In an $LCR$ series circuit,when $L$ is removed from the circuit,the phase difference between voltage and current is $\frac{\pi}{3}$. If $C$ is removed from the circuit instead of $L$,the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit is $(\tan 60^{\circ}=\sqrt{3})$.

  • A
    $1$
  • B
    $\frac{1}{\sqrt{2}}$
  • C
    $\frac{\sqrt{3}}{2}$
  • D
    $\frac{1}{2}$

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Similar Questions

Given below are two statements :
Statement $I$ : In an $LCR$ series circuit,current is maximum at resonance.
Statement $II$ : Current in a purely resistive circuit can never be less than that in a series $LCR$ circuit when connected to the same voltage source.
In the light of the above statements,choose the correct option from the options given below :

Statement-$I$: $A$ capacitor can be used in an $a.c.$ circuit in place of a choke coil.
Statement-$II$: $A$ capacitor blocks $d.c.$ and allows $a.c.$ to pass.

You are given many resistances,capacitors and inductors. These are connected to a variable $DC$ voltage source (the first two circuits) or an $AC$ voltage source of $50 \ Hz$ frequency (the next three circuits) in different ways as shown in Column $II$. When a current $I$ (steady state for $DC$ or rms for $AC$) flows through the circuit,the corresponding voltage $V_1$ and $V_2$ (indicated in circuits) are related as shown in Column $I$. Match the two.

In an $A.C.$ circuit,the current:

$A$ series $LCR$ circuit has $L=0.01\,H$,$R=10\,\Omega$,and $C=1\,\mu F$ and it is connected to an $AC$ voltage of amplitude $V_m = 50\,V$. At a frequency $60\%$ lower than the resonant frequency,the amplitude of the current will be approximately $...............\,mA$.

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