Frequency effect on capacitive reactance: what happens to Xc when the operating frequency of a capacitive circuit is doubled?

Introduction / Context:
Designing filters and timing networks requires a precise understanding of reactance behavior with frequency. Capacitive reactance Xc is inversely proportional to frequency. This question asks about the effect of doubling frequency on Xc, a common quick check in analysis.

Given Data / Assumptions:

• Capacitive reactance formula: Xc = 1 / (2 * pi * f * C).
• Capacitance C is constant.
• Frequency f changes from f to 2f.

Concept / Approach:
Since Xc is inversely proportional to f, scaling f by a factor k scales Xc by 1/k. For k = 2, Xc becomes Xc/2. This relationship is independent of the actual numerical values of f and C: only the ratio matters.

Step-by-Step Solution:
1) Start with Xc = 1 / (2 * pi * f * C).2) Replace f by 2f: Xc(new) = 1 / (2 * pi * 2f * C) = 1 / (2 * (2 * pi * f * C)).3) Simplify: Xc(new) = (1/2) * [1 / (2 * pi * f * C)] = Xc / 2.4) Conclude that Xc is cut in half.

Verification / Alternative check:
Use example numbers: If f = 1000 Hz and C = 1 µF, Xc ≈ 1 / (2 * pi * 1000 * 1e-6) ≈ 159.15 ohm. Doubling f to 2000 Hz yields ≈ 79.6 ohm, half the original.

Why Other Options Are Wrong:
Xc doubling or “no effect” contradicts inverse proportionality. Multiplying by 6.28 confuses constants (2 * pi) with scaling behavior.

Common Pitfalls:
Mixing up inductive and capacitive reactance trends; for inductors, reactance increases with frequency, but capacitors do the opposite.

Final Answer:
Cuts the capacitive reactance in half.