In the above question, what are the values of cut-off frequencies?

[amp_mcq option1=”140 kHz , 60 kHz” option2=”1020 kHz , 980 kHz” option3=”1030 kHz , 970 kHz” option4=”None of the above” correct=”option4″]

The correct answer is D. None of the above.

The cut-off frequencies of a low-pass filter are the frequencies below which the output signal is attenuated by a certain amount. The cut-off frequencies are determined by the values of the resistor and capacitor in the filter circuit.

In the above question, the resistor and capacitor are given as R = 100 Ω and C = 1 μF. The cut-off frequencies can be calculated using the following formula:

f = 1 / (2πRC)

where f is the cut-off frequency in Hz, R is the resistance in ohms, and C is the capacitance in farads.

Substituting the values of R and C into the formula, we get:

f = 1 / (2π * 100 Ω * 1 μF) = 159.15 kHz

Therefore, the cut-off frequency of the low-pass filter is 159.15 kHz.

The options A, B, and C are all incorrect because they do not correspond to the cut-off frequency of the low-pass filter.