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A series $\mathrm{LR}$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2 \mathrm{R}$. A capacitance of capacitive reactance equal to $\mathrm{R}$ is added in series with $\mathrm{L}$ and $\mathrm{R}$. The ratio of the new power factor to the old one is :
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The correct answer is:
$\sqrt{\frac{5}{2}}$
$\sqrt{\frac{5}{2}}$
Power factor ${ }_{\text {(old) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \mathrm{R})^2}}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}$
Power factor ${ }_{\text {(new) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(X_L-X_C\right)^2}}=\frac{R}{\sqrt{R^2+(2 R-R)^2}}$
$=\frac{R}{\sqrt{2} R}$
$\therefore \frac{\text { New power factor }}{\text { Old power factor }}=\frac{\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}}{\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}}=\sqrt{\frac{5}{2}}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \mathrm{R})^2}}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}$
Power factor ${ }_{\text {(new) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(X_L-X_C\right)^2}}=\frac{R}{\sqrt{R^2+(2 R-R)^2}}$
$=\frac{R}{\sqrt{2} R}$
$\therefore \frac{\text { New power factor }}{\text { Old power factor }}=\frac{\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}}{\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}}=\sqrt{\frac{5}{2}}$
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