Question

Learning Goal:

To analyze and design a passive, first-order low- pass filter using a series RL circuit. The analysis and design will be repeated for a series RC circuit.

An electrocardiogram needs to detect periodic signals of approximately 1 Hz (since the resting heart rate of a healthy adult is between 55 and 70 beats per minute). The instrument operates in an electrical environment that is very noisy with a frequency of 60 Hz. It is desirable to have a low- pass filter that will block any noise above 23 Hz. A low-pass RL filter can be built using available 140 H inductors; a low-pass RC filter can be built using available 4.2 µF capacitors; resistors of any necessary value are also available.

Part E - Designing and analyzing a series RC low-pass filter
Using the available 4.2 µF capacitor, what is the value of the resistor needed to make a low-pass filter with a cutoff frequency of 23 Hz?
Express your answer to three significant figures and include appropriate units.

Answer 1

To create a series RC low-pass filter with a cutoff frequency of 23 Hz using a 4.2 µF capacitor, a resistor with a value of approximately 1.555 kΩ is needed.

Step-by-step explanation:

To design a series RC low-pass filter with a cutoff frequency of 23 Hz using a 4.2 µF capacitor, we need to calculate the required value for the resistor. The cutoff frequency (ωc or fc) for a low-pass filter is given by the formula fc = 1/(2πRC), where R is the resistance in ohms (Ω), C is the capacitance in farads (F), and fc is the cutoff frequency in hertz (Hz). Rearranging this formula to solve for R gives us R = 1/(2πfcC).

Substituting the known values (fc = 23 Hz and C = 4.2 µF), the calculation for R is:

R = 1 / (2π x 23 Hz x 4.2 µF)

R = 1 / (2π x 23 x 4.2 x 10-6) = ≈ 1.555 kΩ

Therefore, a resistor value of approximately 1.555 kΩ would be required to create a low-pass filter with a cutoff frequency of 23 Hz using the available 4.2 µF capacitor.