Difficulty: Easy
Correct Answer: both series and parallel RC impedance decrease
Explanation:
Introduction / Context:
RC networks are fundamental frequency-selective elements. Understanding how their impedances vary with frequency is essential for filter design, coupling/decoupling, and timing applications. This question contrasts series and parallel RC behaviors as frequency rises.
Given Data / Assumptions:
Concept / Approach:
As frequency increases, Xc decreases. In a series RC, total impedance magnitude |Z_series| = sqrt(R^2 + Xc^2) falls because the reactive term shrinks. In a parallel RC, the total admittance increases because the capacitive susceptance (1 / Xc) grows with frequency, so the equivalent impedance |Z_parallel| = 1 / |Y| decreases as well.
Step-by-Step Solution:
Compute Xc = 1 / (2 * pi * f * C); increasing f → decreasing Xc.Series RC: |Z_series| = sqrt(R^2 + Xc^2) → decreases as Xc decreases.Parallel RC: Y_total = 1 / R + 1 / (jXc); |Y| increases with frequency, so |Z_parallel| = 1 / |Y| decreases.Therefore, both series and parallel RC impedances decrease with increasing frequency.
Verification / Alternative check:
Limiting cases: as f → 0, Xc → infinity; series RC behaves like largely resistive with large |Z|, parallel RC tends toward R (capacitor open). As f → infinity, Xc → 0; series RC tends to R (smaller than at low f), parallel RC tends to near 0 (capacitor acts short), confirming decreasing trends.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
both series and parallel RC impedance decrease
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