Compute capacitive reactance: a 0.47 µF capacitor across a 2 kHz sine-wave source—what is Xc?
Electronics
Capacitors
Difficulty: Easy
Choose an option
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A170 Ω
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B17 Ω
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C1.7 Ω
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D0.000169 Ω
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E340 Ω
Answer
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