Parallel RLC behavior — does the smaller reactance branch determine the net reactance of the overall circuit?

Difficulty: Easy

Correct Answer: True

Explanation:


Introduction / Context:
In parallel reactive networks, branch currents depend on branch reactances. Understanding which branch dominates the net behavior is important for filter design and tuning around resonance.


Given Data / Assumptions:

  • Two reactive branches in parallel, one inductive and one capacitive, with finite resistive loss neglected for the conceptual argument.
  • Sinusoidal steady state and linear components.


Concept / Approach:

In parallel, admittances add. The branch with smaller reactance magnitude has larger susceptance magnitude and therefore draws more current for the same applied voltage. Consequently, that branch dominates the net reactive behavior, effectively determining whether the total appears predominantly inductive or capacitive away from exact resonance.


Step-by-Step Solution:

Compute branch susceptances: B_L = −1 / XL and B_C = +1 / XC (signs per convention).Total susceptance: B_total = B_C + B_L.If |XC| < |XL| (capacitor reactance smaller), then |B_C| > |B_L| and the circuit is net capacitive.If |XL| < |XC|, the inductor dominates and the circuit is net inductive.


Verification / Alternative check:

The equivalent reactance magnitude of parallel reactances is less than the smaller of the branch magnitudes. Measurements will show current primarily in the smaller reactance branch, confirming dominance.


Why Other Options Are Wrong:

  • Dominance is not restricted to resonance; it holds for any frequency where branch magnitudes differ.
  • Nonzero resistance modifies Q but does not reverse which branch dominates reactively.
  • Both inductive and capacitive branches can dominate depending on frequency.


Common Pitfalls:

Interpreting “determine” as “equal to.” The smaller reactance does not set the exact value; it dominates the net behavior. The exact equivalent must be computed from admittances.


Final Answer:

True

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