Capacitive reactance calculation At f = 500 kHz, what capacitor value C gives a capacitive reactance Xc of 1 kΩ? (Use Xc = 1 / (2π f C)).

Introduction / Context:
Designing RF and filter circuits often requires selecting a capacitor to achieve a target reactance at a given frequency. Capacitive reactance decreases with both increasing frequency and capacitance, which guides component choice in coupling, bypassing, and tuning networks.

Given Data / Assumptions:

• Target capacitive reactance Xc = 1 kΩ.
• Frequency f = 500 kHz.
• Ideal capacitor behavior (no parasitics).

Concept / Approach:
For an ideal capacitor, capacitive reactance is given by Xc = 1 / (2 * π * f * C). Rearranging gives C = 1 / (2 * π * f * Xc). Substitute the numerical values carefully, keeping units consistent in SI.

Step-by-Step Solution:
Given f = 500 kHz = 5.0 * 10^5 Hz, Xc = 1.0 * 10^3 Ω.Compute denominator: 2 * π * f * Xc ≈ 2 * 3.1416 * 5.0 * 10^5 * 10^3 = 3.1416 * 10^9.Compute C: C = 1 / (3.1416 * 10^9) ≈ 3.18 * 10^-10 F.Convert to practical unit: 3.18 * 10^-10 F ≈ 318 pF.

Verification / Alternative check:
Back-check: Xc = 1 / (2 * π * 5.0 * 10^5 * 3.18 * 10^-10) ≈ 1000 Ω, confirming the calculation.

Why Other Options Are Wrong:
2 nF: would yield Xc ≈ 159 Ω at 500 kHz, too small a reactance.3.18 µF: would give Xc on the order of 0.1 Ω, far too small.2 F: unrealistic and would act nearly as a short at 500 kHz.

Common Pitfalls:
Unit mistakes with kHz, kΩ, and pico/mega prefixes.Forgetting to use radians-based constant 2π in the formula.

Final Answer:
318 pF