Discussion
For that break to fall at \( 20 \ \mathrm{Hz} \) instead, the capacitor may be reduced to about
Answer: D
The capacitance needed scales inversely with the frequency wanted:
\[ C = \frac{1}{2\pi (20)(15000)} \approx 0.53 \ \mathrm{\mu F} \]
so about 0.53 microfarads. Two microfarads would push the break lower rather than higher, which is the direction error to watch for: a larger capacitor extends the response downwards. Reading off which way a component change moves a break frequency is worth doing before any arithmetic.
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