Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 k $\Omega$ with C = 2 $\mu F$ . The resonant frequency $\omega$ is 200 rad/s. At resonance the voltage across L is

Option 1)
$4\times 10^{-3}\, V\;$

Option 2)
$\; 2.5\times 10^{-2}\, V\;$

Option 3)
$\; \; 40\; V\;$

Option 4)
$\; 250\; \; V$

Answers (1)

As we learnt in

where $E=\sqrt{V_{R}^{2}+(V_{L}-V_{C})^{2}}$

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$

At resonance, $X_{L}=X_{C}$

$\therefore \; \; Z=R$

Again at resonance, $V_{L}=V_{C}$

$\therefore \; \; E=V_{R}$

Current $I=\frac{E}{Z}$

$\therefore \; \; I=\frac{V_{R}}{R}=\frac{100}{1\times 10^{3}}=0.1\; A$

$\therefore \; \; V_{L}=IL\omega =\frac{I}{C\omega }=\frac{0.1}{(2\times 10^{-6})(200)}$

$\therefore \; \; \; V_{L}=250\; volt$

Option 1)

$4\times 10^{-3}\, V\;$

This option is incorrect

Option 2)

$\; 2.5\times 10^{-2}\, V\;$

This option is incorrect

Option 3)

$\; \; 40\; V\;$

This option is incorrect

Option 4)

$\; 250\; \; V$

This option is correct

Posted by

Sabhrant Ambastha

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters

anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in

anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question