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:
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:
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
Discussion & Comments