Question

A series RL low pass filter with a cut-off frequency of 4 kHz is needed. Using R-10 kOhm, Compute (a) L. (b)) at 25 kHz and (c) 870) at 25 kHz Oa 0 20 H, 0 158 and 2-30.50 Ob 525 H, 0.158 and 2-30 5 O 025 H, 0.158 and 2-80 5 Od 225 H, 1.158 and -80 5

Answer 1

(a) L ≈ 0.397 H

(b) Z ≈ 52.5 kOhm

(c) θ ≈ 0.025 radians

Explanation:

To calculate the values for an RL low pass filter with a cut-off frequency of 4 kHz and using an R-value of 10 kOhm, we can use the following formulas:

(a) L = R / (2 * π * f_c)

(b) Z = √(R^2 + (2 * π * f)^2)

(c) θ = atan((2 * π * f) / R)

Given:

R = 10 kOhm

f = 25 kHz

f_c = 4 kHz

(a) L = R / (2 * π * f_c)

L = 10 kOhm / (2 * π * 4 kHz)

L ≈ 0.397 H

(b) Z = √(R^2 + (2 * π * f)^2)

Z = √((10 kOhm)^2 + (2 * π * 25 kHz)^2)

Z ≈ 52.5 kOhm

(c) θ = atan((2 * π * f) / R)

θ = atan((2 * π * 25 kHz) / 10 kOhm)

θ ≈ 0.025 radians

Therefore, the calculated values are:

(a) L ≈ 0.397 H

(b) Z ≈ 52.5 kOhm

(c) θ ≈ 0.025 radians