A series LR circuit is connected to an ac source of frequency and… - Vidyadip

MCQ

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

A
$\sqrt {\frac{2}{3}} $
B
$\sqrt {\frac{2}{5}} $
C
$\sqrt {\frac{3}{2}} $
✓
$\sqrt {\frac{5}{2}} $

✓

Answer

Correct option: D.

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

d
Power factor $_{(old)}$

$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(2 \mathrm{R})^{2}}}=\frac{\mathrm{R}}{\sqrt{5 \mathrm{R}}}$

Power factor $_{(\text {new })}$

$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(2 \mathrm{R}-\mathrm{R})^{2}}}$

$=\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}$

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

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