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JEE Main 2023 (31 Jan, Shift 2) · Pure R, L and C Circuits

Pure R, L and C Circuits question from JEE Main 2023 (31 Jan, Shift 2)

Q35JEE Main 2023 · 31 JAN · SHIFT 2Easy

Pure R, L and C Circuits

An alternating voltage source V=260sin⁡(628tV = 260 \sin(628t) is connected across a pure inductor of 5 mH5\text{ mH}. Inductive reactance in the circuit is:

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Answer: (d)

  1. From V=260sin⁡(628t)V = 260\sin(628t), ω=628\omega = 628 rad/s.
  2. XL=ωL=628×5×10−3=3.14 ΩX_L = \omega L = 628 \times 5\times10^{-3} = 3.14\ \Omega
  3. So the answer is (d) 3.14 Ω3.14\ \Omega.
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