RC high-pass sizing: Given an 820 Ω resistor, choose C so that the capacitive reactance Xc is one-tenth of R at f = 12 kHz.

Introduction / Context:
Designing a high-pass RC filter often requires setting the capacitive reactance Xc relative to the resistor at a chosen frequency. Making Xc much smaller than R (for example, R/10) pushes the corner to a frequency well below the operating band, improving high-pass behavior at that frequency.

Given Data / Assumptions:

• R = 820 Ω.
• Target: Xc = R/10 = 82 Ω.
• Frequency f = 12 kHz.
• Ideal, linear components.

Concept / Approach:
Capacitive reactance is Xc = 1 / (2 * pi * f * C). Solve for C when Xc is specified. Careful unit handling is essential to get microfarads (µF) correctly.

Step-by-Step Solution:
Xc target = 82 ΩC = 1 / (2 * pi * f * Xc)C = 1 / (2 * 3.1416 * 12,000 * 82)C ≈ 1 / (6.183 × 10^6) ≈ 1.617 × 10^-7 FC ≈ 0.161 µF

Verification / Alternative check:
Back-substitute: Xc ≈ 1 / (2π * 12 kHz * 0.161 µF) ≈ 82 Ω, matching the design goal of R/10.

Why Other Options Are Wrong:

• 81 µF, 161 µF, 220 µF: These are orders of magnitude larger than required and would not yield Xc = 82 Ω at 12 kHz.

Common Pitfalls:
Misreading “ten times less than R” (here it means Xc = R/10), or failing to convert units (Hz, F, µF) correctly. Always solve symbolically first, then convert units.

Final Answer:
0.161 µF