The power factor of an R-L circuit is $\frac{1}{\sqrt{2}}$. If the frequency of $\mathrm{AC}$ is doubled the…
The power factor of an R-L circuit is $\frac{1}{\sqrt{2}}$. If the frequency of $\mathrm{AC}$ is doubled the power factor will now be
- $\frac{1}{\sqrt{3}}$
- $\frac{1}{\sqrt{5}}$
- $\frac{1}{\sqrt{7}}$
- $\frac{1}{\sqrt{11}}$
Solution
The power factor of an R-L circuit is given as,
$\begin{aligned}
& \quad \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}} \\
& \therefore \quad \frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}}=\frac{1}{\sqrt{2}} \\
& \therefore \quad \frac{1}{\sqrt{1+\left(\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right)^2}}=\frac{1}{\sqrt{2}} \\
& \therefore \quad \frac{1}{\sqrt{1+\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)^2}}=\frac{1}{\sqrt{2}} \\
& \therefore \quad\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)^2+1=2 \\
& \therefore \quad \frac{\omega \mathrm{L}}{\mathrm{R}}=1
\end{aligned}$
So, when the AC frequency is doubled,
$\begin{aligned}
\frac{\omega \mathrm{L}}{\mathrm{R}} & =2 \\
\therefore \quad \cos \phi & =\frac{1}{\sqrt{1+\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)^2}}=\frac{1}{\sqrt{1+(2)^2}} \\
\therefore \quad \cos \phi & =\frac{1}{\sqrt{5}}
\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 2)
Practice more Alternating Current questions on Aicharya