Physics · Alternating Current

JEE Main 2024 — 6 April, Shift 1 — Question 42

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II : Statement I : In an LCR series circuit, current is maximum at resonance. Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source.

In the light of the above statements, choose the correct from the options given below :

Option A:
Statement I is true but Statement II is false
Option B:
Statement I is false but Statement II is true
Option C:
Both Statement I and Statement II are true
Correct
Option D:
Both Statement I and Statement II are false

Answer: C

Step-by-step solution

Statement-I :

$I_{m}=\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ at resonance $X_{L}=X_{C}$ Thus, $I_{m}=\frac{V_{m}}{R}$

$\because$ Impendence is minimum therefore $I$ is maximum at resonance.

Statement-II :

$\mathrm{I}=\left(\frac{\mathrm{V}}{\mathrm{R}}\right)$ in purely resistive circuit.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor

When a dc voltage of 100 V is applied to an inductor, a de current of 5A flows through it. When an ac voltage of 200 V peak value is…

Answer: C

Step-by-step solution

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