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In a series LR circuit $\mathrm{X_{L}=R}$ and power factor of the circuit is $\mathrm{P_{1} .}$ When capacitor with capacitance $\mathrm{C}$ such that $\mathrm{X_{L}=X_{C}}$ is put in series, the power factor becomes $\mathrm{P_{2}. }$ The ratio $\mathrm{\frac{P_{1}}{P_{2}}}$ is :
Option: 1
$\frac{1}{2}$

Option: 2
$\frac{1}{\sqrt{2}}$

Option: 3
$\frac{\sqrt{3}}{\sqrt{2}}$

Option: 4
$2: 1$

Answers (1)

best_answer

$\mathrm{Power\: factor=\cos \phi =\frac{R}{Z}}$

$\mathrm{\cos \phi=\frac{R}{\sqrt{R^2+\left|X_L-x_C\right|^2}} }$

$\mathrm{P_1=\cos \phi_1=\frac{R}{\sqrt{R^2+\left(X_L\right)^2}} }$

$\mathrm{P_1=\frac{R}{\sqrt{R^2+R^2}}=\frac{1}{\sqrt{2}} }$

$\mathrm{P_2=\cos \phi_2=\frac{R}{\sqrt{R^2+\left(x_L-x_C\right)^2}} }$

$\mathrm{P_2=\frac{R}{R}=1 }$

$\mathrm{P_1=\frac{1}{\sqrt{2}}}$

Hence (2) is correct option.

Posted by

shivangi.bhatnagar

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