If the power factor is 1/2 in a series RL cir | vedclass

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

Question

(A) The power factor of an $RL$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{1}{2}$.Since $\cos \phi = 1/2$,the phase angle $\phi = 60^\circ$.

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{\pi}\,H$

Vedclass · Answer

(A) The power factor of an $RL$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{1}{2}$.Since $\cos \phi = 1/2$,the phase angle $\phi = 60^\circ$.In an $RL$ circuit,$	an \phi = \frac{\omega L}{R}$.Substituting the values,$	an 60^\circ = \sqrt{3} = \frac{2\pi f L}{R}$.Given $f = 50\,Hz$ and $R = 100\,\Omega$,we have $\sqrt{3} = \frac{2 	imes \pi 	imes 50 	imes L}{100}$.$\sqrt{3} = \frac{100\pi L}{100} = \pi L$.Therefore,$L = \frac{\sqrt{3}}{\pi}\,H$.

If the power factor is $1/2$ in a series $RL$ circuit with $R = 100\,\Omega$ connected to an $ac$ mains $(50\,Hz)$,find the value of $L$.

Similar Questions

An e.m.f. $E=E_0 \cos \omega t$ is applied to a circuit containing $L$ and $R$ in series. If $X_L=2 R$,then the power dissipated in the circuit is

For a series $LCR$ circuit,$R = X_L = 2X_C$. The impedance of the circuit and the phase difference between $V$ and $i$ will be:

In a series $LCR$ circuit,operated with an $ac$ of angular frequency $\omega$,the total impedance is

In an $L-R$ circuit,the value of $L$ is $(\frac{0.4}{\pi}) \, H$ and the value of $R$ is $30 \, \Omega$. If an alternating e.m.f. of $200 \, V$ at $50 \, Hz$ is connected to the circuit,what are the impedance and the current in the circuit?

An e.m.f. $E = 4 \cos(1000t) \, V$ is applied to an $LR$-circuit of inductance $3 \, mH$ and resistance $4 \, \Omega$. The amplitude of current in the circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module