An $L-C-R$ circuit is connected to a source of $A.C.$ current. At resonance,the phase difference between the applied voltage and the current in the circuit is:

  • A
    $0$
  • B
    $\pi / 2$
  • C
    $\pi$
  • D
    $\pi / 4$

Explore More

Similar Questions

$A$ series $LCR$ circuit is connected to an alternating source of emf $E$. The current amplitude at resonant frequency is $I_0$. If the value of resistance $R$ becomes twice of its initial value,then the amplitude of current at resonance will be

In a series resonant $LCR$ circuit,the voltage across $R$ is $100 \, V$ 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......$V$.

The current in resistance $R$ at resonance is

$A$ sinusoidal voltage of peak value $250\, V$ is applied to a series $LCR$ circuit,in which $R = 8\, \Omega$,$L = 24\, mH$ and $C = 60\, \mu F$. The value of power dissipated at resonant condition is $'x'\, kW$. The value of $x$ to the nearest integer is .............

If the frequency of the $AC$ supply is equal to the resonant frequency,which ammeter will show zero reading?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo