Compute capacitive reactance: a 0.47 µF capacitor across a 2 kHz sine-wave source—what is Xc?

Difficulty: Easy

Correct Answer: 170 Ω

Explanation:


Introduction / Context:
Capacitive reactance indicates how a capacitor impedes AC at a given frequency. It is critical for sizing coupling capacitors, setting filter cutoffs, and calculating impedance in signal paths.


Given Data / Assumptions:

  • Capacitance C = 0.47 µF = 0.47 × 10^-6 F.
  • Frequency f = 2 kHz.
  • Ideal capacitor (neglect ESR).


Concept / Approach:
The formula for capacitive reactance is Xc = 1 / (2 * π * f * C). Plug in values carefully, keeping units consistent, to avoid power-of-ten mistakes.


Step-by-Step Solution:

Compute the product: 2 * π * f * C = 2 * π * 2000 * 0.47e-6.2000 * 0.47e-6 = 0.00094; multiply by 2π ≈ 6.283 → ≈ 0.005907.Xc = 1 / 0.005907 ≈ 169.3 Ω ≈ 170 Ω.


Verification / Alternative check:
Sanity check: At audio kHz and sub-microfarad capacitance, Xc should be on the order of hundreds of ohms; result is reasonable.


Why Other Options Are Wrong:

  • 17 Ω or 1.7 Ω: Off by decade(s), likely unit conversion errors.
  • 0.000169 Ω: Physically unrealistic; would imply a near-short.
  • 340 Ω: Would correspond to half the frequency or half the capacitance; not our case.


Common Pitfalls:

  • Losing micro (10^-6) in calculations.
  • Rounding early instead of at the end.


Final Answer:
170 Ω

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