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.

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
$3.16A$

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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.

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