In an LCR series resonant circuit,at resonanc | vedclass

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

Question

(D) In an $LCR$ series resonant circuit,the voltage across the inductor $(V_L)$ leads the current by $90^{\circ}$,while the voltage across the capacit

Vedclass · Accepted Answer

$180^{\circ}$ out of phase

Vedclass · Answer

(D) In an $LCR$ series resonant circuit,the voltage across the inductor $(V_L)$ leads the current by $90^{\circ}$,while the voltage across the capacitor $(V_C)$ lags the current by $90^{\circ}$.Therefore,the phase difference between $V_L$ and $V_C$ is $90^{\circ} - (-90^{\circ}) = 180^{\circ}$.Since they are $180^{\circ}$ out of phase,they act in opposite directions.At resonance,the inductive reactance equals the capacitive reactance $(X_L = X_C)$,which implies $V_L = V_C$.Because they are equal in magnitude and $180^{\circ}$ out of phase,they cancel each other out.

In an $LCR$ series resonant circuit,at resonance,the voltage across $L$ and $C$ will cancel each other because they are:

Similar Questions

$A$ series $LCR$ circuit with inductance $L = 10\,H$,capacitance $C = 10\,\mu F$,and resistance $R = 50\,\Omega$ is connected to an $AC$ source of voltage $V = 200 \sin(100t)\,V$. If the resonant frequency of the $LCR$ circuit is $\nu_{0}$ and the frequency of the $AC$ source is $\nu$,then:

In the circuit shown,the readings of voltmeters $V_1, V_2$ and $V_3$ at resonance are given by

In an $LCR$ circuit,$L = 8.0 \, H$,$C = 0.5 \, \mu F$,and $R = 100 \, \Omega$ are connected in series. The resonance angular frequency is ........ $rad/s$.

In a series $LCR$ resonance circuit,if we change the resistance only,from a lower to higher value:

In the circuit shown,the ratio of the quality factor to the bandwidth is: (in $\text{ s}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module