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Alternating Current question

2023 · 31 Jan · Shift 2 · Q60
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Alternating Current question

2023 · 31 Jan · Shift 2 · Q60

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
An alternating voltage source V=260sin⁡(628t\mathrm{V}=260 \sin (628 \mathrm{t}V=260sin(628t) is connected across a pure inductor of 5mH5 \mathrm{mH}5mH Inductive reactance in the circuit is :
  1. A
    6.28Ω6.28 \Omega6.28Ω
  2. B
    0.318Ω0.318 \Omega0.318Ω
  3. C
    0.5Ω0.5 \Omega0.5Ω
  4. D
    3.14Ω3.14 \Omega3.14Ω
View written solutionFree

Correct answer: D

  1. Given data

The alternating voltage is V=260sin⁡(628t)V = 260\sin(628t)V=260sin(628t) Comparing with the standard form V=V0sin⁡(ωt)V = V_0\sin(\omega t)V=V0​sin(ωt) we get:

  • Angular frequency: ω=628 rad/s\omega = 628\ \text{rad/s}ω=628 rad/s
  • Inductance: L=5 mH=5×10−3 HL = 5\text{ mH} = 5\times 10^{-3}\text{ H}L=5 mH=5×10−3 H
  1. Formula for inductive reactance

For a pure inductor, XL=ωLX_L = \omega LXL​=ωL

  1. Substitute the values

XL=628×5×10−3X_L = 628 \times 5\times 10^{-3}XL​=628×5×10−3

XL=628×0.005X_L = 628 \times 0.005XL​=628×0.005

XL=3.14 ΩX_L = 3.14\ \OmegaXL​=3.14 Ω

  1. Match with the options

The correct option is: D: 3.14 Ω\boxed{D:\ 3.14\ \Omega}D: 3.14 Ω​

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