Final answer:
To design an RL low pass filter with a cutoff frequency of 5 kHz using a 40 mH coil, you would need a capacitor with a value of approximately 0.01 µF.
Step-by-step explanation:
An RL low pass filter is a circuit that allows low-frequency signals to pass through while attenuating high-frequency signals. To design one with a cutoff frequency of 5 kHz using a 40 mH coil, we need to calculate the value of the resistor.
The cutoff frequency (fc) for an RL low pass filter is given by the formula fc = 1 / (2π√(LC)), where L is the inductance and C is the capacitance. Rearranging the formula, we have C = 1 / (4π²f²L), where f is the desired cutoff frequency.
Substituting the given values into the formula, we get C = 1 / (4π² * (5 kHz)² * (40 mH)). Calculating this results in a capacitance of approximately 0.01 µF.