JEE Main 2015PhysicsAlternating CurrentMediumMCQ

JEE Main 2015 — Alternating Current Question with Solution

JEE Main 2015 (04 Apr)

Question

An inductor ( L=0.03 H ) and a resistor (R=0.15  kΩ) are connected in series to a battery of 15 V E.M.F. in a circuit shown below. The key K1 has been kept closed for a long time. Then at t=0, K1 is opened and key K2 is closed simultaneously. At t=1 ms , the current in the circuit will be : Take, e5≈150

Choose an option

Show full solutionCorrect option: A
Correct answer
A0.67 mA

Step-by-step explanation

Case I: K1 is closed for long time, 



for a long time, the inductor acts as a conducting wire,  ⇒ current in the circuit =VR = 1 5 1 5 0 ⇒i0=0.1 A
Case II: K1 is open and K2 is closed



Current in the circuit, i=i0e-tτ;  τ=LR

After t=1 ms=10-3 s

⇒i=i0e-10-3×1503×10-2 = 0.1 e - 1 5 3 =0.11e5=0.1150 A=0.67 mA.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Alternating Current chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2015, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.