The power factor of the circuit shown in the | vedclass

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

Question

(C) In the given $LCR$ series circuit,the inductive reactance is $X_L = 100 \ \Omega$,the capacitive reactance is $X_C = 50 \ \Omega$,and the resistan

Vedclass · Accepted Answer

$1/\sqrt{2}$

Vedclass · Answer

(C) In the given $LCR$ series circuit,the inductive reactance is $X_L = 100 \ \Omega$,the capacitive reactance is $X_C = 50 \ \Omega$,and the resistance is $R = 50 \ \Omega$.The impedance $Z$ of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values,$Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = \sqrt{2 	imes 50^2} = 50\sqrt{2} \ \Omega$.The power factor is defined as $\cos \phi = \frac{R}{Z}$.Therefore,$\cos \phi = \frac{50}{50\sqrt{2}} = \frac{1}{\sqrt{2}}$.

The power factor of the circuit shown in the figure is:

Similar Questions

In an $A.C.$ circuit,the instantaneous values of $\text{e.m.f}.$ and current are $e = 10 \sin(50 \pi t) \text{ V}$ and $I = 10 \sin(50 \pi t + \frac{\pi}{3}) \text{ A}$. The average power consumed in the circuit is $... \text{ W}$.

The power factor of an $R-L$ circuit is $\frac{1}{\sqrt{3}}$. If the inductive reactance is $2 \Omega$,the value of the resistance is:

In an $AC$ circuit,the current flowing is $I=5 \sin \left(100 t-\frac{\pi}{2}\right) A$ and the potential difference is $e=200 \sin (100 t) V$. The power consumption is equal to (in $W$)

An $AC$ voltage $V=20 \sin 200 \pi t$ is applied to a series $LCR$ circuit which drives a current $I=10 \sin (200 \pi t + \frac{\pi}{3})$. The average power dissipated is: (in $W$)

In the circuit shown in the figure,if the value of $rms$ current is $2.2\, A$,the power factor of the box is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module