Alternating Current

JEE Main level, single correct, medium

19 of 190

Question 19 of 190

An alternating voltage source V=260sin⁡(628t)V=260\sin (628t) is connected across a pure inductor of 5mH5mH. Inductive reactance in the circuit is:

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Answer: (A) 3.14Ω3.14Ω

Comparing the given equation of AC voltage with standard expression, V=V0sin⁡(ωt)V={V}_{0}\sin \left(\omega t\right), we get ω=628rads−1\omega =628rad{s}^{-1}.
Inductive reactance is given by, XL=ωL{X}_{L}=\omega L.
=628×(5×10−3)=628\times \left(5\times {10}^{-3}\right)
=3.14Ω=3.14\Omega

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