Given below are two statements:Statement I: M | vedclass

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

Question

(C) The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi$ is the power factor.For a series $LCR$ circ

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true.

Vedclass · Answer

(C) The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi$ is the power factor.For a series $LCR$ circuit, the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.At resonance, $X_L = X_C$, which makes the impedance $Z$ minimum $(Z = R)$ and the phase difference $\phi = 0$. Thus, $\cos \phi = 1$, and power dissipation is maximum. Therefore, Statement $I$ is true.In a pure resistive circuit, the current and voltage are in the same phase $(\phi = 0)$, so $\cos \phi = 1$. This results in maximum power dissipation for a given voltage. Therefore, Statement $II$ is true.Hence, both statements are correct.

Similar Questions

The $Q$ factor of a series $\text{LCR}$ circuit with $L=2 \text{ H}$,$C=32 \mu\text{F}$ and $R=10 \Omega$ is :-

An inductance $L$ having a resistance $R$ is connected to an alternating source of angular frequency $\omega$. The Quality factor $Q$ of the inductance is

The power factor of an $LCR$ circuit at resonance is:

$A$ resonant frequency of a circuit is $f$. If the capacitance is made four times the initial value,then the resonant frequency will become

The self-inductance of the motor of an electric fan is $10\;H$. In order to impart maximum power at $50\;Hz$, it should be connected to a capacitance (in $\mu F$) of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module