In the given circuit the $\mathrm{AC}$ source has $\omega =100\mathrm{rad}{s}^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit? ![JEE Main 2021 Physics, Alternating Current — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Alternating_Current/648b5a6b417cc3fb48d6740f/question_1__q_648b5a6b417cc3fb48d6740f__cdn-question-pool.getmarks.app__340e7beb-a78e-4ff4-9569-2121298e7805-image__0e23fffd40_final_ppt_sync.png)

${Z}_{C}=\sqrt{{(\frac{1}{\omega C})}^{2}+{R}^{2}}$ $=\sqrt{{(\frac{1}{100\times 100\times {10}^{-6}})}^{2}+{100}^{2}}$ $=\sqrt{{100}^{2}+{100}^{2}}$ $=100\sqrt{2}\Omega$ ${Z}_{L}=\sqrt{(\omega L{)}^{2}+{R}^{2}}=\sqrt{(100\times 0.5{)}^{2}+{50}^{2}}=50\sqrt{2}\Omega$ ${i}_{L}=\frac{200}{{Z}_{L}}=\frac{200}{50\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\text{A}$ ${i}_{C}=\frac{200}{{Z}_{C}}=\frac{200}{100\sqrt{2}}=\sqrt{2}\text{A}$ $\mathrm{cos}{\phi }_{1}=\frac{100}{100\sqrt{2}}=\frac{1}{\sqrt{2}},{\phi }_{1}=45^{\circ}$ $\mathrm{cos}{\phi }_{2}=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}},{\phi }_{2}=45^{\circ}$ ![JEE Main 2021 Physics, Alternating Current — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Alternating_Current/648b5a6b417cc3fb48d6740f/solution_1__q_648b5a6b417cc3fb48d6740f__cdn-question-pool.getmarks.app__998ea0dc-cb5f-43a5-8398-0cd23b89328a-image__f04b5af419_final_ppt_sync.png) $I=\sqrt{{I}_{C}^{2}+{I}_{L}^{2}};I=\sqrt{8+2};I=\sqrt{10}=3.16A$ This question is bonus as the correct answer was not available for students to mark as option in original paper by NTA.

In the given circuit the AC source has ω… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021mcqhard

In the given circuit the $\mathrm{AC}$ source has $\omega =100\mathrm{rad}{s}^{-1}$ . Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit?

JEE Main 2021 Physics, Alternating Current — question figure

Official previous-year question

Held on 26 Aug 2021 · Verified 6 Jul 2026.

Options

A
$5.9A$
B
$6A$
C
$0.94A$
D
$3.16A$

Did you get this right?

Solution

${Z}_{C}=\sqrt{{(\frac{1}{\omega C})}^{2}+{R}^{2}}$

$=\sqrt{{(\frac{1}{100\times 100\times {10}^{-6}})}^{2}+{100}^{2}}$

$=\sqrt{{100}^{2}+{100}^{2}}$

$=100\sqrt{2}\Omega$

${Z}_{L}=\sqrt{(\omega L{)}^{2}+{R}^{2}}=\sqrt{(100\times 0.5{)}^{2}+{50}^{2}}=50\sqrt{2}\Omega$

${i}_{L}=\frac{200}{{Z}_{L}}=\frac{200}{50\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\text{A}$

${i}_{C}=\frac{200}{{Z}_{C}}=\frac{200}{100\sqrt{2}}=\sqrt{2}\text{A}$

$\mathrm{cos}{\phi }_{1}=\frac{100}{100\sqrt{2}}=\frac{1}{\sqrt{2}},{\phi }_{1}=45^{\circ}$

$\mathrm{cos}{\phi }_{2}=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}},{\phi }_{2}=45^{\circ}$

JEE Main 2021 Physics, Alternating Current — solution figure

$I=\sqrt{{I}_{C}^{2}+{I}_{L}^{2}};I=\sqrt{8+2};I=\sqrt{10}=3.16A$

This question is bonus as the correct answer was not available for students to mark as option in original paper by NTA.

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism