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JEE Main 202330 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 30 January
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2023 · 30 January · Morning Shift · Physics · Alternating Current

In a series LR circuit with X _ L = R , power factor is P _ 1 . If...

In a series

LR

circuit with

{X}_{L} = R

, power factor is

{P}_{1}

. If a capacitor of capacitance

C

with

{X}_{C} = {X}_{L}

is added to the circuit the power factor becomes

{P}_{2}

. The ratio of

{P}_{1}

to

{P}_{2}

will be :

In a series LR circuit with X _ L = R , power factor is P _ 1 . If a capacitor of capacitance C with X _ C = X _ L is

In a series LR circuit with X _ L = R , power factor is P _ 1 . If a capacitor of capacitance C with X _ C = X _ L is

Options

A $1: 3$
B $1: \sqrt{2}$
C $1: 1$
D $1: 2$

Correct answer

B. $1: \sqrt{2}$

Step-by-step solution

In case of $LR$ circuit:

In a series LR circuit with X _ L = R , power factor is P _ 1 . If a capacitor of capacitance C with X _ C = X _ L is

Given: ${X}_{L} = R$

${P}_{1} = cosϕ = \dfrac{R}{Z} = \dfrac{R}{\sqrt{{R}^{2} + {\left({X}_{L}\right)}^{2}}} = \dfrac{1}{\sqrt{2}}$

In case of $LCR$ circuit:

In a series LR circuit with X _ L = R , power factor is P _ 1 . If a capacitor of capacitance C with X _ C = X _ L is

${P}_{2} = cosϕ = \dfrac{R}{Z} = \dfrac{R}{\sqrt{{R}^{2} + {\left({X}_{L} - {X}_{C}\right)}^{2}}} = 1$

Therefore,

$\dfrac{{P}_{1}}{{P}_{2}} = 1: \sqrt{2}$

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