Question

The power factor of an R-L circuit is $\frac{1}{\sqrt{2}}$. If the frequency of AC is doubled, what will be the power factor?

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{5}}$
• C. $\frac{1}{\sqrt{7}}$
• D. $\frac{1}{\sqrt{11}}$

Answer 1

Answer: (B)

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

Power factor $PF=1/\sqrt{2}$

Power factor of $AC$ circuit can be given as

$P.F=\frac{R}{\sqrt{R^2+x_2^2}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{1+{\left(\frac{x_L}{R}\right)}^2}}$ substituting $x_L={\omega }_L$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{1+{\left(\frac{\omega L}{R}\right)}^2}}$

$\Rightarrow 1+\frac{\omega .L}{R}=2$

$\Rightarrow \frac{\omega .L}{R}=1$

As the frequency is doubled the volume of $\omega .L/R$ also becomes twice

$\therefore P.F.=\frac{1}{\sqrt{1+{\left(\frac{\omega L}{R}\right)}^2}}=\frac{1}{\sqrt{1+(2{)}^2}}$

$=\frac{1}{\sqrt{1+4}}$

$\Rightarrow P.F.=\frac{1}{\sqrt{5}}$