Series RC circuits — does the total impedance vary directly with frequency?

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:This statement probes how the magnitude of impedance in a series RC circuit changes with frequency. Understanding this relationship is fundamental for filter design, Bode plots, and time-constant intuition in electronics. Given Data / Assumptions: Series RC network driven by a sinusoidal source of angular frequency ω = 2 * π * f. Impedance of resistor R is R (independent of frequency). Impedance of capacitor C is Zc = 1 / (j * ω * C). Concept / Approach:The total series impedance is the vector sum of R and the capacitive reactance magnitude Xc = 1 / (ω * C). As frequency increases, Xc decreases. The impedance magnitude therefore decreases toward R, not increases “directly with frequency.” Step-by-Step Solution: Write total impedance: Z = R − j * (1 / (ω * C)).Compute magnitude: |Z| = sqrt( R^2 + (1 / (ω * C))^2 ).Observe dependence: as ω increases, (1 / (ω * C)) decreases, so |Z| monotonically decreases toward R.At low ω → 0, |Z| → very large (dominated by capacitor); at high ω → ∞, |Z| → R. Verification / Alternative check:Sketch a Bode magnitude plot for |Z|. The curve slopes downward with a corner near ω = 1 / (R * C), confirming inverse dependence on frequency due to the 1 / (ω * C) term. Why Other Options Are Wrong: Choosing “True” implies |Z| grows with f, contradicting |Z| = sqrt( R^2 + (1 / (ω * C))^2 ). Common Pitfalls:Confusing impedance of a capacitor (inversely proportional to frequency) with that of an inductor (directly proportional to frequency). Only inductive reactance Xl = ω * L increases with frequency. Final Answer:False

Series RC circuits — does the total impedance vary directly with frequency?

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