An alternating voltage source V = 260sin (628t) is connected across a pure inductor of 5 mH. Inductive reactance in the circuit is:

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$0.5 Ω$

$6.28 Ω$

$3.14 Ω$

$0.318 Ω$

answer is A.

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Detailed Solution

$\begin{array}{l} V = 260 \sin (628) t \\ L = 5 mH, ω = 628 rad / s \end{array}$
Inductive reactance, $X_{L} = ωL$
$X_{L} = 628 \times 5 \times 10^{- 3} = 3.14 Ω$

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