Question

Suppose we have a simple AC circuit with just an inductor in it, and we want a

22.5KΩ22.5 kΩ reactance at a frequency of 505 Hz.
A. L=0.141 L=0.141 H
B. L=0.070L=0.070 H
C. L=0.035 L=0.035 H
D. L=0.282L=0.282 H

Answer 1

B. L=0.070L=0.070 H. To achieve a reactance of 22.5 kΩ at a frequency of 505 Hz, the required inductance is 0.070 H.

Step-by-step explanation:

An inductor in an AC circuit contributes reactance to the circuit, denoted as XL. The reactance of an inductor is given by the formula XL = 2πfL, where f is the frequency in hertz and L is the inductance in henries. To find an inductance of 22.5 kΩ reactance at a frequency of 505 Hz, we can rearrange the formula as L = XL / (2πf). Substituting the values, we have L = 22500 / (2π * 505) = 0.070 H.