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JEE Main 20204 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2020 · 4 September · Evening Shift · Physics · Alternating Current

A series L - R circuit is connected to a battery of emf V . If the...

A series L−RL - R circuit is connected to a battery of emf VV. If the circuit is switched on at t=0t = 0, then the time at which the energy stored in the inductor reaches (1n)\left(\dfrac{1}{n}\right) times of its maximum value, is :A series L - R circuit is connected to a battery of emf V . If the circuit is switched on at t = 0 , then the time at

Options

  1. ALRln⁡(nn−1)\dfrac{L}{R}\ln \left(\dfrac{\sqrt{n}}{\sqrt{n} - 1}\right)
  2. BLRln⁡(n+1n−1)\dfrac{L}{R}\ln \left(\dfrac{\sqrt{n} + 1}{\sqrt{n} - 1}\right)
  3. CLRln⁡(nn+1)\dfrac{L}{R}\ln \left(\dfrac{\sqrt{n}}{\sqrt{n} + 1}\right)
  4. DLRln⁡(n−1n)\dfrac{L}{R}\ln \left(\dfrac{\sqrt{n} - 1}{\sqrt{n}}\right)

Correct answer

A. LRln⁡(nn−1)\dfrac{L}{R}\ln \left(\dfrac{\sqrt{n}}{\sqrt{n} - 1}\right)

Step-by-step solution

A series L - R circuit is connected to a battery of emf V . If the circuit is switched on at t = 0 , then the time at

Potential energy stored in inductor is given by U=12L2U = \dfrac{1}{2}{L}^{2}

∪∝l2\cup ∝ {l}^{2}

UU0=(1I0)2\dfrac{U}{{U}_{0}} = {\left(\dfrac{1}{{I}_{0}}\right)}^{2}

1n=(1I0)2\dfrac{1}{n} = {\left(\dfrac{1}{{I}_{0}}\right)}^{2}

1I0=1−e−RT/L=1n\dfrac{1}{{I}_{0}} = 1 - {e}^{- RT / L} = \dfrac{1}{\sqrt{n}}

t=LRlnnn−1t = \dfrac{L}{R}ln\dfrac{\sqrt{n}}{\sqrt{n} - 1}

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