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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Physics
Chapter
Alternating Current

JEE Main 2016 · 9 April · Online · Physics · Alternating Current

A series LR circuit is connected to a voltage source with V(t) = V...

A series LRLR circuit is connected to a voltage source with V(t)=V0sin⁡(\omegat)V\left(t\right) = {V}_{0}\sin \left(\omegat\right) . After a very large time, current I(t)I\left(t\right) behaves as \left({t}_{0} ≫ \dfrac{L}{R}\right)$:A series LR circuit is connected to a voltage source with V(t) = V _ 0 ( ) . After a very large time, current I(t)A series LR circuit is connected to a voltage source with V(t) = V _ 0 ( ) . After a very large time, current I(t)A series LR circuit is connected to a voltage source with V(t) = V _ 0 ( ) . After a very large time, current I(t)A series LR circuit is connected to a voltage source with V(t) = V _ 0 ( ) . After a very large time, current I(t)

Options

  1. AIf I = ∫ ( tan x + cot x )dx , then I equals to (where, C & is an arbitrary constant)
  2. BIf I = ∫ ( tan x + cot x )dx , then I equals to (where, C & is an arbitrary constant)
  3. CIf I = ∫ ( tan x + cot x )dx , then I equals to (where, C & is an arbitrary constant)
  4. DIf I = ∫ ( tan x + cot x )dx , then I equals to (where, C & is an arbitrary constant)

Correct answer

D. If I = ∫ ( tan x + cot x )dx , then I equals to (where, C & is an arbitrary constant)

Step-by-step solution

When sinusoidal voltage is applied across LRLR circuit, current is also sinusoidal but it lags by some phase.

Current will be in the form of
I=I0sin⁡(\omegat−ϕ)I = {I}_{0} \sin \left(\omegat - ϕ\right)
Graph will be sinusoidal.

Here I0{I}_{0} is the maximum current and ϕϕ is phase difference between V and IV \text{ and } I

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