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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2020 · 7 January · Evening Shift · Physics · Alternating Current

An emf of 20V is applied at time t = 0 to a circuit containing in...

An emf of 20V20V is applied at time t=0t = 0 to a circuit containing in series 10mH10mH inductor and 5Ω5\text{Ω} resistor. The ratio of the currents at time t=∞t = \infty and at t=40st = 40\text{s} is close to: (Take e2=7.389{e}^{2} = 7.389 )An emf of 20V is applied at time t = 0 to a circuit containing in series 10mH inductor and 5Ω resistor. The ratio of

Options

  1. A1.061.06
  2. B1.151.15
  3. C1.461.46
  4. D0.840.84

Correct answer

A. 1.061.06

Step-by-step solution

i=i0(1−e−tL/R)i = {i}_{0}\left(1 - {e}^{\dfrac{- t}{L / R}}\right)
An emf of 20V is applied at time t = 0 to a circuit containing in series 10mH inductor and 5Ω resistor. The ratio of
=205(1−e−t0.01/5)= \dfrac{20}{5}\left(1 - {e}^{\dfrac{- t}{0.01 / 5}}\right)
=4(1−e−500t)= 4\left(1 - {e}^{- 500t}\right)

now at t=∞t = \infty
i∞=4{i}_{\infty} = 4

at t=40st = 40\text{s}
i40=4(1−e−500×40){i}_{40} = 4\left(1 - {e}^{- 500 \times 40}\right)
=4(1−1(e2)10000)=4(1−17.2910000)= 4\left(1 - \dfrac{1}{{\left({e}^{2}\right)}^{10000}}\right) = 4\left(1 - \dfrac{1}{{7.29}^{10000}}\right)
i∞i40≈1\dfrac{{i}_{\infty}}{{i}_{40}} \approx 1 Slightly greater then one.

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