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JEE Main 202331 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2023 · 31 January · Evening Shift · Physics · Alternating Current

An alternating voltage source V = 260 ( 628t ) is connected across...

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:

Options

  1. A3.14Ω3 . 14 Ω
  2. B6.28Ω6 . 28 Ω
  3. C0.5Ω0 . 5 Ω
  4. D0.318Ω0 . 318 Ω

Correct answer

A. 3.14Ω3 . 14 Ω

Step-by-step solution

Comparing the given equation of AC voltage with standard expression, V=V0sin⁡(\omegat)V = {V}_{0}\sin \left(\omegat\right), we get ω=628rads−1\omega = 628rad{s}^{- 1}.

Inductive reactance is given by, XL=\omegaL{X}_{L} = \omegaL.

=628×(5×10−3)= 628 \times \left(5 \times {10}^{- 3}\right)

=3.14Ω= 3 . 14Ω

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