Discussion
A coupling capacitor works into \( 10 \ \mathrm{k\Omega} \) total. For a \( 20 \ \mathrm{Hz} \) break it must be at least about
Answer: C
Rearranging the break frequency relation:
\[ \begin{aligned} C &= \frac{1}{2\pi f R} \\ &= \frac{1}{2\pi(20)(10^{4})} \\ &\approx 0.8 \ \mathrm{\mu F} \end{aligned} \]
giving about 0.8 microfarads. The others misplace a decade. Because the capacitance needed is inversely proportional to both frequency and resistance, low-frequency response into a low impedance is what drives capacitor values up. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.
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