In a series LCR circuit,the inductive reactan | vedclass

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

Question

(C) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$.The impedance $Z$ of the series $LCR$ circuit is calculated as:$Z = \sq

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$.The impedance $Z$ of the series $LCR$ circuit is calculated as:$Z = \sqrt{R^2 + (X_L - X_C)^2}$Given values are $R = 6\, \Omega$,$X_L = 10\, \Omega$,and $X_C = 4\, \Omega$.Substituting these values:$Z = \sqrt{6^2 + (10 - 4)^2}$$Z = \sqrt{36 + 6^2}$$Z = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\, \Omega$Now,calculating the power factor:$\cos \phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$

In a series $LCR$ circuit,the inductive reactance $(X_{L})$ is $10\, \Omega$ and the capacitive reactance $(X_{C})$ is $4\, \Omega$. The resistance $(R)$ in the circuit is $6\, \Omega$. The power factor of the circuit is :

Similar Questions

$A$ sinusoidal alternating current of peak value $I_0$ passes through a heater of resistance $R$. What is the mean power output of the heater?

In an $A.C.$ circuit containing $L, C, R$ in series,the ratio of true power to apparent power is ($Z=$ impedance of the circuit and $R$ is the resistance).

In an $AC$ circuit,$V$ and $I$ are given by $V = 150 \sin(150t) \, V$ and $I = 150 \sin(150t + \frac{\pi}{3}) \, A$. The power dissipated in the circuit is ....... $W$.

On which of the following does the power consumed in an $L-C-R$ series $AC$ circuit depend?

Can the instantaneous power output of an $AC$ source ever be negative? Can the average power output be negative?

Vedclass Test Series

Exam Paper Generator

Online Exam Module