Target Exam

CUET

Subject

Physics

Chapter

Alternating Current

Question:

The power factor of an RL circuit is $\frac{1}{2}$. If the frequency of a.c. is doubled, the power factor will be:

Options:

$\frac{1}{\sqrt{5}}$

$\frac{1}{\sqrt{7}}$

$\frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{13}}$

Correct Answer:

$\frac{1}{\sqrt{13}}$

Explanation:

The correct answer is Option (4) → $\frac{1}{\sqrt{13}}$

Power factor (P.F.) = $\frac{1}{2}$

$P.F.=\frac{R}{\sqrt{R^2+{X_L}^2}}$

$⇒\frac{1}{2}=\frac{1}{\sqrt{1+\left(\frac{X_L}{R}\right)^2}}$

$⇒1+\frac{w.L}{R}=4$

$⇒\frac{w.L}{R}=3$

$∴P.F.=\frac{1}{\sqrt{1+6^2}}=\frac{1}{\sqrt{13}}$