JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 16)
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 :
$6.28 \Omega$
$0.318 \Omega$
$0.5 \Omega$
$3.14 \Omega$
Explanation
$\omega$ = 628 rad/s
$X_{L}=L \omega$
$$ \begin{aligned} & =5 \mathrm{mH} \times 628 \\\\ & =3.14 \Omega \end{aligned} $$
$X_{L}=L \omega$
$$ \begin{aligned} & =5 \mathrm{mH} \times 628 \\\\ & =3.14 \Omega \end{aligned} $$
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