Verified Solution • 2025–26 Session moderate

QJEEM16Ph1e6a: JEE Main 2016 Physics - Class 12 JEEM16Ph1e6a Solution: A series LR circuit is connected to a voltage source with V(t)…

A series LR circuit is connected to a voltage source with
V(t) = V0 sin Ω\Omega t. After very large time, current I(t) behaves as
(t0 >> LR{L \over R} ) :

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/71f41488c7cf842b74e59bef68ab31fe94f63a97c96319353a136b3847b0b8a1.webp
  • B https://ik.imagekit.io/novatra/photos/313891458ded1206c9227f6fac98c73feeded5cf48988320928a5cf8a02c4389.webp
  • C https://ik.imagekit.io/novatra/photos/de18aa50e94f252c6ce47241737f37096499bcd569b8e9b7a66ac3858a12dffc.webp
  • D https://ik.imagekit.io/novatra/photos/07c1cfd18c610d42e73184d31eead0c6b04788f08e30e924fdd89262ac5b1394.webp

Step-by-Step Verified Solution

Current in LR circuit,

I=V0R2+w2L2sin⁡(ωt−π2){\rm I} = {{{V_0}} \over {\sqrt {{R^2} + {w^2}{L^2}} }}\sin \left( {\omega t - {\pi \over 2}} \right)

it will be a sinusoidal wave.

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