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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 18 March · Evening Shift · Physics · Alternating Current

The time taken for the magnetic energy to reach 25 % of its maximum...

The time taken for the magnetic energy to reach 2525 % of its maximum value, when a solenoid of resistance RR, inductance LL is connected to a battery, is :

Options

  1. ALRln⁡5\dfrac{L}{R}\ln 5
  2. Binfinite
  3. CLRln⁡2\dfrac{L}{R}\ln 2
  4. DLRln⁡10\dfrac{L}{R}\ln 10

Correct answer

C. LRln⁡2\dfrac{L}{R}\ln 2

Step-by-step solution

Magnetic energy =12Li2=25= \dfrac{1}{2}L{i}^{2} = 25 %</p><p>ME ⇒ 25 % ⇒ i = \dfrac{{i}_{0}}{2}</p><p></p><p>i = {i}_{0}(1−R−Rt/L)\left(1 - {R}^{- Rt / L}\right) for charging

t=LRln⁡2t = \dfrac{L}{R}\ln 2

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