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

1 Answer

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}

User Dan Andreatta
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