Calculating Cutoff Frequency for a First-Order Low-Pass Filter

This solution explains how to find the cutoff frequency ($f_c$) for a first-order low-pass filter given the resistance ($R_1$) and capacitance ($C_1$) values.

Given Parameters

• Resistance: $R_1 = 1.2 \ k\Omega = 1.2 \times 10^3 \ \Omega$
• Capacitance: $C_1 = 0.02 \ \mu F = 0.02 \times 10^{-6} \ F$

Filter Cutoff Frequency Formula

The cutoff frequency ($f_c$) for a simple RC low-pass filter is determined by the formula:

$f_c = \frac{1}{2\pi R C}$

Step-by-Step Calculation

1. Substitute the given values of $R_1$ and $C_1$ into the formula:

   $f_c = \frac{1}{2\pi (1.2 \times 10^3 \ \Omega) (0.02 \times 10^{-6} \ F)}$

2. Calculate the product of resistance and capacitance (the time constant, $\tau$):

   $R_1 C_1 = (1.2 \times 10^3) \times (0.02 \times 10^{-6}) = 24 \times 10^{-6} \ s$

3. Calculate the cutoff frequency:

   $f_c = \frac{1}{2\pi (24 \times 10^{-6} \ s)} = \frac{1}{48\pi \times 10^{-6} \ s}$

   $f_c \approx \frac{1}{150.796 \times 10^{-6}} \ Hz \approx 6631.45 \ Hz$

4. Convert the frequency to kilohertz (kHz):

   $f_c \approx 6.63 \ kHz$

Conclusion

The calculated cutoff frequency for the first-order low-pass filter is approximately $6.63 \ kHz$.