Discussion
Find the cut off frequency of a first order low pass filter for \(R_1 = 1.2\ \mathrm{k\Omega}\), \(C_1 = 0.02\ \mu\mathrm{F}\).
Answer: A
The cut off frequency of a first order RC low pass filter is the frequency at which the output has fallen to \(1/\sqrt{2}\) of its low frequency value, \(f_c = \dfrac{1}{2\pi R C}\).
\[ \begin{aligned} f_c \\ &= \frac{1}{2\pi \times 1200 \times 0.02 \times 10^{-6}} \\ &= \frac{1}{1.508 \times 10^{-4}} \\ &= 6631\ \text{Hz} \approx 6.63\ \text{kHz} \end{aligned} \]
The other values are 10, 10 and 100 times off, which would come from a slip of a power of ten in the capacitance (0.02 microfarad is \(2 \times 10^{-8}\) F). The answer is 6.63 kHz, option A.
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