Question

An alternating emf of frequency $f$ is applied to a series $\text{LCR}$ circuit. For this frequency of the applied emf.

• A. The circuit is at ‘resonance’ and its impedance is made up only of a reactive part
• B. The current in the circuit at resonance is in phase with the applied emf and the voltage across $R$ equals this applied emf
• C. The sum of the potential differences across the inductance and capacitance equals the applied emf which is ${180}^{\circ }$ ahead of phase of the current in the circuit
• D. The quality factor of the circuit is $\frac{\omega L}{R}$ or $\frac{1}{\omega CR}$ and this is a measure of the voltage magnification (taking V out-across inductor or capacitor) produced by the circuit at resonance as well as the sharpness of resonance of the circuit

Answer 1

Answer: (B), (D)

The quality factor of the circuit is $\frac{\omega L}{R}$ or $\frac{1}{\omega CR}$ and this is a measure of the voltage magnification (taking V out-across inductor or capacitor) produced by the circuit at resonance as well as the sharpness of resonance of the circuit

At the resonance condition , $X_L=X_C$

Impedance $z=\sqrt{R^2+(X_L-X_C{)}^2}$

$\Rightarrow z=R$

If voltage is $E=E_0\sin \omega t$

then, current $I=I_0\sin (\omega t\pm \varphi )$ (as phase does not change across resistor)

Quality factor in $\text{LR}$ circuit is, $Q=\frac{X_L}{R}=\frac{\omega L}{R}$

For $\text{CR}$ circuit , $Q=\frac{X_C}{R}=\frac{1}{\omega CR}$

Sharpness curve,



Hence, options (b) and (d) are the correct answers.