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 an $ac$ circuit,the current lags behind the voltage by $\pi /3$. The components in the circuit are

$A$ circuit when connected to an $A.C.$ source of $12 \; V$ gives a current of $0.2 \; A$. The same circuit when connected to a $D.C.$ source of $12 \; V$,gives a current of $0.4 \; A$. The circuit is

In an $LCR$ series circuit,the source voltage is $120 \ V$,the voltage across the inductor is $50 \ V$,and the voltage across the resistor is $40 \ V$. Determine the voltage across the capacitor.

In a series $LR$ circuit,$X_L=R$,the power factor is $P_1$. If a capacitor of capacitance $C$ with $X_C=X_L$ is added to the circuit,the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be

The following series $L-C-R$ circuit,when driven by an emf source of angular frequency $70 \text{ krad/s}$,behaves effectively as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module