Series RLC above resonance — “In a series RLC circuit, at frequencies above resonance the circuit behaves more capacitive than inductive.” Is this correct?

Electronics RLC Circuits and Resonance Difficulty: Easy
Choose an option
  • A
    True
  • B
    False
  • C
    True only if R is very small
  • D
    True only for DC signals
  • E
    True only for ideal capacitors

Answer

Correct Answer: False

Explanation

Introduction / Context:Knowing how a series RLC behaves relative to its resonant frequency helps in filter and matching design. The sign of net reactance determines whether current leads or lags voltage and which component dominates energy storage.

Given Data / Assumptions:

  • Series RLC (single current path).
  • Sinusoidal steady state.

Concept / Approach:Net series reactance is X = XL − XC, where XL = ωL and XC = 1/(ωC). At resonance, XL = XC and X = 0 (purely resistive). For frequencies above resonance, ω is larger, so XL grows while XC shrinks; therefore XL > XC and X is positive (inductive). Thus the circuit is more inductive above resonance and more capacitive below resonance.

Step-by-Step Solution:

1) Identify resonance condition: XL = XC.2) Increase frequency above f0: XL increases, XC decreases.3) Compute X = XL − XC > 0 → inductive behavior.4) Conclude the statement is false.

Verification / Alternative check:Phasor diagrams show current lagging voltage above resonance (inductive), and leading below resonance (capacitive).

Why Other Options Are Wrong:

Any “True” variants contradict the algebraic combination of reactances for series RLC.

Common Pitfalls:Remembering the rule in reverse; mixing series with parallel behavior.

Final Answer:False

Discussion & Comments
No comments yet. Be the first to comment!
Join Discussion