Question 164 of 190

In a series $LR$ circuit with ${X}_{L}=R$, power factor is ${P}_{1}$. If a capacitor of capacitance $C$ with ${X}_{C}={X}_{L}$ is added to the circuit the power factor becomes ${P}_{2}$. The ratio of ${P}_{1}$ to ${P}_{2}$ will be :

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

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Answer: (B) $1:\sqrt{2}$

In case of $LR$ circuit:

Given: ${X}_{L}=R$
${P}_{1}=\cos \phi =\dfrac{R}{Z} =\dfrac{R}{\sqrt{{R}^{2}+{\left({X}_{L}\right)}^{2}}} =\dfrac{1}{\sqrt{2}}$
In case of $LCR$ circuit:

${P}_{2}=\cos \phi =\dfrac{R}{Z} =\dfrac{R}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}} =1$
Therefore,
$\dfrac{{P}_{1}}{{P}_{2}}=1:\sqrt{2}$