Physics · Alternating Current

JEE Main 2024 — 8 April, Shift 1 — Question 42

A LCR circuit is at resonance for a capacitor C , inductance $L$ and resistance $R$ . Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

Option A:
Zero
Option B:
double
Correct
Option C:
same
Option D:
halved

Answer: B

Step-by-step solution

In resonance $\mathrm{Z}=\mathrm{R}$

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

$\mathrm{R} \rightarrow$ halved

$\Rightarrow \mathrm{I} \rightarrow 2 \mathrm{I}$

I becomes doubled.

Answer key and solution verified before publishing.

Practise Alternating Current

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor