$A$ series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from the circuit,the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit,the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

  • A
    -$1.0$
  • B
    zero
  • C
    $0.5$
  • D
    $1.0$

Explore More

Similar Questions

Which of the following combinations should be selected for better tuning of an $L-C-R$ circuit used for communication?

In a series $LCR$ circuit,$C = 2\,\mu F$,$L = 1\,mH$,and $R = 10\,\Omega$. When the current in the circuit is maximum,what is the ratio of the energy stored in the capacitor to the energy stored in the inductor?

Power delivered by the source of the circuit becomes maximum,when

In a series $LCR$ circuit, $C = 2 \mu F$, $L = 5 \text{ mH}$, and $R = 5 \Omega$. What is the ratio of the energy stored in the inductor to that in the capacitor when the maximum current flows through the circuit (in $:$)?

An $L-C-R$ series circuit with $100 \, \Omega$ resistance is connected to an $ac$ source of $100 \, V$ and angular frequency $300 \, rad/s$. When the capacitance is removed, the current lags behind the voltage by $45^{\circ}$. When the inductance is removed, the current leads the voltage by $45^{\circ}$. The current flowing in the circuit will be ... $A$.

Difficult
View Solution

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