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

Question

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

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Solution

The cut-off frequency for a RC (Resistor-Capacitor) filter can be calculated using the formula:

f_c = 1 / (2πRC)

where: f_c is the cut-off frequency, R is the resistance in ohms (Ω), C is the capacitance in farads (F).

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

R = 1 / (2πf_cC)

Let's calculate the resistance for each given capacitor value:

  1. For C = 0.48 microfarad (or 0.48 x 10^-6 F), R = 1 / (2π * 330 * 0.48 x 10^-6) ≈ 1.01 KΩ
  2. For C = 0.047 microfarad (or 0.047 x 10^-6 F), R = 1 / (2π * 330 * 0.047 x 10^-6) ≈ 10.1 KΩ

Comparing these calculated resistances with the given options, we can see that the combination of R = 1K Ohm and C = 0.48 microfarad is the closest to our calculated values. Therefore, this is the correct combination to design a filter with a cut-off frequency of 330 Hz.

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