
ASSERTION: In a series LCR circuit at resonance condition, power consumed by the circuit is maximum.
REASON: At resonance, the effective resistance of the circuit is maximum.
(A) Both ASSERTION and REASON are correct and REASON is the correct explanation for ASSERTION.
(B) Both ASSERTION and REASON are correct but REASON is not the correct explanation for ASSERTION.
(C) ASSERTION is correct but REASON is incorrect
(D) Both ASSERTION and REASON are incorrect.
Answer
126.6k+ views
Hint: When a series LCR circuit is in resonance, the impedance of the same series LCR circuit is minimum. This happens because the inductance and capacitance are equal in magnitude. As a result of this the current in the series LCR circuit also becomes maximum. We can easily arrive at the correct answer if we remember the relationship between the current in a series LCR circuit and the power consumed.
Complete step by step answer:
When a series LCR circuit comes in the state of resonance, the impedance becomes minimum and thus $Z = R\Omega $. We already know that this happens because the value of both inductance and capacitance becomes equal. Now, as the root mean square current is inversely proportional to the value of impedance $({I_{rms}} = \dfrac{{{V_{rms}}}}{Z} = \dfrac{{{V_{rms}}}}{R})$ thus the value of current increases. Now, we already know that the power of a series LCR circuit is given by the mathematical expression $P = I_{rms}^2R$ thus we can say that the power of a series LCR circuit is maximum during resonance. But this is not because the effective resistance is maximum but rather because the root mean square current is maximum. Thus we can say that the ASSERTION given in the question is correct but the REASON is incorrect.
Hence, the correct answer to the above problem is (C).
Note: It is always important to remember how the value of one variable influences the value of another in questions related to series or parallel LCR circuits in resonance. Many students approach the problem in a wrong manner and just use the expression for power to justify their answer. It is better to start to analyse each and every term given in the question and its influence in other terms. This way we can arrive at the right answer.
Complete step by step answer:
When a series LCR circuit comes in the state of resonance, the impedance becomes minimum and thus $Z = R\Omega $. We already know that this happens because the value of both inductance and capacitance becomes equal. Now, as the root mean square current is inversely proportional to the value of impedance $({I_{rms}} = \dfrac{{{V_{rms}}}}{Z} = \dfrac{{{V_{rms}}}}{R})$ thus the value of current increases. Now, we already know that the power of a series LCR circuit is given by the mathematical expression $P = I_{rms}^2R$ thus we can say that the power of a series LCR circuit is maximum during resonance. But this is not because the effective resistance is maximum but rather because the root mean square current is maximum. Thus we can say that the ASSERTION given in the question is correct but the REASON is incorrect.
Hence, the correct answer to the above problem is (C).
Note: It is always important to remember how the value of one variable influences the value of another in questions related to series or parallel LCR circuits in resonance. Many students approach the problem in a wrong manner and just use the expression for power to justify their answer. It is better to start to analyse each and every term given in the question and its influence in other terms. This way we can arrive at the right answer.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application

Young's Double Slit Experiment Step by Step Derivation

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main
