In a radio tuning $RLC$ circuit,the half-power frequencies are $100 \, MHz$ and $120 \, MHz$. Find the quality factor.

  • A
    $5$
  • B
    $4.5$
  • C
    $6$
  • D
    $5.5$

Explore More

Similar Questions

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:

$A$ $110 \; V, 50 \; Hz, AC$ source is connected in the circuit (as shown in figure). The current through the resistance $55 \; \Omega$,at resonance in the circuit,will be $\dots \; A$.

If the inductance and capacitance are both doubled in an $L-C-R$ circuit,the resonant frequency of the circuit will

In the circuit shown,the $AC$ source has voltage $V=20 \cos (\omega t) \text{ V}$ with $\omega=2000 \text{ rad/s}$. The magnitude of the amplitude current will be nearly

An $LCR$ series circuit with $100 \Omega$ resistance is connected to an $AC$ source of $200 V$ and angular frequency $300 \text{ rad/s}$. When only the capacitor is removed,the current lags behind the voltage by $60^{\circ}$. When only the inductor is removed,the current leads the voltage by $60^{\circ}$. The power dissipated in the $LCR$ circuit will be:

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