Two equal capacitors in series across a 10 kHz sine source: with two 0.68 µF capacitors in series, what is the total capacitive reactance?

Introduction / Context:
Capacitive reactance determines how a capacitor impedes AC. When capacitors are placed in series, their equivalent capacitance drops. This problem connects series-capacitance calculation to reactance at a specified frequency, common in filter and coupling design.

Given Data / Assumptions:

• Two capacitors, each C = 0.68 µF, in series.
• Frequency f = 10 kHz.
• Ideal components; no ESR or parasitics.

Concept / Approach:
For equal capacitors in series, the equivalent capacitance is C_eq = C/2. Capacitive reactance is Xc = 1 / (2 * π * f * C_eq). Compute C_eq first, then apply the reactance formula.

Step-by-Step Solution:

Verification / Alternative check:
Check scale: A few tenths of microfarads at 10 kHz gives tens of ohms reactance—reasonable. If you incorrectly used a single 0.68 µF value, you would get half the correct reactance, which flags a series-capacitance mistake.

Why Other Options Are Wrong:

• 23.41 Ω or 11.70 Ω: These correspond to using 0.68 µF (not the series equivalent) or additional division errors.
• 4.68 Ω: Off by a factor of 10; likely a frequency or unit slip.

Common Pitfalls:

• Forgetting that two equal capacitors in series halve the capacitance.
• Unit conversion errors (µF to F) leading to decade mistakes.

Final Answer:
46.8 Ω