An AC circuit has an inductor and a resistor | vedclass

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

Question

(B) The power factor of an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.For the initial circuit ($RL$ circuit):$Z = \sqrt{R^2 + X_L^2} = \sqrt{R

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The power factor of an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.For the initial circuit ($RL$ circuit):$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10R^2} = R\sqrt{10}$.Old power factor,$\cos \phi = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}}$.For the new circuit ($RLC$ circuit):$Z' = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (3R - 2R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.New power factor,$\cos \phi' = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.The ratio of the new power factor to the old power factor is:$\frac{\cos \phi'}{\cos \phi} = \frac{1/\sqrt{2}}{1/\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} = \frac{\sqrt{5}}{1}$.Comparing this with $\sqrt{5} : x$,we get $x = 1$.

An $AC$ circuit has an inductor and a resistor of resistance $R$ in series,such that $X_L = 3R$. Now,a capacitor is added in series such that $X_C = 2R$. The ratio of the new power factor to the old power factor of the circuit is $\sqrt{5} : x$. The value of $x$ is ...... .

Similar Questions

In a series $LCR$ circuit, the $rms$ voltage across the resistor and the capacitor are $30 \ V$ and $90 \ V$ respectively. If the applied voltage is $V = 50 \sqrt{2} \sin \omega t$, then the peak voltage across the inductor is

In an $AC$ circuit,a resistance of $R$ $\Omega$ is connected in series with an inductor of self-inductance $L$. If the phase angle between voltage and current is $45^{\circ}$,the value of inductive reactance $(X_{L})$ will be equal to . . . . . . .

An inductor of inductance $L$ and a resistor of resistance $R$ are joined in series and connected to a source of frequency $\omega$. The power dissipated in the circuit is

When an inductance $L$ and resistor $R$ are connected in series to $25 \ V$, $50 \ Hz$ supply, a current of $0.5 \ A$ flows in the circuit. The current lags behind in phase from applied voltage by $(\frac{\pi}{3})$ radian. The value of $R$ is $(\cos 60^{\circ} = \frac{1}{2})$ (in $\Omega$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module