Question

The cutoff frequency of a first order low pass filter for \(R_1 = 1.2 \, \text{k}\Omega\) and \(C_1 = 0.02 \, \mu F\) is:

• A. 1.86 kHz
• B. 2.63 kHz
• C. 6.63 kHz
• D. 10.63 kHz

Hint:

For low-pass filters, remember \(f_c = \dfrac{1}{2 \pi R C}\) to calculate cutoff frequency.

Answer 1

The Correct Option is B

The cutoff frequency \(f_c\) for a first-order low-pass filter is defined as:\[f_c = \frac{1}{2 \pi R_1 C_1}\]Using the provided component values, the calculation is:\[f_c = \frac{1}{2 \pi \times 1.2 \times 10^3 \times 0.02 \times 10^{-6}} \approx 2.63 \, \text{kHz}\]
Final Answer: \[\boxed{2.63 \, \text{kHz}}\]