Question

The power factor of an ac circuit having resistance \((\mathrm{R})\) and inductance (L) connected in series and an angular velocity w is, (a) \((\mathrm{R} / \mathrm{cL})\) (b) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$ (c) \((\omega L / R)\) (d) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}-\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$

Short answer

The short answer to the question is: The power factor of an AC circuit having resistance R and inductance L connected in series and an angular velocity ω is (b) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$.

Step-by-step solution

Calculate the impedance of the circuit

For a series R-L circuit, the impedance Z is given by the following formula: \[Z = \sqrt{R^2 + (ωL)^2}\]

Obtain the angle between voltage and current phasors (θ)

The angle θ can be calculated using: \[\tan θ = \frac{ωL}{R}\] Now, we can find the cosine of the angle θ, which is the power factor: \[PF = \cos θ\] Since \[\cos θ = \frac{1}{\sqrt{1 + \tan^2 θ}}\], we can substitute the expression for tan θ.

Calculate the power factor using the angle θ

Now, let's substitute the expression for tan θ and simplify: \[PF = \frac{1}{\sqrt{1 + (\frac{ωL}{R})^2}}\] \[PF = \frac{1}{\sqrt{1 + \frac{ω^2L^2}{R^2}}}\] \[PF = \frac{R}{\sqrt{R^2 + ω^2L^2}}\] So, the correct answer is: (b) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$