One need to design a filter with cut off frequency 330 Hz. Which of the following resistor and capacitor combination is correct. R = 2K Ohm C=0.047 micro farad R = 20K Ohm C=0.047 micro farad R = 1K Ohm C=0.48 micro farad R = 10K Ohm C=0.48 micro farad

The cut-off frequency (f_c) of a filter can be calculated using the formula:

f_c = 1 / (2 * π * R * C)

where:

• f_c is the cut-off frequency,
• R is the resistance,
• C is the capacitance,
• π is a constant (approximately 3.14159).

We can rearrange this formula to solve for the combination of R and C that will give us a cut-off frequency of 330 Hz:

R = 1 / (2 * π * f_c * C)

Let's calculate the resistance for each combination of R and C:

1. For R = 2K Ohm and C = 0.047 micro farad: R = 1 / (2 * π * 330 * 0.047 * 10^-6) = 10.14 K Ohm

2. For R = 20K Ohm and C = 0.047 micro farad: R = 1 / (2 * π * 330 * 0.047 * 10^-6) = 10.14 K Ohm

3. For R = 1K Ohm and C = 0.48 micro farad: R = 1 / (2 * π * 330 * 0.48 * 10^-6) = 1.01 K Ohm

4. For R = 10K Ohm and C = 0.48 micro farad: R = 1 / (2 * π * 330 * 0.48 * 10^-6) = 1.01 K Ohm

From these calculations, we can see that the combination of R = 1K Ohm and C = 0.48 micro farad is the closest to the desired cut-off frequency of 330 Hz.