At what frequency would an inductor of inductance 0.8H have a reactance of 12000^?

asked Dec 18, 2022 157k views

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4 votes

Step-by-step explanation:

The inductive reactance
X_L is given by

$X_L = \omega L = 2 \pi fL$

Solving for f, we get

$f = (X_L)/(2 \pi L) = \frac{12000\:\text{ ohms}}{2\pi (0.8\:H)}$

$\:\:\:\:\:\:\:= 2387.3\:\text{Hz}$

Dan Andreatta answered Dec 24, 2022

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