The power factor of an R-L circuit is frac{1} | vedclass

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

Question

(B) The power factor of an $R-L$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $\cos \phi = \frac{1}{\sqrt{2}}$,w

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}$

Vedclass · Answer

(B) The power factor of an $R-L$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $\cos \phi = \frac{1}{\sqrt{2}}$,we have $\frac{R}{\sqrt{R^2 + X_L^2}} = \frac{1}{\sqrt{2}}$.Squaring both sides,$\frac{R^2}{R^2 + X_L^2} = \frac{1}{2}$,which implies $2R^2 = R^2 + X_L^2$,so $R^2 = X_L^2$ or $X_L = R$.Since $X_L = \omega L = 2\pi f L$,if the frequency $f$ is doubled,the new inductive reactance $X_L'$ becomes $2X_L = 2R$.The new power factor is $\cos \phi' = \frac{R}{\sqrt{R^2 + (X_L')^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

The power factor of an $R-L$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of $AC$ is doubled,the power factor will now be

Similar Questions

In a series $R-C$ circuit shown in the figure,the applied voltage is $10\, V$ and the voltage across the capacitor is found to be $8\, V$. Then,the voltage across $R$ and the phase difference between the current and the applied voltage will respectively be

$A$ series $R-C$ combination is connected to an $AC$ voltage of angular frequency $\omega = 500 \ rad/s$. If the impedance of the $R-C$ circuit is $R\sqrt{1.25}$,the time constant (in $ms$) of the circuit is:

In an $RL$ circuit,the reactance of the coil is $\sqrt{3}$ times the resistance. The phase difference between the voltage and the current is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module