Parallel RL circuit — impedance vs frequency trend: As frequency decreases in a parallel RL circuit (R in parallel with an inductor), the magnitude of the circuit impedance will ______.

Difficulty: Easy

Correct Answer: decrease

Explanation:


Introduction / Context:
Parallel RL networks are common as loads and bias elements. Their input impedance varies with frequency because the inductor's reactance changes with ω. Recognizing the trend helps you predict loading and filter behavior.


Given Data / Assumptions:

  • Topology: R in parallel with L.
  • Linear components at small-signal levels.
  • We examine how |Z_in| changes as frequency decreases.


Concept / Approach:
Inductive reactance X_L = 2πfL. At low frequency (f → 0), X_L → 0 Ω, so the inductor behaves like a short. The parallel of any finite R with nearly 0 Ω approaches 0 Ω, so the total impedance decreases toward zero. At high frequency, X_L becomes large (approaching open circuit), so the parallel pair approaches R. Thus, as frequency decreases from high to low, |Z| decreases from approximately R toward 0 Ω.


Step-by-Step Solution:

Compute X_L = 2πfL.At low f: X_L ≈ 0 → Z_total ≈ 0 (short-dominated).At high f: X_L ≫ R → Z_total ≈ R (resistor-dominated).Therefore, decreasing frequency decreases |Z|.


Verification / Alternative check:
Plot |Z| vs f using Z = (jωL * R) / (R + jωL). The curve drops toward zero as f approaches DC.


Why Other Options Are Wrong:

Increase/constant: contradicts the limiting behavior at f → 0.Unpredictable: the dependence is monotonic and well-defined.


Common Pitfalls:
Confusing series and parallel behaviors; forgetting that inductors look like shorts at DC.


Final Answer:
decrease

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