JEE Main 2019 — Alternating Current Question with Solution
From: JEE Main 2019 (Online) 10th April Evening Slot
Question
A coil of self inductance 10 mH and resistance 0.1 is connected through a switch to a battery of internal
resistance 0.9 . After the switch is closed, the time taken for the current to attain 80% of the saturation
value is: [take ln 5 = 1.6]
Choose an option
Show full solutionCorrect option: D
Correct answer
D0.016 s
Step-by-step explanation
L = 10 × 10–3 H, r1 = 0.1
0.8 = 1 - e-t/2 ; e-t/2 = 0.2
et/L = 5
t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3
0.8 = 1 - e-t/2 ; e-t/2 = 0.2
et/L = 5
t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3
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This is a previous-year question from JEE Main 2019, 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.