Frequency dependence of impedance in a series RL: As frequency increases in a series RL circuit, how does the magnitude of the impedance |Z| change?

Difficulty: Easy

Correct Answer: It increases because XL = 2 * pi * f * L increases

Explanation:


Introduction / Context:
Understanding frequency dependence is critical when predicting how circuits respond to different signals. In a series RL circuit, the reactive part grows with frequency, changing both the magnitude and the phase of the impedance. This directly affects current draw and filter behavior.


Given Data / Assumptions:

  • Series RL, ideal components.
  • Inductive reactance XL = 2 * pi * f * L.
  • Interest is in impedance magnitude |Z|.


Concept / Approach:
Impedance magnitude for a series RL is |Z| = sqrt(R^2 + XL^2). Because XL is proportional to frequency f, increasing f increases XL and therefore increases |Z|. At low frequency XL is small and |Z| ≈ R; at high frequency XL dominates and |Z| grows approximately linearly with f (when XL ≫ R).


Step-by-Step Solution:

Start with |Z| = sqrt(R^2 + (2 * pi * f * L)^2).Differentiate qualitatively with respect to f: |Z| increases as f increases.Consider limits: f → 0 ⇒ |Z| → R; f → ∞ ⇒ |Z| ≈ 2 * pi * f * L.Therefore, the trend is monotonic increase with frequency.


Verification / Alternative check:
Bode plots show magnitude rising at +20 dB/decade beyond the frequency where XL surpasses R. Bench measurements with an LCR meter exhibit the same trend in the measured impedance.


Why Other Options Are Wrong:
Inductors do not become shorts at high frequency; they impede more. Constant magnitude equal to R applies only at f ≈ 0. There is no resonance in a simple RL (no capacitor), so no oscillatory behavior in |Z|.


Common Pitfalls:
Confusing the RL series with RC trends; thinking resonance exists without a capacitive element.


Final Answer:
It increases because XL = 2 * pi * f * L increases

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