JEE Main 2022 — Alternating Current Question with Solution
From: JEE Main 2022 (Online) 30th June Morning Shift
Question
In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be :
(f1 = original resonant frequency, Q1 = original quality factor)
Choose an option
Show full solutionCorrect option: A
Step-by-step explanation
We know,
Quality factor (Q factor)
Now, when and then
Q2 remains same as Q1.
Also, as
When and then new resonating frequency
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 2022, 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.