Question

The power factor of a $CR$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of ac signal is halved, then the power factor of the circuit becomes

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

Answer 1

The correct option is A $\frac{1}{\sqrt{5}}$

The power factor of a CR circuit is given as,

$\cos \varphi =1/\sqrt{2}$

$\varphi ={45}^o$

We have, $\tan \varphi =1/\omega CR$

$\tan 45=1/\omega CR$

$1=1/\omega CR$ ............eq1

Now if the frequency is halved, ${\omega }^{\prime }=\omega /2$

then $\tan {\varphi }^{\prime }=1/{\omega }^{\prime }CR$

$\tan {\varphi }^{\prime }=2/\omega CR$

But $1/\omega CR=1$

hence, $\tan {\varphi }^{\prime }=2$

therefore, new power factor

$\cos {\varphi }^{\prime }=1/\sqrt{2^2+1^2}=1/\sqrt{5}$