Question

The time constant of the given circuit is τ. If the battery is replaced by an AC source having voltage $\mathrm{V}={\mathrm{V}}_0\mathrm{cos\omega t}$, the power factor of the circuit will be?

Answer 1

Step 1: Given parameters

As the circuit is L-R circuit, the

Time contact, $\mathrm{\tau }=\frac{\mathrm{L}}{\mathrm{R}}$

Impedance, $\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{{\mathrm{X}}_{\mathrm{L}}}^2}$

Here,

${\mathrm{X}}_{\mathrm{L}}$ is the inductive resistance or ${\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}$

$\mathrm{\omega }$ is the angular velocity.

$\mathrm{L}$ is the inductance.

Step 2: To find

We have to find the power factor of the circuit.

Step 3: Calculation

We know the formula,

$\mathrm{cos\phi }=\frac{\mathrm{R}}{\mathrm{Z}}$…(equation $1$)

Here,

R is the resistance.

Z is the net impedance.

$\mathrm{\phi }$ is the angle of the impedance triangle.

By substituting the values in the above equation,

$$\begin{gathered} \mathrm{cos\phi }=\frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{\omega }}^2{\mathrm{L}}^2}} \\ =\frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{\omega }}^2{\mathrm{\tau }}^2{\mathrm{R}}^2}} \\ =\frac{1}{\sqrt{1+{\mathrm{\omega }}^2{\mathrm{\tau }}^2}} \end{gathered}$$

Therefore, the power factor of the circuit is $\frac{1}{\sqrt{1+{\mathrm{\omega }}^2{\mathrm{\tau }}^2}}$.