Numerical AC analysis: what is the capacitive reactance Xc of a 0.1 μF capacitor operating at f = 1000 Hz? Provide the closest standard-value option.

Introduction / Context:
Capacitive reactance quantifies how a capacitor resists changes in voltage in AC circuits. Computing Xc quickly is essential for filter design, impedance matching, and estimating current draw at a given frequency.

Given Data / Assumptions:

• Capacitance C = 0.1 μF = 0.1 * 10^-6 F = 1 * 10^-7 F.
• Frequency f = 1000 Hz.
• Ideal capacitor; no series resistance or parasitics considered.

Concept / Approach:
Use Xc = 1 / (2 * pi * f * C). This relation comes from the frequency-domain impedance of a capacitor, Zc = 1 / (j * 2 * pi * f * C), with magnitude Xc equal to the reciprocal of 2 * pi * f * C.

Step-by-Step Solution:
Step 1: Compute the product k = 2 * pi * f * C.Step 2: Substitute values: k = 2 * pi * 1000 * (1 * 10^-7) = 2 * pi * 10^-4 ≈ 6.2832 * 10^-4.Step 3: Compute Xc = 1 / k ≈ 1 / (6.2832 * 10^-4) ≈ 1591.55 Ω.Step 4: Round to the nearest listed option: 1590 Ohm.

Verification / Alternative check:
A quick mental check: doubling frequency halves Xc; at 1 kHz and 0.1 μF, designers often memorize Xc ≈ 1.6 kΩ, matching the computed result.

Why Other Options Are Wrong:

• Less than 1 Ohm: Two orders of magnitude too small.
• 312 Ohm or 690 Ohm: Do not follow from Xc = 1 / (2 * pi * f * C) with given values.
• None of the above: Incorrect because 1590 Ohm is correct.

Common Pitfalls:
Misplacing powers of ten when converting microfarads to farads causes large errors. Always convert 0.1 μF to 1 * 10^-7 F before calculation.

Final Answer:
1590 Ohm.