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Alternating Current question

2019 · 12 Apr · Shift 2 · Q56
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Alternating Current question

2019 · 12 Apr · Shift 2 · Q56

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t = LR{L \over R}RL​ is : JEE Main 2019 (Online) 12th April Evening Slot Physics - Alternating Current Question 141 English
  1. A
    2.7ELR2{{2.7EL} \over {{R^2}}}R22.7EL​
  2. B
    EL2.7R2{{EL} \over {2.7{R^2}}}2.7R2EL​
  3. C
    7.3ELR2{{7.3EL} \over {{R^2}}}R27.3EL​
  4. D
    EL7.3R2{{EL} \over {7.3{R^2}}}7.3R2EL​
View written solutionFree

Correct answer: B

  1. Current growth in an LRLRLR circuit

When the switch is closed at t=0t=0t=0, the current in a series LRLRLR circuit connected to a battery of emf EEE is

i(t)=ER(1−e−Rt/L)i(t)=\frac{E}{R}\left(1-e^{-Rt/L}\right)i(t)=RE​(1−e−Rt/L)

The time constant is

τ=LR\tau=\frac{L}{R}τ=RL​

We need the charge passed through the battery from t=0t=0t=0 to t=τ=LRt=\tau=\frac{L}{R}t=τ=RL​.

  1. Charge as integral of current
q=∫0L/Ri(t) dtq=\int_0^{L/R} i(t)\,dtq=∫0L/R​i(t)dt

Substitute i(t)i(t)i(t):

q=∫0L/RER(1−e−Rt/L)dtq=\int_0^{L/R} \frac{E}{R}\left(1-e^{-Rt/L}\right)dtq=∫0L/R​RE​(1−e−Rt/L)dt q=ER[∫0L/R1 dt−∫0L/Re−Rt/Ldt]q=\frac{E}{R}\left[\int_0^{L/R}1\,dt-\int_0^{L/R}e^{-Rt/L}dt\right]q=RE​[∫0L/R​1dt−∫0L/R​e−Rt/Ldt]
  1. Evaluate the integrals

First integral:

∫0L/R1 dt=LR\int_0^{L/R}1\,dt=\frac{L}{R}∫0L/R​1dt=RL​

Second integral:

∫e−Rt/Ldt=−LRe−Rt/L\int e^{-Rt/L}dt=-\frac{L}{R}e^{-Rt/L}∫e−Rt/Ldt=−RL​e−Rt/L

So,

∫0L/Re−Rt/Ldt=[−LRe−Rt/L]0L/R=−LRe−1+LR=LR(1−e−1)\int_0^{L/R} e^{-Rt/L}dt =\left[-\frac{L}{R}e^{-Rt/L}\right]_0^{L/R} = -\frac{L}{R}e^{-1}+\frac{L}{R} =\frac{L}{R}(1-e^{-1})∫0L/R​e−Rt/Ldt=[−RL​e−Rt/L]0L/R​=−RL​e−1+RL​=RL​(1−e−1)

Hence,

q=ER[LR−LR(1−e−1)]q=\frac{E}{R}\left[\frac{L}{R}-\frac{L}{R}(1-e^{-1})\right]q=RE​[RL​−RL​(1−e−1)] q=ER⋅LR⋅e−1q=\frac{E}{R}\cdot \frac{L}{R}\cdot e^{-1}q=RE​⋅RL​⋅e−1 q=ELeR2q=\frac{EL}{eR^2}q=eR2EL​

Now using e≈2.7e\approx 2.7e≈2.7,

q≈EL2.7R2q\approx \frac{EL}{2.7R^2}q≈2.7R2EL​
  1. Match with options

This corresponds to:

EL2.7R2\boxed{\frac{EL}{2.7R^2}}2.7R2EL​​

So the correct option is B.

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