JEE Main 2019 — Physics Electromagnetism

2019mcqmedium

A circuit connected to an ac source of emf $e={e}_{0}\mathrm{sin}⁡(100t)$ with $t$ in seconds, gives a phase difference of $\frac{\pi }{4}$ between the emf $e$ and current $i$ . Which of the following circuits will exhibit this?

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
$RC$ circuit with $R=1 k\Omega$ and $C=1 \mu F$
B
$RL$ circuit with $R=1 k\Omega$ and $L=1 mH$
C
$RL$ circuit with $R=1 k\Omega$ and $L=10 mH$
D
$RC$ circuit with $R=1 k\Omega$ and $C=10 \mu F$

Did you get this right?

Solution

$\omega =100$ rad/sec

phase difference between $e$ and $i$ is given by

$\mathrm{tan}\theta =\frac{X}{R}\therefore \theta =\frac{\pi }{4}$

so $\mathrm{tan}\theta =\mathrm{tan}\frac{\pi }{4}=1$

in case of $RL$ circuit $X={X}_{L}=\omega L$

And in case of $RC$ circuit $X={X}_{c}=\frac{1}{\omega c}$

For option $(A):$ $R=1000 \Omega$ and

${X}_{C}=\frac{1}{\omega C}=\frac{1}{(100)(1\times {10}^{-6})}=10000 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{C}}{R}\neq 1$

For option $(B):$ $R={10}^{3} \Omega$ and

${X}_{L}=L\omega =({10}^{-3})(100)=0.1 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{L}}{R}\neq 1$

For option $(C):R=1000 \Omega$ and

${X}_{L}=\omega L=(100)(10)=1000 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{L}}{R}\neq 1$

For option $(D):R=1 k\Omega =1000 \Omega$ and

${X}_{C}=\frac{1}{\omega C}=\frac{1}{(100)(10\times {10}^{-6})}=1000 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{C}}{R}=1;\theta ={45}^{o}$

Your note

JEE Main Electromagnetism in other years

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JEE Main 2019 — Physics Electromagnetism

2019mcqmedium

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
$RC$ circuit with $R=1 k\Omega$ and $C=1 \mu F$
B
$RL$ circuit with $R=1 k\Omega$ and $L=1 mH$
C
$RL$ circuit with $R=1 k\Omega$ and $L=10 mH$
D
$RC$ circuit with $R=1 k\Omega$ and $C=10 \mu F$

Did you get this right?

Solution

$\omega =100$ rad/sec

phase difference between $e$ and $i$ is given by

$\mathrm{tan}\theta =\frac{X}{R}\therefore \theta =\frac{\pi }{4}$

so $\mathrm{tan}\theta =\mathrm{tan}\frac{\pi }{4}=1$

in case of $RL$ circuit $X={X}_{L}=\omega L$

And in case of $RC$ circuit $X={X}_{c}=\frac{1}{\omega c}$

For option $(A):$ $R=1000 \Omega$ and

${X}_{C}=\frac{1}{\omega C}=\frac{1}{(100)(1\times {10}^{-6})}=10000 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{C}}{R}\neq 1$

For option $(B):$ $R={10}^{3} \Omega$ and

${X}_{L}=L\omega =({10}^{-3})(100)=0.1 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{L}}{R}\neq 1$

For option $(C):R=1000 \Omega$ and

${X}_{L}=\omega L=(100)(10)=1000 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{L}}{R}\neq 1$

For option $(D):R=1 k\Omega =1000 \Omega$ and

${X}_{C}=\frac{1}{\omega C}=\frac{1}{(100)(10\times {10}^{-6})}=1000 \Omega$

$\mathrm{tan}⁡\theta =\frac{{X}_{C}}{R}=1;\theta ={45}^{o}$

Your note

JEE Main Electromagnetism in other years

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JEE Main 2019 — Physics Electromagnetism

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JEE Main 2019 — Physics Electromagnetism

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