Question

In a series LR circuit $X_L=R$ and power factor of the circuit is $P _1$. When capacitor with capacitance $C$ such that $X _{ L }= X _{ C }$ is put in series, the power factor becomes $P _2$. The ratio $\frac{ P _1}{ P _2}$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{\sqrt{2}}$
• D. $2: 1$

Answer 1

Answer: (B)

In case of L-R circuit
$Z =\sqrt{ X _{ L }^2+ R ^2} \&$ power factor
$P _1=\cos \phi=\frac{ R }{ Z }$
As $X _{ L }= R$
$\Rightarrow Z =\sqrt{2} R$
$\Rightarrow P _1=\frac{ R }{\sqrt{2} R } \Rightarrow P _1=\frac{1}{\sqrt{2}}$
In case of L-C-R circuit
$Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$
As $X_{ L }=X_C$
$\Rightarrow Z=R$
$\Rightarrow P_2=\cos \phi=\frac{R}{R}=1$
$\Rightarrow \frac{P_1}{P_2}=\frac{1}{\sqrt{2}}$