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A series LR circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the old one is :
Option: 1
$\sqrt{\frac{2}{3}}$

Option: 2
$\sqrt{\frac{2}{5}}$

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

Option: 4
$\sqrt{\frac{5}{2}}$

Answers (1)

best_answer

$\text { Power factor (old) }=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}=\frac{R}{\sqrt{R^{2}+(2 R)^{2}}}=\frac{R}{\sqrt{5} R}$

$\text { Power factor (new) }=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}=\frac{R}{\sqrt{R^{2}+(2 R-R)^{2}}}=\frac{R}{\sqrt{2} R}$

$\frac{\text { New power factor }}{\text { Old power factor }}=\frac{\frac{R}{\sqrt{2} R}}{\frac{R}{\sqrt{5} R}}=\sqrt{\frac{5}{2}}$

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Divya Prakash Singh

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