Discussion
A coupling capacitor of \( 1 \ \mathrm{\mu F} \) works into \( 15 \ \mathrm{k\Omega} \). Its break frequency is about
Answer: C
Setting reactance equal to resistance:
\[ \begin{aligned} f &= \frac{1}{2\pi R C} \\ &= \frac{1}{2\pi (15000)(10^{-6})} \\ &\approx 10.6 \ \mathrm{Hz} \end{aligned} \]
giving about 10.6 Hz, comfortably below the audio band. The other options misplace a decade or invert part of the product. Because the frequency is inversely proportional to both resistance and capacitance, a low-impedance point in the circuit always demands a much larger capacitor for the same break.
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