An alternating voltage source $\mathrm{V}=260 \sin (628 \mathrm{t}$ ) is connected across a pure inductor of $5 \mathrm{mH}$ Inductive reactance in the circuit is :
Solution
$\omega$ = 628 rad/s
<br/><br/>$X_{L}=L \omega$
<br/><br/>$$
\begin{aligned}
& =5 \mathrm{mH} \times 628 \\\\
& =3.14 \Omega
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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