For the series LCR circuit connected with a 2 | vedclass

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

Question

(C) The power factor of an $LCR$ series circuit is given by $\cos \phi = \frac{R}{Z}$.First,we calculate the impedance $Z$ of the circuit:$Z = \sqrt{R

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The power factor of an $LCR$ series circuit is given by $\cos \phi = \frac{R}{Z}$.First,we calculate the impedance $Z$ of the circuit:$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$Given $R = 60 \ \Omega$,$X_{L} = 70 \ \Omega$,and $X_{C} = 150 \ \Omega$.$Z = \sqrt{60^{2} + (70 - 150)^{2}}$$Z = \sqrt{60^{2} + (-80)^{2}}$$Z = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \ \Omega$.Now,the power factor is:$\cos \phi = \frac{R}{Z} = \frac{60}{100} = \frac{6}{10}$.Comparing this with the given power factor $\frac{\alpha}{10}$,we get $\alpha = 6$.

For the series $LCR$ circuit connected with a $220 \ V$,$50 \ Hz$ a.c. source as shown in the figure,the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is . . . . . .

Similar Questions

$A$ sinusoidal $AC$ current flows through a resistor of resistance $R$. If the peak current is $I_p$,then the power dissipated is

Average power in the $L-C-R$ circuit depends upon

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

When a voltage $V_{s} = 200 \sqrt{2} \sin(\omega t + 15^{\circ})$ is applied to an $AC$ circuit,the current in the circuit is found to be $I = 2 \sin(\omega t + 45^{\circ})$. The average power consumed in the circuit is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module