Frequency dependence of impedance in a series RL circuit Consider a series resistor–inductor (RL) circuit driven by a sinusoidal source. Does the magnitude of its total impedance vary inversely with frequency, or does it increase with frequency due to the inductor's reactance?

Introduction / Context:
Impedance in AC circuits indicates how much a circuit opposes the flow of sinusoidal current. In a series RL circuit, impedance depends on both resistance and inductive reactance. The statement claims an inverse relationship between impedance magnitude and frequency, which we must examine carefully.

Given Data / Assumptions:

• Series RL circuit containing R (ohms) and L (henry).
• Sinusoidal steady state with frequency f (hertz).
• Inductive reactance: XL = 2 * pi * f * L.
• Total impedance magnitude: |Z| = sqrt(R^2 + XL^2).

Concept / Approach:

For inductors, reactance increases linearly with frequency. As frequency rises, XL grows; as frequency falls toward zero, XL shrinks to zero. Therefore, the impedance magnitude of an RL series circuit should increase with frequency, not vary inversely.

Step-by-Step Solution:

Verification / Alternative check:

Take R = 10 Ω, L = 10 mH. At f = 50 Hz, XL ≈ 3.14 Ω, |Z| ≈ 10.5 Ω. At f = 1 kHz, XL ≈ 62.8 Ω, |Z| ≈ 63.6 Ω. The impedance clearly increases with frequency, contradicting the inverse claim.

Why Other Options Are Wrong:

• “True” and conditional variants are incorrect; there is no resonance in a simple RL series circuit, and the behavior is not inverse at low frequency.
• “True for DC but false for AC” is wrong because at DC, impedance equals R, and for AC it increases with f.

Common Pitfalls:

Confusing RL with RC: in RC, the capacitive reactance decreases with frequency, but that is not “inverse impedance” of the entire circuit; it affects only the capacitive part. Another pitfall is assuming “inverse with frequency” because current may decrease with frequency; current reduction does not imply impedance varies inversely.

Final Answer:

False