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In a series $L R$ 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
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{\sqrt{2}}$
$2: 1$
Solution
In case of $L-R$ circuit
$Z=\sqrt{ X _{ L }^{2}+ R ^{2}} \&$ power factor
$P _{1}=\cos \phi=\frac{ R }{ Z }$
$\text { As } X _{ L }= R$
$\Rightarrow Z =\sqrt{2} R$
$\Rightarrow P _{1}=\frac{ R }{\sqrt{2} R } \Rightarrow P _{1}=\frac{1}{\sqrt{2}}$
In case of $L-C-R$ circuit
$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$
As $X_{L}=X_{C}$
$\Rightarrow Z=R$
$\Rightarrow P _{2}=\cos \phi=\frac{ R }{ R }=1$
$\Rightarrow \frac{ P _{1}}{ P _{2}}=\frac{1}{\sqrt{2}}$
