC).2) Evaluate at ω < ω0: X_C is large; X_L is smaller.3) Net reactance: X = X_L − X_C is negative → capacitive behavior.4) A series network with dominant capacitive behavior resembles a series RC circuit.
Verification / Alternative check:
Bode plots of phase for a series RLC show phase leading below resonance, consistent with capacitive dominance. Above resonance, phase lags due to inductive dominance, confirming the mirror behavior.
Why Other Options Are Wrong:
Series RL corresponds to inductive dominance (true above resonance). “Purely resistive” occurs only at exact resonance in an ideal series RLC. Option C does not answer the “below resonance” case.
Common Pitfalls:
Mixing series and parallel intuitions; remember, series below resonance → capacitive, parallel below resonance → inductive.
Final Answer:
series RC circuit
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