In an LCR series circuit, at resonance, | vedclass

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

Question

(D) In an $LCR$ series circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.At resonance, the inductive reactance $X_L$ is equal to

Vedclass · Accepted Answer

the current and voltage are in phase

Vedclass · Answer

(D) In an $LCR$ series circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$, i.e.,$X_L = X_C$.Therefore, the impedance $Z$ becomes equal to the resistance $R$, which is the minimum possible value of impedance.Since $Z$ is minimum, the current $I = \frac{V}{Z}$ is maximum.At resonance, the phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R} = 0$, which means $\phi = 0$.Thus, the current and voltage are in the same phase.

In an $LCR$ series circuit, at resonance,

Similar Questions

At which point is the circuit inductive?

An $LCR$ circuit is at resonance for a capacitor $C$,inductance $L$,and resistance $R$. If the value of the resistance is halved while keeping all other parameters the same,the current amplitude at resonance will be:

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

$A$ series $LCR$ circuit is connected across a source of alternating emf of changing frequency and resonates at frequency $f_0$. Keeping capacitance constant,if the inductance $(L)$ is increased by $\sqrt{3}$ times and resistance is increased $(R)$ by $1.4$ times,the resonant frequency now is

$A$ sinusoidal voltage having a maximum value of $283 \ V$ and a frequency of $50 \ Hz$ is applied to an $LCR$ series circuit where $R = 3 \ \Omega$,$L = 25.48 \ mH$,and $C = 796 \ \mu F$. What is the impedance at resonance condition (in $\Omega$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module