Calculate the cut-off frequency of an OP-Amp based first order low pass filter for R = 2·5 kΩ and C = 0·05 μF.
MH SET - 2018, SET Exam - 2018,
Answer: A
The cut-off frequency is 1.273 kHz.
For a first order RC low pass filter, active or passive, the cut-off frequency is
\[ \begin{aligned} f_{c} &= \frac{1}{2\pi RC} \\ &= \frac{1}{2\pi (2.5 \times 10^{3})(0.05 \times 10^{-6})} \\ &= \frac{1}{2\pi \times 1.25 \times 10^{-4}} \\ &\approx 1273\ \text{Hz} = 1.273\ \text{kHz} \end{aligned} \]
The other options are the same digits with the decimal point moved, the result of a slip with the powers of ten for kilo-ohms and microfarads: 12.73 kHz and 127.3 kHz are ten and a hundred times too high, and 127.3 Hz ten times too low.
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