A series LR circuit is connected to an ac source of frequency ω and the inductive reactance is equal to 2 R . A capacitance of capacitive reactance equal to R is added in series with L and R . The ratio of the new power

Correct answer: D. (5)/(2) Power factor _ (old) = R R^2+X_ L ^2 = R R^2+(2 R)^2 = R 5 R Power factor _ (new) = R R^2+(X_L-X_C)^2 = R R^2+(2 R-R)^2 = R 2 R New power factor Old power factor = R 2 R R 5 R = (5)/(2)

JEE Main 201325 AprilOnline

Exam: JEE Main
Year: 2013
Date: 25 April
Mode: Online
Subject: Physics
Chapter: Alternating Current

JEE Main 2013 · 25 April · Online · Physics · Alternating Current

A series LR circuit is connected to an ac source of frequency ω and...

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 :

Options

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

Correct answer

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

Step-by-step solution

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}}

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