Frequency behavior of a capacitor: how does capacitive reactance (Xc) vary as the operating frequency increases, assuming capacitance remains constant?

Introduction / Context:
Understanding how a capacitor opposes AC is fundamental for filter design, coupling/decoupling, and impedance matching. The parameter that captures this opposition is capacitive reactance, denoted Xc, which depends on both capacitance and frequency. Designers rely on the frequency dependence of Xc to shape signal spectra and stabilize power rails.

Given Data / Assumptions:

• Ideal capacitor with fixed capacitance C.
• Sinusoidal steady-state operation at frequency f.
• We examine magnitude relationships only.

Concept / Approach:
The magnitude of capacitive reactance is Xc = 1 / (2 * pi * f * C). As f increases, the denominator increases; therefore Xc decreases. At DC (f = 0), Xc tends toward infinity, meaning an ideal capacitor blocks steady current. At very high frequencies, Xc can become very small, allowing significant AC current for a given voltage amplitude (subject to real-world ESL/ESR limits).

Step-by-Step Solution:
Write the formula: Xc = 1 / (2 * pi * f * C).Observe inverse proportionality to frequency.Conclude: increasing f causes decreasing Xc.Select option stating “decreases as frequency increases.”

Verification / Alternative check:
Compute a simple example: with C = 1 µF, at 50 Hz Xc ≈ 3183 Ω; at 1 kHz Xc ≈ 159 Ω; at 10 kHz Xc ≈ 15.9 Ω, confirming the inverse relationship.

Why Other Options Are Wrong:
A misstates the DC behavior (Xc is maximum at DC). C denies the clear mathematical dependence. D is the opposite trend. “None” is invalid because the relationship is well known.

Common Pitfalls:
Ignoring nonideal ESR/ESL that dominate at very high frequencies; confusing reactance with resistance; forgetting phase shift of −90 degrees in ideal capacitors.

Final Answer:
Capacitive reactance decreases as the operating frequency increases