In a series LCR circuit,the voltage leads the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a series $LCR$ circuit,the impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and the phase angle $\phi$ is given by $	an \phi = \frac{X_L

Vedclass · Accepted Answer

$\omega > \omega_0$

Vedclass · Answer

(C) In a series $LCR$ circuit,the impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and the phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.Voltage leads the current when the phase angle $\phi$ is positive,which implies $X_L > X_C$.Substituting the expressions for inductive reactance $(X_L = \omega L)$ and capacitive reactance $(X_C = \frac{1}{\omega C})$,we get $\omega L > \frac{1}{\omega C}$.This simplifies to $\omega^2 > \frac{1}{LC}$.Since the resonant angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$,we have $\omega_0^2 = \frac{1}{LC}$.Therefore,the condition for voltage to lead the current is $\omega^2 > \omega_0^2$,which implies $\omega > \omega_0$.

In a series $LCR$ circuit,the voltage leads the current when (Given that $\omega_0$ is the resonant angular frequency):

Similar Questions

The quality factor of an $LCR$ circuit having resistance $(R)$ and inductance $(L)$ at resonance frequency $(\omega)$ is given by:

In the circuit shown, the ratio of the quality factor $(Q)$ and the band width $(BW)$ is (Given: band width = $R/L$)

$A$ series $LCR$ circuit is shown in the figure. Where the inductance of $10 \ H$,capacitance $40 \ \mu F$,and resistance $60 \ \Omega$ are connected to a variable frequency $240 \ V$ source. The current at resonating frequency is (in $A$)

In an $LCR$ resonant circuit,the phase difference between the current and the voltage is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module