The power factor of a CR circuit is frac{1}{s | vedclass

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

Question

(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.Given $\cos \phi_1 = \frac{1}{\sqrt{2}}$,this implies $\frac{R}{Z} = \frac{1}{\sqrt{2}}$,so $R = X_{C1} = \frac{1}{2\pi f_1 C}$.When the frequency is halved,$f_2 = \frac{f_1}{2}$.The new capacitive reactance is $X_{C2} = \frac{1}{2\pi f_2 C} = \frac{1}{2\pi (f_1/2) C} = 2 X_{C1} = 2R$.The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{C2}^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

The power factor of a $CR$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of an $AC$ signal is halved,then the power factor of the circuit will become:

Similar Questions

When a $DC$ voltage of $200\, V$ is applied to a coil of self-inductance $\frac{2\sqrt{3}}{\pi}\,H$,a current of $1\, A$ flows through it. When the $DC$ source is replaced by an $AC$ source of $200\, V$,the current in the coil is reduced to $0.5\, A$. The frequency of the $AC$ supply is......$Hz$.

$A$ coil has self-inductance $L = 0.04 \ H$ and resistance $R = 12 \ \Omega$. When it is connected to a $220 \ V, 50 \ Hz$ supply,what will be the current flowing through the coil (in $A$)?

$A$ lamp is connected in series with a capacitor and an $AC$ source. What happens if the capacitance of the capacitor is reduced?

In a series $LCR$ circuit,$R=300 \Omega$,$L=0.9 \text{ H}$,$C=2 \mu\text{F}$,and $\omega=1000 \text{ rad/s}$. The impedance of the circuit is: (in $\Omega$)

$A$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module