Question

Time constant of the given circuit is $\tau$. If the battery is replaced by an ac source having voltage $V=V_0\cos \omega t$, power factor of the circuit will be:

• A. $\omega \tau$
• B. $\frac{1}{\sqrt{1+(\omega \tau {)}^2}}$
• C. $\sqrt{1-(\omega \tau {)}^2}$
• D. $None$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{1+(\omega \tau {)}^2}}$

The power factor of a circuit is given by:

$\cos \phi =\frac{R}{Z}$ ...$\left(1\right)(R=$ resistance, $Z=$ net impedance, $\phi$ = Angle of impedance triangle)

Time Contact $\left(\tau \right)=\frac{L}{R}$ (As the given circuit is series $L-R$ circuit)

Impedance $\left(Z\right)=\sqrt{R^3+X_L^2}$ $(X_L=$ Inductive Reactance $=\omega L$)

Using these values in equation 1:

$\cos \phi =\frac{R}{\sqrt{R^2+{\omega }^2{L}^2}}$

$=\frac{R}{\sqrt{R^2+{\omega }^2{\tau }^2{R}^2}}$ (Using $L=\tau R)$

$=\frac{1}{\sqrt{1+{\omega }^2{\tau }^2}}$ which is the power of the circuit