JEE Main 2021 — Alternating Current Question with Solution
From: JEE Main 2021 (Online) 27th July Morning Shift
Question
Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1 is connected to T2. As the current in R = 6 attains a maximum value of steady state level, T1 is disconnected from T2 and immediately connected to T3. Potential drop across r = 3 resistor immediately after T1 is connected to T3 is __________ V. (Round off to the Nearest Integer)
This question includes a diagram. The text above accompanies the figure.
Enter your answer
Show full solutionCorrect answer: 3
Correct answer
3
Step-by-step explanation
What T1 and T2 are connected, then the steady state current in the inductor
When T1 and T3 are connected then current through inductor remains same. So potential difference across 3
V = Ir = 1 3 = 3 Volt
When T1 and T3 are connected then current through inductor remains same. So potential difference across 3
V = Ir = 1 3 = 3 Volt
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 2021, 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.